Canon camera recommendation I am an amateur astronomer looking to upgrade from a simple point and shoot camera looking through a Newtonian telescope to a more sophisticated set-up. I lecture at a local college and I am lucky enough that they may lend me a schmidt-cassegrain telescope (Celestron CPC series telescope).
I am looking to invest into my first DSLR camera and while I know Canon has their own dedicated astronomy camera (EOS Ra) such a camera is out of my budget. I have narrowed it down to four options and I would like to have more expert advice on which I should choose:

EOS 90D
EOS 77D
EOS 250D
EOS M50

I am lost to choose between those options. May you please shed some light? Thank you very much.

Based on how you described your situation, I surmised that you do not have prior experience or equipment for astrophotography through a long focal-length telescope.  There are a few things you did not specifically ask, but should know before you invest in an expensive camera.  Also, it sounds like you would have preferred a dedicated astrophotography camera -- rather than a general purpose camera -- if budget would allow for it.
Altitude/Azimuth Mounts and Field Rotation
The Celestron CPC series scopes are all SCT's... but on an Altitude/Azimuth style mount.  The mount's axis of rotation is not parallel to Earth's axis of rotation and this means that when you perform long-exposure astrophotography, you'll experience an issue called "Field Rotation".  The amount of field rotation you experience depends on where you point the telescope in the sky BUT ... to avoid field rotation, the maximum exposure duration is about ... 20 seconds (tops).
The scope has an f/10 focal ratio (virtually all SCT's are f/10 scopes... but just occasionally f/8 ... that's fairly uncommon).   This means 20 seconds will not be adequate (you'll likely need 5 minutes or longer).
Very large professional telescopes (e.g. 1 meter and larger) are almost always on alt/azimuth mounts, but those telescopes are equipped with field-rotators which rotate the camera while the scope tracks to counter-act the field rotation effect.  The rotation rate is highly dependent on the location of the sky being imaged -- so this is computer controlled.  Field rotators are expensive.
The less expensive way to solve this issue is to add an equatorial wedge.  This tilts the entire alt/az setup over on an angle such that the azimuth axis becomes a right-ascension axis and can be aligned parallel to Earth's polar axis.  In this orientation, the motion of the mount exactly cancels the motion of the Earth and the object is held in position with no field rotation.
Guiding
It is possible to take long-exposure images at short focal lengths (e.g. I've done as much as 8 minutes un-guided at around 500mm focal length).  But even an 8" SCT would have a focal length of around 2000mm.  It isn't possible to take long exposures without an auto-guider ... even on a quality mount.  This requires a separate guide-camera.  The good news is that guide cameras usually aren't very expensive.  A starter camera is around 150 USD (e.g a ZWO ASI120MC-S is around 150).  The Celestron CPC mount is designed to be affordable for visual use.  While you can use it for astrophotography, it is not a high-performance mount ... and may present challenges.
Commonly the guide-camera is attached to a guide-scope ... a separate telescope usually piggy-backed to the main imaging scope (or on a side-by-side mounting plate).  The focal length would ideally be around 1/3rd of the focal length of the main imaging scope (e.g. for an 8" SCT with around 2000mm focal length, the guide-scope would ideally have a focal length of around 700mm). People do push this limit ... some people will push it up to around 1/10th (e.g. a 200mm guide scope).
The guide scope need not be high-quality ... it's staggering how bad the optics can be and still result in exceptionally good guiding quality (this has to do with the algorithms used ... they don't assume the Star is a pin-point source of light).
Another option is to use an off-axis guide adapter.  This are gadgets mounted between the telescope and main imaging camera.  It has a 45° prism which acts as a "pick off" mirror.  It is position just outside the field of view of the main imaging camera and sends light out at a 90° angle and into a guide-camera mounted right next to the main imaging camera.  An off-axis guide adapter has the advantage of sharing the same telescope as the main imaging camera -- which means no differential flexure issues between two scopes resulting in elongated stars.  The down side of this is that the pick-off mirror has to be very small so as to not present an obstruction in the view of the main imaging camera.  And that means you have to try to find a suitable guide-star in that very tiny field of view (sometimes you have to rotation the off-axis adapter to try to find a star.)
Schmidt Cassegrain Optics
The telescope wont have a "flat" field.  Stars near the edges of the field of view will exhibit off-axis aberrations ... which will include some loss of focus and also the addition of coma and possibly a bit of astigmatism.  Celestron does make a line of SCT optical tubes they refer to as their "EdgeHD" line (more expensive).  These telescopes are not coma-free but they do have a flatter field with respect to focus and overall do a decent job of correcting the field of view well enough for use with an APS-C camera.  When using an APS-C camera with traditional non-correct Schmidt Cassegrain optics, expect to see some reduction in off-axis image quality (edges and corners of the frame).
Cameras
The Canon EOS Ra is a full-frame sensor which has been modified for astrophotography use (the primary difference is the band-pass on the filter -- specifically at the Hydrogen alpha wavelength).  Full-frame cameras are a can-of-worms.  They require well-corrected optics because of the larger physical sensor size. A DSLR camera is typically attached to a telescope via a nose-piece that slides into the eyepiece receiver, but has the bayonet-style mount required to mate with the camera body.  This often means either a 1.25" or 2" diameter nose-piece is used.  A 1.25" outer-diameter loses a few millimeters on the inside due to wall-thickness ... but generally offers something like 28mm clear internal diameter.  That is large enough for an APS-C sensor.  But a full-frame camera needs nearly 44mm of clear aperture -- so a 1.25" nose-piece would result in extreme vignetting.  A 2" nosepiece is generally large enough as long as the clear unobstructed diameter is larger than the diagonal measure of the sensor (just about 44mm). Given that the CPC doesn't offer a well-corrected field of view with a 44mm diameter, you would likely be disappointed with the results of such a large camera.
The issue with non-modified DSLR's for deep-space astrophotography is that the built-in IR filter doesn't allow full-pass of light in the visible spectrum.  These cameras trim light transmission in an effort to match the sensitivity of the human eye.  This "trimming" is rather substantial.  A camera designed for traditional photography is trimming around 75-80% of the photons at the Hydrogen alpha (~656nm wavelength) band. This means deep-space nebulae that glow in Ha will require substantially longer exposure times.
Dedicated astrophotography cameras don't have UV or IR filters.  They are full-spectrum cameras with only the limitations imposed by the sensitivity of the Bayer matrix -- and usually are sensitive well into the near IR spectrum.  A hard-cut IR filter is recommended as an accessory.
A second issue is that long exposure astrophotography results in a build up of heat which another source of noise.
I still have and use my Canon 60Da, but I've added a number of dedicated CMOS astrophotography imaging cameras.
To get an APS-C size sensor in such a camera, the popular sensor is the Sony IMX071 (this is the same sensor used in, say, a Nikon D5100 ... except without Sony or Nikon firmware (so the RAW images are true RAW images... not cooked RAW's like you'd get in a Nikon or Sony camera -- where stars go missing because they try to clean up noise and mistake faint stars for noise.)
From ZWO, the camera is called a ZWO ASI071MC-Pro.  This is a color (all ZWO cameras with 'MC' in the name are color.  Cameras with 'MM' in the name are monochrome) CMOS imaging camera designed specifically for long exposure deep-sky astrophotography.  List price in USD is about 1480 USD.  This is a few hundred above the price of a Canon 90D (about 1200 USD).  (Incidentally, monochrome cameras actually cost more because these sensors are manufactured with the Bayer color-filter array on the sensor, and they are factory-modified to remove the filter (a manual process) -- which increases the price.)
The camera is has a built-in cooling system that is able to chill the sensor to as much as 40°C below ambient temperature to reduce noise.
The camera uses a USB 3.x connection but also includes a built-in USB hub that provides two additional USB ports.  This allows connection of an auto-guider camera and/or filter-wheel without having to run separate cabling up to the scope.
Lastly, it has a 256MB memory buffer -- which is why ZWO tag this as a 'Pro'.  Many dedicated astronomy cameras do not include a built-in buffer.  If reading out the sensor across a slow USB 2 bus to the computer, the delay can result in a problem called 'amp glow' (almost looks like a like flare coming out of the edge of the frame).  The buffer allows for instantaneous read-out to the buffer ... and the computer's speed is no longer an issue.  Amp-glow is usually not an issue if the computer has a fast bus and fast storage).
Here is a link to the manufacturer's website:  https://astronomy-imaging-camera.com/product/asi071mc-pro  (Although nearly every astronomy dealer I am familiar with probably carries these cameras ... the more difficult issue is finding it in-stock during the pandemic).
For slightly less money, the same sensor is available from another vendor:  QHYCCD.  The QHYCCD QHY168C is the very same sensor.  It sells for roughly 1400 USD (just slightly cheaper than ZWO) but it's on-board buffer is 128MB.  Also is does not include the USB hub to piggy-back an auto-guider camera or filter wheel.
Here is a link to the manufacturer's website: https://www.qhyccd.com/index.php?m=content&c=index&a=show&catid=94&id=13
If the 1400-1480 USD price tag is out of budget, it is also possible to get a 4/3rd's sensor camera with the same technology ... for just around 1000 USD.
If you inspected either of those links, you might have noticed that these cameras have absolutely no controls of any kind on the body -- not even a power switch.  They are completely controlled by computer -- which is fine for astrophotography since using a computer is preferred even with a traditional camera.  They are powered via the USB bus with the exception of the cooling system (it runs on 12v power ... and usually the 12v power supply is not included.)
Traditional Cameras
It is somewhat less risky to use a traditional camera ... even with the short-comings -- because if you decide that astrophotography isn't for you... you've still got a camera that you can use for other things.
When you use a camera for astrophotography, you generally will not use any automatic features of the camera.   The focus system does not matter -- you will have to manually focus the telescope.  The burst-rate of the camera doesn't matter... you'll be taking long-exposure images generally only in Bulb mode.  Basically ... you just want a decent sensor.
The EOS 250D and the M50 actually have the same sensor.  (A 24.1 megapixel sensor).  the 77D has a 24.2 megapixel sensor (it may actually be the same sensor).  Only the 90D has a higher resolution sensor but that is not necessarily an advantage in astrophotography (often, lower resolution sensors will have less noise on long-exposure imaging).
Affordable alternatives
A highly affordable way to do deep-sky astrophotography is to use a decent (solid) photo tripod and add an equatorial tracking head.  Today the common models are the Sky Watcher "Star Adventurer" head or the iOptron "Sky Guider Pro" head.  These only have a Right-Ascension motor (no equatorial... you just manually re-point the camera.  You could use, for example, a ball-head attached to the tracking head for this).  The counter the rotation of the Earth.  You use a camera with traditional camera lenses and these allow for long exposure imaging and do not require an auto-guider.  If you own a decent photo tripod and camera, then the tracking head is typically in the 300-400 USD price range.
This option is far more forgiving (easier) than starting astrophotography with a long focal-length telescope.
Planetary Imaging
I didn't previously mention this because you did not specifically ask about it.  But planetary imaging does work well through an SCT and does not require auto-guiding or an equatorial mount.  This is because the planets are bright enough that they are best captured via video (e.g. about 30 seconds is typically long enough).  Tiny sensors are more than large enough to handle this -- such as the ZWO ASI120MC-S at around 150 USD.  But if you have deeper pockets... an ASI290MC or an ASI178MC, etc. are even better options (in the 300 to 400 USD price range).
Summary

Be aware of the issues of using an alt/az mounted SCT telescope without an equatorial wedge and the need for a guiding solution.

The Canon EOS Ra is a full-frame camera modified for astrophotography, but pairs well with telescopes that can provide a well-corrected flat field with at least a 44mm diameter.

A dedicated CMOS astrophotography imaging camera with an APS-C size sensor would certainly be a huge advantage and ... for not significantly more than the price of the Canon EOS 90D.  A dedicated CMOS astrophotography imaging camera with a Micro 4/3rds sensor would be even more affordable (less than the price of the Canon EOS 90D)

A tracking head for a photo tripod is significantly easier and more affordable than all the gear acquisition needed to use a high focal length telescope.

Planetary imaging does not require auto-guiding or an equatorial mount ... and the planetary imaging cameras are significantly more affordable.


First observation that the movement of a planet or asteroid in its orbit was affected by another planet? Answers to the question How did Kepler determine the orbital period of Mars? describing careful observations centuries ago got me thinking.

Question: What was the first analysis of observations that directly demonstrated that the movement of a planet or asteroid in its orbit was affected by gravitational attraction by another planet?

I don't mean inferences based on the spacing of orbits, I'm thinking of detailed positional measurements; something like planet/asteroid A was X kilometers or Y arcseconds away from it's predicted position and it couldn't be accounted for if not for the gravitational effects of planet B.

I am not sure if First observation that the Sun and Jupiter (and friends) move around a common barycenter? is a completely separate question, or if this confirmation happened all at once, but I've currently asked it separately.

My best guess: The late periapsis of 1P Halley in 1759.

In 1705, with the mathematical assistance of Issac Newton, Edmond Halley published Synopsis of the Astronomy of Comets which made the prediction that the comet observed in  1682 had previously been observed in 1531 and 1607 was the same object returning about every 76 years or so.  He included a rough estimation of the gravitational effects of Jupiter and Saturn, and predicted it would return in 1758.

From Wikipedia: Halley's Comet

  Halley's prediction of the comet's return proved to be correct,
  although it was not seen until 25 December 1758, by Johann Georg
  Palitzsch, a German farmer and amateur astronomer. It did not pass
  through its perihelion until 13 March 1759, the attraction of Jupiter
  and Saturn having caused a retardation of 618 days. This effect was
  computed prior to its return (with a one-month error to 13 April) by a
  team of three French mathematicians, Alexis Clairaut, Joseph Lalande,
  and Nicole-Reine Lepaute. The confirmation of the comet's return was
  the first time anything other than planets had been shown to orbit the
  Sun. It was also one of the earliest successful tests of Newtonian
  physics, and a clear demonstration of its explanatory power.

This predates the discovery of Uranus(1781) and Neptune (1846).


How to convert theoretical template spectrum from luminosity density to flux density units? I'm working with galaxy spectral templates (e.g., Bruzual & Charlot 2003) which seem to always come with y-axis units of $L_{\odot}$/A and x-axis units of Angstroms. Thus the y-axis is a luminosity density instead of a flux density. In contrast, observationally, we tend to always work with spectra that have y-axis units of flux density ($F_{\lambda}$ in erg/s/cm$^2$/A or $F_{\nu}$ in erg/s/cm$^2$/Hz). Similarly, spectral energy distributions (SEDs) from photometry tend to have $\lambda F_{\lambda}$ or $\nu F_{\nu}$ such that the y-axis is flux, not flux density.

How do I convert a theoretical template spectrum from units of luminosity density ($L_{\odot}$/A) to flux density (erg/s/cm$^2$/A)?

For context, I want to fit spectral templates to an observed SED. The observed SED is for an object at a redshift $z$, so I think I can either convert the templates to flux density units, or I can convert my observed SED to luminosity density units. I feel like working in flux density units is more natural -- plus I'm not sure if multiplying the observed SED y-axis values by $4\pi D^2$ (D is distance of object) and x-axis (wavelengths) by $1/(1+z)$ would be sufficient (e.g., normalization concerns).

Use this equation:
$$F_\nu = \frac{L_\nu}{4\pi D_L^2}.$$
This is the relationship that defines the luminosity distance, $D_L$, in a static Euclidean universe (i.e. not ours). Hogg has a good review on the arXiv of how to handle the relationship in an expanding universe, including when you need to perform something called a "$K$-correction" for spectral densities like the question asks.


Ross 128 (What does the number mean?) I see that there are a number of stars that have a number at the end of them. Ross 128 is an example. What does this number mean?

The number refers to the number in the catalog compiled by Frank Elmore Ross. From the Wikipedia article on Ross

  At Yerkes Observatory he was the successor to the late E. E. Barnard, inheriting Barnard's collection of photographic plates. Ross decided to repeat the same series of images and compare the results with a blink comparator. In doing so, he discovered 379 new variable stars and over 1000 stars of high proper motion. Some of the high–proper motion stars turned out to be quite nearby, and many of these stars (such as Ross 154) are still widely known by the catalog number he gave them.


Can visible wavelength spectroscopy study an exoplanet's chemical composition directly? I know spectroscopy of light in visible wavelengths is very effective for studying and determining the chemical composition of bodies within the solar system and bright objects outside of it. However, can this same method be used with exoplanets?

In looking about some related articles, I've learned that it is possible, if a planet passes between us and its star, to filter out the star's light leaving only what passes through the planet's atmosphere.
But what if there is no such occultation? Can we filter out the star's light in other wavelengths like UV or infrared and look at just the planet via its own emitted radiation (assuming the planet is large and hot as these are what we are most able to detect at this time)?

This can be done by high-contrast imaging.

Currently it is only feasible for a very limited class of systems (e.g. HR 8799) since it is difficult to achieve the required angular resolution (0.1 arcsecs for exoplanets at a distance 10 parsec) with today's adaptive optics.

Future large telescopes such as the European Extremely Large Telescope are designed for direct imaging as well as visible and near-infrared spectroscopy of exoplanets.


Relationship between Periapsis and apoapsis longitude What is the relationship between Periapsis and Apoapsis longitude ? Are they 180° apart ? (ie) Periapsis + 180° = Apoapsis ?

Related question: if apoapsis is not known, what orbital elements are required to compute it ?

The periapsis and apoapsis are 180° apart.
According to Kepler's first law,
an orbit is an ellipse with the central body at one focus.
The periapsis (P), central body (F), and apoapsis (A) all lie on the major axis of that ellipse.


Could gravity waves be used to confirm the existence of a ninth planet? Shortly after the news of the discovery of gravity waves, my physics professor explained the significance of the discovery by comparing it to a medical X-ray machine. He said that gravity waves can be used to detect objects that cannot be seen using telescopes. Therefore, could gravity wave detectors be used to confirm the existence of the theorized ninth planet? Even though it doesn't have a high enough albedo to be seen, it would still be detectable gravitationally, wouldn't it?

No, they can't.

Gravity waves from a small, simple object moving slowly are very, very faint, to the point of being undetectable with current (or foreseeable) technology.

The waves that have been detected come from the merger (a very fast movement in the last orbits) of two black holes (two very big masses). And they were just detected over the noise level.

Do not confuse that with the possibility to detect the gravitational influence of such a planet over other bodies. That is how Neptune and Pluto were detected, indeed.


What happens to the 99.9 % of the sun-rays that do not fall on any planets or any other celestial body? I assume that about 99.9% of the sun-rays that do not fall on any planet or any other celestial body keep on traveling farther and farther unto infinity. Apparently such rays get lost. Keeping in mind the colossal energy Sun has produced since 4.5 billion years I am somehow reluctant to  reconcile myself to the idea that Nature would have allowed wastage of so much energy produced by the Sun. Nonetheless I want to get enlightened whether it got really lost or got utilized. If it got utilized I want to know how it may have got utilized at all and whether any sustainable evidence is available in support of any such finding?

The light from the Sun spreads, at least initially, in an isotropic fashion into the universe.

As it gets further from the Sun, some of that light will interact with the interstellar medium (ISM) and therefore some of the energy emitted by the Sun will be used to excite atoms and molecules or even ionise some atoms. This will be the fate of almost all the light which is emitted from the Sun towards the plane of our Galaxy, which contains sufficient molecular gas and dust to block starlight travelling through it for any distance. We know this happens because we can "see" dark clouds in the Milky Way, that can be penetrated by longer wavelength radiation to reveal all the billions of Sun-like stars that lie behind them. Roughly speaking, about half the visible light from the Sun will be absorbed every 1000 light years when travelling in the Galactic plane, so it is essentially all absorbed within a few thousand light years.

But most of the Sun's light is not travelling in the direction of the Galactic plane, and interstellar and intergalactic space has a very low density of gas and dust. The equivalent extinction number for the intergalactic medium is that light travels many billions of light years with almost no chance of being absorbed (see Zu et al. 2010). This means that most of the light from the Sun will travel to cosmological distances (billions of light years) over the course of the next billions of years. Indeed light emitted from the Sun shortly after its birth has already travelled 4.5 billion light years. We know this has happened and will happen, because we can observe galaxies (the light from which is nothing more than the summation of light from many stars like the Sun) that are 4.5 billion (and more) light years away.

As the Sunlight travels towards cosmological distances, its wavelength is "stretched" by the expansion of the universe, becoming redder and redder. We know this happens because distant galaxies have redshifted spectra. If the universe keeps expanding, then its density will continue to decrease and there is little to stop the radiation from the Sun travelling on forever, with a wavelength that scales as the scale factor, $a$, of the universe.

If we follow a co-moving and co-expanding cube containing the Sun's radiation as the universe expands. The total radiative energy inside that cube diminishes as $a^{-1}$ - that is, the energy content of the universe in the form of radiation from stars (and other sources) becomes energetically less important as the universe expands and appears to be being superseded by the energy contained in the vacuum itself (a.k.a. dark energy).

In conclusion, most of the energy emitted by the Sun is not "used" for anything; it propagates into space, becoming more and more dilute.


Is it possible to be 'still' in space? So I was reading this answer about how galaxies are the fastest moving objects in the universe because space is expanding faster than the speed of light. This got me wondering, would it be possible to be 'still' inside of space so that galaxies are moving away from you and others towards you? If so, wouldn't this allow us to travel from place to place in the universe much quicker? I know that this wouldn't be something that's even close to possible today, but I was just curious if such a thing is possible. Thank you.

The so called expansion of the universe is not as trivial as most people think. What is happening, in fact, is that the distance between two points in space (note that I'm not talking about objects with velocities, but just coordinates in space) increases with time in a manner proportional to a given factor (in this case, the Hubble constant - which is actually not a constant in time, but let's ignore that for now).

This means that galaxies are not exactly moving away from a specific point with a particular velocity (i.e., a particular rest frame), but that the space between them is expanding. It is somewhat hard to wrap your head around the difference between these two situations, but this has to do with the geometry of space, and not with what's inside it.

It is widely accepted that the expansion of the universe is homogeneous and isotropic (the Cosmological Principle), which means that there is no "special" position (rest frame) on the universe, and whatever your velocity and your location, the universe will seem to expand the same way.

The best way to imagine this is to reduce to 2 dimensions expanding in a third dimension. For example, suppose that you live in a galaxy embedded on the surface (a 2D universe) of a balloon (completely unaware of the third dimension), with other galaxies distributed homogeneously. If, for some reason, the balloon is expanding isotropically and, every point of the surface will be moving away from every other point, so will every galaxy from each other. Think about this: is it possible to find a special, "still", point (without resorting to the elusive third dimension) where you will see galaxies moving from you in one side and galaxies moving towards you on the other side?

Our universe works similarly, but we're embedded on a three-dimensional space instead of a two-dimensional one.


Why is [O III] a good density probe in interstellar medium? According to Draine in his book "Physics of the interstellar and intergalactic medium" (page 210/211) and Caltech [O III] line ratio's (amongst others) are a good density probe, but I cannot really understand why that is the case. Can someone explain me how this works? It doesn't have to be numerical or with detailed formulas, just the idea what is going on.

The key to understanding this is the concept of forbidden lines and forbidden transitions.

A forbidden transition is one that cannot occur radiatively via an electric dipole interaction. Instead, it must proceed either through magnetic dipole or electric quadrupole emission, with much lower probability, or the transition may be accomplished through collisional (de)excitation.

A forbidden line is the radiation at a wavelength corresponding to a forbidden transition. In laboratory plasmas these are usually not seen - hence the term forbidden lines - because the transitions are normally accomplished through collisional de-excitation on a timescale much shorter than the radiative lifetime through magnetic dipole/electric quadrupole emission. However, in astrophysical plasmas the densities can be many orders of magnitude lower than even the best laboratory vacuums. In this case, forbidden lines are seen and indeed are often an important means of radiative cooling.

The property of forbidden lines which makes them an excellent density-sensitive tool is that they can be "quenched" if the density becomes large enough to make collisional de-excitation more probable than radiative de-excitation. The strength of the forbidden line is thus sensitive to the electron density (which is what dominates the collisions) between densities where "quenching" starts to become effective and higher densities where the line essentially becomes unobservable.

Taking the specific example of OIII (doubly ionised oxygen) optical line emission. There are three forbidden transitions of interest between the $^1D_2 \rightarrow\ ^3P_2$ (501 nm), $^1D_2 \rightarrow\ ^3P_1$ (496 nm) and $^1S_0 \rightarrow\ ^1D_2$ (436 nm) states.
These transitions produce optical forbidden lines which are quenched at different characteristic densities. Measuring a line strength ratio is important because the ratio will be independent of the abundance of OIII. The ratio will depend on electron density, providing the electron density is $>10^{5}\ cm^{-3}$, and temperature (at lower densities, the ratio is just temperature dependent).

To get the density in this case requires more information - usually supplied by a similar line ratio for something like NII, which has a similar set of forbidden lines, but with differing density and temperature dependence.

A more straightforward route is to use the line ratio for forbidden lines of OII or SII where there are a closely spaced pair of energy levels, both undergoing forbidden transitions to the same lower level. In this case the ratios are density sensitive, but not temperature sensitive, again over a range of densities where one or other of the transitions is in the process of being quenched (e.g. for OII 372.6/372.8nm, $10^2 <n_e < 10^{6}\ cm^{-3}$.)


Will the "fabric of space time" rip from expansion? I'm quite familiar with the fabric of space time theory and as I've been quite lonely these past few weeks, I started thinking that since the universe is expanding and since the "fabric" is the universe, would our nice little sheet eventually rip and mess up space time?

There is no "fabric" to rip.

Spacetime can stretch as much as you want (and in ways you cannot visualize - hence we use mathematics).  There is no way to do anything more than distort spacetime.  It can (at least in theory) be distorted is very odd ways (odd in terms of human experience) but there is no distortion that would "rip" or "tear" it.  It is always there.

Regarding the expansion of the universe, think about it like a having two points some fixed distance apart.  As the universe expands they become further and further apart, but it's not them moving, it's the scale factor of spacetime getting larger.  The universe gets larger because the scale factor increases.  In the past it was smaller and things were closer together.

There is a theoretical idea called "The Big Rip", but basically this just says that eventually the scale factor (which in this model keeps increasing) gets so large that everything is isolated from everything else, down to the level of elementary particles.  But spacetime itself does not rip - it's still intact, it's just that the scale factor has become huge.

There are models of the expansion of the universe that say the opposite - that eventually the expansion stops and the scale factor stops increasing and starts decreasing and we end up with everything in a Big Crunch.  Again no spacetime was hurt in the making of this model. :-)


Are we sure solar cycle is not related to the orbit of Jupiter? Are we still debating the relationship between Jupiter and solar cycle? If not, what is the common or mainstream opinion about this topic? Or in other words, have we disproved the solar cycle is related to the "year" of Jupiter?

I think for people who want to believe the Jupiter link to solar cycles it will be impossible to disprove.  They'll cling to it forever.

I think in a practical sense the "link" is little more than numerology.  The force Jupiter can exert on the Sun's atmosphere and internal structure is a minute fraction of the magnetic forces at work generated by the Sun itself.

A simple model helps.  Place two magnets a foot apart.  Now place a magnet a thousand feet away from both.  The effect of the distant magnet is one millionth that of the two close together.  That gives you a feel for the relationship between magnetic forces within the Sun's atmosphere and Jupiter's distant influence.  And gravity is an even weaker force.

As I understand it the current daring of models is the Double Dynamo, but certainly the dynamo type models ( all relating to electromagnetic fields generated by the Sun ) are the main contenders.

With people who propose the Jupiter link ( often extended to include Saturn ), there is a tendency to do a lot of algebra to find ways to relate various periods without actually justifying the supposed link on the basis of forces at work ( i.e. doing a simulation or a model based on real physics ).  This is why I call it numerology.

The dynamo models actually are based on real physics and make predictions, both past and future, which can be tested.  This also means the details of these theories can be refined in a proper scientific manner to tease out the truth.


Recognising lunar month only by looking the sky Suppose somebody went on an island and don't have sense about time, dates, months. The only thing he has is seeing the sky. Is it possible to recognise which lunar month it is by just looking the sky? I ask this because I know how to see beginning of the lunar month, but how to recognise which month is it? So we can look at the Sun, Moon, the stars. Is there a way to figure that out?

Not easily.

The Lunar Islamic calendar is based on a strict succession of months, with each month based only on the position of the moon, and no intercalated months or months extended to 31 days in order to keep the cycle of months aligned with the solar year. The start of the months are determined by the moon, the names of the months are not defined by any astronomical feature.

So to determine which Islamic month it is you need to know the actual amount of time that has passed. Now the positions of the planets change day by day, and by careful observation of the planets, and given a detailed ephemeris of their positions, one could work out the exact date, and therefore the Islamic month.

This is similar to determining the day of the week by astronomical observations, since the start of the day is based on the motion of the sun, but the name of the day is not defined on any astronomical feature.


Strange bright light in the sky flying fast I was out with my dad on our porch watching the stars with our telescope. after awhile of looking and taking pictures I noticed something strange in the sky. There were two airplanes flying close to each other, but next to those two airplanes was a very bright light (see how the stars are bright? Well that one was a lot bigger and I lot brighter, and to give you an example it was moving as fast as the planes, I showed my dad and he as well saw it, so I knew I wasn’t dreaming, it wasn’t moving straight either, both me and my dad could see it was going in zigzags, not a straight line, it desapeared after awhile behind he trees, we didn’t get to look at it through the telescope since it was moving so fast...what could it be...? Both me and my dad agree it was going as fast as the planes next to it, it was brighter then both the planes and the stars and it could not have been a third plane since broth planes were just so small and it looked like a very big star. FINALY it was not moving in a straight line...so the what would it be...?

This sounds kind of like a Foo Fighter.
There have been various reports of weird glowing orbs following aircraft and zig-zagging around them ever since World War two. In fact both sides assumed they were secret weapons belonging to the other side up until the end of the war.

As to what they actually are exactly, well nobody really knows. The most plausible answer is that these are examples of Ball lightning or some other electrostatic phenomenon. As far as I know, it has never been conclusively replicated in the lab.


Space telescope located in outer solar system I want to pose a question from someone who's astronomy knowledge would fit neatly onto a single piece of paper so please bear this in mind with any comments or answers. I am interested and want to learn. I could not find a sane answer to this via a search engine.

The question: If in theory we placed another Hubble telescope (or similar instrument of at least equal or ideally more power) either on or very near the dwarf Pluto, as in in its orbit if we did not consider a ground mounted instrument, what if anything would or could we gain in astronomical knowledge ?

I do realise data transmission from such a distance would take time, power could be a concern and the fact if anything went wrong it would be beyond any sensible help, not to mention logistics in getting it there in the first place.

I also realise this is a completely hypothetical question too but I am genuinely curious if we would "see more" (so to speak) with such an instrument so far out.

The disadvantages would likely outweight the advantages.

It's cold out there. This makes it easier to keep an infra-red telescope cool
The sun's just a super-bright star.  This means more of the sky is visible and not in the glare of the sun. However you orbit so slowly that there will be a few objects that you won't be able to image because they are behind the sun for years.
There's no Earth to get in the way. The Earth blocks half the sky for Hubble. Putting a telescope on Pluto would have the same problem.
There's no space debris there (yet)

There may be other advantages, but compared to the immense difficulty of putting a probe into an 250 year orbit (you need to get it out there, with enough fuel left to put it into orbit) And the time it takes (you can't take much advantage of slingshots because you need to be in an orbit not just on an escape trajectory). And all the other disadvantages

Saw something interesting, can't just get hubble to take a look, because there is a five hour light time.
No power, sun is too weak so depend on a box of plutonium or similar.
No nice high resolution images, sorry those take too long to download. Low resolution only.
We've never landed anything as complex and fragile as Hubble on any other body, And the landing has to be fully automatic.

And you don't get to see anything new because space is just as clear from Hubble's orbit as it is from Pluto.


What is the brightest star (by absolute magnitude) that we can see by naked eye? Is it Rho Cassiopeiae or Mu Sagittarii?
I see in the Stellarium 0.13.0 that Mu Sagittarii has absolute magnitude -11.43!!

Mu Sagittarii is a star system, not a single star. If that can be included, then Eta Carinae should be included, and it has an absolute magnitude of -12.0. It's a star system about 7,500 light-years from Earth.

It looks like the brightest (absolute magnitude) single star visible to the unaided eye is WR 24 (in Carina Nebula). Its absolute magnitude is −11.1 and apparent magnitude is 6.48, so just barely visible.

source: Wikipedia - List of most luminous known stars

edit: Rho Cassiopeiae's absolute magnitude is -9.5.


Would an analogue of the definition for planets also work for moons? This is a follow up question to What exactly is a "moon"? The conclusions I draw from James K's answer is that the IAU should define what a moon is. They haven't done so yet, but they should.

The obvious way to go about defining "moon" would be to build on the definition of a planet:

Planets must be in orbit around the Sun
Planets must be round due to their own gravity
Planets must have "cleared the neighbourhood"

This immediately suggests the equivalent definition of a moon:

Moons must be in orbit around the host planet
Moons must have [minimum size]. Unfortunately being round doesn't work since not all moons are round (c.f. Mars' moons, unless one reclassifies those as not moons)
Moons must have "cleared the neighbourhood" of their own orbit, i.e. in their particular orbit around the host they are by far the most massive body

Would such a definition work? If so, why hasn't the IAU also defined moons? It seems so natural after all. If not, what's the catch?

In 2006 the IAU had a trilemma.

They could decide that Eris was a planet, and potentially allow for future discoveries of tens of new planets.
They could be inconsistent, declare that Pluto was a planet, but Eris (and Ceres) wasn't
They could come up with a definition of "planet" that would exclude Eris, and consequently also exclude Pluto.

Each option is problematic: We teach eight-year-olds to chant the names of the planets, would we want to make them chant the names of 20 or thirty Kuiper belt objects?  Being willfully inconsistent would lead to further argument. Re-describing Pluto would go in the face of 75 years of tradition. There was no option that would keep everyone happy, but they chose the third.

We don't have a problem with a more relaxed defintion of "moon" we are used to new "moons" being discovered. Eight-year-olds don't need to chant their names.  The casual definition of "if you find it, you can name it" works.

The issues with planets are particular to our expectations that "planets" are important. We have no such expectation with moons.


Telescope selection for concrete uses Suppose I want to build or buy a telescope for the following uses:

Astrophotography (mainly galaxies and nebulae).
Transits of exoplanets and binary systems like Cygnus X-1.
To search for BH+star systems.
To measure the rotation curve of some spiral galaxies

Is that ALL feasible and/or cheap at amateur/semi-amateur level?

(1) is very doable. The others don't seem doable, at least not easily. If you practice astrophotography for a while you'll have a better understanding of the difficulties involved.

It's best (but not mandatory) if you begin doing plain visual astronomy, with something like an 8" dobsonian, for a year or two, to get yourself familiarized with how it all works. Once that's accomplished, the transition to astrophotography should be easier.

For AP (astrophotography) you'll need: a mount, a camera, a telescope. The mount is by far the most important. The camera is very important. Oddly enough, the telescope is the least important.

The accuracy, overall quality, and load carrying capacity of the mount will determine how well the whole system works. Just get the best mount you can afford.

Entry level mount examples: Celestron CGEM, Sky-Watcher EQ6.

Midrange mounts: Losmandy

High-end mounts: Software Bisque Paramount, Astro-Physics GTO

Cameras: for solar system objects like planets or the Moon, you'll need cameras with smaller sensors that can take lots of frames quickly (planetary cameras). For galaxies and nebulae (DSO - deep space objects) you'll need cameras that perform well with long exposure (DSO cameras); the best DSO cameras have cooler elements installed on the sensor.

There are OSC (one-shot color) cameras that don't require filters. There are monochrome cameras that require a filter box with R, G, B filters; you need to take separate exposures through each filter. OSC cameras are easier to use; monochrome cameras tend to provide higher performance.

Planetary camera examples: ZWO ASI 224 MC, ZWO ASI 178 MM

DSO camera examples: ZWO ASI 1600 MM, ZWO ASI 071 MC-Cool

Telescope:

For DSO it is very highly recommended to start with a relatively small refractor with short focal length. Something like the Orion ED80. Add a focal reducer / field flattener too. The long exposures require very high precision tracking from the mount, and a long focal length only creates problems. Keep it short until you learn how to make everything work. You could use a larger instrument like an SCT, but make sure your mount can carry that weight, and make sure you've fine tuned the process so the very long focal length of the SCT doesn't amplify the tracking errors.

For planetary imaging, the bigger the aperture the better. Use an SCT such as a Celestron C8 or bigger (up to C14 if the mount can carry it).

Capture software:

For DSO: Nebulosity (beginner), Sequence Generator Pro (advanced)

For planetary: FireCapture

Image processing software:

For DSO: Deep Sky Stacker (beginner), PixInsight (advanced)

For planetary: AutoStakkert for stacking. Subsequent processing could be done with Registax, or Astra Image.

Guiding:

For DSO, you'll do very long exposures. The mount alone is not precise enough. You need to guide it. This is done by using a smaller telescope (guidescope) installed on the main telescope, using a small camera (guide camera).

The finderscope that's normally used for visual astronomy could be used as a guidescope.

If you have a DSO camera and a planetary camera, you could use the planetary camera as a guide camera for DSO photography.

You'll have to connect the camera to a software that can control the mount and issue corrections to it. The most popular app is PHD2 a.k.a. Open PHD.

When doing DSO imaging, you'll spend most time during your first year basically learning all the subtle aspects of guided exposures.

Learning:

'The Deep-sky Imaging Primer' by Charles Bracken. A bit old but still very relevant.


How to Distinguish between Temperature and Doppler effect using Black-body radiation? I understand that the radiation of a body can be described using the curve for black-body radiation. In the sense that a hotter body will be blue shifted and a cooler body will be red shifted.

The doppler effect gives similar results in the sense that a body is approaching us is blue shifted. While an object that is travelling away from us is red shifted.

How is it possible to distinguish between temperature and doppler effect?

Without any other information, you cannot distinguish between the two effects.

$$ T = T_0 (1 + z) $$

A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$.

For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody.  For the CMB, we can use the fact that atomic combination happens only below a particular temperature.


How do moons get captured? A moon-sized object is running loose in the Solar System, perhaps after a planetary collision.  As it approaches a planet, it's presumably following an approximately hyperbolic path.  If it goes on past, it's still on the same hyperbola, on a curve mirroring its approach (presumably).  How can the planet ever capture it, whatever the body's velocity?  Why doesn't it either collide or go on past?

How can a planet capture a moon?

There are 178 moons in the Solar System, according to the NASA Planetary Fact Sheet, so it seems to be a common event. The following sections will show that moon capture is actually unlikely, but when a planet has one or more moons capture becomes easier.

Initial Conditions

Starting from the initial conditions, the planet is in orbit about the sun, and an asteroid is in a different orbit about the sun.

In order for capture to become possible, the asteroid and the planet must come into proximity. When the asteroid comes inside the Sphere of influence of the planet, the gravity of the planet is the main factor in determining the path of the asteroid.

Possible Outcomes

Relative to the planet, the asteroid will be following a hyperbolic trajectory, and hence has sufficient kinetic energy to avoid capture. A large variety of outcomes may occur, but the ones that lead to capture are those where the asteroid somehow loses enough kinetic energy for its velocity to fall below the escape velocity of the planet while retaining enough energy to achieve a closed (elliptical) orbit. The main (not the only) possible outcomes are

the orbit of the asteroid is perturbed, by a greater or lesser extent, and it continues on its way out of the sphere of influence of the planet.
the orbit of the asteroid is perturbed, and the asteroid impacts the planet surface. That would usually be the end of the process, but current theories on how Earth captured the Moon are that a body named Thea impacted the Earth, and the Moon formed from some of the collision debris.
the orbit of the asteroid is perturbed, and the path of the asteroid intersects the atmosphere of the planet, losing kinetic energy as heat in the atmosphere (similar to aerobraking).
the orbit of the asteroid nears an existing moon of the planet and is accelerated (in the sense that deceleration is just acceleration with the opposite sign) by the existing moon, as used by the MESSENGER spacecraft to slow its speed before orbiting Mercury.

The last two cases admit the possibility of capture.

Possible Capture

After losing energy in the planetary atmosphere, if the asteroid has lost enough energy it may enter a closed orbit around the planet. The problem is that the orbit will intersect the atmosphere again, losing energy each time it does so, until it impacts on the planetary surface. Capture can occur when an existing moon is present and is in just the right place for its gravity to reduce the eccentricity of the orbit of the asteroid.

So, the most likely case where a planet can capture a free asteroid is when there is already one or more moons present. The incoming asteroid must avoid entering the Hill sphere of the existing moon - the region where the moon would dominate the path of the asteroid.

Gravity assist can accelerate an asteroid when the asteroid is passing outside the orbit of the moon, but can decelerate the asteroid is passing inside the orbit of the moon. In this case some of the kinetic energy of the asteroid is transferred to the moon. As is the case with aerobraking capture, gravity assisted capture requires the existing moon to be in just the right place.

Another mechanism

A rather elegant paper published in Nature (mentioned below) shows how two bodies orbiting each other as they approach the planet could have led to one being captured by Neptune. This mechanism could apply in other cases also. This Dissertation (pdf) discusses a similar process for Jupiter.

Irregular bodies

It turns out that irregular shaped bodies can be captured more easily than spherical bodies. Orbiting within the Hill sphere of the planet is not enough for capture to be permanent. Only orbits in the lower half of the Hill sphere are stable. Bodies in higher orbits can be perturbed by nearby planets, and the body can eventually be ejected. But irregular shaped bodies exert minute fluctuations in gravitational attraction on the planet, and actually orbit in a chaotic manor. When other moons or rings are present these chaotic orbits gradually transfer energy to the bodies in the lower orbits, causing the new body to orbit lower, and hence become immune to external perturbation. [citation needed]

Prograde vs retrograde orbits

The same analysis of chaotic orbits, and earlier work also concluded that retrograde orbits are more stable than prograde orbits. Whereas prograde orbits are only stable in the inner half of the Hill sphere, retrograde orbits can be stable out to 100% of the Hill radius. Hence retrograde capture is more commonly observed (this is not the whole story, it is still a matter if research).

Multiple existing moons, rings, and the early Solar System

While the probability of a single moon being in the right place at the right time is low, when there are multiple moons the probability of an initial helpful interaction rises linearly. But the probability of additional interactions rises geometrically, so the more moons a planet has the more likely it is to capture more. The existence of rings also aids capture by exerting a drag on the new moon, taking it's energy and lowering it's orbit, in much the same way that uncaptured gas would do in the early Solar System.

The biggest planets have the most moons

It may be obvious, but the biggest planets have the most moons. This is because they have deeper gravity wells, and sweep in more objects. Even though the probability of capture is low (most objects are just pulled into the planet), a steady trickle have have captured over millions of orbits.

Conclusion

Each capture mechanism requires a fortuitous set of conditions, and so is actually a quite rare event. One mechanism is a that a pair of co-orbiting asteroids become separated when one enters the planetary Hill sphere. The odds for an individual asteroid are improved when the asteroid arrives with low kinetic energy that must be given up to other bodies orbiting the planet, and when there are already many moons or a ring system.

See also

Could Earth's gravity capture an asteroid? - earthsky.org
Dynamics of Distant Moons of Asteroids - Icarus (pdf)
Hypothesis: New evidence on origin of the Moon support cataclysmic collision theory
Neptune's capture of its moon Triton in a binary-planet gravitational encounter - Nature
Planetary Fact Sheet - NASA
The use of the two-body energy to study problems of escape/capture (pdf)
Wikipedia
Aerobraking
Elliptic orbit
Escape velocity
Hill sphere
Hyperbolic trajectory
Gravity assist
Retrograde and prograde motion
Sphere of influence
Thea


Relationship between precession and dating ancient alignments Perusing the topics here. I am convinced that this is the right place to get some answers. Although it does appear that my question should be relegated to the "childrens table".

I am curious to understand more about the topic in archeoastronomy that is used to date ancient alignments utilizing calculations based on precession (which I very basically understand as an imaginary circle drawn in space by the earth's axis-line as the earth wobbles).

For example a "temple" in Peru has been dated as being aligned with the summer SOLSTICE(es)[edited] as of 6000 years ago by unnamed experts in the field. Other unnamed experts date it at 28000 years old - to which the first set of "experts" grudgingly concede is "remotely" possible - with an absolute qualification that there was no one here to build it that far back.(Statements of absolute certainty like this and, without proof make me itchy)

First part of question is: Why the huge discrepancy in the dating? Isn't the concept of precession a proven, quantifiable fact? (OK, that was two questions.)

Second part of question is: How do the first set of experts find themselves entrenched in their dating if they must also grudgingly agree that the second dating is remotely possible? (Both groups agree that neither the Inca nor any known predecessors actually built it.)

Anything which can help me to come to grips with this dating rationale would be great.

I am not versed in astronomy whatsoever. I can barely find Sirius on a clear night. And, math is also a foreign concept as well.
What I DO understand is Ancient Civilizations.
So, I would like to understand this topic more so that I can discuss it with others in the field - without sounding like a complete idiot.

[Edit to clarify before more protests stack up] I acknowledge that I did not recall and therefore omitted the fact that non-solar stellar alignments AT THE TIME of summer solstice are the basis for this alignment-dating. At least that is the theory I am discussing. (There are others) One window frames the Sun at sunrise while another frames some star. The third window... I cannot locate a theory of what it was supposed to be for. I just wanted to understand the basis for DRIFT over time. And, the accepted answer does that very nicely. Thank you everyone

The answer is fairly simple.   It takes about 22,000 years for the Earth to perform 1 spin or move 360 degrees along it's axial precession.   The 22,000 year cycle means that the temple was lined up at 6,000 years ago and at 28,000 years ago, as well as 50,000 and 72,000.

In other words, if we build a statue that points directly at the North Star, In a couple thousand years it will visibly no longer be pointing at the North star, but in 22,000 years, after a full circle, it will be pointing at the North star again.   More specifically the North Star will apparently move in a circle that takes 22,000 years to complete, moving away from and back towards the celestial north pole.

Any skepticism on the 28,000 year date wouldn't have anything to do with alignment as the alignment works just as well with either date, and it works for any period a full circle in the past, like 50,000 years ago.  The skepticism is likely related to doubt that it's that old.

Similar to the hour hand on a clock that takes 12 hours for the hour hand to return to where it started, the Earth takes about 22,000 years to return to the direction the North Pole is pointing.   That means, no time between 6,000 and 28,000 years ago will line up.  Those are the two most recent alignment periods.   It moves in a circle in 3D, more like the picture below than like a clock.

As an astronomical note, I want to point out that 22,000 is somewhat inaccurate.   The Earth's axial precession in and of itself is about 25,771 years, but that's only part of the movement.  There's also a larger, slower apsidal precession which adds to or subtracts from the 25,771 year wobble.   These together affect the alignment of the Earth to the mostly fixed stars, and that period varies, but it was about 22,000 years between 6,000 and 28,000 years ago.   It can take as little as 20,800 years or as many as 29,000.   Average is about 23,000.

Many Native American structures lined up quite well to astronomical alignments, so dating ancient structures by measuring when it lined up a reasonable approach.    As a related, somewhat similar example, the Mayan 2012 prophecy is also based on an astronomical alignment and they were, as I recall, a year or two late, but it was still a very good estimate given when they made it.   Understanding these alignments requires seeing things in 3 dimensions and that can be a little confusing to try to picture.


Motivation for using $\sigma_{8}$? Press-Schechter formalism allows one to universally quantify the number of massive dark matter halos in a given volume -- The Halo Mass Function. The density contrast is given by the rms mass variance in Fourier space, which is the dependent on the linear power spectrum $P_{\rm lin}(k,z)$ in a smooth window $\widetilde{W}(R;k)$
\begin{align}
\sigma^{2}(R,z)
= \frac{1}{2\pi^{2}}
\int_{0}^{\infty} dk\ k^{2} P_{\rm lin}(k,z) \left| \widetilde{W}(k;R)\right|^{2}
\end{align}

The halo count can be scaled via observations in a defined window region of $R = 8\ h^{-1}\rm Mpc$. We can solve for this analytical/numerically and measure it from CMB fluctuations, given by the parameter $\sigma_{8}$. The power spectrum, as well as the halo mass function, is then scaled through the normalization of the analytical variance and the observed variance.

My question is then, why do we choose a smoothing radius of $R = 8\ h^{-1}\rm Mpc$? Is it arbitrarily chosen to compute the number variance in a cluster, or is their more too that? What prevents me from defining the window radius of $R = 6\ h^{-1}\rm Mpc$ (just ignore the fact that the  WMAP and PLANCK results give $\sigma_{8}$ in the radius of $8$)?

The specific radius of $8 h^{−1} \text{Mpc}$ is used because the value of $\sigma_8$ turned out to be close to unity. To quote e.g. from Amendola & Tsujikawa (2010), p. 39:

  If the cells have a radius of $8 h^{−1} \text{Mpc}$, it turns out that $\sigma_R$ is close to unity. Conventionally the normalization of the power spectrum is therefore given by quoting $\sigma_8$.


What lens should I use for my Raspberry Pi? I am building myself a camera that attaches to the back of my Celestron NexStar 4 SE telescope and I am not sure what type of lens to choose to replace the stock lens on the Raspberry Pi camera. I bought this CS Mount and by mistake bought M12 lenses thinking it was a mount for M12 lenses but now I know CS mounts require bigger lenses than M12. So should I instead get a M12 mount and use M12 lenses or should I buy a CS lens and what kind if I am looking to do a high quality video/live stream?

Raspberry Pi Camera: https://www.adafruit.com/products/3099

Thought I would revisit this and answer my own question...no lens is needed at all :)

The telescope itself is the lens, so once that finally made sense to me I then took the lens off the Raspberry Pi camera and I can now use the camera with the telescope


Location on (or in) the celestial sphere First off, I apologize if I am breaking any site-etiquitte. Secondly, this is probably extremely elementary.

For a lab, I was given the question:

  When we are sitting in a planetarium, where are we located on(or in)
  the "celestial sphere"?

To me, this seems like a silly question - because the answer is really dependent on where the operator places you. Is this assumption wrong? Is there a particular place in the celestial sphere that you would be?

I also thought that it might just be the very center of the celestial sphere.

Well, at least you did some thinking and proposed a couple of personal thoughts instead of just asking for the answer to a homework question. Both observations you make are pertinent.

Remember, a celestial sphere doesn't actually exist. It is an imaginary concept, a simplification based upon our perspective of the universe.

A planetarium, whether it is an old-fashioned one that depicts an earth-bound perspective of the universe, or a modern digital projection system that can "fly" you through space to view the universe from a different place, is projecting a perspective of the view of the universe from a particular place, that is, you see the "celestial sphere" as it would appear from that place. In the case of a view from elsewhere in space, it would be a different celestial sphere from our celestial sphere.

So yes, you would be at "the centre" of the celestial sphere. Maybe not the "very" centre in the case of old fashioned projectors with fixed star fields, as they wouldn't account for the 150 million kilometre imprecision caused by the movement of the earth around the sun. But, as this is an insignificant distance in astronomical terms, that doesn't matter a great deal, so "the centre" is probably a sufficient description.


Is the Roche limit used this way? Has the Roche limit affect been observed having an effect on an constructed object gravitational wise to cause damage? Could the Roche limit be used to slow an object?

No. The Roche limit refers to the radius at which a satellite which is held together only by its own self gravity is disrupted by tidal forces. Spacecraft are not such satellites, but are much harder to tear apart.


What is the temperature of an accretion disc surrounding a supermassive black hole? What is the temperature of an accretion disc surrounding a supermassive black hole? Is there plasma in the disc?

It depends on the distance from the central body. This gives the temperature $T$ at a given point as a function of the distance from that point to the center ($R$):
$$T(R)=\left[\frac{3GM \dot{M}}{8 \pi \sigma R^3} \left(1-\sqrt{\frac{R_{\text{inner}}}{R}} \right) \right]^{\frac{1}{4}}$$
where $G$, $\pi$, and $\sigma$ are the familiar constants, $M$ is the mass of the central body (and $\dot{M}$ is the rate of accretion onto the body), and $R_{\text{inner}}$ is the inner radius of the disk - possibly (if the object is a black hole) the Schwarzschild radius $R_s$, in which case we can simplify this a little more. So the temperature in the accretion disk is far from constant.

Whether or not there is plasma depends on the exact nature of the disk, the central object and the region around it. For example, a supermassive black hole may have different matter in its disk than that of a stellar-mass black hole. I should think, though, that black holes in binary systems accreting mass from a companion should have plasma in their accretion disks, and supermassive black holes might also have plasma from nearby stars.


How can we install a radar on radio telescopes like FAST or GMRT? Recently I came across an article mentioning that ever since the failure of Arecibo telescope, we only have one asteroid radar imaging telescope which is Goldstone observatory.
The article also mentions the need of Asian (or any other country) radio observatory to upgrade their telescopes to telescope + radar system to take the place of Arecibo.
This made me wonder how can we upgrade such system instead of starting all over again (building a new telescope that could handle radar imaging as well).

The article is quite informative. It is a summary of "a paper published in the journal Scientia Sinica Information" which appears to be Discussion on the requirements and feasibility of constructing China's near-Earth asteroids radar system. While it is written in Chinese the tables and figures alone are very informative.
They are going for a "planetary scale" or very long range system; out to 0.1 AU or 15 million km for an asteroid is ambitious, and what's really interesting it sounds like their transmitted radar beam will com from "four or five 35-metre (115-foot) diameter radio dishes" rather than one big one.
I've pasted some bits into google translate so far and it looks like they will try to have the option to use the multiple transmitting dishes coherently

(3) X 和 Ka 频段上行天线组阵技术. 在国家高技术发展计划支持下, 我国在上行天线组阵技术研 究方面也已经取得了突破, 成功实现了对地球同步静止轨道通信卫星在 C 频段 (发射频率 6 GHz) 的 3 个天线上行组阵技术验证, 达到了 80% 的合成效率. 后续需要针对频率更高的 X 和 Ka 频段上行天 线组阵, 重点开展上行链路相位延迟变化准确估计技术、时延和相位对齐的精确控制技术和大回路系 统标校技术等研究.

X and Ka band uplink antenna array technology. With the support of the national high-tech development plan, China has also made breakthroughs in the research of uplink antenna array technology, and successfully achieved the C-band communication satellites in geosynchronous geostationary orbit. (Transmit frequency 6 GHz) 3 antenna uplink array technology verification, achieved 80% synthesis efficiency. The follow-up need for higher frequency X and Ka band uplink antenna array, focus on accurate estimation of uplink phase delay changes Research on precise control technology of technology, time delay and phase alignment and large loop system calibration technology.

Rather than build one giant dish for transmitting like Goldstone Solar System Radar which is

...a large radar system used for investigating objects in the Solar system. Located in the desert near Barstow, California, it comprises a 500-kW X-band (8500 MHz) transmitter and a low-noise receiver on the 70-m DSS 14 antenna at the Goldstone Deep Space Communications Complex. It has been used to investigate Mercury, Venus, Mars, the asteroids, and moons of Jupiter and Saturn. The most comparable facility was the radar at Arecibo Observatory, until that facility collapsed. GSSR now stands alone.

...they will use several 35 m transmitting dishes and an even larger number and more widely separated array of receiving dishes.
Table 7, "China asteroid detection multi-bases radar system layout station composition" is a list of transmitting sites. Obviously not all can be coherent and they will be used in a variety of ways.

It's really hard to add a new multi hundred kilowatt transmitter to an existing very large dish like a 70 m DSN or FAST. You can read more about that in answer(s) to What is a Beam Waveguide dish and why does the Deep Space Network use them?. The big 70 m DSN dishes use the focus up between the primary and secondary; it is a real challenge to add more hardware there. The image below shows a DSN 70 m dish, for scale, the red lines in the dish itself are a safe walking path and going up each arm of the secondary reflector are stairways for people, not ants.
I think that the Chinese project is quite ambitious but it's a next step in technology, rather than a makeshift retrofit that would interrupt availability of currently very busy large dishes, and it seems that adding a transmitter to FAST is not an option. From this answer to How will the closure of the Arecibo dish impact deep space communications? (found here):

With the loss of Arecibo, Goldstone's DSS-14 now becomes the world's largest and most powerful radar dish.  (China's 500 meter FAST dish is larger, but has no transmitter and is purely passive.)  Sky and Telescope reports that "Arecibo offered 18 times the sensitivity of other existing facilities, such as NASA's Goldstone receiver."  It also states

Arecibo is also irreplaceable for scientists. Even though it’s technically the second-largest radio dish in the world (China’s Five-hundred-meter Aperture Spherical Telescope, or FAST, recently broke the record Arecibo held for decades), the observatory has unique capabilities, among them its radar. “FAST cannot do radar, it’s specifically incapable of doing active observation,” Springman explains. Because of that, FAST can’t take Arecibo’s place in planetary defense by characterizing asteroids and their orbits.

See also answers to

What will succeed the Arecibo Observatory?
Arecibo: Advantages of Giant Dish?
What are monostatic radar observations, and how will Deep Space Network's DSS-13 be used to observe asteroid 1999 WK4's flyby of Earth?

From Yahoo News China urged by scientists to create biggest radar system to save lives threatened by asteroid strike (similar/same article reprinted in several places, including intellasia.net, toysmatrix.com):

The proposed system would use four or five 35-metre (115-foot) diameter radio dishes in Kashgar, Xinjiang, to send powerful beams into space. Returned signals would be picked up by large antennas across the Chinese mainland in a number of cities, including Jiamusi, Beijing, Tianjin, Shanghai and Kunming.
The radar detection range would be expected to exceed 0.1 astronomical unit – or a tenth of the mean distance between the Earth and the sun.
The project “will fulfil the responsibility of our nation to the mankind,” said the team led by Li Haitao with the People’s Liberation Army’s Beijing Institute of Tracking and Telecommunications Technology in a paper published in the journal Scientia Sinica Informationis on Tuesday.
Continuing to rely solely on information provided by the Americans for a threat that could cause a global catastrophe would have a negative impact on China’s rising status in the international community, the researchers said.
The US had two planetary radars but one of them – the Arecibo Observatory in Puerto Rico – collapsed in December because of ageing, hurricane damage and a shortage of funding for repair and maintenance. This leaves the Goldstone Solar System Radar in a Californian desert as the last line of defence, and things could go wrong.
“A single failure of the system would leave us without an essential tool to ensure our protection. Such a failure occurred as recently as 2019-20, when the Goldstone transmitter was inoperative because of klystron (high power microwave vacuum tube) failures,” said professor Jean-Luc Margot, a planetary radar expert with the University of California, Los Angeles.
“I am glad to hear that China is considering building a radar system to study near-Earth asteroids,” he added.

From this answer to Why is the reflector on this millimeter-wave antenna spinning?:

Below are photos of one of the 70 meter Deep Space Network telescopes for talking to deep space spacecraft. This one is in the Goldstone complex. From the relative sizes in the image, the secondary mirror is close to 8 meters in diameter. Considering the size and mass of the secondary (those are stairs for humans on each leg, and the red lines in the dish are "safe paths for walking") other, more advanced techniques to scan electronically have been developed, but the concept is the same.

above: Photo credit JPMajor, creative commons CC BY-NC-SA 2.0. click for larger.

above: From commons.wikimedia.org click for larger.


The moon has just the right speed not to crash on the Earth or escape into space. What are the odds? My understanding is that the moon was created a long time ago when Earth was hit by a big asteroid.
The debris then agglomerated into the Moon, which happens to be orbiting at the exact speed required to neither crash back into the Earth, nor escape into space.
Having the exact correct speed seems extremely unlikely. Yet, our moon is there, and many other planets have moons.
Are these just the few survivors out of thousands of events that didnt have the « goldilock » speed?

There isn't a "Goldilocks speed" for orbit.  If you put two objects in space, and give them a velocity relative to each other, then provided that velocity is less than the escape velocity (at their relative distance) the two objects will orbit each other.
Those orbits will be elliptical, and it is possible that the ellipse is skinny and "eccentric" enough for the two bodies to collide when they are closest to each other. But for an object that is several hundred thousand km from Earth, there is a quite a wide range of possible elliptical orbits.
So when (and if) the grand collision happened, there was a huge amount of matter that was ejected up into space.  Some probably was moving so fast that it escaped, Some certainly went into orbits that didn't have enough energy and so were small skinny ellipses and the matter fell back to Earth.  But there was a lot that ended up in some kind of elliptical orbit.  This matter was not all in the same orbit, but it started to coalesce, and form into a single ball, under its own gravity.
Other moons weren't formed like this, they either formed at the same time as their planets as a "mini solar system" (such as the four major moons of Jupiter) or they were captured from the asteroid or Kuiper belts).  Initially, the captured moons may have had rather elliptical orbits.
But most moons are in rather circular orbits.  Even if the moon was originally in an elliptical orbit, tidal effects will tend to make the orbit more circular.  A planet and moon system has a certain amount of angular momentum and a certain amount of energy.  The angular momentum can't change, but energy can be converted into heat and since tides dissipate some energy as heat, the orbit will tend to change to a shape that minimizes energy, for a given amount of angular momentum.  That shape is a circle.  (See Is the moon's orbit circularizing? Why does tidal heating circularize orbits?)
So the effect of tides is to give moons the "Goldilocks speed" that keeps them in a circular orbit.


Why is it said that the far side of the moon's crust is thicker? This video states that NASA has determined that the far side of the moon's crust is thicker than the near side's crust. How did they determine this?

Please note I'm not asking for theories or hypotheses on how this came to be, but rather if the difference in thickness indeed is measurable scientifically today.

[So off-topic: If it is in fact true, I favour the theory that the centrifugal force of the earth's gravity along with the fact that the far side is always pointing outwards, is responsible for this and not some other moon smashing into our moon etc. as is shown in that video.]

The near side and far side have very different topography. Compared to the near side, the far side is far more rugged and has far fewer maria. The near side in places has very thin crust (the maria).

Other than these surface features, the biggest indicator of a disparity in crustal thickness is the two kilometer offset between the Moon's center of figure and it's center of mass. This is typically attributed to a thicker crust on the far side of the Moon, consistent with the observations of the different surfaces of the near side and far side.

NASA discovered that the center of figure (geometrical center) and the center of mass are offset from one another in the late 1960s. The simple act of observing how a probe orbits an object gives clues regarding the interior of that object. Scientists know, for example, that Mars has a partially molten core from observing the orbits of the many probes that have been sent to Mars.

In the case of the Moon, NASA sent 16 unmanned probes to the Moon in the 1960s prior to the Apollo 8 mission. The first four of these, Ranger 6, 7, 8, and 9, were impactors. Their impacts the Moon indicated the Moon's center of mass was displaced from its center of figure by 2-3 kilometers. Later missions confirmed this, and also confirmed that the Moon's near side has five large mass concentrations ("mascons").  The Apollo 15, 16, and 17 missions carried laser altimeters that gave very precise measurements of the distance to the Moon's surface. These further improved the estimate of the Moon's center of figure / center of mass offset.

(Aside: Read Bizarre Lunar Orbits for a description of how those lunar mascons change how things orbit the Moon.)


Kuiper belt planet formation There are a lot of objects of different sizes in the Kuiper belt. Are objects in the Kuiper belt stable? Is there any chance in the future that one of the larger objects would accumulate enough critical mass to start the process of becoming a proper planet (i.e., clearing its neighboorhood)?

The Kuiper Belt is fairly large (with inner and outer radii $\sim30\text{ AU}$ and $\sim50\text{ AU}$), but it does not contain much mass. The total mass is likely somewhere between $0.01$ and $0.1$ Earth masses[1], [2], which is essentially a mass range from five times the mass of Pluto to the mass of Mars. Now, it was likely much more massive in the past - its origin and mass loss are tied to the evolution of the orbits of ice giants in the early Solar System (see in particular the Nice model) - but we're talking about the formation of planets in the future. At present, there simply isn't enough mass to go around.

Let's talk about "clearing the neighborhood". The IAU's definition of a planet merely states, as one of its three criteria, that a planet must have

  cleared the neighbourhood around its orbit.

Wikipedia gives several quantitative measures of whether or not an object has cleared its neighborhood. The quantity $\Pi$ (Margot (2015) is given by
$$\Pi(m,a)=\frac{m}{M^{5/2}a^{9/8}}k$$
where $m$ is the mass of the body (Earth masses), $M$ is the mass of the star (solar masses), $a$ is the radius of its semimajor axis (AU), and $k\simeq807$. A body is a planet if $\Pi>1$.  If the mass of this hypothetical planet, $m_H$, is $0.01$ Earth masses, then no orbit in the range $30\text{ AU}<a<40\text{ AU}$ gives me $\Pi(.01M_{\oplus},a)>1$. If $m_H$ is $0.1$ Earth masses, then almost all $a$ in the same range give me $\Pi(0.1M_{\oplus},a)>1$. That said, it is extremely unlikely that a body could amass even a small fraction of the total mass in the Kuiper Belt - remember the inner and outer boundaries!

The quantity $\Lambda$ (Stern & Levison (2002)) is given by
$$\Lambda(m,a)=\frac{\mu^2}{a^{3/2}}\times\left[\text{Terms accounting for the properties of the body's orbit}\right]$$
where $\mu$ is the ratio of the mass of the body to the mass of the star. The author's list a value on the order of $\Lambda=3\times10^{-3}$ for Pluto; if we assume that the orbital terms are about the same for this body, we find $\Lambda(0.01M_{\oplus},a)\sim0.075$ and $\Lambda(0.1M_{\oplus},a)\sim7.5$, with the cutoff being $\Lambda=1$. Again, it is not possible for the body to accumulate the total mass of the Kuiper Belt, so we can safely assume that $\Lambda\ll1$.


What types of star do we know are hostile to life (like ours)? Certainly all stars have enough energy for a planet to host life. But a more magnetically active star is not preferred because solar ejection could strip planets of their atmosphere. What's the right balance?

Which stellar category of a star is suitable for hosting life excluding G2V class (our Sun's category)?

We so far have a sample size of 1. We know that a largish star, with a surface temperature at about 4.5 billion years of 5800K can be stable enough to support life for long enough for intelligence to evolve. We are fairly sure that very large stars don't last long enough.

With only one data point to extrapolate there is a tendency to think that stars that are like the sun offer the best environment for life. We are fairly sure that very large stars don't last long enough. It seems reasonable to exclude highly variable stars, or ones whose outbursts would strip the atmosphere from any planet. Intuitively these would seem to be less likely sites for life. But there is no strong evidence in the form of multiple discoveries of extra-terrestrial life to support this.

In short we don't know what stars can support life.


First meteor(s) of Lyrid meteor shower? So I am a newcomer to Astronomy - only aged 15 - and am currently looking upon the sky in hopes of witnessing several meteors. This is my first time ever - maybe - seeing them. I just went inside for food and drink, come out and i saw a trail (this trail was not here before I went in). I do have a few doubts, one it was seen below the moon, when it was supposedly meant to be near Vega - however I read about them having a tendency to be different to predictions. Also, considering it is so close to the Moon and I can't currently see Vega from this position I highly doubt it.

please view the below image in the hopes of identifying the object (please excuse the camera quality, I am currently setting up the DSLR and this was photographed on my Nexus.)

The feature in the photograph ressembles a condensation trail.

Meteors usually move rather rapidly, mostly perish within a few seconds or less.
In rare cases they may leave long trails; the meteor would have been very bright (brighter than the moon) in those cases, moving much faster than an aeroplane.


Why is it always planets orbiting stars? In our solar system, there are 8 planets orbiting a star, the Sun.
And I understand that there are about 500 confirmed solar systems out there.

But why is it always planets orbiting stars? Why can't it be several stars orbiting a planet, or a star orbiting a star? Why is a star by definition stationary as opposed to planets which are moving.

Of course, in that case, it wouldn't make much sense to call it a solar system, but still.

"Why is a star by definition stationary as opposed to planets which are moving." This just isn't true. Both star and planets orbit a point in the system known as the barycenter (a.k.a. the center of mass of the system).

Because stars are much more massive, this barycenter is much closer to the center of the star than it is to the planets. Hence it appears to the casual observer that the less massive object orbits around the more massive object. In fact both objects orbit the barycenter. Like so:

The following diagram (seen on the wikipedia page above) shows the motion of the solar system barycenter with respect to the center of the Sun. Note that the barycenter is close to the center of the Sun, but often spends time outside the visible surface of the Sun. It follows this complicated path, mainly because of Jupiter but all the other planets make smaller contributions too.


Latest sunrise/earliest sunset out of phase with solstice? Everybody knows that the shortest and longest days of the year occur on the solstices.  However, examination of sunrise/sunset tables shows that the maxima and minima of sunrise/sunset times are offset from the solstices by about 10 days.

According to timeanddate.com for my location (Portland, OR), for the winter solstice, the earliest sunset is about December 10 (16:27), and the latest sunrise is around January 1 (07:51).

This is counterintuitive... what's the explanation?

The point here is that sunrise and sunset does not only move towards or away from noon, but noon itself moves around 12:00. There is a difference between "civil time" (the one counted by wall clocks) and "solar time" (the one marked by sun clocks). This difference is called the "Equation of Time" and its geometrical shape is called the Analema.

So in order to have the earliest sunset you need a combination of both a short day and a noon moved before 12:00. Same reasoning for latest sunrise.

You can see more about the Equation of Time on Wikipedia: http://en.wikipedia.org/wiki/Equation_of_time


How do we find the exact temperature of a star? This is a very basic question, but I am a little confused. As far as I know, the temperature of a star is analyzed based on the color of the light it emits. So, if a star is moving away from us, then the light emitted by it will be redshifted(or if it is stationary with respect to us and the light undergoes gravitational redshift), then how do we know the exact temperature of that star or any other object because it is possible that we observe red light but actually the star might be emitting yellow light.

This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do of course have line of sight velocities which give their spectrum a redshift (or blueshift). It is a reasonably simple procedure to correct for the line of sight velocity of a star, because the redshift (or blueshift) applies to all wavelengths equally and we can simply shift the wavelength axis to account for this. i.e. We put the star back into the rest-frame before analysing its spectrum.

Gerald has talked about the blackbody spectrum - indeed the wavelength of the peak of a blackbody spectrum is inversely dependent of temperature through Wien's law. This method could be used to estimate the temperatures of objects that have spectra which closely approximate blackbodies and for which flux-calibrated spectra are available that properly sample the peak. Both of these conditions are hard to satisfy in practice: stars are in general not blackbodies, though their effective temperatures - which is usually what is quoted, are defined as the temperature of a blackbody with the same radius and luminosity of the star.

The effective temperature of a star is most accurately measured by (i) estimating the total flux of light from the star; (ii) getting an accurate distance from a parallax; (iii) combining these to give the luminosity; (iv) measuring the radius of the star using interferometry; (v) this gives the effective temperature from Stefan's law:
$$ L = 4\pi R^2 \sigma T_{eff}^4,$$
where $\sigma$ is the Stefan-Boltzmann constant. Unfortunately the limiting factor here is that it is difficult to measure the radii of all but the largest or nearest stars. So measurements exist for a few giants and a few dozen nearby main sequence stars; but these are the fundamental calibrators against which other techniques are compared and calibrated.

A second major secondary technique is a detailed analysis of the spectrum of a star. To understand how this works we need to realise that (i) atoms/ions have different energy levels; (ii) the way that these levels are populated depends on temperature (higher levels are occupied at higher temperatures); (iii) transitions between levels can result in the emission or absorption of light at a particular wavelength that depends on the energy difference between the levels.

To use these properties we construct a model of the atmosphere of a star. In general a star is hotter on the inside and cooler on the outside. The radiation coming out from the centre of the star is absorbed by the cooler, overlying layers, but this happens preferentially at the wavelengths corresponding to energy level differences in the atoms that are absorbing the radiation. This produces absorption lines in the spectrum. A spectrum analysis consists of measuring the strengths of these absorption lines for many different chemical elements and different wavelengths. The strength of an absorption line depends primarily on (i) the temperature of the star and (ii) the amount of a particular chemical element, but also on several other parameters (gravity, turbulence, atmospheric structure). By measuring lots of lines you isolate these dependencies and emerge with a solution for the temperature of the star - often with a precision as good as +/-50 Kelvin.

Where you don't have a good spectrum, the next best solution is to use the colour of the star to estimate its temperature. This works because hot stars are blue and cool stars are red. The colour-temperature relationship is calibrated using the measured colours of the fundamental calibrator stars. Typical accuracies of this method are +/- 100-200 K (poorer for cooler stars).


When was the distance to a star measured for the first time without using parallax? This is somewhat of a follow-up to When was the parallax of a star first measured? Once the distance to the nearest stars was determined, it was possible to discover physical properties of stars (such as the relation between intrinsic luminosity and color, or the relation between luminosity and period for Cepheid variables), then use the knowledge of those physical properties to find the distance of stars with no visible parallax.
When was the distance to a star (other than the Sun) determined for the first time with a method other than parallax?

From Wikipedia

In his 1698 book, Cosmotheoros, Christiaan Huygens estimated the distance to Sirius at 27664 times the distance from the Earth to the Sun (about 0.437 light years, translating to a parallax of roughly 7.5 arcseconds).

and

It was also in this book that Huygens published his method for estimating stellar distances. He made a series of smaller holes in a screen facing the Sun, until he estimated the light was of the same intensity as that of the star Sirius. He then calculated that the angle of this hole was {\displaystyle 1/27,664}1/27,664th the diameter of the Sun, and thus it was about 30,000 times as far away, on the (incorrect) assumption that Sirius is as luminous as the Sun. The subject of photometry remained in its infancy until the time of Pierre Bouguer and Johann Heinrich Lambert.[124]


After a supernova, why don't new stars coalesce with higher heavier element content? May be a beginner's question but I couldn't find the answer anywhere.
Given a supernova event and the cloud left by it, why does a new star coalesce with a disproportionally higher hydrogen(and helium) content compared to the matter left on it's vicinity?
Shouldn't gravity form bodies(maybe stars still) with all, or most of, the heavier elements and leave the hydrogen and helium in the vicinity instead?
I would also appreciate any book indications on that matter.

Supernovae are extremely energetic events. The remnants left behind by a supernova are ejected at much greater than escape velocity. Instead of coalescing due to self-gravitation, the remnants disperse and eventually spread throughout vast regions of the galaxy.


Why is the exoplanet HD 131399Ab so hot? Recently, a fascinating triple-star system with an exoplanet has been discovered (arxiv).

I'm very surprised about the temperature of the exoplanet: it is roughly 850 kelvin, even though its orbit is ~82 AU away from the main star (~1.5 $M_☉$), and >250 AU from the other two stars. The planet itself has ~4 jupiter-masses.

Why can such a planet be so hot, for instance compared to Jupiter? I would have expected it to be very cold because it loses a lot of energy via radiation (that's actually the reason why the group discovered it via direct imaging), and it does not get much energy from its stars.

It's a giant planet and it's young. p.3 of the paper you reference gives a likely age of 16 million years.

The heat arises from gravitational contraction and it would be a similar temperature even if it were further away from the stars.

When a ball of gas contracts, the virial theorem establishes that half the gravitational potential lost is radiated away and half heats up the gas.

Jupiter would also have been much hotter in the past and even now emits more radiation than it receives from the Sun. The rate of contraction (and hence the planet's luminosity and temperature) decreases as the planet ages.

Direct imaging surveys for giant planets deliberately target young stars for this reason.


Do all the objects in the universe exert force on all other objects? Do all the objects in the universe exert force on all other objects? Like a type of gravity; also, how much does it decrease as it gets farther away?

Yes - this is the formula:

$$F = G\frac{m_1m_2}{d^2}$$

Using this equation, we can say that all atoms in the universe exert force upon eachother. One carbon-12 atom has a mass of $1.660538921(73)\times10^{-27} kg$. That's a crazy small mass.

Now let's say that these two atoms are 100,000,000 light years apart. That's $9.461\times10^{23} m$, which is a very long distance.

Now, if we plug these values into our equation, we get that the force is: $1.709191430132 \times 10^{-59} N$

That's a very, very small amount of force. But it's still force.


Why do proto stars on the Hayashi track get dimmer as they contract? Why do proto stars on the Hayashi track get dimmer as they contract?

My expectation is that they would get hotter and brighter the more they contact. Why is this wrong?

The Hayashi track has an almost constant effective temperature. The pre main sequence star (not protostar) is powered by the release of gravitational potential energy caused by its contraction.

If the star is getting smaller at constant effective temperature, then of course it's luminosity ($\propto R^2 T^4$) decreases.

So perhaps your question is why do PMS stars contract at constant temperature? In other words, why is the Hayashi track almost vertical in the HR diagram? The answer to this is derived in some detail on the relevant Wikipedia page and arises from the fact that low mass PMS stars are fully convective and have atmospheric opacity dominated by H- ions.


How to calculate the angle formed between 2 planets? I am interested in calculating the angles formed between planets like Saturn opposite to Neptune - 180 degrees, PlanetX opposite/trine/square to PlanetY. How to find such degrees?

I use Stellarium software to find positions of planets. Here I am adding positions of Saturn, Jupiter and Sun.

Stellarium shows information about selected objects, can it be used to calculate angle formed, for example, angle formed between Saturn and Jupiter taking Sun as reference point?

The planets are moving along elliptical paths, and we are viewing them from another planet that is also moving along an elliptical path, this makes the motion that the planets make in the sky (relative to distant stars) seem to loop and move with an inconsistent rate. The planets don't all orbit in the same plane, and so may be above or below the ecliptic (the apparent path in the sky of the sun)

So there is no shortcut to finding the angle between two planets: You have to calculate the sky position of the planets and then find the angle between those positions.

The first step is called calculating an ephemeris. The details are beyond the scope of this answer, but NASA have a good ephemeris calculator at https://ssd.jpl.nasa.gov/horizons.cgi. Alternatively, software like Stellarium will be able to calculate the position of a planet on any given date in the near future or past.

When you have the position of the planets as RA and Dec, and you have converted the units to decimal degrees, you can calculate the angle $A$ between them using

$$\cos(A) = \sin(\mathrm{Dec_1})\sin(\mathrm{Dec_2}) + \cos(\mathrm{Dec_1})\cos(\mathrm{Dec_2})\cos(\mathrm{RA_1} - \mathrm{RA_2})$$

Further details and a calculator is found at http://www.gyes.eu/calculator/calculator_page1.htm