Why is cosmological event horizon closer to us than the particle horizon?

The particle horizon marks the region from within which we may have received light. It started out at zero, because light from nowhere had had the time to reach us, and increased as time went by because light from increasingly larger distance reached us. Due to the expansion it will always increase in physical size, but due to the accelerated expansion, there is a limit to the region of the Universe that we will ever be able to receive light from. Today, the particle horizon is 46 Glyr (billion lightyears) away, and the limit — in the far future — is the part that today is 63 Glyr away (but in the far future will be much farther away).

The event horizon can be thought of as sort of the opposite. At any time $t$, it is the distance from which we will be able to receive light. If the Universe contained no dark energy, there would be no limit to how distant a galaxy could be and we would still be able to see it — just wait long enough. However, the accelerated expansion means that an increasingly smaller part of the Universe will be able to emit light toward us with us receiving it, before the expansion carries the photons away faster than they can travel toward us.

At Big Bang, the event horizon consisted of particles that today are 63 Glyr away — that's what I meant by "the opposite of the particle horizon".

The following figure from Davis & Lineweaver (2004) may help explain. It show basically three versions of the same thing, namely how the sizes of various horizons evolve with time. The top panel shows time as a function of physical size, while in the middle panel, the $x$ axis show comoving coordinates, i.e. the coordinates that expand along with the Universe. The lower panel shows not time on the $y$ axis, but conformal time, which can be thought of as the time it would take a photon to travel a distance if we "froze" the Universe.


Red Moon a Characteristic of all Total Lunar Eclipses?

No, not all total lunar eclipses will turn the Moon deep red. Most of them do, but not all.

If you were standing on the Moon during the eclipse, you'd see the Earth passing in front of, and obscuring, the Sun. But the Earth will never become fully dark, even when the Sun is fully covered. A bright ring will always surround the Earth. Why?

That ring is sunlight refracted by the atmosphere. It's there because the Earth has an atmosphere. You could say it's all the sunrises and sunsets of the Earth, all seen at once. It's this light that continues to illumine the Moon during the eclipse.

But why is the Moon red, instead of some other color? This is because the blue end of the spectrum is scattered more easily in all directions (same mechanism that explains why the sky is blue on Earth), whereas the red part of the spectrum is scattered less easily and moves on a straighter path along the refraction lines. The bright ring around the Earth, as seen from the Moon, is probably red, because most of the blue in it has been scattered away.

Now, if the Earth's atmosphere is full of dust particles from huge volcanic eruptions, the bright ring is a lot weaker. That makes the Moon during the eclipse a much darker shade of red. Sometimes the Moon is a very dark, dull grey during the eclipse, no red hue at all - so dark in fact that it's hard to see in the sky while the eclipse is full. This has happened some decades ago after the Pinatubo eruption.

http://en.wikipedia.org/wiki/Lunar_eclipse#Appearance

http://eclipse.gsfc.nasa.gov/LEcat5/appearance.html

EDIT:

A measure of the brightness and color of a lunar eclipse is the Danjon scale. Eclipses are rated between 0 (almost invisible, black or very dark grey) and 4 (bright orange with bluish rim).

http://en.wikipedia.org/wiki/Danjon_scale

http://eclipse.gsfc.nasa.gov/OH/Danjon.html

My estimate is that the last eclipse was a 3.

I can't find a list of recent eclipses rated on a Danjon scale, but here's a list of 20th century eclipses, with a measure of the magnitude of the umbra.

http://eclipse.gsfc.nasa.gov/LEcat5/LE-1999--1900.html

As you can see, the magnitude varies quite a bit.

Bottom line: each eclipse is a bit different. A majority would have some kind of orange, copper or red tint. A minority are too dark to see any color. Unusual colors (outside of the red-yellow interval) are very rarely possible. Brightness and hue varies from one instance to another, since it depends on Earth's atmosphere, which is a system that changes greatly over time.


Calculating 2004 BL86 position using software

I'm going to describe the steps that I followed to show 2004 BL86 in Stellarium in my notebook:

Open the Configuration Window by pressing F2
Select the Plugins tab and from the left list select Solar System Editor
Click in the configure button at the bottom
Go to Solar System tab in the opened window
Click the button Import orbital elements in MPC format...
Click in the drop-down menu called Select bookmark... and select MPCORB: near-Earth asteroids (NEAs)
Click the button Get orbital elements
In the search field, type 2004 BL86 and select it once it appears below
Click the last button called Add object
Close the windows and then press F3 to open the Search window
Start typing 2004 BL86 and once it is showed below, click in the magnifying glass to the right, that will get you to the position of 2004 BL86

Please let me know if that answers your question. I'm using Stellarium version 0.13.1... and sorry if my English is not very precise :)


Does gravity repel when dark energy is involved?

In addition to not knowing, there's other reasons to suspect gravity wouldn't reverse.   There's no known force that reverses over distance (OK, the strong force, kinda/maybe but over very short distances).  Also, the expansion of space explains things that we seem to observe or suspect occurred like galaxies moving away from us faster than the speed of light and the rapid expansion that's suspected to have happened very shortly after the big bang.   Gravity can't do either of those things.

There's also the problem of how would gravity know that things are so far apart that they should start repelling, where nearby galaxies are still too close but more distant galaxies the repulsion kicks in?   It doesn't make a whole lot of sense that gravity would account for both attraction and expansion of space.

That's not to say it's impossible.  We don't know how gravity works or what it is, so, maybe.  Never say never, until something is understood.

@eshaya's answer, "We don't know" is probably the more straight forward answer to this question, but I wanted to point out some problems with repulsive gravity theory as well.

In addition (and I don't understand precisely why), but the gravitational waves discovery has ruled out some theoretical gravity/dark energy proposals.


The difference is based on the different efficiency of the processes.
We can describe the luminosity by:

$$L = \eta m c^2$$

where $\eta$ is the conversion efficiency, and describes how much matter can be converted into luminosity (photons).
Main sequence stars (if you mean this by "normal operation") extract energy from matter by nuclear fusion.
The efficiency of nuclear fusion in stars is of the order of $\eta = 1\%$.

The other competitive mechanism, much more efficient than nuclear fusion, is conversion from gravitational potential.
In core-collapse SNe, this is the mechanism that brings to the explosion.
Gravitational potential energy is defined as $U=-\frac{GMm}{r}$ where m is the mass of a particle, at a distance r, from a large object of mass M; G is the gravitation constant.

If you consider the SN scenario, $r$ is the distance of a particle that is falling from the external layer towards the center (center at $r=0$). So the particle goes from $r_1$ to $r_2$ with $r_2 < r_1$. Then the system lost part of the potential energy, converted to kinetic energy.
All the external matter collapse in a very brief time (order of seconds), from the external shell to the compressed nucleus, bringing more compression and then generating energetic shock in the explosion.

The virial theorem says that half the change in gravitational energy stays with the star, while the other half is emitted as luminosity.
The efficiency coefficient for gravitational conversion is $\eta\sim10\%$, i.e., 10 times larger than the nuclear fusion efficiency.

From this reference, you can check that the ratio between the invisible against visible emission is of the order $\sim10^7$ (I don't know what kind of reference is that, but the authors are totally trusty).

Something more here and here.


Does the gravitational attraction near the surface of dense celestial objects diverge from inverse square?

Leckner's paper deals with the effect of induced polarization on the spheres. Electrons are redistributed, making the force different from what one would expect. The gravitational counterpart is tidal distortion: since the gravitational field is non-radial when you have two heavy masses close to each other, matter will move to make the surface an equipotential surface. This means that the gravitational acceleration at the surfaces will not be constant at all locations.

Doing an analytic solution of how two ellipsoids attract each other and deform seems to be tractable (e.g. see this question) but algebraically very tedious and likely involves lots of special functions. See the addendum below for an approximate numeric model.

Black holes produce another complication: since spacetime nearby is curved and expanded the meaning of the distance in the inverse square law becomes problematic. The Paczyński–Wiita potential is an approximation of the potential, and it deviates from the $U=-GM/r$ as $U_{PW}=-GM/(r-R_S)$ (where $R_S$  is the Schwarzschild radius). It makes the force increase faster than the classical potential as we approach $r=R_S$.

Addendum: I did a numeric exploration of the force between two ellipsoidal, self-gravitating masses with centres of mass separated by a given distance. To find the shape I started with spheres and adjusted the semi-major axis (while preserving volume) so the potential along the surface became more equal at the poles. After a few iterations this gives a self-consistent shape. Then I calculated the force (the derivative of the potential) due to this shape on the other mass.

The result is indeed that the force increases faster than $1/r^2$
as the bodies approach each other, since they elongate and eventually merge (a bit before this they will deviate from my ellipsoidal assumption). If one multiplies the force by the squared distance the product should be constant for pure $1/r^2$ forces, but it starts to increase as they approach enough. Note that this is a non-rotating model: with rotation the numbers will change and the ellipsoids will become tri-axial, but I suspect the qualitative behaviour stays the same.


Saturn observation : how to calculate requirements for Parallax method

Are you treating $\Theta$ as the only unknown, i.e.  you believe you know the baseline distance exactly?    If so, then  you compare your angular sighting of a distant star vs. Saturn with the same for the other observer.    The devil is in the error analysis.  Your two angles will be absurdly close to each other in magnitude, which means a very small error/uncertainty in value leads to a large error in the calculated distance $D$ to Saturn.

You should find some first-year-level text on measurement & statistical error analysis to get a better feel for the difficulty here.


Hamiltonian of general three body problem

The condition $$\sum_{i=1}^{3}m_iq_i'=0$$
of this paper generalizes to
$$\sum_{i=1}^{n}m_iq_i'=0,$$
meaning the barycenter should be at the origin of the frame.

Equation (1) generalizes to
$$H=\frac{1}{2}\sum\frac{|p_i'|^2}{m_i}-\sum_{i=1}^n\sum_{j=1, j\neq i}^n\frac{m_im_j}{|q_i'-q_j'|}.$$
The first part is the total kinetic energy of the system relative to the barycenter, written in terms of momenta: $$E=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{p}{m}\right)^2=\frac{1}{2}\frac{p^2}{m}.$$
The second part is the sum of all self-potential energy between each pair of (spherical) objects, with Newton's gravitational constant $G$ set to 1.

See also n-body problem on Wikipedia.

The 3-dimensional 3-body (or n-body) problem is like the planar version. Just take 3-dimensional vectors for $p_i'$ and $q_i'$.


Direction of movement sun with respect to the NGP

Backer and Sramek 1999
observed that relative to more distant radio sources, the galactic center
Sgr A*
had a heliocentric proper motion
of about -6 milliarcseconds per year in galactic longitude.
Reid et al. 1999
reported a similar result.
Looking from NGP to SGP,
galactic longitude
increases counterclockwise, a decrease in longitude is a clockwise motion, and the Sun revolves clockwise around the galactic center.


What is the significance of the velocity resolution in spectral line observations using radio interferometry?

Short version: velocity resolution is the smallest velocity difference you can measure between two moving objects, using a given spectrum.
More details:
As you probably know (based on your implicit use of the formula in your question), we can measure velocities by using the Doppler shift.  To do that, we need to measure a feature (an absorption or emission line) in an object's spectrum, and then determine how much it is shifted from the known "rest" wavelength or frequency.
Now imagine that there are two gas clouds along the line of sight when you take a spectrum, both emitting light in some spectral line.  (This is pretty common when observing neutral hydrogen in the plane of the Milky Way, for example.)  If they have the same velocity, then their emission spectra will lie exactly on top of each other, and you won't be able to see that there are two clouds there.  If they have slightly different velocities, then the different Doppler shifts of the two lines will give them slightly different wavelengths.  You can measure the two different velocities if the shift between them is larger than the resolution of your spectrograph, i.e. if the finite width of spectral features imposed by your optics and detector doesn't smear them together too much.
You can probably see from this that a smaller velocity resolution is better.  A resolution of 1 km/s means that you can distinguish features as close as that, while 10 km/s means that a 1 km/s difference in two features would be undetectable.  It is similar to spatial resolution in this sense.  A spatial resolution of 1 arcsec is better than a resolution of 10 arcsec, since it means you can distinguish finer details.
Note that a given velocity resolution does not mean that that is the smallest Doppler shift velocity you can measure for a single object.  That is because you can generally find the central wavelength of a given spectral feature (which is what you need for a Doppler shift measurement) very precisely, even if the feature is broad.  For example, the spectrographs used to measure radial velocities in searches for extrasolar planets (e.g. HARPS) typically have spectral resolution on the order of 1 part in 100,000, i.e. the resolution is 1/100,000 of the observing wavelength.  If you put that in the Doppler shift formula, that translates to a velocity resolution of 1/100,000 of the speed of light, or about 3 km/s.  Yet those instruments routinely measure velocities of just a few meters per second, which is 1,000 times smaller than their nominal velocity resolution.
As a final technical note, in your question you calculated the velocity width of one channel in the spectrum, but the velocity resolution would be twice as large as this, because of the need to Nyquist sample the spectrum.  One way to think about that would be to imagine a series of spectral features that were separated by exactly the velocity width of one channel.  If that were the case, then every channel would have a feature, and your spectrum would just appear flat. (Even if you just had two features in consecutive channels, they would blend to appear as one.)  But if the features were separated by twice that velocity, then only every other channel would have a feature and you could see that there were distinct features there.  So the spectral resolution is always at least twice your channel width, but could be limited by other parts of your instrumentation to be larger than that.


Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at only the "simple" versions of these equations.

You say that the you've heard that the source of the two is because of receiving the signal and reflecting it, which is entirely true. Let's consider the simpler case of a police officer using radar to track a car's speed (just so we don't have to drag in the more complicated concepts of orbital motions of satellites, dispersion relations, etc.) You have a police officer standing by the side of the road with a radar gun, shooting radar waves of some frequency $f_o$ at the car as it recedes from him (note I'm going to use frequency but it is completely analogous to wavelength, assuming a simple dispersion relation). That wave travels to the car and the first peak of the wave hits the car. The car then has time to move slightly forward by the time the next peak hits the car. From the car's perspective, because of his relative motion to the police officer, the frequency of the wave hitting the car has been Doppler shifted to a new frequency, $f_1$. This new frequency is given by the Doppler equation

$$\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)$$

Note that this equation is the same as what you presented, except that it uses frequency instead of wavlength. So now, our car has been hit by these frequency shifted waves. It will now reflect those waves back at the police officer. The process of reflection effectively makes the car an emitter of the frequency $f_1$ but since it's obviously moving with respect to the police office, the police officer receives a Doppler shifted frequency, which I'll call $f_2$. Since the speed between the car and the police officer is still $v$, we know that the Doppler equation for the frequency shift of the reflection is given by

$$\frac{f_2}{f_1} = \left(1+\frac{v}{c}\right)$$

Now the police officer receives back a frequency $f_2$, but he originally emitted $f_o$, so he wants to know the velocity which caused the shift from $f_o$ to $f_2$. Fortunately, this can be determined by combining the two shifts above.

$$\frac{f_2}{f_o} = \frac{f_2}{f_1}\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)\left(1+\frac{v}{c}\right)$$

$$\frac{f_2}{f_o} = \left(1+\frac{v}{c}\right)^2 = 1 + 2\frac{v}{c} + \frac{v^2}{c^2}$$

At this stage, we make the assumption that $v^2/c^2 \ll v/c$ and this second order term can be safely ignored. We then wind up with

$$\frac{f_2-f_o}{f_o} = \frac{\Delta f}{f_o} = \frac{2v}{c}$$

What the police officer measures by using the radar gun and applying the Doppler equation is $2v$, but of course we know this to be twice the speed of the car, so of course to get the actual speed of the car, the speed measured by the radar gun must be divided by two. As you can hopefully see, this 2 comes from the double receive and reflect of the radio waves by the car. The same exact principle holds whether we're talking about radar guns and cars or satellites and planets.

Note: As I said at the start, thing's are vastly more complicated than I've presented. For example, we assumed that $v^2/c^2 \ll v/c$. In reality, to get a more correct answer, you wouldn't want to drop this term and you'd find that $v_r \ne 2v_p$ as you've indicated in your answer, and there's actually more to it than that. This is all further exacerbated by the fact that the doppler equation I started with already had some assumptions built in. The fully correct equation is more complicated than that.


I saw a bright orange light floating slowly across the sky ,occasionally slowing as if looking for something
it looked like a star, so was it a star,

For the sake of teaching some scientific reasoning skills, I'll point out that just because you think it looks like a star does not mean it is a star.
In fact the motion of the object suggests it was not a star.  Stars do not move except as part of a general motion of the night sky dues to Earth's rotation.  What you describe is not that.
When we try and work out what something is, we use all the evidence.
What would it be like ?  Perhaps a helicopter with lights.


How did Ole Christensen Romer measure the speed of light?

Ole Rømer did not measure a change in the frequency of light. He measured an apparent change in the orbital period of Io, one of Jupiter's moons.

The orbit of Io can be measured very accurately by observing when it enters or leaves the shadow of Jupiter. When the Earth is moving away from Jupiter, Rømer noted that the orbit of Io appeared to be very slightly longer than when the Earth was moving parallel to Jupiter.

Suppose the orbital period of Io is $p$ seconds. If you are not moving and you observe an eclipse at time $t$, the next eclipse will be at time $t+p$. However, if you are moving away from Io, the next eclipse will be seen at $t+p+x$, where $x$ is the time that light takes to travel from your position at time $t$ to your position $p$ seconds later.

The delay over a single orbit was too small for him to measure (about 30 seconds). But the delay was cumulative, and over about 30 orbits of Io, the orbit was delayed by about a quarter of an hour. He noted:

  [This] appears to be due to light taking some time to reach us from the satellite; light seems to take about ten to eleven minutes [to cross] a distance equal to the half-diameter of the terrestrial orbit.

Of course, Rømer didn't know the actual size of the half-diameter of the terrestrial orbit (the astronomical unit) However putting the modern value of about 150 million km this gives a value of the speed of light at about 220000km/s, which is not so far from the modern value of about 300000km/s


Determine distance of asteroid based on two observations from different locations

Draw a diagram.
You get a quadrilateral EACB.  You should be able to determine the four angles in the quadrilateral.
Draw the diagonal EC, which is the length to be determined.
You know that ACE + BCE = ACB  (the known angle at the asteroid)
And you can use the law of sines twice, once in each of the triangles EAC and EBC
These should give you three equations, with three unknowns:  length EC,  and angles ACE and BCE.  Given three equations and three unknowns, a solution should be possible. You should find the distance is a little more than the Earth-moon distance.
Note that angles at the centre made by the diagonal: AEC and BEC are not 46° 27' and  22° 34', but the sum of those angles is the sum of 46° 27' + 22° 34'


Can we (theoretically) spin the black hole so strong that it will be broken apart by centrifugal force?

Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?

For a Kerr-Newman (rotating, charged, isolated) black hole of mass $M$, angular momentum $J$, and charge $Q$, the surface area of the event horizon is given by
$$A = 8M\left[M^2 + (M^2-a^2-Q^2)^{1/2} - Q^2/2\right]\text{,}$$
where $a = J/M$. An extremal black hole occurs when $M^2 = a^2 + Q^2$. Beyond that, if the black hole is even more overspun or overcharged, is an "overextremal" Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.

Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.

Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any completely general proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.

For example, the Hawking temperature of the black hole is $T_\text{H} = \kappa/2\pi$, where
$$\kappa = \frac{\sqrt{M^2-a^2-Q^2}}{2M\left(M+\sqrt{M^2-a^2-Q^2}\right)-Q^2}$$
is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.


The radial velocity curve of a star in a binary system (with another star or a planet) is defined through 6 free parameters
$$V_r(t) = K\left(\cos(\omega + \nu) +e \cos \omega \right) + \gamma,$$
where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity, $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $\nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $\tau$, the orbital period $p$ and the eccentricity $e$.

To proceed you estimate what all these parameters are - i.e. an initial guess.

Then, for each time $t_i$ of a data point in your RV curve you:

Calculate the mean anomaly
$$M(t) = \frac{2\pi}{p}(t - \tau),$$
Solve "Kepler's equation"
$$M(t) = E(t) - e \sin E(t)$$
numerically (its a transcendental equation, you could use Newton-Raphson or similar) to give $M(t_i)$, the eccentric anomaly.
Use
$$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$
to calculate the true anomaly $\nu(t_i)$.
Calculate $V_r(t_i)$

You then calculate some figure of merit (e.g. chi-squared) for how closely the model and data agree and go through an iterative process to adjust the parameters and optimise the fit of model to data.

A more sophisticated discussion can be found in this paper by Beauge et al.

If you have the RV curves of both stars, then you can fit them both simultaneously. Obviously, they have $p$, $e$, $\gamma$ and $\omega$ in common, but their RV amplitudes $K_1$ and $K_2$ will be different. The ratio of $K_1/K_2$ gives you the ratio of the two stellar masses.

If you only have one RV curve you are limited to estimating the mass function of the binary system.
$$ \frac{M_2^{3} \sin^{3} i}{(M_1 + M_2)^2} = \frac{p K_1^{3}}{2\pi G},$$
where $i$ is the inclination of the orbit with respect to the line of sight.
This can only give you a lower limit to $M_2$ unless $i$ is known.

Taking your specific case study. If you know $M_1$ and $i$ (this could be the case for a transiting exoplanet, or maybe a binary featuring an eclipsed black hole candidate), then the primary radial velocity curve gives you $K_1$ and hence $M_2$. If the masses and $p$ are known then Kepler's laws give the orbital separation.

There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is Systemic Console.

There is no fundamental difference between analysing the RV curves of stars with exoplanets and stars with unseen (stellar) companions.


Can a rogue planet neither orbit around a star nor a galaxy?

They would be enormously difficult to observe, but with a big enough gravitational assist a rogue planet certainly could be ejected from a galaxy and become a "double rogue" so to speak, or intergalactic rogue planet is probably a better term.  Rogue (or intergalactic) stars have been observed.  Here and Here.  The same gravitational forces that create intergalactic stars absolutely should create inter-galactic rogue planets and in fairly significant numbers.  The merging of 2 galaxies should create millions of them.


Were K2-141b and WASP-63b selected as the first two exoplanet atmospheres for JWST to look at during its first year of observations? If so, why?

The list of JWST approved programs for Cycle 1 (12 months long, starting after the 6 month long commissioning phase following launch) is available here. If you unfold the Exoplanets and Disks science category, you will see the programs, along with the links to proposal abstracts and target lists.
Looking at the proposal titles only, the exoplanet targets for atmospheric studies  include:

HD 189733b (2 proposals)
WASP-121b
Gliese 486b
GJ1214b
TOI-421b
55 Cnc e
HD 80606b
WASP-166b
K2-141b (2 proposals)
WASP-127b
Kepler 51d

The Espinoza et al. proposal is proposal number 2159 is one of two to study K2-141b. So K2-141b and WASP-63b are just two of the exoplanets that will be studied by JWST in the first cycle, with others likely in following cycles.
In the overall summary of the JWST Cycle 1 proposal process that was made available to the community, exoplanets accounted for 928 out of the 3500 hours allocated to small programs (slide 7). The exoplanets to be studied are chosen by the individual investigation teams that submit the proposals based on what particular aspects of the atmosphere they are trying to study. These proposals are evaluated by the various exoplanet subpanels of the time allocation committees and by Space Telescope Science Institute (STScI; who operate JWST and HST) staff for scientific merit, technical feasibility, capabilities of the proposing team and optimal use of JWST, a process which is common to almost all competed observing facilities. (The science panels are supposed to only consider the scientific merit, the technical feasibility is handled by STScI staff; this process has been improved by STScI for recent HST and all JWST Cycles by employing Dual Anonymous Peer Review where the panel doesn't know the identity of the proposers and the proposers don't know the membership of the review panels)


How does NASA's ASTHROS stratospheric telescope compare to its James Webb space telescope?

I found this link that might seem interesting : https://www.nasa.gov/feature/jpl/nasa-mission-will-study-the-cosmos-with-a-stratospheric-balloon
The article in the link doesn't clearly specify the actual instruments or their functionality, but as this is an official article, I thought it might be useful.
The only info that NASA has officially released about ASTHROS is the cooling systems :

Because far-infrared instruments need to be kept very cold, many missions carry liquid helium to cool them. ASTHROS will instead rely on a cryocooler, which uses electricity (supplied by ASTHROS' solar panels) to keep the superconducting detectors close to minus 451.3 degrees Fahrenheit (minus 268.5 degrees Celsius) — a little above absolute zero, the coldest temperature matter can reach. The cryocooler weighs much less than the large liquid helium container that ASTHROS would need to keep its instrument cold for the entire mission. That means the payload is considerably lighter and the mission's lifetime is no longer limited by how much liquid helium is on board.

Here is a link about the JWST : https://en.wikipedia.org/wiki/James_Webb_Space_Telescope
This link also does not specify the difference between JWST and ASTHROS, but it compares the JWST with the Hubble Space Telescope. Here are some comparions:-

The JWST has an expected mass about half of Hubble Space Telescope's, but its primary mirror, a 6.5 meter diameter gold-coated beryllium reflector will have a collecting area over six times as large, 25.4 square metres (273 sq ft), using 18 hexagon mirrors with 0.9 square metres (9.7 sq ft) obscuration for the secondary support struts.[22]

The JWST is oriented toward near-infrared astronomy, but can also see orange and red visible light, as well as the mid-infrared region, depending on the instrument. The design emphasizes the near to mid-infrared for three main reasons: high-redshift objects have their visible emissions shifted into the infrared, cold objects such as debris disks and planets emit most strongly in the infrared, and this band is difficult to study from the ground or by existing space telescopes such as Hubble. Ground-based telescopes must look through the atmosphere, which is opaque in many infrared bands (see figure of atmospheric absorption). Even where the atmosphere is transparent, many of the target chemical compounds, such as water, carbon dioxide, and methane, also exist in the Earth's atmosphere, vastly complicating analysis. Existing space telescopes such as Hubble cannot study these bands since their mirrors are insufficiently cool (the Hubble mirror is maintained at about 15 °C or 288 K) thus the telescope itself radiates strongly in the infrared bands.

The article goes on to compare the JWST with other telescopes (mainly the cooling systems, aperture and wavelengths) so you can compare the currently being planned JWST telescope with other already launched ones like IRT, ISO and Spitzer.
The above WIKI article is amazingly detailed and may help.
The article linked above contains the wavelength range of ASTHROS as follows :

Managed by NASA's Jet Propulsion Laboratory, ASTHROS observes far-infrared light, or light with wavelengths much longer than what is visible to the human eye. To do that, ASTHROS will need to reach an altitude of about 130,000 feet (24.6 miles, or 40 kilometers) – roughly four times higher than commercial airliners fly. Though still well below the boundary of space (about 62 miles, or 100 kilometers, above Earth's surface), it will be high enough to observe light wavelengths blocked by Earth's atmosphere.

It also defines the targets for ASTHROS, listed as follows :

ASTHROS will make the first detailed 3D maps of the density, speed, and motion of gas in these regions to see how the newborn giants influence their placental material. By doing so, the team hopes to gain insight into how stellar feedback works and to provide new information to refine computer simulations of galaxy evolution.

It will also for the first time detect and map the presence of two specific types of nitrogen ions

A third target for ASTHROS will be the galaxy Messier 83. Observing signs of stellar feedback there will enable the ASTHROS team to gain deeper insight into its effect on different types of galaxies. "I think it's understood that stellar feedback is the main regulator of star formation throughout the universe's history," said JPL scientist Jorge Pineda, principal investigator of ASTHROS. "Computer simulations of galaxy evolution still can't quite replicate the reality that we see out in the cosmos. The nitrogen mapping that we'll do with ASTHROS has never been done before, and it will be exciting to see how that information helps make those models more accurate."

Finally, as its fourth target, ASTHROS will observe TW Hydrae, a young star surrounded by a wide disk of dust and gas where planets may be forming. With its unique capabilities, ASTHROS will measure the total mass of this protoplanetary disk and show how this mass is distributed throughout. These observations could potentially reveal places where the dust is clumping together to form planets. Learning more about protoplanetary disks could help astronomers understand how different types of planets form in young solar systems.

Wavelength Range of JWST explained as :

The JWST will observe in a lower frequency range, from long-wavelength visible light through mid-infrared (0.6 to 28.3 μm), which will allow it to observe high redshift objects that are too old and too distant for the Hubble to observe.[8][9] The telescope must be kept very cold in order to observe in the infrared without interference, so it will be deployed in space near the Earth–Sun L2 Lagrangian point, and a large sunshield made of silicon-coated and aluminium-coated Kapton will keep its mirror and instruments below 50 K (−220 °C; −370 °F).

Also, here are some targets of the JWST : https://www.nasa.gov/feature/goddard/2017/icy-moons-galaxy-clusters-and-distant-worlds-among-selected-targets-for-james-webb-space


Is the longitude assigned to sunspots spherical or projected?

The transformation from projected geometry to heliographic coordinates is straightforward.
As Peter Meadows explains, this can be done either mathematically, or graphically using a latitude/longitude grid template.
Since the solar equator is inclined about 7° to the ecliptic, different Stonyhurst disks are used at different times of year.

W T Thompson 2006 section 2 describes the two main heliographic coordinate systems.
Latitude in both systems is measured from the solar equator.
The central meridian of longitude in Stonyhurst coordinates is fixed facing the Earth.
The Carrington coordinate system rotates rigidly with a sidereal period of 25.38 days, approximating solar rotation at a typical sunspot latitude of ±26°.

The NOAA Space Weather Prediction Center's daily Solar Region Summary provides spherical coordinates in both forms.
Here are excerpts of two reports four days apart:

SRS Number 129 Issued at 0030Z on 09 May 2019
Report compiled from data received at SWO on 08 May
I.  Regions with Sunspots.  Locations Valid at 08/2400Z
Nmbr Location  Lo  Area  Z   LL   NN Mag Type
2740 N08E14   307  0160 Cao  04   03 Beta
2741 N05E52   269  0160 Hsx  03   01 Alpha

SRS Number 133 Issued at 0030Z on 13 May 2019
Report compiled from data received at SWO on 12 May
I.  Regions with Sunspots.  Locations Valid at 12/2400Z
Nmbr Location  Lo  Area  Z   LL   NN Mag Type
2740 N08W40   308  0070 Hax  03   02 Alpha
2741 N06W02   270  0210 Hsx  03   02 Alpha

According to their README,
Location is a latitude and longitude relative to the central meridian, and Lo is a Carrington longitude.
A typical sunspot region moves about 13°/day from east to west in Stonyhurst longitude and less than ±1°/day in Carrington longitude.

GOES X-ray event report columns 29-34 also appear to be Stonyhurst spherical coordinates.
For example, sunspot region 9042 was associated with several X-ray events in June 2000.
If you filter the report by that region number, you can see that its progress from E90 to W90 longitude is roughly linear in time.
In projection, the plot of east-west position vs. time would be sinusoidal, flattening at the limbs.


Generating random galaxy catalogs for correlation functions
... the random catalog can be just be a uniform distribution of points in the same volume as the real data.
However, real-world galaxy catalogs are not homogenous boxes, and have very irregular shapes and observationally induced redshift and angular distributions. Even some mock-catalogs are endowed with these properties to better represent real data. In these cases, how is a random catalog best produced? ...
If the answer is that some nonuniform distribution should be used, is there a usual method for doing so? The literature I have found (e.g. for the SDSS published data, and for galaxy mocks mimicking SDSS's constraints) does not discuss this in any way. Note, if this is a highly involved process beyond a brief summary and a source or two, I do not expect a full answer here. Rather, I am trying to determine if there is an existing set of standard methods which are documented but have eluded me thus far.

For SDSS see:

"Large scale correlations in galaxy clustering from the two degree field galaxy redshift survey" (2006), by N. L. Vasilyev, Yu. V. Baryshev and F. Sylos Labini.

"The SDSS Galaxy Angular Two-Point Correlation Function" (12 Mar 2013), by Y. Wang, R. J. Brunner, J. C. Dolence. The include their SQL query to the SDSS catalog archive server:

1 INTRODUCTION
One of the most powerful and simplest probes of the galaxy distribution is the two-point angular correlation function, which quantifies the excess probability above a random distribution of finding one galaxy within a specified angle of another galaxy. For the case of a Gaussian random field, the two-point angular correlation function and its Legendre transform pair provide a complete statistical characterization of the galaxy clustering (see, e.g., Peebles 1980). Even for the case of non-Gaussianity, the two-point angular correlation function provides a simple and important statistical test of galaxy formation models (Tegmark et al. 2004). The two-point angular correlation function has been studied at bright magnitudes from the data releases from the Sloan Digital Sky Survey (SDSS) such as the Early Data Release (EDR; Connolly et al. 2002).
...
To accurately calculate the galaxy two-point angular correlation function, we must first minimize potential systematic effects in the galaxy catalog used to measure the correlation function. The systematics of the SDSS EDR were thoroughly studied by Scranton et al. (2002). To minimize the systematic effects of seeing and Galactic extinction, they determined that the SDSS EDR galaxy sample had to be masked to exclude regions with seeing greater than 1."75 and reddening > 0.2 magnitudes. Given the importance of minimizing the impact of systematic effects on the galaxy two-point angular correlation function and the significant changes that were made in the SDSS data processing pipeline between the SDSS EDR and the SDSS DR7, we have repeated many of the tests presented in Scranton et al. (2002) by using the SDSS DR7 data. In this paper we present the methods used to contain these systematic effects, the results of these systematic tests, the actual galaxy two-point angular correlation function for the SDSS DR7, and our massively parallel implementation that rapidly calculates correlation functions for large data sets.

Also: "The galaxy correlation function and power spectrum" (.PDF).
Wikipedia's "Correlation function (astronomy)".
CalTech: "Measurements of Clustering" (Use [Next] or [Contents] buttons to view all the pages).
There are fixups applied to the raw data to correct for redshift and angular diameter distance.

For example catalogs and a description of the procedures see Shaun Cole's website: "Mock Galaxy Redshift Catalogues" where software and full details of the N-body simulations and the contruction and properties of the mock catalogues is provided for both 2dF Galaxy Redshift Survey (2dFGRS)  and SDSS galaxy redshift surveys.
Halo bias relation to dark matter and N-body simulations are discussed in "BAM: Bias Assignment Method to generate mock catalogs" (15 June 2018), by A. Balaguera-Antolínez, Francisco-Shu Kitaura, Marcos Pellerejo-Ibañez, Cheng Zhao, and Tom Abel.
Also check out the (under construction, and too busy working and writing papers) website: "The Carnegie-Irvine Galaxy Survey (CGS)" - with missing papers available on arXix:
"The Carnegie-Irvine Galaxy Survey. I. Overview and Atlas of Optical Images"
"The Carnegie-Irvine Galaxy Survey. II. Isophotal Analysis"
"The Carnegie-Irvine Galaxy Survey. III. The Three-Component Structure of Nearby Elliptical Galaxies"
"The Carnegie-Irvine Galaxy Survey. IV. A Method to Determine the Average Mass Ratio of Mergers That Built Massive Elliptical Galaxies"
"The Carnegie-Irvine Galaxy Survey. V. Statistical study of bars and buckled bars"
"The Carnegie-Irvine Galaxy Survey. VI. Quantifying Spiral Structure"
"The Carnegie-Irvine Galaxy Survey. VII. Constraints on the Origin of S0 Galaxies from Their Photometric Structure"


How bright was Scholz when is passed near the Sun 70,000 years ago?

I just found the answer to my question at this place:
http://astronomynow.com/2015/02/18/suns-close-encounter-with-scholzsstar/

Here is the answer:

  Currently, Scholz’s Star is a small, dim red dwarf in the
  constellation of Monoceros, about 20 light-years away. However, at the
  closest point in its flyby of the solar system, Scholz’s Star would
  have been a 10th magnitude star — about 50 times fainter than can
  normally be seen with the naked eye at night. It is magnetically
  active, however, which can cause stars to “flare” and briefly become
  thousands of times brighter. So it is possible that Scholz’s Star may
  have been visible to the naked eye by our ancestors 70,000 years ago
  for minutes or hours at a time during rare flaring events. The star is
  part of a binary star system: a low-mass red dwarf star (with mass
  about 8% that of the Sun) and a “brown dwarf” companion (with mass
  about 6% that of the Sun). Brown dwarfs are considered “failed stars;”
  their masses are too low to fuse hydrogen in their cores like a
  “star,” but they are still much more massive than gas giant planets
  like Jupiter.


Why N$_2$ is a non-absorbing species in the spectrum of the Earth?

As your question is based on the plot you posted, I suggest you to look for a lower wavelength range of the atmospheric electromagnetic absorption. A quick search in google gave me this paper, which says:

The importance of molecular nitrogen as the most abundant species in the Earth's atmosphere is evident. The strong absorption bands in the range 80–100 nm shield the Earth's surface from theextreme ultraviolet (XUV) part of the solar radiation.

As you can see, the absorption bands of diatomic nitrogen resides in the <1000 nm, which is in the extreme left of your plot, shadowed by its scale (1 micrometer = 1000 nanometers).


What have we learned from the observation of most distant galaxies

Nearby galaxies are seen in their old age. Distant galaxies are, on average, the same age as the local ones. But due to the finite speed of light, the farther you look, the younger you observe them. That means that the most distant galaxies are observed in their infancy, and they look much different from the old ones.
Galaxy evolution
This is one reason to observe galaxies at all distances: We get a statistical picture of the galaxies through different epochs, and hence probe their evolution. From such observations, we see e.g. how they grow with time, how their star formation increase in the early Universe until the Universe was roughly four billion years (Gyr) old and the decreased (Madau et al. 1996), and how the metals and dust built up over time.
Galaxy formation
Observing the most distant ones is not just a matter of breaking a previous record in distance, or time. When we look at the most distant (spectroscopically confirmed) galaxy to date, GN-z11, we see 13.4 Gyr back in time. What's the big deal if we find another one that is 13.5 Gyr back in time? Well, since the Universe at that time was only a few hundred million years (Myr) old, that would give us quite good constraints on the minimum timescale of galaxy formation. Can a galaxy form in only 300 Myr or 200 Myr, or does it need 400 Myr?
Physical properties
As I said above, the most distant galaxies look vastly different from the local ones. For instance, disks hadn't formed yet, both because it takes some time to settle into a disk, and because galactic mergers were more common at that time, which disturbed the galactic dynamics. The physical properties of the interstellar medium were also different, e.g. because it was more metal-poor, since metals are formed by stars over time. Dust, which is made from metals, was also sparse. But how long time does it take for stars to form metals, and dust? Previously it was thought that most dust was produced in AGB stars which die peacefully and hence have good conditions for making dust (high density and cool gas). But these stars typically live for billions of years, and observations of the most distant galaxies showed us that they contained dust as well. So people started wondering if supernovae, which are the violent deaths of massive and hence short-lived stars, could do it, even though it was thought that the gas was much too hot too form dust right after the explosion, and much too dilute after it had cooled down. But indeed it seems they can (e.g. Kozasa et al. 2009).


What is the most appropriate way to estimate the helium composition of a star?

Nearly all the helium in the photosphere of the sun comes from the helium in the interstellar gas that collapsed to form the sun. That helium was produced shortly after the Big Bang (in about the first 20 minutes) while the universe was hot and dense enough for hydrogen to fuse to helium.  That produces a universe in which ordinary matter is about 25% helium and 75% hydrogen. (by mass) Over the eons, the gas is enriched in helium somewhat by previous stars to about 27% helium.

The outer layers of the sun still have this 25%-75% composition. Some of the helium has settled under gravity, reducing the composition of the photosphere to your 24.85% helium. The core is enriched in helium by fusion reactions. It isn't constant. The outer core is about 30% helium. The inner core has as much as 65% helium.  The average composition is about 28% helium, only slightly more than what it started with.

Proxima Centauri is fully convective, which means that helium from the core gets mixed up through the whole star. But it is such a dim candle that it has hardly produced any more helium than when it formed: about 27%.

So for any star, no matter how bright or dim (with a few exceptions) the helium composition is about 25%, or a little more.


Is there a term for asteroseismology as applied to giant planets?

Is there a specific term for this technique (asteroseismology) as applied to giant planets instead of stars?

No it seems there is not.

Instead, people just use normal literal constructions.

"Seismology of Giant Planets""
"Jovian Seismology" and "giant planet seismology" or (your planet here) seismology in general
"Planetary normal-mode seismology"

A good example of the Fourier imaging technique is SYMPA:

SYMPA, a dedicated instrument for Jovian seismology I. Principle and performance
SYMPA, a dedicated instrument for Jovian seismology
II. Real performance and first results

JOVIAL is next-gen implementation, also from Earth:

JOVIAL; Jovian Oscillations through radial Velocimetry ImAging observations at several Longitudes.
Jupiter’s interior: from gravimetry to seismology

note: after posting I realized that one of the linked papers is the same as in the OP"s question. I'll leave it here for completeness.


Do stars have "radio photospheres"? Are they different from their optical photospheres?

Stars certainly do have radio emission from their photospheres, since blackbodies emit at all wavelengths. But that’s not usually what is detected, because it’s so faint.  Doing some simple numbers, emission from a star like the Sun peaks at a wavelength of about 500 nm (= $5 \times 10^{-7}$ m), and the long-wavelength side of the Planck function, in the Rayleigh-Jeans approximation, goes as $\lambda^{-4}$, so at a wavelength of 5 cm (= $5 \times 10^{-2}$ m) the emission will be $10^{20}$ times fainter.  With nearby or very luminous stars it is possible to detect this in some cases (e.g. with ALMA which is both very sensitive at operates at somewhat shorter wavelengths), but more typically the radio emission detected is from other processes like synchrotron emission.
Whether the radio photosphere is at a substantially different radius depends on the opacity at that wavelength, but Reid & Menten (1996) Radio Photospheres of Red Giant Stars suggests that for some giants it may be a lot bigger than the optical photosphere.
From the abstract:

[...]These observations suggest that long period variables have a “radio photosphere” at about two stellar radii, where the stellar radius is defined by line-free regions at optical wavelengths. The radio photosphere is just inside the SiO maser shell, and the limited variability of the radio emission suggests that stellar shocks are mostly damped within this region. The density and temperature of the radio photosphere are estimated to be >~ 1012 cm-3 and $\approx$ 1400 K, respectively. For these physical conditions, free electrons, obtained predominantly from the ionization of potassium and sodium atoms, provide the opacity through interactions with neutral H atoms and H2 molecules.


Is there an optimum orbit for a Hipparcos/Gaia-like parallax observatory

There is several reasons why the L2 point is an optimum point for the satellite.

First you have a stable orbit with low gravitational perturbation and which is not too far away. So, you won't have a huge travel time. And the parallax is good for what the satellite is supposed to do.
Secondly in a Lagrange point you can easily point out the same point. In orbit, you will have some grey area.
Eventually, I think you are doing a mistake when you say : "Assuming no cost". In my opinion, if we are not sending a satellite like Gaia around another planet, it is because we want to avoid risks and sending the satellite, around Mars for example, adds a ton of risk.

That's why we use the L2 point to place a satellite in regards to all the parameters. But, you can choose another location if you want to maximize something in particular.


Can a planet's gravity rip out its moon's atmosphere?

It's probably rare that moons have atmospheres to begin with. Planets don't generate solar flares. Planets emit thermal radiation as a result of their heat and during formation and collision planets can get significantly hot.   Earth is believed to have glowed red hot following the theoretical giant impact and that heat actually may have helped give the Moon its lopsided crust.  See the Earthshine part of this article.  The Moon, which also likely glowed red hot after it formed, was unlikely to have ever had an atmosphere, but in theory, a hot planet could thermally strip a nearby moon if that moon had an atmosphere.

A planet with a surface temperature of a couple thousand degrees could heat the near side of its moon more than the star that they both orbit, perhaps giving off enough heat to cause the loss of the moon's atmosphere through thermal or Jeans escape which is basically a product of upper atmospheric temperature and escape velocity.  It might not even be that uncommon for moons to receive more heat from the planets they orbits than the star they both orbit when the planet is undergoing sufficient bomboardment.

A second way that a planet might strip a moon's atmosphere is if the moon was close to the planet's Roche limit.  The atmosphere is less gravitationally bound and even if the moon comfortably holds itself together outside the Roche limit, its atmosphere would extend miles above its surface and be less gravitationally bound where gradual atmospheric stripping would be possible.

Moons rarely have atmospheres to begin with, but ice moons at the right temperature can generate atmospheres through outgassing.   Pluto, for example, has a tenuous atmosphere from the thawing of frozen nitrogen and other ices on its surface and Pluto's low gravity means this atmosphere is very dispersed and it can extend quite far from the planet, far enough that Pluto and its moon Charon may exchange atmosphere (grain of salt warning with that article, to say they share an atmosphere is an interpretation that I'm not 100% comfortable with, but the article does make the point that an atmosphere can be tenuously held by a comparatively low gravity object like Pluto, and as a result, affected and even shared with a nearby object in a close orbit).

A third theoretical way a planet could strip a moon of its atmosphere would be through magnetism.   Jupiter's enormous magnetic field generates a radiation belt that extends beyond its inner moons, certainly beyond the inner 3 of its 4 Galilean moons.   The high temperature of the charged particles in a planet's radiation belt could strip a moon of its atmosphere in a similar way that the solar wind can strip planets of theirs.

You just asked about gravity and I gave you three possible scenarios; radiation belts and thermal radiation aren't gravity but they could theoretically strip a moon of its atmosphere.  The Roche limit scenario is essentially driven by gravity, so the answer is yes, it's possible.  It's not actually the gravity but the tidal forces, which is a product of the gravity that can strip the atmosphere of a moon close to its planet's roche limit.

Moons with atmospheres are probably pretty rare and moons that orbit close to their planet's Roche limits are also fairly rare.  So it might be a rare scenario where the two situations coincide.


What is a singularity?

A singularity is a point in the universe where matter is infinitely dense. The singularity is at the centre of the black hole, and is often hiding behind an event horizon. So basically, a singularity is a point in space where a lot of matter is squeezed together in a very small space. When you have a singularity matter does not go to infinity. Density is what goes to infinity. Density and matter are different things.

A singularity is often created in a supernova explosion. A supernova is where a star is caused when there is so much matter in the centre of a star (at the end of its life) that the star cannot support its own gravitational force, and it collapses in on itself and explodes.

The second way a supernova can happen is in a binary star system - where if there is a white dwarf, it steals matter from the other star and eventually it accumulates so much matter that it explodes.

A singularity is caused by the big stars, not small ones. Small stars that explode in a supernova create something called neutron stars.

So what is a black hole?

A black hole is a area in space with a gravitational field so strong not even light can escape. Black holes don't have the colour black, they don't have a colour because we can't see any light coming back from them. So no, light is not absorbed. Black holes can also eventually dissipate, due to a mechanism called Hawking Radiation

There are multiple types of black holes including:

Supermassive black holes (often found at the centre of galaxies)
Stellar black holes
Primordial black holes
Rotating (Kerr) black hole

Finally, there are 3 main areas inside a black hole.

This diagram can show us many things, for example, how black holes don't have a gravitational pull which goes across the whole universe.

On a final note, we can work out the area where even light cannot escape by using the Schwarzschild radius. Seen as you said this answer shouldn't include relativity, I won't include any maths either. The Schwarzschild radius, in summary, is where the event horizon is in a black hole.

Summary

Sorry for providing a awkwardly long answer, I was having trouble understanding what you didn't understand, and I thought that if I explained a singularity - then I'd surely help you, sorry if I've just said things you already know. I've provided some Wikipedia links if you want to have a look at some of the terms I left in italics. You might also want to look at black hole thermodynamics

Schwarzschild radius
Ergosphere
Hawking radiation
White dwarf
Black hole thermodynamics


How to determine the age of an old galaxy?

When the article speaks of the "age", what it means is that we are seeing that galaxy as it was a long time ago. We are seeing it as a very young galaxy. The galaxy is very far away, so light has taken a long time to reach us from the galaxy. The light from the galaxy is 11 billion years old. We measure the age of the galaxy by finding its distance from us.

To find the distance we use the Hubble Law, which states that the further away a galaxy is, the faster it moves away from us, due to the expansion of the universe. And we can measure the speed that the galaxy is moving away from us because when an object moves very fast, the light from that object becomes shifted towards longer wavelengths (red-shifted). By measuring light that is known to have a fixed wavelength when not red shifted, we can find the speed of the galaxy accurately, and then use the Hubble Law to find the distance, and hence the age of the galaxy. (source)

A1689B11 has a redshift z=2.54 which corresponds to a light travel time of 11.1 billion years

Gravitational lensing makes the galaxy brighter, so it can be observed at these immense distances, it isn't directly used for finding the distance, but without gravitational lensing, the galaxy wouldn't be visible.


Estimation of average rock and asteroid mass associated with different stars

The composition of the gas from which stars and their planetary systems form is reasonably well known. About 1-2% of this gas is in the form of chemical elements heavier than Helium (the so-called metallicity of the gas).

A fraction of these "metals" - the iron, silicon, oxygen etc. is capable of forming dust and then accumulating to form "rocky" material.

So to answer your last question first, given the available reservoir it seems most unlikely that even 1% of the baryonic (normal) matter in the universe could be rocks, even if none of it were gathered into luminous stars and galaxies.

Forming stars are surrounded by circumstellar material from which planets and other rocks form. Observations of young stars suggest these discs can be as massive as 10% of the stellar mass, but more usually 1% or less. This means that as a fraction of the whole formed star system the rocks will be at most 1% of 10% (ie 0.1%) of the stellar mass. For a Sun-like star that means there's less than 330 Earth masses of rocky material around it.


How are telescopes protected from blinding light and satellite glares?

Random bright objects can occasionally appear in astronomical images. Examples include asteroids and satellites. However, these objects are not moving at the sidereal rate across the sky, and thus unless the telescope is set to track them, they leave a trail across the image.

This is unlikely to cause damage and CCD detectors are fairly robust to saturation of exposures in any case. However, it may ruin the image.


Are you sure it's true that all hydrogen will eventually be processed?   I can see that star formation rate should eventually slow to a point that there would be essentially no more light being created, but that's not the same thing.  Right now, most of the hydrogen in the universe is in the space within clusters of galaxies, and some of it falls into galaxies and some gets ejected, so you might imagine that all of that will eventually get processed.  But there is also hydrogen in between galaxy clusters, in the "cosmic web", and since the universe is expanding quite rapidly, it seems quite possible to me that much or most of that gas will never fall into a galaxy, but instead just end up stranded in between the expanding clusters.

What's more, our models now include the "cosmological principle" that treats the universe as homogeneous, but that's just the universe we can see.   We have good evidence that it extends way beyond what we can see, so we have no way to know (or even to expect) that the homogeneity is a truly global property of the entire universe (i.e., not just the "observable" universe).  On scales we cannot probe, the structure might be fractal, meaning that there could be pockets somewhere doing very different things, and there might always be stars forming somewhere.  Another possibility is that there could be a "landscape" of universes, which is hard to say if they count as the same universe or completely different ones, but it does mean there could be in existence hydrogen elsewhere in the "landscape" even after our universe has none.   So it's a tricky question to answer if "all the hydrogen" will eventually be processed, but I suppose you could look at all the hydrogen in the observable universe-- but you still have to wonder if the gas between the clusters ever gets into a galaxy.


If the planets only attract, then how can they stay in their orbit?

Gravity does not repel, but there is another "force" at work here. The angular momentum of the original protoplanetary disk is conserved. In other words, just because the cloud condensed into planets doesn't mean it is going to stop spinning.

When we orbit artificial objects, we speed them up until they reach a velocity that sustains orbit--so again there is another force at work, just not gravity. Gravity keeps objects in orbit (rather than letting them fly off into space). Imagine if you spin a yo-yo around in the air. The tension of the string is like gravity, keeping the yo-yo in orbit, but if you let go of the string then the other "force" (its momentum) will cause it to fly off.

(Technically, momentum is not a force. It is the product of the mass and velocity of the system--a conserved quantity, but the object required a force to bring it up to that velocity. In the case of a planet, it is the astronomical forces that acted on the protoplanetary disk. In the case of the yo-yo, it was your applied force to spin it.)


Is there an official list of objects in the sky?

Lists of astronomical objects are called "catalogues" and other answers have already mentioned the NGC catalogue of galaxies and nebulae. However, you asked about quasars and black hole x-ray binaries.

For quasars, there are several catalogues. None claim to be complete, as new discoveries happen all the time. The MILLIQUAS catalogue is a compendium of 460,222 type-I QSOs and AGN. It also contains about a million "Quasar candidates". (sources that may be quasars, 80% confidence)

For Black hole binaries, the difficulty is proving that the object is indeed a black hole. This usually requires calculating the mass of the accreting object, which is not always easy. The X-ray catalogues are classified by the mass of the normal stellar component. The High Mass x-ray binary catalogue notes objects that are "black hole candidates". It is in no way a definitive list, since no such list exists.


How are the three character IAU Minor Planet Center's Observatory Codes assigned? Why are some letters so popular?

I asked the at the minor planet center how the codes were decided, and the answer was

  Historically, the observatory codes were assigned ascending by longitude toward east (from prime meridian): 360 degrees were divided by numbers. When three digit numerical codes were not sufficient, letters plus two numbers were used again in bands toward the east. Some bands are already full (e.g. H**). New observatory code is assigned in the available letter+number band (based on its longitude).


How can a Grade-11 Student avail TESS data and learn to process?

I'll use the TESS satellite as an example, which you might use to look at transiting exoplanets for example.

Go to https://heasarc.gsfc.nasa.gov/docs/tess/data-access.html this gives you a list of links to other useful TESS websites, data access and analysis tools. There are links here to get raw and processed light curves or even to get light curves for stars with known exoplanets. This page also explains what data are available.
What I want to do is look at some lightcurves. So you go to the MAST portal. First thing is to decide what target you want to find. There's a box at the top where you can type something in. Not every star in the sky will have data. There are limitations to the coverage and the types of stars for which data are acquired and you probably need to familiarise yourself with that if you are doing something systematic.
So I try HD 36705 (AB Dor, this is a star I know - about 6th magnitude in the constellation of Doradus. Up pops an image of the sky (on the right of the screen). HD 36705 is the bright star in the middle. In the centre panel you see a list of observations by whatever satellites are in the archive (not just TESS, but IUE, Kepler etc.) This star has lots of TESS observations listed. Each row is a dataset. You want the TESS lightcurves, which are some way down. You can identify them because they are labelled as "TESS", they have a little icon next to them which indicates you can view the data (see below), they are labelled with a target name of "149248196" and you want the ones which have an Observation-ID that ends in "-s".
These are lighcurves. You can quickly view them using the internal viewer, by clicking on the little lightcurve icon.
 The light curve viewer allows you have a basic look at the data, but you need to download it to do any analysis.
Click on the download icon  and put it on your own computer. This is a zip file. Unzip it. The good stuff is here:
inflating: MAST_2020-02-05T1007/TESS/tess2018292075959-s0004-0000000149248196-0124-s/tess2018292075959-s0004-0000000149248196-0124-s_lc.fits
This is the light curve in fits format. So you need a fits viewer to do anything with it or you need to import it into your own (python) code using something that reads fits files. To get started I would download a java program called topcat. This is free and works on windows, linux or mac.
Load the light curve file into topcat. Plot "SAP_FLUX" against "TIME" and that is your basic lightcurve (see below). You can zoom in, manipulate the data, fold the data, subtract or add things to the data, lots of cool things. Plus make plots, anotate them, save them in various formats etc.

It took me 15 mins to write this answer while downloading the data and doing those plots. You can do it!


Let's call them "hematite spherules".
In principle, things on the Marsian surface can last a long time, billions of years https://sci.esa.int/web/mars-express/-/55481-the-ages-of-mars and https://science.sciencemag.org/content/343/6169/1247166.full state that a mudstone sample was deposited before 1.6Gy.
Unfortunately, the answer so far is we do not know. As longs the formation processes of the spherules aren't known, everything is open. If they formed in mid-air during a meteorite impact they may just be a few years old (https://www.sciencedirect.com/science/article/abs/pii/S0032063314000245) (*). But if they formed in a similar process as those on earth they may be older than 3 billion years (https://advances.sciencemag.org/content/4/12/eaau0872.full).


What is the distance of Earth from the 7 planets, moons and the Sun at a point of time?

There are plenty of sky simulators where you can calculate that information yourself. e.g. Celestia, KStars

Just install one of them and it will tell you position in sky and distance to most known bodies in Solar System, for any given time.


Recent research regarding if the local group of galaxies will remain bound to the Virgo Cluster?

The most recent paper I have found is Hoffman, Y., Lahav, O., Yepes, G., & Dover, Y. (2007). The future of the local large scale structure: the roles of dark matter and dark energy. Journal of Cosmology and Astroparticle Physics, 2007(10), 016. They also found that the Local Group gets separated.


What units are used for the Stefan-Boltzmann law?

The Stefan-Boltzmann constant $\sigma$ is not a dimensionless quantity, it comes with units. So whatever units you use, you must ensure that the value you use for the Stefan-Boltzmann constant is consistent with them.
So using the value expressed in terms of SI units:
$$\sigma = 5.670\,374\,419\ldots \times 10^{-8}\,\rm W\,m^{-2}\,K^{-4}$$
you would either have to work with radius, luminosity and temperature in metres, watts and kelvins, or convert $\sigma$ to the units you are actually using.
For example, if you want to work in terms of solar radii and luminosities you have to account for the conversion factors $L_\odot = 3.828 \times 10^{26}\,\rm W$ and $R_\odot = 6.957 \times 10^8\,\rm m$, giving:
$$\sigma = 7.169\ldots \times 10^{-17}\ L_\odot\, R_\odot^{-2}\, \rm K^{-4}$$


Circular velocity in axisymmetrici potential

What you do is predict the orbital velocity for a more realistic hypothesised mass distribution. There is an easy analytic solution for a disc. More complicated distributions require a numerical solution.

If you get far enough outside an arbitrary mass distribution, then that solution will asymptotically look similar to $v_{\rm circ} = \sqrt{GM/r}$.

Astronomers get into trouble when explaining the evidence for dark matter from galaxy rotation curves to the layperson. They often use the analogy of the solar system, where the orbital speed goes as $r^{-1/2}$. The trouble is that that a Galaxy is not like the solar system because the mass is distributed and in general cannot be assumed to be concentrated at a point and Newton's shell theorem does not apply.

In practice what you can do is posit a more realistic distribution of matter based on star counts and the gas that can be observed. You can then predict (using a computer model) what the rotation velocity should be at any point if that is all the mass that exists. This model will converge to the simple $r^{-1/2}$ law once you get well outside the bulk of the mass in the model. The model can be compared to the observed velocities of gas, globular clusters, planetary nebulae and other velocity tracers - some of which do indeed exist well outside the bulk of the visible galaxy matter. What you find is that they move much faster than predicted, indicating that considerably more unseen matter must exist inside their galactic orbits. You can then add that (dark) matter in various ways. Non-spherically symmetric and spherically symmetric dark matter profiles (of the same total mass) would produce different orbital velocities at different radii for precisely the reason you identify. By looking at tracers at different radii you can map out the distribution of unseen mass. Once happy with your model you can integrate the density of seen and unseen matter to estimate the total mass.

The bottom line is that the $\sqrt{GM/r}$ is just a "back of the envelope" demonstration of the effect. Real research uses more accurate models or works in a regime far outside the bulk of the visible matter where the "back of the envelope" approximation works.


Is it possible to throw a camera attached to a rope within the event-horizon of a black hole and then pull it back in?

According to the observer holding the rope, the camera would never make it past the event horizon. It would appear to slow down to a standstill just above the horizon (though would effectively disappear because of redshift). At some point the rope would snap, but the camera would still not appear to cross the horizon.

According to an observer with the camera, they would cross the horizon. But if they were lowered slowly, the rope would snap before they got there, since infinite force is needed to "hover" just above the event horizon.

They would then find themselves inexorably falling inwards in the same way that you and I are inexorably drawn into the future and can't revisit the past.

That different observers in different frames have different versions of events is not unusual in relativity.


Is there a term for a horizontal moon crescent

No formal astronomical term, but it is sometimes said to be "on its back"

Examples from "weather lore"

  When the moon lies on its back
  Then the sou-west wind will crack.

  When the moon is on its back,
  She sucks the wet into her lap.

  It's sure to be a dry moon if it lies on its back
  so you can hang your hat on its horns.


What are some of the best astronomy gatherings in the USA?

NEAF is pretty big (and probably the biggest).  There's also Southern California Astronomy Expo (SCAE).

For observing, there is the Texas Star Party in west Texas.  There is the Winter Star Party in the Florida Keys (that site for that event was severely damaged by a hurricane a couple of years ago, had to relocate up-state, but are now back in the keys.  This past year the had to limit attendees because the camp wasn't fully restored.

A little closer to home for you (in Pennsylvania)... is the Cherry Springs Star Party (there is also the Black Forest Star Party at the same park ... different months).


Why is the height of a satellite greater than the apogee of its orbit?

You are right!

Apogee is higher than Perigee, except when the orbit is perfectly circular when they're equal.
Semi-major axis will be half-way between the two.
572 km doesn't make sense.

As pointed out in comments, expressed like this they are the distances from Earth's center minus a fixed equatorial radius of the earth, in this case about 6378.137 kilometers. From the numbers in Wikipedia for peri and apo you'd call this a 567 x 515 km orbit. It's a simple way to express roughly the size and shape.
The linked de.Wikipedia page says
Umlaufzeit:       95,3 min[1]
Bahnhöhe:         572 km
Bahnneigung:      97.7°
Apogäumshöhe:     567 km
Perigäumshöhe:    515 km

[1]  Bahndaten nach [TES](https://www.n2yo.com/satellite/?s=26957). N2YO, 18. November 2017, abgerufen am 18. November 2017

I don't know what Bahnhöhe means, google translates it as "track height", so I've asked What exactly would a distance labeled Bahnhöhe mean in the context of a satellite's orbit around the Earth? in German SE.
While the data might be static, the mentioned n2yo link says:
NORAD ID:           26957
Int'l Code:         2001-049A
Perigee:            513.2 km
Apogee:             564.0 km
Inclination:        97.7 °
Period:             95.3 minutes
Semi major axis:    6909 km

and looking it up in Celestrak gives the two line element set or TLE as
TES
1 26957U 01049A   20269.88561907  .00000195  00000-0  15464-4 0  9995
2 26957  97.7413  31.7408 0036778 246.8588 112.8763 15.11565287 39121

The TLE is also shown in n2y0.
Here's a link direct to Celestrak for this object. They are always changing, the orbits of Earth satellites are never really fixed.
Atmospheric drag tends to lower them and Earth's lumpy gravity causes them to precess, just for starters.
Many satellites have small thrusters to raise their orbit after atmospheric drag lowers it, so we can't expect numbers in static Wikipedia data fields to be accurate.


Night-Day Cycles in S-Type Planetary Systems

Although this question would probably be more suited for the worldbuilding.se, you're asking for scientific references, so I guess this is ok to ask here.

What you describe is a hierarchical system, akin to Sun-Jupiter-Anymoon. I assume the planet orbits star $S_1$ directly, and $S_2$ is the larger star, and the planet will be abbreviated with $P$.

I don't know any references, but just from geometric considerations I'll state a formula here that should work in a quasi-2D setting (what I mean by this, later).
The surface illumination function $I(t)$ should be a simple sum of the two-star contributions $$I(t)=I_1\,\sin(\omega_0 t) + I_2(t)\, \sin(\omega_1\,t), $$ where $\omega_0$ is the spin rotation frequency of P around itself. The amplitudes of the sinii are $I_1$, the constant amplitude related to the constant distance of $S_1$ to $P$, and $I_2(t)$ being the irradiation amplitude coming $S_2$. As example, $I_1$ could be a number like the familiar $1360 \, kW/m^2$ for the sun.
I wrote $I_2(t)$ as a function of time, because distance from the planet to $S_2$ varies over time, as it orbits $S_1$. Therefore, as simplest approximation, I would assume that this is also a sinus, hence $$I_2(t)=I_{2,flct}\; \sin(\omega_1\,t)+I_{2,avg},$$ with $I_{2,avg}$ being the average irradiation at $S_1$'s orbit around $S_2$, and $I_{2,flct}$ is given by the minimum and maximum distance of the planet towards $S_2$, and finally $\omega_1$ being the orbital frequency of P around $S_1$.

Once you have some numbers for the constants, you can use python, or any other routine, in order to investigate the behaviour of your day/night cycles. Here it is important that your routine can replace the negative contributions in $I(t)$ with zero values, otherwise the negative values of $I_1$ could 'eat' the positive values of $I_2(t)$ and instead of perpetual day you'd get a perpetual night from the formula, when P would be directly between $S_1$ and $S_2$. In python you would do this via some command like "I_updated = [0 if i < 0 else i for i in I_raw]", where "I_raw" would be the values from the individual $S_1$ or $S_2$ contributions, before summing them up to $I(t)$.

If you want to know 'which sun is up' at any given moment, you can plot the sinii separately.

This quickly scribbled formula ignores possible 3D-effects on the illumination, such as a possible Kozai resonance in your system, but also misrepresents 'shadowing' of $S_2$ by $S_1$. Therefore the formula represents a very slightly inclined orbital plane w.r.t $S_1$.

Please observe that your system, as you want to build it, might not be stable for too long:
For the hierarchical structure of your orbits to exist, the stars must have a mass ratio of at least $m_2/m_1\approx 23$. If $m_1$ is a sun-like G0 star, then $m_2$ will be an O9, which has a lifetime of only 4 million years on the main sequence.
However, if you assign $m_2$ to be the G0 star, then $m_1$ could be a M2 dwarf, and that should work out fine. Also remember, that in your chosen configuration, there's always one part of the planets orbit around $S_1$, namely when it's exactly between $S_1$ and $S_2$, where the whole planet experiences perpetual day, and depending on the choice of stellar type, might get quite cooked.

For more on three-body motion, I'd recommend diving into Murray & Dermott: Solar system dynamics.


They are captured.

The formation of supermassive black holes is rather uncertain. They may have formed from core collapse of very ancient stars  (and then grown) or they may have some other mechanism of formation in the early universe. However we do know that the first formed early on in the history of the universe. They are as old as the galaxy, or perhaps even older.

On the other hand the stars that orbit the black hole are of regular age.
They are not particularly old. Therefore the black hole formed long long before the stars that orbit it. And so those stars must have been captured by some mechanism.

Star formation doens't normally happen in supernova remnants. A supernova can provide the shock that starts stellar formation in molecular clouds.


Questions about terminology used for Mars and its moons

Deimos is, like our own moon, tidally locked to Mars, so one hemisphere of Deimos faces the planet, and the faces away. The Sub-Martian hemisphere faces Mars.

Since Deimos is locked, there is a point on its surface that always faces in the direction of orbital motion, that point is the leading apex. It is an apex in the sense of a "leading point"

A meridian is a line of longitude, running from the North to the South pole. One of Deimos's meridians goes through the centre of its hemisphere that faces away from Mars, this is the anti-Mars Meridian. As you say, the "meridian on the side of Deimos facing away from Mars.

The image here shows a map of Deimos:

The leading face is the left side of the image, and the trailing face is on the right. The central 50% is the sub-martian hemisphere and the anti mars meridian is the line on the left and right edge, where the 3D image of Deimos has been cut to make a 2D map


What constellation does sun occupy by location and date?

You can see for yourself for any date and location using the free Stellarium program (http://stellarium.org/).

The constellation that the Sun is in as seen from Earth does not depend noticeably on the precise location of the observer on Earth.  It depends mostly on the time of year.  Here are the approximate days of the year at which the Sun enters a new constellation:

Aries: April 18
Taurus: May 14
Gemini: June 21
Cancer: July 20
Leo: August 10
Virgo: September 16
Libra: October 31
Scorpius: November 23
Ophiuchus: November 29
Sagittarius: December 17
Capricornus: January 19
Aquarius: February 16
Pisces: March 12

The "approximate" part is that the Sun may be up to one day earlier or later than the quoted dates, just like the start of the seasons may be one day earlier or later than the average.  (Actually, it is the calendar that is early or late with respect to the phenomenon.)  Also, because of the precession of the equinoxes, these dates shift by about 1 day every 70 years.

See http://aa.quae.nl/en/antwoorden/sterrenbeelden.html#6.


How did ancient astronomers know to order the planets from the closest to the farthest from the Sun?

According to the Cambridge Concise History of Astronomy (p 33 of my edition), essentially the Greeks took the (not unreasonable) view that the planets that moved more slowly were further away and were orbiting on larger spheres.

That's obviously not the same as suggesting they knew "the right distances" to the planets, merely the order. They did develop a mathematically sound means of estimating the distance to the Moon, though their observational data let them down on a correct estimate.

They also thought the Sun was in the fourth sphere around the Earth (to explain the observed behaviour of Mercury and Venus).

Not all Greeks thought the Sun revolved round the Earth and those who argued for a heliocentric universe said the reasons the stars did not move as the Earth did was because they were very, very far away.


Scientific possibility of stationary moons for a writer of fiction

The possibility you mention is feasible if the moon is in the equivalent of a geostationary orbit, where the orbital period of the moon is exactly the rotation period of the planet and it orbits in the equatorial plane of the planet.

Such a moon would always keep the same face to the planet if it were close enough to the planet (which in turn depends on the spin period of the planet) such that its spin is tidally locked to the same orbital period.

The presence of significant tides may be required to keep this configuration stable, without a tidal elongation of both objects it is likely that it would drift away from synchronous rotation. This in turn probably needs the moon to have a significant size/mass compared with the planet.

The tides on such a planet would essentially be "fixed". i.e. the tidal forces would be constant (with the exception of the possibly quite significant tidal forces due to the star it was orbiting). That would nean no water sloshing around due to tides. The "sea level" at point nearest to, and opposite, the moon would be higher than if there were no moon at all.

Something like this has been asked in another Astronomy SE question (Can a natural satellite exist in a geostationary orbit?) and, as pointed out there, Pluto and Charon are an example of the system you are looking for.


How did the Earth come to be in orbit around the sun?

The picture was so much cleaner 20 to 25 years ago. I'll present that nice clean picture first. Stars form from the gravitational collapse of huge clouds of interstellar gas. Those gas clouds inevitable have some net non-zero angular momentum. This forces the gas cloud to change shape from being more or less spherical to being disk-shaped. (Why? That's a different question. Ask it.)

While this protoplanetary disk continued to feed mass to the growing protostar, it also set the stage for the formation of planets. The gas cloud was mostly primordial hydrogen and helium, but it also contained heavier elements thanks to stellar fusion and supernovas in the billions of years that preceded the formation of our solar system.

Those heavier elements behave quite differently than do hydrogen and helium. They have chemistry. The planets started as microscopic clumps of mass of these heavier elements, bound together chemically. These microscopic clumps occasionally collided, eventually forming larger clumps of mass. These larger clumps in turn collided with one another, forming even larger clumps of mass. Eventually the clumps became large enough that they interacted gravitationally, making them grow even larger. This process continued, eventually forming protoplanets, and then planetary embryos, and finally planets.

Temperature in the protoplanetary disk were high near the forming protostar but dropped precipitously with increasing distance from the protostar. At some point, volatiles such as water, ammonia, methane, and carbon dioxide become as solid as rock. This is the ice line, aka the snow line or frost line. Asteroids inside of Ceres' orbit tend to be rocky. Asteroids outside of Ceres' orbit tend to be icy.

Planets that form outside the ice line can grow very quickly and then they can grow very, very large. The stuff that comprises the protoplanetary disk orbits the growing protostar at something other than the rate suggested by Kepler's laws thanks to the pressure of all that stuff in the disk. Thanks to the square-cube law, larger objects aren't as subject to that pressure. Those larger objects orbit at a Keplerian rate. Planets that form outside the ice line grow very quickly and then sweep up gas and ice because they are orbiting at a different speed than the immediate surroundings. The result is gas giants such as Jupiter and Saturn and further out, ice giants such as Uranus and Neptune. Planetary growth is a much more difficult process and much slower process inside the ice line. That's why Mercury, Venus, the Earth, and Mars are rocky and so much smaller than Jupiter, Saturn, Uranus, and Neptune.

That's the pretty picture. The not so pretty picture:

Why are Mercury and Mars so much smaller than Venus and the Earth?
Simulations suggest that the rocky planets should all be more or less the same size. That is not the case in our own solar system, let alone elsewhere.
How could Uranus and Neptune have formed?
Simulations cannot recreate Uranus and Neptune at their current distances from the Sun. The material in the protoplanetary disk should have been too sparse at those distances to form large planets.
Much, much worse, what's the deal with all the weird exoplanets scientists have found?
Scientists have found Jupiter-sized objects orbiting very close to their sun, Neptune-sized objects orbiting where the simple model would have only rocky planets forming, and planets in highly inclined (and sometimes retrograde) orbits that don't make sense.

These simulations (which have become to be very good) and the plethora of exoplanets have pushed the theory of how planets form back into the “that’s funny” stage. ("The most exciting phrase to hear in science, the one that heralds new discoveries, is not “Eureka!” but “That’s funny...”", a quote widely attributed to Isaac Asimov.)


What kinds of technologies are required to directly image exoplanets

When the light from two telescopes is combined to make an interferometer, the resolution of the combined instrument is equal to the resolution of a single telescope with a diameter equal the the distance between the two (or more) smaller telescopes. This is the technology you need to view exo-planets to any resolution you want.

This has been incredibly successful in radio astronomy. Long baseline radio astronomy gives the resolution of a radio telescope the size of the diameter of the Earth. Optical telescopes in space can be even bigger. Arrays of relatively small telescopes with specially calculated spacings can do even better. Big mirrors are still better in order to get enough light on a sensor to be useful.

The twin Keck telescopes in Hawaii can form an interferometer but funding has dried up at the moment. It has an effective size of 280 feet. I don't know why there is no funding, but at this point it may be likely that a space-based interferometer would work so well that finishing the ground based instrument is not cost effective. Or the time required to set up and get data is a poor trade-off compared to other demands for telescope time. Also, longer IR is better for the planet imaging and space is a lot better place for that.

For some reason, this is a hot topic among PhD candidates, but a delay in getting pictures of these objects is not going to mean much for the rest of us.


Are there ways other than the collapse of a star which have been observed to form black holes?

There are currently three classifications of black hole and these are stellar, super massive and miniature.

Stellar

Stellar black holes are formed by a massive star collapsing and its mass being great enough to form a black hole of moderate size

Super-massive

These black holes can have the mass equivalent of a billion suns, these most likely exist at the centre of galaxies and are probably as old as time itself.

Miniature

Miniature black holes are theoretical and are thought to have been created at the early stages of the universe. these require masses less than that of our sun and are thus very unlikely.

Other

"But Rhys" I hear you cry, "these are all from stars you fool!"

And yes, yes they are, this is because black holes are formed from mass, and lots of mass (except miniatures), coincidentally large amounts of mass tends to gravitate towards each other and form stars long before it forms black holes.

Therefore theoretically, a black hole does not even need stars, it just needs lots and lots of mass, but having that much mass without forming a star is no easy task.

Short answer, you can make a black hole from anything as long as it has enough mass, but good luck stopping this from forming a star beforehand!


Will new stars stop forming at some point of time?

Cosmic GDP has already crashed, as Peak Star was ~11 billion years ago.

According to Sobral et al's prediction, the future star production by mass will give only 5% of the stars in the universe today, "even if we wait forever." More theoretical predictions, such as this one, suggest that nebulae will run out of hydrogen on the order of $10^{13}$ years, while star formation will occasionally happen due to collision of brown dwarfs until somewhere on the order of $10^{14}$ years.

Of course, hydrogen itself may have a finite lifetime. The half-life of a proton is experimentally known to be longer than $10^{34}$ years, but it may still be quite finite.


The spectrum is of course heavily weighted towards red wavelengths, but your answer is mainly to do with the physiology of the eye, not astrophysics.
This site has folded stellar spectra through the appropriate response and arrived at "orange-ish" for M-dwarfs.
A "white-hot" piece of metal is not really white. Sunlight appears white - a blackbody spectrum at around 5800 K. The composition doesn't play a great role, though the spectrum of heated pure metal is probably closer to a blackbody than is an M-dwarf. I suspect that when people talk about "white hot metal", what is going on is the eye being overwhelmed by a large flux of photons. The intrinsic spectrum is clearly not that of a "white" object (such as sunlight). However, it could be that the emissivity is a bit lower in the red part of the visible spectrum so that when at least some green/blue light is produced (recall that the Wien tail is an exponential fall on the short wavelength side) that the eye's color sensory cells may be triggered sufficiently to fool it that the light is white. Either way that's physiology, not astrophysics.
Postscript: Most people can make out the orange-ish colour of M-type stars like Betelgeuse and Antares. These are (super)giants, but their spectra are not so different from M-dwarfs.


It moved in a straight line with sparkling track.What could it be?

It is most likely a jet contrail. Not all contrails are persistent, and a high altitude jet can be seen over a great distance.

A fireball may leave a trail, but if it does, then it is likely to persist for as long or longer than a jet contrail. A fireball appears to move much faster than a jet, crossing the sky in a few seconds. It also appears to be very bright.

The image and description both point to this being a jet.


Is gravitational energy released when a body contracts?

It is correct. Contracting stars heat up due to their own contraction. This justifies the initial temperature raise in a star, before fusion begins in earnest.

Kelvin was not 100% wrong, it's just that fusion, over the life time of the star, makes much, much more energy than contraction. The collapse heats up the star at the very beginning, then fusion takes over and makes essentially all the heat you observe.

This is also true of gas giant planets. Even a tiny amount of contraction may raise their temperature measurably. E.g., Jupiter radiates more energy than it receives from the Sun - this is due to contraction, at a rate of 2 cm / year (diameter).

TLDR: Gravitational energy becomes heat.


Why are blackholes generally not considered as solid spheres with an escape velocity FTL?

The idea of a hard surface on a black hole is very much a minority theory, bordering on a fringe theory. The reason is that general relativity is fairly well tested, and does not just predict event horizons around black holes but also the Buchdahl bound on how small objects in hydrostatic equilibrium can be. Still, it is always worth testing the theory, which is what the paper proposes.
If you drop an object from infinity onto an object of radius $R$ the Newtonian kinetic energy at that moment would be $GM/R$ J/kg, much of which would be released as a flash. This is why accretion onto white dwarfs and neutron stars produce a lot of radiation beside the accretion disk: for white dwarfs about 0.0003 of the mass-energy would be released, while for a neutron star 0.147 of the mass energy would be released. For a black hole-radius object this crude estimate produces 50% efficiency of converting mass into energy.
However, the energy of that flash is reduced by the gravitational redshift. At the Buchdahl limit the redshift is z=2, halving the energy received (for neutron stars you only get 74% of the energy). But for a black hole horizon general relativity  gives infinite redshift, making the flash disappear altogether.
It does not make sense to see black holes as solid objects since the structure of spacetime inside the event horizon makes the inward direction behave like time does outside: you can only move inward, just as you can only move forward in time. Thinking of the singularity as a material object with mass is very misleading: it is not even point-like. Hence talk about the density of a black hole or imagining it as a sphere of something dense tends to end in confusion. Similarly talking about escape velocity is also tricky: it does not define the horizon as people usually say. The horizon is defined by being a trapped surface that can only be crossed in one direction.


How far away is the light that would reveal the Big Bang?

The particle horizon is the point where light arriving at us now would've had to have been emitted at the big bang (though note this is hypothetical as the Universe did not become transparent to light until it was several 100,000 years old).. How far away it is depend on the model you use (not to mention there are several different distance measures in cosmology), but it's proper distance is reckoned to be about 46.9 billion light years.

A note of caution though, when calculating the particle horizon cosmic inflation is usually ignored. What is light arriving (again hypothetically) from usually called the particle horizon therefore would be actually light that was emitted from just the end of the inflationary epoch. If you were to take a model with a big bang singularity and inflation the "true particle horizon" would in fact be much further away.


Is it moving straight to the west (i.e. it keeps the same distance from the horizon) or does it 'move with the stars'?

Have you checked an online planetarium like The Sky Live? Based on the current information, it could just be Sirius, the brightest star in the night sky, or even Venus which is currently in her Evenstar period (but planets generally don't twinkle).