Difficulty in using the telescope as a projector

Yes this is a bit more of a challenge! The first scope in the photo is long and thin and before the big square cardboard shadow-maker is put on it can be easily pointed at the Sun by moving the telescope until its shadow becomes a small circle, the minimum outline of the scope. If it's off a little bit that long tube's shadow will immediately change shape.
But with a short squat form factor your telescope won't be so easy. Start with the smallest magnification largest focal length eyepiece and at least try to use the shadow method of looking at the shadow that your telescope makes behind it (on something flat and held perpendicular to the Sun's direction, and move the scope around trying to make the most circular profile shadow. For this method, instead of a finder scope you can use an empty long thin tube to make a shadow.
Wait, you have a finder scope!
Eyepiece projection will work just as nicely for your finder scope as your main scope, so BEFORE you use it for the Sun, check that the finder is aligned to the main scope EITHER AT NIGHT or using TERESTRIAL OBJECTS on the ground. DON'T POINT IT AT THE SUN if there's any chance that you will look through it.
Those have to be distant objects so parallax doesn't lead to misalignment.
Once your finder scope is aligned to your main telescope, use eyepiece projection to get a nice centered image from it. At that point your main scope should be really close.


How much greater is the star density in our galactic bar?

In other barred galaxies which are vaguely similar to the Milky Way, the contrast in (projected) stellar surface density between the bar and the inter-bar region at the same radius (e.g., along the bar's minor axis, perpendicular to the bar) is typically a factor of at least two; in particularly strong bars it can be as high as six (see, e.g., Figure 5 in Ohta et al. 1990). Similar contrasts are seen in N-body models of disk galaxies that form bars.

It's much harder to figure this out for the Milky Way, because we're not looking down it from above. The best attempt to derive a model of the bar's 3D stellar density from star counts and distance estimates that I know of is Wegg et al. (2015). From the face-on projected view of their model (their Figure 14), I would guesstimate the maximum contrast as a factor of 4 or so.

Figure 14 of Wegg et al.: projected face-on view of stellar density for the Milky Way (full model in right-hand panel).

The 3D density (which is maybe what you're really asking about) in the inner part of the bar is not quite as great as this suggests, because the inner part of the bar is vertically thick, forming a "boxy/peanut-shaped" bulge (this would correspond to the red region in the figure above). So the contrast would be a little less compared to the (less thickened) inter-bar region. But the outer part of the bar is roughly as thin as the rest of the disk, so the projected surface density contrast would mean a similar contrast in 3D stellar density.


TL;DR: 1000 K (according to differentiated model of Pluto)

According to the density value of Pluto, astronomers proposed three types of structural models:

Undifferentiated or "cold" model: rocks mixed with water-ice
Differentiated or "hot" model: rocky core and water-ice mantle
Rocky core only (temperature high enough to boil off water-ice.

There are two types of rocks to consider i.e. CI rock- carbonaceous chondrite($\rho$ = 2.66g/cm3) for undifferentiated models which is $\approx$13% by wt. eposomite(MgSO4) and Prinn-Frigley rock($\rho$ = 3.262 g/cm3) for differentiated models which is sulfide-rich rocks like troilite(FeS).
McKinnon and Parmentier(1986); McKinnon and Mueller(1988) correctly predicted the structural model of Pluto to be differentiated one. During initial formation of Pluto, radiogenic heating was the only source of heat, hot enough to melt the outside part of the undifferentiated core consisting of hydrated CI-rocks to form liquid matter of what is known today as water-ice in the mantle (water-ice melting temperature was 251 K at 0.21 GPa). During that time, 80% of core was at above 500 K . The anhydrous silicate core at the center was still rigid resisting the outside flow (the temperature of core was 500 K). Convection through heating allowed steady-state heat flow to form water-ice in the mantle, some of which erupted into the surface. Although the gravitational potential energy was modest, the process become self-accelerating.
This process was also accelerated by the "accretion process" due to a large-body impact that resulted in the formation of Charon. This collision heated up the core by 100 K and also resulted in several differentiation of Pluto that made the surface level ice layer mixed with organic compounds.
Present day scenario: Pluto is a rock-rich world with rock mass fraction $\gtrsim$ 60%. Core temperature is 1000 K, surrounded by 250-300 km thick mantle shell of predominately water-ice and mixed liquid rock. Surface temperature is 40 K. Core pressure is 1.1-1.4 GPa. Rock fraction (rock/rock+H2O+ice) for differentiated Pluto is 0.69-0.79(mainly PF-rock). There is also an extra thermal boundary layer between the ice-mantle and the ice-rock layer just outside the core. There might also be an organic layer in between the core and mantle but it hasn't been proven yet. The water-ice mantle is differentiated into several layers based on its transitions. You can check the details and calculation in ref.2

Fig. 1: Temperature of Pluto at several Pressure value

Fig. 2: Temperature of the mantle and outside layer

Fig.3: Differentiated model of Pluto
References:

http://articles.adsabs.harvard.edu//full/1988LPI....19..764M/0000764.000.html
Pluto and Charon, Alan Stern, David James Tholen, University of Arizona Press, 1997
https://arxiv.org/ftp/arxiv/papers/1510/1510.06604.pdf (This paper discusses the possibility that Pluto might have an iron core)


Current estimates put Apophis's diameter around 325 m and its 2029-04-13 approach about 38000 km from the center of the Earth. I figure an angular size <= 2 arcsec, almost starlike even if you manage to track it in a telescope.
Ephemerides show it at apparent magnitude 3.5 or so just before closest approach - visible but not outstanding to the unaided eye at night - and spending about 36 hours within 1 lunar distance of the Earth.


Special Relativity as Applied to an Interstellar Starship?

In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.

For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.

It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.


The singularity itself is, by definition, zero dimensional (well...you could argue one dimensional as it exists over time, but it has zero spacelike dimensions).

The event horizon is a two dimensional sphere, just like the surface of a beach ball - whether it can be said to have any thickness is unknown, there might be a thickness down at the Planck scale, but that's currently unknown.


Can I see "my own hand" in the void space?

Yes, and no. You will be able to see your hand's silhouette as it blocks incoming light from other sources, but you won't be able to see any features on it. For example, illuminance of starlight overcast moonless night sky is quoted at $0.0001\ lux$ which means that in absence of a brighter light source, human eye simply isn't that sensitive to such low illuminance, even if your hand reflected all of the incoming light (it doesn't). What is the minimum lux rating that a human eye can resolve is not as straightforward to answer though, as it varies greatly by specific wavelengths. This is explained in more detail for example in this Human eye sensitivity and photometric quantities (PDF) document, or perhaps Wikipedia on Scotopic vision. But in a nutshell, no, you wouldn't be able to see any features on your hand in outer space, unless closer to a brighter visible wavelength light source. You'll still be able to infer that it's still there and all fingers on it accounted for, though.


Has the existence of earth-sized rogue planets been confirmed?

The list on Wikipedia looks current. It doesn't list planets that have only been detected as one-off microlensing events. This can detect Jupiter sized objects (with considerable uncertainty in mass). But observing a microlensing doesn't allow for study, so a microlensing is not considered to be enough evidence for naming or cataloging a known planet. So the short answer is no Earth sized rogue planets are known.

This is entirely due to the difficulty in detection. An Earth sized rogue planet would cool on its surface to a few degrees above absolute zero, and be undetectable with current telescopes. Microlensing events with small rogue planets would also be smaller and harder to detect above noise.

An Earth sized rogue planet has a perfectly good inertial frame, (it is the bound planets that are in a rotating, non-inertial frame), nor does a frame of reference have anything to do with implosions. Dark matter is doesn't grind anything since the main property of Dark matter is that it doesn't interact with regular matter.

In all likelihood, there are billions of rogue planets in the galaxy, and some are smaller, and some are Earth-sized and "rocky". They could have been formed around a star and then ejected.


Does Jupiter rotate at the same speed at every depth?

No. The deep interior of the planet rotates as (nearly) a rigid body (Guillot et al. 2018), while the outer part of the planet undergoes differential rotation. Therefore the interior and exterior rotation cannot match at all latitudes simultaneously.


What would the sky have looked like when galaxies first started to form?

In early times of galaxy formation you would have seen a sky similar to this Hubble Extreme Deep Field image. A sky filled with small, irregularly shaped, young galaxies and galaxy predecessor fragments (sub-galaxies). Galaxies probably formed bottom-up from smaller fragments.

Further back in time it would have been dark. Hydrogen clouds would begin to contract, as predecessors of first stars (population III) and clusters of stars.

Even further in the past (redshift between 300 and 400) the sky would have been glowing red by what we now call cosmic microwave background.


Open source alternative for RegiStar (align, rotate, and combine images into single frame)

I've had success with Hugin. It's a panorama creator, but it can be used to rotate, resize, and relocate images into a coherent whole. It has a few different automated profiles, but sometimes they are a bit off with star stacks. You can manually select the guide stars though.

A really good program for after that is GIMP and the plugin G'MIC to stack and remove noise.

Unfortunately it's not a single package like RegiStar, but it's free and can do quite a lot!

Before

After


How bright can white dwarf stars glow as they accrete matter?

There might be a problem with your calculation, because by the time a white dwarf made of helium gets up to 0.5 solar masses, not only are the surface layers hot enough to fuse hydrogen, they are actually hot enough to fuse helium!  That's what will happen to the Sun when it ceases to be a red giant.  You are talking about carbon white dwarfs, but they are similar to helium white dwarfs.  So hydrogen fusion at the surface of a white dwarf is a common result-- it is what causes a phenomenon called a "classical nova."  So you are on the right track-- but what you are talking about is a well-known area of study.


How to polar align an equatorial mounted telescope when you are close to the equator?

Mount a camera on the tripod and take a long exposure. You should be able to determine the concentric circles of the stars paths. The center of the circle is the North pole.


Can virtual particles be the source of dark matter and dark energy?

The problem with associating virtual particles with either dark matter or dark energy is that, quite literally, virtual particles aren't real.  By that I mean, they are intended to change the effects of energy, like how the energy is transported and where it goes, but they shouldn't alter the amount of energy.  As such, virtual particles are just a kind of picture, a way to view the terms in a particular way of writing the equations.  Many physicists reject that picture, they think it gets taken too literally, while others embrace it wholeheartedly, but it shouldn't appear in the stress energy that is used in general relativity.

But of course, the problem is that general relativity may be wrong, especially since it does not seem very consistent with quantum mechanics.  So the idea is, maybe there is some way to correct general relativity where virtual particles do have a more direct role.  Then the people who tend to dismiss their relevance would have to reconsider.  But so far it's not at all clear that this will be the right way to fix GR, because you run into significant problems.

The problems you run into with dark matter being virtual particles is that most of the effects of dark matter are seen on completely Newtonian scales, like the gravity in a galaxy.  On that scale, one should not need GR, and no one is suggesting that we need to fix Newtonian gravity by including virtual particles, that would not be consistent with everything Newton did.  For example, we see no effects of dark matter within our own solar system, so why would we need to fix Newtonian gravity on the galactic scale but not the solar system scale?  We would need to do more than include virtual particles, we'd need a whole new kind of gravity (sometimes called "MOND", modifications to Newtonian dynamics").  If you are going to do MOND, it opens a full range of possibilities, there's really no reason to be constrained by the virtual particle concept.

The problems you run into if you try to associate virtual particles with dark energy is that the answer you get, if you proceed in the most straightforward way possible, is that the dark energy should either be zero (if you don't think virtual particles count in the energy of the vacuum), or something more than 100 orders of magnitude too large (if you think they do count).  So how do you take that virtual particle concept and whittle down its effects from something more than 100 magnitudes too large to something tiny but not zero?  It's not at all clear this is the best starting point!


Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does?

Even though I'm a professional astronomer (but not an expert on this field), I don't have a good answer. AFAIK, this question is still open. Actually, the fact that most planets in the Solar system rotate in the same sense as they orbit the Sun is an important constraint on possible planet formation scenarios (of which still several are presently discussed by scientists).

For example, the idea that planets form out of dust collected at the centre of vortices in the proto-planetary disc can be almost ruled out, because retrograde vortices are much more stable and long-lived than prograde ones (Jupiter's red spot is a retrograde vortex), and hence planets should be retrograde, but aren't.


Why does the eclipse in this video look annular?

It is quite clear from the film (and the awful soundtrack) that a total eclipse was observed. The Sun's corona and prominences cannot be seen during an annular eclipse. Nor is there a sudden fading or drastic brightening at the beginning and end of an annular eclipse.

The accompanying article talks about "totality" and removing safety glasses to look at the Sun during totality, and that the period of totality was 1 min 53 seconds; the (frankly, annoying) narrator shouts the word "totality" about 23 times.

The eclipse was observed towards the end period of totality (what I mean by that is the shadow is tracking west to east across the Earth's surface. That means the Sun was above the western horizon as seen by the plane, and at 5:35pm, would be 10 degrees above the horizon. That means that in order to observe this event the plane must have been closer to the moon than a point on the surface of the Earth where totality would have been observed at the same time. That means the period of totality would have been slightly longer and the moon would have had a slightly larger angular size as seen from the plane.

If I had the exact latitude and longitude I could tell you just how much longer the eclipse was. Messing about with Stellarium, I can get a reasonable match of the eclipse time at W157 58 35, N31 22 00. At 10,000m the eclipse is total and lasts about 70s. At sealevel at the same latitude it lasts maybe a few seconds longer (it's difficult to be sure). But the sea level region of totality observed at the same time would have been somewhat further east (but still along the path where totality was observed), where the period of totality would have been shorter.


Gravity is no different from any other force. Doesn't matter it comes from a black hole, a star, a planet, or a lump of cheese floating in outer space. It's just a force, like any other.

Put a very powerful rocket engine on the pebble. In a short time it would accelerate the pebble near the speed of light. In the next moment it should exceed the speed of light. Why not?

Because that's not how it works. As you accelerate more and more, time dilation and space contraction also get stronger, so it gets harder and harder to get even closer to the speed of light, no matter what's pushing or pulling the pebble, be that a giant rocket engine or the enormous gravity of a large black hole. Same thing. The faster the pebble goes, the more warped space and time become, the harder it is to increase the speed.

It cannot even reach the speed of light because then time dilation, etc., would become infinite, effectively preventing it from reaching speed of light.

You cannot cheat relativity.

Note: This is not a rigorous explanation but it's close enough and should provide some intuitive understanding of what's going on.


Is it plausible that Mars could have been one of Jupiter's natural satellites at one time?

I'm going to say no, for a few reasons, but if anyone wants to give a more detailed answer, feel free.

1) Planets and moons that form inside the frost-line have much less water and other ices than those that form outside.   Mars likely once had water and oceans but if it had formed outside the frost-line it should have a lot more water and a lower density.   The frost line moves away from the sun as the sun grows more luminous and planets move, so there is a degree of imprecision with this argument, but I believe this on it's own is a pretty good reason why it's not likely.

2) A moon that escapes Jupiter's orbit should enter a near Jupiter orbit initially, and at some point later, it could receive a gravity assist and be propelled either in towards the sun or away, but it's unlikely that Mars could have obtained it's relatively circular orbit had it originated as one of Jupiter's moons, unless Jupiter was much closer to the sun when Mars escaped.

3) it's difficult to see how the asteroid belt could have stayed around if Mars originated outside the asteroid belt and moved inside it.   Mars passing through the asteroid belt would have caused significant disruptions to it.


Has the great Andromeda Galaxy ever collided with any galaxies?

Galaxies grow through cosmic time by accretion of the surrounding matter. Some of its mass increase happens through smooth accretion of gas, but much also happens through merging with small clumps of dark matter, gas, and stars, called satellite galaxies. This is called "minor merging".

If merging galaxies are similar in size, it's called major merging. Most large galaxies have gone through of the order of one major merger through its life (Man et la. 2014), but whether or not it happened for Andromeda, we cannot know for certain.

However, if you take a look at Andromeda, you'll see that it has quite a large bulge, i.e. the central, reddish part of the galaxy. This region is dominated by stars with a more "chaotic" velocity patterns. That is, their paths aren't "rotation-dominated" (lie in the disk), but more "dispersion-dominated" (leave the disk). To achieve such velocities, the system has to be disturbed by infalling matter, so a large bulge is probably a signature of a major merger. Investigating the velocity patterns of Andromeda, Dorman et al. (2015) find that it has had a much more violent history than the Milky Way, and in fact models by Davidge et al. (2012) and simulations by Hammer et al. (2010) of the history of Andromeda suggest that 8-9 billion years ago, Andromeda went through a major merger.

Note however that a large bulge can also be explained, at least in cosmological simulations, by a more smooth and "cold" accretion, i.e. gas that doesn't get shock-heated (Dekel et al.)


How do I remove fungus from a telescope mirror?

Several points need to be made.

1. Performance

The performance of the whole optical stack is incredibly resilient w.r.t. small spots on the primary mirror. The mirror might look visually very, very dirty, but the performance of the whole instrument will remain essentially the same. Even if you chip the mirror at the edge, it shouldn't matter. Take a marker and scribble a dozen lines on the mirror - it doesn't make a large difference for performance, not greater anyway than what the 4 vanes of the secondary mirror already do (we can start a separate discussion about diffraction artifacts from spots and lines on the mirror - but the bottom line is, don't worry about it).

With that in mind, you should be conservative when deciding to clean the mirror. Let it accumulate some amount of dirt, the scope will be okay.

2. Permanent damage

That being said, mold or other living things may chemically corrode the mirror. Again, a few spots of corrosion do not impact performance, but you don't want to allow this process to continue.

Dust will generally not corrode the mirror. Salt (from ocean breeze) will usually corrode it in time. With mold it's hard to say, but it's best if it's removed.

3. Prevention

Mold grows because the mirror is wet. The mirror fogs up usually when you bring the scope back into a warm room from outside; the mirror is cold, and water in the warm air condenses on it.

Procure a cheap, small fan, and point it at the rear end of the scope while it's warming up after you bring it back in. The air current will reduce water condensation on the mirror. Turn the fan off after 30 ... 60 minutes.

Also, keep the OTA (the big tube of the telescope) capped off while not in use. This will reduce dust deposits, and therefore mold will find less food to grow. At least cap off the top end and store the scope in a vertical position; that will reduce dust deposits greatly. The bottom end is less important.

Avoid storing the scope in a place with large day/night temperature variations. When the mirror is colder than air, condensation may form. This can happen spontaneously with scopes stored in a shed outside. The mirror fogs up in the morning, when air is getting warmer but the scope is still cold. This is much less of an issue with small scopes, but it's something to keep in mind.

You could cap off both ends while in storage, and place a bag of silica gel inside the OTA (don't drop it on the mirror). That will suck up all humidity inside the OTA, but you have to bake the silica gel in the oven every so often to keep it working effectively. You could buy silica gel online, it's very cheap. Avoid temperature variations and the gel is not necessary.

4. Clean up procedure

This is the procedure recommended by Optic Wave Labs:

https://www.youtube.com/watch?v=9Y8xFnXFVGQ

I make mirrors but I don't coat them myself (yet) - however, I've found that procedure to work very well on both coated and non-coated glass.

A few minutes of initial soaking in water makes everything easier later on.

After soaking it, add a few drops of detergent and rub very gently with your clean fingertips or with a soft, clean, lint-free cloth. Don't put pressure on the mirror, just try to detach the dirt from mirror in the horizontal direction and let it wash off. Don't drag dust particles across the glass, just detach them and let them go; this is why fingers are better than cloth - because you can feel large particles coming off.

When done, wash off the detergent with lots and lots and lots of water.

Using tap water for most of the cleaning procedure is absolutely fine. If you can drink the water, it's fine for the mirror too. Don't use hot water if the mirror is cold, there's a very small chance of cracking it, especially with cheap, mass produced glass.

When you're done cleaning, rinse the mirror with distilled water. Again, a few water spots on the mirror don't matter to performance, but the mirror looks better if it's spot-free. Tap water tends to leave water spots if you don't follow up with distilled water rinsing, and most people find the spots not pleasing from an esthetic perspective (but performance is same). So just splash a bottle of distilled water on it at the end and you'll be fine. No touching after rinsing.

Some dirt or mold may not be removed by the usual water-based procedure. In that case, use isopropyl alcohol, the highest concentration you can find (over 90% is fine). Rub gently with your clean, non-greasy fingertips or a lint-free cloth. If it doesn't come off quickly, leave it alone. Alcohol has probably killed the mold anyway.

Acetone might be used for things that resist even alcohol. But again, you have to use your judgement: a few little spots won't affect performance, so is it worth the trouble?

In any case, rinse with distilled water at the end, then let it dry off in an inclined position (carefully to not tip the mirror over on its face). Put a towel on the table in front of the mirror, so if it does topple over, at least it falls on something soft. Do not let it dry in a horizontal position, you'll get water spots and dust.

If there's some permanent damage in a few little spots, don't worry about it. The mirror will work essentially the same.

It is recommended that you do the water-based procedure every 1 or 2 years, depending on how dirty the mirror is. If dust contamination is moderate, leave it alone.

5. Post-cleanup

After everything is done, re-collimate the telescope carefully. But this should go without saying, since collimation is part of the regular maintenance of newtonian telescopes.


Assuming 2007 OR10 is in tidal lock with its satellite, what is its mass?

We can use Kepler's Third Law, the Law of Periods, to solve this problem. The law can be stated mathematically as:

$T^2 = \frac{4\pi^2}{G(M+m)} a^3$

Where

$T$ is the orbital period,

$G$ is the gravitational constant,

$M$ is the mass of the primary,

$m$ is the mass of the secondary, and

$a$ is the semimajor axis.

Solving for $(M + m)$ with the given values returns a system mass of approximately $8 \times 10^{22}$ $\mathrm{kg}$.

Other data are necessary to isolate the mass of the satellite from the primary.


There are probably four different processes, any or all of which may be described as "spiralling into a black hole"

The effects of general relativity. as @AtmosphericPrisonEscape has indicated, once you get very close to a Black hole Newtonian physics ceases to be a good approximation and must be replaced by general relativity (GR). When you do the calculations, you find that all circular orbits within 3 Schwarzschild radii of the black hole are unstable. Any object orbiting closer than that will definitely spiral in.
Gravitational radiation GR tells us that any two orbiting objects will emit gravitational waves, which carry away energy and angular momentum, slowly reducing the separation between the objects. This effect is, however, incredibly small, unless the objects are very massive and orbiting very close to one another (which means they must also be very dense). Thus this is mostly relevant for black holes and neutron stars in close orbits.
Turbulence in an accretion disk When a lot of gas or dust (which may be the remnant of larger objects pulled apart by tides) is orbiting a black hole, it tends to form a disk, called an "accretion disk". Basically the disk is oriented according to the average angular momentum of the stuff falling in, and anything orbiting in a different plane will collide with lots of the material in the disk and quickly join the disk. This is the same mechanism that caused the early proto solar system to form a disk, and indeed Saturn's rings (I know I'm oversimplifying somewhat).  Anyway, to remain in orbit, the inner parts of such a disk must be rotating faster than the outer parts, and this "shear" in the flow of gas tends to cause turbulence, which, it turns out, tends to transfer angular momentum outwards through the disk. This results in material at the inner edge of the disk getting slowly closer to the Black Hole until it reaches the Innermost Stable Circular orbit, at which point effect number 1 takes over.
Dynamical "friction"  If there are a lot of larger bodies (stars, asteroids, etc.) orbiting a large central mass, they will, from time to time, pass near each other, and  change each others orbits, transferring angular momentum between them. It turns out that averaged over many encounters, the effect of this is to drive larger bodies inwards and smaller ones outwards (possibly ejecting the smaller bodies from the system completely) so that it acts as something like friction on the larger bodies.


Total apparent magnitude of eclipsing binary system

Without information about stellar radii, I think it's reasonable to assume
$R_A \approx R_B$.
Then your equation becomes

$$ m_p - m_s = -2.5 \log \frac{F_A}{F_B} $$

and you can compute $k$ and the non-eclipsed total magnitude.


How is it known that Pillars of Creation are destroyed?

Much of the evidence for the destruction of the Pillars of creation lies in observations of dust clouds in the region (Flagey et al. (2009)). The morphology and structure of a cloud in the dust are suggestive of a shock wave, which seems to be indicative of a recent supernova. The shock waves from supernovae do often interact with both the interstellar medium and local gas clouds. The original observations supporting this hypothesis were made by the Spitzer Space Telescope.

It should be noted that the SST uses infrared light, and as the dust wasn't seen in the visible portion of the spectrum, we wouldn't necessarily have realized from the Hubble photos that there were any effects from a hypothetical supernova.

Some dissenters believed that something completely different is responsible for the behavior of the dust - possibly the winds of massive stars. Now, more recent x-ray observations (see e.g. Guarcello et al. (2012)) have failed to see either the x-ray spectra expected of O-type stars or the x-ray emission that would be indicative of a supernova, leading some to conclude that the Pillars of Creation have not yet been destroyed.


Which redshift is used to determine the Hubbleconstant?

The term "Hubble flow" refers to the homologous expansion of space and the resulting recession of all galaxies from each other (if they're not close enough to be gravitationally bound). This effect causes the "cosmological redshift", i.e. the redshift that light from distant galaxies attain as it travels through space.
In addition to this motion away from each other, galaxies have a so-called peculiar velocity, i.e. a motion through space. This motion adds an additional Doppler shift to the cosmological redshift, either to larger or smaller wavelengths, depending on the direction.
Whether or not you see these two types of redshift as fundamentally different, is not trivial I think. In most textbooks, they are described as two different things, the former having to do with the dynamics of the fabric of space, and the latter having to do with motions of emitters and observers. But in fact it may not be so different. For instance the Welsh cosmologist Geraint Lewis argues that, in some sense, the cosmological redshift can be interpreted as the sum of infinitely many infinitesimally small Doppler shift (Lewis 2016). On the other hand, the American physicist Sean Carroll argues that the notion of expanding space is nevertheless an extremely useful concept (Carroll 2008).


why was the probability of one in 10 billion per planet considered highly pessimistic by a recent study on aliens?

The main reason for this is the new data we aquired with the Kepler telescope and our findings about the goldilocks zone (where temperature is right for water to be liquid on the planet).

In the last decade a lot of discoveries, like the highly likely liquid water on the moon europe (which is way outside of the goldilock zone), the findings of amino acids in stelar clouds, etc, lead to the understanding that the chemical ingrediants and the chemical processes for life are possible in way more places than we thought.

Adding to this, the data from the Kepler telescope suggests that around 25% of all solar systems have a potentially habitable planet in some way (which makes that around 60-100 billion potential habitable planets in the milky way alone)


Have we ever observed a large meteor hit the Moon?

Yes, we have, and the impacts occur very frequently according to "Bright explosion on the moon" (NASA):

  For the past 8 years, NASA astronomers have been monitoring the Moon for signs of explosions caused by meteoroids hitting the lunar surface. "Lunar meteor showers" have turned out to be more common than anyone expected, with hundreds of detectable impacts occurring every year.

On March 17, 2013 a significant impact occurred that could have been observed from the Earth.

Updated information about lunar impacts is found at the Marshall Space Flight Centre's Lunar Impacts page.  Below is a map of the impacts that occurred on the lunar surface between 2005-2008, all of which, according to the article "Amateur Astronomers see Perseids hit the Moon, could have been seen from the Earth (with a telescope):

There have been instances throughout our history when, so is believed, people on the ground have witnessed lunar impacts. One of the more famous ones was in 1178 AD, when monks witnessed fire coming from the moon; However, according to the article "Historic lunar impact questioned", the accuracy of this historical account is questionable.


Does the Moon capture radiation pressure from the sun causing momentum from photon propulsion?

If we assume the 15 CM per year is accurate (which I'm not 100% sure it is), then it's possible to answer this.  In a loose sense, yes, the pressure should move the moon away from the sun faster, but more accurately and the real answer is that doesn't happen because there are 3 bodies in play and when you have 3 body gravitation, things get a lot more complicated.

First, just for fun, a source for the 15 CM.

  The moon away each year by photon momentum.The moon is the perfect
  color,size and The moon has a mass of 7.35 x 10²² kilograms. It is
  only about 60 percent as dense as Earth so it should be effected by
  the photons from the sun causing movement.And that movement should be
  combined with earths 15cm giving a new total distance a year right

If we only look at equal force per surface area, which is what you're describing, it's not density, it's the ratio surface area to mass.  The Moon's surface area is about 7.4% of Earth's (about 1/13.5) and it's mass about 1/81st of Earths.  Source.  81/13.5 = about 6.  That means given equal pressure, the moon should accelerate 6 times as much which corresponds over small distances to 6 times the distance or 90 CM per year - BUT, that's if you ONLY take into account the pressure.

The effect of the sun losing mass has equal effect on both the Earth and the smaller Moon.

And a 3rd factor to consider is the tidal bulge on the sun caused by the Earth-Moon system, which is very small but all these effects are small.   The Earth-Moon system creates a tiny bulge on the surface of the Sun and because the Sun rotates ahead of the Earth-Moon, that tiny tidal bulge has a tiny pull on the Earth-Moon system that slowly accelerates them and pushes them slowly away from the sun.  That effect is equal for both the Moon and Earth too.

All 3 of those are factors in the 15 CM per year estimate and only one of them has 6 times the effect on the Moon than the Earth.   The planets Venus and Jupiter might also be factors in that 15 CM per year estimate too, but lets leave that alone for now.

If we consider the 3 body problem it gets very mathy, but I'll just talk about how it applies to solar pressure.   Lets start with a picture.

When the Moon is waning its moving towards the sun and any pressure from the sun slows the moon down, (a tiny bit) and that slow-down moves the Moon closer to the Earth which speeds it up even more - funny how that works, pushing to slow something down in orbit and it goes faster - but that's how it works, cause potential energy converts to kinetic as the orbit drops, kinda like how falling makes things go faster.

When the Moon is waxing, it's moving away from the sun and any pressure from the sun speeds it up which moves it away from the earth and that in turn, slows it down.  So that's your answer in a nutshell.  The solar pressure doesn't push the moon away from the sun so much as it pushes the moon into a lower and then higher orbit around the Earth depending on where the Moon is in it's orbit around the Earth and the various positions in both the Moon's and Earth's elliptical orbits.  The overall effect is very difficult to calculate and I suspect it could go either way depending in part on the timing of large coronal mass ejections hitting the moon and in part on any possible resonance between the eccentricity of the Moon around the Earth and the Earth around the sun, er, I think.

What is safe to say is that the Earth-Moon system obeys the same surface area to mass ratio the Earth alone because that's simple momentum which needs to be conserved.    The Earth-Moon system has a combined surface area of 1.074 Earths and a combined mass of 1.012 Earths, so the Moon being in the Earth-Moon system makes the Earth move away due to solar pressure about 6% faster.  15 CM to 15.9 CM.  Not a very big change, and it's possible that the 15 CM estimate is based on the Earth-Moon barycenter not the Earth itself.

So, the effect is more of a wobble in the Moon's orbit around the earth than an actual force that pushes the Moon away from the sun faster than it pushes the Earth away.

All this, is, of-course, unimportant compared to the much bigger solar tidal force on the Earth-Moon system.   As the Moon gets closer to the sun (new moon in the picture above) The sun effectively pulls the Moon a little bit away from the Earth and when it's further from the sun (full moon), The Moon is effectively pulled towards the earth.  The net effect of this solar tide is a measurable wobble in the Moon's orbit around the earth.  The effect on the Moon's orbit around the earth due to photon pressure and coronal mass ejection pressure - basically insignificant.

Hope that's clear.


My query for stars north of +75° and brighter than magnitude 4 returns only 6 stars, brightest of which is γ Cephei.
Polaris may be too bright for Gaia.
Even if it doesn't saturate the detector, a bright star will probably have larger astrometric errors than a star between magnitudes 6 and 12.
DR2 coverage of magnitude 5 and brighter stars is better than DR1 but still rather sparse.
Section 3.1 of this paper plots parallax coverage vs. magnitude and explains why some stars have positions but not parallaxes.


When was the distance to a star measured for the first time without using parallax?

From Wikipedia

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

and

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


Cancelling out earth rotation speed, Altazimuth mount

Given ra, dec, lat, lon in radians, and d in number of fractional days
since '1970-01-01 00:00:00 UTC', the azimuth of a star is:

$-\cot ^{-1}(\tan (\text{dec}) \cos (\text{lat}) \sec (\text{d1}+\text{lon}-\text{ra})+\sin (\text{lat}) \tan (\text{d1}+\text{lon}-\text{ra}))$

and the elevation is:

$\tan ^{-1}\left(\frac{\sin (\text{dec}) \sin (\text{lat})-\cos (\text{dec}) \cos (\text{lat}) \sin (\text{d1}+\text{lon}-\text{ra})}{\sqrt{(\cos (\text{dec}) \sin (\text{lat}) \sin (\text{d1}+\text{lon}-\text{ra})+\sin (\text{dec}) \cos (\text{lat}))^2+\cos ^2(\text{dec}) \cos ^2(\text{d1}+\text{lon}-\text{ra})}}\right)$

where d1 is: $\frac{\pi  (401095163740318 d+11366224765515)}{200000000000000}$

(you will need to resolve the ambiguity in the arc(co)tangent, but
this isn't difficult).

These formulas aren't as daunting as they seem, since, for you, lat,
lon, ra, and dec will be fixed, and the only thing that changes is d.

Hope this helps, but I'm worried that it just demonstrates how complicated these formulas are.


Does astronomical observing "count" as remote sensing?

As suggested by, e.g., the Wikipedia page (or this page at the US Geological Survey), "remote sensing" is used pretty much exclusively for airborne or satellite-based observations of the surface of the Earth (including the ocean floor in some cases).
This makes sense if you consider that "remote" is used in contrast to observations made locally (or "on-site"). E.g., if you walk around on the ground and make observations, those are "local" observations; if you fly a plane and make observations of the same place, those are "remote" observations.

Since astronomical observations are overwhelming of places you simply can't make local observations of, it's unnecessary to specify them as "remote" observations, just as no one call looking up at high-altitude clouds or distant mountain peaks "remote sensing". (And it seems more than a little silly to claim that humans have been doing "remote sensing" for millions of years, even though humans and proto-humans have indeed been looking at the Sun, the Moon, planets, and stars "remotely" for about that long.)

(I have occasionally seen satellite observations of planetary surfaces -- e.g., imaging by probes orbiting Mars -- referred to as a type of remote sensing; this is clearly based on the direct analogy with Earth remote-sensing satellites. But that's about as far as you can push the term.)

You're free to consider astronomy to involve "remote sensing", based on an overly literal and overly logical interpretation of the term, if you like; in practice, no one working in either astronomy or remote sensing does.


Is the conversion from proto-star to main sequence an event or a process?

Astronomers distinguish a prototstar from a star based on whether the object is visible. A protostar is hidden by the gas cloud that surrounds it. Protostars aren't visible. At some point in their evolution (and where this occurs depends on mass and metallicity), a protostar will start clearing the surrounding cloud of gas. This process happens very quickly from an astronomical point of view. (Aside: From a human point of view, this is anything but quickly.)

In the case of very massive stars, the newly emerged star is already on the main sequence. Very massive stars are "stars" (objects with a stable mass, stable size, and fusing hydrogen into helium) well before astronomers can see them. In contrast, very small stars spend hundreds of millions of years of evolution between being a "star" (visible to astronomers) and being on the main sequence. The star is a pre-main sequence star during this long span of time.

Intermediate mass stars also spend some time as a pre-main sequence star. How much time depends on mass and metallicity. In the case of a star with a mass of ~1 solar mass and metallicity comparable to that of the Sun, the time spent as a pre-main sequence star is on the order of tens of millions of years. In contrast, the time spent as a protostar is very short, tens of thousands of years. The time taken to clear the star system of gas (the transition from protostar to PMS star) is shorter yet, hundreds to a few thousand of years.

There is no point at which one can say fusion starts. It's a probabilistic thing, with probability increasing sharply with density and temperature. What about objects at the low end of density and temperature? This is the boundary that separates red dwarfs from brown dwarfs.

There is very little difference between the smallest red dwarfs and the largest brown dwarfs. The smallest red dwarfs are quite cool. They fuse hydrogen into helium via the p-p chain, but rather infrequently. The largest brown dwarfs also fuse hydrogen into helium via the p-p chain, but here the fusion reactions are so very infrequent that the brown dwarf gets cooler and cooler. In a trillion years, we might be able to point to a marked boundary between old red dwarfs and old brown dwarfs. Right now, that boundary is rather arbitrary.


How does the Hubble Space Telescope "Drift and Shift"?

The "drift-and-shift" (DASH) mode was described in a paper by Momcheva et al. (2016), and works the way you supposed: a sequence of nearby images is taken using blind offsets without guide-star recalibration: move to the first field and take an image (possibly first acquiring a guide star), move to a nearby target field and take another image (no guide star acquisition), repeat...

Guide-star "calibration" means correcting the internal model of "where in the sky HST is pointing" by locating a bright star with well-known coordinates, tweaking the pointing of HST to put the guide star in the right position within the telescope's focal plane (so that the center of the focal plane is now pointing where the observing proposal specifies), and setting up a monitoring process where the position of the guide star is continually tracked during an observation to make sure the telescope pointing is not drifting (and correcting it when it is). The whole acquisition-and-setup process introduces an overhead of about six minutes; since a typical HST observing window is only about 50 minutes (before the spacecraft's orbit carries it around to the other side of the Earth and it can't see the target any more), trying to change target fields multiple times would rapidly eat up your actual observation time.

(In some [most?] cases, it might make sense to do guide-star acquisition for the very first field, to which HST has likely just made a very large slew, to ensure the initial pointing is accurate and that the (smaller) offsets to new fields will be reasonably accurate.)

Not having a guide star does impact the stability of the telescope pointing during an observation: guiding using just the gyros means a drift of about 0.001 to 0.002 arc seconds per second of time. So long exposures are not a good idea, because they'll be smeared by the telescope drift: an exposure of 100 seconds means a typical drift of several pixels during that time. But since Riess et al. were taking very short exposures (just a few seconds), this wasn't a problem for them.


Has Hawking Radiation Ever Been Observed?

According to this MIT Technology Review "First Observation of Hawking Radiation" and an associated article "Hawking radiation from ultrashort laser pulse filaments" where the authors have simulated Hawking radiation in the laboratory, using ultrashort laser pulse filaments.

So, in part answer to your question, yes Hawking Radiation is believed to have been observed, but in a laboratory environment.

According to "Observing Hawking radiation in Bose-Einstein condensates
via correlation measurements" (Fabbri, 2012), direct observation of the Hawking radiation is (in their word): hopeless.

However, once again simulations are possible:

  Hawking effect, which depends only on kinematical properties of wave propagation
  in the presence of horizons, is present also in nongravitational contexts,
  for instance in stationary fluids undergoing supersonic flow.

From these articles, it would seem that direct observations of Hawking radiation is not feasible currently; however, we are able to produce analogues of the phenomena in the laboratory.


The temperature of an exoplanet is dependent on a number of factors, including:

The distance between the exoplanet and the parent star.
The type (energy output) of the parent star.
The ability of the exoplanet's atmosphere to distribute the heat energy.

Recent discoveries, such as reported in the article "Sizzling Planet Makes Some Stars Look Cool" (Than, 2007), have found that some exoplanets are hotter than some types of stars.

For example, the exoplanet  HD 149026b  has a star-facing side of over 2000C (3700F), this planet reflects so little light that it would appear like a "glowing piece of charcoal". It is believed (from the article) that the planet is tidally locked and the 'dark side' of the planet is believed to be substantially cooler.


You have gone wrong because the radius of Andromeda is not 12250.40 arcminutes.
It is difficult to be more helpful. Possibly the units are arcseconds?

Neither do I see why you have inserted a factor of 0.5 in your second equation.
or why you have inserted $300^2$ into the third equation, when $v$ is given in km/s, so it should be $(3\times 10^5)^2$.


How can white dwarf form Oxygen ? (Temperature problem)

Some oxygen is produced during CNO cycle processing of hydrogen, starting with carbon nuclei. Oxygen is also produced by alpha capture onto carbon nuclei at temperatures well below 350 million K.

Both of these occur in and around the cores of low mass stars before they become white dwarfs. Neon production is not required.


Convolve a SED with a filter. Is convoluting the mathematical operation?

To get the flux of an SED through a particular filter, you actually multiply the the SED by the filter's response. Talking about convolution in this context is a bit of a misnomer.

Basically, for each wavelength, you look at what fraction of the light will go through the filter, and you sum up the values you get at those wavelengths.
The division by $\int T_{b}(\lambda)d\lambda$ is just a normalization term.


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

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

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

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

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

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


Why is sunset on 21st of June not the latest?

It's explained in this Wikipedia article. Basically, the length of a day is defined by both the Earth's rotation itself and Earth orbiting around the sun. Because the Earth's orbit around the Sun isn't a circle but an ellipse (making it moving faster when it's closer to the Sun), and the Earth rotates in a plane different than it orbits the Sun, an apparent solar day isn't always 24 hours; in June, it's slightly longer. That means every day, "noon" (the time of day when the Sun is at its highest point) happens 'later' (compared to the time on your watch), so sunset (and sunrise) are also later. The first few days after June 21st, this effect is more significant than the effect of the Sun moving south (causing the day* to become shorter and the sunrise falling earlier). The same effect causes the earliest sunrise (in your location) to be on June 15th, so before the solstice.

*: day as in day vs. night, the period between sunrise and sunset, so not 24 hours


Rendering stars in 3D space - AbsMag to OpenGL scale values

"Accurate star size" is a problem. Obviously you do not have the resolution on your screen for accurate angular resolution of stars (they would require an absurdly fine resolution), and the dynamic range of brightness is also too low -- stars range over many orders of magnitude in brightness, far more than a screen can show. Deeply annoying. What one might want to do is to make each star have a brightness and screen size that is proportional to their actual brightness in the sky to give the same feeling as the sky would give. This is still very tough, since screens have different gamma correction. To top it off, the human eye actually has a pretty logarithmic response to light (this is why magnitudes and gamma correction make sense).

Here is a rough idea. A star with AbsMag of $M$ has an actual luminosity of $$L = L_\odot 10^{0.4(M_\odot - M)}$$ where $L_\odot$ is the luminosity of the sun and $M_\odot$ is the absolute magnitude of the sun. Things get easy if we just count luminosity in terms of $L_\odot$, making the sun one unit of luminosity.

A spot of radius $r$ on the screen with luminance $l$ radiates power as $P=\pi r^2 l$. That luminance is due to the gamma-corrected luma value $V$ the computer displays: $l = K V^\gamma$.

Assuming the radius changes with luminosity too as $r(L)$, I would try $r(L)=r_0 L^a$ where $a\approx 0.6$ (but this is guesswork) and the sun has radius $r_0$ pixels.

So trying to put this together, we get the pixel brightness as $$V = V_0 (L/ r(L)^2)^{1/\gamma} = V'_0 L^{(1-2a)/\gamma}$$ where $V_0$ and $V'_0$ are the pixel brightnesses used for the sun in this model. Basically this squashes the actual luminosity with $a$ (representing using bigger spots for brighter stars, not requiring as intense pixels) and $\gamma$ (to correct for the screen and the eye). What values to use will largely be trial and error unless you want to try to use screen photometry equipment.

So the full formula converting from absolute magnitude to pixel value would be: $$V = V_0 10^{0.4(4.83 - M)(1-2a)/\gamma}.$$


How long will it be until the nearby groups of galaxies are receding at speeds faster than light?

This will take about 2 trillion years.

Given everything we know now, it's not too hard to predict the eventual fate of the universe. Of course there may be some changes as our knowledge advances, but I think the general course of events will occur as we expect them to. Based on our current knowledge, if you run through the math, you find that after about 2 trillion years from now, the expansion of the universe will have made it so that all by the closest galaxies will no longer be visible to us. You do ask two separate questions, but they're more or less the same question. They might not occur precisely at the same time, but predictions of the future aren't precise enough to distinguish such a small time difference.

By the way, you can read more about the future of the universe on Wikipedia, and specifically this section as it relates to your question.


How can we tell how many exoplanets a star has?

What you are describing is a basic signal processing problem. The doppler shift that one observes is due to the motion of the star in the system around the system's centre of mass. The star will be influenced by the gravitational pull of each of the planets in that system, each of which exerts a gravitational pull that increases with the mass of the planet and decreases with the orbital radius of the planet.

The overall motion of the star will be the sum of the effects of all the planets. Importantly, the effect of each planet will have its own amplitude and will be periodic with a period equal to the orbital period of that planet.

Let's imagine that each planet is in a circular orbit (elliptical orbits are more complicated, but the principle is the same). Each planet would cause a circular motion in the star about the centre of mass of the planet-star pair, leading to an observable doppler signal which has the form of a sine wave with a period equal to the orbital period of the planet. The amplitude of that signal will increase with the mass of the planet and increases with decreasing orbital radius.

The overall signal is the sum over all the planets in the system. Fortunately the decomposition of this signal back into its individual components is a well-trodden problem in physics, electronics and many other fields and is known as Fourier Analysis. Whether you can successfully recover the original signals from each of the planets depends on how long you observed the system (ideally you want to observe for longer than the longest orbital period) and the amplitude of the signals compared with the noise in your observations.

In general it is easier to recover high-mass planets with short orbital periods and more difficult to recover low-mass planets with long orbital periods.

The image below might be helpful. It shows the track of the solar system centre of mass compared with centre of the Sun over a period of several decades. Notice how the Sun executes a complex trajectory (with respect to the solar system centre of mass) that is mainly caused by the orbit of Jupiter, but then there are smaller, superimposed, signals caused by the smaller planets. In principle, if you observed for longer than the period of Neptune and had a detector which gave perfect measurements, you could reconstruct how many planets there were in the solar system, what their orbital periods were (and then from Kepler's 3rd law, what the planet-star sepration was) and what their masses were (multiplied by the inclination of their orbits with respect to the line of sight of observation, which is generally an unknown in doppler measurements).

In terms of what we could currently see if we observed the Sun as a star: basically we would (assuming we observed for 20 years) detect Jupiter quite easily with a doppler amplitude of about 13 m/s. We would also see that there was a drift in Jupiter's signal due to the influence of Saturn, but we would have to observe for >Saturn's orbital period in order to confirm the presence of Saturn, its orbital period and mass. The inner planets produce an amplitude that is too small to be visible using the technology currently available. e.g. The Earth would produce a doppler wobble of amplitude $<8$ cm/s, but the current precision of doppler measurements is limited to about 50 cm/s.

The doppler amplitudes in m/s due to each of the planets (assuming we view them edge on) are:

Mercury <0.01
Venus    0.08
Earth    0.08
Mars    <0.01
Jupiter  12.5
Saturn   2.6
Uranus   0.28
Neptune  0.26

Thus with current technology, only Jupiter and Saturn are detectable.

Below I simulate what the doppler signal due to these two planets would be. I hope you can see that the overall signal consists of the superposition of two sinusoidal signals with different periods and amplitudes. Myriad computational tools are available to do the fourier decomposition to establish these.


Why is the asteroid belt shaped like a triangle?

It's not. The image doesn't show the main asteroid belt. It shows the Jovian Trojans (in green) , and the Hilda Asteroids (in red).

The Hildas are a dynamical group of a few thousand known asteroids in elliptical orbits that are locked in a 3:2 orbital resonance with Jupiter, and reach aphelion coinciding with the regions near Sun-Jupiter Lagrange points L4, L3, and L5 in succession.

"Animation showing Hildas with 3 indicated orbits" from "Asteroid (and Comet) Groups", Petr Scheirich, Retrieved 2020/02/14.

In accordance with Kepler's Second Law, objects in elliptical orbits move slower near aphelion than perihelion, and as a result, when you just look at the Hildas, they bunch up near the Lagrange points, and the pattern they appear to show at any one time resembles a triangle.

A more complete view of all the Asteroids inside of Jupiter looks like this.


It seems you saw a Tumbling satellite. They don't have a constant brightness. This is a result of satellite's rotation. Learn more: http://www.satobs.org/tumble/tumbleintro.html


Is scattering efficiency of dust related to albedo?

There will be a relationship but not a simple one. If we model a uniform slab of material by a complex dielectric constant $n + ik$ then we can calculate three things; reflection, absorption and transmission. If we model a dust particle as a uniform sphere of the same, then the far field electric amplitude scattered from an incident plane wave on it will depend on the complex dielectric constant. The reason we can't just use reflection, absorption and transmission is that those are the amplitudes squared, and so phase information is lost, so you can't easily work your way backwards.
There is also the angular distribution to worry about, which will be different for each size and also for each complex index of refraction. So two identically shaped and sized dust particles; one from a high bulk albedo material and one from a low bulk albedo material could end up with a variety of different relative scattering efficiencies.
A cloud of either one might attenuate light from a star, depending on the details, and one might scatter light more strongly at some specific angle than the other.
There might be some rules of thumbs for certain kinds of dust particles made of materials most likely to be observed and discussed by astronomers, but I don't think there are any first-principles-based generalizations that always apply.


Why does the precession of the earth affect other planets?

Why would precession affect the motion of the other planets?

First things first: That's an unreferenced portion of a wikipedia article.

That said, a perfectly spherical body acts exactly like a point mass in Newtonian mechanics. A non-spherical body does not. The Earth's equatorial bulge has a significant effect on satellites in low Earth orbit. Sun synchronous orbits would be impossible without that bulge. For example, a sun synchronous orbit about Venus with its very slow rotation rate and hence very small equatorial bulge is not possible (reference).

The Earth's equatorial bulge adds a quadrupole (inverse quartic, or $1/r^4$) term to the inverse square law relation for point masses, and this inverse quartic term depends not only on distance but also on latitude. The added term is
$$3 J_2 \frac {GM_\oplus a^2}{r^4}\left(\frac 3 2 \cos^2 \lambda - 1\right)$$
where $J_2$ is the Earth's second dynamic form and $\lambda$ is the geocentric latitude.

Since this term drops off as $1/r^4$, it becomes very small beyond the Moon's orbit. But it still exists beyond the Moon's orbit. However, I've yet to see a paper that does account for (or says we need to account for) Earth's $J_2$ effect on the other planets.


What's the name of the point where you are closest or furthest from the ecliptic?

I don't know of a standard term for that quantity, but I'd call it the ecliptic latitude of the zenith.
With north positive and south negative, this would reach a minimum at 6:00 and a maximum at 18:00 local sidereal time.
These correspond to local solar noon or midnight at the solstices and occur 2 hours earlier per month.

The vectors perpendicular to the ecliptic are called the ecliptic poles. The zenith only coincides with one of these on the Arctic or Antarctic circle, at sunrise (winter-spring) or sunset (summer-fall).


How to calculate the mean molecular weight of the Sun

For your second question, it depends upon whether you are talking about composition by number, or composition by mass (the numbers you give are roughly the composition by mass).  If you want to get really specific, you can add in the other elements too.

To answer the first, you add them up proportionally.


How are the days relating to right ascension around a polar star chart decided?

The date on the charts shows the meridian (north south line) at 8pm local time on that date.  Over a period of a month, the RA of the meridian will move 2hr (so that it gets back to where it started from after 12 months), and hence over 16 days, as you ask about, it will move 1.05 hours.  So you can either say that's close enough to 1 hr, or worry about the small discrepancy - that's assuming the 1hr you cite is actually 1.00hrs.


I assume you are referring to the rotation curve of galaxies, i.e. the speed of matter (stars, gas clouds, etc.) as a function of its distance from the center. This speed will depend on the matter distribution, that is, how the galaxies' constituents are distributed throughout the galaxies.

Keplerian rotation

If (almost) all of the mass $M$ were located in the center, then a star at a distance $R$ would have a speed $V = \sqrt{GM/R}$. This is called Keplerian rotation, because this is how planets in the Solar system move, which were described by Johannes Kepler. In this case, the mass inside a radius $R$ is (almost) constant, i.e. $M'(<R) = M$.

Flat rotation

Most disk galaxies have a large concentration of visible mass in the center (the "bulge"), so one might think that Keplerian rotation would be a good description of the stars' speeds. But this is not what is observed. In fact, the speed is seen to rise steeply the first few kpc, and then stay constant out to distances where no light is detected. This will happen if the mass $M'(<R)$ inside $R$ increases with $R$, i.e. if the matter is distributed throughout the galaxy. This is one of several pieces of evidence that 1) there exists some matter that we cannot see, in addition to what we can see, 2) that this matter dominates the kinematics of the galaxies, and 3) that it resides in a halo that extends beyond the visible matter. We call it dark matter.

Solid body rotation

If in addition to a given matter distribution the distances between the individual parts are fixed, the object we are discussing would describe a solid body. Of course this can't be the case for a galaxy, but it will be the case for a rocky planet. In this case the velocity of a given point will depend linearly on its perpendicular distance to the axis of rotation, i.e. $V \propto R$.

Density profile of the matter distribution

Excactly how the matter is distributed in the galaxy is not easily deducted, and varies from galaxy type to galaxy type. An early description of the matter density as a function of distance from the center was the so-called NFW profile, but other models may be better, e.g. a Sérsic profile or an Einasto profile.

So, to summarize, the term "matter distribution" refers to density as a function of location, i.e. $\rho(\mathbf{r})$, for some object. This can be for any object — solid, liquid, gaseous, or consisting of several components that are held together by gravity — but in any case it is true that its form determines its rotation, and the three different rotation forms you mention are often discussed in the context of galaxies.

The three different cases are illustratred below. Keplerian, solid, and flat rotation are drawn in green, red, and blue, respectively. For comparison, a typical spiral galaxy velocity curve is drawn in magenta.


Based on your description, that might be the International Space Station:

  The ISS is visible to the naked eye as a slow-moving, bright white dot because of reflected sunlight, and can be seen in the hours after sunset and before sunrise, when the station remains sunlit but the ground and sky are dark.

So if what you saw was not long after sunset, and if the object was quite bright and moving slowly across the sky, that might be the answer.

If you want to check for yourself, you can input your location on this webpage and change the search period to see if the ISS passed above your heads on that day.

Other possibilities include other satellites (not very bright but also moving smoothly in the sky), or—more rarely—particularly large shooting stars.


What was the "optical illusion" that led to erroneous metal concentrations in stellar atmospheres in the galactic center?

The journal paper is Thorsbro et al. (2018).

The facts are somewhat mundane. The atmospheres of cool M-giants are not well understood in detail. The infrared lines of neutral Scandium that had previously been used to claim massive overabundances in stars near the Galactic centre are problematic.

The paper by Thorsbro et al. shows that the same "anomalies" are present in other M giants nowhere near the Galactic centre. Instead they argue that the details of line formation, especially the formation low excitation lines formed high in the atmosphere, like the IR transitions of neutral Scandium, are particularly vulnerable to poorly modelled NLTE (non local thermodynamic equilibrium) effects. If analysed with atmospheric codes that assume LTE, then you get the "illusion" of a much higher abundance.


Can young surface of Venus be caused by early oceans instead of global resurface event?

Our best guess is that if Venus had an ocean, it did 4 billion years ago. Then it had the runaway greenhouse effect due to its thick atmosphere. The runaway greenhouse effect would have boiled away the oceans if the planet ever had oceans.

There are several guesses about why its surface is so young (300 million to 600 million years old). One is that the planet get so hot the entire surface melts. It has no plate tectonics and it has a super thick atmosphere to trap the heat. All that heat melts the surface every half billion years or so. Another reason is that it has many very large volcanoes. (167 volcanoes over 100 km across.) That's far more very large volcanoes than Earth has. Those volcanoes resurface the planet and keep it young.


How can I use my current camera whose zoom lens can't be removed behind an f/6 refractor for as an astrophotography?

The AT60ED is a nice small refractor. However, its objective lens simply projects an image at the focal plane. Adding a field flattener mimimizes curvature that degrades sharpness at the edges of the image. The flattener requires a specific back-focus, or distance (about 57 mm for the AT60 FF) between its (T2, 42 mm thread) flange and the image plane where the sensor should be located.

If your camera has a non-removable lens, its sensor cannot be placed directly at the focal plane. You cannot just attach your lens to the telescope or field flattener to take an image.

I believe that the only alternative for a camera with lens is to use eyepiece projection, also called digi-scoping. An eyepiece magnifies the image at the refractor's focal plane and projects it, usually so the eye can then focus the image on the retina. Instead, if a camera is attached to the eyepiece using a suitable adapter, its lens can then focus the image.

Eyepiece projection also has disadvantages, including a small field of view, vignetting, and often balance issues due to the camera weight and adapter attached to a small eyepiece, rather than directly to the focuser of the telescope. However, lots of people use eyepiece projection just by holding their cell phone camera at the exit pupil of the eyepiece.

Edit:  Added some comments regarding deep-sky imaging.

Also, deep-sky imaging presents additional challenges. The Moon and planets can be photographed with short exposures. However, images of even the brightest nebulas and galaxies require long exposures. Astrophotographers usually take a large number of individual images, called sub-exposures, each typically lasting tens of seconds to many minutes. These are then aligned and stacked to improve signal to noise by averaging. Total integration time can often be several hours or more, depending on the brightness of the object.

A telescope mount that can accurately track a celestial object, compensating for the Earth's rotation, is necessary to prevent motion blur over these long exposures. The cost of a mount suitable for even a relatively short focal length refractor such as the AT60ED will be several times the cost of the telescope, and much more than a camera capable of prime focus astrophotography.


How uniformly is cometary debris distributed along the entire orbit of the comet?

There are more or less two possibility to get debris in a sufficient amount: collisions and breakup/ejection.

Collisions

This is mostly the case for asteroids. At the point of impact you get a cloud of debris which basically moves as if the colliding bodies combined into a bigger one. However, for small objects, the fragments in the cloud are not gravitationally bound to each other and thus can separate over time. If larger objects survive the initial collision, they can undergo further impacts at different orbital positions leading to the seasonal meteor showers.

Breakup/Ejection

This is mostly the case for comets. Particles get ejected from the original body by heating of the surface, or other effects. This is only possible for small grains, which are massively influenced by stellar radiation pressure. The result is an effectively smaller gravity of the star and thus a different orbit. Being on different orbits, theses particles naturally have to separate.

Conclusion

In both cases you get a physical separation over time. Being that all of the quantities involved are on a continuum, you effectly have a smeared distribution of fragments over an orbit. Depending on the special case, you can get a distribution over the whole ellipse, but a part of it seems more likely.

EDIT: As you asked about catalogues, I would recommend you looking at https://www.ta3.sk/IAUC22DB/MDC2007/ plus at http://adsabs.harvard.edu/abs/2006mspc.book.....J for a scientific read.


If a photon were 1 Planck length away from the event horizon of a blackhole, would it be able to escape the blackhole's gravitational pull?

It depends on the direction the photon travels.

If the photon is directed straight away from the black hole, it will escape (but will be very redshifted).

Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. (See slide 6 in this presentation)

See also  this interactive web page; try it for a fixed position looking along the "orbit". The higher up in the picture the closer you get to the horizon. I have found that one can get past the $1.5R_s$ limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 $R_s$ is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.


Which planets have abundant amounts of photoluminiscent matter?

Sorry, but none of the planets have substantial amounts of phosphorescent chemicals.

Perhaps the closest to what you are thinking is the aurora, which can be seen on all the planets with atmospheres (to a greater or lesser degree) notably on the Gas Giants and on Earth. However, while the mechanism of aurora is similar to florescence, it is not phosphorescence.

No planets outside of the solar system have been directly imaged beyond a dim dot, so there is no information about exoplanets.


Why would the interplanetary medium atomic hydrogen Lyman‐α emission fall as 1/r rather than1/r^2?

The brightness considered here is the brightness of scattered Lyman $\alpha$ as viewed in the upstream direction, i.e. away from the Sun.

The authors assume that the density of the atoms responsible for scattering the Ly$\alpha$ — the neutral hydrogen atoms — is 1) roughly constant throughout the Solar System and 2) is optically thin, meaning that the radiation field is not attenuated considerably with distance from the Sun.

Since the intensity of a point source falls off as $1/r^2$, the scattering emissivity in a given point should then also fall off roughly as $1/r^2$.

But when viewed from inside the Solar System and in the upstream direction, the brightness is the integral of all scattering events from that point to infinity, and since $\int 1/r^2 \propto 1/r$, you have your quoted brightness.


Using a refractor telescope 5" aperture, 900mm focal length for terrestrial viewing

Yes, potentially.  There are a few limitations:

It will be easier to use in alt azimuth mode, which in practice means setting the polar axis to 90 (ie straight up).  Not all equatorial mounts can do this.

With the standard astronomical diagonal, images will be inverted left to right.  You can get diagonals for terrestrial viewing that will fix this.  If you don't use a diagonal, the image will be upside down.

The field of view will be quite restricted if you use a typical eyepiece.  For example a 20mm plossl would give a magnification of 45 and a field of view of about 1 degree.  To get more useful field of view, you would need longer focal length eyepieces, with with a wide angle design.  These are more expensive than ordinary eyepieces.

Unless your scope is an apo, you may see quite noticeable chromatic aberration, particularly at the edges of objects with a bright background (eg birds).


How do I create a galaxy stellar mass function?

Your approach is completely correct, just note three things:

Logarithmic distribution

First, since the distribution of masses is logarithmic in nature (as is most other things), be sure to bin them logarithmically. Otherwise you will oversample (undersample) the bins at the low-(high-)mass end.

Comoving densities

Second, to be able to compare mass functions at different redshifts, ones uses the comoving volume rather than the physical volume, such that the expansion of the Universe is factored out. The two are related as $V_\mathrm{com} = V_\mathrm{phys}(1+z)^3$.

Damn you, little $h$!

Finally, observers and modellers tend to use a slightly different definition of the unit volume. Whereas observers usually use $\mathrm{Mpc}$ for distances, and hence $\mathrm{Mpc}^{-3}$ for number densities, if your galaxies come from a cosmological simulation where the cosmological parameters can when tuned at will, it is custom to factor out the Hubble constant $H_0$. In simulations, masses and distances are then measured in $h^{-1}M_\odot$ and $h^{-1}\mathrm{Mpc}$, respectively, so number densities are measured in $h^3  \mathrm{Mpc}^{-3}$. Here $h\equiv H_0\,/\,(100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1})$.

This is probably a reminiscence of the time when the Hubble constant was rather uncertain. Nowadays, in my opinion there is no need to do this, but since everybody does, it's difficult to go against the stream. For a discussion of this issue, see Croton (2013).

Python code

Since you've tagged the question with python, I wrote this little snippet that should do the work (I randomly chose $10^5\,\mathrm{Mpc}$ as your survey volume; note also that in this example I don't factor out $h$):

import numpy as np
import matplotlib.pyplot as plt

M     = np.loadtxt('Mstar.dat')         #Read stellar masses in Msun
logM  = np.log10(M)                     #Take logarithm
nbins = 10                              #Number of bins to divide data into
V     = 1e5                             #Survey volume in Mpc3
Phi,edg = np.histogram(logM,bins=nbins) #Unnormalized histogram and bin edges
dM    = edg[1] - edg[0]                 #Bin size
Max   = edg[0:-1] + dM/2.               #Mass axis
Phi   = Phi / V / dM                    #Normalize to volume and bin size

plt.clf()
plt.yscale('log')
plt.xlabel(r'$\log(M_\star\,/\,M_\odot)$')
plt.ylabel(r'$\Phi\,/\,\mathrm{dex}^{-1}\,\mathrm{Mpc}^{-3}$')
plt.plot(Max,Phi,ls='steps-post')


If Jupiter and Saturn were considered stars, which of their moons would qualify as planets by 'clearing the neighbourhood'?

I've noted before that the IAU naming critera are guidelines rather than laws of nature.  If applied to Jupiter's moons the four Galilean satellites would probably be "planets" (they don't share their orbits with anything else of comparable size and are large enought to be in hydrostatic equilibrium)  There is a mean-motion resonance 4:2:1 between them but this is a more or less symmetric relationship without one body dominating another.
The other moons are tiny by comparison.  There are 4 major bodies orbiting Jupiter.
For Saturn, Titan certainly dominates its orbit (Hyperion is much smaller.)  Dione and Rhea also seem to have a value of mu that are over 100 (though determining the value is tricky because not all moons and moonlets have been discovered) Enceladus and Tethys and Iapetus also seem to qualify,  As you note, Mimas is too small and borderline spherical. The other moons are just so much shrapnel by comparison.
So there would probably be the six pre-1800 moons, minus Mimas.
It is possible that the rules might be bent, the Rings are no permanent features, so they might be ignored for the sake of planetary determination. Remember the rules exist to serve us.  Humans make the rules. Some can be bent, some broken.


What is it that distinguishes one atmospheric layer from another?

What is it that defines where one layer stops and another starts?

Temperature. More specifically, it's whether temperature rises or falls with increasing altitude.

In the troposphere, temperature generally decreases with increasing altitude, at an average rate of 6.4 °C/km (the environmental lapse rate). This decrease stops at the tropopause, the boundary between the troposphere and the stratosphere. The ozone layer is in the stratosphere, making temperatures in the stratosphere increase with increasing altitude. While this boundary is a bit fuzzy, it is still very real. It takes an incredibly strong thunderstorm (think hurricanes, storms powerful enough to spawn tornados, and very tall and strong storms in the Inter-Tropical Convergence Zone) to penetrate that boundary.

Temperatures in the stratosphere stop rising at the stratopause, the very fuzzy boundary between the stratosphere and the mesosphere. Like the troposphere, temperatures in the mesosphere fall (and fall very sharply) with increasing altitude. Thermodynamics makes the boundary between the stratosphere and mesosphere very different (and not nearly as clear) as the boundary between the troposphere and the mesosphere.

The boundary between the mesosphere and the thermosphere is similar to the boundary between the troposphere and stratosphere. Temperatures rise with increasing altitude in the thermosphere. This change from falling temperatures in the mesosphere to rising temperatures in the thermosphere makes for a rather stable boundary.

Something else happens near that boundary between the mesosphere and the thermosphere. Long-lived gases are fairly well-mixed in the troposphere, stratosphere, and mesosphere thanks to turbulence. Gases in the thermosphere and exosphere act more like a bunch of individual particles rather than a gas. Unlike the dense layers below, gases in the thermosphere and exosphere are differentiated. The makeup tends toward lighter and lighter particles (e.g., helium and hydrogen) with increasing altitude. Eventually, all one finds are hydrogen and helium. These are the gases that escape from the atmosphere.


Do planets form around lone stars, multiple stars, or both? Do we know this yet?

There are currently 3424 planets identified (including those in the Solar System and unconfirmed exoplanets). Of these, 3310 are confirmed exoplanets.

Out of the 3424, 180 are in binary (or multiple) star systems. Of those, 28 are circumbinary. I am not certain if any circumternary planets have been identified, though I know some circumternary disks have.

It is more difficult to identify circum-n-ary planets than those that only revolve around one star, so these numbers might not precisely represent the actual proportions.

Source:

Open Exoplanet Catalogue