Pulsar, or X-ray navigation is being actively tested right now. Currently an X-ray telescope mounted on the International Space Station completed a test where a small navigation computer was able to solve for it's own position and orbit in space relying only timing information from X-rays produced by a group of pulsars.
I've written more about it in the question Is NICER/SEXTANT the first civilian “spacecraft” to determine it's own position in space without GPS or uplinked data? Spoiler alert, hold your cursor over the box:

 Yes, it is the first

You can also read about an investigation for X-ray communication in How will NavCube (actually) be important for the XCOM testing and demonstration? I believe this will be a significantly longer term technology development path.
NICER, NASA’s Neutron star Interior Composition Explorer, will of course also be used for pure Astronomy research on neutron stars.
Will this replace GPS? Not really, No.
GPS serves the Earth's surface, and near Earth orbit (NEO) very nicely and efficiently. However, once you get a little farther, to MEO and GEO, it gets much more difficult to receive several simultaneous GPS signals because the satellites concentrate most of their power directly on to the Earth's disk, with only some side-lobes "leaking" past the terminator and into space.
See the questions and of course their answers:

Will Glonass, Galileo, or BeiDou-2 satellites provide better cis-lunar navigation than GPS?
Has GPS been used beyond GEO? (duplicate)
How far up have satellites used a GNSS for positioning, and how does the precision degrade with altitude?

So for spacecraft in farther Earth's orbits it may be possible to augment or replace GPS or GNSS in general, and for deep space, there is currently no way to autonomously navigate today, so it will be the first.

above: NICER X-ray timing telescope/concentrator array, from About NICER.


Are there any works discussing planetary bodies being forms of life?

Look up Gaia hypothesis, James Lovelock, and Lynn Margulis.


What effects do other planets have on the solar system?

The gravitational pull of all the planets and the sun, and the rest of the galaxy and the universe, all play a part, but gravitational effects fall off with distance.

For Earth's orbit, the Sun is far and away the single biggest influence. Jupiter perturbs our orbit slightly, but with it or without it we have a simple elliptical orbit round the sun. We can measure the perturbation in our orbit from all the planets, but the effects are minor.

When it comes to gravitational effects on small bodies passing one of the planets, however, the effects can be dramatic - an asteroid passing close to Jupiter will be moved far from its original path. The same would be true when passing close to Saturn or in fact any planet - just to a lesser degree.

So while you can't say that Jupiter protects Earth from anything, especially not the Oort Cloud, you also can't say that it draws objects in. Anything out in the Kuiper belt or Oort cloud will be orbiting the barycentre of the solar system (which is within the Sun) and will keep doing that unless knocked out by something  like a collision with another object, or the pull of other Kuiper belt objects like Pluto, which can give a much greater tug.


Magnitude, satellite flare and the Heavens Above app

The astronomy "magnitude" scale works backwards: smaller numbers indicate brighter objects.  Back in the days before precision measurements of brightness, stars were categorized by eye, with the brightest being "stars of the first magnitude".  When more precise measurement became possible, this scale was retained, and extended into the negative numbers for very bright objects like Venus, the Sun, and a few of the brightest stars.

The Iridium satellites have enormous, mirror-like antenna arrays.  When one of them is angled correctly, it will reflect sunlight straight at you, producing an incredibly bright flare visible even in broad daylight.  Very few other satellites have large reflective surfaces other than their solar panels, and solar panels are kept pointed straight at the Sun, so they never generate flares.

To give some points of comparison, the Iridum flare listed in your screenshot, at magnitude -5.2, is comparable to Venus at its brightest, visible during daylight if you know where to look.  The other satellites listed, with magnitudes in the 2-3.5 range, have brightnesses between that of Polaris, and that of the dimmer stars making up Ursa Minor.


Type II supernovae as a distance indicator

The open supernova catalogue calculates absolute magnitudes from the observed apparent magnitude, luminosity distance and redshift decrements.  They note that they don't account for spectral energy distribution shape, so absolute magnitudes may be approximate if there is significant absorption by dust.
So the catalogue isn't recording a measured or modelled absolute magnitude derived independently of the distance, (which is required if the SN are used as standard candles), but a calculated absolute magnitude derived from the apparent magnitude and the redshift.


Is the Cosmos a super set of the Universe and any hypothetical multiverse(s)?

"Cosmos" is the English cognate of the ancient Greek word which refers to the entirety of existence. It is equivalent to the Latin-derived word "universe;" the only difference between the two is the language of origin.


Why is molecular hydrogen (H2) so difficult for astronomers to detect?

The light that we receive from space contains a lot of information.  Specifically, because of quantum nature of molecules and atoms, theses small species absorb light at very specific wavelengths.Each molecule, atom or ion has a unique set of absorption features.  Think of it as the molecule own fingerprint.  Hence, if we look at a star or a hot gas, we can deduce which species are present by looking at the very specific emission lines present in our spectrum of light.  Similarly, if the light of a star or of a galaxy passes trough a dust cloud, then we can deduce which species are present in that cloud from the absorption lines present.
Molecular hydrogen doesn't really have absorption lines in the infrared, visible and UV light and so it doesn't emit nor absorb very much light and so we don't see it in our spectrum, i.e. whether there is hydrogen or not, the light that we receive looks the same.


Why is only one side of the Moon visible from Earth?

The reason for this is what we call tidal locking:

Tidal locking (or captured rotation) occurs when the gravitational
gradient makes one side of an astronomical body always face another,
an effect known as synchronous rotation. For example, the same side of
the Earth's Moon always faces the Earth. A tidally locked body takes
just as long to rotate around its own axis as it does to revolve
around its partner. This causes one hemisphere constantly to face the
partner body. Usually, at any given time only the satellite is tidally
locked around the larger body, but if the difference in mass between
the two bodies and their physical separation is small, each may be
tidally locked to the other, as is the case between Pluto and Charon.
This effect is employed to stabilize some artificial satellites.

    Fig. 1: Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit the Earth. (Source: Wikipedia)

Fig. 1, cont.: Except for libration effects, this results in the Moon keeping the same face turned towards the Earth, as seen in the figure
on the left. (The Moon is shown in polar view, and is not drawn to
scale.) If the Moon were not spinning at all, it would alternately
show its near and far sides to the Earth while moving around our
planet in orbit, as shown in the figure on the right.

                         Fig. 2: Lunar librations in latitude and longitude over a period of one month (Source: Wikipedia)

Libration is manifested as a slow rocking back and forth of the Moon
as viewed from Earth, permitting an observer to see slightly different
halves of the surface at different times.
There are three types of lunar libration:

Libration in longitude results from the eccentricity
of the Moon's orbit around Earth; the Moon's rotation sometimes leads
and sometimes lags its orbital position.

Libration in latitude results
from a slight inclination between the Moon's axis of rotation and the
normal to the plane of its orbit around Earth. Its origin is analogous
to how the seasons arise from Earth's revolution about the Sun.

Diurnal libration is a small daily oscillation due to the Earth's
rotation, which carries an observer first to one side and then to the
other side of the straight line joining Earth's and the Moon's
centers, allowing the observer to look first around one side of the
Moon and then around the other—because the observer is on the surface
of the Earth, not at its center.

All quotes and images from Wikipedia on Tidal locking and Wikipedia on Libration.


What's the differences between one-dimensional spectrum and two-dimensional spectrum?

When you place a spectrograph slit on a source, the spectrum recorded can be thought of as lots of images of the slit at different wavelengths.

Ordinarily, you would sum up this spectrum along the direction of the slit images to give you a one dimensional spectrum. If however, you leave the image as recorded, then you have a two dimensional spectrum - intensity as a function of wavelength along one axis and as a function of position along the slit in the other.

Two dimensional spectra are used when we expect the spectrum to vary with position along the slit. Examples might include a spectrum recorded across a galaxy, or a spectrum of a binary star with the slit placed across both components.

An example is shown below. The inset image shows a broad-band image of V458 Vul - a classical nova that is surrounded by (not visible) shells of ionised material. The authors of this particular study lined up a spectrograph slit as shown in the inset and then obtained the two two-dimensional spectra shown in the main image. What you have to imagine is that each position on the slit produces a horizontal spectrum at a vertical position that corresponds to its position on the slit. Therefore we see a bright spectrum across the middle corresponding to the central source, but there are then "knots" of emission at particular wavelengths that are some distance away from the central star.

Slitless two dimensional spectroscopy is also possible using integral field spectrographs. Fibers record spectra over a two dimensional area. This can also be referred to as two dimensional spectroscopy.


The formula you are using is for a point mass, but a hole through the Earth experiences less gravity than a point mass, as the comment points out.  One can then ask, what if the Earth's mass were all at the center, why wouldn't the mass fly through and go to a distance r on the other side?  It's because a mass like that is a singularity, one that doesn't matter unless you fall right through it.  If the Earth's mass was really concentrated at a point, then falling from r would still tend to just miss that point, since it's just a point anyway, and the huge forces would turn the orbit around, such that the object fell from r, went to zero, turned around, and flew back out to r.  That's a very squeezed ellipse, with a=r/2, as desired.  The motion that goes straight through and out to the opposite r, on the other hand, is not an ellipse with the mass at one focus, so it violates Kepler's laws.  Again, the violation of Kepler's laws is that the mass is singular in a way that matters if you fall through a hole in it.  Interestingly, it doesn't matter how concentrated the mass is toward the center-- if there's a hole in it that the object falls through, then the object does not obey Kepler's laws.  It will only obey Kepler's laws if the orbiting object does not enter the mass.


How do I see Deep Sky Objects in my telescope?

To see Deep Sky Objects you'll definitely have to head out of whatever town you're in by at least 5km, but the further the better. Next, you'll have to have a good knowledge of where to look for them or a guide. You should be able to see a fair few once you've done that, especially if you've already seen the Orion Nebula.


Stanley, there really isn't a very clear definition and this is still a keenly argued point.

Definitions include:

Browns dwarfs burn deuterium. In models this happens if they are more massive than about 13 times Jupiter. The weakness of this that we think isolated brown dwarfs could condense from a gas cloud that are less massive than this; and young brown dwarfs won't have got around to fusing deuterium.

Planets must form from the disk around a star. This is ok, but: brown dwarfs may also form from the disk and it is also possible for planets to be tidally stripped from their stars and be found alone in space.

Planets must have a rocky core. This used to be thought definite, but now we think maybe sometimes planets can collapse from a gas instability in the disk in some circumstances, without the need for a rocky/icy core. It is true that brown dwarfs should not have a rocky core. However as an observational definition this is fairly hopeless since we can't even tell yet if Jupiter has a rocky core.

A flavour of the controversy can be gleaned from reading between the lines of the IAU statement on the definition of planets vs brown dwarfs.


Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit?

because I thought the conservation of angular momentum didn't have to do with the rotation of the objects.

The angular momentum of the system is conserved.  As rotation can contribute to angular momentum, it does affect the equation.  You can ignore it for point-like objects.  In the case of the earth-moon system, the earth's rotation is a significant portion of the total momentum.

A not-too-wrong version would be to assume an axis centered on the earth.  Then the total angular momentum of the system could be found by summing the individual momenta for:

the earth's rotation
the moon's rotation
the moon's orbit

Assuming the moon is tidally locked to the earth and has a nearly circular orbit, this gives you only two variables.  You can then calculate that, if you slow the earth's rotation, the only way to keep angular momentum of the system constant is to increase the angular momentum of the moon by an equal amount.  This would correspond to a larger orbit.

Further if you calculate the KE from both configurations, you find the second one has less energy.  So there must be an energy loss to transition between them.


Are stars beyond the asteroid belt less constant?

No, stars that are seen through the asteroid belt are not any more variable than stars in any other part of the sky.

All stars vary in brightness on measurable timescales, although some have been measured to have such small variations that they are recorded as "standard" stars and for most purposes can be considered to have a constant brightness and thus be used as reference for other brightness measurements.

What you are referring to, as noted, are occultations. There are many hundreds of thousands of known asteroids in our Solar System, and billions of stars in our galaxy, and so there could be hundreds of predicted occultation events every day, and there are many people interested in occultation events. For one thing, we can calculate the distance and speed of an asteroid quite accurately but we can't directly photograph it to determine its size. But when an asteroid passes in front of a star, we can measure the duration of the occultation and set precise limits on its size.

But for all the asteroids in the Main Belt, the space between them is huge and for any one star the chances of an occultation are quite (vanishingly) small, so the stars seen through the asteroid belt are not any more variable than anywhere else in the sky.


Comparing Light Collection Between Pupil and Telescope?

You have the idea right! The amount of light collected is indeed proportional to the area of the collecting surface.

Now, if you want to take it a bit further, try to think about the following:

Binoculars have one lens for each eye, does this mean you get twice the amount of light in the end?
Reflecting telescopes almost always have a secondary mirror which results in part of the light being masked and not collected by the primary mirror. Now imagine you have an 8-inch telescope with a 2-inch secondary mirror, can you calculate again the light collection ratio compared to the naked eye?


Use the Optical Spectra Query Form. You'll want to set PRIMTARGET to "GALAXY". Here's the list of possible parameters it can return http://skyserver.sdss3.org/dr10/en/help/docs/QS_UserGuide.aspx#Spectroscopy. Note, there is a query limit of 500,000 rows, so you'll have to break up the query since you'll have more spectra than that.


How to evaluate the fit of an isochrone to a stellar population?

Finding the best-fitting isochrone, a.k.a. isochrone fitting, is a standard approach to determine the age of globular clusters. This problem can be solved with a least-square method, where the data to be fitted are the points on the color-magnitude diagram and the fitting curve is the isochrone. Since there are many ways to apply this method, I will first outline the general approach and then I will provide you with a real case application.

Outline of the method

Let's first break down what is isochrone fitting via least-square method. You have your CMD with all your photometric data scattered around. You assume your sources belong to a coeval stellar population (almost true for many globular clusters), and so you want to exploit a theoretical isochrone to find the age.  So you built a set of isochrone with different ages, fixing carefully all the other parameters (e.g. [Fe/H] and [$\alpha$/Fe]). Then you define an error function and you find the isochrone that minimizes it. The simplest choice for the error function is
\begin{equation*}
\chi^2=\sum_i(C_{i,\textrm{obs}}-C_{i,\textrm{exp}})^2
\end{equation*}
where the sum is over all the sources in the CMD, $C_{i,\textrm{obs}}$ is the color of the $i$-th source, and $C_{i,\textrm{obs}}$  is the color of the isochrone at the magnitude of the $i$-th source. Heuristically, $(C_{i,\textrm{obs}}-C_{i,\textrm{exp}})$ is the horizontal distance of the $i$-th point from the isochrone and we assume the best-fit isochrone is the one that minimizes the sum of the distances.

This $\chi^2$ could be the parameter you're looking for. Similar parameters can be obtained using the vertical distance or the minimum distance instead of the horizontal distance.

Application to 47 Tuc

Let's see a real application of the $\chi^2$ minimization technique. I will use some plots that I've made for didactical purposes, obtained from the photometric data of 47 Tucanae in the HUGS catalogue (Piotto et al. 2015, Nardiello et al. 2018). I've used the HST ACS/WFC bands $F606W$ and $F814W$, which are similar to the $V$ of Johnson and $I$ of Cousins.

First point: usually we do not use all the sources at once, and there are several reasons for that. The main one is the dispersion on the CMD, which can lead to a bigger uncertainty on the final age measure. For example, your sources can have a huge dispersion on the CMD due to photometric errors, or they can have an intrinsic dispersion (e.g. due to differential reddening, binaries above the main sequence, multiple populations, and so on). You want to get rid of those outliers, you want to narrow the distribution of sources in the CMD to have a more robust determination. Second point: instead of using all the sources at once, we can bin them and average their position on the CMD. In this way, we have to deal just with a few fiducial points that represent the whole distribution.

So we bin the sources in magnitude, and for each bin we calculate the clipped mean of the colors.  in finding the standard deviation of colors $\sigma$, discard all the sources outside e.g. $3\sigma$, eventually iterate this selection some times, and at the end calculating the mean color. In this way we get rid of the outliers and obtain the fiducial points. The results is shown in the image below: in black, the CMD of 47 Tuc; in grey, the final clipped selection, in red, the clipped mean color in bins of width $\Delta m_{F606w}=0.1$, i.e. the fiducial points. The error bars on the fiducial points are the standard deviations in both dimensions.

$\hskip2in$

With these fiducial points, we can easily apply a least-square method. I've downloaded a set of isochrones from the Dartmouth Database, with fixed chemical abundances and different ages ($\Delta\textrm{age}=0.2$ Gyr). I've then calculated for each isochrone
\begin{equation*}
\chi^2=\sum_i(C_{i,\textrm{obs}}-C_{i,\textrm{exp}})^2
\end{equation*}
where the sum is over all the fiducial points, $C_{i,\textrm{obs}}$  is the color of the $i$-th fiducial point, and $C_{i,\textrm{exp}}$ is the color of the isochrone at the magnitude of the $i$-th fiducial point. The results are in the image below. In the panel above, you can see the $\chi^2$ for each age. In the panel below, the red crosses are the fiducial points, while the colored points are the points extrapolated from each isochrone at the magnitude of every fiducial point (colors in the two panels are matched). Now you see that, for a given isochrone, in the $\chi^2$ we are summing the horizontal distances between the colored points and the red crosses.

$\hskip2in$

Conclusions

  After generating the required isochrones, what should be a parameter for the goodness of fit for the isochrone?

It depends on the method you are using. Assuming you're using a least-square method, the parameter you're looking for is the sum of the squared residuals (e.g. my $\chi^2$).

  Is there something like a 'best fit line' that also exists for general curves?

From the context of the question, I suppose you're asking if there's something beyond linear regression, i.e. something beyond finding the best-fitting line. If so: yes, as you can see from this wikipedia page. In fact, least-square methods are not limited to lines, neither linear functions, and the method I've outlined is an example.

Remark

It is well known that the determination of the absolute age of globular clusters carries considerable uncertainties (e.g. Bolte & Hogan (1995)). Therefore, to do meaningful science is necessary to analyze the error budget (e.g. how the uncertainties in distance, metallicity, and reddening propagate to the final uncertainty, or what is the entity of the photometric errors) and to take actions to mitigate these uncertainties (e.g. rejecting sources with high errors, using clipped averages, transposing theoretical isochrones into observational CMD and not observational data into theoretical CMD to control error propagation).


NGC 2516 is somewhat richer and more massive than the Pleiades. A careful look at the mass function suggests there is about $1000 M_{\odot}$ in stars $\geq 0.3 M_{\odot}$, within the central 0.9 square degrees (with about a 15% uncertainty), but there is probably a few hundred solar masses beyond this (Jeffries et al. 2001). This agrees with the figures of Dachs & Kabus (1989), who essentially extrapolated from the numbers of high-mass stars using a typical cluster mass function.
New Gaia data will probably provide a more complete census in the halo of the cluster. There will also be some additional mass in lower mass stars and brown dwarfs, but this is unlikely to add more than 10% to the figure.
The lower masses quoted in your question are way too low. In an incomplete spectroscopic survey of the central square degree, a recent paper by
Jackson et al. (2020) found about 500 low-mass members ($0.4-1.2M_\odot$), that alone would add up to about $400 M_\odot$, before considering the many high-mass stars.


The idea of removing material from the sun for either fuel or to lengthen its lifespan is called "star lifting". Various hypothetical methods have been suggested, including heating parts of the atmosphere or squeezing it using magnetic fields. My own tentative suggestion is to combine these with chaos control to manipulate turbulence into making coronal mass ejections in desired directions.

The problem is not just that the sun is hot (5777 K at the surface), but that the stuff that is easy to scoop up with a huge magnetic field is a fairly sparse plasma - you don't get that much material despite the vast scales. Of course, you may want to compress the plasma into a smaller and more convenient ball, but now the ideal gas law tells you that it will heat up proportionally - to get a cool hydrogen ball you will need to bleed off all that heat slowly.

Which is why I think Dyson spheres are probably the easiest way to start.


Given telescope diameter and focal length only, how can I find the size of a produced image?

To solve this problem, you can replace the telescope imaging system with a simple pinhole (conceptually) and just draw the rays that pass through the center of the entrance aperture. The Moon is about 29 to 34 arc minutes wide or about 0.0085 to 0.01 radians.
If your telescope has a focal length of 200 cm, then at the focal plane it will be 1.7 to 2.0 cm wide. Since the angles are small you don't need to use trigonometry, just use a small angle approximation:
$$\sin(\theta) \approx\ tan(\theta) \approx \theta$$
when $\theta$ is in radians. So the size of the Moon will be about $\theta F$ where $F$ is the focal length.

In other words, ignore the diameter of the aperture, objective lens or mirror. Just use the pinhole camera model for imaging systems. If you have a multi-element telescope (e.g. Cassegrain or refractor with a Barlow lens) make sure to use the effective focal length of the system at the image plane for $F$. A Cassegrain might have a 40 cm focal length objective but a 120 cm effective focal length for the whole system. You might have a 100 cm focal length refractor but with a 2x Barlow lens you effective focal length will be 200 cm.

Source


I don't think this will be a great answer because these are just speculations. I cannot find concrete evidence or sources discussing why Jupiter (the planet) became associated with the "king of gods" and not Venus, but here's my go at citing some potential reasons.

To start, I want to point out that ultimately, we have to trace this back to the Greeks and Babylonians. Sure Jupiter (the planet) is named after the Roman god Jupiter, but really the Roman mythology traces pretty distinctly to Greek mythology and Greek mythology has strong ties to Babylonian mythology. The Greeks were the ones who associated Zeus with the planet Jupiter (and they were likely taking cues from the Babylonians) and the Romans just tagged along with that idea, but changed the god's name. So I think your question then transforms into, Why didn't the Greeks/Babylonians decide to associate the planet Venus with their chief god (being that it is the brightest) and instead chose the planet Jupiter?

Brightness of Venus and Jupiter

First, let's just confirm that Venus is indeed brighter. If you look up the Apparent Magnitude system of Wikipedia, you see a description of how astronomers measure visible brightnesses of astronomical objects (and coincidentally enough, this system was created by a Greek). In this system, smaller and more negative numbers indicate brighter objects. From the table on that page, we can see that the apparent magnitude of Venus varies from $-4.89$ at its brightest to $-3.82$ at its dimmest (this variation is caused by the different configurations of Earth, Venus, Sun positions). For Jupiter, we see that at its brightest it has an apparent magnitude of $-2.94$ and a minimum of $-1.61$. Clearly, even at its brightest, Jupiter is far outshone by Venus at its dimmest. Note that none of this takes into account light pollution or other atmospheric effects which would have been negligible to the Greeks and Babylonians anyway. So we can say emphatically yes that Venus is distinctly brighter than Jupiter, but importantly, we note that Jupiter is still brighter than every other star in the sky $-$ the brightest being Sirius with a magnitude of $-1.47$.

Confusion about Venus

Okay, the Greeks and Babylonians were sitting around looking at these planets and wondering what they were. Likely they started thinking they were big, great deities and started attributing gods to them. I can't say this is a certainty, but one possible reason Venus didn't get the top god was that initially the Greeks and Babylonians weren't even sure Venus was a planet like the rest of the planets were. Because Venus is so close to the Sun, we only ever see it at sunrise and sunset. Otherwise, it is in the sky during the day and dwarfed by the brightness of the Sun. This meant that it took the ancient peoples a long time before they even realized the bright object they saw in the evening was the same as the bright object they saw in the morning. To quote Wikipedia on this:

  Observations of Venus are not straightforward. Early Greeks thought that the evening and morning appearances of Venus represented two different objects, calling it Hesperus ("evening star") when it appeared in the western evening sky and Phosphorus ("light-bringer") when it appeared in the eastern morning sky. They eventually came to recognize that both objects were the same planet. Pythagoras is given credit for this realization.

Potentially, the ancient Greeks didn't even consider Venus to be a planet initially. The word planet is Greek and means "wandering star". The ancient Greeks identified objects in the night sky which were different and seemingly more important that the normal stars. However, initially, they don't seem to have grouped Jupiter and Mars in with the same type of object as Venus (and possibly Mercury). Jupiter's motion around the sky may have seemed more grandiose and full of purpose, hence why they attributed it to being their biggest and most important god.

Love and Beauty

As I said, the Babylonians, as well as the Greeks before they figured it out, didn't consider Venus to be a planet, but rather two separate and strange "stars". However, once they figured out the two "stars" were really one and the same and Venus was a planet just like Jupiter, they decided to attribute a proper god or goddess from the pantheon of gods to it. Jupiter was already taken so they had to pick someone else.

Despite that, I don't think the argument that Venus is brighter automatically means it should be the most important god. Note that Venus is named after the Roman goddess Venus (of course) who is also the Greek goddess Aphrodite. This is the Greek goddess of love and beauty and she herself was supposed to be exceedingly beautiful. Venus, the planet, is a beautiful sight to behold and it seems appropriate that the most radiant and beautiful planet be named after the most radiant and beautiful Goddess.


If I had a horseshoe orbit of two moons...

It's not the size of the Moons so much as the proximity to the sun that makes a 2 moon system (and especially a 2 moon horse-shoe orbit) unstable.   Size matters too, but the proximity to the sun matters more.    The sun's tidal force perturbs the Moon's orbit around the earth quite a bit.   If Earth was much further from the sun, it could have several moons, not just one and some a fair bit more distant and the big moon's orbit would be nearly circular due the inverse tidal forces, which tend to circularize an orbit.

Making the moon's smaller helps some, but what you really need, if you want that orbit to be relatively long-lasting, is to push the Earth away from the sun at least a few more astronomical units.

If we ignore the solar-tidal effects, then Earth could have 2 moons in a horse-shoe orbit to each other.    As to what it would look like, the first thing to consider is that everything in the sky rises in the east and sets in the west and that's because the Earth rotates.    If you watch each of the two horseshoe moons individually, they would move much like our moon, rising in the east, moving slowly against the fixed stars, and setting in the west.

The second thing to consider is that the Moons, the planets and the sun move against the fixed background stars, but this movement is slow unless the Moon is very close to Earth.  Our Moon, moves a full 360 degrees across the fixed stars once every 27 days (sidereal orbit).

The motion of the two moons individually would be very similar to our moon now in terms of new, waxing, full, waning, and repeat.

What the horseshoe orbit would do is you'd see one moon slowly catching up to the other, perhaps touching it from the point of view of earth, but never passing it.  it would get close, then it would reverse direction relative to the other moon and in a couple/few weeks, it would approach from the other direction, slow down and reverse.    Each moon individual would behave much like our moon but relative to each other there would be this weird game of attraction-repulsion between them.

I believe (but don't quote me on this), from Earth you'd never see the Moons' cross, but you'd perhaps see them touch.   The nature of the horeshoe orbit prevents crossing and passing, you'd just see them move closer, then move further away from each other.   If you had a really good eye, you might be able to notice the smaller moon change size as it moves closer and further from the Earth as the moon's get close, but that difference would be subtle.

This is the best diagram I've seen for what happens as the moons get close.  Unfortunately it's not in English, but it's not hard to follow:

https://upload.wikimedia.org/wikipedia/commons/9/99/JanusEpimetheus.gif

The horseshoe back and forth, assuming some degree of stability, it should have a regular period, just as the full-moons would have a period.   The optimal time for moon watching would be when those 2 periods coincide and you have a double full moon.   The horseshoe period and the full moon period (currently 29 days - sidereal orbit) should have nothing to do with each other but they could in theory be coincidentally in some kind of resonance.

As pointed out in the other thread, even if the Earth was far enough away from the sun to have two moons in this pattern, it's probably not truly long-term stable.   It might last for a few hundred orbits, or a few thousand.   It's not what is generally considered a stable orbit.

  Would both have the same cycle?

This is a bit of a trick question.   On average, yes.  The orbital periods would need to be the same for the horseshoe to work when averaged out, but individual orbits would need to vary by a little bit, so the individual orbits wouldn't have the same cycle.

Corrections are welcome if I got any of that not quite right.


What's so interesting about Coma-type galaxy clusters?

I think the answer to this isn't that Coma-type galaxy clusters are interesting, it's that galaxy clusters are interesting. Chiang et al. (2014) talks about the importance of studying protoclusters as a means of understanding how galaxies evolve over time.

  Galaxy clusters are extreme products of structure formation. They are ideal laboratories to study galaxy assembly, quenching, and sub/super-halo galaxy environments. It has become clear that to obtain a complete picture of cluster formation, we also need to find and study their progenitors at high redshifts that  were  still forming.

In their paper, they discuss the discovery of a large number of new protoclusters. They bin their protoclusters by mass and designate them as "Fornax-type" ($M_{z=0} = 1-3\times 10^{14}\ M_{\odot}$), "Virgo-type" ($M_{z=0} = 3-10\times 10^{14}\ M_{\odot}$), or "Coma-type" ($M_{z=0} > 10^{15}\ M_{\odot}$). Clearly, they're using present-day clusters as analogs to describe what these protoclusters may turn into as they continue to evolve. The fact that a protocluster is considered a Coma-type cluster, is only based on the mass. Astronomer's aren't so much interested in studying Coma-type protoclusters specifically, they're just studying protoclusters and the high mass protoclusters are considered Coma-type protoclusters.

It's very possible that you were preferentially seeing references to Coma-type clusters because Virgo- or Fornax-type clusters are harder to find (being of smaller mass) and so there is an observational bias that preferentially makes Coma-type protoclusters easier to find and subsequently publish about.


Galactic revolutions relative to the age of the universe

Gravitationally bound dynamical systems evolve towards their equilibrium states on a dynamical timescale of $\sim (G \rho)^{-1/2}$, where $\rho$ is their average density.

If we assume the Galaxy started of with $10^{12}$ solar masses in a sphere of radius $10^5$ pc, then the dynamical timescale is 1 billion years.

If we consider an object at the "edge" of such a system, Newtonian gravity and a bit of rotational mechanics, we find that the fastest possible bound orbital period is also of order $(G\rho)^{-1/2}$. The Sun is not at the edge and the average density inside the solar orbit is much higher and so the orbital period can be a bit shorter than this (a factor of four).

Thus what we see is exactly what is expected for a $>10$ billion years old universe - a relatively settled dynamical system.

You could extend this argument to the timescale for major galaxy mergers. The distance between galaxies is of order ten times their size and thus the mean density of galaxy systems is 1000 times lower than for an individual galaxy. This increases the relevant dynamical timescale to $1 \times 1000^{1/2} \sim 30$ billion years. That the Milky Way and Andromeda will collide some 17-18 billion years post big-bang is also therefore not an unexpected event.


Why does the Moon's terminator look "wrong" in this image?

The reason that moon image looks wrong is because it is wrong. It is not a real image of the moon $-$ at least the terminator is not real.

The original article you cite has a link just below their image indicating the source of their image of the moon. That source is the night sky planner, hosted by JPL. You'll find the same image on that website, albeit slightly darker (it seems the NPR people lightened up the image a bit).

If you do some more digging, you'll see that the night sky planner got its image of that moon from someone else. Within the html code, they have the image defined as:

<img alt="the moon" src="http://api.usno.navy.mil/imagery/moon.png">

Clearly you can see that the image was taken from the United States National Observatory.

After a bit of digging, I found out what exactly is going on here. The purpose of this site is to show you the current phase of the moon. To do this, they take a single image of the full moon and artificially shade out a region to make it appear as the current phase of the moon. You can see their process here and how it was done by some guy named R. Schmidt. They've broken down the moon phases into 181 images which you can download here, if you're interested.

As you can see, the terminator on that image is wrong because it is not a real image of the current phase of the moon, but rather a computer generated "shading" of the full moon to indicate the current moon phase.


Is there a formula for absolute magnitude that does not contain an apparent magnitude term?

Yes. You derive the absolute bolometric magnitude by reference to the Sun. The Sun has a bolometric magnitude (i.e. integrated over all wavelengths) of about 4.75. Thus the difference in bolometric magnitudes is given by the log of the ratio of luminosities like so
$$ M_{\rm bol} = 4.75 -2.5 \log_{10} \left(\frac{L_*}{L_{\odot}}\right)\ ,$$
where $L_{\odot} = 3.8\times 10^{26}$ W is the luminosity of the Sun and $L_*$ is the luminosity of the star.
To get the absolute visual magnitude then you need to subtract a bolometric correction, which to first order, depends on the temperature of the star (it also depends to second order on the composition and gravity).
$$ M_V = M_{\rm bol} - BC_V\ .$$
A table of bolometric corrections vs temperature can be found in the 6th column of this authoritative table, which is in common use by astronomers/astrophysicists.


Hypothetical scenario, two nearby black holes

This is actually a very nice example of why black holes are such extreme objects, they even break the basic rules arithmatic:

Black hole math: $$1 \text{ black hole} + 1 \text{ black hole} = 1 \text{ black hole}$$

(this is a little insider science joke)

A black hole is a region of space from which nothing, not even light, can escape. To achieve that it must have very strong gravity, so a lot of mass in a small region of space. So we may call it a "hole in spacetime", but it still has a mass.

In physical sense black holes can be described by only three quantities:

Its mass M (how much stuff was used to make it)
Its total angular momentum, J (how fast is it spinning)
Its electric charge, Q (does it have an electric/magnetic field)

All these three properties can change if the black hole is fed with 'stuff'. If you add matter the black hole mass increases. If that matter has (aligned) angular momentum, the black hole will spin faster and if the matter is charged, then the black hole charge changes accordingly.

When dealing with merging black holes, it is the mass property that is important. If I repeat the black-hole-math for its mass you get

$$ 1 \text{ black hole mass} + 1 \text{ black hole mass} = 2 \text{ black hole mass} $$

So a merger of two identical black holes just gives a single black hole with twice the mass.

Such black hole mergers are probably less hypothetical than you might think. They are considered to be likely method of creating supermassive black holes and are also considered as a potential source for gravitational wave emission.


Why does the neutral hydrogen velocity have this characteristic behavior in the galactic plane?

The reason is detailed in depth in this pdf, which contains the following diagram:

Some key quantities:

$R_0$: Distance from the observer to the center of the Milky Way
$R$: Distance from target gas to the center of the Milky Way
$V_0$: Velocity of the observer with respect to a certain reference frame
$V$: Velocity of target gas with respect to the same reference frame
$l$: Angle between $R_0$ and $\mathbf{SM}$, the distance to the gas

When measuring the radial velocity, we look for the projection of $V$ onto our line of sight. Some geometry results in the relation
$$V_r=V_0\left(\frac{R_0}{R}-1\right)\sin l$$
for the radial velocity $V_r$, assuming $V\approx V_0$ (which is not always true!). The $\sin l$ term, in addition to the asymmetrical rotation of the Milky Way, leads to the peculiar data you asked about.


Assuming proton decay is false, what's the ultimate fate of a neutron star or white dwarf?

This is a classic question in physical eschatology, seeing what happens if we extrapolate current understanding of astrophysics forward. The classic papers are (Dyson 1979) and (Adams & Laughlin 1997).
Obviously, over very long timescales white dwarfs cool down, crystallize. and become "black dwarfs". This is fairly well-established from observation and modelling, although the final stages have not been studied much.
If weakly interacting dark matter can be captured, dense objects would acquire internal halos: in this case, if the dark matter is a mix of particles and antiparticles there would be some annihilation, heating the object for a long time. If it is non-annihilating in principle it could build up until the object imploded into a neutron star or black hole. This is highly dependent on the dark matter model, so this should be regarded as conjectural.
However, it is fairly well established that galaxies dissolve due to gravitational interactions over long timescales, and this will dump such objects into the central black hole before they plausibly could undergo collapse, or eject them into intergalactic space where they would no longer acquire dark matter.
The fate of intergalactic black dwarfs and neutron stars without proton decay depends on what other modes of decay and change are possible. In white dwarfs pycnonuclear fusion would continue until all fusible elements had fused. Dyson estimated the timescale until everything is iron to $10^{1500}$ years, although there are environmental effects in white dwarfs that likely speed things up. This can actually make heavier white dwarfs (above 1.2 solar masses) collapse into supernovas on a timescale of $10^{1100}$ years (Caplan 2020).
Dyson noted that over timescales of $10^{65}$ years matter behaves like a quantum fluid due to tunneling. But this doesn't change structure of remaining objects much. A more important issue may be tunneling into black hole states where a small part of the object tunnels together to form a small black hole that evaporates. Adams and Laughlin estimate timescales of $10^{45}$ years for neutron stars and $10^{336}$ years for white dwarfs to evaporate this way.
Even if this does not happen, there is an argument to be made that thermodynamic fluctuations eventually dissolve bound objects since this minimizes the Gibbs free energy $E-TS$: at a finite temperature (which is the standard assumption for accelerating expansion in $\Lambda$CDM) if there is enough space the entropy $S$ can be maximized by separating particles despite some binding energy: tunnelling will eventually dissolve everything. This is similar to the discussion about Herzfeld's paradox of the spontaneous ionization of hydrogen atoms. The mere presence of other matter "outside the lab" normally stabilizes bound systems but in the very far future isolated systems become destabilized. This assumes that (1) temperatures will stay finite (i.e. our understanding of horizon radiation and continued accelerated expansion are right), (2) there are no other limits on dissociation, (3) the Gibbs energy argument is valid in this context. All three can be debated.
Overall, the trend seems to be that entropy maximization will tend to dissolve objects into isolated particles while gravity either just keeps them together or causes implosion into black holes through some pathway, followed by evaporation.


The term for when the sun is at its highest is "solar noon".

https://en.wikipedia.org/wiki/Noon#Solar_noon It is the moment the sun crosses the meridian. The meridian is an imaginary half-circle that goes from the northernmost point on the horizon to the highest point directly above you and then to the southernmost point on the horizon. The sun is always at its highest point in its daily path when it crosses the merdian.

You're right that in earlier centuries, each town would set its clocks at noon for when the sun was at its highest point. Those of us who are familiar with astronomy and history would call that type of timekeeping "apparent solar time". You can't guarantee an exact 24 hour day with "apparent solar time", so some people used "mean solar time" instead. (Historical note: Having each town choose its local noon became a nightmare for railway companies that wanted to set train schedules. So the railway companies urged the creation of 4 time zones in America and encouraged each small town to use "standard time" instead of "apparent solar time".

And as rgettman already mentioned, the term for the sun's height above the horizon is "altitude". Astronomers use the term "elevation" for a person's distance above sea level.


Changes of the velocity of the Moon due to impacts from other objects?

Our best current understanding of the formation of the Moon has it forming quite close to Earth.  Because even then its orbit around the Earth took more than one day (Earth was spinning faster, also), the effect of tides has always been for the Earth to transfer some of its rotational energy to the Moon. This slows the Earth's rotation  and makes the Moon move outwards into a higher (and slower) orbit.

It's been around four billion years since the Moon last suffered a major impact, so collisions have not played a significant role since then.

Expanding (a lot) on this:

How do we measure Lunar ages? Two ways: Cratering and isotope dating. The cratering technique was developed by William K. Hartmann of the University of Arizona and makes use of the amazingly simple fact that when two craters overlap, you can tell which is older. Combine that with the observed fact that while the cratering rate varies with time, it changes smoothly and the distribution of craters is essentially random. Using that and a lot of painstaking analysis of photography, the order of events on the Moon can be determined quite closely (E.g., this crater, then this mare, then those craters, etc.) as well as getting rough estimates of the actual time.

Isotope dating of Lunar meteors (bits of Lunar rock blasted into space by impacts) and Apollo samples lets us put precise times to various bits of the sequence resulting in a reliable chronology going back nearly to the Moon's formation.

Arguably there were four stages in the Moon's history:

Stage 1: Formation of the Moon by Giant Impact

The Moon forms out of the debris in orbit after a Mars-sized impactor strikes the Earth about 4.5 billion years ago during the formation of the Solar System.  This was the final giant impact involving either the Earth or the Moon.

Stage 2: The Early Years

When formed, the early Moon was around 25,000 miles from Earth and the huge tides raised by the Moon in the Earth and by the Earth in the moon rapidly (by astronomical standards, anyway) further out.  The Moon (and Earth) continued to be hit by big rocks (dinosaur-killer-sized and bigger) but not by things in the many hundreds of miles in diameter.

Stage 3: The Late Heavy Bombardment

We can tell from cratering patterns and from dating Lunar samples, that 4.1 to 3.8 billion years ago there was an upsurge in big rocks (dinosaur-killer-sized) hitting the Moon. We also see evidence of this on other planets, though not on Earth since so little remains of the Earth's crust from that era.) We don't know why the LHB happened, though there are lots of theories.  (There are also scientists who think the LHB was not as heavy and more spread out, but they are currently in a minority.)

Stage 4: Retirement

Since the end of the LHB, the Moon has had few new craters created and has simply slowly receded from Earth (enjoying a serene and fairly uneventful old age.)  There are still craters being formed now and then -- the freshest big one, Tycho, is only 108 million years old!  Another recent one, Aristarchus, is around 400 million years old.

These are typical of the last few billion years: Very occasional craters caused by rocks in the 10-50 mile rage. They make a big hole, but happen rarely (ca. even hundred million years) and have had no significant impact (pun intended) on the Moon's orbit.


Retrieving input effective radius from Sérsic profile

The conceptual problem is that the half-light radius is the radius of a circle enclosing half the total light, assuming that the object is circular with a radial intensity profile equal to the Sersic function.

So you need to integrate over circular rings, each of which has an area of $2 \pi r \, dr$ and an intensity (or surface brightness) of $S(r)$:
$$L(< R) = \int_{0}^{R} 2 \pi r \, S(r) \, dr$$
where $S(r)$ is the Sersic function evaluated at radius $r$.

Instead, you're just doing the integral of a line of fixed width:
$$\int_{0}^{R} S(r) \, dr$$

In Python terms, the (pre-calculated) function you integrate should be

I = 2 * np.pi * x * Sersic(x)

not

I = Sersic(x)

(Alex is also correct that you have almost no resolution near the value of r_eff that you are trying to find.)


What is the Inner Horizon of a rotating black hole?

This is an attempt at a completely non-mathematical description although I'll use the proper terms for things.  Caveat: it may be wrong: my memory of this is rather distant now and I haven't spent much time trying to sort out the details.  I would welcome corrections (and/or a better answer!).
This answer also started with the intention of being a brief sketch and has turned into something much bigger.  It is as a result very thinly referenced: sorry for that.  I will try to make it better.
Some definitions
The first thing to think about is what sort of paths through spacetime do massive objects take?  The answer is that they travel on what are called 'timelike curves', and, more precisely future-directed timelike curves.  A future-directed timelike curve is simply a trajectory that some massive object, like a person, say, could take through spacetime (you actually need to be fussy about this and insist that the curve is smooth and probably a few other things).  Things which are massless, like light, can travel on 'null curves' and there are also 'spacelike curves' which nothing travels on: null curves represent the boundary between timelike and spacelike curves in some sense.
Future-directed timelike curves are often called 'worldlines', because they are the curves corresponding to the paths objects like people and rockets take through spacetime.
So you can think of spacetime as being full of these timelike curves: each curve represents some possible future for an object starting from a given point in spacetime (and also, going the other way, some possible past for an object which ended up at this point in spacetime).  A 'point in spacetime' is an event: its both a position in space and time.  Spacetime is made of events.
There are special subsets of these families of curves called geodesics, which are the curves that things which experience no acceleration travel on.  A timelike geodesic is the curve a stone would follow (if you threw it in a vacuum and there were no other forces on it...).  Geodesics don't really matter below.
Timelike curves have lengths, and the length of them is the time experienced by an object following them, which is called the 'proper time'.  Null curves have zero length (yes, this is fine).  Geodesics are extrema of length: the geodesic (or geodesics) between two points are the longest curves between those two points.  This is like the definition of a straight line in ordinary geometry but longest rather than shortest.  It's also how the twin 'paradox' works, of course.
Schwarzschild
So first of all, let's look at a Schwarzschild black hole: one that is not rotating.  It has one special surface, which is the event horizon.  The important thing about the event horizon is that any future-directed timelike curve which passes through it will meet the singularity, and will do so after only finite proper time.  What this means is that, once you've passed the horizon there is no way back through it, and at some time after passing through it (time from your point of view, so proper time) you will hit the singularity.  (Caveat: in GR, in real life you presumably meet whatever quantum thing actually replaces the singularity).  This holds for all timelike curves, not just geodesics: you can have a rocket which is as powerful as you like and it doesn't help (although the acceleration from it will kill you).
Kerr
OK, the Kerr solution is more complicated: it has three special surfaces, two of which are horizons.
The static limit
The first special surface is what's called the 'static limit'.  To see what this is you need to know about a weird thing in GR called 'frame dragging'.  Frame dragging is a feature by which rotating objects in GR (such as planets, stars, anything) tend to 'drag' the spacetime around them so that objects near them 'want' to rotate with the large rotating thing.
Well, for a Kerr BH there's a region where, in order not to rotate with the object, you would need to be going at the speed of light.  Or, in other words, all timelike curves within this region get wrapped around the BH.  Within the static limit you can't stay still: you must rotate with the BH.
But the static limit is not a horizon (it does touch the outer horizon at the poles): you can pass into and out of it.  The region between the static limit and the outer horizon (next) is called the 'ergosphere' of the BH.  (Alternatively, the static limit may be called the ergosphere and the region between it and the horizon the 'ergoregion'.)
The outer horizon
The next special place is the outer horizon.  This is just an event horizon as we commonly know it: once you've passed through it, there's no going back.  Any future-directed timelike curve which passes the outer horizon will meet ... something ... in finite proper time.  Once you've passed the outer horizon then you're gone from the point of view of any observer outside the BH (and I think that, as with Schwazschild, it takes an unbounded amount of their proper time for this to happen although it makes no difference in practice).
The outer horizon is, I think, always called the outer horizon.  It is common to label regions of spacetime:

region I is outside the outer horizon
region II is between the outer and inner horizons
region III is inside the inner horizon

So once you're inside the outer horizon you are travelling through region II, and you spend only a finite time in it.  After that, the 'something' you meed is not, this time, the singularity, it is ...
The inner horizon
The inner horizon is, in a way, the nearest thing to the Schwarzschild singularity for the Kerr solution: it's the place that you will get to, in finite proper time, once you pass the outer horizon.  But it's not very like the singularity because it's not a singularity: it's just another one-way door through which you pass.
The inner horizon is a 'Cauchy horizon' (see below).
Once you're inside the inner horizon, you're in region III.  And as before there's no going back.  But region III is very strange indeed.  If you take it seriously, it has two extremely strange features:

not all future-directed timelike curves hit the singularity (I think perhaps almost none do);
there are closed future-directed timelike curves in this region.

Both of these lead to very awkward situations: a closed timelike curve (CTC) is a time machine: it is a worldline which you can move along into your own past.  So that's a bit awkward.  Fortunately these CTCs are all inside not one, but two horizons.
Except: what happens to the worldlines which miss the singularity?  Well, if you take this seriously, what you have to do is to do what's called 'extending' the spacetime of the Kerr solution, and once you do this you have two possible fates for these things:

they can pass through the singularity, which is ring shaped, and get into some very strange, asymptotically-flat spacetime, which I don't understand at all;
they can miss the singularity altogether and then pass out, through a horizon, into a new copy of region II from which they inevitably pass, though another horizon, into a new copy of region I;

And now, in this new copy of region I there is another copy of the Kerr solution (or, really, the same solution again), and they can pass in through another horizon into yet another copy of region II, and thence into another region III and so on.  There is an infinite chain of these things, in both directions.
A Penrose diagram of Kerr
One way of seeing this is to use a Penrose diagram, which is a picture of spacetime which has been conformally transformed to make it compact.  What that means is two things: a conformal transformation is one which can change lengths but doesn't change angles, and making spacetime compact means that points at infinity are brought in to a finite distance, so you can have a picture of an entire, infinite, spacetime on a finite bit of paper (well, OK, you have to suppress two space dimensions in the usual way).
Here is a diagram I drew of the Kerr solution (any errors in this are mine).  This only includes two copies of it: you need to remember that there is an infinite chain of these things.  Time goes upwards.

So, here is what is going on in this diagram.

At the bottom is region I, for which the two outer diagonal edges are past & future null infinity, $\mathcal{I}^-$ and $\mathcal{I}^+$ respectively (not marked).
The outer horizon is the line between I and II.
The inner horizon is the line between II & III, and you can see that once you've got into II you are going to hit this.
The singularity is the dotted line.  If you pass through it you end up in the unmarked region to the left or right of it, and this is the bit of the thing that I just don't understand at all.
If you otherwise miss it (which you almost always do) you end up passing through another horizon into a new II region.
And from that you pass out through another horizon into a new I region, which has its own  $\mathcal{I}^-$ and $\mathcal{I}^+$, also not marked.
And if that's not enough for you you could then travel into yet another II and do the whole thing over again.  Personally at this point I decide I've had enough and just go off to live my life in the new I region.

As I said above, I think this is (now more) correct, but any it may (still) not be.  Note also that the static limit, which is in region I, isn't shown here at all: it just has the horizons.
Probably a toy
But I think it's commonly accepted that this is pretty clearly all just a theoretical toy, because this only works for a solution which actually is a Kerr BH: what people call an 'eternal black hole'.   Real physical BHs are not like this, because they originate from collapsing objects: no BH can be older than the age of the universe.  Therefore I believe that the inner structure of real BHs is probably not like this at all.
In particular I mentioned above that the inner horizon is a Cauchy horizon.  A Cauchy horizon is something which, loosely, divides spacetime into regions where there are closed spacelike curves (no problem with these!) and closed timelike curves (big problem with these!), and serves to 'censor' the nasty region (the one with the CTCs) from the nice region (the one with only closed spacelike curves).  However, I think it is the case that Cauchy horizons have two unfortunate features:

they're not stable under plausible assumptions, so tiny perturbations turn them into something else (not sure what) which would make them unphysical;
if you cross one you experience an infinite flux of energy which is clearly also unphysical since this could only happen for eternal BHs.

Both of those things, in particular the lack of stability, make it seem unlikely that the internal structure of real spinning BHs is Kerr.  Least of all would this seem to be true for BHs which originate from collapse, which is probably all BHs, since whatever internal structure there may be does not yet exist from the perspective of an external observer.
However this all well beyond my competence, and I am simply not sure what the current state of things in this area is.
References

Some lecture notes on the Kerr solution which used to be at https://www.roma1.infn.it/teongrav/leonardo/bh/bhcap3.pdf but no longer are.  I believe these may be a chapter from this book which I found via Leonardo Gaultieri's home page: I have not read the book but the notes looked very good.
On Crossing the Cauchy Horizon of a Reissner-Nordstrom Black-Hole, S. Chandrasekhar and J. B. Hartle, Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences 384, no. 1787 (1982) – I think this is the thing you'd want to read to understand Cauchy horizons, and I may once have read it, but it's behind a paywall so I can't check now.  Certainly Chandrasekhar did a lot of interesting work on this sort of thing.
Wikipedia page on Kerr solution is probably reasonable.
Wikipedia page on Penrose diagrams likewise.

I wish I had some better references, sorry.


Expansion of Space in the Very Distant Future

Maybe someone can give chapter and verse on this, but my understanding is that bound objects are unaffected by the expansion for the case of dark energy represented by a simple cosmological constant.

But a "big rip" will occur in a finite time if the dark energy equation of state $P = w \rho$ is characterised by $w < -1$. In this scenario, the expansion accelerates so quickly that the scale factor becomes infinite in a finite time and structures must disintegrate when the size of the observable universe becomes smaller than that structure.

The present state of play is that $w$ appears to be very close to -1 (a cosmological constant). But given that dark energy is not understood, who can really say whether it might change in the future...


How does the illumination of the sky from the sun compare to that of the moon?

The question is not altogether well defined. One needs to specify what wavelength is being considered and at what position in the sky - since a twilight sky has a very significant gradient of sky brightness.

A paper by Patat et al. (2006) gives the brightness of the sky in the UBVRI photometric bands, at zenith, with the Sun at a variety of altitudes (including below the horizon). Figure 5 from this paper is reproduced below, it shows the zenith sky brightness measured at the Paranal observatory in the V band (approximately 550 nm). Curves for the other bands are qualitatively similar but with different numbers on the y-axis. The vertical dashed lines show where the Sun is 12 or 18 degrees below the horizon (known as nautical or astronomical twilight respectively).

Now to complete an answer to the question we need to know the brightness of the night sky at full Moon. Again, this is not completely defined, since it depends how close you are to the Moon.

The "dark" sky at Paranal is approximately 21.6 mag/arcec$^2$ when the moon is down. That is the horizontal dashed curve in the picture. Another article by Patat (2014) discusses the contribution of the moon as a function of wavelength and angular distance from the Moon. A reasonable number for the full moon at 90 degrees (i.e. the full Moon close to the horizon) is that the sky brightness is increased by about 1.5 mag in the V band (but by as much as 3 mag in the B band). Obviously, if you get closer to the full Moon then the sky becomes even brighter than this.

Using this number and the plot above, we can estimate a night sky brightness of 20.1 mag/arcsec$^2$ (equivalent to a full Moon at 90 degrees) is reached in the V band when the Sun is about 12.5 degrees below the horizon (i.e. very close to nautical twilight).

If you do a similar estimate for other bands, then you arrive at results that are very similar - perhaps 12 degrees below the horizon for the B band and 14 degrees below the horizon for the I band, presumably because moonlight does not have a drastically different spectrum to sunlight (it is a bit redder).

Caveats (at least): The answer will also depend on how far away the Moon is from the Earth, the altitude of your observing site, whether there is any cloud in the sky, what the density of scattering particles are in the atmosphere and the altitude of the full Moon.


Assumed a Venus year is 0.615 198 Earth years.

0.615 198 yr / (1 - 0.615 198) = 1.598739 yr = 583 days 11 hours 24 minutes for one period in mean, or roughly 19.5 months.

That's 1.598739 periods for Earth and 2.598739 periods for Venus:
1.598739 * 365.256 days = 583.95 days = 2.598739 * 224.705 days

The relative accuracy should be somewhere near 1e-5.

Since both orbits are elliptical the actual period oscillates around that mean value.


What would the effects on or around Earth if Betelgeuse went supernova?

tl;dr - The main measurable effect may be minor climate cooling, but in day-to-day life, the only difference would be that we see a cool, bright explosion in the sky, and eventually, Orion becomes "incomplete".

The effects would likely be quite minimal. What Will Happen When Betelgeuse Goes Supernova? by  Corey S. Powell, former editor in chief of Discover and reproduced from their Quora answer addresses this exact question quite well. The supernova itself would be quite bright in the sky for a while, and it would certainly be something we'd all want to get nice photos of before it goes away, but the actual "shock wave" would take much longer to reach us, and given the distance between us and Betelgeuse, it will have largely dissipated.

The article describes a period of climate cooling a couple million years ago which corresponds with evidence of a nearby supernova, but we have yet to establish a firm, causal link. As of now, we don't expect to really measure any effects, also partly since the shock wave from the supernova would take much longer to reach us than the light. So even if it we saw it go supernova in the next 20 seconds, we would not be able to observe the effects in our lifetimes.

As for our upper atmosphere, as you mentioned, the radiation and material from the supernova would not nearly be strong enough to do anything like destroy ozone (or as the article mentions, ionize the atmosphere), so I imagine the upper atmosphere would stay relatively the same (this assumes the lack of causal link between supernovae and climate cooling - if a link is established definitively, then perhaps that cooling could have some effect on the upper atmosphere - that goes more into meteorology and weather than astrophysics)


Did Earth "tilt" seasonally before it was accepted that the world was not flat?

This is not really a physics or astronomy question, but I reckon that when you model the Earth as flat, the stars and the Sun are merely minor objects and their motion is governed by heavenly spirits rather than regular orbits. So, rather then the flat Earth tilting, the Sun's trajectory changed with the seasons.

Btw, what I find more concerning is how the (infinitely extended) flat Earth can be reconciled with the Sun rising in the East after setting in the West: how did it get there?

Btw2, I don't think that elites in classic times (Greek/Egyptian/Middle East Cultures ~1000Bc and more recent) seriously considered the Earth to be flat. All the constructions of epicyles etc for explaining the data assume a spherical Earth in the centre of the cosmos. Flat is much more arcane than Geocentric.


Could a binary system of two planets with oceans reflect each other?

The surface of the ocean isn't smooth, so there would not an image formed.

Look at how the Earth's oceans reflect the sun.

(credit: NASA/ESA)

Obviously the sun is reflected by the ocean, but the rough surface of the ocean scatters the light enough for no image to be formed. It would be the same if there were a planet being reflected: No detail would be visible.


though the discovery of such a polished large body is odd, even if it is there it would not be much useful because it will then show us the historic times of 19th century which archaeologists themselves can easily discover using evidences.


Your question is one that bothered cosmologists for a long time, how do you get clumps of matter out of a smooth initial condition, in the time allowed?  It was long understood that a perfectly smooth matter distribution could not be maintained forever, because gravity creates instabilities-- the smallest overdensity in one region would tend, over time, to pull in more matter and increase the overdensity.  But this takes time, and the CMB measurements were coming in surprisingly smooth-- so how did galaxies and stars form in less than a billion years, given that very uniform initial density?

The answer seems to be dark matter.  Dark matter increases the mass and therefore the gravity, so it reduces the time it takes for instabilities to develop and overdensities to grow.  So the simple answer to your question is that even though cosmology assumes a smooth initial density, it does not say the density is exactly smooth, and indeed the CMB tells us there are tiny density variations (you can look at the tiny variations seen by the PLANCK satellite at https://en.wikipedia.org/wiki/Planck_(spacecraft)#/media/File:Ilc_9yr_moll4096.png, for example-- the scale shows variations of 1 part in ten thousand).  Then gravitational instability increases the overdensity, but needs stronger gravity to do this in the time allowed, so that appears to be provided by dark matter.


Why won't the Milky Way colliding with Andromeda affect the Solar System?

Outside my field, but collecting information from the Wiki article, I can make an attempt nevertheless:

You're right that we will probably pass through Oort-like clouds of other stars. But the largest of the planetesimals of the Oort Cloud is measured in kilometres, whereas the distance between these objects is measured in AU, so the total cross section of object compared to that of the cloud is still miniscule. Smaller objects are larger in number, but still it's a very low probability.

Gravitationally, the clouds also won't affect us. The total mass of the Oort Clouds is probably of the order of 10 Earth masses, i.e. much smaller than the Sun.

The outer parts of the Oort Cloud is only loosely bound to the Sun. I suppose the best evidence that we won't have any major collisions is that then we would already have similar objects bombarding us from our neighboring stars' clouds.


Are many exoplanets synchronously tidally locked like Mercury?

GJ 581d and GJ 667c are candidates for showing spin-orbit resonance.
This paper mentions GJ 3634b and 55 Cnc b as two further candidates.

Spin-orbit resonances different from 1:1 are expected to be rather common among rocky planets orbiting close to a star. Direct observational evidence is difficult to obtain; results are based on model calculations.


What is the radiant intensity of the celestial sphere?

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump.

A plot of $\nu I_\nu$ (proportional to energy density per logarithmic frequency interval), taken from Hill et al. (2018)

The CMB is approximately isotropic blackbody radiation with a temperature of 2.73 K and therefore a specific intensity of
$I_{\nu} = \frac{2h\nu^3}{c^2}\frac{1}{\exp[h \nu/k_B T] -1} $$

The integral of this over all frequencies is
$$ u = \frac{8\pi^5 (k_B T)^4}{15 (hc)^3} = 4.2 \times 10^{-14}\ \ {\rm J\ m}^{-3} $$

In terms of power, we just multiply by $c$ to get $1.3\times 10^{-5}$ Wm$^{-2}$.

Note that you can't harness this unless you have a receiver that is colder than the CMB.

You can do a rough calculation for the contribution of optical background light by noting that the dark sky at a good, dark astronomical site is around 22 mag per square arcsecond in the V band. Only about 0.1-0.2 mag of this could be attributed to light pollution at the best astronomical sites (see for example Benn & Ellison 2007). Noting that the zero point for the V-band magnitude scale is $3.6 \times 10^{-12}$ Wm$^{-2}$ per angstrom for $V=0$ and taking the optical band as 3000 angstroms, and noting there are $5.34\times 10^{11}$ square arcseconds in the sky, then a point in space will receive $10^{-5}$ Wm$^{-2}$ (or perhaps about half this, because a significant fraction of the dark sky brightness is caused by airglow).

This roughly agrees with what is shown in the plot. Contributions at mid-infrared and at shorter wavelengths than optical do not contribute appreciably.

It might be thought that discrete sources or the Milky Way might contribute more, but I think that isn't the case. The Milky Way surface brightness is only about a factor of three above the dark sky brightness, but of course occupies a tiny fraction of the total sky.

A way to see that discrete stars don't contribute much is to note that the optical flux from the entire dark sky is equivalent to about 1000 zeroth magnitude stars or $10^7$ stars at 10th magnitude. Both numbers are higher by orders of magnitude than the actual numbers of Galactic stars at these brightnesses.


Could gravitational waves near merging black holes collapse to a black hole themselves?

You might like to look at some of the simulations from the SXS collaboration. Essentially the black holes and the gravitational waves are aspects of the same thing -- the curvature of space. In particular, the way black holes seem to merge in these simulations is that the two black holes come very close and space around them becomes more and more curved (and in more complex ways). Some of that curvature spreads out in the form of gravitational waves, becoming less intense as it gets further away, but some of it, very close to the two black holes reaches the "critical" level of curvature at which an event horizon appears (extending the event horizons of the two black holes) which then "rings down" becoming spherical and form the merged black hole.


Since this seems to be a kind of niche question I will answer and leave some comments for future souls with the same problem.

To whom it may interest that may be searching for the same problem.
The solution is the basic $M = m + 5( \log (parallax) + 1)$. Parallax in arcsec.

The reason I'm going with K instead of V is due to a more independent relationships with the stellar metallicity

Parallaxes can be from catalogues like Hipparcus or Gaia.

If anyone is searching and gets here with the goal of using $M_K$ in the $\log (M / M_\odot) - M_K$ relation from Delfosse et al. (2000) keep in mind that in the paper the $\log$ relation is in $\log_{10}$ !!

Delfosse et al. uses "Johnson-Cousins-CIT", for that I followed equation (12) on Carpenter, J. M. 2001, AJ, 121, 2851.

If you want the $\log g$ I recommend the relation in:

Bean, J. L., Sneden, C., Hauschildt, P. H., Johns-Krull, C. M. and Benedict, G. F. (2006), ‘Ac-
curate m dwarf metallicities from spectral synthesis: A critical test of model atmospheres’, The
Astrophysical Journal 652(2), 1604.


It would be more accurate to call them surveys than studies.   A study looks at an object in detail.  A survey counts and categorizes objects.

Rogue planets are enormously difficult to see, and only a few very large ones have actually been observed directly.  For the largest rogue planets, it's unclear if they actually are ejected planets as opposed to failed stars.

This article discusses rogue planets in pretty good detail.

In 2011 the MOA survey was published, estimating a high number of Jupiter sized Rogue planets, perhaps outnumbering stars.

More recently the OGLE Project estimated about 1/10th that many rogue Jupiters, which is more in line with statistical predictions.   Smaller planets are more likely to get ejected from their solar-systems than larger ones.    Rogue Juiters outnumbering stars seemed unlikely and my understanding is that most scientists are more in line with the OGLE estimate than MOA.

Both estimates rely on microlensing which amounts to very small and temporary changes when rogue planets drift between the Earth and a star.  There's a fairly large margin of error for both surveys.

NASA's WFIRST telescope should answer a lot of questions on rogue planets.   It's not expected to be launched till the mid 2020s though.


If the Universe is infinite, why isn't it of infinite density?

I would like to point out that any information (even quantum mechanical) travels at most at the speed of light. (There are some problems with entangled systems etc, but we can assume most systems to not be entangled, which is a fair assumption, imo). So, if you consider the age of the universe to be finite, any point will have only finitely many overlapping wavefunctions (those within the range of visibility, and the probability can be very very small at most points). So that means that there is never an infinite amount of atoms in the visible region at any finite time, meaning that the density never reaches infinity (or does so only at t=inf).


Assuming your question is a reference request:

I now about one resource that has a very broad range of spectral data, the

NIST Atomic Spectra Database Lines Form

This has worked great for me.

This page lists a couple of different alternatives. For the near IR case you are interested in, several collections are listed here.


Where does the energy of light go, when it red-shifts?

The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify any inertial frames and certainly not ones that encompass a cosmologically significant volume.

What this means is that if you make a local "box" small enough that it is not affected by the expansion of the universe, then energy conservation will apply. But of course in such a box, a photon would enter and leave with the same energy because the box is unaffected by the expansion of the universe and so there would be no redshifting of the photon.


In which direction does the Milky Way rotate?

The Milky Way has arms that form due to density waves. Like the majority of spiral galaxies, the arms are trailing. Individual stars orbit in circles (roughly), neither towards or away from the centre.

If you consider a common map of the Milky way (imagined from a point North of the Earth, Celestial North is not the same as Ecliptic North, which are both about 60 degrees off from Galactic North)

The stars in the galaxy would be moving in a clockwise fashion.

If you were to view from the other side, it would be as if you had made a mirror image of the galaxy, so the motion would be counterclockwise. If you look at the back of a clock, then from the wrong side, the hand would move counterclockwise. However the rotation of the galaxy is still with its arms trailing.

There are some galaxies that rotate with arms leading: NGC 4622 is one example.


The light of which star is next to reach the earth?

I think you may be thinking of Olbers' paradox
. This supposes that if the universe were infinite in time and extent, and stars were more or less randomly placed, then every line of sight would end at a star, and so the sky would be as bright as the surface of a star.

The paradox can be resolved by supposing that the universe is not infinite in time, but had a beginning.

We don't see stars suddenly appearing as their light reaches us. What we see is stars in all stages of their development. When we look the nebulae where stars are born we see clumps of gas in the process of collapsing to form stars. These are called protostars. Now the process of stellar formation takes a long time: about a million years for a medium sized star. The start of a star is not a sudden "turning on". Instead, as the core of the protostar shrinks and heats up, nuclear fusion starts; slowly at first, then increasing until it is able to prevent further collapse and heating.

Protostars usually are dim in visible light, they can be found with infra-red telescopes.

Since the process is slow and there isn't a sudden turning on, your questions cannot be directly answered. However, there is an object HOPS383 which is 1400 light years distant, which between 2004 and 2008 had a "growth spurt" and became visible in infra-red for the first time. It can't be seen in visible light yet - so in a sense this protostar answers your question 1 and 2, in different wavelengths.


Did Hubble see a pair of UFOs? (Identify this artifact)

I'm going to guess that these are reflection artifacts of one or two nearby bright stars. The fact that they have about the same length as the stellar streaks, suggests they are smeared by the telescope's tracking of the comet. The fact that they are oriented with a slightly different angle than the star streaks may be due to the reflection shifting slightly as the telescope is moved.

In fact, Figure 6.9 on this page from the WFC3 Handbook (the instrument used for the comet images) shows a clear examples of "figure-eight shaped ghosts" due to bright stars in the lower-right quadrant, described thus:

  When a point source is positioned in the lower right quadrant of the
  UVIS detector, out-of-focus reflections between the CCD and windows
  appear along a diagonal from the source towards the upper left, well
  removed from the source. These figure-eight shaped ghosts gradually
  move outside the field of view as the target moves out of the lower
  right corner. They contain a few percent of the flux of the target.
  The ghosts of several bright stars are visible in the exposure of 47
  Tuc shown in Figure 6.9.

Figure 6.9 -- note the clear example paired elliptical rings ("figure-of-eight shaped ghosts") on the left part of the upper half of the image.

(Note that the comet images were taken with a sub-array of the WFC3 UVIS detector, which is probably why you don't see the star that produced the ghosts.)

You see ring-like ghost images in ground-based images, too; I've been examining an image taken by some collaborators with a telescope in the Bavarian Alps, and it's got several examples (fortunately not on top of the galaxy we're interested in!).


Do the heliospheres of stars at the center of globular clusters interact?

Similar to what @Barrycenter said in his comment, There's a wide range of distances between 100 or 1,000 times closer on average and as close as binary stars.

Alpha Centauri A and B get as close as 11.2 AU.   The Sun and Alpha Centauri are roughly 270,000 AU apart so at 1,000 times closer, the close stars in a globular closer would still be (ballpark), 250-300 AU from their nearest neighbors.  In a crowded ball of stars you're not likely to get binary systems that far apart, so most of the interaction that you're looking for would be in close binary systems.   Neighboring stars in a globular cluster would mostly be too far apart to have much interaction.

There's also a world of difference between the corona and the heliosphere.  The Sun's corona extends about 12 solar radii or about .05 AU.   The heliosphere, while, it's one, not a sphere and two, probably has some variation in it's boundary, extends roughly 100 AU (121 by this estimate), over 2,000 times further than the corona.

Binary systems like Alpha Centauri A & B probably have a shared heliosphere and/or some form of heliopause between them where their two solar winds meet.   It would be interesting to study that in more detail.  I'm not sure how much is known about the heliopause between two binary stars, especially given that they orbit each other.   I'm not sure what form it would take - perhaps that would make a good question.

Binary stars need to be much closer than 11.2 AU to have corona to corona interaction.  Stars that close are probably quite rare but they do happen.   Here's one example of a binary star system so close they're actually thought to be touching.


Subject Size
In angular dimensions, planets are tiny compared to the Moon.  The Moon is roughly 30 arc-minutes (about 1/2° ... varying slightly from apogee to perigee).  In real life they are, of course, MUCH larger than the Moon... but they are also MUCH farther away so they appear much smaller.
In comparison to the Moon's approximately 30 arc-minutes size, Mars is nearing it's closest approach (making it appear larger than usual) but is a still a mere 23 arc seconds.  Saturn is 17 arc-seconds across.
This means you are trying to resolve fine detail on an extremely tiny area and this is pushing the limits of what your telescope handle.
Collimation
Other factors such as optical collimation (alignment of the optical axis of the primary and secondary mirrors).  Portable Newtonian reflectors tend to need somewhat frequent collimation.  Collimation is easy to perform and will improve optical quality.  Also, if the temperature of the telescope mirror has not had time to acclimate to the ambient temperature, the mirror itself will reduce the quality of what you can see (e.g. if a telescope were stored in a warm house but taken outside on a cool evening).  Smaller telescopes will acclimate faster because the mirror has less mass.
Seeing Conditions
Seeing conditions will also have a large factor.  Weather fronts such as cold fronts, warm fronts, or the presence of the jet stream will impact seeing quality.  There is little you can do about this.  (Large telescopes can use 'Adaptive Optics' but this isn't a common thing in amateur astronomy.)
The best analogy I've come across to describe seeing conditions is to imagine placing a coin in the bottom of a pool or fountain.  If the surface of the water is perfectly flat -- no waves -- then you can easily see the coin.  With magnification you can see details on the coin.  But if a second person starts tapping the water surface to create waves... you can see the coin ... but not details.  The coin is optically distorted by the waves.  The atmosphere does the same thing when we use telescopes.  It need not be physical "waves" but can be things such as mixing warm & cool air in the upper atmosphere.
Infinity Focus
Everything in space is effectively focused to "infinity" for your telescope.  If anything is focused, then everything is focused.  This means if it makes things easier for you, focus your instrument on the moon and achieve the sharpest focus possible.  Then re-point the telescope to a planet and should not need to re-focus the telescope.  You will likely want to do this using your 6mm eyepiece (because changing eyepieces usually requires re-focusing the instrument.  There is such a thing as "parfocal" eyepieces -- which all share the same focus and you can swap eyepieces without refocusing.  Most eyepieces are not parfocal with other eyepieces.
Maximum useful magnification
There are some formulas for working out the maximum useful magnification of an instrument.
One such (rather simplistic) approximation is that the maximum magnification is 2x the aperture when measured in millimeters.  As you have a 130mm aperture, this would imply that 260x is the maximum magnification.  HOWEVER (huge caveats) this assumes flawless optics in flawless seeing conditions -- which almost never happens.  A more practical limit is roughly 1x of your aperture.  This becomes subjective as to when you decide that the magnification is making things worse instead of better.
It turns out mathematically... that the focal length of an eyepiece needed to provide magnification that matches 1x of your telescope's focal length in millimeters... will always be the same as the focal ratio of the telescope.
Since you have an f/5 telescope, a 5mm eyepiece will provide 130x magnification.
Both your 20mm and 6mm are fine and should not over-magnify the image ... you can use a 2x barlow with the 20mm, but you will not likely enjoy the views of the 6mm eyepiece combined with the 2x barlow.
There are more technical ways to discuss maximum focus such as the very simple Dawes' Limit or using Rayleigh criterion.  These methods discuss maximum resolving power of an instrument based on aperture and wavelength of the light.  But even these methods have a low-bar.  That is to say... their standard is whether two points of light with some amount of angular separation could be resolved as two points of light and not just one merged point of light.  They wont look sharp ... they'll just be blurry points that have noticeable separation.
One conclusion that can easily be drawn by looking at the formulae for maximum resolving power of an instrument... is that the bigger the aperture of the instrument, the greater the detail it will be able to resolve (which is why professional astronomers want to strive for larger and larger telescopes).
On a personal note, my first telescope had a 3.5" aperture.  I could see that Saturn had "rings" and I could see that Jupiter had "bands".  Then I got a 5" telescope... and on nights with better seeing, I could just barely make out that the ring system on Saturn had a dark band in it (the Cassini division) and that Jupiters belts sometimes had texture.  Over the years I would eventually get larger telescopes and, unsurprisingly, larger aperture telescopes allowed for viewing greater amounts of detail.  Have a reasonable expectation as to what you should be able to resolve with a 130mm (roughly 5") instrument.
Atmospheric Dispersion
When looking at anything low in the sky you may notice some "color fringing".  If a planet has a slightly blue edge on one side and a slight red edge on the opposite site, then this is typically a result of the atmosphere acting like a prism ... it's separation the light spectrum because the object is very low in the sky.  I live at fairly northern latitudes and this is true of observing outer planets during the summer months (when Earth's northern pole is tilted away from the outer solar system and toward the Sun ... from my latitude.  But southern hemisphere observer's are tilted toward the outer solar system, so they see these same planets very high in the sky where atmospheric dispersion doesn't impact them as much.  Basically outer planets appear high in the sky when observed during your hemisphere's "winter" season and low in the sky when observed during your hemisphere's "summer" season.
This dispersion will tend to blur the surface details of any planet.  It isn't really the fault of your instrument (there is a device called an Atmospheric Dispersion Corrector which is basically an adjustable prism and it is used to reverse the dispersion created by the atmosphere).


What is the distance from Alpha Centauri to Barnard's Star?

To find the distance from one star to another, we need three things for both of the stars: their right ascensions, declinations, and the distance from Earth to those stars.
So, let's get those things:
From the Wikipedia page on Alpha Centauri:
$RA = 14^h\:39^m\:36.49400^s$
$DEC = -60^{\circ}\:50'\:0.23737''$
$R = 4.37\:\rm{ly}$ (you gave 4.366, some other sources give 4.367... I'm going to stick with 4.37)
and for Barnard's Star:
$RA = 17^h\: 57^m\: 48.49303^s$
$DEC = +04^{\circ}\: 41'\: 36.2072''$
$R = 5.958 \: \rm{ly}$ (again, you gave a slightly different value, I'm sticking with Wikipedia for now)
where RA is right ascension, DEC is declination, and R is radial distance from Earth to the target star.
Now, by themselves, it is relatively difficult for us to obtain an actual distance. What I would do is convert these to rectangular coordinates, and then it's a matter of using the 3-d distance formula.
First, however, we need to convert RA and DEC into units like radians or degrees.
For right ascension, we can use the general formula:
$degrees = 15  (h + \dfrac{m}{60} + \dfrac{s}{3600})$
and for declination:
$degrees = deg + \dfrac{m}{60} + \dfrac{s}{3600}$
(when the declination is negative, however, multiply all terms in the formula by -1)
So, for Alpha Centauri AB, we have:
$RA = 15  (14 + \dfrac{39}{60} + \dfrac{36.49400}{3600}) \approx 219.902^{\circ}$
$DEC = -1  (60 + \dfrac{50}{60} + \dfrac{0.23737}{3600}) \approx -60.833^{\circ}$
and for Barnard's Star, we have:
$RA = 15  (17 + \dfrac{57}{60} + \dfrac{48.49303}{3600}) \approx 269.452^{\circ}$
$DEC = 04 + \dfrac{41}{60} + \dfrac{36.2072}{3600} \approx 4.693^{\circ}$
Now, to convert from spherical to rectangular coordinates, we have to define which of RA, DEC, and R can be assigned to $r$, $\theta$, and $\phi$. R should be $r$ - that's pretty straightforward. Since RA can be thought of as "celestial longitude", we'll assign it to $\theta$, and thus declination will be $\phi$.
To clarify, I'm defining $\phi$ as the angle from the xy-plane - so a $\phi$ of $\dfrac{\pi}{2}$ would mean pointing straight upwards. I know some sources define $\phi$ as the angle complementary to that angle (so, $\dfrac{\pi}{2}$ -
the angle from the xy-plane), but for astronomical purposes, I think the definition I'm using is more intuitive and easier to work with.
We can then use the conversions:
$x = r\cos{\theta}\cos{\phi}$
$y = r\sin{\theta}\cos{\phi}$
$z = r\sin{\phi}$
So, for Alpha Centauri AB:
$x = 4.37  \cos{219.902^{\circ}}  \cos{−60.833^{\circ}} \approx -1.634\: \rm{ly}$
$y = 4.37  \sin{219.902^{\circ}}  \cos{−60.833^{\circ}} \approx -1.366\: \rm{ly}$
$z = 4.37  \sin{−60.833^{\circ}} \approx 3.816\: \rm{ly}$
and for Barnard's Star:
$x = 5.958  \cos{269.452^{\circ}}  \cos{4.693^{\circ}} \approx -0.057\: \rm{ly}$
$y = 5.958  \sin{269.452^{\circ}}  \cos{4.693^{\circ}} \approx -5.938\: \rm{ly}$
$z = 5.958  \sin{4.693^{\circ}} \approx 0.487\: \rm{ly}$
And now, finally, we can use the distance formula for 3-d:
$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$
So, the distance between Alpha Centauri AB and Barnard's Star is:
$d = \sqrt{(-1.643 + 0.057)^2 + (-1.366 + 5.938)^2 + (3.816 - 0.487)^2} \approx\mathbf{5.873\,ly}$
Well, that was certainly tedious - but it's a process that you can standardize to pretty much any star, or really, any two astronomical objects:
First, convert RA and DEC to degrees.
Second, assign R, RA, and DEC to the spherical coordinates $r$, $\theta$, and $\phi$.
Third, convert spherical coordinates to rectangular coordinates.
Lastly, use the distance formula with the two sets of $x$, $y$, and $z$ coordinates.
Hope this helps. :)


What would happen to the Sun's evolution if it collided with another yellow dwarf?

If the Sun collided with another star about the same mass, then its mass would be slightly less than 2 solar masses, as some material would be ejected away from the merger. This would result in an A-type star, as the merger's mass is about 2 solar masses. A good example of a 2 solar mass star is Fomalhaut A, which is an A3V  star. Therefore, this merger would be somewhere around Fomalhaut's position on the HR diagram.
If the mass of a star is doubled, then its lifespan is decreased by eightfold, meaning that the merger would live for about $\dfrac{10}{8}= 1.25 \text{ billion years}.$ With that said, the transition to the red giant phase would not be prolonged, but shortened.
When this merger becomes a red giant, its core will have less mass than the Chandrasekhar limit, and become a white dwarf. At its largest, it will exceed 256 solar radii, up to about 400 radii, according to Wikipedia. χ Cygni is a red giant that is about 2 solar masses, which is a good reference to the aging merger. This star will become a white dwarf, so that is further proof that the Sun-merger will have a shorter lifespan as well as not have enough mass to exceed the Chandrasekhar limit.


How do I find the signs that planets are in during any given date?

You can use the software Stellarium, which is available on most plateforms, and in-browser. With this software, you can show the constellations and the planets at any time and date, and see where the planets are in the sky.

For instance, here you can see that Mars is in Libra and Mercury is in Scorpius.

At the dates you specified, the positions of the planets were the following