Partial answer about the retrograde motion only:
Yes, the Sun has yearly retrograde motion as seen from Mercury
I've written a simple script to generate the orbital motion of Mercury based on its eccentricity. I've used reduced units with $GM = a = 1$ and period $T = 2 \pi$.
Mercury's rotational period is 2/3 of its orbital period.
I've plotted the rotational angle of mercury minus the angle of the Sun as seen from Mercury's position. To address retrograde motion, it's not necessary to consider from where on Mercury you are looking,
There are tiny dips below zero every year.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc = - x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

halfpi, pi, twopi, fourpi, sixpi = [f*np.pi for f in (0.5, 1, 2, 4, 6)]
to_degs, to_rads = 180/pi, pi/180

# Mercury
# https://space.stackexchange.com/questions/43213/what-is-the-analytical-closed-form-solution-of-the-two-body-problem-to-verify-it/43237#43237
# https://solarsystem.nasa.gov/planets/mercury/by-the-numbers/

e = 0.20563593
a = 1
peri = (1 - e) * a
v_peri = np.sqrt(2/peri - 1/a)

X0 = np.array([peri, 0, 0, v_peri])
times = np.linspace(0, sixpi, 1001)
dt_years = (times[1] - times[0]) / twopi

answer, info = ODEint(deriv, X0, times, full_output=True)
xy = answer.T[:2]
x, y = xy
r = np.sqrt((xy**2).sum(axis=0))

theta_rot = np.mod((3/2) * times, twopi)
sun = np.zeros(2)[:, None] - xy
theta_sun = np.arctan2(sun[1], sun[0])

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    angle = theta_rot - theta_sun
    dangle = np.mod(angle[1:] - angle[:-1]+halfpi, twopi)-halfpi
    dangle_dt = dangle / dt_years
    angle = np.mod(angle+halfpi, twopi)-halfpi
    plt.plot(times/twopi, angle)
    plt.ylabel('angle (rad)')
    plt.subplot(2, 1, 2)
    plt.plot(times[1:]/twopi, dangle_dt)
    plt.plot(times[1:]/twopi, np.zeros_like(dangle_dt), '-k', lw=0.5)
    plt.ylabel('dangle/dt (rad/year)')
    plt.xlabel('years')
    plt.show()

if True:
    plt.figure()
    plt.subplot(3, 1, 1)
    plt.plot(x, y)
    plt.plot(x[:1], y[:1], 'ok', ms=2)
    plt.plot([0], [0], 'oy', ms=8)
    plt.gca().set_aspect('equal')
    plt.subplot(3, 1, 2)
    plt.plot(times/twopi, x)
    plt.plot(times/twopi, y)
    plt.plot(times/twopi, r, '-k')
    plt.subplot(3, 1, 3)
    plt.plot(times/twopi, theta_rot)
    plt.plot(times/twopi, theta_sun)
    plt.show()


Can you still see Polaris even if you are in the south pole?

Currently Polaris is at a declination of a bit over 89 degrees, which means that no one south of 1 degree south latitude can see Polaris. That's almost all of the Southern hemisphere, let alone the South Pole.

Polaris won't be the North Star forever, thanks to axial precession. In about 13000 years or so, Polaris will have a declination of about 46 degrees or so (twice the 23 degree axial tilt). Polaris will thus be visible in 13000 years or so as a wintertime star to all of Africa, all of Australia, and most of South America, but none of Antarctica.

After millions of years, proper motion may make Polaris visible over Antarctica. But then again, being a yellow supergiant, its unlikely that Polaris will be visible anywhere (without a telescope). It will instead be dead.


How would one estimate the rotation period of a star from its spectrum?

It takes a spot "bump" half the rotation period to traverse from the blue side of the line profile to the red.
You can see that the bump has maybe moved across 20% of the profile in one night, so the rotation period would be estimated as $\sim 10$ days.
Looking up the period of Sigma Gem - it is 19.6 days. So my method is not very accurate by eye, but you would need to simulate the star plus spot spectrum more accurately to take account of limb darkening and the spectral resolution of the spectrograph.


The probability of observing a planet transit is approximately $(R_p + R_s)/a$, where $R_p$ and $R_s$ are the planet and stellar radius respectively and $a$ is the semi-major axis of the planet's orbit. This assumes that planetary orbits are circular and randomly oriented with respect to our line of sight to the star (for which there is little or no counter-evidence).

In the case of Trappist-1, it has an estimated radius of $R_p=0.11R_{\odot}$, the planetary companions have radii and semi-major axes of (in units of Earth radii and au respectively)  (1.09,0.011), (1.06, 0.015), (0.77,0.021), (0.92,0.028), (1.05, 0.037), (1.13, 0.045), (0.75, 0.059).

Using approximation above, then the transit probabilities are 0.051, 0.037, 0.026, 0.020, 0.015, 0.012, 0.009 respectively.

Now these probabilities cannot be treated as independent of one another, since it is quite likely (though this is still the subject of research) that planets will naturally tend to inhabit the same orbital plane (the Trappist-1 planetary orbit inclinations are within 0.2 degrees of each other - flatter than the solar system). Thus you are much more likely to see second and subsequent transits in a system with one transiting planet than you are in a star picked at random. On the other hand, a system with lots of planets that are not quite in the same orbital plane gives an increased chance of seeing one of them transit.

Either way, you can see from the probabilities above that seeing one and only one transiting system of this type from 20 targets, even if all of them had similar planetary systems,  is not at all unexpected.

Radial velocity studies of such systems are difficult, but currently being attempted. Trappist-1 has an apparent J magnitude of 11.4, which is very challenging for current telescopes and instrumentation to get high resolution spectroscopy.


Average number of exoplanets in a system

Best answer is that we don't know. The techniques that are used for finding exoplanets are incomplete - that is they are capable of finding planets around stars in some circumstances but not in others.

For example, finding stars using the transit detection technique is more sensitive to planets that are (i) big, (ii) close to their parent star and (iii) orbiting their star with an edge-on orientation from our point of view. The doppler techniques are sensitive to (i) massive exoplanets, (ii) those that are close-in to their parent star, (iii) favour detection when the line of sight lies close to the orbital plane.

Each detection method (and each observing strategy and telescope/satellite) has their own particular selection bias, but overall the picture is that the census of exoplanets is more complete for close-in massive planets, and very incomplete for planets that are either small or orbiting with periods much longer than 10 years. Then at very large separations and very long orbital periods (think 100s of years), direct imaging surveys are again capable of spotting massive exoplanets. Microlensing observations are also capable of detecting even small planets at large distances from their star.

The result of all this is that any figure for the observed fraction of stars with exoplanets or the number of exoplanets per star is a lower limit.
The exception is where researchers try to account for the various selection effects and give a figure for the fraction of stars with certain types of planet in certain types of orbit around certain types of star.

At the moment it looks like the modal number of exoplanets per star is at least 1. In other words it seems increasingly likely that most stars have exoplanets of one sort or another.

Some facts and figures. From Kepler transit data, Petigura et al. (2013) say that 22% of Sun-like stars have "Earth-sized" planets orbiting in their habitable zone (my italics), but their plots show that the overall results for planets with orbital periods in the range 5-100 days (where Kepler is most sensitive), are that nearly 60% of stars have planets from Earth- to Jupiter- sized in this range. Cassan et al. (2012) use microlensing survey data to claim that 17% of stars have "Jupiters" at 0.5-10 au from their parent star, and that Neptune-sized or "super-Earth" sized objects orbit a further (approximately) 50% and 60% respectively. The combination of these figures also suggests multiple planets are not rare and indeed the title of their paper is "One or more bound planets per Milky Way star from microlensing observations".

The frequency of multiple planet systems is also hard to assess. All the techniques are capable of detecting certain types of multiple system, but equally all are capable of missing multiple planets even when they are there.
Kepler did detect many multiple transiting systems, so they are not rare. A study by Tremain & Dong (2012) estimated a minimum fraction of 20-30% of stars that have one transiting planet will also have others. This is on top of the 15% of Kepler planetary systems where multiple planets can be directly observed (i.e where the orbital planes are very close together).


Is there an antonym for 'transit', i.e. the transit of a celestial body through the meridian

The transit of a star when it reaches its highest point is called the upper culmination and when it reaches its lowest point is called the lower culmination. I think both are referred to as meridian transits.


Why isn't Earth tidally locked to the sun?

Really, it's just because the tidal locking timescale is so long for Earth:
$$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$
where $a$ is semi-major axis, $m_s$ is the mass of the secondary object, $m_p$ is the mass of the primary, and $R_s$ is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see
$$\frac{a_1}{a_2}\approx380,\quad \frac{m_{s_1}}{m_{s_2}}\approx80,\quad \frac{m_{p_1}}{m_{p_2}}\approx333000,\quad \frac{R_{s_1}}{R_{s_2}}\approx3.67$$
where $_1$ denotes the Sun-Earth system and $_2$ denotes the Earth-Moon system. To make up for these differences (and to have Earth be tidally locked by now), assuming similar Love numbers and dissipation functions for Earth and the Moon, we would need Earth's initial spin to be substantially smaller than the Moon's initial spin by many, many orders of magnitude, and this just wasn't the case.


What is the proper interpretation of a negative parallax?

Did you read this section of the documentation?  It suggests there are ways to deal with it, but I have not examined the paper it refers to.

  • For closely aligned sources (separated by 0.2–0.3 arcsec), which are only occasionally resolved in the Gaia observations, confusion in the observation-to-source matching can lead to spurious parallax values which are either very large or have a negative value very far away from zero in terms of the formal parallax uncertainty quoted in the catalogue. These sources tend to be faint and located in crowded regions and are also associated with unreliable (large) proper motions (Gaia Collaboration et al. 2018b). Guidance on how to clean samples from spurious parallax values is provided in Lindegren et al. (2018).

It also says there's a small bias, but it shouldn't be universally removed.

  • The systematic errors in the parallaxes are estimated to be below the 0.1 mas level (Lindegren et al. 2018) but the following systematics remain. There is an overall parallax zeropoint which, from an examination of QSO parallaxes, is estimated to be around −0.03 mas (in the sense of the Gaia DR2 parallaxes being too small). The estimated parallax zeropoint depends on the sample of sources examined (Arenou et al. 2018) and the value above should not be used to ’correct’ the catalogue parallax values....


Is the Big Bang inflation caused by Dark Energy

The expansion caused by a cosmological constant (a particularly simple version of dark energy) is not exactly Hooke's law, it is an expansion that maintains a fixed Hubble constant. So this means $\mathrm{d}a/\mathrm{d}t = Ha$, for scale parameter $a$ and constant $H$. The solution is exponentially growing in time.  Hooke's law (without the minus sign) would have the acceleration $\mathrm{d}^2 a/\mathrm{d}t^2$ being proportional to $a$, which permits more complicated solutions, but you're right it certainly has similarities  if one picks out the exponentially growing solution.

What's more, you are right that some treatments of inflation look like that of a cosmological constant, in that they are also exponential in time, but with a much much larger H constant (like 50 orders of magnitude larger).  So if it's the same thing, then for some reason the cosmological constant must have dropped vastly in value.  It's thought to be for a rather different reason, though this is all speculation.  The common speculation is that the "false vacuum" went through a phase change during inflation, leading to the vacuum we have now, and the vacuum we have now has a very different cosmological constant.  Maybe you can think of it as a phase change in the cosmological constant itself, I don't know.  So there could be some connection, but they can't be just the same.


Why is the outline of a crescent moon brighter?

In the image you are now refering to the only significant brightness difference is between the mare and the "highlands".

The retina enhances edges. So you'll see edges enhanced, even if they aren't.

By image processing edges can also be enhanced to make contours clearer. This is not the case for the referred image.

To better assess, which part of the perception is real, and which is illusion, here the illusion part:


The laws of physics through out the universe

Let's start in the middle:

  What is the furthest radius we can prove from earth, with absolute certaintity, that the laws of physics do not vary?

Zero. Proofs are found in mathematics and court rooms, and are impossible in natural science. The best we can do is have falsifiable theories. This holds for every description of reality - there's no "proof" even for the Laws of Gravity.

So, what could we observe that would tell us that physical constants or relationships between physical quantities are different in other parts of the universe, or at other times during its existence?

Gravity: For galaxy clusters, we have independent mass measurements from several different sources that agree within their (admittedly large) error bars. Gravitational lensing, velocity dispersion of the member galaxies and X-ray temperatures are all in agreement. So the laws of gravity seem to work even at redshifts up to 0.5 or even higher.
Atomic physics: We observe highly redshifted objects. The wavelength of the light emitted by these objects is made longer by the expansion of the universe. Observing redshifted spectral lines of different chemical elements (or molecules) tells us that atomic physics worked the same when and where this light was emitted. If the transition levels between electron orbits had changed over time, we would get different redshifts for the same objects depending on what element's spectral line we observe.
Nucleosythesis: Shortly after the big bang, the temperature lowered such that protons and neutrons were no longer created and destroyed constantly. A free neutron has a half live of about 8.5 minutes before it decays into a proton and an electron. Our theories predict that we'd get a helium (2x proton, 2x neutron) content in the universe of about 25%. (The rest of the "normal" matter being essentially all hydrogen), and that is indeed what we observe. Now, the helium content is dependent both on the matter density at the time this took place and the half live of the neutron. From other observations (BAO come to mind) we are fairly certain that we got the matter density about right. Which leaves only a small wiggle room for the half live of the neutron, and hence for changes in the weak force.
We've covered gravity, electromagnetism, and the weak force. I don't know any good test for the strong force.

For a change of natural laws over time, we can look at the isotope distribution in rocks here on earth. We should be able to tell whether the decay rate of various elements was different at earlier times by looking at how many of each of their decay products are around.

To summarize, we cannot say with "absolute certainty", but what we observe seems to indicate that natural laws are the same throughout the universe.


How do I get the change in angular diameter value?

You can see from the paper you linked that they followed the procedure you outlined in option (2). Figure 4 of that paper shows the $\theta (t)$ for a few stars. They specifically state towards the end of section 6

  Figure 4 shows in the top panel the relation between the angular diameter predicted by the SB-relation versus radial variation, and in the bottom panel the angular diameter as a function of phase.

The SB-relation they're talking about is the one you cite in your question. So your process should be to take your list of $V-K$ colors over time and convert them to angular size using the surface brightness relation, then fit a function to that (via cubic spline, least squares, etc.) and determine the total angular variation over time.

I will note also that the $\theta$ vs $V-K$ relation you've cited here is equation 10 in the linked paper which is not a result of that paper. Their result is listed in equation 9 with equations 10 and 11 being results from other, previous work for the reader to compare. You should sure of which equation you want to use and cite the appropriate source.


A lot of the water ice on mars is covered with dust.  NASA's Phoenix mars lander only had to scrape off some of it to find the layer of ice underneath  as seen here: https://www.nasa.gov/mission_pages/phoenix/images/press/sol_020_024_change_dodo_v3.html


"What" is straightforward: L4 and L5 are two points on an orbit that are 60 degrees ahead and 60 degrees behind the planet, and so move as the planet orbits.  If the planet is less than about 1/25 of the mass of the central object, then a combination of centrifugal and Coriolis forces will cause the L4 and L5 points to be places where a third body can have a stable orbit.
Asteroids that have orbits at (or near) the L4 and L5 points are known as "Trojans" Jupiter has hundreds of Trojans that share its orbit, even Earth has one 2010 TK7 (and perhaps more).
"Why" is harder.  It seems that this is chosen to ensure that the "planet" is significantly smaller in mass than the central "star". So that we can clearly claim that "the planet orbits the star" and not "there are two roughly equal-sized bodies in orbit around each other".  Such configurations would be a binary system and not a planet in orbit.
The significance would be in the nomenclature.  In a binary (or multiple) system, the objects are named "on the same level" (for example Alpha-Cen A, Alpha Cen B and Alpha Cen C)  Planets, however, are named at a lower level, with lower case letters (Alpha Cen Cb) There isn't really another reason for specifying an upper limit to the size of second body.
It also seems to be a proxy for the mode of formation (this is hinted at by the parenthetical comment). Two roughly equal bodies will have formed by gravitational collapse of a cloud of gas that split in two as it collapsed to form two objects mutually orbiting each other.  Whereas classical "planets" form in the disc surrounding a proto-star as a secondary process. But rather than use the unobservable "mode of formation" as a definition, the IAU (in its infinte wisdom) chooses the astronomically meaningful L4/L5 instability


Light pollution and apparent magnitude of objects in solar system

The answer is yes: dust is highly constrained to the ecliptic plane and so viewing from above the plane would result in less light that needs to be subtracted from observations of the solar corona.  The zodiacal light is quite strong in the direction of the sun because the dust strongly scatters in the forward direction. Near the Sun, its surface brightness is somewhere around 21 - 23 mag/arcsec$^2$ from our viewpoint in the ecliptic, and this would be an additive contamination to any measurement of the Sun's corona.  From the ecliptic pole, this light foreground would be much less.  Of course, a careful observer would know to also measure the zodiacal light contribution and to subtract it off, so both observers would get the same brightness for the corona roughly.  The observer at the pole would have lower noise for a comparable instrument and observing time.


Finally, after searching quite a deal, I found this paper which would aptly answer the question.

The paper from 1999 published by Pietrzynski and Udalski in Acta Astronomica lists the Cepheids in the star clusters of Magellanic Clouds.


How do stars or galaxies get their spin?

You could start from the premise that there was no net angular momentum in the universe at all; but it would still be the case that everything of interest was spinning.

On the scales of stars and planets there are (at least) two important mechanisms that result in individual systems having angular momentum. The first is turbulence. If you take a parcel of turbulent gas from a giant molecular cloud it will always possess some angular momentum, even if the total cloud does not. As the parcel collapses to form a star/planets conservation of angular momentum $J$ and dissipative interactions result in an ncrease in spin rate and collapse towards a planar geometry.

Second, stars form in clusters. There is interaction between stellar systems early in their lives. Again, the cluster may have little net J, but groups of stars can, relative to their own centre of mass frame.

On bigger scales (galaxies) the second of these explanations becomes more important. The interaction and accretion of galaxies is what gives individual galaxies a spin, even if the clusters they are born in have much less or even no net angular momentum.

As an example of how turbulent velocity fields lead to gravitational condensations containing angular momentum you could do worse than study the star formation simulation performed by Matthew Bate and collaborators. These simulations start off in clouds with zero net angular momentum, yet produce a host of stars with swirling accretion disks, binary systems of all shape and sizes etc. An example journal paper can be found here: http://adsabs.harvard.edu/abs/2009MNRAS.392..590B  Here is a web page where you can download the animations and study them at length  http://www.astro.ex.ac.uk/people/mbate/Cluster/cluster500RT.html

Turbulent clouds are by their nature random and stochastic in terms of their motions. Often the velocity field is defined in terms of a power law dependence on spatial scale. The formation of vortices is a characteristic of turbulent media. They can be produced in the absence of external forces. The vortices contain angular momentum.

It is also worth noting that not all galaxies have an appreciable spin. Spiral galaxies do, but many elliptical galaxies have little net rotation. See https://physics.stackexchange.com/questions/93830/why-the-galaxies-forms-2d-plane-or-spiral-like-instead-of-3d-ball-or-spherica


How can the Event Horizon Telescope image Sgr A* when it's not visible from all sites at one time?

I don't think it is necessary for all the telescopes to view the target simultaneously if you can make the assumption that the source you are observing is only varying on timescales longer than it takes the source to be viewable by all the telescopes. Sure, the maximum baseline you can get will be defined by the most distant pair of simultaneously observing telescopes, but that doesn't stop you adding together images taken at other times.

This is very probably the reason why M87 is in the fact the first image presented by the EHT. The timescale for significant variability around this (big) black hole will be of order a few time $2GM/c^2$, which is a few days for the M87 black hole. Imaging Sgr A* is going to be more difficult (or at least more blurry) because the variability timescale is minutes.

It looks like (from the data reduction paper) that the first steps involve treating every baseline in a pair-wise way (as I suggested above) and then combining them across the network, with whatever baselines were observing at the time, to give data at the surprisingly (to me) high time-resolution of 10s.


Can the Milky Way be seen with the naked eye? Does this apply to any galaxy? If yes, then how and when?

There are a few nearby galaxies that can be seen with the naked eye, but clear & dark skies are essential. Most notable for northern-hemisphere viewers is the Andromeda Galaxy also known as M31, best viewed in late summer and fall; from the southern hemisphere the Magellanic Clouds, irregular satellites of our own Galaxy, are famous landmarks. These all look like small fuzzy patches, rather unexciting until you consider that they‘re on the order of a million light-years away.


Is "Giga-hertz Astrophysical Transient Emission" distinct in any way from "Fast Radio Burst"?

I am going to float this as a tentative answer. I'll appreciate feedback, either validating it or letting me know it's wrong, in which case I'll delete it. If it is part of a better answer, please run with it and I'll also delete this.

X-ray navigation XNAV is being look at as a source of navigational data for spacecraft in deep space that does not require measurements and telemetry from a deep space facility on Earth. While a spacecraft could not easily receive radio pulses from a pulsar without a very large antenna to overcome the thermal noise in the receiver's front-end, a suitable X-ray detector could potentially build up timing data from several pulsars.

Since a spacecraft is usually moving without thrust under gravitational forces from known bodies, an instantaneous fix is not needed and data could be collected for several minutes or even longer.

While radio pulses from pulsars are generally periodic, there are transient phenomena that can be related to infall of ionized material interracting with the magnetic field causing bumps and wiggles of individual radio pulses, which may have a bearing on X-ray time structure as well.

And so I am speculating that these transient deviations in a Pulsar's otherwise regular behavior would be of relevance to technology development for both deep space exploration and business development.

However, while intermittent download of data from spacecraft could be considered Transient GHz Emissions, it would not be considered Astrophysical.

NOTE: The variability is in the production of electromagnetic radiation in a pulsar's incredibly intense magnetic field (radio) and at the surface (X-ray) produced during the infall of matter along the field lines towards the magnetic poles. The underlying rotation of the neutron star and it's field are still quite regular.


Why are Cassegrain telescopes shorter than Newtonian reflecting telescopes?

Basically, the light has a longer way to go after hitting the secondary mirror in a Cassegrain than in a Newton telescope.

In a Newton telescope (top image), the light hits the primary mirror and goes through the telescope tube once. In a Cassegrain telescope (bottom image), the light hits the primary mirror and goes through the entire length of the tube twice.

The focal length of a telescope is the distance between the telescope's primary mirror and the point where the light rays come together in focus. This is how far the light has to go after hitting the primary mirror before reaching the eyepiece.

Since the light goes a longer distance within a Cassegrain telescope than a Newton telescope of the same length, the focal length of the primary mirror can be made longer.
Conversely, for two telescopes with the same focal distance, the Cassegrain will be shorter than the Newton telescope.

You can imagine the Cassegrain telescope folding the focal length in two, whereas the Newton telescope leaves the focal length stretched out flat.

In this image, light has a longer path to go in the bottom (Cassegrain) telescope than in the top (Newton) one. Even though they are the same size, the bottom telescope has a longer focal length.

This would make it seem that a Cassegrain can have a focal length twice the length of the tube, whereas a Newton can have a focal length roughly the length of the tube.
In fact, Cassegrain have an extra trick up their sleeve to increase the focal length, and can have a focal length equal to five times the length of the tube.

The secondary mirror of a Cassegrain telescope is convex, and has its own focal length. The effective focal length of the primary mirror + the secondary mirror can thus be made longer than that of the primary mirror alone.


Where can I watch the announcement from LIGO about gravitational waves?

The event will be live streamed on youtube here: https://www.youtube.com/user/VideosatNSF/live

The will also be tweeting: https://twitter.com/ligo

A video of the conference will also be available afterwards if you miss it. It starts at 10:30 EST, which is 3:30 GMT.


What this video is talking about is Hawking Radiation, as you've linked. Hawking Radiation is a proposed hypothetical (by no means verified or proven) way for a black hole to radiate its energy into space. The basic idea is that a black hole is nothing but mass/energy compressed to an infinitesimal point, which is radiating its energy into space over time. For large black holes (such as solar mass or bigger), this radiation process is tiny and the time taken to leak all the black hole's energy into space (and thus for the black hole to "evaporate") is exceedingly long. For tiny black holes however, the time to radiate all the black hole's energy is exceedingly short.

You can calculate how long it will take for a black hole of mass $m$ to evaporate (and release all its mass/energy) with the equation

$$t_{\text{ev}} = \frac{5120\pi G^2 m^3}{\hbar c^4} = (8.41\times 10^{-17}\:\mathrm{s}\:\mathrm{kg}^{-3})\:m^3$$

For $m = 5\:\mathrm{g}=0.005\:\mathrm{kg}$, you get $t_{\text{ev}} \simeq 4\times10^{-19}\:\mathrm{s}$. Now that means in this tiny amount of time, the black hole will radiate all of its mass/energy away and completely evaporate. But the output of all the energy in 5 g of mass is a huge output. Putting out 450 Terajoules of energy in $10^{-19}\:\mathrm{s}$ is basically just an explosion. You can determine the total energy output from the famous equation

$$E=mc^2$$

Just plug in $m = 0.005\:\mathrm{kg}$ and $c = 3\times 10^8\:\mathrm{m/s}$ and you'll get $E=4.5\times 10^{14}\:\mathrm{J} = 450\:\mathrm{Terajoules}$.

So in short, hypothetical calculations (not even theory at this point) suggest that a tiny black hole with the mass of a nickel would immediately explode out in a huge amount of energy. Whether such a black hole can form, or if such an evaporation would/could occur is still highly debated and ultimately unknown at this point.


The metal content of the Solar system is completely dominated by the Sun. The Sun contains $\sim1\%$ of 'metals' (in astronomical language anything but hydrogen and helium is a 'metal'), but all the other bodies of the Solar system combined have less mass than that. So even if they were only made of metals (but the outer planets are mostly made of H and He) the Sun would still dominate the metal budget.
The Sun does not dominate the angular momentum of the Solar system, which is dominated by the orbital angular momentum of Jupiter. The angular momentum due to the spin of the Sun is rather modest. This is easy to estimate: the orbital angular momentum of a planet is $\sim m\sqrt{GM_\odot a}$ which increases with semi-major axis $a$, while the spin of the Sun contributes $\beta \omega M_\odot R_\odot^2$ with $\omega$ the Solar spin frequency, $R_\odot$ the Solar radius, and $\beta\sim0.1$ a factor depending on the inner structure of the Sun. The ratio of the latter to the former is
$$
\beta \frac{M_\odot}{m} \frac{R_\odot^2}{a^2} \frac{\omega}{\Omega}
\sim0.01
$$
For Jupiter, the first factor is $\sim10^3$, the second $\sim10^{-6}$, and the third $\sim180$ is the ratio of the Solar spin frequency (once in 25 days) to the orbital frequency $\Omega=\sqrt{GM_\odot/a^3}$ of Jupiter (once in 12 years).


How to find maximum and minimum right acsension and declination based on the telescope's location?

In the Southern hemisphere, the maximum declination you can see is
90-L where L is your latitude. The minimum is -90, since you can see
the south celestial pole.

As the earth rotates and revolves, you can see any right ascension
within those declinations.

The only exception is that you ordinarily can't see stars that are too
close to the Sun. However, the Sun moves enough between October and
March that all right ascensions will be visible at night sometime
between October and March.


Have there been studies of "old photons" to see just how constant things like Planck constant has been?

You can not check if a dimensional constant has changed because you can always reverse that change by a smart change of coordinates (system of units). Despite that, since the current Physics assumes the immutability of certain constants, you can verify this assumption by testing the change of an adimensional constant.

One of the most common adimensional constants that is used to search for constant variability if the fine-structure constant (a fundamental physical constant characterizing the strength of the electromagnetic interaction between elementary charged particles)
$$\alpha=\dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{\hbar c}\,.$$

The Wikipedia pages about Time-variation of fundamental constants and fine-structure constant have nice explanations about this.


What is it like at the edge of the universe?

The edge of the Universe in space-time can only be the moment of its creation (or end if the Universe is closed). What it would be like at that time depends on what you would consider close to the edge. Your question is about the edge of the expansion so I will presume that you mean during the Inflationary era (until about $10^{-32}\textrm{s}$ after the creation of the Universe).

There would be no objects as matter as we know it today did not exist (according to the big bang model). The major interacting forces (gravity, strong and weak nuclear force, and the electromagnetic force) were unified at the time of the Inflationary era. There would be no protons or neutrons as the strong interaction has not decoupled yet from the other forces. The temperature would be extremely high $>10^{27}\textrm{K}$.


With gravitational lensing, is it possible to mathematically compute the correct image of the galaxy that is being distorted?

That would be possible theoretically. However, in practice, the lensed images are highly non-linearly distorted and due to the limits in resolution, they are often limited to a thickness of a few pixels at best. Due to the non-linear transformations, this would cause the reconstructed original image to be even blurrier.

Having said that, I don't see any real need of reconstructing the original image. You can get most parameters (like a rough estimate of the size and type of the background galaxy) from the distorted image anyway.


Can radio-based telescopes produce visible light images?

Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble?

Of course, either via false color or pseudocolor. A false color radio image maps various frequencies (typically three) in a multispectral radio image to visible colors. A pseudocolor radio image maps various intensities in a single spectrum gray scale radio image to visible colors. The same techniques are also used for gamma, X-ray, UV, IR, and microwave imagery. We can't see in those frequencies, so something needs to be done to the captured imagery to make them visible to us.

If the question is asking whether a radio telescope can be used to capture visible light, the answer is no. I'll start with the receiver. It's designed to capture long wavelength radiation. Photons in the visible spectrum will have minimal, if any, impact on a radio telescope receiver.

An even bigger problem with some radio telescopes is depicted below.

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The above is the Haystack Radio Telescope at the Haystack Observatory in Westford, Massachusetts. The radome protects the telescope against weather, but also hides the antenna from optical light. A couple of other radio telescopes at the Haystack Observatory show yet another problem:

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The fixed zenith telescope antenna in the foreground and the fully movable telescope antenna in the background are made of a wire mesh, making them see-thru in the visible portion of the spectrum. This wire mesh construction is quite common for long wavelength radio antennae because it allows the wind to blow through it. Shorter wavelength radio antennae are more likely to be solid, but those are oftentimes painted.

A wire mesh with wires separated by centimeters is very smooth to a meter long radio wave, as solid surface smooth at the millimeter level to a centimeter long radio wave. A surface with millimeter-scale roughness would make for an extremely lousy optical mirror. There is no reason to make a radio telescope antenna smooth to the visible portion of the spectrum, and there are many millions of reasons (i.e., many millions of dollars) not to do so.


How can we predict a Big Crunch when all galaxies are moving further apart?

In a homogeneous and isotropic Universe (even if recent observations challenge this hypothesis), you can derive the Friedmann equations, which describe the evolution of the Hubble constant with time:
$\frac{\dot{a}}{a} = H(t) = \frac{8 \pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3}$   (with $c=1$) (Equation $1$)

where $a=a(t)$ is the scale factor, $\dot{a}$ its derivative, $G$ the gravitational constant, $\rho$ the matter density, $\frac{k}{a^2}$ the spatial curvature (a parameter that describes the metric of the Universe), and $\Lambda$ the cosmological constant (an integration constant added by Einstein).
It could be useful to rewrite the equation as:

$H^2 = \frac{8 \pi G}{3}(\rho + \rho_{\Lambda}) - \frac{k}{a^2}$

where $\rho_{\Lambda} = \frac{\Lambda}{8 \pi G}$ is the "density of cosmological constant".

We can also expand the matter density as $\rho = \rho_{matter} + \rho_{radiation}$.

So we have a "total" density $\rho_{tot} = \rho_{matter} + \rho_{radiation} + \rho_{\Lambda}$. The destiny of the Universe depends on this amount.

In case of $\rho_{tot} > \rho_{crit}$, or equivalently a closed Universe ($k=+1$), the equation $(1)$ becomes:

$\dot a^2 = \frac{8 \pi G}{3}\rho a^2 -1$

Which points out that the scale factor must have an upper limit $a_{max}$ ($\dot a^2$ must be positive).
This in turn means that the second derivative $\ddot a$ of the scale factor must be negative, when approaching $a_{max}$, that is the scale factor function inverts its behavior:

Look at here and here if you want to go deeper.

@Bardathehobo This figure shows what I mean when I say that a currently accelerating Universe can still crunch. This is because we are basically ignorant upon the dark energy issue.


Surprisingly your linked Wikipedia article does not link to an explanation of how to calculate it, and neither does this.
The equation for a synodic period is based on subtracting two frequencies, the same way a radio down-converts to an intermediate frequency by mixing two frequencies nonlinearly and using the difference in frequencies as the new signal:
$$f_{\text{diff}} = f_> - f_<$$
where the > and < indicate the larger and smaller frequencies.
In orbital mechanics we usually don't talk about orbital frequency, but instead orbital period where $T = 1/f$, so:
$$\frac{1}{T_{syn}} \ = \ \frac{1}{T_<} - \frac{1}{T_>} \ = \ \frac{T_> - T_<}{T_> \ T_<}.$$
Use that if the two rotations are in the same direction which is the case for Earths rotation around its axis and around the Sun.

could anyone illustrate it or point me in the right direction?

Put the sidereal day for $T<$ and a year for $T>$ and you'll have your answer.
But what if they are rotating in opposite directions?
In that case you can define one as having negative frequency or a negative period, and replace the minus sign with a plus sign.


You and your phone will be destroyed before you get to the black hole, but
If we pretend that you and your phone can survive being in the vicinity of a black hole ... And let's suppose this is a regular black hole of about 5 to 10 solar masses, and you are falling in, you are not trying to orbit it.

As you fall towards the black hole, at first there is no problem. As you get closer to the event horizon you will accelerate, but until the last second or so the relativistic effects are quite small. Your phone signal will be redshifted, so the receiver will have to adjust the frequency that they are listening in on.

The last second before you fall in will last forever. Your signal will become redshifted further and further, your apparent rate of speaking will slow down and the last second before you pass the Event horizon will, to the outside observer, last forever.

You however are doomed. You pass the Event horizon and reach the singularity in a fraction of a second.


Why there are three equations governing blackbody emission?

Historically, two people (or groups of people) independently came up with different equations to model the blackbody equations in different parts of the spectrum. Rayleigh-Jeans law (classically derived) is valid for longer wavelengths and Wien's law (not Wien's displacement law) is valid for shorter wavelengths.

The Planck Distribution approaches the two laws in its limits, as in, for shorter wavelengths it is approximately equivalent to Wien's law and for longer wavelengths, it is approximately equivalent to Rayleigh-Jeans law. However, the quantum mechanically formulated Planck's law is accurate at all wavelengths, so it is the one that should always be used. One can use the other laws, for example, for brightness temperature definitions in radio astronomy, they use the RJ law for convenience, but that's because the wavelengths are long enough, and the approximation of the Planck's law gives the same result. (https://en.wikipedia.org/wiki/Planck%27s_law#Approximations)

Wien's displacement law only relates the peak wavelength to the temperature, which is a different law completely, though Planck's law (differentiating the function to find the maximum) also gives the same result. Though in this case, the displacement law is not approximate. It is accurate and can be used whenever you want.


can we use connect a long pipe with space station and use vacuum for fuel transfer?

A pump can lift water about thirty feet. That's the height where the column of water in the pipe weighs the same as an equivalent column of air (and that's why deep wells use submersible pumps: they push the water up from below, instead of pulling from above and relying on atmospheric pressure to provide the lift). While fuel would be less dense than water, it's still much more dense than air, and the pressure of the atmosphere simply doesn't provide enough force to raise that column of fuel very far.


Telescope on drones to escape light pollution?

Yes, but not in the exact way you think of.
To avoid light pollution it is better to go sideways.  If you are in a light-polluted city, there would still be light pollution at an altitude of 400ft (the maximum operating altitude of drones) And, moreover, amateur drones are not powerful enough to carry a telescope of any useful size. So you will find that driving a few miles out of town will be much more effective, and much cheaper.
On the other hand, water vapour in the atmosphere absorbs infra-red light. If you can take a "drone" up to 40000 ft, you would have a good location for infra-red astronomy.  You would just need a big "drone". . . A 747 would be big enough. SOFIA is just that: a telescope mounted on a jumbo jet to get above the water vapour in the lower atmosphere.
So, it's not practical to use an amateur drone to get above light pollution. It is practical to put a telescope on a plane but only for quite specialist observations.


The function you're requesting — i.e. the number density $N$ of DM halos above a given mass $M_\mathrm{h}$ — is called the cumulative halo mass function (cHMF). It is obtained by integrating the halo mass function (HMF) from a given mass to infinity. The HMF, in turn, is thus the function that describes the (differential) number density of DM halos of a given mass.

In other words,
$$
\boxed{N(>\!\!M_\mathrm{h})
= \int_{M_\mathrm{h}}^\infty \!\!\!dM_\mathrm{h}'\, \frac{dN}{dM_\mathrm{h}'}.}
$$

Halo mass function

So, the problem is to determine the HMF, i.e. $dN/dM_\mathrm{h}$. This was first calculated, analytically, by Press & Schechter (1974) assuming spherical collapse of structures from an initial, smoothed density field. It may be written as
$$
\frac{dN}{dM_\mathrm{h}} = \frac{\rho_{\mathrm{m,0}}}{M_\mathrm{h}}
                    \left| \frac{d\ln\sigma}{dM_\mathrm{h}} \right|
                    f(\sigma),
$$
where
$\rho_{\mathrm{m,0}}$ is the present-day average mass density of the Universe,
$\sigma = \sigma(M_\mathrm{h},z)$ is the rms fluctuations of the (smoothed) density field, and
$f(\sigma)$ is the "multiplicity function" (note that because the number density decreases so fast with halo mass, for computational purposes often the HMF is expressed as $dN/d \ln M_\mathrm{h}$ rather than $dN/dM_\mathrm{h}$).

If you want, I can give you more details about how to calculate $\sigma(M_\mathrm{h},z)$ and $f(\sigma)$. You can also find details on this in sec. 2.1 of Laursen et al. (2018), but note that there's an error in Eq. 1. If you're happy with using a "black box", you can get both the HMF and cHMF for your favorite cosmological parameters using this online HMF calculator.

Only in the Press-Schechter formalism can an analytical form of $f(\sigma)$ be derived, and it was subsequently found that it over(under)predicts the collapsed fraction at the low-(high-)mass end (see e.g. Governato et al. 1999); in general one must obtain it by fitting halo abundances in cosmological $N$-body simulations.

Defining a halo

This is what N. Steinle describes in his answer, but actually you don't need to assume a density profile of the halos. So, how do you count halos in a simulation? There are several ways, arguably the two most popular ones being the spherical overdensity (SO) and the friends-of-friends (FoF) method.

Spherical overdensity

In the SO method, you first calculate the center of mass (CoM) of a clump of particles in the simulation, and then you calculate the average density $\bar{\rho}$ of particles in successively larger spheres centered on the CoM. As you increase the radius of the sphere, $\bar{\rho}$ drops because you include less and less dense regions. When the density has reached a certain factor $\Delta$ times the average density $\rho_\mathrm{m}(z)$ in the Universe you stop, and the total mass of all particles inside that sphere is then the halo mass. The overdensity factor $\Delta$ is usually chosen to be around 200, but other choices exist, e.g. 500 which then gives slightly smaller masses (because you stop counting earlier).

A variant of this method is using ellipsoids rather than spheres, allowing for more realistic results for elongated structures.

Friends-of-friends

In the FoF method, you start at an overdensity and count all particles that are "linked" together, meaning they are within some chosen distance (the "linking length") of each other. This method may give more realistic results for very non-spherical structures, but also tends to include particles in filaments streaming into the galaxies, which perhaps shouldn't be considered a part of a galaxy.

The figure below (from Klypin et al. 2011) shows halos from the Bolshoi simulation identified with the two methods; SO (red points) and FoF (blue points), compared to an extension of the P-S HMF (solid black line) to allow for ellipsoidal structures (Sheth & Tormen (1999, 2002). At all masses, FoF halos are seen to be more massive than SO halos.

Redshift range and area

Re-reading your question, I think maybe you're interested not in the number density, but in the absolute number in a volume spanned by a redshift range $dz$ and area $dA$. If that is the case, then you simply multiply your $N(>\!\!M_\mathrm{h})$ by the cosmological volume given by $dz$ and $dA$. Let me know if you also want to know how to calculate that.


What's the minimal argument I can give to reject The Geocentric Universe?

Part One: Moons of other planets.
The discovery of the Galilean moons of Jupiter in 1609-1610 proved that not everything had to orbit directly around one signel object.
The discovery of the Galilean moons showed that objects could revolve around an object which revolved around another object.  There was no way for all known objects to revolve directly around each other.
Either:

The Sun, the Moon, and the planets revolved around Earth, and the 4 Galilean moons revolved around Jupiter, thus only indrectly orbiting the Earth.

or:

Earth, the Moon, and the planets orbited around the Sun, and the 4 Galilean moons revolved around Jupiter, thus only indrectly orbiting the Sun.

or:

The Sun, the Earth, the other planets, and the 4 Galilean moons, all revolved around Jupter, and the Moon revolved around Earth, thus only indrectly orbiting Jupiter.

Decades later, Titan was disvoered in 1654, and four other large moons of Saturn were discovered in 1671 to 1684.
So now there were three objects in the solar system which certainly had objects orbiting them, Earth, Jupiter, and Saturn, and there was a big astronomical question whether everything ultimately orbited the Earth or the Sun.
Part Two:  Phases of the Planets.
When telescopes were first used to observe the planets starting in 1609-1610, the phases of the planets began to be observed, phases caused by the changing angles between the Sun, the planets, and Earth.
It became noticed that the phases of Mercury and Venus went all the way from thin crescents to 100 percent full, and that their angular diameters were larger when they were thin cresents and smaller when they were full. Half the times they appeared close to the Sun they appeared small and full, and half the times they appeard close to the son they appeared large and cresent shaped. That was consistent with them orbiting around the Sun at a distance less than the distance tebween Earth and the Sun.
But the phases of Jupiter and Saturn were different.  They ranged from slightly more than half full to totally full.   Jupiter and Saturn always appeared smaller and less full the closer they were to the direcion of the Sun but were never less than half full.  They never appeared larger and less than half full when they were close to the direction of the Sun.  So they must have orbited around either the Earth or the Sun at distances greater than the distance between the Earth and the Sun.
Part Three: Parallax.

Diurnal parallax

Diurnal parallax is a parallax that varies with rotation of the Earth or with difference of location on the Earth. The Moon and to a smaller extent the terrestrial planets or asteroids seen from different viewing positions on the Earth (at one given moment) can appear differently placed against the background of fixed stars.[12][13]
The diurnal parallax has been used by John Flamsteed to measure the distance to Mars at its opposition and through that to estimate the astronomical unit and the size of the solar system.[14]

https://en.wikipedia.org/wiki/Parallax#Distance_measurement[1]
john Flamsteed lived from 1646 to 1719.

Solar parallax
After Copernicus proposed his heliocentric system, with the Earth in revolution around the Sun, it was possible to build a model of the whole Solar System without scale. To ascertain the scale, it is necessary only to measure one distance within the Solar System, e.g., the mean distance from the Earth to the Sun (now called an astronomical unit, or AU). When found by triangulation, this is referred to as the solar parallax, the difference in position of the Sun as seen from the Earth's centre and a point one Earth radius away, i. e., the angle subtended at the Sun by the Earth's mean radius. Knowing the solar parallax and the mean Earth radius allows one to calculate the AU, the first, small step on the long road of establishing the size and expansion age[20] of the visible Universe.
A primitive way to determine the distance to the Sun in terms of the distance to the Moon was already proposed by Aristarchus of Samos in his book On the Sizes and Distances of the Sun and Moon. He noted that the Sun, Moon, and Earth form a right triangle (with the right angle at the Moon) at the moment of first or last quarter moon. He then estimated that the Moon–Earth–Sun angle was 87°. Using correct geometry but inaccurate observational data, Aristarchus concluded that the Sun was slightly less than 20 times farther away than the Moon. The true value of this angle is close to 89° 50', and the Sun is actually about 390 times farther away.[18] He pointed out that the Moon and Sun have nearly equal apparent angular sizes and therefore their diameters must be in proportion to their distances from Earth. He thus concluded that the Sun was around 20 times larger than the Moon; this conclusion, although incorrect, follows logically from his incorrect data. It does suggest that the Sun is clearly larger than the Earth, which could be taken to support the heliocentric model.[21]

Although Aristarchus' results were incorrect due to observational errors, they were based on correct geometric principles of parallax, and became the basis for estimates of the size of the Solar System for almost 2000 years, until the transit of Venus was correctly observed in 1761 and 1769.[18] This method was proposed by Edmond Halley in 1716, although he did not live to see the results. The use of Venus transits was less successful than had been hoped due to the black drop effect, but the resulting estimate, 153 million kilometers, is just 2% above the currently accepted value, 149.6 million kilometers.
Much later, the Solar System was "scaled" using the parallax of asteroids, some of which, such as Eros, pass much closer to Earth than Venus. In a favourable opposition, Eros can approach the Earth to within 22 million kilometres.[22] Both the opposition of 1901 and that of 1930/1931 were used for this purpose, the calculations of the latter determination being completed by Astronomer Royal Sir Harold Spencer Jones.[23]
Also radar reflections, both off Venus (1958) and off asteroids, like Icarus, have been used for solar parallax determination. Today, use of spacecraft telemetry links has solved this old problem. The currently accepted value of solar parallax is 8".794 143.[24]

https://en.wikipedia.org/wiki/Parallax#Solar_parallax[2]
Part Four: Stellar Parallax.
The heliocentric theory implies that as the Earth orbits the Sun, the angles to stars should change. Because of the vast distances to the stars, astronomers failed to detect the parallaxes and distances to the stars for centuries.
Advances in instruments and techniques led to the first successfull measurements of stellar parallax in the 1830s, when the parallaxs and distances of Alpha centauri, 61 Cygni, and Vega were first measured.  Alpha Centauri is the closest star system to the Sun, 61 Cygni is the 15th, and there should be a few hundred star systems closer - and easier to measure their parallaxs - than Vega.
Part Five: Conclusion.
Any amateur astronomer who has telescopes, instruments, and observing technigues as good as those of professional astronomers of 17th and 18th centuries, or the 19th century up to the 1830s, can duplicate those observants which give stronger and stronger evidence that the Earth orbits the Sun instead of the Sun orbiting the Earth.


How does time work beyond the cosmic event horizon?

First, let's clear up a few misconceptions:

The Hubble sphere

The speed of light as an upper limit is valid in special relativity (SR). In general relativity (GR), which must be used to describe the expansion of the Universe, although locally (i.e. where SR is a good approximation) you cannot exceed the speed of light, there is no limit to the relative velocity of two objects.

By Hubble's law ($v=H_0 d$), all galaxies that are more distant from us than the "edge" of the so-called Hubble sphere $d_\mathrm{H} = c/H_0 \simeq 14.4\,\mathrm{Glyr}$ (billion light-years) recede from us faster than $c$. This is no hindrance for us to observe them; we regularly observe galaxies much farther away (see e.g. this and this answer).

The event horizon

The cosmic event horizon (EH) is the largest distance a galaxy can be at, and still emit a light signal that may reach us in a finite time. It is currently roughly 16 Glyr away, i.e. a bit farther than the Hubble horizon. We also regularly observe galaxies farther away than the EH, but we see them in the past (as we see everything else), and the light they emit today will never reach us (not because space expands, but because the expansion rate accelerates).

The particle horizon

The particle horizon (PH) is the largest distance from which light has had the time to reach us since the Big Bang. Due to the expansion of the Universe, this is currently 46.3 Glyr away. Galaxies at this distance recede from us at 3.2 times the speed of light. We don't see any galaxies here because we look back in time to just after the Big Bang where they hadn't yet formed (and because the Universe at that time was foggy), but the most distant galaxy yet observed, GN-z11, is 32.2 Glyr away and recedes at $2.2c$.

Time dilation

However, you're right that the galaxies' recession from us makes their time go slower, as seen from us. Not by the SR factor $(1 - v^2/c^2)^{-1/2}$, but by the GR factor $(1+z)$, where $z$ is their observed redshift.

Galaxies receding from us at $v=c$ have a redshift of $z\simeq1.5$, so when we observe physical processes in such a galaxy — e.g. the decline time for the brightness of a supernova — it's slower by a factor $\simeq2.5$. GN-z11 is at redshift $z=11$, so things there happen slower by a factor 12.

The farther you look, the slower things happen, and if we could look arbitrarily far, you could in principle see the event right after the Big Bang in slow motion. You cannot see "zero o'clock", which would have an infinite redshift. And you can never, not even in theory, see anything that's farther away than it's light has had the time to travel, so not even in theory does it make sense to try to assign a time dilation factor to these events.

As far as we know, there are galaxies outside the PH. As time goes, the distance to the PH increases, both because space expands, and because light fro more distant region will reach us. But as soon as that light is able to reach us, the object that emitted it is inside the PH, and thus has a finite time dilation factor.

But remember that the time dilation is only from our point of view. If you could magically teleport yourself there right now, time would pass normally for you there. So nothing particular happens there.


What does this figure imply about the luminosity of pre-main-sequence stars?

It shows that the mass of the (proto)star is the major factor determining bolometric magnitude.

The brightness of the proto-star increases a little as it contracts and becomes significantly hotter (and the radius decreases), The star moves toward the left of the diagram. As fusion in the core becomes significant the star reaches the maximum point on the curve, this stabilises the core and the star settles into the main sequence, it cools slightly and becomes slightly less bright.


Is a planetary system star's referred to as their sun?

The technical name (in English) for the Sun is not Sol, which is just Latin for sun. The technical name for the Sun is the Sun. Another body in the sky has a similarly boring name, the Moon. There's one more boringly named object: in the Solar System: The Earth. Note the use of "the" (a definite article) and the use of capitalization to indicate a specific object.
The problem with Sun, Moon, and Earth is that we have been using these names (or their predecessors) for thousands of years. For example, sun, sol, ἥλιος (helios), and a bunch of other names for that very bright object in the sky whose presence distinguishes day from night all derive from the same proto-Indo-European word sóh₂wl̥.
As an end note, from http://curious.astro.cornell.edu/about-us/159-our-solar-system/the-sun/the-solar-system/4-what-are-the-names-of-the-earth-moon-sun-and-solar-system-beginner,

You may read or hear people using Luna for the Moon, or Terra or Gaia for the Earth, or Sol for the Sun, but in English-speaking countries, these are poetic terms, often seen in science fiction stories, but not used by astronomers in scientific writing. In some countries where Romance languages are spoken, these terms are the official names.
It's also interesting to note that most astronomers do not call our galaxy the Milky Way in technical writing--they call it the Galaxy.


Does the Honey moon have a precession problem?

Is there a problem here though? For any given month won't the full
Moon oscillate between low and high above the horizon at midnight on
an 18.6 year cycle?

No. The full moon is always opposite to the sun (within 5.2 degrees, the angle between the moon's orbit and the ecliptic). So, when in summer the sun is high in the sky, the full moon will be low, similar to the height of the winter sun.
You are correct about the oscillation with the 18.6 year cycle, but that oscillation amounts to only +/-5.2 degrees, while the summer/winter difference results in +/-23.5 degrees. So (in northern latitudes) the typical summer full moon will be 47 degrees lower than the typical winter one, and even the highest possible summer full moon will still be 37 degrees lower than the lowest winter full moon.


Usage of $\sim$, $\approx$, $\simeq$, and $\cong$ in observational astronomy?

$\simeq$ and $\approx$ both mean "approximately equal to". I don't think $\cong$ is used so often, but if I read it, I would interpret is as the same as the two others.

$\sim$ in principle means "of the order of", i.e. correct to within an order of magnitude. However, astronomers tend to be rather slobby sometimes (we call all elements except H and He for "metals", we call it "gas" even when it's ionized and should be called "plasma", we omit factors of order unity if we can get away with it, we mix up terms like "flux", "flux density", and "intensity", and so on), and thus $\sim$ is very often used to mean "approximately equal to". It is even sometimes used as "proportional to", but (usually) only when there's no risk of misunderstanding (e.g. $L_\mathrm{UV}\!~\sim\!~\mathrm{SFR}$, meaning that the UV luminosity is proportional to the star formation rate, omitting a factor of $10^{27}$).

Journals usually have rather strict standards for style and nomenclature, so they might not accept a $\sim$ for a $\propto$, or even for a $\simeq$. But if you read a conference proceeding or lecture notes or some other non-refereed text (even a paper on arXiv waiting to be refereed), you best be prepared for anything.


Relationship between Dark Energy and Dark Matter

First of all, if there were no dark matter (DM), you wouldn't ask this question, since structures — including galaxies, stars, planets, and you — wouldn't have had the time to form in the early Universe before it had expanded too much for gravitational collapse to occur. But let's use magic and make the galaxies anyway:

The CMB (specifically the power spectrum of the CMB) shows that the total density $\rho_\mathrm{tot}$ of mass/energy in the Universe is extremely close to the critical density $\rho_\mathrm{c}$. That is,
$$
\frac{\rho_\mathrm{tot}}{\rho_\mathrm{c}} \equiv \Omega_\mathrm{tot} \simeq 1.
$$
The "$\Omega$" is a common way to express densities; as a ratio to the critical density. The CMB also gives some constraints on the total amount of mass (DM + "normal" matter, i.e. baryons), but it is better at constraining the ratio of baryons-to-DM. Together with the matter density $\Omega_\mathrm{M}=\Omega_\mathrm{b}+\Omega_\mathrm{DM}$ obtained from observations of supernovae and, in particular, baryonic acoustic oscillations, we then obtain the amount of DM. This fraction is roughly $\Omega_\mathrm{DM}=0.26$ (Planck Collaboration et al. 2016).
If there were no DM, then $\Omega_\mathrm{tot}$ just wouldn't be $1$, but rather $\Omega_\mathrm{tot} - \Omega_\mathrm{DM} \simeq 0.74$.
Similarly, if there were no DM, there would be less matter to counteract the expansion of the Universe. That means that we wouldn't observe the same relation between the brightness and the distance of supernovae, from which we infer the presence of dark energy (DE). Instead, the brightnesses would be somewhat lower, because the supernovae would be farther away due to the faster expansion.

In other words, you are right that if there were no DM, and if we observed the same thing as we do, then there would be a contradiction. But if there were no DM, we  wouldn't see what we see. Therein lies the resolution.


You are making the mistake of thinking that there was a single spot in space that everything expanded from. Rather, the expansion was not of things in space, it was space that expanded. It was an expansion OF space, not IN space. There wasn't "space" around the singularity.


Why does the sensitivity to GWs drops off inversely proportional to the distance?

EDIT I'm leaving the original, highly upvoted answer below, but I've had a fundamental rethink about this, prompted by questions from Keith McClary and a helpful clarification from a Physics SE question.

The original answer I gave is the reason that we can detect gravitational waves (GWs) at all. Their coherent nature as single oscillators, means that despite their comparatively low powers, they can be detected right across the universe. In comparison, electromagnetic sources are usually the superposed light from countless incoherent emission sources. On average this has a destructive interference effect that reduces the power (intensity) received; and due to the rapidly changing signal, it is usually intensity that is measured.

However, the answer to the present question is actually just how "sensitivity" is defined.

In order to detect a source, we must identify it in a background of noise. This is done by defining a signal-to-noise ratio. The signal is the product of your source strength (more on that in a minute) and how long you observed it for. The noise is a property of your instrument. The sensitivity of the instrument then is something like the minimum (signal $\times$ observation time) that will produce a significant detection.

In astronomy it is conventional to express the signal in terms of power received since, due to the arguments given above and in the original answer, power (intensity) is generally what is measured. The sources of noise are therefore also defined in terms of a power and the sensivity has units of something like Watts $\times$ seconds, or more conventionally, W/Hz.

In gravitational wave astronomy, because it is the amplitude that is directly detected, GW astronomers express their source signal in terms of amplitude (which is proportional to the square root of detected power) and their sensitivities are expressed in terms of gravitational wave amplitude (which is dimensionless) divided by $\sqrt{Hz}$ as a result.

i.e. we are not comparing like with like. Doubling the sensitivity of a gravitational wave detector is actually like quadrupling the sensivitity of an electromagnetic wave detector. Thus there is no fundamental difference here, the apparent difference in behaviour is merely a result of how sensitivity is defined. The reason for the different definition is as per my original answer below.

Original answer

The difference is that usually when we detect sources of electromagnetic waves, we are detecting intensity, which obeys the inverse square law.

In contrast, we are detecting the amplitude of gravitational waves, and amplitude only scales as the inverse of distance

Why the difference? Sources of gravitational waves are coherent oscillators. A  merging binary produces a single coherent wave train with an amplitude that can be defined and measured. By contrast, when we look at a distant star or galaxy in electromagnetic waves we are seeing the incoherent contribution from countless accelerating particles and atoms and all we can detect is the resultant summed intensity. There is no coherent electromagnetic wave with an amplitude that can be measured.

This difference in behaviour is fundamentally because whilst there are positive and negative electric charges, which require contrived circumstances in which to behave coherently (e.g. in a laser), gravitational waves are produced by accelerating masses, and since there is only one sign of "gravitational charge", the individual parts of a gravitational wave source are able to act in concert quite naturally to produce a coherent waveform that has a wavelength larger than the body itself.

An excellent discussion of these points can be found on the first page of the review article by Hendry & Woan (2007).

In principle, if we were looking at a single coherent source of electromagnetic waves then we can detect the amplitude (for example by the force it exerts on charged particles), and then the sensitivity would just reduce as the inverse of distance. At optical frequencies the electric field varies so rapidly that this cannot be done, but it is possible at radio frequencies. Unfortunately the coherence length and coherence time (the time over which the phase of the wave is predictable) are so short that this is rarely practical in laboratory, let alone astronomical, sources.


Possible to use tachyons to explore black hole?

Let's look at the validity of tachyons first. Travelling faster than light, as said by userLTK, would create time travel, which would create the famous paradox, the tachyonic antitelephone. The tachyonic antitelephone is a violation of causality, and thus most likely, tachyons would not exist.

Now, take a look at another assumption. Wikipedia is right almost all times, but sadly, it has some wrong information. Such is with tachyons. It gives us some contradictory statements of tachyons too. It is wrong to assume that tachyons always move faster than light. A tachyon, is a particle with an imaginary number as its rest mass. This, however, does not mean it can travel faster than the speed of light, nor that there's any conflict between their existence and the special theory of relativity.

The main idea here is that the typical intuition we have about particles -- them being billiard ball-like objects -- just fails in the quantum world. It turns out that the correct classical limit for quantum fields in many situations is classical fields rather than point particles, and so you must solve the field equations for a field with imaginary mass and see what happens rather than just naively assume the velocity will turn out to be faster than light.

Leandro M. wrote on physics stack exchange, that

  a good way to think about tachyons is to imagine hanging several
  pendulums on a clothesline, one after the other. If you disturb one of
  them, some amount of force will be transmitted from one pendulum to
  the next and you'll see a traveling disturbance on the clothesline.
  You'll be able to identify a "speed of light" for this system (which
  will really be the speed of sound in the string). Now you can make a
  "tachyon" in this system by flipping all the pendulums upside down:
  they'll be in a very unstable position, but that's precisely what a
  tachyon represents. Nevertheless, there's absolutely no way that you
  could send a signal down the clothesline faster than the "speed of
  light" in the system, even with this instability.

This means that tachyons don't actually move faster than the speed of light, and your assumption is wrong.

Furthermore, StephenG wrote in a comment that tachyons essentially can't exist in a black hole.

user106422 wrote in the linked question on physics stack exchange that

  in a nutshell, creating a tachyon itself would lead to some form of
  a spontaneous phase transition in which the tachyonic mode no longer
  exists. So from a rigourous QFT + GR setting, I don't think tachyons
  exist the way it is portrayed in simpler QM models. So no, tachyons
  will not escape the horizon simply because the theory will not allow
  for them to exist.

Thus, no, even if they exist, they don't really travel faster than the speed of light, and even if they do, that can't really "travel" through a black hole. So, it is not possible to use tachyons to explore a black hole.


Mountains on the moon during solar eclipse?

The beads of bright light can be seen thanks to the Moons rugged topography. They're called Baily's beads in honor of an English astronomer who explained the phenomenon, Francis Baily.

In the photograph you posted, you can also see Solar prominences. It's plasma ejected into the corona. Its temperature is lower than that of the coronal plasma and it emits more visible light.


That's an interesting question, but as the comments say might be a better place to ask this question in the Philosophy SE site, but the question about the universe being englobed maybe by another much bigger 'space' is something we might never be able to prove right or wrong. This 'existential' question might refer to the holograph theory, the multiverse theory. The thing is, at the moment the only possible explanation of whats beyond the 'observable universe' boundary is nothing, neon, an infinite multi-dimensional plane of nothingness.


Jupiter with a mobile phone and Celestron Astro FI 102mm Maksutov

Congratulations on your purchase.  The first pictures dont' show anything much. Just a out-of-focus blur.
The last one shows Jupiter and three of its moons.  I've overlaid the image onto a simulated image from stellarium (at about 10pm BST, the moons move pretty quick so you need an exact time!):

You can clearly see which moon is which, and why Europa is invisible (it's in line with Jupiter)  Look back tonight and the moons will have moved.  Ganymede will have crossed to the other side, Europa will be visible. Take a series of observations and you can rediscover Kepler's third law.
Jupiter itself is overexposed, which is why you can't see any detail. There is quite a trick to getting the exposure right. Astrophotography is a rabbit hole that goes very deep.
You don't want a "zoom". You've got a couple of eyepieces.  You don't see any extra detail with a more powerful eyepiece, you just magnify the blur.
But well-done.  You've imaged a solar system body with a telescope.  Galileo would have been proud.


Might the universe be really small yet seem infinite from spherical reflections?

The simple answer to your headline question is: No.

That kind of correlation between different angles has been investigated, and no significant correlation beyond 1 degree (first acoustic peak), e.g. large-scale circular structures in the CMB, has been found.

Details about the investigation of non-trivial topologies within the limits of recombination, see this Planck paper.

Closer-by structures (like galaxy clusters) visible by telescopic observations are correlated to gravitational lensing of the CMB, as described in this Planck paper, or by the integrated Sachs-Wolfe effect, as described in this Planck paper, since galaxy clusters tend to be correlated with dark matter halos.

Repeating pattens (from mirroring) would be evident from 3d maps (like this one for the local 300 Mpc environment) of the mass distribution or telescopic observations.

Multiple images of distant objects are well-known, e.g. as Einstein crosses, or Einstein rings, caused by gravitational lensing.
This can include mirroring in extreme cases; it's well-known how this would look like.


Does Einstein's theory of gravity say anything about the direction of space

There are numerous ways to try to explain space-time by analogy, and the flat plane is one of them. Another is the inflating balloon. The are all only analogies. Either of your pictures above is equally valid, however the one "the right way up" is simpler to understand.
For humans, who can easily conceive of 3 dimensional space, an example that works in 3D is essential. If you have ever tried to visualise more than 3 space dimensions, you'll find it very tricky.


You're using the wrong terms. Engineers use yaw, pitch, and roll to describe the orientation of a vehicle. Some erroneously call these rotations Euler angles. Astronomers and physicists use true Euler angles, a rotation about the Z axis of some reference plane, followed by a second rotation about the once-rotated X axis, followed by a third rotation about the twice-rotated Z axis. Note that the Tait-Bryan angles (aka Cardano angles) use a sequence of rotations about three distinct axes. Euler angles use only two axes.

The first rotation is the planet's axial precession angle. The second rotation is the planet's axial tilt, or obliquity. The third rotation represents the planet's daily rotation. The rates at which the precession and obliquity change are much smaller than the quickly-changing third angle.


How does this telescope work, and what is it for?

I'm pretty sure this is one of the four auxiliary (or "outrigger") telescopes for the European Southern Observatory's Very Large Telescope Interferometer, part of the VLT complex at Paranal Observatory in Chile. It's used in conjunction with the four main 8 m ("unit") telescopes in an optical/infrared interferometric system, so its purpose is to send images of the sky in the optical or near-IR to be combined with images from the unit telescopes and the other auxiliary telescopes (via underground tunnels) in the interferometric system. (The octagonal gray frame is kind of a giveaway, and the fact that this video is ESO-based helps confirm this.)

https://www.eso.org/public/teles-instr/paranal-observatory/vlt/auxiliarytelescopes/

(Note that, per the link above, the primary mirror in each auxiliary telescope has a diameter of 1.8 m, while the secondary mirror is only 0.14 m -- very small, as you guessed.)

The object you identify as "B" is the tertiary mirror, which is flat, elliptical, and tilted at a 45-degree angle so that light from the secondary mirror is redirected off to the side. It sits in a mount (supported by the "B vanes") located just in front of the primary mirror, so there is no need for a hole in the primary mirror. You're probably used to versions of this (Nasmyth) design where the tertiary mirror is located behind the primary mirror, requiring a hole in the primary. Since the tertiary mirror in a Nasmyth design needs to sit along the elevation rotation axis of the telescope, whether it goes in front of or behind the primary depends on the overall telescope balance.

(Sometimes there is a hole in the primary, and the tertiary sits on top of a small "tower" that protrudes up through the hole from behind the primary; this would help disguise the existence of the hole. That's how the tertiary mirrors in the 8 m unit telescopes of the VLT work, though I don't think they have support vanes the way this one does. The existence of the support vanes and the nature of the B' reflection suggests there really isn't a hole in the primary here.)

The "dome" is an example of a "clamshell" design, in which two halves fold down out of the way; you can see one of the two hinges in the front, near the bottom right of your images. If you google for images of these telescopes, you can see them in the daytime, folded-up configuration.

Here's an image of one of the telescopes when it was still in the factory (before the mirrors were installed), which may help you understand the design better:


We didn't only visit the Moon once! Neil Armstrong, Buzz Aldrin and Michael Collins were the first manned mission to land on the Moon with the famous Apollo 11 landing, but 5 later Apollo missions also landed safely on the Moon and did various exciting things.
This Wikipedia article gives a brief list of all the manned and unmanned missions that have investigated the Moon, and this NASA article gives a nice history of the Apollo program with all the manned missions. (So far the US are the only country to land astronauts on the Moon.)

As for why there haven't been more visits since Apollo, there are a couple of points:

1) We have continued investigating the Moon, through remote observation, close encounters and landings, just not with humans on board. The future of human space flight is quite a debated question, since humans are very fragile, have lots of requirements in terms of food, space, health etc. and are not really very efficient at many tasks. Automated investigation of space is by far cheaper and easier. The main advantages humans have are intuition and adaptability. We can change plans, make snap decisions and respond much quicker than a robot controlled by people far away on Earth.

2) Turn the question around: why would we go back to the Moon? Why spend money and time investigating this rather boring lump of cold rock when we could go to places we have absolutely no data on, or very shaky theories, or places that have environments hugely different to our own? If we want to test scientific theories, we often want extreme environments, not just something  a little bit different to home. And if, as your comment says, we are looking for other life or future homes, these things are very rare so we will have to look a lot further away than the Moon to find something that interesting!


How would the night sky look from inside a globular cluster?

Globular clusters occupy an interesting place in the spectrum of composite stellar systems. As you point out, they are highly concentrated populations of stars, and seem to lack any dark matter component, unlike more massive dwarf galaxies.

Binary interactions become very important in simulating globular clusters, and interestingly enough (maybe unsurprisingly), the one example of a discovery of a planet found in a globular cluster has been around a binary star system (see: PSR B1620-26 b; this circumbinary planet was found orbiting a pulsar and a white dwarf.). This is not to say there are not other examples, however, this was the easiest for me to come across. I would be interested to know how common this situation is, and in addition, how stable it is given the potentially highly chaotic environment it lives in. These speculations don't answer your question, but I thought it interesting enough to bring up as evidence in favor of your question not being an unreasonable one to ask.

From the wiki page:

  Globular clusters can contain a high density of stars; on average about 0.4 stars per cubic parsec, increasing to 100 or 1000 stars per cubic parsec in the core of the cluster.[26] The typical distance between stars in a globular cluster is about 1 light year,[27] but at its core, the separation is comparable to the size of the Solar System (100 to 1000 times closer than stars near the Solar System).[28]

This seems to indicate to me that location within the globular cluster would matter quite a bit. If at the core the average distance between stars is about three thousand times closer than our nearest neighbor is to our sun (my estimate to give some perspective: a few lightyears to Proxima Centauri divided by 100 is about 3000AU (about 100 times further than Pluto from the sun)), then stable orbits may be shifted inward, or simply may not exist due to two-body interactions.

However, if life were to exist (an assumption we're going to make for the purposes of your question), one would see a very different night sky. According to this paper, the number density profile of stars within the globular cluster M92 follows a Wilson Profile fairly well, which has the form:

$$ f_{W} = A\{ e^{-aE} - e^{-aE_{0}} [1 - a(E-E_{0}) ] \} $$
where $E \le E_{0}$. E is the specific energy of the star:

$$ E = v^{2}/2 + \Phi(r) $$

and where $\Phi(r)$ is the mean-field gravitational potential, determined from the Poisson equation. For each family of models, the constants A, $E_{0}$, and a in the above distribution function define two dimensional scales (a typical radius and a typical mass or velocity) and one dimensionless parameter, the central depth of the potential well (related to the concentration parameter) (all of the information is taken from the paper I've linked).

It seems to be the case that globular clusters are not "simple stellar populations", in that they are usually made of multiple generations (sources: 1,2). However, globular clusters generally consist of population II stars and are older stellar systems when compared to other star clusters. I bring all of this up because in addition to the number density of stars, the stellar distribution of stellar types would certainly be an important factor in how the night sky would look. If you lived a thousandth of a lightyear from a blue supergiant, you could imagine that that would make a huge difference in what you would see on a day to day basis. At the same distance, a supergiant star is on the order of $10^{5}$ times the luminosity of our sun (and therefore is $10^{5}$ the flux, since $L \propto f$ holding $D_{L}$ constant). At the same distance as our sun, the magnitude of a star with $10^{5}$ times the flux would have an apparent magnitude of about -38 (I used Rigel as my test case; this produces a star in our sky which is 12 magnitudes brighter than our sun). Moving this to the average distance between stars at the center of a globular cluster we would get an apparent magnitude of:

$M = -6.43 $ from $m=-38$ at distance of $d=1$AU.
$m=-26.43$ from $M=-6.43$ at a distance of $d = .00326 ly$ (the new average distance between stars at the center of a globular cluster)

In other words, a blue supergiant at an average distance between stars within a globular cluster would appear to be as bright as our sun is to us! This is absolutely crazy. Depending on where it is in relation to the sun, it could effective cause two days, or potentially one day which is greater than half the time it takes your planet to rotate once. I would image that this would certainly interfere with observing in optical (and shorter wavelengths).


Can the Hubble Sphere ever extend over the cosmic event horizon?

No, the Hubble sphere can never extend over the cosmic event horizon; it can only asymptotically approach it.
The Hubble sphere is the region within which galaxies (and other stuff) recede from us slower than light. The event horizon is the region outside which no light can ever reach us. If the Hubble sphere extended beyond the event horizon, it would mean that there exist a region in spacetime that recede subluminally, but which we could never see. But we have always been, and will always be, able to see light from subliminal regions.
However, a galaxy can cross the event horizon, such that it is no longer able to emit light that we can see. This is most easily seen in comoving coordinates, i.e. the coordinates that expand along with the Universe. In these coordinates, galaxies have fixed positions, but the event horizon is always shrinking.
In the spacetime diagram below you see time as a function of comoving coordinates. We are where the blue lines cross. The yellow region is the part of spacetime from which we may receive a signal, bounded by the event horizon (orange), and slowly approaching the Hubble sphere (purple). Galaxies at fixed coordinates follow lines parallel to the black dotted line, and may hence eventually cross the horizon.
For example, today the distance to the horizon is 16.5 Glyr. A galaxy at this distance may emit a photon now that we may detect in a almost infinitely distant future, redshifted to the extreme radio regime.
But tomorrow, it's too late.


One calls Mercury a planet because it doesn't share its orbit with any other bodies of comparative size. The fact that it is in its orbit means that that orbit hasn't been cleared by another body (for if it had been cleared by another planet, then Mercury would have been cleared too.)
In practice one doesn't look at "How" an orbit is cleared, only if there is one body that dominates the orbit, or if there are many with no one body dominating.  So Mercury is clearly much larger than any other bodies in roughly 88 day orbits.  But, (for example) Ceres does not dominate its orbit, as it shares its orbit with Vesta, Pallas and Hygeia and many other asteroids.


What is the centre of rotation of Pluto?

I know that Pluto and Charon have been found to circle round each other, hence the reason Pluto is no longer a planet

This is not correct.

It is considered a dwarf planet because of a technical definition employed by the IAU in deciding what is a planet and is not.  This technical definition relates to an object's ability to "sweep" it's orbit clear of asteroids and other small debris.  The decision has nothing to do with Charon.

The centre of rotation is calculated using Newtonian physics.  I think what you want is the Center Of Mass and related topics.


Can BY Draconis variable stars sustain any form of complex life?

Life could easily form underground, where flares don't reach.  Never underestimate the power of 10^(googol) possible locations in the universe for living things to come into being.   No reason to limit to organic, or CO2 / O2 connected, or photon-dependent, life forms.


Here is an overview of different scales we may look at the universe. On scales beyond it's almost homogeneous, and we get at the border of the visible universe. Many detailed images are available, e.g. from the Hubble Space Telescope.

If you need a three-dimensional description of the universe, your program will probably need to read portions of star, cloud, star cluster or galaxy catalogs, similar to a planetarium software.

The general structure of the universe, and its contents may be summarized the following way:

The observable universe has a diameter in the order of 10s of billions of lightyears; it depends on which of the several notions of distance is chosen. For distances below a few hundred million lightyears the notions get similar.
The universe is about 13.8 billion years old, started as a hot and dense state (Big Bang), expands, and is almost 3-dimensional Euclidean plus one time dimension on large scales; that kind of space plus time is called a Minkowski space-time. Its average temperature is now near 2.7 K; we can detect a background of this temperature as cosmic microwave background (CMB); the CMB is thought to be the redshifted afterglow of its state as hot plasma at 380,000 years after the Big Bang.
The universe mainly consists (according to current models) of dark energy (about 70%); that's invisible; dark energy is thought to be equally distributed, although this is going to be investigated in more detail.
Most (about 63%) of the matter is cold (slow) and "dark", but not equally distributed. Dark matter is thought to form large halos around galaxies and galaxy clusters. It's invisible and interacts via gravity with ordinary matter. It can be inferred from gravitational lensing and the velocity distribution of stars within galaxies.
Much of the remaining matter consists of hot (fast) neutrinos; they are very hard to detect, but they can be detected, and they can be inferred from nuclear decay measurements.
Part of the energy of the universe consists of photons (e.g. radiowaves, light, x-rays, gamma rays, magnetic and electric fields).
A minor part (about 5%) of the universe consists of atoms and ions (simplified called baryonic matter); that's the kind of matter we know from our immediate experience.
About half of the baryonic matter in the universe is thought to be the intergalactic medium made of a rarified plasma, consisting of hydrogen atoms, protons, electrons, helium ions. It's thought to form huge filaments around voids of roughly 100 million light years diameter.
The filaments connect and embed galaxy superclusters. Our local galaxy group is part of the Virgo supercluster.
Galaxy superclusters are made of galaxy clusters (diameter up to about 30 million lightyears), like the Virgo cluster, and galaxy groups (diameter up to about 10 million lightyears), like the Local Group. Sometimes an intermediate grouping (between supercluster and cluster) like a galaxy cloud is defined.
Galaxy clusters contain up to about 1000 galaxies, galaxy groups less than 50. Galaxies may have sizes about 100,000 lightyears across, and consist (besides a halo of dark matter) mainly of interstellar gas and dust. After galaxy collisions this gas may be consumed by star formation or removed from the galaxy. A galaxy typically contains billions of stars. Interstellar gas of different temperature is distinguished.
Galaxies can contain stars across the Hertzsprung-Russel diagram, neutron stars, black holes, sometimes supernovae, and rougue planets, probably also smaller rocks ejected from planetary systems. Stars can be grouped to binary, ternary or larger systems of stars up to open and globular star clusters. Globular clusters are usually old, located in the halo of a galaxy, and contain up to about a million, mostly old stars, may be with a black hole (IMBH) in their center. Open clusters are mostly young and in the galactic disk. A couple of different galaxy types are distinguished, e.g. elliptical and spiral galaxies. Dense molecular clouds may be the location where stars form and start as open clusters.
The distance between stars is usually a few lightyears; in globular clusters it's well below a lightyear. In the galactic center (bulge) stars are also closer to each other. The diameter of a star can vary from about 10 kilometers (neutron stars) to several astronomical units (red supergiants), the mass from about 13 Jupiter masses for brown dwarfs (or by a stricter definition from 0.075 solar masses for red dwarfs) to  several 100 solar masses.
A galaxy center is mostly populated by a supermassive black hole (SMBH), with about 0.1% the mass of the galaxy, the mass of the SMBH can vary from galaxy to galaxy.
Galaxies can collide and temporarily form irregular galaxies.
Planetary systems around stars seem to be common; they are thought to form from protoplanetary disks shorty after the central star has formed. The outermost part of planetary systems may reach out half a lightyear, to the limits of the Hill sphere of the central star; it depends on the distance and mass of neighbouring stars. Planets can form outside the Roche limit of a star; small asteroids can survive closer to the central star for some time. A planetary system can contain planets, dwarf planets, moons, asteroids, trojans, comets, dust, magnetic fields, stellar wind.

That's certainly still incomplete, and most of the notions can be subdivided to much more detail.