Air molecules per cubic foot at a certain altitude above the surface

The kinds of molecules / atoms and the density of those molecules / atoms as a function of altitude in the Earth's upper atmosphere depends very much on the time of day and the activity of the Sun.
Time of day: The Earth's upper atmosphere has a marked diurnal bulge. The density variation at a constant altitude can vary by an order of magnitude (a factor of ten) just due to the difference between night and day.
Solar activity: One poof from the Sun can make the Earth's upper atmosphere swell by not just one order of magnitude as is seen with the diurnal bulge but multiple orders of magnitude. Satellites in low Earth orbit need to use a good deal more propellant when the Sun's activity is high as compared to now when the Sun is rather quiescent.

To answer your question, there is no one answer to your question. This is part of the reason (perhaps the key reason) why predicting when a satellite in a degrading orbit will re-enter is imprecise. There are multiple models of the Earth's upper atmosphere that vary from fairly simple to ridiculously complex.
On the simple side, I prefer to use the US Naval Research Laboratory's mass spectrometer and incoherent scatter atmosphere model, or NRLMSISE-00 for short. One needs to provide latitude, longitude, date, time of day, and three parameters that indicate recent and current solar activity. And that's on the simple side.


Would an object placed at L1 be seen in front of the Sun from Earth?

Objects are not place at the Sun-Earth L1 or L2 Lagrange points. They are instead placed in pseudo orbits about these points. These pseudo orbits intentionally avoid being directly in line with the Earth and the Sun for two key reasons. One reason is that these points are directly in the line between the Sun and the Earth. An object at the Sun-Earth L1 would be hard to see, while an object at the Sun-Earth L2 point would be in the Earth's umbral shadow.
A much more important reason is the incredible amount of propellant that would be needed to maintain their position at those unstable Lagrange points. The pseudo orbits, while also unstable, are much less unstable and require considerably less propellant for stationkeeping.


How to determine the mass of a runaway star?

OK, I'll assume you are talking about this star or some other high-mass "runaway star".

Most massive stars are born in rich clusters and are often part of multiple system. Is not unusual for such stars to get ejected from these regions are high velocities, either as a result of dynamical interactions in multiple systems or when they are companions to stars that explode as supernovae.

The steps to estimate the mass are: (i) Obtain very detailed optical and UV spectroscopy. From this you estimate the spectral type of the star, it's effective temperature and its (current) mass-loss rate. (ii) You combine this with a known distance (the star in the link was in the Large Magellanic Cloud, so the distance was known fairly accurately) to estimate a luminosity. (iii) You compare the luminosity, temperature and mass-loss rate with the predictions of a specialised high-mass stellar evolutionary model, which includes mass-loss and rotation. The comparison yields an estimate of the mass and age of the star.

The details are found in Evans et al. (2010). It is probably fair to say that the estimate of $90\ M_{\odot}$, is uncertain by a few tens of solar masses.


It's indeed a coincidence. The human limits on observables cannot have some deeper explanation.
In fact, we can claim many things like this, without having some sort of deep meaning.
The other thing is they are not "exactly" equal. Their ratios just happens to be in the same order. There are also uncertainties in all these measurements and the ratios are never equal to each other. If they were exactly equal in some sense then we may thought there is some deeper meaning.


What is the accepted theory as to why Uranus' axis is tilted so severely?

Nasa.gov speculates that
the most likely possibility is that an object with a mass close to that of Earth's collided with Uranus, causing it to rotate on its side from then on.
A recent test by Space.com suggests that "Planet Uranus Got Sideways Tilt From Multiple Impacts". These findings suggest that two or more smaller collisions probably occurred, asserting that the early solar system was probably more turbulent than most would think.


What are the prerequisites for a meteorite to reach the ground?

Your first resource here can be the impact effects calculator, It can show you when it is likely that a high velocity impact will occur. Somewhat surprisingly, faster objects are less likely to hit the ground at hyper-velocity, as they are more likely to break up and explode at altitude. They will likely form multiple meteorites, as the Chelyabinsk meteor did

If you are interested in meteorite formation, rather than craters, then most impactors will in some part survive the fall, provided they are rocky rather than icy. Objects smaller than 5cm are mostly cometary in origin. Larger than that and they are more likely to be rocky.


What is the distance between the solar systems towards the center of galaxy?

We don't really know what the statistics are for the frequency of solar systems is as a function of stellar density and many such systems in dense regions may be disrupted by close encounters with another star.

What I surmise you want is an estimate of the density of stars as a function of Galactocentric radius.

In the solar neighbourhood, there are 378 known stars within a distance of 10pc of the Sun, according to the RECONS project. This gives a stellar density $n = 0.09$ per cubic parsec, which includes stars of all masses and white dwarfs. However, this includes multiple systems. If we ask what the density of stellar systems is, then it more like $n = 0.075$ per cubic parsec.

The average distance between stars is tricky to calculate because of the issue of multiple systems. About half of stars are in multiple systems, with a companion at distances of anything from a fraction of an AU to $10^5$ AU, though the distribution peaks at around 1000 AU.

The average distance between systems is about $n^{-1/3}$, which is about 2.36 pc (or 7 light years).

The density of stars increases dramatically towards the Galactic bulge and then again towards the centre. However, since we would not actually detect all the very low mass stars that we can see in the solar neighbourhood, to some extent stellar density estimates rely on an assumption that the mass distribution of stars is similar throughout. There are huge uncertainties here; a lot of detail is discussed in the review by Genzel et al. (2010).
The consensus seems to be that there are of order a million stars in the central parsec, giving an average separation of 0.01 pc, or 2500 AU.

This density falls by several orders of magnitude in the first 10pc. The bulge of the Galaxy has a radius of about 2 kpc and a mass, mostly in stars, of about $2\times 10^{10} M_{\odot}$. If the average star has a mass of $\sim 0.3 M_{\odot}$ (as in the solar vicinity), then the stellar density in the bulge is about 2 per cubic parsec, with an average stellar separation of 0.8 pc (or 3 light years).


Why do planets in red dwarf star systems tend to have a higher probability to get tidally locked?

A rough back-of-the-envelope way of seeing what's going on...

The tidal force is due to the difference in gravitational force, so follows an inverse cube law:

$$F_\mathrm{tide} \propto M_\ast R^{-3}$$

where $M_\ast$ is the stellar mass and $R$ is the distance. So at the same distance from a less massive star, the tidal force will indeed be weaker. But what's usually of interest is the effect on planets in the habitable zone. As a first approximation, define the habitable zone as the location where a planet receives the same amount of insolation from its star as the Earth does from the Sun. The inverse square law then gives:

$$R_\mathrm{HZ} \propto L_\ast^{1/2}$$

where $L_\ast$ is the luminosity of the star. For main sequence stars, we can approximate using a power-law mass–luminosity relationship:

$$L_\ast \propto M_\ast^{k}$$

where the exponent $k$ is traditionally given as 3.5, although it may vary between about 2 and 4 in the mass range of interest. Combining these gives:

$$F_\mathrm{tide,\ HZ} \propto M_\ast^{1-3k/2}$$

The exponent is negative for $k > \frac{2}{3}$, which is the case for the main-sequence mass–luminosity relationship. Therefore the tidal force in the habitable zone increases with decreasing stellar mass: the smaller distance to the habitable zone for these stars provides a stronger effect than the decreased mass.


Metalicity and age of bulge stars vs halo

Your intuition is largely correct: the key is that the proto-bulge region had a deep enough potential well so that the supernovas couldn't expel the remaining gas, and so new stars could form out of the gas (enriched by the supernova ejecta) in a continuing cycle. In the low-mass, isolated protogalactic clouds which probably contributed to the halo, the initial round of supernovas ejected most of the gas (including the original gas that hadn't yet formed stars) -- thus, little opportunity to form more stars (out of higher-metallicity gas) on a continuing basis.

It's really the total mass in a given region that matters. For example, the density of stars in the central regions of a globular cluster is pretty high, but the mass of the cluster as a whole isn't enough to keep most of the gas when its massive stars go supernova.

(Side note: the traditional terms "Population I" and "Population II" aren't used all that much any more, since age and metallicity can vary continuously and aren't always strongly associated.)


Is it possible to do moon sighting in advance for 5 years with 100% accuracy?

Note: this answer was posted under duress; though I mentioned in a comment under the question that I was composing an answer, several users have decided to close the question out from under me. Therefore I've put this together a little hastily. It's late here and I'll come back in the morning to address any questions or requests for clarification, and probably add some more information.

  Is it scientifically possible to predict (simulate...) in future with 100% accuracy as to the location of the moon?

Nothing is 100% accurate or possible. But it is highly likely that the apparent position of the Moon from a given geographic position on Earth can be predicted to very high precision over the next five years. Each aspect:

Geometry
Geography
Time
Meteorology/atmospherics

is predictable based on well-understood scientific principles which have been tested against a substantial extent of data.

Without a specific specification in the question, I can present these aspects of the problem and give an example of the precision to which they can be predicted.

Geometry

We have records of bits and pieces of the Moon's motion for several thousand years from accounts of eclipses, and several hundred years with much more precision with respect to time and to other stars from records of lunar occultations of stars as seen through telescopes.

Combine this with laser ranging measurements of the moon which fit theoretical predictions to centimeter accuracy over a period longer than forty years, and it seems that the relative positions of the Earth and Moon are probably going to continue to be within a few centimeters for the next five years as well.

You can read further about that in
 - Progress in Lunar Laser Ranging Tests of Relativistic Gravity
 - Lunar laser ranging: the millimeter challenge

Now the position of the Earth relative to the Sun is a different matter. If you want to read about how these calculations are done you can have a look at this answer to How to calculate the planets and moons beyond Newtons's gravitational force? These days the movements of the Sun, Earth, planets, and other bodies with enough mass to affect their motions substantially are know extremely well.

Let's look at one of the major JPL Development Ephemerides. In the abstract of The Planetary and Lunar Ephemerides
DE430 and DE431 there are a few sentences that give us some idea of what's discussed in this pretty in-depth report:

  The planetary and lunar ephemerides DE430 and DE431 are generated by fitting
  numerically integrated orbits of the Moon and planets to observations. The present-day
  lunar orbit is known to submeter accuracy through fitting lunar laser ranging data with
  an updated lunar gravity field from the Gravity Recovery and Interior Laboratory (GRAIL)
  mission. The orbits of the inner planets are known to subkilometer accuracy through fitting radio tracking measurements of spacecraft in orbit about them.

Geography

The visibility of a crescent moon from a given spot on Earth needs more than the positions of the center of masses of the Moon, Earth, and Sun. We need the position of the viewer. These days we can get GPS coordinates of the site, but those coordinates are fixed to the Earth, and the Earth rotates around its axis and its axis nutates and wobbles.

That motion is pretty small on the scale of five years, you can see the image below showing the motion of the Tropic of Cancer each year.

Time

The rotational speed of the Earth increases and decreases in a somewhat unsteady way, and this is why the placement of leap seconds looks a bit uneven. However, the fact that we are talking about one second makes it clear that the uncertainty over five years is likely to be a handful of seconds.

Still, this may be the single biggest risk in the accuracy of the prediction five years into the future!

There could be a major seismic event, magma movement, or big chunk of ice that falls into the ocean. Specific details about how these can affect the rotational speed of the Earth (and the exact direction of its axes) need to be addressed in detail in separate questions, but this could be an upset.

Of course we would know about it, and a request for an updated five year report could be issued. The time error wouldn't suddenly jump to some big number, it would slowly start accumulating, and there would be time to recalculate before it became significant.

Atmospherics and Meteorology

This is the hardest one. Refraction from the atmosphere can move the apparent position of objects significantly, up to a fraction of a degree. But the amount of refraction is fairly well predictable. It depends a little bit on the Weather because index of refraction's deviation from 1.0 depends almost linearly on the density of the air, and slightly on the humidity. But this is a very small correction on top of a small correction.

Lunar Libration image from here

Lunar Laser Ranging images from here

 Source

Source


How can you find the distance to a galaxy using right ascension and declination?

You can't use the RA and dec.  These tell you the direction that the galaxy is in, relative to the stars.  It is possible and normal for two galaxies to have the same (or very similar) RA and dec and be at completely different distance.
You can estimate the distance using the Hubble law $v= H_0 d$, if $v$ is measured in km/s and d is in mega-parsecs the value of H_0 is approximately 70 (measuring this value accurately is a major problem in cosmology)
So $D= 2727/70 \approx 39$ So we estimate the distance to the galaxy to be 39 million parsecs  (multiply by 3.26 to convert to millions of lightyears).  This can only be an estimate, as the actual value of $H_0$ is not know with certainty, and the galaxy can have particular motion relative to the Hubble flow (for example, Andromeda is actually blue shifted, but it isn't a negative distance from us, as this value would suggest!)


How are gravitational lensing and dark matter related?

Gravitational lensing of background galaxies and quasars is used to probe the amount of gravitating matter in foreground objects such as clusters of galaxies.

Here is an example, taken with the Hubble Space Telescope.

The foreground object is a cluster of galaxies and the lensed background objects, some of which are marked, are galaxies and quasars that are much further away.

The shapes of these lensed objects can be used to deduce the gravitational potential in the foreground cluster. This potential is due both to the galaxies that can be seen, hot intra-cluster gas (that can be seen in X-rays), but chiefly is due to dark matter that pervades the cluster. Thus a combination of optical, X-ray and lensing observations can map the distribution of dark matter and hot gas in the cluster.

An example is shown below for Abell 1689, where the diffuse, blobby emission is a false-colour representation of the density distribution of dark matter inferred to be in this cluster from the many lensed images of background galaxies that can be seen.

http://hubblesite.org/news_release/news/2010-37

A particularly important example that is a landmark in dark matter research is the Bullet Cluster. This is actually two clusters that appear to have collided and passed through each other about 150 million years ago. The distribution of galaxies in the clusters can be seen in optical images; the hot gas can be traced by X-ray observations and the dark matter is traced by looking at the lensed images of distant background galaxies.

The key image is shown below - it illustrates the distribution of optical galaxies, the X-ray-emitting hot gas (pink/red) and the inferred distribution of dark matter (blue).

As the clusters move through each other, the hot gas interacts and piles up into bow-shocks that are displaced from the distribution of optical galaxies. On the other hand, the dark matter, which is thought to consist of non-baryonic, non-interacting particles remains centred on the visible galaxies and each dark matter "cloud" has apparently passed through the other with no interaction occuring.

The displacement between the hot gas and the majority of the gravitating matter, that is dark, is very strong evidence for a non-baryonic particle description of dark matter, rather than some of the alternatives to dark matter like Modified Newtonian Dynamics.


Term for a momentary geometric pattern formed by astronomical objects

Momentary events in general (comets, supernovae, solar flares, etc.) can be lumped under the general heading of "transient phenomena". 'Transient asterisms' may be a candidate term for the concept you are describing.

Then again, we now know that on a large enough time scale all asterisms are eventually transient in nature. So one might justifiably take the point of view that the adjective "transient" is redundant. So perhaps calling the patterns you describe simply as "asterisms" is not as bad of a candidate term as you might suspect.


Distance and orbital period for terrestrial binary planets

The formula for orbital period is $$T= 2\pi\sqrt{\frac{a^3}{G \left(M_1 + M_2\right)}}$$

For your example, $a=19000000$ metres, and $M_1 = M_2 = 6e24$

Which gives an orbital period of just over 5 hours.

Whether such an arrangement is likely depends on the nature of the the formation of the binary planet, and how the orbital period and rotation rates are affected by tides.


Unidentified color changing object in sky

Probably a satellite flare

It sounds like a satellite flare.

Coming, showing up clearly for a few moments, moving, then going away, are characteristic of a satellite flare, like Iridium flashes.

Check Heavens Above or any other similar site to see if any predicted flares match with your location and time.

The changing colours can be explained with "Twinkling".


How to convert celestial Coordinates with a radius, to max and min celestial coordinates?

If you were working with a rectangular coordinate system as shown in the figure below (that also had no bounds), then it is correct that the right ascension would be $RA=RA_p\pm d$ (points 1 and 3) and the declination would be $Dec=Dec_p \pm d$ (points 2 and 4) where d is the radius of the circle.

The sky is spherical, so the lines of RA are not a constant separation. (It also has limits such as declination less than $\pm 90$.) Therefore you need to use an equation for spherical trigonometry. The angle d between two points i and j is

$$\cos(d) = \sin(Dec_i)\sin(Dec_j) + \cos(Dec_i)\cos(Dec_j)\cos(RA_i - RA_j)$$

For points 1 and 3 on the circle, $Dec_i=Dec_j=Dec_p$ in the above equation, and you can solve for the difference in right ascension $RA_i - RA_j$ to be added to or subtracted from $RA_p$. For points 2 and 4, $RA_i=RA_j=RA_p$ and the above equation simplifies to $Dec=Dec_p \pm d$. Of course, keep in mind that declination is between -90 and +90. For any other point on the circle, you can assume either an RA or Dec (within the bounds of the above 4 points) and solve for the other unknown.


If there would be air at room temperature between Earth and Moon, sound would take about 28 days from Earth to Moon and echoed back, with helium instead of air 9 days, and with hydrogen about 7 days.

Calculated from http://www.phy.mtu.edu/~suits/SpeedofSoundOther.html and http://en.wikipedia.org/wiki/Moon

You could also reflect Earth light from Pluto back to Earth, 11 times this double Earth - Pluto distance to get about 5 days of travel of light, much depending of the orbital position of Pluto.


What is the most asymmetric known planet?

That depends on what you mean by asymmetric, on what counts as planet, and on what parts of a planet are to be considered / excluded. The dominant feature of many planets is that they aren't quite spherical due to their rotation. If that counts then Saturn is the winner. Saturn's poles are 5900 km closer to Saturn's center of mass than are points on Saturn's equator.
That oblateness is however exactly the shape an object such as Saturn is expected to take on. It means that Saturn is in an equilibrium shape. Moreover, it has a very definite axis of symmetry, the rotation axis. So that's probably not what is meant in the question as being asymmetric. All of the giant planets are quite symmetric by this standard. We need to look to the terrestrial planets.
One concept that might be of aid is the center of mass - center of figure offset of each of the four terrestrial planets. Do the oceans count? Do the icecaps count? Excluding the oceans but not the ice leads to COM-COF offset of about 2100 meters. Including both leads to a COM-COF offset of about 800 meters, about the same as Mercury and Venus. Mars however has the Earth beat by far. Various estimates place the Mars COM-COF offset between 2.5 km and 3.3 km.
Another approach is to look at how lumpy the planet's gravitational field is. Mars once again is the winner in this regard.

I have a simple technical question: an asymmetric body gets a torque in a gravitational field, is the rotational energy/acceleration to be subtracted from the regular energy/acceleration? In other words: does an asymmetric planet get less radial acceleration than a perfect spherical one?

No. In a system of particles bound by central forces such as self-gravitation, Coulomb forces, and chemical bonds, the acceleration of the system's center of mass is the net sum of the external forces acting on the individual particles, divided by the total mass.


Do all stars have an Oort cloud or is it a rare occurence?

Awesome question, especially since we know so little of the answer.

Nobody knows for sure how the Oort Cloud formed - I'll put that out there right now - but the current hypothesis is that it was originally part of the Sun's protoplanetary disk. All of the ice and rock coalesced into small bodies - proto-comets, if you will. While these bodies were much closer in to the Sun than they are today, they were tossed far out by gravitational interactions with the gas giants. Other interstellar comets could also have been captured by the Sun, adding to the population.

So why is the Oort Cloud spherical? After all, the protoplanetary disk was just a flat disk. Why were the orbits of the objects perturbed? Well, the Oort Cloud objects are only loosely bound to the Sun - relatively, that is. They can be influenced by passing stars or other objects. It appears that galactic-scale tidal forces, combined with the influence of passing stars, molded the Cloud into its current spherical shape.

So what does this all tell us? Well, we know other stars have protoplanetary disks, right? Some also have exoplanets - gas giants like Jupiter. They are also subject to tidal forces and the passing of nearby stars. So, theoretically, there's no reason why other stars shouldn't have Oort Clouds.

So can we find them? The answer is, most likely, no. Here's why. According to Wikipedia,

  The outer Oort cloud may have trillions of objects larger than 1 km (0.62 mi), and billions with solar system absolute magnitudes brighter than 11

An absolute magnitude of a solar system object of 11 is very dim. Now, the object's apparent magnitude is how it would look from a given distance; the absolute magnitude is how it looks from a distance of 1 AU (in the case of Solar System objects, this quantity is denoted $H$). Oort-cloud objects are 2,000 - 50,000 (or more) AU away-- so these objects, to us in the same solar system, have an apparent magnitude much fainter than 11.

The point of that poorly-explained interlude is that these objects are faint. Very faint. And objects in Oort Clouds around other stars would appear even fainter. Using the distance modulus, we can calculate the apparent magnitude of an object if the distance to that object and its absolute magnitude are known:

$$m-M=5(\log_{10}d-1)$$

(from here)

where $m$ is apparent magnitude, $M$ is a scaling of $H$ normally used for stars, and $d$ is the distance in AU.

Given an Oort Cloud object $x$ light-years away, you can figure out how bright (or dim) it would appear, given that 1 light-year is 63241 AU. Try this with the distances of nearby stars, and you'll realize how dim objects in these stars' Oort Clouds would be.

As a final note: We don't know for sure if other Oort Clouds exist. From what I've been able to find, we don't have sufficiently powerful telescopes to observe these hypothetical Clouds, and so we don't (and may never) know if they exist.

I hope this helps.

This paper was instrumental in this answer. Start at page 38 for the relevant information. This page, too, has some good information.

As I found from a link from an answer to this question on Physics, we've found Kuiper-Belt-like disks around other stars. This means it is certainly plausible for these stars to have Oort Clouds, too. And exocomets have been detected, which is another good sign.


What is the effect of this expanding universe on the energy density of it?

I am no expert and I am sure there will be a more rigorous answer later, but here my 2 cents:

It appears to me that the term "energy" and energy conservation becomes a bit more complicated in general relativity and cosmology, taking into account more abstract forms as the energy of the gravitational field or space curvature. As you seem to be more interested in the implications I just want to focus on the common forms of energy, i.e., radiation, matter and dark energy.

For a given expanding volume of space only the energy from matter (dark and normal) is dilutes as expected. You could say that with respect to expansion its energy is conserved. This means its energy density decreases with $\sim a(t)^{-3}$, where a is the scale factor (or just the diameter) of the universe.

Radiation additionally looses energy via cosmological red-shift, so its energy density decreases even faster with  $\sim a(t)^{-4}$.

Finally you have the dark energy density. What makes it special is, that its density does not depend on $a(t)$. One can think of it as the energy associated with space itself. So if the universe expands new space and dark energy "appears" and its density stays constant.

So over time the average (matter and radiation) energy density decrease and approaches the energy of empty space itself (or vacuum energy).

Note, however, that this is only true for the average densities. In regions of space, that are held together by gravity, as our local group, the matter density will not decrease. On the contrary, these region are still contraction due to gravity.

So to answer your question on the effect on all the celestial bodies:
Those which are gravitationally bound, won't notice much but others will drift apart never to see each other again.


First time visit to dark sky stargazing site: what do I need?

If you have yet to plan your adventure and the moon is not you main target, be sure to set your date nearest a "New Moon". During this phase the moon will not be visible at night, reducing the light pollution that much more. This will increase the amount of stars you should be able to see. Peak of any meteor shower event would be a bonus aswell.

Red light is also important as you suggested, but keep in mind a cellphone is just as bad as a flashlight. There are free apps available for many devices to reduce your brightness levels even further. Your eyes can take roughly 20 to 30 minutes to adjust to the darkness after exposure to light. This will significantly impact your stargazing experience.

Wether its your first or hundredth time, naked eye observation is always an enjoyable experience. Being that you shall be going to a region with rather dark skies, you should see a suprisingly bright milky way. There should be so many stars in the sky, one could get "lost" looking for often easily  recognized constellations. You may even see meteors, satellites or even the ISS.

That being said, binoculars are not neccesary but are a great tool for the avid or begginer gazer. With a steady hand and a decent pair of binoculars it is possible to see the rings on saturn and moons of jupiter. Many galaxies and starclusters can also be enhanced with the aid of the binoculars. Be sure to lean on something though or it will be way to shaky to see clearly. If you have the space and a half decent pair, I would bring them. If not, no worries the pea soup of stars in the sky should keep you busy.

There are many free resources online that can help you find what your looking for if you have any celestial targets in mind. If you have a mobile device, an astronomy/planetarium app is a very useful tool for any stargazer.

For your first time, the most important thing I can suggest to bring is a comfortable reclining chair or blanket and an open mind. There are so many amazing things to learn about the universe and there is no better place to do it than under the stars.

Also be sure not to forget the bug repellent. Can't see much of the stars if your hiding in your tent. Best of luck and hope you get great weather.


What percentage of galaxies rotate with trailing arms?

  I am looking for the percentage of galaxies that rotate with trailing arms.

  ...

  Do you know of any newer publication on this question? More recent than Pasha & Smirnov (1982) where the sample is quite small?

I think the best answer is that exact numbers are very difficult to determine.

Take for example page 5 of "Arm classifications for spiral galaxies" (1987) by D. M. Elmegreen and B. G. Elmegreen, cited over 200 times.

Affiliation: IBM Thomas J. Watson Research Center, Yorktown Heights, NY

Publication: Astrophysical Journal, Part 1 (ISSN 0004-637X), vol. 314, March 1, 1987, p. 3-9.

  Page 5:

  "II. THE CATALOGS

  a) Selection Criteria

  Table 2 includes all 708 spiral galaxies listed in the Second
  Reference Catalogue of Bright Galaxies - [3rd] Edition link (A. de Vaucouleurs, G. de Vaucouleurs, and H. G. Corwin 1976), with declinations $\delta \gt -35°$, inclination-corrected radii at 25 mag arcsec$^{-2}$, R$_{25}$, greater than $1^\prime$, and inclinations less than 60°; 654 galaxies could be classified, 54 galaxies which met the criteria but were too faint, had no details (were overexposed on the POSS or were too early in Hubble type to distinguish structure) or were too inclined for reliable classification are also included in Table 2, but with designations f n, or i, respectively, instead of arm classes.

  ...

  b) Self-Consistency Checks

  Most galaxies in Table 2 were classified at least 3 times, separated by intervals of several months, as a consistency check. They were classified from both paper and glass copies of the POSS. In the de Vaucouleurs sample, ~10% of the galaxies were given different arm classes during independent examinations on the paper prints. An 8% variation occurred when the glass images were subsequently examined.

  ...

  For the high-resolution galaxies listed in Table 3, again ~10% of the arm classes changed upon reclassification. In these cases the large amount of detail presented confusion; irregular features were given more or less weight at different times of classification. Arm classifications become increasingly difficult for very small, faint galaxies. The cutoff of $1 ^\prime$ radius for the galaxies in Table 2 represents a minimum practical size for classification from the POSS.

As you can see, even with expert examination multiple times of the same data, determination of exact numbers with any accuracy is impossible at the present time. A guess is "most of them", an exception example is: NGC 4622 (also called a backward galaxy).


How much lighter would someone be on the Jupiter-facing side of Io?

On the leading or trailing points of IO, you are orbiting Jupiter at the same speed as the center of mass of Io, so you experience its gravity only
On the equator, facing Jupiter, you are 1821.6 km closer to Jupiter, but still orbiting with the same period. So there is a bit of Jupiter's gravity that is not countered by Io's orbital speed.
How much?
Gravity from Jupiter is more by: Assume 1kg at Io core, vs 1kg at io-jupiterface
G(core) = 0.71813N
G(jupiterside) = 0.7244N
Differnece = 0.00627N
In addition, you have less centrifugal force from the orbit, to be being in an orbit that is "smaller" by 1821.6km but has the same period.
Centrufugal at Io(core) = .71813N (as expected, at Io's center of mass, its orbital centrifugal force matches Jupiter's attraction, hence an orbit)
Centrifugal at IO(jupiterface) = 0.7151N
Thus the difference in perceived weight of a 1kg mass is
(.71813-0.7151) + (0.7244 - 0.71813) = 0.0093N
Which is a difference of 1.295% , or 1 part in 77


How is the total mass of all the Solar System's small asteroids and meteoroids estimated?

Your question body is different from your question title and it seems you really want to ask what you did in the question body so I'll address that.

Short Answer: The simple power law which applies for larger asteroids and comets actually doesn't extend that well to smaller bodies and shouldn't be trusted too much in that range.

Long Answer:

You're right to be suspicious of using a simple power law to relate asteroid size to quantity at small sizes. Especially because there are known deviations from that power law at certain radii. As wikipedia notes:

  The number of asteroids decreases markedly with size. Although this generally follows a power law, there are 'bumps' at 5 km and 100 km, where more asteroids than expected from a logarithmic distribution are found.

Who's to say there aren't other bumps or deviations at radii that we can't observationally confirm. Our confidence in any power law's applicability for small asteroids and comets basically comes from modelling.

This power law you're asking about is essentially something referred to as an "Initial Mass Function" (IMF). A lot of work has been done to define an IMF for stars, relating the mass of stars to the number of stars at that mass. This stellar IMF follows a power law more or less, but strongly deviates for very low masses. At the surface, it would be reasonable to assume an IMF for asteroids and comets likewise deviates.

The way you answer this, is through models. See for example Cuzzi, Hogan, and Bottke (2010) (that link is to the formal journal paper, but for some reason some plots are obscured so you can also look at their "unofficial" arxiv version). In this paper, the authors attempt to construct a representative asteroid/comet population by modeling their formation via accretion of protoplanetary dust grains into asteroids, comets, KBO, and planetesimals. They make certain assumptions about the initial gas disk around the Sun and bake in physics of how these dust grains formed into larger conglomerates. Their end goal is to produce an IMF that identifies how many asteroids/comets exist at a given size. They back up a lot of their modelling with data from observations so they stick to reality as much as possible.

I'll let you peruse their results since they actually produce many different IMFs using a variety of different assumptions and starting conditions. However, I think one can summarize their results in a way which addresses your main question. As they state it:

  Asteroids were born big

What they mean is that asteroids (and comets), rather than forming by successive growth via accretion of many small particles, seem to just clump together into larger bodies, completely bypassing the formation of many smaller meter-sized asteroids and comets (that's not to say there aren't meter-sized objects). You can actually see this in their various plots (fig. 4 for example). Much like the stellar IMF, the power law drastically changes once you get to small sizes and the power law which applied for larger sizes is no longer applicable. At which size the general power law no longer becomes applicable depends upon certain tunable parameters. Cuzzi et al. shows many different results with their tunable parameters set to many different possible values.


Why dust is optically thin in Far Infrared wavelengths?

The term "optically thin" means that the optical depth is small. The optical depth is a measure of the opacity of a medium, in this case dust, experienced by light traveling through that medium, and is defined as
$$
\tau \equiv n \, r \, \sigma,
$$
where $n$ is the density of the particles in question, $r$ is the distance traveled through the medium, and $\sigma = \sigma(\lambda)$ is the cross section of the particles, which is dependent on the wavelength $\lambda$ of the photons. If $\tau\ll1$, the medium is said to be optically thin, while if $\tau\gg1$, it is said to be optically thick. The fraction of the light that is extinguished by the journey through the medium is $e^{-\tau}$, so the two cases indicate when the light is mostly transferred and mostly extinguished, respectively.

Cosmic dust is composed of particles spanning a large range of sizes. Photons interacting with the dust are either scattered in another direction, or absorbed, depending on the albedo of the dust. But in both cases, a photon has a larger probability of interacting with a dust grain which is comparable to its wavelength. Because of the size distribution, an ensemble of dust grains hence has a characteristic extinction curve. In the figure below (modified from Laursen et al. 2009), I plotted the (functional fits to the observed) extinction curves of dust in the Small (dashed line) and Large (solid line) Magellanic Clouds$^\dagger$:

The extinction is here given in terms of "cross section per hydrogen atom", but you can just think of it as an average dust grain cross section. The colors show the ultraviolet (purple) and infrared (red) regions of the spectrum.
You can see that the extinction, or the cross section, is largest for UV photons, and as you go to longer wavelengths, the cross section decreases sharply and is very small in the far infrared ($\lambda \gtrsim 30 \, \mu\mathrm{m}$; the exact definitions depend on your field of interest).

This is what your statement refers to. But in fact it's badly phrased, because the optical depth is also a function of density and distance, so for instance for a distance $r = 1 \, \mathrm{cm}$ through a dust cloud, the optical depth is $\ll1$ for all photons.

$^\dagger$Extinction curves in other galaxies look similar. In particular, the extinction along most sightlines in the Milky Way features the same bump around $\lambda \simeq 2175$ Å. The origin of this bump still isn't well understood, but may be partly due to graphites and/or PAHs.


What does Sa and Sab mean for spiral galaxies?

Capitalization is critical here. (As is knowing which classification system is being used.) The traditional spiral stages (going back to Hubble) are Sa - Sb - Sc; this was a sequence from more centrally concentrated (more dominant central "bulge") + smoother disk + more tightly wrapped spiral arms (Sa) to less centrally concentrated (smaller or even nonexistent bulge) + less smooth disk + more open spiral arms (Sc). Later, S0 (for lenticular galaxies, which are not spirals but seem to have significant or dominant disks) was added at the beginning and Sd (and sometimes Sm) were added as stages beyond Sc.

Hubble also specified that clearly/strongly barred galaxies should be labeled SB0 - SBa - SBb - SBc (with a later extension to SBd), creating a parallel sequence to the nominally unbarred sequence S0 - Sa - Sb - Sc - Sd.

De Vaucouleurs emphasized the existence of intermediate stages: thus, Sab is in between Sa and Sb, Scd is in between Sc and Sd, and so on. (There is also a stage intermediate between S0/lenticular galaxies, which are not spirals, and the Sa spirals: "S0/a".) He also applied this to bar classifications, arguing that there were weakly barred galaxies in between unbarred and strongly barred galaxies. To do so, he changed the Hubble classifications so that "SA" mean unbarred, "SAB" mean weakly barred", and "SB" meant strongly barred. These precede the lower-case Hubble-sequence labels, so you end up with SA0, SAB0, SB0 - SA0/a, SAB0/a, SB0/A - SAa, SABa, SBa - SAab, SABab, etc.

To add to the confusion, de Vaucouleurs also recognized that some spiral galaxies had an unclear bar status (especially true for edge-on galaxies). His famous and widely used RC3 catalog uses "S" for those cases, so "Sa" in RC3 actually means "Hubble stage a, unknown bar status". This, of course, looks exactly like the traditional Hubble classification (continued by other classifiers such as Allan Sandage) meaning "Hubble stage a, unbarred), which has confused people over the years.

To directly address your question: the "a" and "ab" mean what you said in your comment: "Sa galaxies tend to have significant bulges, and tightly-wrapped and relatively smooth spiral arms. Sab galaxies are similar to Sa galaxies, but show more obvious resolution of the arms." Loosely, people will often use "Sa", etc., to refer to spirals regardless of their bar status, as in your quoted text (or in something like, e.g., "Sa galaxies tend to have lower star-formation rates than Sc galaxies."). If it's a specific classification of an individual galaxy, then it might mean "also unbarred" or even "bar status unclear".


Will intergalactic travel be impossible in roughly 15 billion years?

Yes, over time the accelerating expansion of the universe (assuming the $\Lambda$CDM model is right) will separate gravitationally bound clusters of galaxies from each other with exponentially growing distances. However, this will not happen in 15 billion years.

This excellent paper analyses the expansion in detail. What takes about 17 billion years is for the distances to increase by 2.71 times. They predict that typical clusters will grow isolated in about 120 billion years, with the Local Group becoming isolated in 175 billion years from the other nearby groups.

In this "era of isolation" the only reachable galaxies are the ones in one's local cluster. However, they will tend to merge over time into one big elliptical galaxy (like the Milky Way and Andromeda are likely to do in a few billion years). Extragalactic astronomy also becomes really boring since the remote clusters are redshifted so much they cannot be observed.

If one just travels outwards from Earth in some direction with a high velocity one will eventually find oneself in a last cluster beyond which there is nothing reachable.


Is it possible to see Callisto with the naked eye when it's at its greatest elongation from Jupiter?

This has been an area of controversy for quite some time now, with no clear resolution (if you'll pardon the pun). There's been interest in supporting or refuting claims that observers were able to see the moons with the naked eye before Galileo. There are, of course, two major impediments (which you might well know about): brightness and angular separation from Jupiter.

The human eye has a diffraction-limited resolution of (roughly) one arcminute, whereas Callisto can reach separations from Jupiter of about 10 arcminutes - so, arguably, resolving it is not an issue, as it is with some other moons. However, we do have to deal with the problem that light from Jupiter could dominate any light we see from the moons (even if Ganymede and Callisto are too close to be separated and appear as one source). It's the same issue we get when observing faint sources near bright sources using a telescope, such as direct imaging of exoplanets.

If you do want to give this a shot, you might want to block out Jupiter's light with some foreground object, while making sure not to block out the moons. Choose a dark night when Callisto is at its greatest separation, and you might have a shot.

But not great odds, of course. I'd be surprised if you succeed. On the other hand, get yourself a good pair of binoculars and you won't have a problem.


How do we measure the age of the universe?

Once we look far enough away to get past "local effects" we see everything moving apart, spreading out in all directions, like a set of dots on a balloon being inflated, or like currants in a loaf of bread rising in the oven.

We can use the laws of physics to follow those movements back in time to work out how far apart all the galaxies were at various dates in the past. If we do this we find that 13.8 billion years ago they were all in the same place (there are a lot of subtle details to get right in this calculation, but a lot of people have worked on it over a long time). That much matter and energy packed into that small a space warps both space and time so that it is not really possible to talk about time "before" that event. Hence that is taken as the origin of the universe.


How does the dense atmosphere of Venus affect the planet's rotation?

I don't know all the details, but I can give a partial answer, with links to further reading if interested.

Tidal locking, and orbits slowing down due to a tidal bulge and tidal forces and/or a denser or greater gravitational gradient on one side of the tidally locked object are all pretty straight forward, and I don't think that part needs to be explained too much.  The tidal bulge has mass and that mass creates drag on the rotation (or speeds it up if the bulge is behind the object causing the tides).

And it's worth adding that rocky worlds are gravitationally lumpy, where the heavier side (gravitational gradient), tends to, over time and in tidal locking, face the object it orbits.   The near side of the Moon, for example, is slightly denser on average than the far side.  See here for a better explanation and here for gravitational map of the moon (the Earth side is on the left in the picture below).

Venus' atmosphere, particularly it's upper atmosphere, behaves very differently.   There's still an atmospheric tidal bulge, but because of the heat from the sun, the hot gas on the sun side of Venus upper atmosphere has lower density, so the tidal bulge may have less mass, not more mass, even with greater size.   Atmosphere doesn't behave like a typical tidal bulge orbiting close to a star because the hot atmosphere is less dense and the night-side of Venus' upper atmosphere is more dense.  The (by mass) true tidal bulge is on the far side of Venus, away from the sun, not the near side.

Another effect is that the heat from the sun causes something like a jet-stream on Venus, a very powerful jet stream.   The wind speed on Venus' upper atmosphere is faster than any hurricane on Earth and the upper atmosphere tends to rotate in the same direction around Venus.   This solar driven rotation (may be) a factor in Venus' surface having a slow retrograde rotation where, without an atmosphere it might be tidally locked.

It's difficult to imagine an upper atmosphere wind could "spin" a planet, and it would only do so very very slowly, (Venus' atmosphere is about 1/10,000th it's total mass, not too far off from the ratio of Earth's oceans to Earth's total mass) and the tidal slowing of a planet is very gradual, at Venus distance taking perhaps 100 million years to slow a rocky body of that size to tidal locking.     So the rotation may be a balance between Venus' solar driven jet stream and it's gradual tidal locking.

At least, that's what I came up with doing some reading on this.    From Wikpedia:

  The rotation period of Venus may represent an equilibrium state
  between tidal locking to the Sun's gravitation, which tends to slow
  rotation, and an atmospheric tide created by solar heating of the
  thick Venusian atmosphere

and link to the source, which is in French but English translation is available.   (see footnote 100 and 101)

On Venus "jet-stream" like upper atmosphere, note that while the jet-stream on Earth is driven by surface temperature variation perhaps more than solar and by Earth's Coriolis effect, and it varies by seasons.   Venus doesn't have surface temperature variation to speak of and essentially no Coriolis effect due to it's slow rotation, so it's jet-stream tends to be much more along latitude lines and much less curved than Earth's.    That's not to say that there's no variation.   Venus' wind seems to operate on a 5 day cycle and it appears to have grown stronger in the years since we've been observing Venus.  Neither of those is well understood.

Article here

  At the very top of the cloud layers on Venus, wind speeds reach 355
  km/hour (or 100 meters/second). This is the same the jet stream here
  on Earth. As you descend through the cloud layers, though, the wind
  speeds pick up. In the middle layer, the winds can reach speeds of
  more than 700 km/hour. That’s faster than the fastest tornado speed
  ever recorded on Earth.

  But then as you descend further down through the clouds, the
  thickening atmosphere slows the winds down, so that they act more like
  currents in the ocean than winds in the atmosphere. Down at the
  surface, the winds only move at a few km/hour. That’s not much, but
  the thick atmosphere can still kick up dust and push around small
  rocks.

and

  The winds on Venus travel in a westerly direction, the same backwards
  direction that Venus rotates. Seen from above, Venus rotates in a
  clockwise direction. This is backwards from the other 7 planets, which
  rotate counter-clockwise.

another article here.

and, related, quote from Wikipedia and source article.

  Venus's rotation has slowed down by 6.5 min per Venusian sidereal day
  in the 16 years between the Magellan spacecraft and Venus Express
  visits

So, short summary, Venus atmosphere may play a key role in it's reverse rotation, or, maybe it's rotation was reversed by an impact a long time ago and it's still slowing down from that impact.   The impact theory might make the most sense, but impacts large enough to spin a planet are quite rare so, it shouldn't be assumed.   I don't believe it's known 100% whether Venus' atmosphere caused it's backwards rotation, but it seems reasonable and possible.   My personal favorite idea, though it may not be likely, is the gyroscope flipping hypothesis.  (Posting video cause it's really cool, but it has nothing to do with your question).

https://www.youtube.com/watch?v=1n-HMSCDYtM


How are space telescopes stabilised to a perfect standstill?

Pointing

In some ways it's easier to keep the direction that a telescope is pointing stable in orbit in space than it is on Earth.

There's no wind Why is the opening in the Anglo-Australian Telescope's dome so small?
You aren't on a spinning planet requiring you to constantly move the telescope with electric motors and gears
there's no sagging due to gravity

There are tiny torques due to differential solar photon pressure on different parts of a space telescope, but these are small, very reproducible and can be learned, predicted, and compensated.

The James Webb Space Telescope has a giant sunshield that keeps the infrared telescope cold, and this can produce substantial torque. So it has a second reflector used to passively null out most of the torque. You can read more about it in answers to How will JWST manage solar pressure effects to maintain attitude and station keep it's unstable orbit?

In low Earth orbit (LEO) telescopes like Hubble will also experience tiny torques due to tidal forces; for example, one end of Hubble is in a higher or lower orbit than the other end. Without damping, this could cause it to slowly oscillate see this answer. Again, this will be small, very reproducible and can be learned, predicted, and compensated.

If you can somehow stop something in space from rotating, it will simply stay stopped and point in a stable direction, at least for a short time. If you can predict the torques due to solar pressure and gravitational tidal forces, you can apply tiny controlling torques to minimize those effects and it will stay pointed longer.

How?

Sensitive gyroscopes to measure tiny rates of rotation (inertial stabilization)
Watch the stars at the edge of the telescope's field of view, using them as guide stars

Both are shown in the image below, taken from Hubble Space Telescope Pointing Control System.

The objects labeled "Gyros" are gyroscopes used to measure tiny rates of rotation of the telescope and send those measurement to the attitude control system for the telescope. Systems can use real spinning wheels, or fiber optic rings with light going in both directions, or other techniques, but the result is the same.

Currently Hubble is working without all three gyroscopic sensors on-line because it has redundant systems: You can read more about that in Hubble Space Telescope Reduced-gyro control law design, implementation, and on-orbit performance

The objects labeled Fine Guidance Sensors use CCDs at the edge of the Hubble's field of view to track stars:

  The FGSs use starlight captured by the telescope’s mirrors to find and maintain a lock on guide stars to ensure that the spacecraft’s attitude does not change. One FGS can also be used as a scientific instrument to determine a star’s position with high accuracy. The level of stability and precision that the FGSs provide gives Hubble the ability to remain pointed at a target with no more than 0.007 arcsecond of deviation over extended periods of time. This is the same as holding a laser beam focused on Franklin D. Roosevelt’s head on a dime roughly 200 miles away — which is about the distance from Washington, D.C., to the Empire State Building in New York City — for 24 hours.

Feed this and other bits of information from other star cameras into the computer, combine it with the predicted torques from tidal forces and photon pressure, and the computer decides on a tiny correction torque necessary to keep it pointed stably.

That message then goes to the objects labeled as Reaction Wheels. These are four heavy, spinning wheels whose angular momentum is balanced such that it normally applies no torque to the telescope.

When the computer senses that a little torque is needed to keep the telescope pointed in the right direction, or to move it to a new field, motors on the spinning reaction wheels will speed them up or slow them down a little bit, and that will apply a compensating torque to the telescope.

Vibrations

Vibrations and oscillations of a space telescope's structure can also be a problem. These are addressed through very careful engineering and special materials. Structures are design to minimize vibration and oscillation by damping, isolation of moving parts (pumps, solar panels, etc.) and by use of light weight and stiff materials like silicon carbide to move resonant frequencies as high as possible where they are most easily damped and isolated.

Silicon carbide is a very popular material in newer space telescopes and is found in the optical system of too man of them for me to remember, but here is mention of it in GAIA's optical bench answer and in New Horizion's LORRI telescope answer.

above: Gaya's Silicon Carbide Optical Bench, with the two objective mirrors of it's twin telescopes pointing 106.5° apart. From Spaceflight 101, image credit: ESA/Astrium.

above: The LORRI telescope is described for example in the ArXiv preprint Long-Range Reconnaissance Imager on New Horizons

below: From Hubble Space Telescope Pointing Control System


Since the Julian day is defined as a continuous count of seqential days starting at Jan 1 4013 BC, and there is no limit to the number of integers available for  the count, there is no specified maximum.

If you are asking when the Julian Day Number will no longer be useful, possible answers are when there is no one left to count it, or when it is replaced by another concept or definition. Note that there are already several shortened versions, such as Modified JD and Truncated JD, used when JDN became too large to fit in smaller computer word sizes with the desired precision.


Can we change Earth's orbit with modern technology?

No, we cannot.
It's relatively simple math to show that we are a very long shot from changing Earth's orbit by anything significant at all: consider the kinetic energy of Earth in its current orbit, and do the same math for that in the desired orbit:
$ E_{kin} = \frac{m_E}{2}v^2 \approx \frac{m_E}{2}\frac{GM_S}{a}$
where $m_E = 6\cdot 10^{24}$kg is Earth's mass, $M_S=2\cdot 10^{30}$kg is the solar mass and $a=150\cdot 10^9$m is the distance from Earth to sun. I used the approximation in calculating orbital velocity $v = \sqrt{\frac{GM_S}{a}}$ that the Earth mass, is much smaller than the Sun's mass. So current orbital velocity is (using numbers as above) $29822$m/s.
Now, changing oribital distance outward by 1%, thus by $1.5\cdot 10^9$m, to $151.5\cdot 10^9$m yields a speed of $29674$m/s - and in energy that subsequently gives a difference of $2.64\cdot 10^{31}$J to move Earth's orbit about 1% to further from the Sun.
Annual world primary energy production currently is around $10^{20}$J - so we "just" need to produce 100.000.000.000 times the yearly energy production of the entire world to change Earth's orbit about 1%. And that implies that we have a technique to transfer energy 100% into Earth orbital velocity - something which cannot exist either due to simple thermodynamic argument.
Edit to add: Another way to argue is the rocket way: typical speed of conventional rocket exhausts are a few km/s, which is less than needed to leave Earth's orbit - that's why you need multi-stage rockets to get anything into orbit. But if you want to move the whole Earth with a rocket motor, you need a propellant which is fast enough that it directly leaves Earth's gravitational field - or you will not be able to create any effective change of momentum and only move around mass in the Earth system itself. So you need to make some substantial technological advances here, too - or some entirely different technique.


Can impact craters on the moon act like giant radio telescopes?

Interesting idea. I think the answer is both yes and no -- yes with a manufactured dish but no in the crater's raw state.

The Arecibo telescope sits in a natural crater, but adds a dish which has a couple of important things required by a radio dish:

a radio-reflective surface
a specific curvature, classically parabolic, but also shaped
low surface roughness typically in the order of mm to µm for radio frequencies we're interested in. See http://en.wikipedia.org/wiki/Ruze%27s_Equation
a receiver close to the reflective surface (focal length), a satellite would probably be beyond the focal point
probably some some other things, but that's probably enough.

Edit: Added #4


Why don't (or can't) stars be more than 325 or so times the mass of the sun? What limits their size?

There is basically an upper limit to the mass of a star because their luminosity is so great that the radiation pressure prevents the accretion of further mass.
However, the upper limit depends on the composition of the accreting material. This is because the effect of the radiation depends on the opacity of the material - stuff that is more metal-rich is more opaque. The upper limit you refer to is a bit too high for compositions similar to the Sun (I think more like $200 M_\odot$ would be the limit for that). For stars born in the distant past that were metal-poor, or even born from primordial material with no metals, the upper limit could be much higher.
For primordial material, with no metals at all, the physics that determines the upper limit is likely to be fragmentation. Primordial gas is quite transparent to radiation, but this also means it is poor at radiating away the heat from gravitational contraction. In principle, much larger bodies can only begin to contract towards being very massive stars if some of the gas is ionised or if there is hydrogen in molecular form. However, this cooling also reduces the Jeans mass, allowing the collapsing cloud to fragment into a cluster of smaller objects before a supermassive $(>10^6 M_\odot)$ star/black hole could form.
Nobody knows for sure how supermassive black holes formed. It seems likely that they grew from seed intermediate black holes that formed from very massive primordial stars - perhaps thousands of solar masses - or even by direct collapse to black holes of mass $10^4-10^6 M_\odot$ without ever forming a stable star.


Summary

There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now.

Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no guarantees it's completely accurate, but the numbers seem to pass the sanity check so I think we're good.

Caveat the First: The numbers for stars other than the Sun are based on data with a great deal of uncertainty, such as the number of stars in the universe and the average size of a star. The numbers above could easily be off by a factor of 10 in either direction, and are merely intended to give a rough idea of how empty space is.

Caveat the Second: The numbers for the Sun and the Milky Way are based on the assumption that you are standing (or floating) at a random point on Earth. Anyone outside the tropics will never have the Sun over their head. People in the northern hemisphere are more likely to have Milky Way stars over their head, with the best odds being people near 36.8° N, because at that latitude straight up passes through the galactic center once a day.26

Note: You can mostly ignore everything in this answer and just look up the solid angle of the Sun to get the same result. All the other stars are really far away and very spread out. The difference in solid angle subtended is five thousandths of a percent more when we add the rest of the universe to the Sun.

Background

Let's try to get a somewhat realistic, hard number. To do that, we'll need some assumptions.

As pointed out in Michael Walsby's answer1, if the universe is infinite (and homogeneous2), there is only an infinitesimal chance of there not being a star overhead, which normal math treats as exactly zero chance. So let's presume the universe is finite.

Presumptions

Specifically, let's presume the universe only consists of the observable universe. (Look up the expansion of the universe3 for further information.)
Further, let's presume the contents of the observable universe are measured at their current (presumed) positions, not the position they appear to be. (If we see light from a star from 400 million years after the universe began, we would measure it as being about 13.5 billion light years away, but we calculate that it's likely closer to 45 billion light years away due to expansion.)
We'll take the number of stars in the observable universe to be $10^{24}$. A 2013 estimate4 was $10^{21}$, a 2014 estimate5 was $10^{23}$, and a 2017 estimate6 was $10^{24}$, with each article expecting the estimate to increase as we get better telescopes over time. So we'll take the highest value and use it.
We'll take the size of the observable universe7 to be $8.8\cdot10^{26} \text{m (diameter)}$, giving a surface area8 of $2.433\cdot10^{54} \text{m}^2$ 9, and a volume10 of $3.568\cdot10^{80} \text{m}^3$ 11.
We'll take the average size of a star to be the size of the Sun, $1.4\cdot10^{9}\text{m (diameter)}$ 12. (I can't find any sources for average star size, just that the Sun is an average star.)

Model

From here, we're going to cheat a bit. Realistically, we should model each galaxy separately. But we're just going to pretend the entire universe is perfectly uniform (this is true enough as we get farther away from Earth in the grand scheme of the cosmos). Further, we're going to start counting far enough out to ignore the Milky Way and Sun entirely, then add them back in later with different calculations.

Given the above presumptions, we can easily calculate the stellar density of the observable universe to be $\delta = \frac{10^{24}\text{stars}}{3.568\cdot10^{80} m^3} = 2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ 13.

Next, we need to calculate the solid angle14 subtended by a star. The solid angle of a sphere is given by $\Omega=2\pi\left(1-\frac{\sqrt{d^2-r^2}}{d}\right)\text{ sr}$ 15, where $\Omega$ is the solid angle in steradians16 (sr), $d$ is the distance to the sphere and $r$ is the radius of the sphere. Using $D$ as the diameter, that converts to $\Omega=2\pi\left(1-\frac{\sqrt{d^2-\left(\frac{D}{2}\right)^2}}{d}\right)\text{ sr}$. Given the average diameter presumed above ($1.4\cdot10^{9}\text{m}$), this gives an average solid angle of $\Omega=2\pi\left(1-\frac{\sqrt{d^2-4.9\cdot10^{17}\text{m}^2}}{d}\right)\text{ sr}$ 17.

At this point, we could set up a proper integral, but my calculus is rather rusty, and not very sharp to begin with. So I'm going to approximate the answer using a series of concentric shells, each having a thickness of $10^{22}\text{m}$ (about a million light years). We'll put our first shell $10^{22}\text{m}$ away, then work our way out from there.

We'll calculate the total solid angle of each shell, then add all the shells together to get the solid angle subtended by the entire observable universe.

The last problem to fix here is that of overlap. Some stars in the farther shells will overlap stars in the nearby shells, causing us to overestimate the total coverage. So we'll calculate the probability of any given star overlapping and modify the result from there.

We'll ignore any overlap within a given shell, modeling as if every star in a shell is at a fixed distance, evenly distributed throughout the shell.

Probability of Overlap

For a given star to overlap closer stars, it needs to be at a position already covered by the closer stars. For our purposes, we'll treat overlaps as binary: either the star is totally overlapped, or not overlapped at all.

The probability will be given by the amount of solid angle already subtended by previous shells divided by the total solid angle in the sky ($4\pi\text{ sr}$).

Let's call the probability a given star, $i$, is overlapped $P_i$, the solid angle subtended by that star $\Omega_i$, and the number of stars $n$. The amount of non-overlapped solid angle subtended by a given shell, $k$, is then $\Omega_{kT}=(1-P_1)\Omega_1+(1-P_2)\Omega_2+\ldots+(1-P_n)\Omega_n\frac{\text{ sr}}{star}$. Since we've said stars in a shell don't overlap each other, $P_i$ is the same for all $i$ in a given shell, allowing us to simplify the above equation to $\Omega_{kT}=(1-P_k)(\Omega_1+\Omega_2+\ldots+\Omega_n)\frac{\text{ sr}}{star}$, where $P_k$ is the probability of overlap for shell $k$. Since we're treating all the stars as having the same, average size, this simplifies even further to $\Omega_{kT}=(1-P_k)\Omega_k n\frac{\text{ sr}}{star}$, where $\Omega_k$ is the solid angle of a star in shell $k$.

Calculating Solid Angle

The number of stars in a shell is given by the volume of the shell times the stellar density of said shell. For far away shells, we can treat the volume of the shell as being its surface area times its thickness. $V_\text{shell}=4\pi d^2 t$, where $d$ is the distance to the shell and $t$ is its thickness. Using $\delta$ as the stellar density, the number of stars is simply $n=\delta V_\text{shell}=\delta 4\pi d^2 t$.

From here, we can use the calculation for solid angle of a shell (from Probability of Overlap, above) to get $\Omega_{kT}=(1-P_k)\Omega_k \delta 4\pi d^2 t\frac{\text{ sr}}{star}$.

Note that $P_k$ is given by the partial sum of solid angle for all previous shells divided by total solid angle. And $\Omega_k$ is given by $\Omega_k=2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right)\frac{\text{ sr}}{star}$ (from Model, above).

This gives us $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\text{ sr}$. Given that each shell is $10^{22}\text{m}$ away, we can substitute $d_k$ with $k 10^{22}\text{m}$. Likewise, $t$ can be substituted with $10^{22}\text{m}$. And we already calculated $\delta=2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ (from Model, above).

This gives us
$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k 10^{22}\text{m})^2-4.9\cdot10^{17}\text{m}^2}}{k 10^{22}\text{m}}\right) 2.803\cdot10^{-57}\frac{\text{stars}}{\text{m}^3} 4\pi (k 10^{22}\text{m})^2 10^{22}\text{m}\frac{\text{ sr}}{star}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right) 2.803\cdot10^{-57} 8\pi^2 k^2 10^{66}\text{ sr}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\,2.213\cdot 10^{11}\,k^2\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right)\text{ sr}$

From here, we can just plug the numbers into a calculation program.

$\Omega_{T}=\sum_{k=1}^{k_\text{max}} \Omega_{kT}$

Where $k_\text{max}$ is just the radius of the observable universe divided by the thickness of a given shell. Thus $k_\text{max}$$=\frac{4.4\cdot 10^{26} \text{m}}{10^{22} \text{m}}$$=4.4\cdot 10^4$$=44000$

$\Omega_{T}=\sum_{k=1}^{44000} \Omega_{kT}$

Results

Because of the large numbers involved, it's difficult to just run this in a program. I resorted to writing a custom C++ program using the ttmath library18 for large numbers. The result was $2.386\cdot 10^{-11}\text{ sr}$, or $1.898\cdot 10^{-12}$ of the entire sky. Conversely, there's about a 1 in 500 billion chance you're standing under a star right now.

Note that we ignored the Milky Way and the Sun for this.

The C++ program can be found at PasteBin25. You'll have to get ttmath working properly. I added some instructions to the top of the C++ code to get you started if you care to make it work. It's not elegant or anything, just enough to function.

The Sun

WolframAlpha helpfully informed me the Sun has a solid angle of about $6.8\cdot 10^{-5}\text{ sr}$, or about 2.8 million times more than all the stars in the universe combined. The solid angle formula above gives the same answer18 if we provide the Sun's 150 gigameter distance and 0.7 gigameter radius.

The Milky Way

We could get an approximation for the Milky Way by taking its size and density and doing the same calculations as above, except on a smaller scale. However, the galaxy is very flat, so the odds greatly depend on whether you happen to be standing in the galactic plane or not. Also, we're off to one side, so there are far more stars towards the galactic center than away.

If we approximate the galaxy as a cylinder with a radius of $5\cdot 10^{20}\text{ m}$ (about 52000 light years) and a height of $2\cdot 10^{16}\text{ m}$ (about 2 light years), we get a volume of $1.571\cdot 10^{58}\text{ m}^3$ 20.

Current estimates of the galaxy's radius are closer to 100000 light years21 22, but I'm presuming the vast majority of stars are a lot closer than that.

There are estimated to be 100 to 400 billion stars in the Milky Way21. Let's pick 200 billion for our purposes. This puts the density of the Milky Way at $\delta = \frac{200\cdot10^{9}\text{stars}}{1.571\cdot 10^{58}\text{ m}^3} = 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3}$ 22, or about 4.5 billion times denser than the universe at large.

This time, we'll take shells $10^{17}\text{ m}$ thick (about 10 light years) and go out from there. But we need to re-organize the math into a spherical form, so we'll presume the galaxy has the same volume, but is a sphere. This gives it a radius of $1.554\cdot 10^{19}\text{ m}$ 24, or 155.4 shells. We'll round to 155 shells.

$\Omega_{T}=\sum_{k=1}^{155} \Omega_{kT}$

Using our formula from above (Calculating Solid Angle), we can start substituting numbers.

$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k\cdot 10^{17}\text{ m})^2-4.9\cdot10^{17}\text{ m}^2}}{k\cdot 10^{17}\text{ m}}\right) 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3} 4\pi (k\cdot 10^{17}\text{ m})^2 10^{17}\text{ m}\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2\cdot 10^{34}\text{ m}^2-4.9\cdot10^{17}\text{ m}^2}}{k 10^{17}\text{ m}}\right) 1.273\cdot 10^{-47} \frac{\text{stars}}{\text{m}^3} 8\pi^2 k^2 10^{51}\text{ m}^3\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\cdot\,1.005\cdot 10^6\,k^2\,\left(1-\frac{\sqrt{k^2\cdot 10^{34}-4.9\cdot10^{17}}}{k 10^{17}}\right)\text{ sr}$

Plugging this into the program gives $3.816\cdot 10^{-9}\text{ sr}$, which is $3.037\cdot 10^{-10}$ of the total sky. The odds you're standing under a star in the Milky Way are about 1 in 3.3 billion.

Solid Angle Totals

Solid angle is:

Sun, $6.8\cdot 10^{-5}\text{ sr}$
Milky Way, $3.816\cdot 10^{-9}\text{ sr}$
Universe, $2.386\cdot 10^{-11}\text{ sr}$
Total, $6.800384\cdot 10^{-5}\text{ sr}$ (the extra digits are basically meaningless, adding about five thousandths of a percent to the Sun's solid angle)
Milky Way plus Universe, $3.840\cdot 10^{-9}\text{ sr}$ (about 0.6% more than just the Milky Way)

References

1 Michael Walsby's answer to this question, is there a star over my head?. https://astronomy.stackexchange.com/a/33294/10678
2 A Wikipedia article, Cosmological principle. https://en.wikipedia.org/wiki/Cosmological_principle
3 A Wikipedia article, Expansion of the universe. https://en.wikipedia.org/wiki/Expansion_of_the_universe
4 A UCSB ScienceLine quest, About how many stars are in space?, from 2013. https://scienceline.ucsb.edu/getkey.php?key=3775
5 A Sky and Telescope article, How Many Stars are There in the Universe?, from 2014. https://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/
6 A Space.com article, How Many Stars Are In The Universe?, from 2017. https://www.space.com/26078-how-many-stars-are-there.html
7 A Wikipedia article, Observable universe. https://en.wikipedia.org/wiki/Observable_universe
8 A Wikipedia article, Sphere, section Enclosed volume. https://en.wikipedia.org/wiki/Sphere#Enclosed_volume
9 A WolframAlpha calculation, surface area of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=surface+area+of+a+sphere%2C+diameter+8.8*10%5E26+m
10 A Wikipedia article, Sphere, section Surface area. https://en.wikipedia.org/wiki/Sphere#Surface_area
11 A WolframAlpha calculation, volume of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=volume+of+a+sphere%2C+diameter+8.8*10%5E26+m
12 A nineplanets.org article, The Sun. https://nineplanets.org/sol.html
13 A WolframAlpha calculation, (10^24 stars) / (3.568⋅10^80 m^3). https://www.wolframalpha.com/input/?i=%2810%5E24+stars%29+%2F+%283.568%E2%8B%8510%5E80+m%5E3%29
14 A Wikipedia article, Solid angle. https://en.wikipedia.org/wiki/Solid_angle
15 Harish Chandra Rajpoot's answer to a geometry.se question, Calculating Solid angle for a sphere in space. https://math.stackexchange.com/a/1264753/265963
16 A Wikipedia article, Steradian. https://en.wikipedia.org/wiki/Steradian
17  A WolframAlpha calculation, 2*pi*(1-sqrt(d^2-(1.4*10^9 m/2)^2)/d). https://www.wolframalpha.com/input/?i=2*pi*%281-sqrt%28d%5E2-%281.4*10%5E9+m%2F2%29%5E2%29%2Fd%29
18 Website for ttmath. https://www.ttmath.org/
19 A WolframAlpha calculation, 2*pi*(1 - sqrt(d^2 - r^2)/d), where d = 150 billion, r=0.7 billion. https://www.wolframalpha.com/input/?i=2*pi*%281+-+sqrt%28d%5E2+-+r%5E2%29%2Fd%29%2C+where+d+%3D+150+billion%2C+r%3D0.7+billion
20 A WolframAlpha calculation, pi * (5*10^20 m)^2 * (2*10^16 m). https://www.wolframalpha.com/input/?i=pi+*+%285*10%5E20+m%29%5E2+*+%282*10%5E16+m%29
21 A Wikipedia article, Milky Way. https://en.wikipedia.org/wiki/Milky_Way
22 A Space.com article from 2018, It Would Take 200,000 Years at Light Speed to Cross the Milky Way. https://www.space.com/41047-milky-way-galaxy-size-bigger-than-thought.html
23 A WolframAlpha calculation, (200*10^9 stars) / (1.571*10^58 m^3). https://www.wolframalpha.com/input/?i=(200*10^9+stars)+%2F+(1.571*10^58+m^3)
24 A WolframAlpha calculation, solve for r:  (4/3)*pi*r^3 = 1.571*10^58 m^3. https://www.wolframalpha.com/input/?i=solve+for+r%3A++%284%2F3%29*pi*r%5E3+%3D+1.571*10%5E58+m%5E3
25 My C++ program code on PasteBin. https://pastebin.com/XZTzeRpG
26 A Physics Forums post, Orientation of the Earth, Sun and Solar System in the Milky Way. Specifically, Figure 1, showing angles of 60.2° for the Sun, and 23.4° less than that for Earth. https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/


Arago Spot in the Shadows of Celestial Bodies

The Arago spot phenomenon requires the light source to be point-like and very bright. The sun is not a point-like light source, as it has an apparent diameter of about 0.5 degrees

The only body large enough to cast a full solar shadow is the moon.  Arago's spot has not been observed during a total solar eclipse.


How to adapt small Dobsonian to use with a tripod?

First, a tripod won't help you point to a particular location in the sky.  For that you would need a correctly-aligned equatorial mount with setting circles, or a mount with digital encoders (you can get these for dobsonians) that when correctly configured can convert the telescopes position to a position on the celestial sphere.

It is possible to convert a dobsonian to a different kind of mounting.  If your telescope doesn't already have a tripod bracket, you would probably need to buy tube rings such as these, and put one on either side of the circles on the sides of the tube that sit in the rocker box, so that the center of gravity of the scope was between the rings.  Then you'd probably need a plate connecting the rings and providing a mount point between them that you could attach to a tripod or a more complex mount (like an equatorial).

My suspicion is that rather than trying to adapt the dobsonian to a different mount, you'd be better served either by buying after-market digital encoders for your dobsonian, or just getting a whole new telescope with the kind of mounting you want.


Does the twin paradox work in an almost empty universe?

Velocity is relative, but acceleration is not: you can feel it (and it feels just like gravity). But as I said in the question comments, acceleration is a red herring.
The real issue is that the travelling twin occupies two (or more) distinct reference frames. That means their worldline (their trajectory through spacetime) is not a straight line, and any inertial observer measuring that worldline will agree on that, although different observers will disagree on the lengths of the segments of that trajectory, depending on their speed relative to the traveller.
So the important factor is that the worldline has a bend in it, and so it cannot be transformed into an unbent line. It doesn't make much difference whether it's a perfectly sharp bend (your Dirac delta function), or a smooth gradual curve.

There's a variant of the twin paradox that requires no acceleration, but it has 3 participants, A, B, and C. A is based on Earth, C is based on Alpha Centauri (which we can assume is at rest relative to Earth).
B's rocket takes off. It's moving at constant speed $v$ (relative to Earth) as B passes A, and at that moment  they use radio to synchronise their clocks to zero. B then travels towards Alpha Centauri.
One year later (according to B's shipboard clock), C sees B approaching. C starts their rocket and heads towards Earth. C's rocket has a speed of $v$ relative to Alpha Centauri. (According to the relativistic law of composition of velocities, B & C have velocity relative to each other of $2v/(1+v^2)$, using natural units, where the speed of light $c=1$). Just as C passes B, C synchronises their clock to B's clock. That is, C "clones" the value of "1 year" from B's clock the instant they pass one another.
One year later, according to that clock, C arrives at Earth, and as C passes A, they compare clocks. C's clock says "2 years". A's clock says "9 years".


Yes, Pluto is still a dwarf planet. According to the IAU website, it still fits the criteria for a dwarf planet, fails to meet the criteria for a planet, and still carries the "dwarf planet" label, whatever its future status may be. I'm sorry I can't provide a longer or more detailed answer, but this is really a yes-or-no question.


How to estimate uncertainty of measurements of equivalent widths?

Yes there is a problem. You seem to be trying to derive an uncertainty in the measurement of EW by doing repeated measurements of the same data?

This can only give you the uncertainty associated with your measurement technique (i.e. where you define the limits of the line and how you set the continuum level) - the systematic error you might call it (although there can be other systematic errors inherent to EW measurements, like whether you subtracted the sky or scattered light in your spectrograph correctly for example).

What it does not do is evaluate the uncertainty in the EW caused by the quality or signal-to-noise ratio of the data itself. You might assess this using some rule-of-thumb formulae for a Gaussian line, e.g.
$$\Delta EW \simeq 1.5 \frac{\sqrt{fp}}{{\rm SNR}},$$
(eqn 6 of Cayrel de Strobel 1988) where $f$ is the FWHM of the spectral line (in wavelength units), $p$ is the size of one pixel in wavelength units and SNR is the signal-to-noise ratio of the data in an average pixel. Or you could take a synthetic spectrum and add some artificial noise to it with the appropriate properties and measure the EW of several randomisations of the same spectrum, taking the standard deviation of your EW measurements to indicate the EW uncertainty for a particular level of signal-to-noise ratio.

If this statistical uncertainty is not negligible, then you would then need to add it to any systematic uncertainties associated with your analysis of the spectrum. As far as the latter is concerned then your suggested method does give some indication of what that error might be, though I suspect it will overestimate the 1-sigma uncertainty.


Comments converted to community wiki

Ceres' density (2.08 g/cc) suggests it has abundant water/ice below its presumably mostly dry surface. That and the shiny spot (salts left after subsurface water has sublimated) suggests it wouldn't too difficult. Low gravity, keep the dig in shade to prevent the ice from sublimating in direct sunlight.

We could not do it now, but with improved technology and continued space exploration, it should be very possible. Ceres should be a comparatively easy place to build a space elevator too. It seems a pretty good mining option. Only potential issue might be the occasional asteroid impact.

Ask the asteroid mining companies "Planetary Resources" and "Deep Space Industries". Actual and serious companies. See the websites www.planetaryresources.com and www.deepspaceindustries.com


How can 16 Psyche be the core of a protoplanet?

If this hypothesis is correct, the body from which 16 Psyche formed was a 500km planetesimal (about half the diameter of Ceres) It would have been destroyed in a collision with another body. The other body would have had to have been large enough (300km) to complete fragment the asteroid. The problem with this hypothesis is that no other pieces of this planetesimal have been found (they should still have similar orbits as Psyche).

Another hypothesis is that Psyche suffered from a series of smaller impact, that fractured and disrupted the parent body, but did not destroy it. This would give Psyche a composition similar to the stony-iron mesosiderite meteorites.

The Psyche mission is intended to answer these questions.


Does the cosmic microwave background change over time?

The CMB patterns do indeed change over time, although statistically they remain the same, and although it will not be noticeable on human timescales.

The CMB we observe now comes from a thin shell with us in the center, and with a radius equal to the distance that the light has traveled from the Universe was 379,000 years old and until now. As time passes, we will receive CMB from a shell with an increasingly larger radius. As that light has traveled farther through space, it will, as you say, be more redshifted, or "cooler". But it will also have been emitted from more distant regions in the early Universe that, although statistically equivalent, simply will be other regions and hence look different.

The patterns that change the fastest are the smallest patterns we can observe. The angular resolution of the Planck satellite is 5-10 arcmin. Since the CMB comes from a redshift of ~1100, the angular diameter distance defining the physical distance spanned by a given angle — is ~13 Mpc, so 5 arcmin corresponds to a physical scale of roughly 19 kpc in physical coordinates, or 21 Mpc in comoving coordinates (that is, a structure spanning 5 arcmin today were ~19 kpc across at the time of emission, but have now expanded to a size of ~21 Mpc, with 1 Mpc = 1000 kpc = 3261 · 10³ lightyears).

Assuming an isotropic Universe, if the smallest observable parcels of gas were 19 kpc across perpendicular to our line of sight, they are also on average 19 kpc across along our line of sight.

So the question of how fast the CMB changes comes down to how much time did it take light to travel 19 kpc when the Universe was 379,000 years old. This is not simply 19 kpc divided by the speed of light, since the Universe expands as the light travels, but it's pretty close. Hence, it took light roughly 62,000 years to traverse such a patch.

Since we see events at redshift $z$ time dilated by a factor $1+z$, we will have to wait $62\,\mathrm{kyr}\times1100$, or roughly 70 million years (assuming that Planck will not get replaced by better instruments within that time which is, um, dubious).

So you're right, you could make a 3D image of the CMB, but since the patterns are much larger than a light-year, you don't have to take a new picture every year.


What explains the existence of energy/matter if it cannot be created or destroyed?

That's a very complicated question!

First, let's remember that Moses didn't bring the Law of Conservation of Energy down from Sinai on stone tablets -- it's something that we've observed to be true of the universe today, and which our theories use to make very accurate predictions. But this doesn't mean that it was always true, and particularly that it was true early enough in the Big Bang that our current theories don't apply.  We know that our current theories fail during the Planck Era, for example.

Second, Noether's Theorem says that energy conservation exists because the Lagrangians of mass-energy are time-symmetric. Time symmetry is a very good symmetry today in regionals of space small compared with the entire universe, but there may well have been eras in which it wasn't.

In fact, there probably were such eras, just as time-symmetry is not a good symmetry even today at very large scales.  (The universe is expanding, and on the face of it, this causes a non-conservation of energy.)

Thirdly, if you try to save the law of conservation of energy by calculating the total energy of the entire universe including the gravitational energy, one reasonable set of arguments (energy calculations in General Relativity can be very difficult, since gravitational energy tends to be non-local) yields a total energy of zero! The mass-energy in the universe is exactly balanced by the gravitational potential energy.  If this is correct, the total energy has always been zero.

But this is all speculative!  Remember, that real physics is an experimental science and if you can't test a theory it may be interesting, but it isn't physics.


What is the application of pseudorandomness in astrophysics?

The primary advantage of pseudorandom numbers is speed (as Cristiano mentioned) and convenience: they're what are available in almost any software package or library, whereas true random number generators are not. As long as you use a pseudorandom number generator (PRNG) with good statistical qualities and are not worrying about true cryptographic randomness, this is perfectly acceptable.
In addition to Monte Carlo simulations, pseudorandom number are used for many different applications, including all sorts of physical simulations (cosmological, stellar evolution, planet formation, etc.), statistical analyses (bootstrap resampling), initial conditions for fits, updating parameters in genetic algorithms, sample selection, etc.
A personal example: I've worked on two different studies where we were making classifications of different aspects of galaxies, based on their individual isophote plots or radial surface-brightness profiles. Rather than worrying about the possibility of subtle biases based on possible knowledge of individual galaxies ("Oh, it's NGC 1097, I'd expect that galaxy to be..." or "Well, the list is alphabetically ordered, so the middle galaxies will have NGC prefixes, and those will mostly be really luminous galaxies, so..."), we set up a blind analysis system, where the galaxies were presented to us in a random order, without names (with the actual identifications automatically recorded where we couldn't see them until the whole thing was over). The randomness of the order was, of course, really pseudorandom order.
This is how things like Galaxy Zoo work, for example: the individual galaxies that citizen scientists classify are (pseudo)randomly selected. (And for one of the projects I mentioned above [Kruk et al. 2019], we set up the classification process using the "Zooniverse" framework that Galaxy Zoo developed).
A separate advantage is that if you save the number used as the seed for the pseudorandom number generator, you can use that to regenerate the exact set of pseudorandom numbers used in the analysis/simulation/whatever, which can be very useful for debugging purposes, and also for allowing others to replicate your results. (E.g., you run some kind of relatively short/simple simulation, and derive some results and figures from that. How can someone else trust that your results and figures accurately represent the simulation? Provide them with the code and the PRNG seed, and then can generate the same simulated/random-sampling results and check. Plus, they can vary the seed and make sure you didn't cheat -- or fool yourself -- by reporting atypical results. Here's a Github site for a recent paper of mine that includes both the code for a survey simulation and the seed number I used for the figure in the paper.)


Is dust the only reason why the Martian sky is so bright?

It is very likely due to dust particles. The color of Martian sky is described to be yellow-brown "butterscotch" color during daytime (sometimes pinkish red) and blue during rising and setting of sun (very different from what happens on Earth). The typical color is because as the dust particles in the air are large relative to the wavelength of visible light, it absorb blue light and effectively act as mirrors by scattering the remaining wavelengths. Also, the presence of 1% by volume of magnetite in the dust particles intensify the color. Twilight lasts a long time after the Sun has set and before it rises due to that.
At times, the Martian sky takes on a violet color due to scattering of light by very small water ice particles in clouds.
For more information, read this excellent paper: The color of the Martian sky and its influence on the illumination of the Martian surface
Additional sources:

http://www.webexhibits.org/causesofcolor/14C.html
http://www.mkrgeo-blog.com/what-is-the-colour-of-martian-sky/

For more on Mars Twilight see:

At Mars' equator how long does twilight last?
How bright is the Martian landscape during the long (up to two hour) twilight?
How much more quickly does night come on Mars compared to Earth? Will it be easier to see Mercury there?
How bright is twilight on Mars compared to Earth?
Mars twilight photos


What are the sources of antimatter particles in the magnetosphere?

Matter and antimatter particles are always produced as a pair. This particle physics' process is called pair production. If antimatter is detectable, it means that the
antiparticle of a matter-antimatter-particle-pair was able to survive long enough to be detected, for instance due to relativistic effects, as explained e.g. by the Hyperphysics page on muons:

The measurement of the flux of muons at the Earth's surface produced an early dilemma because many more are detected than would be expected, based on their short half-life of 1.56 microseconds. This is a good example of the application of relativistic time dilation to explain the increased particle range for high-speed particles.

You will now correctly say that a muon is not antimatter, but I just used the muon as example that some particles (or antiparticles) may be detected even if they have a short half-life.
So, what are the sources of antimatter in the magnetosphere? Strictly speaking, the source (in the sense of "location") of pair creation will probably be in the magnetosphere itself, as said by atmosphericprisonescape:

Anywhere where highly energetic reactions take place whose center-of-mass energy surpasses the rest-mass energy of 2*times some particle, this particle and its anti-particle counterpart can be created.

I suppose the real question is: How is their enough energy for frequent pair production in the magnetosphere?


Do other planets in our solar system experience eclipses or is this unique to Earth?

Eclipses are common, but solar eclipses like the ones our moon causes would be unusual since it takes a combination of factors, the angular size of the moon and sun are nearly the same from the viewpoint of Earth so that the moon (depending on where it is in its slightly ellipitcal path around the Earth) can cover the sun nearly exactly.

NASA has a nice picture of an eclipse as seen from the Mars Curiosity rover showing an eclipse by a moon with an angular size less than that of the Sun.

Phobos eclipsing the Sun


Why nuclear fusion is a controlled process in stars?

Because in most stars, the pressure where nuclear burning takes place is proportional to temperature. In order for the nuclear reaction rate to increase, the temperature and hence pressure must increase. This would cause the gas to push out the layers above it, then to expand and cool.

Conversely, if the nuclear reaction rate fell, so would the temperature and pressure. The smaller pressure would allow the outer layers to move inwards, compressing the gas, heating it up and speeding up the nuclear reactions.

In this way the nuclear reaction rate (for hydrogen burning) can be kept almost constant. The small changes (increase) that occur during a star's main sequence life are driven by the increasing mean atomic mass as hydrogen turns to helium. This means you need a gradually higher temperature to maintain the same pressure.

Where the temperature and pressure are decoupled - for instance in the degenerate helium core of a low(ish) mass star near the red giant tip, where electron degeneracy pressure is independent of temperature, then runaway nuclear reactions can be initiated - the "helium flash".


How large must an object be to be seen through a telescope?

A few clarifications.

Telescopes in general operate at very large distances - "at infinity" is the term used in optics parlance.

A bright enough object can be seen from any distance, no matter what its size is. All that matters is that:

It's bright enough to produce an impression on whatever sensor
you're using (or your eye)
The background is dark enough to produce sufficient contrast

But then it would be just a bright but tiny spot.

I believe what you're really asking for is: what is the combination of factors that shows the object as bigger than a simple dot? In that case, it's two factors: aperture of the telescope, and angular size of the object.

Assuming a flawless telescope, its aperture is what determines its resolving power. The resolving power is the angle at which two dots can be separated by the telescope. The formula is:

resolving power = 1 / (10 * aperture)

where resolving power is in arcseconds, and aperture is in meters. Examples:

aperture         resolving power
10 cm            1 arcsec
20 cm            0.5 arcsec
1 m              0.1 arcsec

As long as the object's angular size is bigger than the resolving power, it will appear bigger than a dot.

That's all. In astronomy, we don't speak of an object's absolute size, we only speak of the angular size. But that should be enough. As soon as you have the angular size, and say the distance, then you could deduce the absolute size, it's a simple matter of trigonometry.

absolute size = distance * tangent(angular size)

E.g., this is the size of an object of 1 arcsec angular size, situated at 384,000 km (the orbit of the Moon):

http://www.wolframalpha.com/input/?i=384000+km+*+tangent%281+arcsec%29

It's 1.8 km (in case the link above is unavailable).

In other words, that's the minimum absolute distance resolved by a telescope 10 cm in aperture, for objects on the Moon. Any two dots closer together than 1.8 km, placed on the Moon, are seen as one dot in a 10 cm telescope.