So where are these measurements of galaxies moving faster than light?

They're redshift measurements. Check out the Wikipedia redshift article. It's good stuff.

  "we can actually observe galaxies that are moving away from us at >c"

It's true. You might think it cannot be, but it can.

  Um, I think I missed the groundbreaking headline that said scientists have measured a galaxy moving away from us at speeds faster than light.

It's not groundbreaking at all. It's simpler than you think.

  So the first answer seems to contradict this other answer https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers "If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon".

Yes, that's a bit unfortunate, because he also says this: "galaxies with red shifts greater than ~3 were and are receding from us faster than light".

  It seems like common sense that at a certain distance, space is expanding faster than light, thus preventing light beyond a certain distance from reaching Earth which means we can't directly measure galaxies traveling faster than light. But who's right?

The guy who says we can see galaxies that were and are receding from us faster than light. As per Rob's comment, see Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Tamara Davis and Charles Lineweaver. Note this: "We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light." See page 8: "Amongst those who acknowledge that recession velocities can exceed the speed of light, the claim is sometimes made that objects with recession velocities faster than the speed of light are not observable [App. B: 9–13]."

Also have a read of the Wikipedia ant on a rubber rope article. The rubber rope is being stretched as the ant walks along it: "At first consideration it seems that the ant will never reach the end of the rope, but in fact it does (although in the form stated above the time taken is colossal). Whatever the length of the rope and the relative speeds of the ant and the stretching, providing the ant's speed and the stretching remain steady the ant will always be able to reach the end given sufficient time. Once the ant has begun moving, the rubber rope is stretching both in front of and behind the ant, conserving the proportion of the rope already walked by the ant and enabling the ant to make continual progress".

There's a section on the metric expansion of space. It says this: "By thinking of photons of light as ants crawling along the rubber rope of space between the galaxy and us, we can see that just as the ant can eventually reach the end of the rope, so light from distant galaxies, even some that appear to be receding at a speed greater than the speed of light, can eventually reach Earth, given sufficient time. However, the metric expansion of space is accelerating. An ant on a rubber rope whose expansion increases with time is not guaranteed to reach the endpoint.[3] The light from sufficiently distant galaxies may still therefore never reach Earth".


There are two components to take into account for this question

the earth rotation rate on itself
the orbiting period around the sun

A day back then was shorter than today's earth day. The earth was rotating quicker in the past, but its rotating period around the sun didn't change much because of Kepler's law.
I cannot say how it was 5 billion years ago (anyway, did the earth had something you could call a surface to track the day from then?), but you could find data derived from fossils analysis for the last billion years, e.g. here from Nasa's SpaceMath, on how long a day was back then - I realize now the figures below assume the orbiting period of the earth around the sun (~8766 hours) stayed constant over this period of time. This assumption is wrong as explained here - however stating it "has been close to its current value for the last 2-3 billion years", so below figures are an approximation

Period
Age(years)
*Days/year
*Hours/day
ratio

Current
0
365
24.0
1.000

Upper Cretaceous
70   million
370
23.7
0.988

Upper Triassic
220  million
372
23.5
0.979

Pennsylvanian
290  million
383
22.9
0.954

Mississippian
340  million
398
22.0
0.917

Upper Devonian
380  million
399
22.0
0.917

Middle Devonian
395  million
405
21.6
0.900

Lower Devonian
410  million
410
21.4
0.892

Upper Silurian
420  million
400
21.9
0.913

Middle Silurian
430  million
413
21.2
0.883

Lower Silurian
440  million
421
20.8
0.867

Upper Ordovician
450  million
414
21.2
0.883

Middle Cambrian
510  million
424
20.7
0.863

Ediacarin
600  million
417
21.0
0.875

Cryogenian
900  million
486
18.0
0.750

From which I would roughly interpolate: 600 days earth self-rotation per year full rotation around the sun 2 billion years ago

Edit
We are saying here that earth was rotating quicker in the past:

Detailed  studies  of  fossil  shells,  and  the  banded  deposits  in  certain  sandstones,  reveal a much different length of day in past eras! These bands in sedimentation and  shell-growth  follow  the  lunar  month  and  have  individual  bands  representing  the  number  of  days  in  a  lunar  month.
By  counting  the  number  of  bands,  geologists can work out the number of days in a year, and from this the number of hours  in  a  day  when  the  shell  was  grown,  or  the  deposits  put  down.  The  table  above shows the results of one of these studies.


What will a future cold and "dead" Jupiter look like?

As earlier answers indicate, Jupiter will slowly shrink until it is some 10-20% smaller with elements slowly settling depending on density and solubility.
One can calculate a rough Kelvin timescale for the ratio between binding energy and the blackbody power of the surface to estimate how quickly it cools down (this number will be nudged by the above considerations a bit, so it is an order of magnitude estimate):
$$\tau_{Kelvin}=\frac{3GM^2}{20\pi\sigma R^3 T^4}.$$
For Jupiter this timescale is 54 billion years.
As the universe cools eventually it becomes a ball of layered solid degenerate matter, with a frozen hydrogen crust. This will take longer than $\tau_{Kelvin}$ due to various mild heating processes like crystallization heating, tidal heating from the moons, possibly dark matter annihilation, and other low-energy processes that keep objects warmer than the CMB.


How should I find a telescope for research time?

As an undergraduate, you are a member of a university. Your first step then should be to talk to your Professors.

As a guide to what might be expected, look at the University of Arizona telescope proposal page They accept proposals from members of Arizona University. You have to have a clear aim and plan. This is why you need to talk to your professors first.


Can we measure comets in a similar way to exoplanets?

The thing about detecting exoplanets by eclipses is that the eclipses are repeatable.  You have the same star, whose light curve dims in the same way with each eclipse, which gives you information about the planet's orbital period and its apparent size relative to the star.  The Kepler mission observed many fluctuations in stellar brightness that were not associated with exoplanets, but instead with changes in the instrument, fluctuations in the actual brightness of the star,  probably some other explanations that I can't think of off the top of my head, and actual mysteries.

Basically, if you can only observe it once, then you can't claim to explain it.

Occultations of distant stars by planets or comets don't give you a light curve: the star goes from "on" to "off". So a single occultation doesn't give you any information about the size, speed, or direction of the object doing the occulting.  And the occultations are sufficiently rare that I think you'd be extremely unlikely to find a "line" of them by chance --- though that would make a fun data-mining project for an astronomy student.

What you can do, if you have an object that you already know about, is to make many observations of the same occulatation.  That's the mission of the International Occultation Timing Association.  If you click through to their web page you'll see this image:

Each line here shows the observed brightness of a particular star observed from a particular location at the same time.  You can see clearly that some observers saw the star disappear very briefly, some for longer, and some (who might have been predicted to be in the very center of the path) saw no occultation at all.  Assembling all these individual observations, properly synchronized, shows clearly the shadows of the two parts of a binary asteroid, each of order 50 km in diameter.

Note that 90 Antiope is a main-belt asteroid.  The angular diameter decreases linearly as the distance increases, so an object the same physical size orbiting out by Neptune (perhaps fifteen times farther) would need fifteen times the density of observing locations to create an image with the same detail.


Would a killer asteroid shattered into thousands of pieces produce the same devastation?

TL;DR version: Too big and way, way too late.

The dispersal can't be done, even at the lower end of that 3-20 km scale. Holsapple claims 5 kilojoule/kg are needed to disrupt and disperse a solid 1 km asteroid asteroid, with energy scaling with radius1.65. Disrupting and dispersing a solid 3 km diameter asteroid with a density of 3 g/cc would require a deeply buried two Tsar Bomba weapon. The energy needed to disrupt and disperse a 20 km diameter solid asteroid is beyond anything humanity can achieve.

This disruption and dispersal will not be uniform. It will instead create a few big chunks, perhaps 1/3 the diameter of the original body, a larger number of intermediate sized chunks, and thousands upon thousands of little chunks. Those big chunks are still civilization killers. The intermediate-sized chunks will make Tunguska look small, and the thousands upon thousands of little chunks -- we recently saw what a little chunk can do, over Chelyabinsk.

What if it's a rubble pile or a comet? While they're not quite as dense, they need much, much more energy to disrupt and disperse. Rubble piles and comets are very good at absorbing impact energy. Disrupting and dispersing a one km diameter comet is beyond the scope of human technology.

The time scale, two days, is extremely short. A decade of advanced notice is considered to be very short notice when it comes to diverting a one km impactor. Even larger ones require even more advance notice. A century's advance notice might be enough time to deal with a 20 km diameter object.

Holsapple, "About deflecting asteroids and comets," Mitigation of Hazardous Comets and Asteroids: Vol. 1, Cambridge Univ, 2004.


Calculating orbits using observational data

After researching a bit more, I think I can answer my own question (or at least share what I have found). When Kepler derived his laws using Tycho Brahe's observational data of Mars, he followed the procedure below:

first, the orbital period of Mars must be known. This value was already known in the time of Kepler (aprox. 687 days) $^1$, and can be derived as follows: first, one takes notes of the synodic period of Mars (which "[...] is the time that elapses between two successive conjunctions with the Sun–Earth line in the same linear order."$^2$). This can be done by calculating the elapsed time between two consecutive observations in which Mars is in the same apparent position in the sky. Then to calculate the orbital period of Mars, one can use:$\dfrac{1}{T_{M}} = \dfrac{1}{T_{E}} - \dfrac{1}{T_{M\mathrm{syn}}}$,$^3$ where $T_{M}$ is the orbital period of Mars, $T_{E}$ is the orbital period of Earth and $T_{M\mathrm{syn}}$ is the synodical period or Mars.
knowing the orbital period, Kepler then selected some of Brahe's observations that are exactly 687 days apart from each other (that is, one Martian year apart). This means that regardless of Mars relative position in the sky, the planet will be in the same position in relation to the Sun:

using triangulation, Kepler then calculated the distance of Mars to the Sun in terms of the distance of Earth to the Sun (1 AU). He did this for several positions of Mars.$^1$
while any three points define a circle, four or more do not necessarily do so. Kepler noticed that the collected data of distances between Mars and the Sun could be fit into an ellipse with the Sun in one of the focuses.


Earth rising and setting from moon's perspective

You're quite right.  The Earth is (nearly) stationary in the Moon's sky.

(I say "nearly" because the Moon is in a slightly elliptical orbit around the Earth, but rotates perfectly smoothly. This means that the Earth's motion through the Lunar sky is
 a bit faster when the Moon is at perigee and a bit slower when it's at apogee. Because the Moon is tide-locked, on average they match perfectly, but during the course of the sidereal month the Earth appears to move back and forth by a few degrees.  The effect isn't huge, but it's easily observable with a telescope.  See the Wikipedia article on
"libration" for an OK discussion and a really great animated gif illustration of libration as viewed from Earth.)

Is it possible that you're reacting to the amazing movie from Apollo 8 showing Earthrise over the Lunar horizon? It's been int he news a lot given that this is the 50th anniversary of the flight.  If so, what you saw there was not from the Moon's surface, but from orbit around the Moon and the Earthrise happened as Apollo 8 came around from behind the moon (where the Earth is not visible in the sky) and the Earth first became visible around the edge of the Moon.


Well, I did some digging and found a helpful chart here. The image itself is located at this link.
Wikipedia defines the S band as the section of the electromagnetic spectrum from 2 to 4 GHz. To interpret the respective values the graph displays at these frequencies, we need to convert these two figures to wavelengths:
$\lambda_1 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{2\:000\:000\:000\:Hz} = 0.15\:m$
$\lambda_2 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{4\:000\:000\:000\:Hz} = 0.075\:m$
And, we might as well find the wavelength at 3 GHz, since it's at the middle of the S band and might help us (hint: since the graph is on a log scale, and this calculation comes out to a nice power of 10, it definitely will help us):
$\lambda_m = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{3\:000\:000\:000\:Hz} = 0.1\:m$
So, we can take a look at the graph at 0.1 m. The units seem to be in $W\: m^{-2} \: (c/s)^{-1}$, and I'm assuming that $(c/s)$ is just $Hz$, which would make these units have the same units as Janskys, though, as Connor Garcia mentions in his comment, Janskys are $10^{-26}\: \rm{W\: m^{-2}\: Hz^{-1}}$. So, the quiet sun in the S band lies around $10^{6}\: \rm{Jy}$, and the "disturbed" or "active" sun lies around $10^{7}\: \rm{Jy}$.
And as for the second part of your question, the graph also shows some blackbody curves. You asked for 5800 K, but the closest one on the graph is at 6000 K - hopefully that's okay. As expected, the Sun seems to closely match that line for a good portion of the wavelengths on the graph, but seems to deviate in the S band, which is interesting, to say the least. So, from the blackbody curves in the graph, the sun is brighter than a typical blackbody at ~6000 K in the S band, both in its quiet and disturbed states. This difference is around one order of magnitude in the quiet state, and around two orders of magnitude in the disturbed state.
And lastly, you asked about the difference between the "brightness" of the sun in its active and disturbed states in the S band. From visual inspection of the graph, it seems like the difference in the S band is about one or two orders of magnitude.


How is it ensured that a parabolic shape forms while polishing a telescope mirror?

While a spherical shape is fairly easy to abtain, a parabolic shape definitely isn't!

It's a combination of special polishing strokes, and feedback via control and testing.

https://stellafane.org/tm/atm/mirror-refs/strokes.html

Normal polishing strokes (center over center, 1/3 diameter amplitude) tend to make a sphere. When all goes well, the surface of the mirror tends to become spherical, or very close to it. This is not different from grinding, only more precise. When two surfaces are rubbing against each other in a ball-and-socket type of joint, any irregularities will be removed and both surfaces will "want" to become spherical, or close to it.

When making telescope mirrors, the so-called "normal stroke" is what accomplishes that in the best way.

Now here's the essential detail: no matter what other curve you're trying to obtain (parabola, hyperbola, ellipse), for the purpose of making telescope mirrors all these curves are very close to a circle, within a fraction of a micron.

So a typical process has an initial goal of making a sphere (cross-section is a circle). This is not mandatory, but it's easier to first make a sphere, and then apply "corrections" to it. You can definitely make a parabola directly, but the process is more complex.

For telescope mirrors, the parabola that's needed is basically a circle that's deeper in the center, and more shallow towards the edge. So, if you start with a spherical mirror (circular section) and use polishing strokes that tend to deepen the center in a gradual way, you can get pretty close to a parabola. This is known as a "parabolizing stroke".

Keep in mind that as you apply that stroke, the center keeps deepening. At some point the correction will become deeper than a parabola and the mirror will become hyperbolic. That's sometimes what you actually need (if you make a hyperbolic mirror for a Ritchey-Chretien telescope). Or if you stop too early, the mirror is elliptic, which is used in other designs. The parabola is the thin edge between the elliptic and hyperbolic domains.

You start with a mirror where the cross-section is a circle (spherical mirror). You begin to apply the W stroke. The circle immediately becomes a very shallow ellipse, which keeps deepening. At some point, for a very short time, that ellipse becomes a parabola. As you keep doing the W stroke, the parabola then changes quickly and now you have a hyperbola which just becomes more and more deep.

The key here is testing. You apply the parabolizing stroke for a short while, then stop and test the mirror. There are many testing techniques; the Couder mask is simple enough and very popular; or you could use an interferometer if you have (or can build) one. Is it still elliptic? Keep parabolizing. Has it become hyperbolic? Reset to a sphere and try again. Or adjust the stroke if either the edge or the center are being corrected too fast.

Apply parabolizing strokes. Test. Modify technique if necessary. Repeat. That's all there is to it.

Note: I am using jargon in the text above. "Deeper" really means "more strongly curved". "Shallower" really means "less strongly curved". The difference between circle and parabola is that the parabola is more strongly curved in the center, while the curvature at the edge is less strong. The circle has constant curvature everywhere, by definition.

This doesn't necessarily mean the center of the mirror is physically more depressed; while doing any strokes, the whole surface keeps being polished down all the time. But the relative curvatures of center vs. edge may change.

Deeper vs shallower are terms that reflect the physical process, whether the tool applies more or less pressure on a certain area of the mirror (so polishing goes faster or slower there). But the geometric attribute that the process actually changes, that we are interested in, is the radius of curvature. Where the tool applies more pressure, the radius of the curvature decreases rapidly (the surface becomes more strongly curved).

Deeper and shallower are practical terms related to process. More or less strongly curved are theoretical terms related to geometry. When you make mirrors, after a while these terms become linked in your mind.

It helps if you're familiar with the conic sections:

https://en.wikipedia.org/wiki/Conic_section

red = circle

orange = ellipse

green = parabola

blue = hyperbola


Meteors made of differing substances. Will they all burn up?

I can give a partial answer to this.   In general, larger objects are more likely to break up than smaller ones, and comets are more likely to break up than meteors.   A feather would likely burn up too, here's why:

30,000 KM above the surface, or roughly 4.7 earth radii, the escape velocity at that height would be the square root of that or roughly 2.17 times less or 46% of the escape velocity from the surface, so a feather dropped from that height would be traveling at nearly 54% of earth's escape velocity when it hit the atmosphere, or about 6 KM per second.  That's about 7 times as fast as a bullet and I'd wager, that would burn up a feather pretty quick.

Speed of space collisions is faster than free-fall velocity can reach cause anything that hits the earth from space is also in free fall and you can add to that any relative velocity between the objects.   The slowest meteors hit the earth at about 11 KM per second, not surprisingly, the Earth's escape velocity.

http://www.amsmeteors.org/fireballs/faqf/#12

Now, if you drop a feather from the highest hot air balloon, then it probably would float down to the earth, accelerating faster at first in the thinner atmosphere, then more slowly as the air got thicker.

Small point to add, angle of approach matters too.   A glancing blow and the object can effectively bounce off the atmosphere, not burn up in it.


As you say, expansion is proportional to distance. And in fact, compared to the size of the (observable) Universe, Andromeda and our Milky Way are close together. Indeed, farthest objects are now more than 20 billion Ly from us, whereas Andromeda is only 2.5 million Ly away. That is a 1/10000 factor.

The space expansion between them is not enough to set them apart, because of their general motion. On small scale (until galaxy cluster scale), the motion is primarily ruled by gravity. Galaxy cluster are in fact the largest gravitationally bound systems we know.

  They typically contain no more than 50 galaxies in a diameter of 1 to 2 megaparsecs

Since Andromeda is the closest galaxy, it is inside "our" galaxy cluster, and stays bound to ours by gravity.


Both are solar observatories, and so have different requirements from conventional telescopes.

The sloping sides may allow for smoother air flow around the building, as heating during the day causes air to rise, which affects the resolution of the image. Also the pyramid is structurally strong (useful when you have a building on top of a mountain) and attractive, echoing she shape of the mountains. The Nepali observatory uses the flat walls to mount photovoltaic cells.


What if the black hole in the center of the galaxy grew faster?

this is not really an answer to your question, but to the assumption you made.

The BH in the galactic center cannot grow that much. There are two problems. First, there is not enough food close by. Any potential food (gas clouds, stars, and dark matter) will have some non-zero angular momentum preventing it from coming close enough. Loosing this angular momentum is difficult (it cannot be radiated away like energy), the only way is to exchange it with other objects either via impact (of gas clouds) or gravitational interactions with other objects (not the BH).

The second problem is that a feeding BH is surrounded by an accretion disc of hot gas. In that disc, angular momentum is slowly transported outwards and mass inwards via viscosity (that viscosity most likely originates from turbulent magnetic fields that become unstable -- the magneto-rotational instability). This process inevitably heats the accretion disc to very high temperatures ($10^{6-9}$K) such that it emits a wind of raditation and particles (similar to the Solar wind, but much much stronger). If the BH feeds too much, this wind becomes so strong that it pushes away any further infalling material (potential food).

This second process limits the growth of any BH to double in mass in no less than about $10^6$ years, I think (I'm not too sure--if you want a precise value, consult the literature), even if the first problem was no issue.


Those are probably just "hot pixels" that don't move on the sensor like the real stars do, so that the stacking spread them out in an arc.
I don't think the software processes and treats each star separately, they can't be "skipped stars" mingled with "recognized stars". Notice that they are all 1 pixel wide, whereas stars are much wider.


Planets just look like dots! Beginner telescope, hoping for some advice

Most likely you will not see much detail except that they are a really tiny disc and don't wobble like stars.
The moon has an apparent diameter of 30' (half a degree) in the sky. Mars' apparent diameter varies between 3.5" and 25" (so about between 1/1000° and 1/120°). Thus at closest approach Mars is still about 60x smaller than the moon, currently at ~<10", it's about 1/180 compared to the Moon). How much detail do you see at such tiny fraction of the Moon in your telescope?
You can expect to see on Mars maybe some very slight colour variation between equator and poles. On Jupiter with an apparent diameter of ~50" (thus twice Mars' apparent size) you might see some slight colour changes with the cloud bands in its atmosphere and on Saturn you might see that it has rings.
All the images with details on the surface are both, taken with a camera and selected for (missing) atmospheric distortion and post-processed to enhance the features, and taken with a larger telescope where you can use a larger magnification.
But even with a larger telescope (longer focal length AND larger apperture), you WILL need to have good seeing; there are nights where even with a clear sky, also a larger telescope will not resolve features due to the seeing, thus the influence of Earth's atmospheric disturbances.


Is it possible to detect gravitational lensing behind the Moon?

The angular deflection caused by the lensing of a distant background object by the Moon is given by
$$\theta \simeq 4 \frac{GM}{Rc^2},$$
where $M$ is the mass of the lensing object and $R$ is the closest projected distance of the ray from the centre of the mass.

For the Moon, the maximum deflection would occur when the ray just grazes the limb of the Moon and would equal 26 micro-arcseconds.

At present the most accurate instruments for measuring precision positions are VLBI radio observations of point sources, where 10 micro-arcsecond relative positional precision can be "routinely achieved" (Reid & Honma 2014). To use this technique you need bright, radio point sources to be close in position to the Moon. This is certainly possible and has been done many times using the Sun as the lensing object, although the claimed accuracies of a short time-series of measurements on single sources is an order of magnitude lower and so probably wouldn't work for the Moon (e.g. Titov et al. 2018 ).

The Gaia astrometry satellite has likely end-of-mission positional precisions of about 5 micros-arcseconds for bright stars, but it is unclear what the precision is for a single scan (which would be needed because the Moon moves!). However, to counterbalance that, one could also average the lensing effect over many stars surrounding the Moon and of course average this for different stars with the Moon observed at different times over the entire Gaia mission.

Unfortutunately, although in principle this could be done, because of the way that Gaia scans the sky, it is always pointing away from the Moon and so there will be no observations towards the Moon that would be capable of revealing these small deflections (e.g. see here).

Conclusion: Measuring the gravitational deflection of light by the Moon is just out of reach of current observational techniques.


How much competition there is for jobs in astronomy compared to other fields of science?

The competition for permanent positions in astronomy is very tough.  The field as a whole produces roughly ~200 Ph.Ds per year, but there are usually only a handful (say ~10) tenure-track positions that open up every year.  So perhaps ~5% of Ph.Ds end up in tenure-track positions in astronomy.  There are more permanent positions in astronomy that aren't tenure track positions, but not too many more, probably enough to support about ~20% of Ph.Ds.

As for non-permanent positions in astronomy, there are plenty of those.  The funding situation for postdoctoral positions is such that, while there is still a great deal of competition for open positions (stemming partially from the fact that almost everyone applies to almost every open job --- I exaggerate, but only slightly), there are enough open positions, that almost all astronomy Ph.Ds who want a postdoc will be able to get one.  But those positions only last for an average of three years.

However the job prospects for astronomy Ph.Ds who decide to leave the field of astronomy (like myself) are quite good.  Those holding an astronomy degree have one of the lowest unemployment rates of anyone (around 0.3%).

Here are a few links to some papers on the state of the job market in astronomy:

http://arxiv.org/abs/0712.2820
http://arxiv.org/abs/0805.2624
http://arxiv.org/abs/1512.02223


What is the shape (along the plane, not up-down) of stellar orbits in flat spiral galaxies

This is an interesting question and so often interesting questions cannot easily be answered with any current knowledge, but this one can be answered to some degree. I'm going to run through the basics of orbital theory and describe how they can apply to galaxies and how it differs from Keplerian systems. You should have a reasonable understanding of Newtonian physics (after all orbits derive precisely from Newton's laws) and a strong knowledge of math. If you don't have these things, just jump to the end of each section where I'll try to summarize the important points behind the math.
A quick note on math notation I'll be using. A dot over a symbol indicates a time derivative (e.g., $\dot{a}$) and un-italicised, bold symbols are vector quantities (e.g., $\bf F$). Let's get down to business.
The Orbital Equation of Motion
Consider a mass $m$ as some position $\bf r$ and moving with a motion described by $\bf \dot{r}$. This mass experiences a force $\textbf{F}(r)$ which is only a function of radial distance, $r$, from the center of the coordinate system. The goal here is to determine the equation of motion that can describe the orbit of the mass due to this force. This equation can then be used to solve for $r(\theta)$. By Newton's law the equation of motion can be initially defined to be
$$\textbf{F}(r) = m\textbf{a} = m\Big(\ddot{r} - r\dot{\theta}^2\Big)$$
Note that in this case $r$ is simply the radial component of $\bf r$ and $\theta$ is the azimuthal angle of the body in a spherical coordinate system. I'll leave it to you to determine how to break down the acceleration into the two components above, under the appropriate coordinate system. Let's try to remove our $\theta$ dependence so that we have only a function of $r$. This can be achieved by using angular momentum conservation. The angular momentum per unit mass is given by $\ell = r^2\dot{\theta}$ so that $\dot{\theta} = \ell/r^2$. This gives
$$\textbf{F}(r) =  m\big(\ddot{r} - \ell^2/r^3\big)$$
This is now a differential equation that allows us to solve for $r(t)$, but we want $r(\theta)$ so we need to do some conversion. Let's re-parameterize by defining $u \equiv 1/r$ (the reason will become clear in a bit) and determining $\ddot{r}$ in terms of $u$ and $\theta$.
$$\frac{\mathrm{d}}{\mathrm{d}t}(r) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{1}{u}\bigg)=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac{\dot{\theta}}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}=-\ell\frac{\mathrm{d}u}{\mathrm{d}\theta}$$
Note the substitution of $\ell=r^2\dot{\theta}=\dot{\theta}/u^2$. Now differentiate again to determine $\ddot{r}$.
$$\frac{\mathrm{d}^2}{\mathrm{d}t^2}(r)=-\ell\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg)=-\ell\frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}\theta}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg) = \ell\dot{\theta}\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}=-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}$$
Putting this into our expression for the equation of motion and making the transformation that $r=1/u$ finally gives
$$\textbf{F}(1/u) =  m\Big(-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} - \ell^2 u^3\Big)$$
Writing in a more convenient form we finally arrive at
$$\boxed{\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{\textbf{F}(1/u)}{m\ell^2u^2}}$$
Remember that, $m$ is the mass of the body, $u(\theta) \equiv 1/r(\theta)$, $\ell$ is the angular momentum per unit mass, $\bf F$ is a purely radial force acting on the body, and $r$ and $\theta$ are the radial and azimuthal coordinate positions of the mass.
Punchline: The final result here is a general equation of motion for a body orbiting according to some arbitrary force. This could be gravity, electromagnetic, a spring force, or anything else we decide. It is purposefully derived under general and nonconstrictive assumptions and hopefully you can see that it can be used to understand the orbital motion of a star orbiting in a disk galaxy. The goal with this equation should be to plug in your force (whatever that may be) and solve for $u(\theta)$. From there it's easy to determine $r(\theta)$.
Keplerian Motion
Before we go about looking at orbital motion in a galaxy, let's look at standard Keplerian motion so we have something to compare to. Keplerian motion derives from assuming our mass $m$ is orbiting a single, point-like mass $M$ and under the influence of simple gravity. In that case, we can write our force as $F(r) = kr^{-2}$ and thus $F(1/u) = ku^2$, where $k\equiv GMm$ is a constant, defined here for mathematical simplicity. Note that $G$ is the gravitational constant. The general orbital equation, under this force, now becomes
$$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{k}{m\ell^2}$$
This is a standard second-order, nonhomogeneous differential equation with a constant forcing function. If you know your Diff EQ, you should know the solution almost immediately.
$$u(\theta) = \frac{k}{m\ell^2} + A\cos (\theta - \theta_0)$$
In this equation, $A$ is some unknown constant, and $\theta_0$ represents the starting angle of the orbit, which we can arbitrarily choose to be zero. Our ultimate goal is to get $r(\theta)$ so let's do that. I'm going to do a few steps in one though and leave you to work out the intervening math. I'll fill in for $k = GMm$ and define the angular momentum as $L = \ell\mu$ where $\mu$ is the reduced mass of our system. I will also state, without proof, that $e = A(m\ell^2/k)$ where $e$ is the eccentricity of the orbit.
$$\boxed{r(\theta) = \frac{L^2/GM\mu^2}{1+e\cos (\theta)}}$$
Punchline: We've found a final equation which represents the orbital motion of a mass under the influence of a gravity due to a point-like mass $M$. If you know your stuff, you'll see that this equation precisely describes the conic sections, depending on the value of $e$. If $e = 0$, you get circular motion (since $r(\theta)$ becomes a constant). If $0 < e <1$, you get elliptical motion, $e=1$ is parabolic motion and $e > 1$ is hyperbolic.
As expected, Keplerian motion (i.e., having a central force such that $F\propto r^{-2}$) resulted in conics, which is precisely Kepler's first law. Kepler's second and third laws more or less derive from the same assumptions. It stands to reason then, that any system where $F \not \propto r^{-2}$ does not follow any of Kepler's laws. The orbits are not perfect conics (e.g., ellipses, circles, etc.), they do not sweep out equal areas in equal times, and the standard $P^2 \propto a^3$ certainly does not apply.
Orbital Motion in a Galaxy
Your question correctly describes the situation for stars (or anything really) orbiting in a galaxy. Stars are not orbiting central, point-like masses. They are embedded within both the baryonic and dark matter that comprises the galaxy and are orbiting through it. It is a well known concept in physics that spherically symmetric mass distributions do not have a net gravitational pull on objects interior to that distribution, which means that for stars in a galaxy, the mass affecting its orbit is the mass interior to its radius. If that radius changes, the mass changes!
The central force on our star is still going to be gravity, but the mass acting on it will be all the mass interior to some radius, denoted by $M_r$. We can see that $F(r) = GM_r(r)m/r^2$. If we want to determine the force acting on our star (and thus the exact orbit, via the differential equation above), we first need to figure out what the mass interior to some radius is. This can be achieved by using the mass continuity equation.
$$\frac{\mathrm{d}M_r}{\mathrm{d}r} = 4\pi r^2 \rho(r)$$
You can essentially figure out all the mass interior to $r$ by integrating up all the mass density as a function of $r$. Here of course, you need a good equation for $\rho(r)$. A simple, but physically unrealistic density profile is the Singular Isothermal Sphere (SIS), whereas a more realistic, but mathematically complex equation might be the NFW profile or the Einasto profile.
I've now laid out all the steps you need to figure out the orbital motion in a galaxy, but I have to say, it's not pretty. We can look at part of the simplest case though, that of the SIS.
Single Isothermal Sphere
For the singlar isothermal sphere, you have that $\rho(r) = v^2/(4\pi Gr^2)$, where $v$ is the rotational velocity of your star. This profile relies on a crucial fact of disk galaxies $-$ their rotational profile is flat! This was well established, for example by Ruben et al. 1978. I've reproduced a figure from this paper below which shows the rotation curve for several galaxies. The important point here is that this shows $v$ is constant and does not depend on radius! (Assuming we're not near the galactic bulge or center. That's a whole different beast.)

With this crucial piece of information, we can solve for $M_r$ by integrating $\rho(r)$ (which I'll leave up to you). The result is that
$$M_r = \frac{v^2r}{G}$$
This means your force is given by
$$F(r) = \frac{v^2rm}{r^2} = \frac{v^2m}{r}\Rightarrow F(1/u) = v^2mu\propto ku$$
You can see here that unlike the Keplerian case, our force is proportional to $r^{-1}$ rather than $r^{-2}$. You can do this process with other density profiles (such as the NFW or Einasto I listed above), but you'll wind up with the same result.
If you're so inclined, you could choose to plug this into the orbital equation of motion above and solve it out, but you're now working with a non-linear differential equation and things can get messy fast.
Punchline: I'm not sure if this actually answers your question or not. I've led you partially down the rabbit hole, but hopefully you can appreciate how complex it quickly gets. All the above work was using broad assumptions and simplifications. I suppose the short answer to all of this is that stars orbit galaxies in a complex, yet closed orbit which is not easily described precisely (even for our own galaxy) via calculable equations. We can approximate and do our best to work through the math, but in the end its an approximation. In the crudest approximations though, you may as well consider an orbit such as our star to be circular and be done with it.


why does pluto have a high orbital inclination

I think the best answer is that there is no particular reason.

In any planetary system, most orbits tend to stay close to a common plane - close, but not exactly there. Even the big planets are slightly out-of-plane with each other, but the differences are tiny.

As you move away from the bulk of the planetary system's mass, smaller bodies tend to be more random. They are more easily deviated from encounters and near misses. Their origins might be different from the rest. If they get close to the outskirts of the system, that's a more unruly place, and orbit-changing events are more likely - especially for small bodies. Finally, the whole distribution of orbital planes has a bit of a stochastic (random) component anyway, and there will always be outliers that are quite different from the bulk.

In theory, even some larger planets could be significantly "out of plane" (much more than they are right now), but that's unlikely and it tends to be smoothed out over large periods of time.

So, Pluto is different because it's smaller and it's not really a planet, but don't attach too much significance to it.

Try to install Universe Sandbox and play with some of the included scenarios. Big systems bound by gravity can be pretty complex, and completely orderly configurations are rare.


Can a probe orbit Pluto given Charon's gravity?

Avoiding hard math, which I'm not very good at, the simple answer is yes, provided there's enough distance.   Pluto/Charon have 4 moons orbiting them a bit further away, all in relatively stable orbits.   Source

Here's distances to scale.  - same source.

Because the ratio of gravitational field strength to size is exponential (Power of 1.5), the apparent closeness of Pluto and Charon to the 4 small moons looks unusual, but it's perfectly acceptable for objects of that size.

Lagrange point stability requires a mass ratio of about 26 to 1.  (.0385 to 1 per source).  The Earth has satellites in unstable Lagrange points, so it's certainly doable, it's just not technically "stable".

The hardest part about getting a satellite to orbit Pluto is that Pluto's sufficiently small that any ship that approaches it would need to slow down significantly on it's own to get captured into a Pluto Orbit.   That's why New Horizon was a flyby, not an orbit.


What are evidences of stationarity of rotation of galaxies

If stars were actually about to escape, they would do so on a so-called dynamical timescale:
$$
t_\mathrm{dyn} \sim \sqrt{\frac{R^3}{8GM}},
$$
where $G$ is the gravitational constant, and $R$ and $M$ are the radius and the mass of the galaxy, respectively. For a Milky Way-sized galaxy, this is equates to roughly 30 million years$^\dagger$. But disk galaxies are observed over time spans of billions of years. If they had for some reason been gravitationally bound and then "unleashed", they would have to be so at times synchronized such that we, today, in all directions, observe them in a state that has a very short time span compared their lives throughout the history of the Universe. That is, distant galaxies (which are young because we look far back in time) were unleashed early in their lives, and nearby galaxies (which we see in the "old age" because we look less far back in time) were unleashed recently. This not only seems unlikely, but would also put us in a special place in the Universe.

Note also that the paper you link to is over 20 years old, non-refereed and has no citations (and that the author in general has very few citations). Of course this by no means implies that he's wrong, but it does give a hint that it's a good idea to be cautious of the conclusions of the paper. Since that paper was written, much more independent evidence for dark matter has been obtained.

$^\dagger$Since the roughly constant rotation speed of disk galaxies with galaxy mass (a few 100s of km/s) implies that $R$ grows approximately linearly with $R$ (and not with $R^3$ as one might expect), smaller galaxies — which exist in much larger numbers than large galaxies — have even shorter dynamical timescales.


What is the cause of all of these sharp, concentric rings around bright stars in this HST image?

The diffraction pattern at the focal plane created by a circular aperture is called an Airy Disk or Airy Pattern.  Both the outer opening and the inner hole plus secondary contribute to the exact function.  This is usually not easily observed with ground based telescopes because the seeing fluctuations due to air turbulence smears it out.  These images use filters that are a few 100 Angstroms wide, so although not monochromatic, they are narrow enough to clearly see the pattern.  The spectral shape of the star's light within the filter makes the effective width of frequencies more narrow.


Why is the Sun less massive than other G2V stars?

It isn't. You've just got  dodgy table from wikipedia.
A more modern (and well-used) version is here.
It lists
G1V  1.07 5880
G2V  1.02 5770
G3V  1.00 5720
This is an average relationship. The closest and most consistent relationship will be between spectral type and effective temperature and indeed the Sun is normally attributed a spectral class of G2V and $T_{\rm eff} \simeq 5780$ K.
The reason for any small discrepancy at the hundredth of a solar mass level with the mass in the table could be because the Sun has a slightly different composition to the average star that defines this table. The age of the Sun is unlikely to be a major factor. According to most models, the temperature of the Sun increases by just 100 K between a billion years old and around 8 billion years old and then cools a little bit. On the other hand, stars like the Sun do lose a little mass as they get older, mainly through a stellar wind. The Sun though is probably at around the average age for an early G star in our Galactic neighbourhood.


How is the density parameter estimated via cosmic microwave background anisotropies?

An appropriate explanation might begin by explaining that the cosmic microwave background (CMB) was formed in the expanding universe about 400,000 years after the big bang. An expanding gas cools, and when the gas in the universe cooled to about 3000 K, a transition tok place whereby the protons and electrons, hitherto free particles, combined to form hydrogen atoms. These atoms were almost transparent to the hot radiation field and allowed it to escape, free to travel the universe until we detect it 13.6 billion years later. In the meantime, the cosmic expansion has stretched the wavelength of the radiation from the optical/infrared to the microwaves we see now as the CMB.
At the time when the radiation was "released" there would have been instabilities in the gas - regions of slightly higher or lower density (compressions and rarefactions). For a number of reasons, these will leave a small imprint in the radiation field at that time, which is now observed in the form of tiny temperature variations in the CMB at different positions in the sky.
The "spectrum" of the temperature variations (basically, by how much the temperature varies as a function of angular separation on the sky) depends primarily on two things. (a) The scale of the compressions and rarefactions in the universe when the radiation was formed and (b) by how much and in what way the universe has expanded since then.
Factor (a) is directly dependent on how dense the universe was when the CMB was formed, which depends on how dense the universe is now and a cosmological model of how the universe has expanded. (b) Also depends on the rate of expansion, which can be expressed in terms of the Hubble constant, but also depends on the  topology of the universe - whether it is "flat" or "curved", since this can magnify or diminish the apparent size of the fluctuations.
In the end then, the appearance of the CMB now is dependent on both its formation conditions, the history of the universal expansion and the geometry of the universe. Each of these can be parameterised using standard cosmological models that include the current density parameter and Hubble parameter. These are related to conditions at any previous epoch in much the same way that the position and speed of a projectile can be used to predict is position and speed at any point in the past (or future).


What is the current accepted theory for the fate of hot Jupiters?

This is a fairly loaded question in that it depends heavily on what a "hot Jupiter" actually is defined to be. What is "hot"? What is a "Jupiter"? In reality, there's a continuum of planetary masses and distances from their parent star, and in the literature you'll commonly see references to "hot Neptunes", "hot Saturns," etc.

The predominant theory as to how giant planets form is that they first coalesce from rock and ice beyond the ice line, the distance from the parent star at which water becomes solid. This distance is approximately where Mars lies today in our solar system. What's surprising about "hot gas planets" is that they are found within this ice line, significantly within. This implies that after they formed their cores, they migrated closer to their host stars via some currently undetermined process (for which there are several good candidates, but for now let's assume that the existence of hot planets shows that at least one of these processes operates quite regularly).

And what about the word "hot"? Well, for the planets that are closest to their parent stars, there is known to be a radius anomaly: The radii of these planets are significantly larger than models of giant planet structure irradiated by their host stars would predict. So I would define "hot" planets as gas giants whose radii are larger than what would be predicted by the standard models.

Now that we got some of the definitions out of the way, there's the question of survival. When giant planets are close to their parent stars, they become tidally locked. As a consequence, there is very little energy tidally dissipated on the surface of the giant planet, the shape of the planet is fixed and there are little internal motions. However, the giant planet also raises a tide on its host star as well, and because it takes a lot of angular momentum to change the spin of an object with 1,000 times more mass, the host stars are almost never going to be tidally locked to their closest planet.

The rate at which energy is dissipated within the star is highly uncertain, and this uncertainty is typically swept into a fudge parameter "Q," the quality factor, with lower quality factors reflecting more dissipation. "Q" is measured for certain bodies in our own solar system (i.e. Earth and Jupiter) and in some stellar binaries, but is highly variable from body to body, ranging from about 10 for the Earth to 10^8 for some stars.

Whether a planet survives to be observed today depends on how long the orbital decay time, which is determined by Q, compares to the age of the system. For some systems, such as WASP-12b and WASP-19b, which feature highly inflated hot Jupiters, Q is estimated to be small enough to cause them to fall into their host stars in a surprisingly short time (< 10^7 years).

One other possibility is that the gas surrounding the rock/ice core is blasted away by the tremendous amount of heat deposited into the planet. This leaves you with a relatively low-density planet that's somewhat devoid of iron, as the cores of giant planets form further from their host stars than the rocky planets. There are a few candidate close-in, Neptune-mass objects that may have been produced as a result of them losing the bulk of their atmospheres in this way (Example: GJ3470b).

As for our own solar system, the formation of a hot Jupiter would have likely destroyed the inner solar system as it migrated close to the Sun, owing to the fact that it would violently perturb the inner planets' orbits. Additionally, the Sun would likely be enhanced in metals owing to the accretion of metal-rich material from this giant planet. While it's potentially possible that there was a hot Jupiter in our solar system before the other planets formed, it currently seems unlikely.


Why can falling stars only be seen moving down towards horizon not up away from horizon?

The falling star or shooting star has nothing to do with a star. If shooting stars were actual stars, we'll be burning in their atmosphere, rather than the other way around.

When a meteoroid, comet or (small) asteroid (or space debris in some cases) enters the Earth's atmosphere at high speed (typically in excess of 20 km/s), aerodynamic heating produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This is called a meteor or shooting star (or a falling star).

In general, the meteors travel in a (approximately) straight path. We usually see only the large ones travel down towards the horizon while the small ones are burnt up long before that (giving the impression of coming from the horizon).

Image from Meteors? Even More Remarkable by Bruce Maccabee at brumac.8k.com

Consider the meteor path in the image above. it can be seen that,

The observer will see the meteor only after it passes the initial altitude $H_{i}$. If he sees the meteor at this point, and it burns up in the atmosphere, the impression is that the meteor comes from the horizon.
If the observer notices the meteor after it passes the initial height and tracks it through its travel in the atmosphere, till is exits at the final altitude $H_{f}$, it will appear as if the meteor is moving down towards the horizon.

Another factor is the thickness of the atmosphere itself. The thickness of the atmosphere increases in path of the meteor from $H_{i}$ to $H_{m}$, and decreases from there to $H_{f}$. As the 'glowing' of the shooting star is directly dependent on the density of the atmosphere, its visibility trails of near the observer's horizon. This is another reason for the meteor looking like its falling towards the horizon.

This can be seen clearly in case of meteor showers which appear to radiate from a point.

Image from Lyrids Meteor Shower 2014 by Steve Owens at darkskydiary.wordpress.com

Here, the meteors appear to come away from the (radiant, which is above the) horizon.


Could a heavy exoplanet squeeze water into being liquid and 4° warm?

Hardly. Maybe. Probably. It depends.
Liquid water at atmospheric pressure is densest at 4°C - but that is a function of pressure and temperature.
The phase of a material (solid, liquid, gas, critical) depends on both, pressure and temperature - and so does the density which depends on both, linearily within one phase, discontinuous at phase boundaries. The phase diagramme of water which describes the phase you find for any combination of temperature and pressure is surprisingly complex for water and thus one of the least well-established ones.
Maybe one can answer your question already with a clear 'yes', if you consider a sub-surface ocean an acceptable answer. Enceladus' interior is somewhat heated by tidal interaction with Saturn. From Cassini observations we know that under about a 30km thick ice sheet on Saturn's moon Enceladus we have an ocean of liquid saline water - a source for the geysir activity on that moon.


Black hole at Galactic Center of the Milky way and Bar of stars?

Lets get a sense of scale. The bar is about 30000 light years long, and contains the central bulge of the galaxy, a roughly spherical mass of stars with a diameter of 10000 ly.  It's big.

The supermassive black hole is pretty big for a black hole, but only about 80 light seconds across. Compared to the size of the bar, it is tiny.

The supermassive black hole is found at the centre of the bulge, at the centre of the bar, and entirely within it.

Sometimes a star does fall into the black hole. The black hole is currently inactive, that means that there is not much matter falling in every year. Each year, roughly 1/10000 of the mass of the sun falls into the black hole. This is mostly matter from the solar winds of nearby stars.


Can a planet exist with the same mass as the earth, but have a different diameter?

Yes, a planet can exist with the same mass as Earth, but have a different diameter - if it has a different density. The lower the density, the larger will be the diameter. In terms of an exact formula, we have:

$$d(M, \rho) := 2 \cdot \sqrt[3]{\frac{3M}{4\pi \rho}}$$

The trick is that density is specifically a function of composition: hence, the extent to which the diameter can vary depends essentially completely on what value you have in mind for when you ask for a composition as being "similar". Dramatically different densities, and hence diameters, will only be possible with a dramatically different composition - e.g. a heavy gas envelope will provide a significantly reduced density, as would a significant mass percentage of water (think of this hypothetical Earth-mass planet as being like a big version of certain moons of Jupiter and Saturn, such as Europa and Enceladus. In fact, some percentage of discovered extrasolar planets are believed to be of this type: though you should not think of it as having liquid ocean all the way down, because after about 100 km of depth or so on comparable-to-Earth gravity, water freezes due to compression.).

If by "similar" composition we mean that, say, the planet is still to have a solid, hospitable surface, and to have all the same materials as on Earth but just perhaps in somewhat different abundances, one "easy" way might be a reduction in the mass of iron. A lot of the Earth's density is supplied thanks to its relatively large central iron core: compare it with the Moon, which has little to no iron core - the density for Earth is 5.5 g/cm^3, while that for the Moon is about 3.3 g/cm^3, althuogh one should also be mindful of compression effects due to gravitational crushing of the interior. A small iron core would give one a density in the intermediate range between these two.

With such a smaller iron core, to give one Earth mass (about 6000 Yg), it must be exchanged for a greater quantity of less-dense rock, hence the planet's diameter will be larger. Conversely, we could imagine having more of an iron core, similar to Mercury, and then the diameter would be smaller (and the surface gravity larger) as we should have less rock to avoid going over the one Earth-mass prescription.


How was this interstellar fireball observed and its trajectory analyzed?

"US Government sensors is a euphemism for "nuclear weapon warning early warning satellites", which is why we won't know details of the actual sensor: It is a military satellite and secret.

This particular event occurred in the Bismark sea, North of Papua New Guinea, it probably could have been seen from Papua or parts of Indonesia, if the sky was clear (which it rarely is, in the middle of the tropical rain band), but if it was seen, it wasn't reported to the American meteor society (reports from that time)

The satellites are designed to look for rocket launches, and to track the missiles as the climb through the atmosphere to determine the likely target. Thus they are able to get accurate position and velocity vectors of meteors. If you know the position and velocity vector of the meteoroid just prior to impact, it is an exercise in Newtonian gravity to work out the orbit of the meteoroid around the sun. (It is a three body problem as the meteoroid is moving in both the Earth's and the Sun's gravitational field)


The darkness does not have a speed, as it's just the space the light travels through.  It's always there until the brief instant a photon whizzes through.

It would be analogous to asking what speed the road is travelling at, rather than the car.


Why is it that night has a almost a bluish tint to it instead of red?

The blue tint is an illusion caused by the wavelength sensitivity shift when switching from rods to cones as the light intensity decreases below certain level.

It is called the Purkinje effect.

Objectively, moonlight is not blue. In fact, it is even more yellow than sunlight! (comparing the same elevation, because obviously the true color can be distorted by the atmoshpere, too). The moonlight is not bright enough to fully enable the photopic vision and our red color sensitivity suffers, producing the blue moonlight illusion.


Will you know the date and location from where the image is taken?
If you are sucessfully using the big dipper already you probably using this info.
There is a "degeneracy" between date and time.  The sky rotates once a siderial day (23hrs 56 min) and once a siderial year (which is 20min longer than tropical year: sun to same position). So the sky tonight at say 20:00 is the same as the sky at 20:04 the previous day. In two weeks that's 1 hour. So you need the date.

At the same time (say Universal Time == London time) at each longitude the observer is facing a different direction. So 15 deg long is equivalent to 1 hour in time. So you need longitude. Others have alluded to latitude. You see stars diffrently at different latitudes.   For example at the equator the pole star is on the northern horizon, and invisible in the sothern hemisphere.

The core relationships is between right ascension (like "longitude" for stars), siderial time and hour angle. The hour angle is the angle between the meridial and position of star. Or the siderial time is the right ascension of stars crossing the meridian. You could work this out from an wide field image using a service like https://nova.astrometry.net/ and then convert "siderial time to local time". There's a stack overflow answer plus others.


"What was that spot in the sky?" - How to identify satellites

This will be hard to track down. I'd suggest looking at something like SATVIEW Time Machine. I've entered the location and time you provided (I had to deviate by a few minutes to find something so maybe your time is not 100% accurate) and I can see there were a few satellites flying overhead. With enough work, you may be able to figure out which one was most likely the one you saw based on the Alt-Elev information detailed in the table.

There's plenty of other online satellite tracking websites out there, but keep in mind, most of them aren't going to have everything. And possibly, what you saw is in no database because it's a classified satellite in which case you probably won't find any record of it.


Can we detect the wobble of neighbouring stars around their barycenters?

Yes it is. The Gaia astrometric satellite is and will be doing exactly that. The subtle movement of the photocentre of a target star due to any orbiting planets can be separated from the proper motion and the annual parallax due to our motion around the Sun.

The technique is most sensitive to massive planets that are widely separated from  parent stars that are close to the Sun and have low mass themselves, since these provide the biggest angular shift in the star's position with respect to the barycentre. However, the separations cannot be too wide or the orbital period of the star/planet system will be too long for the motion to be detectable on the timescales of the mission. In practice therefore, Gaia will be most sensitive to planets with orbital periods of 1-10 years (or separations of 1-5 au). Simulations suggest that tens of thousands of exoplanets may be revealed in this way (Perryman et al. 2014).

ESA have even provided a little animation to show how it works.


When was Hubble tension first noticed? When was this term first used?

Hubble tension refers to the incompatibility between different measurements of the value of the Hubble constant. These measurements are incompatible up to more than $5 \sigma$. This incompatibility arises between what we measure "nearby" and what we measure further away, and indicates that there might be some physics we don't understand yet.

Now there has always been some disagreement between the different values for $H_0$ -- in the early 20th century, estimates ranges from 50 to 550 km/s/Mpc -- but the error bars on those measurements were very large. So in the early days of the Hubble constant, although the proposed values varied wildly, there was no tension (well, there were quite a few arguments, but no Hubble tension yet), since it was accepted that the constant was the same on all scales.

Values of the Hubble constant from 1920 to 1962

The most likely value for $H_0$ mostly shrank in the middle of the 20th century, as bias in previous measurements were discovered. By the 60's, most agreed that $60 < H_0 < 130$. But around 1975, there started to be a divide between those who though that the constant was around $55$ km/s/Mpc and those who thought it was around $100$ km/s/Mpc. For instance, de Vaucouleurs found $H_0 = 50$  km/s/Mpc in 1970 and $H_0 = 100\pm 10$ km/s/Mpc in 1977. But error bars were still large, and it was expected that the values for $H_0$ would converge.

Values of the Hubble constant from 1975

Up until very recently, the difference between measurements could reasonably be explained by the size of the error bars. But as measurement errors decreased, the range of possible values did not. This is what led to what we call the Hubble tension.

What really made the Hubble tension clear were the measurements of the Cosmic Microwave Background (CMB) done by Planck in 2013. These were the best observations of the CMB that indicated that $H_0 = 67.74±0.46$ km/s/Mpc, in complete disagreement with the value around $73$ or $74$ km/s/Mpc derived from type Ia supernova.
The Planck observations made the discrepancy between these values statistically significant.
Around the same time, measurements of the Baryonic Acoustic Oscillations (BAO) confirmed the tension between the values obtained at low redshift (with the standard distance ladder) and the values obtained at high redshift.

The Hubble tension was noticed before it was formally called that way. See the following articles that describe this tension without using the expression "Hubble tension":

Planck and the local Universe: Quantifying the tension, 2013
THE 1% CONCORDANCE HUBBLE CONSTANT, 2014
Cosmological implications of baryon acoustic oscillation (BAO) measurements, 2014
The Hubble Constant, 2015

In fact, a search on arxiv, ads or google scholar for "Hubble tension", or "Hubble-parameter tension" before 2014 yields no results. It seems that in 2015, this exact phrase started to really be used (for instance here).

In the last five years, more measurements of $H_0$ have been made, confirming the tension between the "high-redshift" values and the "low-redshift" values, and solving the Hubble tension is a hot topic of research.


Difference in the baryon acoustic oscillation measurements

I think the most important reason is that galaxy clusters more strongly adhere to the BAO shape than individual galaxies do. Galaxies tend to have more dispersion than clusters, making the BAO signal from galaxies smeared out and thus harder to detect. This is alluded to in the Veropalumbo et al. 2014 paper introduction.

  As tracers of the biggest collapsed structures, they are more strongly clustered than galaxies

Another potential reason is that the redshift of a cluster is easier to determine than the redshift of a single galaxy. There are multiple galaxies in a cluster, giving you multiple chances to measure a redshift for that cluster. If you can't measure a redshift for a single galaxy, you can't use it in your analysis. And of course all galaxies in a single cluster are going to have the same redshift to a good approximation. The Estrada et al. 2008 and Hong et al. 2012 both indicate $\sigma_z = 0.01$.

Note also that more often than not, these studies use the photometric redshifts rather than spectroscopic redshifts. Photometeric redshifts are generally less reliable as they rely on correlations (that isn't to say spectroscopic redshifts are perfect though). If you can measure a few photometric redshifts in a cluster or else measure the photometric redshift for the bright, central galaxy with a high signal to noise, you can reduce your redshift errors, compared to measuring it for a single galaxy.

Lastly, in order to use a particular catalogue to measure the BAO signal, the catalogue needs to be both deep (i.e., measures everything down to a specific redshift/magnitude limit) and wide (i.e., measures across a large field of the sky). If your catalogue doesn't meet both of these conditions, it can't be used.

Clusters are easier to spot than individual galaxies. They can be found (as the Estrada et al. 2008 paper describes) by first finding the large, luminous galaxies which generally reside in the center of the cluster and are fairly easy to find. The cluster itself is then identified. Looking at individual galaxies may suffer from magnitude limiting effects or other biases, meaning that your analysis may not include all galaxies that exist at a given redshift. If that is the case, your two-point correlation function will not be accurate and you might miss or inaccurately characterize the BAO signal. I would expect this problem is mitigated by using the easier-to-identify clusters.

Just to point out that cluster samples are likely to be more or less complete, the Estrada et al. 2008 paper is quoted as saying

  Tests on mock catalogs indicate that the MaxBCG sample should be $\gtrsim 90\%$ pure and complete for clusters with $N_{200}\ge 10$.

The requirement of a wide catalogue is also pretty restrictive. The BAO is inconveniently large so not many deep galaxy catalogues are going to be wide enough to even see the BAO. The SDSS is one of the few galaxy catalogues that comes to mind that is wide enough and they have already found the BAO in that. Cluster catalogues are usually built up not from a specific sky survey as galaxy catalogues usually are, but from individual observations or else conglomerated from many galaxy surveys. As such they can be much wider.


Is there a link between the amount of swirling and vorticity in a gas giant's atmosphere and its distance to its Sun?

Distance from the Sun is not a major driver for vorticity of gas giants in our Solar System
According to Yadav et al., Jupiter indeed has a higher vorticity than Saturn:

There is a stark contrast in the number of vortices between the two
planets: about 200 with 1000 km or larger diameter on Jupiter, while
only 10 to 50 on Saturn.  Note that while both planets also have
smaller vortices, the disparity in the numbers still persists.
Vortices are also present at or very close to the rotational poles:
Saturn has a cyclonic polar vortex at each pole , while Jupiter has a
cluster of cyclones at each pole

The authors claim the difference in the vorticity is mostly due to the differences in atmospheric make-up of the two planets (and their magnetic fields) and don't mention distance from the Sun as a factor.  The below image is from their paper:

The difference in vorticity between Saturn and Neptune is less clear. Vortices may be less visible on Neptune as there is less color contrast in the atmosphere. Neptune certainly has many vortices including its own Great Dark Spots.  It also has the highest winds observed in the Solar System.
The vortexes on Neptune are also modeled based on what we know about the atmosphere.  Hadland et al. model methane cloud formation in their paper: EPIC Simulations of Neptune’s Dark Spots Using an Active Cloud Microphysical Model.  Here is a picture of the Great Dark Spot (center) as taken by Voyager 2 in 1989.

You can contrast that image against this more recent image of Saturn from Cassini:

It certainly isn't clear to me which planet is more turbulent.
Reading the aforementioned papers, the vorticity of Jupiter, Saturn, and Neptune are due to:

The atmospheric makeup
The quantity of heat transfer in convective currents
The Coriolis Effects on the atmosphere due to the spin rate of the planets

Note that the ratio for power emitted to power absorbed by the Sun are 2.5 for Jupiter, 2.4 for Saturn, and 2.7 for Neptune, which means the convection currents are mostly driven by geothermal energy loss, rather than solar heating (as opposed to Earth's atmospherics).
Hot Jovian exo-planets are indeed expected to have "extremely swirly, complex weather patterns!"
Though the Sun doesn't play the primary role for our distantly gas giants, it probably plays an enormous role in the weather patterns of Hot Jupiters!  Most of the Hot Jupiters will be tidally locked to their stars since the time to tidal lock is proportional to $a^6$, where $a$ is the semi-major axis of the orbit of the planet.  Planets with a small $a$ tidal lock very quickly.
Hot Jupiters in tidal lock with their star have a much slower rotation rate (on the order of days) than our Jupiter's rotation rate (~10 hours).  That means the contribution to vorticity due to Coriolis Forces will be proportionally less.  However, heat transfer in the upper atmosphere will be much more dramatic since much of the heat absorbed on the "day side" of the planet will be emitted from the "night-side" of the planet. Skinner and Cho (2021) perform numerical simulations in their paper Numerical Convergence of Hot-Jupiter Atmospheric Flow Solutions and highlight that their solution:

features a dynamic, zonally (east-west) asymmetric jet with a copious
amount of small-scale vortices and gravity waves.

Here is one of the beautiful simulation images from their paper:


Correlation of planet sizes with star sizes?

I think there are a number of studies that look at these statistics; I'll try to dig some out. In the meantime I can give you some further things to ponder.

First, let's assume a null hypothesis that the statistics and mass distribution of planets were independent of stellar mass - what would we observe?

Well one selection bias you don't mention is that the likelihood of transit detection depends on the ratio of planet to star size. This means you are less likely to see small planets around F-stars than M-stars. Or to put it another way, you expect the fraction of large planets to increase with stellar mass - as you have found.

It is quite likely that the material available to form planets is correlated with the stellar mass - in other words there is a correlation between stellar mass and protoplanetary disc mass. In which case it could be quite difficult for smaller stars to have the requisite protoplanetary disc density to form giant planets before the disc disperses.


What inaccuracies are in this Stellarium depiction of the sky?

Let's get the ball rolling! Hopefully some of these are what you are looking for.
Point light approximation

I take it the size of the objects are illustrative of apparent magnitude and not the visible size

This is important, and it does not only appear to the planets. Everything in this image, except the Sun, the Earth, the Milky Way, and to some degree Venus are going to be point lights to the naked eye.
Computer monitors or paper are not very good at illustrating point lights. They have by far too few nits.
Instead, brightness is approximated by giving all the stars some visible area, proportional to their brightness. It's a useful approximation, but I would definitely count it as an inaccuracy compared to the naked eye.
Relative size of the Sun and the Earth
The Moon is famously the same relative size in the sky as the Sun. However, in this image, we are viewing the much larger Earth from the same distance, yet it still appears the same size as the Sun. The size of either or both has to be wrong
Sun glare
This image shows glare from the Sun. This will vary a lot depending on your eyes. And if the light form the Sun is blocked out like you propose, none of the solar glare in this image will be visible.
Light from the lunar landscape
Unless you also block out the surface of the Moon, the reflected light is going to severely impact the view.
Smooth edges of objects
Real life has no smooth anti-aliased edges. Cutoffs by planetary discs and stars are very hard, unless an atmosphere is present.
City lights on the dark side of the Earth?
Apollo astronauts could not see this, but admittedly, it was because of limiting observational conditions. Stellarium like does not simulate this
Image compression artefacts
These screenshots have two kinds of compression artefacts introduced by the JPEG format. Needless to say, these are not present in real life.

Banding, due to the limited number of dark colours available in the  YCbCr transform used by JPEG:

Macroblocks of 8x8 pixels, due to the discrete cosine transform

Colour space
All the way from the colour handling in Stellarium, to the YCbCr space of JPEG, to how your monitor is calibrated, decisions are made about colour. The end result may be far removed from how they really appear. Notably, Stellarium is likely considerably increasing the saturation of the colours of the Milky Way.
Black level
The black in this image is merely grey. Even the brightest parts of the Milky Way will appear darker than what can be found in the screenshot.
Field of view
"Personally I don't see in rectangles" as a comment says. The human field of vision is not rectangular, and it is almost twice as wide as the 90 degrees in this crop.
Brightness and colour sensitivity is also very different between the centre and edges of human vision.
mac status bar: this is not visible in real life :)


I'm treating this as a homework question. You have made a clear attempt, so it is not off topic, but you should clarify the source of the problem.

You will first need to establish if the rotation of the Earth is to be considered (it makes a big difference, but in a homework, sometimes the Earth stops spinning!) You have then correctly found the two possible answers, either 24 or 12 hours.

You can also consider the motion of the Earth around the sun (as the Earth only takes 23hr, 56min to spin once, relative to the stars). What this means is that, after 12 hours, the Earth has turned more than 180 degrees, and there is no orbital plane that passes over the person, and through the origin, and is perpendicular to the equator. However the error is only about 1 degree, so probably can be neglected for the purposes of homework.

For the second part, the first picture is certainly wrong, as the line of sight (in green) passes through the Earth. The second shows the line of sight (now in red) being tangent to the Earth's surface. You will need to use orbital mechanics to find the distance from the Earth's centre to the satellite, and find out the distance from the Earth's centre to its surface (green lines), giving two sides of a right-angled triangle. And the rest is maths.

There are a few details that may affect the answer: you should consider if the motion of the person will affect the answer to part (b) (and if not, why not). You should know that the Earth is not spherical. You should consider whether the optical refraction of light by the Earth's atmosphere should be considered.  However, in a homework problem, such issues can often be ignored, as their effects are small.


Did cosmological inflation occur at speeds greater than $c$?

The simple answer to your question is "yes" - the universe expanded at much greater speeds than $c$ during the inflationary epoch. This period of time was very quick but very dramatic, lasting from about $10^{-36}$ to $10^{-32}$ seconds. The universe expanded, in this very short period, by a factor of $10^{26}$. That's pretty incredible, when you think about it.

Inflation was originally proposed to solve, among other things, the horizon problem - that is, why the universe is isotropic and homogenous (on a large scale). This would mean that all the parts of the universe were in "causal contact" at one point in time. Inflation is the explanation for this.

Now, what does this all have to do with the CMB? Well, the temperature of the CMB is the same throughout the universe - a toasty 2.7 Kelvin. For the temperature to be uniform, all the regions of the universe would have had to be in causal contact in some point in time; hence, inflation explains the uniform temperature of the CMB.

However, the CMB was not around during the inflationary epoch. Far from it. It formed a lot later, when the universe was at the ripe old age of $379,000$ years. But the reason that it was formed (photon decoupling) equally throughout the universe is because the conditions were roughly equal, because of inflation. The CMB was, and still is, everywhere. It was never around during inflation, and as such was not otherwise affected by it.

I hope this helps.

My sources for the times:

Inflationary epoch

NASA


What reason is there to doubt the existence of the hypothesised planet Tyche in the far distant solar system?

Generally it would be that peers of Matese and Whitmire looking at the statistics and viewing that the results are not conclusive enough to indicate the existence of Tyche.  There can also be alternative reasons to explain the results mentioned in the paper other than a Jupiter sized planet in the Oort Cloud. As was mentioned in the comments, it is almost impossible to prove the non-existence of something.

At this time, the results from WISE are the best bet for locating Tyche if it exists.  As it will not be visible in the visible light spectrum but more likely in the infrared and the WISE mission scanned the entire sky which means that it won't have been missed due to looking in the wrong place.

http://www.jpl.nasa.gov/news/news.php?release=2011-060

http://www.nasa.gov/mission_pages/WISE/main/index.html#.UnJb3xCJ5fs

There isn't any hard science to show that Tyche doesn't exist.  It is skepticism that this is the explanation for observations to date.  These are the same reasons that Nemesis isn't thought to exist, the data isn't strong enough to support its existence.


To calculate a "luminosity light curve" from a time series of V-band photoetry, you need two things.

You need to know the distance. The distance to Betelgeuse is uncertain and that means the absolute value of the luminosity you get will also be shifted systematically up or down by whatever distance you adopt.

The V-band only contains a small fraction of the luminous flux from Betelgeuse (most is in the red and near infrared). To convert a V-band magnetude to a "bolometric magnitude" requires you to add a bolometric correction, which is dependent on the spectrum of the star. In principle, if the star is varying, in the V-band, that might be because the bolometric correction is varying (i.e. the luminosity is constant, but the distribution of flux with wavelength is changing). Or it could be that the bolometric correction is constant and the total luminosity is varying. You can't tell from the V-band data alone.

In summary - you can't do it.
The best you can do is assume a constant bolometric correction and turn your V-band light curve into a bolometric magnitude light curve and hence into a luminosity light curve. But then, since the absolute luminosity is uncertain because of the distance uncertainty, you may as well just work with the V-band flux.
To do any better requires nearly-simultaneous photometry in more bands, especially in the red and near-infrared.


Is there any definitive proof of stars being "born" in nebula formations?

In science there is no such thing as an irrefutable fact, there is just the best theory we currently have that explains all the data and makes sense. That is very different from an assumption.

A theory is something that you can test - make models and predictions and check that they match up to real life. An assumption is something that you don't have evidence for but you think it should make sense, so you choose to believe it.

Our current understanding of star formation and stellar nurseries is an excellent theory - it has been around for a long time, built up of many parts that have each been tested and confirmed, it matches observations in many ways. It's not perfect and there are still things we don't know (otherwise scientists would have no jobs) but in high school level textbooks we can reasonably present this as a 'fact' because we are pretty certain it won't change any time in the next 50 years.

An assumption would be like me assuming you are reading a high school textbook. It fits the facts (the level of detail in your question) but I haven't tested it (asked you about the book).


Relationship between galactic rotation and stellar lifetime

We know how long stars take to orbit in a spiral galaxy, because we know how big galaxies are (we can measure the size by knowing the distance using various methods such as Cepheid variables, and type Ia supernovae) and we know how fast the stars are moving. (We can measure the velocity of stars using the doppler method)

We know how long stars last for because we have a good model of how stars work. We can use this model to predict that large, hot stars will burn themselves out in a short time. We can verify this model by looking at the numbers of blue stars, red supergiants and supernovae.

From the first part we know that stars take about 250 million years to complete an orbit of the Milky way (which is typical for similar sized galaxies) From the second part, we know that giant blue stars live for only tens of million years, and so could not complete a full orbit of the milky way.

On the other hand, smaller stars like the sun live for billions of years and so make multiple orbits.


Not sure if these kinds of questions are allowed here, but I may as well answer it.

Universe Sandbox 2 is mostly realistic. Its accuracy degrades as you increase the flow of time. The slower the time, the more accurate the calculations. This
means that if you want to simulate the Solar System over a long period of time, it won't be very accurate. If you want to simulate day-long events, like planets crashing, it might be better.

However, Universe Sandbox 2 uses 300-year old Newtonian physics, rather than general relativity. Newtonian physics just requires n-body mechanics, so it is much easier to implement. General relativity requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep.

Instead of simulating N number of bodies, they would be simulating a huge number of points. They would start with some initial data of the shape of spacetime and then see how it evolves according to the Einstein equations, which are 10 highly non-linear partial differential equations. Accurate general relativity simulations require supercomputers.

Still, Newtonian physics can get the job done for the most part. The developers seem interested in incorporating some GR concepts, like gravity propagating at the speed of light, spinning black holes and ways to better approximate GR.


Why do we think that there is no two-solar-mass black hole?

As you said in your question we only know about Black Holes creating in Super Novae, which lead to relatively big BH. The reason they can't be smaller in these cases it that they are created when the core of the star reaches a critical mass that overflows both the electron and the neutron degeneracy. Lower masses can be hold by those degenerate mass states and won't collapse any further.

Theoretically BH with masses less than the solar one could exists if there was some method capable to compress it's mass reaching radii smaller than their Schwarzschild radius.


Herschel, the discoverer of the planet named it Georgium Sidus, "George's star" after his patron, King George III of Great Britain.

The name was not popular outside of Great Britain, and there were various other proposals. It was Johann Elert Bode who proposed Uranus, the Latin form of the Greek god of the sky.  It fits with the existing planets having the names of Roman Gods, and just as Saturn was the father of Zeus, so Uranus was the father of Saturn.


To start with, such two orbits are extremely unlikely. First of all, any object orbiting a star and deviating strongly from a circular orbit has a high risk of crashing into something. As you notice, there are very few planets in our solar system, and all of them except Pluto have a very regular orbit. Planets which have intercrossing orbits would have crashed into each other and fallen apart on the planetesimal stage, much earlier than becoming a planet.

Then, if you suppose that such an event does occur, the consequences will depend on multiple factors. For one, the composition of the planets - if the two planets are terrestrial (rocky), one would expect them to fragmentate and lose a large amount of their mass. Their cores would maybe fuse into one planet, and the debris would form one or several satellites. The exact outcome would be dependent on their speed, their density, and the angle of their collisions. A similar (but not the same) scenario happened during the late stage of Earth formation, when an object close to Earth's size crashed into it to create the Moon. If the sizes are the same and the formation is complete, chances are that most of the matter would fly away and the planet that would be left would be tiny.

It would be a different story if the two planets are gaseous. They have almost no solid matter in them, so chances are that the planets will fuse whilst losing (proportionally) significantly less matter than terrestrial planets. Still, there would be losses.

Lastly, if a terrestrial and a gaseous planet meet, the outcome is quite easy to predict. Gas planets are much larger than rocky planets, so the rocky planet would simply be engulfed. It will maybe cause a disturbance on the surface of the gas giant.

To wrap it up let me just note that such a collision would almost definitely change the orbit of whatever is left, so that it would either get pushed into the sun or away from it, causing it to float eternally in outer space. Have a nice day.


How harmful is Saturn's magnetic field for human life?

Wikipedia says that "The saturnian radiation belts are generally much weaker than those of Jupiter" mostly due to the rings and other orbiting bodies absorbing the ions, I think. I have't yet found anything quantitative though


How do observational astronomers manage streaks and other artifacts from objects in Earth Orbit?

Simple - as you say, in order to get lots of dynamic range in deep astronomical images, you generally need to split your exposure time down into a number of sub-exposures.

When you combine these images, you don't just add them up, you "median stack" them (take the median on a pixel by pixel basis) or perform a slightly more sophisticated flux-weighted averaging that kicks out the outliers.

This process gets rid of any moving objects in the final image as well as cleaning up all the cosmic rays.

The problem with this reflector that has been launched is that it will be the equivalent of a very bright star. It will totally saturate any single image it appears in and might actually be bright enough to damage, or at least leave a persistent image, in some infrared detectors.


Observations of a perigee coincident with a full Moon

The time between full moons is called the the synodic month, and the time between perigees is called the anomalistic month. The anomalistic month is shorter than the synodic month by about two days, (27.6 days and 29.5 days)

Since the anomalistic month is two days shorter, and there are nearly 28 days in a month, it takes about 14 months for the two cycles to come back into alignment. Thus there is a perigee at full moon every 14 (not 13) months.

More accurately the beat period between the two cycles is 411.78443 days. See wikipedia: https://en.wikipedia.org/wiki/Full_moon_cycle


Rising & Setting time of a star on a specific date

You'll need to convert local sidereal time to local time on the specific date, using the longitude of your location. This can be done with the formula described here:

  $$
UT = \frac{LST - \frac{L}{15} - 6.697374558 - 2400.051336*T - 0.000025862*T^2)}{1.0027379093}
$$

  where $UT$ is the time in Universal Coordinated Time, $LST$ is the Local Sidereal Time, $L$ is the longitude, and $T$ is:

  $$
T = \frac{JD-2451545.0}{36525.0}
$$
  where JD is the Julian Date at noon UTC.  Once I've calculated the time in UT, I then add or subtract 24 until the result is within 0 - 24 hours.  Then I can convert from UT to local time.


Why can't moon light (reflected sun light) turn the sky blue?

The simple answer is that it does, but it's not bright enough to be visible to the naked eye. Earth's atmosphere scatters the moon light just like sunlight.

The full moon (like the sun) fills about 1/2 of 1 degree of the sky, the entire sky being 180 degrees, give or take, so the full moon fills less than 1 part in 100,000 of the night sky, so there simply isn't enough blue light to be visible over the brighter stars even with the brightest full moon. Our eyes are very good at seeing variations in brightness, but not that good. . . . and, for what it's worth, the night sky has always appeared to have a dark bluish tint to me, but that might just be my brain playing tricks on me because logically I know it's there. I'm not sure whether it's actually visible.

With a good sized telescope, moonlight scattering acts as a form of light pollution. Telescope users know that you get better visuals when there's no moon.

Source.


Position of specific stars in the sky as a function of time

It's difficult to answer this question with any confidence, since it is so wide ranging, and the original poster has not (yet) provided any clarification.

However, it seems likely that the book "Astronomical Algorithms" by Jean Meeus, published by Willmann-Bell, will provide the answer(s).  Failing that, the publisher's website contains several other reference works on this topic. I have no affiliation with the publisher or author.

Finally, as the author of the question is a beginner, I wonder whether in fact a simple introduction to astronomy, of which there are hundreds, might be a better place to start.  Jumping in by trying to model the positions of stars looks hard, particularly as there are plenty of good applications already available, some of which are free.


How to find the local azimuth of the highest point of the ecliptic

Edit:
Getting the exact value winds up being simple, but it was only later in the day that the answer occurred to me. All we need to do is ask for the altitude and azimuth of the Ecliptic’s south pole! The ecliptic itself will be highest above our horizon in exactly that direction, at an altitude that is exactly 90° above its pole.
Thus the answer is given by the coordinates of a single vector:
from skyfield import api, framelib
from skyfield.positionlib import Apparent

ts = api.load.timescale()
t = ts.utc(2021, 2, 16, 22, 52)
bluffton = api.Topos('40.74 N', '84.11 W')

f = framelib.ecliptic_frame
r = api.Distance([0, 0, -1])
v = api.Velocity([0, 0, 0])

p = Apparent.from_time_and_frame_vectors(t, f, r, v)
p.center = bluffton

alt, az, distance = p.altaz()

print('Altitude of highest point on ecliptic:', alt.degrees + 90.0)
print('Azimuth of highest point on ecliptic:', az.degrees)

The answer:
Altitude of highest point on ecliptic: 67.54977465183501
Azimuth of highest point on ecliptic: 162.61593086793783

— is far more exact, and faster to compute, than my first try. Enjoy!
Original answer:
While it’s likely that someone will soon jump in with a closed-form solution that uses spherical trigonometry — in which case it’s likely that Skyfield or PyEphem will only be used to determine the Earth orientation — here’s a quick way to get an answer within about a degree:

Generate the ecliptic as 360 points one degree apart across the sky.
Compute the altitude and azimuth of each one.
Choose the highest.

The result agrees closely to what I see if I open Stellarium, turn on the ecliptic, and choose a field star right near the point where the ecliptic reaches the highest point in the sky.
import numpy as np
from skyfield import api, framelib
from skyfield.positionlib import Apparent

 = api.tau
ts = api.load.timescale()
eph = api.load('de421.bsp')
bluffton = api.Topos('40.74 N', '84.11 W')

t = ts.utc(2021, 2, 16, 22, 52)

angle = np.arange(360) / 360.0 *
zero = angle * 0.0

f = framelib.ecliptic_frame
d = api.Distance([np.sin(angle), np.cos(angle), zero])
v = api.Velocity([zero, zero, zero])

p = Apparent.from_time_and_frame_vectors(t, f, d, v)
p.center = bluffton

alt, az, distance = p.altaz()

i = np.argmax(alt.degrees)  # Which of the 360 points has highest altitude?
print('Altitude of highest point on ecliptic:', alt.degrees[i])
print('Azimuth of highest point on ecliptic:', az.degrees[i])

The result:
Altitude of highest point on ecliptic: 67.5477569215633
Azimuth of highest point on ecliptic: 163.42529398930515

This is probably a computationally expensive enough approach that it won’t interest you once you or someone else does the spherical trigonometry to find an equation for the azimuth; but at the very least this might provide numbers to check possible formulae against.


If you have, or can borrow binoculars, that is probably a much better way to try to find the comet than your telescope. If it's set up like the one in the image in your question (that I've added) you don't have a computer read-out for RA/Dec or any way to position your mount accurately so that the polar axis points at polaris. The bright skies in LA will make doing so manually a bit of a challenge, and lack of previous experience doing so will make it even more of one.

Go to in-the-sky.org and then to Planetarium view. Go to the bottom right and type 46P into the Find box and it will find your object's name and then display it for you. Turn on the Alt/Az grid instead of the RA/Dec grid and you can see that 46P is about 60 degrees up at 8 PM your time. Lay on your back and take some time to look around with your binoculars.

It should set your location to Los Angeles automatically, but double check that.

You can see that there are some very bright stars lined up to make your viewing easier. First is Betelgeuse the bright corner star in Orion (do not say Betelgeuse three times in a row!) then Aldebaran the eye of the bull, then the comet.

At about 4th magnitude you should be able to see it as a green fuzzy star, unlike anything else. You can also zoom in using in-the-sky.org to see the local stars. The comet is going to be moving relative to the stars, so don't make zoomed-in print-outs and try to use them six hours or a day later.

Set up anything you have around the spot where you'll be laying down, to block lights from neighbors or the street, turn your own lights off. You are going to be looking East-Southeast about 2/3 of the way up to the zenith in the evening, and 46P will remain high in the sky most of the night.

Have fun!

Oh, the telescope! Once you've gotten good at finding 46P with binoculars and have some feel for the other stars nearby, then try the absolute lowest magnification you have on your telescope and just look around slowly. Loosen up your mount so you can move the tube by hand, and scan around until you see the blue fuzzy blob again.