Could an object block out the Sun from Earth's view for 3 hours?

Is it possible?   Not really.  Not with your criteria.   Could some mad scientists make it happen, like if we make the Moon bigger?   Er, maybe.

Types of eclipses.

Nasa's eclipse page (scroll down for maps over 20 year periods).

2001-2020 map below.

The map has a notation for hybrid eclipses (see here), not all that relevant to the question.

Total solar eclipses (blue lines on the map), are quite small because the umbra or fully shaded region that the Moon casts on Earth is small, maybe 150 km in diameter on average, 267 km max.  All planets and objects that orbit the sun cast an Umbra that extends behind them.   The Moon (diameter 3,474 km), is always casting an umbra somewhere in space, being about 1/400th the diameter of the sun, it's umbra is about 1/399th the distance between the sun and the moon, or roughly 375,000 km on average.   Eclipses happen on Earth when the Earth is lined up below the Moon's Umbra.  Mostly they don't line up which is why eclipses are rare, but the Umbra is always somewhere in space, not all that far from the Earth, following the moon's path.

Rogue Planet

Your scenario mentioned planet-like.  Now, if it's a planet from our solar-system, that would imply disturbing the orbits of Earth/Venus/Moon, which you specified you didn't want.   There's also real danger if you were to have a planet like object in a near Earth orbit that passes from time to time, close enough to cast it's Umbra over Earth.   For safety reasons "just passing through" is the way to go.

A rogue planet on a hyperbolic trajectory past the sun that, by very rare coincidence, passes between the Earth and the Sun, but close enough to cast it's Umbra over Earth would be very rare to say the least, but a theoretically possible scenario to cast a total eclipse over Earth, but it would need to pass very close and at just the right angle to block the 1/2 of 1 degree of arc that the Sun takes up in the sky.

How close?

Mars is roughly twice the diameter of the Moon, so would need to pass within about 750,000 km to the earth, and line up just right, to cast a tiny umbra that moved across the Earth.

Venus, about 3.5 times the diameter, about 1.3 million km

Jupiter, about 41 times the diameter of the moon, would need to pass within about 15 million KM to cast an umbra on the Earth of similar size to the typical Moon's eclipse Umbra.

And those are the minimum distances.  You'd need the theoretical planet to be pretty well inside those limits to have an Umbra of significant size and for a 3 hour eclipse, bigger is better.

What would a rogue planet passing that close look like?

Consider that it's passing close enough that it looks larger than the Moon, at least for a 3-plus hours.   How close it would need to be depends on the size, as noted above.

The minimum velocity for an object on a hyperbolic trajectory just passing through the solar system but near Earth distance would be Earth's escape velocity, or the orbital velocity of the Earth (108,000 kph) times the square root of 2, and if it approached on the same plane and in the same direction as the Earth, the relative velocity between this Rogue planet and Earth, the minimum relative velocity would be about 45,000 kph, give or take.   That's absolute bare minimum as a theoretical rogue planet passing inside the Earth's orbit would usually be traveling a fair bit faster than that.

So, for a planet like that to remain between the Earth and the Sun for 3 hours, it would need to be about 3 times 45,000 km in diameter, or, nearly the size of Jupiter.

A small rogue planet couldn't cast an Umbra over Earth anywhere close to 3 hours because it would pass over the Earth much too quickly.   A Jupiter-sized Rogue planet could, theoretically, if it was closer than closer than 15 million km, cast it's umbra over Earth for 3 hours or so, but that would be close enough to change the Earth's orbit and, at the very least, give the Earth significant climate change by pushing the Earth a few million km either closer or further from the sun.    Not an ideal scenario.

So there's no scenario that does what you ask.  You'd need a planet the size of Jupiter, with the mass of Neptune or perhaps less to leave the Earth/Moon undisturbed.    A Neptune mass planet with a large ring system . . . just maybe.   But given how rare your scenario is to begin with, I'm going to say basically no.   You could do it, but it would change the Earth's orbit, or you could do it with a smaller planet but for much less than 3 hours.  In any case, it would be a preposterously rare scenario no matter what.

What if we move the Moon?

With rogue planets, I just looked at the relative velocity between the planets, which is close enough for a near-sighted umpire.  When you look at the Moon's umbra, the Moon's motion is much slower, so you also have to take into account the Earth's rotation too.

The Moon (Synodic period), takes 59 minutes to cross the sun, that's tip to full cover, and another 59 minutes back to tip.  (based on a 29.5 day synodic orbit, 29.5 days to cover 360 degrees, the sun, roughly 1/2 of 1 degree in the sky).   So, 158 minutes of partial eclipse, and a record, of 7 minutes total eclipse in one spot on Earth (7 minutes, the record for a total eclipse is possible when the Moon is slightly larger than the sun, close to it's perigee).

So if our mad scientists pull the moon closer to the Earth, the period of time for a total eclipse increases but the rate that the Moon moves across the sun also increases, so you're effectively increasing the 7 minutes of total but decreasing the 158 minutes of partial.   If you pull the Moon all the way into a geostationary orbit, some 10.5 times closer, the Moon would span an enormous 5.25% of the sky, but it wouldn't move in the sky while the sun would and the sun, which crosses about 1 degree of sky in 4 minutes would pass behind the enormous geo-stationary moon in just 21 minutes.   You can't create a 3 hour eclipse by pulling the Moon closer.

This is where the Earth' rotation speed comes in.  The Moon moves over the Earth in it's orbit at about 3,683 kph, and relative to the sun, but because the Earth is in orbit to, the effective velocity relative to the sun (Synodic not sidereal) is about 8% less.

The Moon's umbra effectively moves with the Moon's relative velocity - at least, close enough.

The Earth rotates in the same direction that the Moon moves, peak velocity at the equator of a bit over 1,600 kph.   In effect, the rotation of the Earth tries to keep up with the movement of the Umbra during an eclipse, but the Umbra moves West to East about twice as fast as the Earth turns West to East.

So the solution for your 3 hour eclipse?   Make the Moon bigger and push it further away so it orbits the Earth more slowly and it's Umbra moves closer to the Earth's rotation speed, making a 3 hour eclipse, at mid day, close to the equator . . . possible.

If we replace the Moon with Venus (3.5 times the diameter of the Moon) and we push it some 3 times further away (which would be pushing the limits of the stability of the Earth's hill sphere), then you'd be a lot closer, as our new "Venus Moon" would move across the sky and across the sun still to fast, but it would move more slowly and closer to Earth's rotational velocity leading to eclipses that stayed in the sky much longer.  If you did that, you might get your 3 hours of total eclipse.  You wouldn't want to push Venus too far, cause you'd want it a big bigger than the Sun in the sky, but 3 times the Moons distance, it would be about 15% larger and close to the limit of orbital stability.

Other silly ideas.

Now if you were to Move the Planet Uranus into Earth's L1 Trojan point, which . . . well, the Trojan point would be further away due to the larger planet's mass dominating so . . . that wouldn't work and because L1 points aren't stable, it would also be highly dangerous.   Moving on.

If you were to move Uranus to where the Earth is and make the Earth a satellite of Uranus . . . we might get longer than 3 hour eclipses every time we were on the far side of Uranus making an orbit.   Seems a lot of work though just for an eclipse. . . . but that would work too.

OK, I'll stop now.


How far from the Sun can an Earth TCO (temporarily captured orbiter) go after leaving?

No answers after a week, so I will put a toe into the water an make a rough estimate. I think a thorough answer would require both a significant 3D orbital dynamics simulation with a million particles, and a way to define dynamically a threshold for what passing bodies would be considered "just barely temporarily captured" and which *not quite temporarily captured".
There should in fact be some good papers to cite that report this kind of simulation. I'll try to look around; I think there is a good opportunity for one or two more answers here.
Also see Have there been any documented mini-moons since 2006 RH120? and all of the sources linked there!
An example of an object in a medium-term stable orbit in the Earth-Moon system might be found in a halo orbit around the Earth-Moon L1, or L2 points. A spherical envelope-back calculation using spherical cow milk for ink suggests that in the rotating frame, it cycles in a 100,000 km orbit every two weeks around the Lagrange point, or about 0.5 km/sec in that frame. In a non-rotating Earth-centered frame, the Moon itself is moving at about 1 km/sec.
Since these temporary orbits are weakly bound, I'm going to say that an object entering the Earth-Moon system at roughly 1.5 km/sec can hang around for a while, then exit with roughly that velocity.
Looking inthe Sun's inertial frame now, The Earth-Moon system is moving at about 30 km/sec. That 1.5 km/sec is now looking fairly small, only about 5%. So if it exited prograde or retrograde, it would end up in a heliocentric orbit with a semi-major axis no more than 10% different than the Earth's. TCO's are NEOs that take the time to flirt with the Earth-Moon system, then continue on as NEOs to perhaps flirt again another day.

EDIT: I just visited my mini-moon question, found Gravnik 2012 and realized the plots are there! Here are some of them. It turns out everything I said is right. That's not a surprise, because I believe I read this paper a few years ago.
1 billion initial conditions, 10 million particles actually integrated.
The population of natural Earth satellites Mikael Granvik, Jeremie Vaubaillon, and Robert Jedicke, Icarus, Volume 218, Issue 1, March 2012, Pages 262-277  https://doi.org/10.1016/j.icarus.2011.12.003

Abstract
We have for the first time calculated the population characteristics of the Earth’s irregular natural satellites (NESs) that are temporarily captured from the near-Earth-object (NEO) population. The steady-state NES size–frequency and residence-time distributions were determined under the dynamical influence of all the massive bodies in the Solar System (but mainly the Sun, Earth, and Moon) for NEOs of negligible mass. To this end, we compute the NES capture probability from the NEO population as a function of the latter’s heliocentric orbital elements and combine those results with the current best estimates for the NEO size–frequency and orbital distribution. At any given time there should be at least one NES of 1-m diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286 ± 18) d. We find a small preference for capture events starting in either January or July. Our results are consistent with the single known natural TCO, 2006 RH120, a few-meter diameter object that was captured for about a year starting in June 2006. We estimate that about 0.1% of all meteors impacting the Earth were TCOs.


What is the mnemonic reason behind b being galactic latitude? (in the Galactic Coordinates frame)

Länge und Breite are the German words for longitude and latitude. Thus 'b' seems like a natural choice for latitude and 'l' for longitude. A century ago when traditional choice of variable names were chosen by vote of feet, a reasonable amount of research papers were still published in German or people with some form of fluency in German. At the same time it's distinct enough from the variable names used in other coordinate systems which also have longitude and latitude (like the terrestrial where $\lambda$ and $\theta$ are commonly used), so the cause for confusion is reduced - even when you don't know German.


Here is a diagram of his setup:

See the Wikipedia page for a discription.


Could Mercury be the creator of Earth's Moon?

There's a principal in science that a strong hypothesis doesn't depend on numerous improbable factors.  I read that somewhere, I think it was Phil Plait who created the bad astronomy website, but I can't find it right now.

For Mercury to have done what you claim, a 2nd planet would have needed to form between Earth and Jupiter, OK, maybe that happened, in fact, the young solar system was probably full of protoplanets.

This planet would then have needed to change it's orbit.  This usually takes a more massive planet to toss the smaller one, but there's two problems here.  Could Mars push Mercury all the way into Earth's orbit?    Probably not.   Could Jupiter?   Sure, but then, where did Mercury get all it's Iron if it formed between Mars and Jupiter.  It's becoming problematic.

Planet on planet collisions are rare, obviously not impossible.

Then there's the problem of how did the collison result in Mercury flying off, losing some material, the moon forming and Earth rapidly rotating.  Planet on planet collisions tend not to be glancing blows because planets are too big, and gravitational wells are pretty deep.   I don't want to say it's impossible but that's an odd scenario which seems highly unlikely.

And then, Mercury changes it's orbit again, perhaps gravitational assist from Venus and it ends up where it is.

A hypothesis like that depends on that many rare events all happening isn't not going to be a popular one.   Building a hypothesis on a single improbable event is fine.   The giant impact theory began like that.

A good hypothesis also has evidence behind it.   Mercury looks like it formed about where it is now, clearing out it's orbital neighborhood and being made up of denser material that was more likely to orbit closer to the sun.    There's very little reason to stand behind the Mercury impacted Earth idea and there's no evidence to support it.


Would DCBH form supermassive stars first before BH formation

This article summarises the research:

How does a cloud of gas become a black hole? According to a 2017 study, such a transformation requires two galaxies with very different personalities: one of them a cosmic overachiever that's forming lots of baby stars and the other a low-key heap of starless gas.

As new stars form in the busy galaxy, they blast out a constant stream of hot radiation that washes over the neighboring galaxy, preventing the gas there from coalescing into stars of its own. Within a few hundred million years, that starless gas cloud could accrete so much matter that it simply collapses under its own weight, forming a black hole without ever producing a star, Basu found.

I assume that some fusion does happen during the collapse, but presumably it never catches up with the increasingly intense gravitational pull of the densening gas cloud, so never supports the gas against gravity


Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?

Perhaps I've got it, or them.
These may be what I was thinking of, an optical intensity interferometer based on the Hanbury Brown and Twiss effect effect, but down-converting via heterodyning with lasers.
However: instead of the heterodyne down-conversion process happening in an optical crystal producing radio waves that are electrically amplified, mixing and detection happens in a semiconductor crystal acting as a photodiode and the output signal being photoelectron current. So this is not exactly what was described in the question. Since this is more likely to be the right way to do the measurement, probably I mis-remembered the conversion to radio followed by electronic amplification.

Possibility of Measuring the Width of Narrow Fe II Astrophysical Laser Lines in the Vicinity of Eta Carinae by means of Brown-Twiss-Townes Heterodyne Correlation Interferometry
The paper has a nice explanation of the effect and the measurement they are proposing.
There is light amplification and photons are bosons, etc.

Abstract
We consider the possibility of measuring the true width of the narrow Fe II optical lines observed in spectra of the Weigelt blobs in the vicinity of Eta Carinae. The lines originate as a result of stimulated amplification of spontaneous emission of radiation in quantum transitions between energy levels showing inverted population (Johansson & Letokhov, 2002, 2003, 2004). The lines should have a subDoppler spectral width of 30-100 MHz, depending on the geometry of the lasing volume. To make measurements with a spectral resolution of R > 107 and an angular resolution better than 0.1 arcsec, we suggest the use of the Brown-Twiss-Townes optical heterodyne intensity correlation interferometry. The estimates made of the S/N ratio for the optical heterodyne astrophysical laser experiment imply that it is feasible.

and more importantly from the introduction:

The Brown-Twiss correlation interferometry (Hanbury Brown & Twiss, 1956;
Hanbury Brown, 1974), modified by heterodyne detection using a CO2 laser as
a local oscillator (Johnson et al., 1974) for the 10 µm region, is well suited to achieve very high angular and spectral resolution simultaneously. We will call
this method the Brown-Twiss-Townes (BTT) technique. In the range 0.9-1.0
µm this can be done today by means of two spatially separated telescopes
equipped with two heterodyne photoreceivers, e.g. avalanche diodes, and a
tunable semiconductor laser diode transporting its radiation via an optical
fiber (Sect. 3). A general approach of this method has already been considered and discussed in a non-astronomical journal (Lavrinovich & Letokhov,
1976; Letokhov, 1996). In the present paper, we focus on the use of the BBT
correlation heterodyne interferometry to study the Fe II laser lines in the range
0.9-1.0 µm and the appropriate signal-to-noise (S/N) ratio estimates (Sect. 4).

And from Section 3:

To master new wavelength regions and achieve high spectral resolution levels, the intensity interferometry method was modified to become heterodyne interferometry (Johnson et al., 1974). This technique uses a local monochromatic laser oscillator to produce beats between the light wave of the star of interest and the coherent laser wave of the local oscillator. The method can be considered intermediate between intensity interferometry and direct interferometry. Townes and co-workers made successful observations in the 10-µm infrared window of the atmosphere using a CO2 laser as a local oscillator (Johnsson et al., 1974; Townes, 1977). The baseline used consisted of a pair of auxiliary telescopes spaced a few meters apart at Kitt Peaks solar telescope.

M. A. Johnson, A. L. Betz, and C. H. Townes, 1974 10-μm Heterodyne Stellar Interferometer

A spatial interferometer for 10-μm wavelength which uses two independent telescopes separated by 5.5 m, heterodyne detection of the infrared radiation, and path equalization by a variable-length rf cable, has given interference fringes from radiation of the planet Mercury. Continuous fringe observations during 4000 sec indicate remarkable stability in the optical-path difference through the atmosphere and the two telescopes, fluctuations between 20-sec averages being about 1/6 of the 10-μm wavelength.

D. D. S. Hale et al (2000) The Berkeley Infrared Spatial Interferometer: A Heterodyne Stellar Interferometer for the Mid-Infrared

A detailed description is given of the Infrared Spatial Interferometer (ISI), developed at the Space Sciences Laboratory of the University of California at Berkeley, which is a high spatial resolution interferometer for mid-infrared wavelengths. The instrumentation, its capabilities and performance, data analysis, science program, and future plans are all discussed. The system's use of heterodyne detection, analogous to that of a modern radio interferometer, is also compared with the homodyne or direct methods more commonly encountered in the visible and infrared. The ISI has been operating productively on Mount Wilson for the past 10 years measuring materials immediately surrounding stars and their changes as well as some stellar diameters. The new spectral capabilities described here, a recent increase in baseline length, and the upcoming expansion to a closure-phase imaging array provide important additional types of measurements.

FIG. 1. Conceptual block diagram of the  ISI optics, circuitry, and heterodyne detection system, including laser phase lock and path length compensation systems. Telescope 1 and its trailer, with associated optics and circuitry, is enclosed within the dashed line. Telescope 2 is represented by equipment outside the dashed line. The trailer for telescope 1 is the primary operating location, with the central computer, delay line, correlator, and data acquisition equipment. Telescope 2 and its trailer contains path-length compensation equipment in addition to the necessary optics and detection hardware. When a third telescope is installed, the controls, delay lines, data acquisition, correlators, and other equipment not needed in each single telescope, will be housed in a central station.

See also

Conceptual Design and Applications of HgCdTe Infrared
Photodiodes for Heterodyne Systems, part 2 (big)
Conceptual Design and Applications of HgCdTe Infrared Photodiodes for Heterodyne Systems


I'll be honest. I looked up Galactic Habitable Zone on Wikipedia. The idea was introduced in 1986 and expanded upon in the book "Rare Earth" by Brownlee and Ward. Essentially the idea is that in the bulge there is too much activity (supernovae, etc.) and in the halo of the galaxy there is not enough heavy elements (carbon, etc.) for life to exist. This was in reference to complex life "as we know it" (Check out another Peter Ward book entitled "Life As We Do Not Know It"). Stellar habitable zones are based on the presence of liquid water. This definition would greatly increase the theoretical "Galactic Habitable Zone". I think we need to look at a region and calculate all of the pluses and minuses of that specific region then keep our minds open.


While such orbits can be proven to exist mathematically, they are unstable, and so no real planet has such an orbit. Such an orbit is analogous to balancing a pencil on its point. Mathematically there is a point of balance, but it is still impossible to do.


What (actually) defines an Aten-class near-earth asteroid?

An Earth-crossing asteroid has either
an aphelion larger than Earth's perihelion ($Q \gt q_\oplus$, 0.983 AU),
or a perihelion smaller than Earth's aphelion ($q < Q_\oplus$, 1.017 AU).
For those which have both, the semimajor axis $a$ determines whether it is an Aten or an Apollo - subject to change by a close encounter with Earth.

This Minor Planet Center
blog post
is consistent with the JPL table.
I think the diagram used the wrong inequality for Aten aphelia.


You can see the constellation Orion on this photo:

(Screenshot from Stellarium)
Aligning the original photo with this screenshot, I found the highlighted stars to be the following:


Can a star orbit around multiple planets or a planet with massive moons?

There's theoretical ways to do it but it's so unlikely as to probably not exist.   Assuming you want a system where the planets are in stable orbits around each other.   The basic difficulty is the 3 body problem or n body problem.  More on it here and here.

For example, a massive planet could (in theory) have a single super-massive moon of similar mass to the planet, though that would probably be considered a 2 planet system.  It's not possible for one planet to have two super-massive moons because that wouldn't be stable.   In general, planets are many times the mass of all their moons combined, similarly stars are many times the mass of all their planets combined and when that stops being true the system is no longer stable.   It's very difficult to generate sufficient mass by adding moons to a system, or by adding enough planets to get more massive than the sun, unless the planets crash into each other but past a certain mass, they'd stop being planets and become a kind of star when enough planets combined.

If we look at gravitational systems, like our solar System, something like 99.7% of the mass is in the sun, so the sun dominates and everything orbits the sun.   A few of the larger objects have moons, and curiously, but only cause they're in relatively calm orbits far away from other planets, a few asteroids orbit each other, but the system is very structured around the sun with much smaller stable orbital zones around the planets.

Pluto also has a rather curious orbital system of it's own, likely caused by an impact, where Pluto and Charon are comparatively large and 4 tiny moons orbit around them.

Source:

But these kind of structured systems are only possible if you have a significant mass differential.   When you have 3 or more bodies of similar mass and similar distance you get a high degree of mathematical chaos and instability.   There are creative mathematical tricks to make it work, but none of them are stable or likely.

This is what 3 similar mass bodies look like, and in such a system, with constant changes, the most likely scenario is that one of the bodies eventually gets ejected.  (source, N-body problem above)

There are star systems with several stars, but they are either unstable or contain significant differences in mass.   The gravitational structure applies for large planets the same as it does for stars, and there's an article about that here.

You can create stability by having 2 objects orbiting each other and a 3rd massive object quite distant.  (Alpha Centauri is that kind of set-up though Proxima Centauri is quite a bit smaller, but it's the same system).

You can even create the heirarchy where you have 2 objects orbiting each other and then 2 more, orbiting each other, but distant so the 2 co-orbitals orbit around each other, and if you do this enough times, you can kind of create enough planetary mass where a more massive star could distantly orbit the entire thing, but it gets very structured and very distant.   It's not what I'd consider a normal orbit.

You could also cheat and have several planets in a wildly unstable general proximity orbit and have a star some distance away orbiting the chaotic mess in the middle, but it wouldn't be stable for long.


How to reduce star trailing when taking long exposure photographs?

Sounds like you should get a new mount. :-)

Being more serious, even expensive mounts have periodic tracking errors (due to manufacturing tolerance of gears and moving equipment) that need to be corrected. These are the basic ways to do the corrections:

Add a guide scope to the side of your scope, use an eyepiece with cross-hairs, and watch a star while you take a photo. When the scope drifts, you push buttons to move the scope to keep the star centered.
Do the same thing with a guide scope but use a dedicated guide camera that provides the corrections to the tracking.
There are also guide cameras that you can insert just in front of the camera. They use a small "pick-off" mirror to look at a star in the field (off axis guider). The guide camera provides the corrections to the tracking.

Of course, all of these methods require a mount that can be moved slowly and smoothly in both right-ascension and declination to keep the guide star centered. Also, the mount needs to be an equatorial mount and accurately aligned. (More expensive mounts can be alt-azimuth mounts and use a gadget to rotate the camera during the exposure.)


What is the density of stars in our region of the galaxy, up to say, 1000 light years distant?

Here's a crude way of estimating the answers:

The density of stars in the Milky Way disk decreases exponentially as you go out in radius and also exponentially as you go out in height above/below the disk midplane. The corresponding scale lengths are about 3 kpc (3 kiloparsecs = about 10,000 ly) radially and about 300 parsecs (about 1,000 ly) vertically ("scale height"); this means that if you go out in radius by about 10,000 ly, or up/down in height by about 1,000 ly, the density decreases by a factor of e (2.718...). Changing radius by about one tenth of the scale length or less, or changing height by about one tenth of the scale height or less, only changes the local density by about 10%, which isn't enough to worry about. That corresponds to about 300 parsecs (1000 ly) in radius, or about 100 ly in height.

So the for case of a 100-ly radius, you can assume the density everywhere is about the same as the local density suggested by the table you linked to. That density is 60 stars / ($(4/3) \, \pi \, 16.3^3$) = 0.0033 stars per cubic ly. For a sphere of radius 100 ly, you would then expect about 14,000 stars.

Naively, a sphere with a radius of 1000 ly would then have about 14 million stars. But a sphere that big extends above and below the disk midplane enough so that it samples about 1 scale height. So you can't just re-use the local density. Without getting into the proper approach of integrating the density functions in 3D, we can make a crude estimate as follows: If we used the density at one scale height (1/e times the local density), then the sphere would have only about 4.5 million stars. If we split the difference, that's about 10 million stars, which is probably not too terribly wrong. (I've been ignoring things like stars belonging to the "thick disk" and stars belonging to the halo, but they won't contribute a whole lot; also, the Sun isn't exactly at the disk midplane, but about 60 or 70 ly above it...)


Can a tectonically inactive planet retain a long-term atmosphere?

Yes, a tectonically inactive planet can retain a long-term atmosphere.
You make the connection that a lack of plate tectonics on a planet indicates a "dead" core and thus said planet has no magnetosphere. As such, I'm going to interpret your question as, can a planet without a magnetosphere retain an atmosphere long-term? As proof, I offer up Venus.
Venus' Magnetic Field
Venus is a planet without a magnetosphere generated by a core. It is thought that the cause for this is Venus' slow rotation rate (of nearly 243 days) and a lack of convection, allowing for bulk motion in the core. As I'm sure you know, you need moving charge to make magnetic fields and Venus' core just isn't moving. As such, we see that Venus is a tectonically dead planet - it's surface is roughly 500 million years old, whereas Earth's surface gets recycled every 100 million years or less due to our plate tectonics.
Now, Venus is not entirely devoid of a magnetic field. Ironically enough, it's lack of a magnetosphere allows for the generation of a magnetic field by it's atmosphere. Because the Sun's radiation is more or less directly hitting the atmosphere, Venus has a strong ionosphere. When you get lots of charged particles moving around in an atmosphere, you get a magnetic field. But on the whole, this field is very, very weak compared to a true magnetosphere such as we have on Earth.
I found this source which talks a lot about this concept and why Venus doesn't have a magnetosphere. Check it out to get a lot more in depth detail.
Venus' Atmosphere
So, Venus has no appreciable magnetosphere (or plate tectonics). Why does it have an atmosphere? And boy does it have an atmosphere. The surface pressure on Venus is estimated to be $\sim93\:\mathrm{atm}$.
In short, the answer is that the inundation of solar wind against an atmosphere is not necessarily the major contributing factor to atmospheric loss. It can be, but not always. For example, Mercury, another planet with a weak (but non-zero) magnetosphere, has no atmosphere (if it ever did) because it is so close to the Sun that the solar wind likely blew away that atmosphere long ago. Venus on the other hand is far enough away that the solar wind just can't strip the atmosphere. Here I'm going to quote wikipedia directly (emphasis mine).

A lack of magnetic field does not determine the fate of a planet's atmosphere. Venus, for instance, has no powerful magnetic field. Its close proximity to the Sun also increases the speed and number of particles, and would presumably cause the atmosphere to be stripped almost entirely, much like that of Mars. Despite this, the atmosphere of Venus is two orders of magnitudes denser than Earth's. Recent models indicate that stripping by solar wind accounts for less than 1/3 of total non-thermal loss processes.

Atmospheric Loss
If the solar wind isn't the contributing factor in atmospheric loss, what is? The answer to that is a process known as Jean's Escape. To put it simply, for gas particles in the atmosphere to escape into space, they need enough energy to climb out of the planet's gravity well. Some particles will have that energy and thus escape into space. Over time, the atmosphere bleeds off, little by little (this is happening for Earth too!).
The factors that contribute to the rate at which a planet loses its atmosphere are such things as the the planet's mass and radius, and the mass of the atmospheric particles. Let's look at Venus. It is comparable in mass and size to Earth and so has a reasonably appreciable gravity well. For something to escape Venus it must be traveling at $10.4\:\mathrm{km/s}$ (compared to the Earth's $11.2\:\mathrm{km/s}$). But, for Venus at least, the important factor there is that the atoms and molecules in its atmosphere are heavy. It is almost entirely ($\sim97\%$) carbon dioxide which has a mass of $\sim44\:\mathrm{amu}$. That means, the chances of such a massive particle getting the energy to escape is pretty small.
Out-gassing
Just one more point to add to this. One may argue that possibly the atmosphere is/can be replenished continuously, but that won't work here because we're assuming the planet is tectonically dead. You can't really have out-gassing on a planet with no active surface.
Conclusion
There are many factors that determine atmospheric escape. Different planets will lose their atmospherics for different reasons. However, it is entirely possible for a planet, under the right conditions, to maintain an atmosphere long-term, despite lacking a global magnetosphere. As we can see from Venus, the conditions are generally that the planet should be sufficiently far from the star, its atmosphere should be sufficiently dense and comprised of heavy particles, and the planet itself should be large enough to have an appreciable gravity well. If these conditions are all met, a planet may retain an atmosphere without having a magnetosphere to protect it.


Is there any difference among exoplanets in composition due to host star?

The chemical makeup of space dust -- the stuff  which agglomerates into planets -- is dependent on the history of supernovas and neutron stars and the like which created the atoms in a given region of space.  This is unlikely to bear any relationship to the size or age of the star around which said items revolve.


Gravity assists such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.

  Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?

The orbital speed of Earth around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$

it has a kinetic energy of

$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$

So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass  $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.

The total momentum of the system must be conserved. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts i and f are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.

$$MU_i^2 + mv_i^2 = MU_f^2 + mv_f^2$$

which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:

$$MU_i + m\vec{v}_i = MU_f + m\vec{v}_f $$

Solving these equations, the solution is

$$v_f = \frac{(1-m/M)v_i + 2U_i}{1-m/M} $$

If Ceres approached Earth at $v_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.

So, the final momentum of Earth is

$$MU_f = MU_i - mv_i - mv_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$

  In fact, Earth's linear momentum will only decrease by $mv_i + mv_f =
 1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by
  $0.019~\mathrm{m~s}^{-1}$.

  Approximating a circular orbit (using $r=GM_{sun} / v^2$), Earth's
  orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190
  km out of 150 million!

Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, it hasn't stopped some from trying!


How to calculate Longitude from Right Ascension?

In other "words", the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$):
$$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$


Are Dwarf Galaxies a Subset of Galaxies?

The IAU has not defined "galaxy", therefore you should not use the word  (Joke)
In all seriousness, language is a tool.  You are trying to dance around definitions.  There are many types of galaxies: spiral and elliptical, barred and unbarred, flocculent and unflocculent, small, medium and large.  They are all galaxies, of one kind or another.  The IAU has chosen to define "Planet" and "Dwarf planet" in a particular way, so that children don't have to learn the names of 46 solars system bodies.  They haven't chosen to give a definition of "galaxy", nor "cluster" nor "nebula".  But it doesn't matter.  You use language to communicate. You are not in thrall to the IAU. So you can use the word "galaxy" as they do to mean "galaxy".
Yes: dwarf galaxies are galaxies.


So, I agree with the other answers, that as we look further away from our Galaxy, we see galaxies as they were in the past. Therefore, by and large, these galaxies are younger than ours. That is, we see them as they were when they had an age that could be at most 13.7 billion (the age of the universe) minus their distance in light years.

Therefore to find the oldest objects in the universe, we must look locally. Probably the most ancient galaxies that we can study, are those that are thought to be the building blocks of larger galaxies. These are the "dwarf spheroidal galaxies", several of which have been identified as companions to the Milky Way and M31, and are probably remnants of a larger population of such objects. Their stellar populations appear to consist of mostly intermediate age stars through to extremely ancient stars.

Other ancients objects include the extremely metal-poor stars of the halo population of our Galaxy and a number of very metal poor globular clusters.

Estimates for the ages of such objects are uncertain, but are likely in the range 12-13 billion years old and so only a short time after the big bang.


Was there ever a Jupiter transit or Saturn transit?

Wikipedia actually has many pages on this subject. Transits of Jupiter, viewed from Saturn are rare. The calculations are done from the centre of the planet, rather than a moon, but the principle is clear: Transits of Jupiter are rare.

The next, non-grazing transit of Jupiter visible from Saturn is in 7541. From Uranus in 2714, and from Neptune in 2188.

But probably the most photogenic transit will be the transit of Saturn from Uranus in 2669.

Saturn will also transit from Neptune in 2061, the only transit which we could live to see. Uranus will not transit from Neptune until 38172, and transits from Pluto or other KBO are even more rare, due to the high orbital inclinations.


How to get from the earth to another planet/ the moon? What happens on the way?

Short answer is easy.   Long answer, there's lots of maths involved, and I'm out of practice with my maths, but there's a few basic parts that your students should be able to follow.   It's a little confusing, not too bad.

We can ignore the Earth's motion around the sun for the most part but not the Earth's escape velocity.

Start with Newton's cannonball thought experiment.  That will get you into space, and that's an example of orbital velocity.  With a fast enough launch, the cannonball enters into and stays in Earth's orbit.

2nd, explain the difference between escape velocity and orbital velocity.   Escape velocity is greater than orbital velocity and the cannonball that orbits the earth hasn't really escaped the earth, it's still in the Earth's gravitational sphere of influence.   You need escape velocity (or very close to it) to get to the moon, about 11.2 KM/s.  You can get away with a bit less cause the Moon hasn't fully escaped the Earth either.

Astronauts can't get shot out of a cannon at high speeds cause it's too many G forces, so what rocket launches do is accelerate over a few minutes.  (you could probably look up the precise amount of G forces), but within G forces that humans can tolerate.   If they experience 4Gs, they're accelerating the equivalent of 3Gs (or 29.4 meters per second), at least at first, as they move away from the Earth, the gravitational acceleration towards the earth drops and after only a few minutes they can shut down the launch and drift towards the moon.   Most of the trip to the moon is done in this way, just drifting through space.   You don't want to go too fast cause you'd just need to slow down as you approach the moon again and the moon has no atmosphere to assist the deceleration like the Earth does.

The ships momentum traves through the combined gravitational field of, primarily, the Earth, Sun and Moon, and the Maths get pretty complicated, so the the launch needs to be quite precise.

The Moon also orbits the Earth at about 2,300 miles per hour and that orbital speed has to be matched as well as, since the rocket will often enter a partial orbit around the moon, that orbital velocity around the Earth needs to be matched too.

Once this is done, landing on the moon requires a simple enough deceleration to have a gentle enough touchdown to not harm any equipment.  (this probably isn't as easy as it sounds).

An orbit around the moon isn't strictly necessary but a partial orbit was used to assist landing on the desired spot.   A partial orbit around the Earth was also used in the launch to the moon.

They accelerate away from the Earth in 2 different times, first to leave the earth and get into Earth orbit.  That's the big one, then from a different spot (and not even an hour later) to accelerate out of Earth's orbit towards the Moon.

See diagram.

Source

At least, that's a pretty basic summary.   I welcome correction if I got anything wrong or missed anything important.


about H$_\alpha$ and H$_\beta$ lines in astronomical objects' spectra

In a normal stellar atmosphere, where the temperature decreases with height, I think this is not possible.

H-alpha absorption arises from the $n=2$ to $n=3$ transition; H-beta from the $n=2$ to $n=4$ transition. Thus both transitions are governed by the number of atoms in the $n=2$ lower level.

The absorption coefficient for a transition can be written in proportionality terms by (in local thermodynamic equilibrium)
$$ \alpha_\nu \propto  \nu B_{lu} n_l ( 1 - \exp[-h\nu/kT]),$$
where $B_{ul}$ is the Einstein absorption coefficient, $n_l$ is the population of the lower energy level and $\nu$ is the photon frequency corresponding to the transition.

Using the well-known relationship between the Einstein emission and absorption coefficients and values for the emission coefficients found here, we can estimate a ratio of absorption coefficients for a given temperature.

$$ \frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = \frac{g_{n=4}}{g_{n=3}} \frac{A_{H\beta}}{A_{H\alpha}} \left(\frac{\nu_{\alpha}}{\nu_{\beta}}\right)^2 \left(\frac{1 - \exp(-h\nu_{\beta}/kT)}{1-\exp(-h\nu_{\alpha}/kT}\right),$$
where $A_X$ are the Einstein emission coefficients and the statistical weights $g_n$ are given by $2n^2$.

Thus if we take $\nu_\alpha = 4.57\times10^{14}$ Hs, $\nu_{\beta}=6.17\times10^{14}$ Hz, $A_{H\alpha} \simeq 10^{8}$ s$^{-1}$, $A_{H\beta} \simeq 3\times 10^{7}$ s$^{-1}$, then
$$\frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = 0.29 \left( \frac{1 - \exp(-29642/T)}{1 - \exp(-21956/T)}\right)$$

So when $T$ (in Kelvin) is small, the ratio is about 0.3. When $T$ becomes large then the ratio increases, but above $T \sim 12,000$ K, all the Hydrogen is ionised.

Thus in thermodynamic equilibrium is it very difficult to manufacture a circumstance where the optical depth in the H$\beta$ line is larger than that in the H$\alpha$ line.

[I'd be grateful if someone can tell me whether the ratio I calculated above is correct, since I could not find it in any reference and need to work it out from scratch].


Has the microwave background radiation ever been blocked in order to prove it's really from the sky?

This is kind of old, but I've been reading a bit about CMB and stumbled on this question. If you go back to the original discovery of the CMB, they basically pointed their antenna at the sky and measured the microwave signal. They found one that was "isotropic, unpolarized, and free from seasonal variation" having run the antenna from July 1964 to April 1965. To calibrate the instrument they had a liquid helium cooled reference that they could switch into the receiving circuit. Blocking the CMB signal would be difficult because it's at 2.7 K, and most anything you would block it with would be a larger source (i.e. warmer). Instead, they used a liquid He reference source as a calibration, and then accounted for all other sources of microwave signal. The biggest "other" source was from looking through the atmosphere. You can read about this here:
http://adsbit.harvard.edu/full/seri/ApJ../0142//0000420.000.html
https://www.wmrc.edu/resources/docs/ch1.pdf

To really see what they did, you'd have to read more about that antenna, and there are references in the paper. The original publication is just a letter and so it is terse and refers to previous results instead of offering a detailed description.

...but if you're in orbit, it's easier to make something cold. For FIRAS on COBE, they actually had a device called XCAL (External Calibrator, I guess) that fit into the CMB receiving antenna like a trumpet mute. It looks like they held it at 2.7K and used it to calibrate by moving it into and out of the microwave receiving horn. That's pretty close to blocking the measurement signal. Here's the reference:
https://www.researchgate.net/publication/2669262_Calibration_of_the_COBE_FIRAS_instrument


Does a gravitational wave loses energy over distance?

tl;dr: Yes. Feynman's beads on a string argument

The other answers skirt what I think is the issue that the OP is asking about. In a lossless medium a spherical wave packet itself, caused by a disturbance, will not "loose energy" itself. If you integrate over a large volume you get a constant energy versus time. Of course the flux per unit area will decrease by the same rate that the area increases, but that's a trivial result.

But, when gravitational waves pass matter, they really do deposit energy in the matter, and thereby become attenuated. The effect is very very tiny for a given amount of matter, but it is non-zero and the universe is pretty big.

In this answer to the question Transfer of energy from gravity back to other “more familiar” forms of energy? I've said:

  This is from the Wikipedia article mentioned.
  I think this is really worth reading, as it not only describes some of the physics associated with the accepted answer and it's citation of Feynman's argument about a bead a stick, but perhaps gives a little insight into the development of the field as well.

    Feynman's argument

    Later in the Chapel Hill conference, Richard Feynman — who had insisted on registering under a pseudonym to express his disdain for the contemporary state of gravitational physics — used Pirani's description to point out that a passing gravitational wave should in principle cause a bead on a stick (oriented transversely to the direction of propagation of the wave) to slide back and forth, thus heating the bead and the stick by friction. This heating, said Feynman, showed that the wave did indeed impart energy to the bead and stick system, so it must indeed transport energy, contrary to the view expressed in 1955 by Rosen.

    In two 1957 papers, Bondi and (separately) Joseph Weber and John Archibald Wheeler used this bead argument to present detailed refutations of Rosen's argument.(5)(6)

    (5) Bondi, Hermann (1957). "Plane gravitational waves in general relativity". Nature 179 (4569): 1072–1073. Bibcode:1957Natur.179.1072B. doi:10.1038/1791072a0.

    (6) Weber, Joseph & Wheeler, John Archibald (1957). "Reality of the cylindrical gravitational waves of Einstein and Rosen". Rev. Mod. Phys. 29 (3): 509–515. Bibcode:1957RvMP...29..509W. doi:10.1103/RevModPhys.29.509.


Moon's revolution around the Earth and it's rotation around it's axis?

Yes. Exactly once. This is how we get the light and dark side of the moon. The reason for it being exactly once is due to something called tidal forces. Wikipedia explains this quite well: https://en.wikipedia.org/wiki/Tidal_locking.


Assumptions necessary for the strong form of Olber's Paradox

For simplicity, let's pick a standard star, of radius $R_\odot$ (you can easily generalize this for stars of any radius by picking $R_\odot$ to be the lower bound of star radius).

Let's split the universe into cubes. It would be much easier to use concentric shells, but later on it will be practical to have everything split into components of standard sizes. Each cube is of side $\ell$, and volume $\ell^3$. Within each cube, there is a probability distribution $P(N_\odot = n)$ to have $n$ stars within that cube. We're not requiring anything of that probability distribution, except

\begin{equation}
E[N_\odot] = \rho_\odot \ell^3
\end{equation}

There is an average stellar density $\rho_\odot > 0$ that is the same everywhere.

Now trace your line of sight from $0$ to infinity. This will go through an infinite number of such cubes. We can simplify the problem by having $\ell \approx R_\odot$, so that we don't have to worry about whether the line passes through a star. The fact that this line will encounter no star is equivalent to the experiment of each of those cubes to have zero stars. As each such probability is independent, this is

\begin{equation}
\prod_{n = 1}^\infty P(N_{n, \odot} = 0)
\end{equation}

We can set this as a Bernoulli trial. Zero stars is a failure with probability $p$, more than zero star is a success with probability $q$. The probability of zero star cannot be $1$, since we assumed a non-zero density. Therefore, the probability becomes

\begin{equation}
\lim_{n \to \infty} p^n = 0
\end{equation}

While it is not impossible, the probability of that line never hitting a star is of measure $0$, and the probability of hitting one is of measure $1$.


statistical techniques for estimating distribution of mass

Matter is indeed distributed in knots (clusters), filaments, and sheets (e.g. Bond et al. 1996). This is predicted from theories and numerical simulations of structure formation under gravitational collapse, and is directly observable in large galaxy surveys by eye, such as The 2dF Galaxy Redshift Survey:

Two-point correlation function

To quantify this clustering, usually a correlation function is used, either in 2D ($x$ and y position on the sky), or 3D (if redshifts are available for the galaxies, giving the $z$ dimension). The (two-point) correlation function $\xi(r)$ gives the excess probability of finding two galaxies at a distance $r$ from each other, relative to a random (Poisson) distribution.
If $\bar{n}$ is the mean number density of galaxies, then the probability $dP$ of finding a pair of galaxies at a separation $r_{12}$ in volume elements $dV_1$ and $dV_2$ is
$$
dP = \bar{n}^2 \big(1 + \xi(r_{12})\big) dV_1 dV_2.
$$
The power spectrum $P(k)$ of the galaxies is related to the two-point correlation function by
$$
\xi(r) = \frac{1}{2\pi^2} \int dk\,k^2\,P(k)\,\frac{\sin kr}{kr}
$$
Observationally it is found that the two-point correlation function is given by a power-law:
$$
\xi(r) = \left( \frac{r}{r_0} \right)^{-\gamma},
$$
where $r_0$ is the correlation length (of the order $5\,h^{-1}\mathrm{Mpc}$) and $\gamma=1.7$–$1.8$.

Note that the value of $r_0$ depends on galaxy type; brighter and redder galaxies have a longer correlation length, i.e. they are more strongly clustered (e.g. Wang et al. 2008).

Excursion set theory

Using a formalism know as excursion set theory, the mass fraction in knots, filaments, sheets, respectively, can be predicted analytically. The idea of this theory is to model the collapse of structure in an expanding Universe as a stochastic process, where an overdensity fluctuates and forms a structure if it crosses a given threshold. See e.g. Shen et al. (2006).

Dynamical classification

Forero-Romero et al. (2009) proposed a dynamical classification of the cosmic web, based on evalution of the deformation tensor
$$
T_{\alpha\beta} = \frac{\partial^2 \phi}{\partial r_\alpha \partial r_\beta}
$$
of the gravitational potential $\phi$, counting the number of eigenvalues above a certain threshold at each grid point in a cosmological N-body simulation, where the case of zero, one, two or three such eigenvalues corresponds to void, sheet, filament or a knot grid point.

Other methods

I quick googling leads me to other methods I don't really know anything about, e.g. "the Shapefinder diagonistic" (Prakash 2016), and large-scale halo ellipticity–ellipticity and ellipticity–direction correlations (Lee et al. 2008).


How distorted will our galaxy be if we are viewing it from several thousand light years away?

The light travel time of 100,000 years is quite small compared to the time it takes the Milky Way's spiral arms to complete an appreciable fraction of one rotation. The arms have a pattern angular speed of $\Omega_{\text{p}}=28.2\pm2.1\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$ (Dias et al. 2019), so they should complete one full rotation on the order of $\tau=2\pi/\Omega_{\text{p}}\approx218\text{ Myr}$.
An observer a distance $h$ above the center of the galaxy would observe light coming from a radius $R$ to have to travel an additional time
$$\Delta t=\frac{1}{c}(\sqrt{h^2+R^2}-h)=\frac{R}{c}\left(\sqrt{\left(\frac{h}{R}\right)^2+1}-\frac{h}{R}\right)$$
which is a decreasing function of $h/R$. Its maximum is at $h=0$, where it would be on the order of 100,000 years for stars on the outer edge of the disk. Clearly, $\Delta t\ll\tau$, so there would not be an appreciable distortion, per se, because of the additional distance. While it's true that the objects you'd see would indeed have moved some amount by the time the light reached you, it wouldn't be significant.


Why is the Moon smooth near the equator and bumpy near the poles?

There are a few factors that create this effect:

First in the visual light images, the view around the equator is seen back lit, with almost no shadow, the few around the pole is seen with low lighting, and low lighting brings out features of the terrain. The Polar mosaics have an unreal quality, since they are made of images all taken when the sun is locally at its maximum above the horizon, so the light appears to be coming from all directions at once.

The first mosaic and the topographic map have a projection that is "conformal" this means that circular craters appear circular on the map, but it does introduce very large distortions of scale this, combined with shading which has been added to the map, makes regular-sized craters near the poles appear huge. The shading scales with the map, so craters near the pole have much more dramatic shading than those near the equator, that is an artistic effect, not real.

Finally on one face of the moon, that which faces us, and is in the middle of the pictures, there are large "maria". These are relatively flat lava plains. There are no maria at the poles. Most are on the near side of the moon: see Why are most lunar maria on the visible side?

So this is a combination of low angle of light at the poles, the way that google has projected and shaded their map and the distribution of lava plains on the moon. The poles are not more bumpy than the equatorial region in general. They are more bumpy that the lunar maria, but the apparent effect you describe is an illusion.


These are the Langrangian Points http://en.wikipedia.org/wiki/Lagrangian_point -you'll find an explanation of the maths at a variety of sources if you search using that.


Puzzling quotes from astronauts about earth size

The Earth is 4 times the diameter of the Moon. The Earth viewed from the Moon will therefore appear to have 4 times the angular diameter of the Moon viewed from the Earth.

The Moon is easily obscured by a thumb at arm's length (by a factor of 3-4).

Now bring your thumb closer (because you can't fully extend your arm in a bulky space suit) and put on the biggest pair of ski gloves you can find.

It is not a great stretch of the imagination to think your thumb, in a spacesuit, would easily obscure something 4 times the size of the Moon.


Can a neutron star ever be less than about 1.44 solar masses (the Chandrasekhar limit)? Why not?

A number of neutron stars in binary systems have measured masses below $1.44 M_\odot$ (e.g. a pulsar of mass $1.251 \pm 0.021 M_\odot$, McKee et al. 2020). I think the current lowest mass contender is $1.174 \pm 0.004 M_\odot$ (Martinez et al. (2015). See the plot below with a pictorial representation of the current state of neutron star mass measurements from Horvath et al. (2020). Conveniently, the vertical line is I think at about 1.4 solar masses.

The Chandrasekhar mass for a ball of iron at the centre of a core collapse supernova is more like $1.15 M_\odot$ because ionised iron has 2.15 mass units per electron, rather than the 2 of carbon or oxygen, and because the electron energy required to neutronise iron is much lower than for carbon and oxygen.
The theoretical lower limit to the neutron star mass is about $0.1-0.2M_\odot$, but none have been observed below $1M_\odot$ and there isn't any known astrophysical mechanism to produce them.
A much fuller answer to this last part, which I won't cut and paste here can be found at Physics SE. In brief, the lower limit arises because the adiabatic index of material in the interior falls below 4/3 (because of inverse beta decay,  the formation of neutron-rich heavy nuclei and the disappearance of some free neutrons) and the material is too compressible to form a stable star with a negative binding energy.


You have correctly identified a niche for photometers. Another point in their favour used to be that they were much more sensitive in the U-band than CCDs, but I think that newer CCDs can almost match or surpass the U-band response of photometers.

CCDs take quite a time to readout. The faster you read them out, generally speaking, the higher the readout noise. By comparison, the output from photometers can be very fast indeed with little impact on the noise characteristics. One of the beauties of photometers is also that they give you an instant readout of how many photons have been detected, requiring little analysis to provide real-time monitoring of signals.

However, things move on. So now there are CCD instruments that can work on extremely fast timescales. The primary example of this is a UK-built and operated instrument called ULTRACAM, which allows data-taking at 100 Hz and which uses a special "drift" mode of readout to achieve this.

Back to your main question - what are the scientific objectives of such instruments? On the link I provided above you will see a series of press releases that describe some of the science. These range from measuring the transits of Trans-Neptunian-Objects in front of background stars; examining detailed time series of flares on magnetic stars; investigating the eclipses caused by white dwarfs passing in front of companion stars; looking at the flickering output from accreting black holes; attempting to get phase resolved light curves of the optical emission from rapidly rotating pulsars and much more.

A rule of thumb is, that if you are able to obtain a time resolution $\Delta t$, then you are probing size scales of $\leq c \Delta t$, where $c$ is the speed of light. Thus if you can get a resolution of 0.01s, this corresponds to size scales of $\leq 3000$ km.

As to whether a photometer would be good on an 8-inch telescope, I'm not sure. When I was a student I used a photoelectric photometer on a small telescope to (i) study the scintillation of stars as a probe of conditions in the upper atmosphere - e.g. Stecklum (1985) - which has timescales of order $0.01-1$ seconds. (ii) To monitor "flare" stars with a time resolution of about a second in an effort to get real-time information so that spectra could be obtained at various points during a flare.


What is the farthest blueshifted object to our galaxy?

My impression is that, aside from the very rare galaxy that is moving towards us, only stuff inside the local galactic gravitational field can be blueshifted.  Written perhaps more clearly,  from spaceanswers ,

  Andromeda is not the only galaxy to be moving towards us. With the
  help of galaxy surveys, astronomers have found that around 100
  galaxies are moving towards us. Compared to the numbers of galaxies
  that we know of (hundreds of billions), blue-shifted galaxies are
  seemingly quite rare. Those that are moving towards us are either part
  of our Local Group, which means that we are gravitationally connected
  to each other, or they are found in the Virgo Cluster which everything
  in our Local Group is moving towards. The galaxies M90, M86 and M98
  are all in the Virgo Cluster and all show blue shifts.


Do white dwarfs produce stellar wind? If not, why is that?

The strong gravity of the white dwarf star makes it difficult for stellar wind to escape. Unglaub (2008) shows that for surface gravity $\log g > 7.0$ hydrogen-helium winds are not possible for solar or sub-solar metallicity white dwarfs and only a weak metallic wind would be possible (note that in astronomy, "metal" tends to refer to any element heavier than helium). There is a group of hot white dwarfs with temperatures above 60000 K called "hot wind white dwarfs" showing ultra-high excitation absorption lines. According to Reindl et al. (2019), these features arise from wind material trapped in the magnetosphere of the white dwarfs.


Yes the Sun has an atmosphere.

Disclaimer: I'm not sure if you meant this, but your question implies the Sun is a planet. It, of course, is a star and not a planet. Just wanted to make that clear.

What is an atmosphere?

When you ask if the Sun has an atmosphere, you're actually asking a tricky question. What do you mean by atmosphere? How do you define the boundary of the Sun, above which anything is considered an atmosphere? This is pretty easy for planets like Earth since they have a nice solid surface. But the Sun is a giant ball of plasma heated to thousands of degrees. There's no easy or clear division between the "surface" and "atmosphere". Any discussion of the Sun's atmosphere involves defining what we mean by the surface of the Sun.

Optical Depth

That being said, astronomers have come up with (arbitrary) ways to define the surface of the Sun. One common metric is to use optical depth. Optical depth is a unitless number that defines one's ability to "see" through a gas (or plasma). An optical depth of 1 or higher means the gas is opaque and cannot be seen through. An optical depth less than one means the gas is transparent and can be seen through.

However, when you have something the Sun or even fog, the optical depth varies with the distance you're looking into that object. I'll talk about fog since it's familiar, but the same idea applies to the Sun's atmosphere. Say you're standing in a forrest and its very foggy out. There's a tree 1 meter away from you that you can see. You could measure your optical depth, $\tau$, of the fog between you and tree and might find that $\tau = 0.15$. Since $\tau$ is less than one, that implies you can see the tree, but the value of $\tau$ also implies how well you can see it. If $\tau = 0$, there's nothing between you and the tree to impede your ability to see it. Let's say there's another tree that's 5 meters away. Now there's more fog between you and the tree and while you can still see it, it is harder to see it. The optical depth of the fog between you and the tree 5 meters away might be $\tau = 0.75$. It's still less than one, implying the tree is visible, but because there's more fog between you and the tree, the optical depth is higher. Finally, there may be a tree 10 meters away with so much fog between you and the tree that the optical depth is $\tau = 1.5$. You can't see this tree because there's too much fog in the way. Hopefully you now realize that anything which is at a distance where $\tau > 1$ is not visible to you. That effectively defines a "surface" around you precisely when $\tau = 1$. Anything beyond that point is not visible and anything closer is visible.

If you're talking about the Sun, you can look at the Sun, but you'll only see light which originates from a point where $\tau < 1$. There are countless photons bouncing around inside the Sun, but you can't see them because they're in an opaque part of the Sun. Astronomers use the optical depth as a metric for defining the "surface" of the Sun.

Keep in mind the above description is highly simplified, almost to the point of being wrong. The optical depth is a useful metric for defining a surface, but it doesn't imply there's an exact radius for the surface or even that the surface is constant for every wavelength. There are a lot of other factors that make this much more complicated than I'm describing here. Hopefully you get the general idea though.

The Sun's atmosphere

For the Sun, the atmosphere would be anything above the surface. Nominally the surface is defined as the point where the $\tau = 2/3$ (despite what I said above, and for reasons I won't go into here). The atmosphere above this surface is complicated and difficult to study. The atmosphere, just above the surface, is violent, turbulent, filled with outbursts and magnetic fields, and extremely hot. Below are some pictures of this region of the atmosphere.

 Left: Image of the corona during a solar eclipse. Right: Image of the corona from SOHO. An occulting mask has been placed over the Sun.

The atmosphere of the Sun extends far beyond that though. In fact, Earth is currently moving through the Sun's atmosphere. It's very tenuous out near Earth, but still exists. The Sun's atmosphere hitting our planet is the reason for aurora. Beyond the lower portions, the atmosphere is generally referred to as the Solar Wind. This solar wind actually extends far out, beyond Pluto even. Exactly how far is difficult to define, but estimates place our Sun's atmosphere extending out to about $\sim230\:\mathrm{AU}$. At that point is the bow shock, where our Sun's atmosphere slams into the interstellar medium surrounding us.

Our own Sun's atmosphere is hard to study on a large scale since we're inside it, but we have been able to observe this bow shock around other stars, as shown below.

LL Orionis bow shock in Orion nebula. The star's atmosphere collides with the nebula flow. Hubble, 1995


What's a good step up from 7x50 binoculars?

For higher magnifications, you are going to get into the territory where a tripod becomes a must. I mention this because you didn't say anything about a tripod, so the implication may not have crossed your mind. Unfortunately, the cost of a tripod will eat into your budget.

I am inclined to think that your best option would be a telescope. (Unless you also want to view terrestrial objects, then a cheap spotting scope might be what you want.) Maybe your best bet, to get a good one, would be to ask around on forums like cloudynights or iceinspace that have some support from astronomers local to you, let them know you are looking to buy, and maybe you can get one second-hand off someone who is trading up. You might get a better telescope for your money than trying to buy new.

If you want a smaller length telescope, you may be put off something like a newtonian reflector or another simple 2-mirror reflector. Still, try one out (maybe you can attend a local star-party and see some up close, even if there isn't a telescope store handy). A 114mm newtonian on a tripod would meet almost all your requirements including budget, even purchased new. It will be a bit less than 2' long; slightly bigger newtonians like 130mm start getting longer than 2'.

Otherwise, what might fit the bill is a small cassegrain-style reflector - these usually also have a corrector lens, making them a catadioptric type telescope - such as a Maksutov-Cassegrain. The extra manufacturing cost of making the lenses pushes up the price of these well past the simple reflecting telescopes, however.

Refracting telescopes with an objective of 100+mm are probably in your "too big" basket, and you're also not likely to find one with that budget.

You use different eyepieces to provide different magnifications using the same telescope. If you buy a telescope as a kit, it will come with at least one eyepiece, and it's likely the magnification provided will be more than 10x (divide the focal length of the telescope by the focal length of the eyepiece to find out).


The Earth's equator is inclined to the plane of the solar system (called the ecliptic) by 23 degrees. In other words, the Sun's position, as seen from the Earth, varies between 23 degrees North (at the northern hemisphere summer solstice) and 23 degrees South (at northern hemisphere winter solstice), appearing to cross the Earth's equator twice (at the equinoxes).

The tropic lines mark the maximum apparent solar move north or south.

The Moon's orbit is inclined by about 5 degrees to the ecliptic so moves over a somewhat wider range of the sky as seen from the Earth. The Moon orbits the Earth every 28 days or so, so moves over that range each month.

At full Moon the Moon is opposite the Sun in the sky, so if the Sun was close to the tropic, then the Moon would be around the opposite tropic when full, and so on.


Is the total energy of the CMBR diminishing with time?

Yes. The energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^{3}$. Since the total energy is the density times the volume, the total energy scales like $a^{-1}$. Note that $a$ is called the "scale factor" and $a = \frac{1}{1+z}$, where $z$ is the redshift, assuming $z=0$ now.


GALEX magnitudes use the AB system, which means that $F_{0} \approx 3631$ janskys for both NUV and FUV, by definition.


Were effects of a planetary magnetic field reversal observed on other planets than Earth?

There are such evidence in the case of Mars. Observations from Mars Global Surveyor show evidences of crustal magnetization. In particular, this magnetization has extensive, east-west trending linear features in Terra Cimmeria and Terra Sirenum. These are probably reminiscent of magnetic features associated with a reversing dipole.

There are no such evidence in the case of Venus: Venus has no intrinsic magnetic field today (contrary to the common belief, its inability to maintain a dynamo activity is either due to its chemical and physical conditions unable to form a solid core, either du to an early complete core solidification; it is not due to its slow rotation), and could not have maintained a remanent crustal magnetization, because its crust would have been to hot during its dynamo activity (above the Curie point).

Sources:

Acuna et al. (1999)
Luhmann & Russell (1997)


A stellar association is a loose cluster of stars, that formed at the same time from the same molecular cloud, and so have the same proper motion. Unlike open clusters, they are not gravitationally bound together, so the stars in a stellar association will gradually separate, forming a moving group of stars.

An example is the Scorpius–Centaurus Association


Did Cassini return a photo of Saturn's rings shown from closer to Saturn?

Yes, on one of the final orbits it took some pictures of the rings while crossing the ring plane:

More details of that image are here, and this page show some still images.
Here’s another one:

though it’s hard to interpret without reading the description.
All the science done during the “Grand finale” orbits is described here.


What can I see using a celestron nexstar 127SLT?

First of all, the high end of magnification is generally not as important as several other factors. However, considering that no additional information was given in the question beyond "reflecting", I will make some assumptions and only give you a very general estimate.

The book "Fundamental Astronomy: H. Karttunen et al.", page 51, formula (3.5) gives a relationship between the diameter of a telescope's aperture and it's useful magnification.This relation is as follows ($e$ is the resolving capacity of the human eye, $D$ is the aperture diameter, $ω$ is the maximum magnification and $λ$ is the wavelength of visible light, used in the formula for the telescope's resolving capacity:

$$θ ≈ 1.22×λ/D ≈ λ/D,$$

$$ω_{max} = \frac{e}{θ} ≈ \frac{eD}{λ} = \frac{(5.8 × 10^{-4} × D)}{(5.5 × 10^{-7} m)} ≈ \frac{D}{1 mm}.$$

That's a key concept for you, because any additional magnification above that can be thought of as "empty magnification". It exists mathematically, but the image is so blurry that you will not see any additional detail, and probably be quite disappointed in your purchase if you paid a lot of money for that magnification!

Let's try backtracking from 500x.

If this is the true maximum useful magnification this telescope can achieve, then its aperture diameter would be about 50cm. This aperture allows for $D^2/d^2$ brighter stars to be observed ($d$ is the diameter of the human pupil). When this is calculated and converted to magnitudes, it turns out you should be able to see stars up to magnitude $9+6=15$.

However this is a very rough estimation and you certainly will not be able to see magnitude 15 stars due to atmospheric extinction, seeing and other factors ... my estimate is you'll see stars about the 12th magnitude on a good night.

If the 500x telescope you are looking at has a much smaller aperture than 50 cm (and probably that's true) then that 500x is not going to be so useful in reality.

Take some time and read further about telescopes. This site has many existing questions and answers on the subject.


What zodiac constellation will the Vernal Equinox be directed towards in ~6500?

Due to Earth's axial precession, the vernal equinox gradually travels westward, completing a cycle in roughly 26,000 years, i.e. about 1º every 72 years. Currently, the vernal equinox crosses the ecliptic in Pisces, near its border with Aquarius. In the year 6500, i.e. in 4500 years, it will have moved a further
$$
4500\,\mathrm{yr} \,\, \times \,\, \frac{360º}{26\,000\,\mathrm{yr}} \simeq 62º,
$$
that is, to the point that currently has a right ascension of roughly 20 hours. From the map below (from Wikipedia), you can see that it will just have moved out of Capricornus and into Sagittarius.

As @barrycarter made me aware, my first drawing was wrong. Today, the vernal equinox in this map is a vertical line in the right-most edge of the map, going from Polaris in Ursa Minor and through Cepheus, Cassiopeia, etc., down to the Octans. However, due to the precession this map will become distorted, and Polaris will no longer be the Northern Star; instead the axis of Earth will travel roughly along the green line, in the year 6500 lying somewhere in the middle of Cepheus. The Vernal Equinox will lie roughly along the red line, going from Cepheus through Cygnus, Vulpecula, Aquila to Sagittarius.

I drew this line using Stellarium, setting the date to 6500-01-01, switching on equatorial grid of that date and today, and using the orthographic view. My line on the southern sky may be a somewhat imprecise, but on the northern it should be fine.


How to calculate the statistically significant level for a guassian line profile?

It is difficult to say, there isn't a description in the paper as to how they arrive at those numbers. Given that the equivalent widths of the lines have signal to noise levels that mean they are only greater than zero at significance levels of 2-3, then it isn't clear these are meaningful numbers at all.
However, here is how it might have been done.
You fit a continuum model to the spectrum, get a minimum chi-squared $\chi^2_1$ (or maximum [log] likelihood value); you then fit a continuum plus absorption line model and get a new, lower value, of chi-squared $\chi^2_2$ (or higher value of max log likelihood)
You then get the difference in chi-squared $\Delta \chi^2 = \chi^2_1 - \chi^2_2$ (or form the likelihood ratio) of the two models, which should be distributed as chi-squared with the number of degrees of freedom being equal to the number of new parameters in the more advanced model (i.e. 3 for a line centre, width and depth). You then check a chi-squared table for the corresponding p-value and significance.
In other words, if $\Delta \chi^2 = 6.25$, that corresponds to p=0.1, or if $\Delta \chi^2 = 11.35$ for p=0.01 etc.
In your comments you say:

First i use powerlaw to fit the continue get reduced chi-square1 = 1.00277 / dof=179. Then i fit with powerlaw+gauss, the reduce chi-square2 = 0.88091 / dof=176. The Δχ2 = 24.455. Is that right?

Yes. A chisquared test for 3 d.o.f. yields p=0.00002012
To convert to a number of Gaussian sigma, it is the stuff of standard statistics that
$$ n\sigma = \sqrt{2} {\rm erf}^{-1} (1 - p)$$
which in the case above suggests a significance of $4.26\sigma$.


Does photon capture in a black hole (or any other "very compact" object) reduce the luminosity/photon flux from its self isotropic emission?

Let's answer the question from the point of view of a stable neutron star.
The luminosity is
$$L_0 = 4\pi R_0^2 \sigma T_{0}^4$$
in the inertial frame of the neutron star surface (assume a non-spinning object).
For a distant observer, the radius of the neutron star is (e.g. see Haensel 2001)
$$R_{\infty}= R_0 \left(1 - R_s/R_0\right)^{-1/2}\, ,$$
the temperature is
$$T_{\infty} = T_0 \left(1 - R_s/R_0\right)^{1/2}\ ,$$
and the luminosity
$$L_{\infty} = L_0\left(1 - R_s/R_0\right)\ .$$
$R_s$ is the usual Schwarzschild radius.
I think the fact that $L_{\infty} < L_0$ is the effect you are talking about.


Is the dark energy between the moon and Earth measurable in any capacity?

Yes and no.

First, the answers here should give you some understanding of dark energy and its effects.

The net effect of dark energy is tiny over small distances. If it is given by a cosmological constant, then its effects on gravitationally bound systems is independent of time (hence 'constant'). Observational evidence puts us at least close to the constant case. Which means that even if we are not actually in the constant case, then it will be many billions of years until there would be a measurable difference.

We can, however, estimate the mass equivalent of dark energy in a given area by measuring the apparent amount in the universe itself.

As estimated here:

  If we take the WMAP value for critical density at $$\rho_{c,0} = 9.47 \times 10^{-27}\, kg/m^3$$
  and presume that dark energy makes up about 73% of that, then the effective density of the dark energy would amount to just over 4 hydrogen atoms ($m = 1.67 \times 10^{-27}$ kg) in a cubic meter of space. If we take $5.9 \times 10^9$ m as a mean radius of Pluto and calculate the volume of a sphere of that radius, then the dark energy in that sphere would be equivalent to just under 6000 kg distributed throughout a space representing the solar system. The density of the asteroid Ida has been measured to be about 2.7 g/cm$^3$. So all the dark energy in the solar system would amount to only about 2.2 m$^3$ of the material of Ida, or a sphere of about 0.8 m radius. This is a hundred times smaller than Ida's tiny moon Dactyl!

Of course, that assumes dark energy is essentially the same on small scales as it is large scales, which is unknown. Nevertheless I think it is pretty clear that it would be extremely difficult to observe dark energy directly within the Earth-Moon system.

See also Stan's comment.

Our understanding of dark energy is minimal at best, to boot, giving us no solid ideas on how to detect it in the first place. We infer its existence from the large scale expansion of the universe, but we're otherwise short on solid ideas on its nature.


Would this effect occur if an ice planet was put into orbit or trajectory?

Welcome to Stack Exchange! I'm a little confused about your question, but I'll answer it the best to my ability.

First of all, I would like to state that such scenario is near impossible because getting the planet there would be extremely difficult.

Second of all, when the ice is being "blown", it would have to be perfectly aligned with the Earth not just in its position in orbit, but also in its angle of orbit.

Third of all, the ice would have to travel probably hundreds of kilometres, meaning that it would either 1) not make it by getting in orbit around the Sun / melting or 2) get lost in space.

The heat of the Sun can barely move things at all - I assume you're sort of thinking of a Light Sail, but that's light, and ice would be too heavy anyway.

Getting the ice off of there would also be challenging because its connected to the planet, and can't just be "lifted" by heat. That's why the Earth's surface doesn't float / scrape away by the Sun's heat.

Overall, I do not think it would be possible.

Hope that helps!


If universe is expanding why is the distance between planets not increasing?

Distances between planets and stars do not expand as long as they are bound together by gravity sufficiently. This is true even at the scale of an entire Galaxy.


I think you may be mixing up "Energy" and "Power" a little here. Power is the rate of Energy output per time. If Power is the speed, then Energy is the distance travelled.

Quasars look like stars, from Earth, they give out a light, and while they do vary a little (or a lot) you would not expect to see new ones forming or existing ones fading away.

Gamma Ray Bursts are a short lived. While a quasar will emit more energy over it's long life. Gamma Ray Bursts are more powerful. They are formed mostly as a result of energy released in gravitational collapse during a hypernova. Moreover this energy is released in a narrow beam.

Imagine comparing a laser beam with a equivalent power light bulb.  If the laser is pointing right at you, it appears much brighter than the bulb.

So the answers:

GRB are more powerful than quasars, but only last a short time.
GRB are form a beam of radiation.
So if a GRB and a quasar are equally distant, the GRB would appear brighter.


In general, you'd have to project the light from the sun (Assuming we're talking about objects in our solar system) onto the 3D surface and project it onto a 2D surface. This can be accomplished via a ray tracing algorithm.

This procedure will give you the fraction of the surface illuminated.

Now if you are concerned about the fraction of the illuminated surface that is reflected in a specific directly -- so that it is "viewable" from that direction -- you have to do the ray tracing in 3D from the source (Sun) to the object and on to the viewer. There packages you can google that will do this type of ray tracing if you can model the objects in 3D...


How can an accretion disk form around the central star of a planetary nebula?

The surrounding gas has already escaped, mostly during the thermal pulsation phase of the AGB progenitor. A common envelope phase just accelerates that process.

The envelope is relatively easy to remove (at only a few km/s as you say) because it escapes from a position a long way (several au) from the centre of the star where gravity is weak and is driven by radiation pressure acting on dust that can form in these cool outer regions.

In contrast, the gas stripped by tidal disruption of the planet is close to the core of the star, which is only a few thousand km in radius. The escape velocity there will be thousands of km/s and the gas will be fully ionised and comparatively transparent to radiation and hence radiative acceleration.


The value is $\beta \simeq -1$ (e.g. see  Cardelli et al. 1988). The extinction would not be well fitted by a power law, except over a very narrow wavelength range.


How long is the Sun closer to Earth then Mars is?

Let us first consider the Earth a fixed point and Mars moving around the Sun on a circular orbit with angular velocity equal to the relative angular velocity ($\omega_\bigoplus - \omega_♂)$
The distance between Mars and the Earth can be described as the square root of $R_E^2+R_M^2-2R_ER_Mcos(\theta)$, and when this is equal to $R_E^2$:  $cos(\theta) = \frac{R_E^2-R_E^2+R_M^2}{2R_ER_M} = 0.75$, meaning this happens when the elongation of Mars is around 0.72 and -0.72 radians (40 and -40 degrees). For smaller angles than 40, the cosine is bigger making the distance smaller (same goes for angles between $-40$ and $0$). Therefore, Mars is closer for $80/360 = 22$ percent of the synodic period of Mars, and for the remaining 78% or around 600 days.


How (the heck) does Astrometry.net work?

I've been trying to figure out the technical details of astrometry.net for quite some time. As others already pointed out, the main input to the whole process is a list of stars. I will not go into details on how astrometry.net does it, just note that you can either use its internal simplexy algorithm or use SExtractor. In the end you need a list of coordinates for stars (plus optional flux/intensity/brightness and background).
The .xyls files are used to store these input values (small example):
 x-coord(px)   y-coord(px)        flux  background
 1008.911987,   557.925659,  10.556271,  32.320175,
 1449.509277,   643.280212,   6.580036,  27.963276,
  185.978119,  1253.869751,   5.525373,  27.713015,

Background is approximated before/during star extraction (SExtractor has a function for that).
Now the algorithm goes into creating quads. The cited paper always talks about 4 stars that form a quad. But by my own experiments they seem to be just triangles. Maybe they should have better named them "Asterism" (the code has a lot of very bad naming, as we soon will discover, took me hours to figure out).
So we have a triangle and want to know if a similar triangle exists anywhere in the selected astrometry.net index. So this is where the real magic happens IMO. The papers talks about geometrically hashed lookup. IMO the word "hash" is a bit misleading here as it has not much in common with real hashes. Real hashes should normally give very different results for small changes in the input (e.g. checksum or hash-table bucket distributions). I would rather say that they perform a geometric transformation where the result must be a one-dimensional value (e.g. a double value).
Abstractly speaking we want to map numerous properties to a single value. The value should vary very little if the overall properties also vary little, so from looking at the delta of both translation results we can deduct how similar the two objects were. One such property could e.g. be the angle between two lines.
From here it should be "obvious" how one can use this approach to drive a search. In this simple one-dimensional case we could e.g. use a binary search. For astrometry.net this problem gets technically more complicated, as we need to search for two "hashes". This is done with a KD-Tree. The one used by astrometry.net is optimized for the pre-built index files, so they can be accessed very fast and without much (memory) overhead. Basically a KD-Tree can optimize the question "give me the closest point(s) to X/Y".
In the picture below I tried to visualize how we can search for similar triangles by reducing the question to two numbers. We basically search for the normalized blue vectors (or ones that are pretty close). It should be obvious that this eliminates any rotation and scaling in the question asked, so any similarly shaped triangle will match, regardless of orientation or size. IMO it simply boils down to the fact that the shape of a triangle is defined by two parameters, e.g. not sure why they didn't use two angles.

In reality we have to make the lookup multiple times to also search for flipped and/or inverted variations. After a similar triangle is found (this will happen a lot), the process goes into the verification step. The resulting triangle will give the program the hint on how all other stars must be translated to match the given triangle. With that it can try to match all other stars with its know star catalog.

Query images may contain some extra stars that are not in your index catalogue, and some catalogue stars may be missing from the image.

These can be seen as distractors and conflicts? in the debug output of the program:
verify: logodds -1.38629, 0 matches, 0 conflicts, 1 distractors after 0 field objects.
verify: logodds 333.123, 99 matches, 0 conflicts, 115 distractors after 213 field objects.
108 matches, 424 distractors, 2 conflicts (all sources)

I believe these numbers are not "really" accurate as the checker will probably bail out once it thinks it is impossibly a match (e.g. after certain distractors are found without any matches). Anyway, this is more or less to whole process in detail. I left out some of the magic, e.g. there are a lot of probability checks for speed. Also the way potential quads are chosen from all stars is quite elaborate in reality (and should probably match closely to how this was done when the index files were created).
Side note on indexes: As they contain x/y positions for known stars and stars wander a little over time (a few faster, most very slow), the index can get outdated and start to not match (no idea if this would be in a decade or a millenium). Regular star catalogs give x/y position at a given epoch time plus x/y-speeds to calculate the actual position in any given point of time. IMO with modern CPUs this can be done in seconds for millions of stars (so time independent indexes could be doable, although probably not for the initial triangle matching, but surely for the verify phase). Also with the new ESO Gaia data releases there shouldn't be any gaps anymore (as noted by astrometry.net as small holes in the USNO-B1.0 catalog). But it seems the official available indexes haven't been updated yet to use the new gaia catalog.
Disclaimer: This knowledge was acquired mostly by reading and testing with the astrometry.net source code. So any conclusions I made could be wrong. But I would say it all makes sense if put together. Below I'll give a few more details into the actual implementation inside astrometry.net.
Edit: After reading https://iopscience.iop.org/article/10.1088/0004-6256/139/5/1782 I came to the conclusion that real quads with four stars are probably used in indexes with smaller view angles (zoomed in). I used a picture taken with a regular 55mm DSLR lens for my tests. This would make sense and it basically is exactly what an n-dimensional tree is made for (the question for the closest neighbors now involves 4 parameters).
Each index_t contains two KD-Trees, namely codekd and starkd. The tree codekd contains the information of all "quads", while starkd contains regular star coordinates for later verification.
The "hashing" bits can interestingly be found in "solver.c" in the function "check_inbox".
static void check_inbox(pquad* pq, int start, solver_t* solver) {
    int i;
    double Ax, Ay;
    field_getxy(solver, pq->fieldA, &Ax, &Ay);
    // check which C, D points are inside the circle.
    for (i = start; i < pq->ninbox; i++) {
        double r;
        double Cx, Cy, xxtmp;
        double tol = solver->codetol;
        if (!pq->inbox[i])
            continue;
        field_getxy(solver, i, &Cx, &Cy);
        Cx -= Ax;
        Cy -= Ay;
        xxtmp = Cx;
        Cx = Cx * pq->costheta + Cy * pq->sintheta;
        Cy = -xxtmp * pq->sintheta + Cy * pq->costheta;

        // make sure it's in the circle centered at (0.5, 0.5)
        // with radius 1/sqrt(2) (plus codetol for fudge):
        // (x-1/2)^2 + (y-1/2)^2   <=   (r + codetol)^2
        // x^2-x+1/4 + y^2-y+1/4   <=   (1/sqrt(2) + codetol)^2
        // x^2-x + y^2-y + 1/2     <=   1/2 + sqrt(2)*codetol + codetol^2
        // x^2-x + y^2-y           <=   sqrt(2)*codetol + codetol^2
        r = (Cx * Cx - Cx) + (Cy * Cy - Cy);
        if (r > (tol * (M_SQRT2 + tol))) {
            pq->inbox[i] = FALSE;
            continue;
        }
        setx(pq->xy, i, Cx);
        sety(pq->xy, i, Cy);
    }
}

The pquads are the potential "asterisms" we are currently creating, and this function's job is to set the "query" parameters via "set[xy]" on the bottom. Those are the actual values later looked up in the codekd tree.
A pquad is basically a line between two points (also called backbone-stars) called fieldA and fieldB (which are actually indexes to get an x/y position). Additionally it must have at least one additioanl xy point (exactly one in our triangular case). The pquad also contains sintheta and costheta (set in check_scale).
    double dx, dy;
    dx = field_getx(s, pq->fieldB) - field_getx(s, pq->fieldA);
    dy = field_gety(s, pq->fieldB) - field_gety(s, pq->fieldA);
    pq->scale = dx*dx + dy*dy;
    pq->costheta = (dy + dx) / pq->scale;
    pq->sintheta = (dy - dx) / pq->scale;

As we see both code parts translate and scale the vector AC in relation to the line AB.
I hope this info is useful to somebody, even if it got a bit long!


How the position of this black hole known?

The article is referring to the black hole known as V4641 Sagittarii. It was originally thought to be the closest known black hole to Earth, at 1,600 light-years (0.5 kiloparsecs). It's a binary system which emits X-rays. Original distance measurements were made based on observations during and following an X-ray outburst in 1999. Analysis of the jets emitted and of the inclination of the system led to initial estimates of $0.4$-$1.7\text{ kpc}$ (see Hjellming et al. (2000)), with lower estimates of around $\sim0.5\text{ kpc}$ being favored - the distance that was then commonly quoted.

Later spectroscopic measurements in more detail appear to show these early observations to be incorrect, as Orosz et al. (2000) found. By observing the orbits of the two objects in the system, their masses were determined, which led to an estimate of the companion star's radius. The temperature of the star was determined by examining its spectrum, and its luminosity was computed from this and the radius. By using the color index and taking into account interstellar extinction, the group was able to calculate the distance to the companion star, and thus to the black hole. They found $7.40\leq d\leq12.31\text{ kpc}$, values closer to the currently accepted distance.

An absorption study (Chaty et al. (2001)) found values somewhere in the middle, with $3\leq d\leq8\text{ kpc}$. However, all groups agree that the actual distance to V4641 Sagittarii is much further than 1,600 light-years. The closest known black hole is now thought to be A0620-00, also known as V616 Monocerotis, at around 2,800 light-years (about $0.85\text{ kpc}$) from Earth.

General black hole detection is discussed in the answers to How are black holes found?. The takeaway for this case - a stellar-mass black hole in a binary system - is that effects from the companion star must be observed and measured. Fortunately, there are typically plenty of ways to choose from.


How much the moon movement can influence an equatorial mount setup?

For most purposes, the difference between the Moon's motion across the sky and star's motion (the sidereal rate) is negligible. The Moon appears to move about 1/2 degree per hour relative to the stars, so while the stars travel 15 degrees per hour east to west, the Moon travels approximately 14.5 degrees per hour. Unless your mount is aligned accurately and has an accurate motor, you probably will not notice this difference.

Some telescope drives do have a "lunar rate" so that it will track the Moon more accurately. But these approximations do not take into account the fact that the Moon is also moving north or south in declination as it orbits the Earth; the lunar rate only compensates for its west to east motion. (Maybe a goto scope will track the Moon more accurately since it theoretically could be calculating the position of the Moon and pointing to a changing position.)