Beginner looking for extra eyepieces for telescope

First of all, any eyepiece of the correct diameter will work with your telescope. Typical diameters are 1.25" and 2".

The eyepieces you have listed tell you the focal length of the lens. This relates to the magnification - the focal length of the telescope divided by the focal length of the lens will tell you the magnification. So the 20 mm eyepiece will give you lower magnification than the 4 mm eyepiece. Thus to work out the magnication you will need to know the size and focal length of the telescope for starters. A Google search showed there are at least a couple of models but your owner's manual will give you this information.

Now a caveat: Forget (at least a bit) about magnification. Many telescope brands will [try to] sell the magnification, because this is a big selling point. BUT higher magnification might just turn a small fuzzy blob into a bigger fuzzy blob.

The most important features of a telescope are the aperture (diameter) and lens/mirror quality. Bigger aperture means more light means easier to see faint stuff. And a quality lens means the difference between a fuzzy blob and a crystal-clear image.

You will probably find that the mirrors in your telescope are reasonable quality; the lenses are usually where the sellers save on cost. I am confident that if you buy a good quality lens in the 10-15 mm range and one in the 20-25 mm range you will see very nice, clear images of the Moon, the planets out to Saturn, nebulae and star clusters.


How fast is a comet moving when it crosses Earth's orbit?

Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.
What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy
$$
E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)
$$
which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)
$$
v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.
$$
The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy
$
2\langle T\rangle + \langle W\rangle=0.
$
For a (near-)circular orbit (such as Earth's), $r$ is constant and we have
$
v^2_{\rm Earth} = GM_\odot/r.
$
Thus, at $r$=1AU
$$
v_{\rm escape} = \sqrt{2} v_{\rm Earth}
$$
as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.
Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since
$$
E = -\frac{GM_\odot}{2a}\qquad\qquad(**)
$$
with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies
$$
r_{\rm apo}\le 2a = 2{\rm AU}.
$$
Such comets don't exist AFAIK. Most returning comets have much longer periods than 1 year.
What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get
$$
\frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.
$$
From equations ($*$) and ($**$), we can then find
$$
v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.
$$
In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.
Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.


Does all of time exist and if so where is it?

Your question is difficult to answer exactly, because it touches the question of consciousness, which is problematic in general. But the answer is basically this:

Theory of relativity describes the space-time as four-dimensional space that exists as a whole. You are described by a world-line, which contains all states you go through from birth to death. There is many slices, which describe "you" in different times. What you can see is given by what information reached your sensory inputs (your senses) and how they are processed by your brain. The brain creates a model of reality based on your sensory inputs, which is stored in space-like slice of your world-line in the space-time, which we call present. This is what "you" can see. So the reason basically is in how the brain processes the data it is getting.

Along you world-line, there is many versions of you at different ages and each version feels its own present and the versions more in the future also have memories of the past (because the brain stores the past sensory inputs). There is no single "present time" in the theory of relativity, only the space-time as a whole.

But you should't forget that the theory of relativity is still only a very well working model. Even within the theory of relativity, the question whether space-time exists as a whole, or whether there is "present state", which evolves in time, is rather philosophical than scientific, because there is no observable difference between the two.


Can I look at the sky and find the day of the week?

It is possible to determine the day of the year using the distance of the sun from the vernal equinox (you have to account for the equation of time etc, but it is possible). It is also possible to determine the year if your measurements are precise enough, using the current position of the vernal equinox (which changes due to precession). The ambiguity of the factor of 26000 years can be removed by noting the positions of other stars (for example, Barnard's star). Knowing these two, you should be able to find the day of the week.

About your second question, Astronomers like to deal with Julian date, which is a continuous day counting system, and you can determine the day of the week from the Julian date, though only if given a reference, for example Jan 1, 2000, was a Saturday. As such, a week and days of a week are man made (https://en.wikipedia.org/wiki/Week).


How big is one arcsecond at various distances?

An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$

Examples:

One arcsecond on the moon is 1.87km
One arcsecond on the sun is 727km
One arcsecond on Mars (when it is closest to Earth) is 237km
One arcsecond at distance of one parsec is one astronomical unit (AU), by definition
One arcsecond on Alpha Centauri is 200 million km
One arcsecond on the Andromeda galaxy is 100 trillion km

The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.


You have two questions here, which are somewhat mutually exclusive. I'll start with the second.
Are there observations of something like non thermal filaments in other galaxies other than our?
The paper you linked pretty plainly spells out that only a handful of nonthermal filaments (NTF) have been found, and all of them have been found near our galactic core. To quote the paper:

Observationally, the NTFs are unique to the GC

Specifically they mention that there were only 9 known NTFs previous to this paper:

nine previously identified isolated NTFs—the most recent NTF discovery is described by Reich (2003)

and this paper reports that they found another 20:

New high-resolution, wide-field 90 cm VLA observations of the Galactic Center (GC) region by Nord and coworkers have revealed 20 nonthermal filament (NTF) candidates.

So it seems the answer to your question, at least according to this paper, is that no, there are no observations of NTFs in other galaxies at this point in time.
Are there non thermal filaments outside our Galaxy?
I interpret this to mean, could there exist NTFs in other galaxies, and not so much have we actually seen NTFs in other galaxies. I think the answer to this, despite being a total guess with no proof to back it up, is a resounding yes. There's no reason to think our galaxy and it's structure/evolution is unique. Very likely NTFs exist in many other galaxies which formed under similar processes.
This article by the NRAO details an attempt to explain where these NTFs originate from and the proposed solution is that they are a result of starburst activity. To quote from the article:

Yusef-Zadeh determined that a number of the non-thermal filaments seemed to connect to concentrated areas of thermal emission, which identify pockets of star formation. Galatic Center Combined radio image from the Very Large Array and Green Bank Telescope. The linear filaments near the top are some of the nonthermal radio filaments (NRFs) studied by the researchers. Other features, such as supernova remnants (SNRs) and the area surrounding our Galaxy's supermassive black hole (Sgr A) are shown.
"What this showed us is that two seemingly disparate processes, thermal and non-thermal radio emission, can be created by the very same phenomenon," said Yusef-Zadeh. "In this case, that phenomenon is pockets of starburst activity."

If this is indeed the source of NTFs, then any galaxy which has or did have starburst formation would likely have NTFs as well.


Simple experimental evidence that Earth revolves around Sun

The answer is ironic: Without good instruments, there is no evidence. The people who thought that the Sun went around the Earth were perfectly correct as far as the actual evidence went until the early 1700s and mid-1800s when two lines of evidence opened up that showed that the Earth moved.
Aberration of Starlight
Wikipedia has a correct but over-complicated explanation. The easiest way to think about it is to imagine yourself at a stop sign in a car in the rain, and the rain is falling straight down. When you start moving, the rain's apparent direction of fall changes so that it appears to be falling from ahead of you and slanting down towards you. That's aberration.
In the early 1700s, the stars were discovered to be shifting position, and in 1727, James Bradley correctly identified it as abberation of starlight due to the motion of the Earth around the Sun. (For any star in the ecliptic, the Earth is moving towards it at some time of the year and away from it six months later.)
Parallax
Wikipedia's article on parallax is better, and I refer you to it for details. Basically, if you hold your finger up before you and look at it with your left eye closed, and then with your right eye closed, it appears to jump with respect to the background -- the wall beyond or the trees outside or whatever. Switch back and forth between your eyes quickly to see it clearly.
As the Earth circles around the Sun, nearby stars also appear to shift their position relative to the more distant stars.  A key point here is that there were good scientific reasons to suppose that the stars were much smaller than the Sun. Seen through a telescope, stars showed disks and if they were like the Sun, their distance could be deduced from those disks. And they were close enough that if the Earth really went around the Sun, parallax should have been observed. But it wasn't and the lack of any noticeable parallax was a strong empirical argument against Heliocentric theories.
In reality of course parallax exists, but the parallax of all stars is small, because they are much further away than was estimated from their disks. (The visible disks were actually diffraction disks and not true disks at all -- but it was not until nearly a century later that diffraction began to be understood.)  Friedrich Bessel first measure the real parallax of a star in 1838.


Impact of Atmospheric Water vapour on Optical Observations

Yes, water vapour interferes even stronger with astronomical observations than $\rm CO_2$.

In the optical, water easily forms droplets which scatter light. But also in the gaseous phase water has strong absorption bands, some of those reach into the optical, while most of those however lie in the infrared, making water a greenhouse gas at the same time.
Let's have a look at the absorption spectrum of various gaseous species in the atmosphere:

Here we see with data what would otherwise take too many words to explain. Take note that the observable spectrum for humans is at the wavelengths $\rm \lambda = 0.4-0.8 \mu m$. The many absorbtion bands of water make clear why this interferes with astronomical observations:
In the same way that outgoing thermal radiation from Earth is stopped at going into space, also infrared coming from space is absorbed in the atmosphere and thus not reaching telescopes.

To deal with this there is a number of ways to cope:

Go high up above all the water vapour, to a dry place like Chile. Or a dry place like space. Then the absorption spectrum looks much more benevolent, and that's part of the reason we have all the major observatories in Chile, Mauna Kea, etc.
Try to observe in one of the absorption bands, but fly above all the water vapour. This is what the two airborne-based observatories SOFIA and its predecessor do/did.
Observe through the infrared-'windows' that let light pass through, from the ground. This is a bit exotic science wise nowadays - AFAIK this is a method mostly used in history, as today it is hard to justify building a massive telescope when most of your prospective signal gets absorbed. This was employed at Mt. Wilson Observatory, but also many others.


Could there be another planet between Mercury and the Sun?

It is very unlikely. While the hypothetical planet Vulcan between Mercury and the Sun was ruled out many decades ago searches for Vulcanoid asteroids would very likely have detected anything that might qualify as a intra-Mercurial planet.


Why is carbon so rare on the Moon and on Mars?

There is not much doubt that the abundance of carbon in the protosolar nebula was not abnormal. We can tell that by looking at the carbon abundance in the atmosphere of the Sun - it has an abundance of 8.43 on a logarithmic (base 10) scale where hydrogen has an abundance of 12 (Asplund et al. 2009). This is typical for stars in the solar neighbourhood.

However carbon in the solids of the solar system - i.e. in the Earth, moon, meteorites etc. is underabundant with respect to this by something like a factor of 10. Nitrogen is similarly underabundant in the Earth and oxygen to a lesser extent (a factor of two). (Lodders et al. 2009)

The reason for this is that C,N,O are all capable of remaining in the gaseous phase at the kinds of condensation temperatures that are typical for forming most of the "rocks" that make up the Earth and other solids in the solar system. In particular, carbon monoxide is readily formed at high temperatures and cannot easily be captured into terrestrial planets and other rocky materials.


Is the age of the solar system equal to the age of meteorites?

The radiometric dating of meteorites tells you when certain radioactive species were incorporated into that body. It is extremely difficult to see how that material could have got there at a later time, so the age you get is the age of formation of that body.

Meteorites do not form in stars. They require considerably cooler conditions, but they also require a certain density to stick together. Those conditions are found in a protoplanetary disk. So you might well ask, well maybe the meteorites were formed around another star and then somehow they have ended up in orbit around our Sun.

This is quite unlikely. Very similar ages have been determined for some tens of meteorites and that age happens to be just a touch older than the oldest rocks that can be found on the Earth and the Moon. There is no big age spread among meteorites. That age also coincides with the age for the Sun deduced from physical models (though this age is less accurate because of some remaining uncertainties in the physics, but it is not hundreds of millions of years out).

The obvious inference is that these meteorites were among the first solid bodies formed in the protoplanetary disk that ultimately formed the Earth and Moon and hence represents a lower limit to the age of the solar system. It depends where you wish to start the clock; the formation of the first solid bodies seems as good a place as any and models for disk formation and evolution suggest that these solid bodies do form within about a million years of the molecular cloud that formed the Sun beginning to collapse.


There are thirteen modern constellations in the Zodiac. In modern astronomy, a constellation is a specific area of the celestial sphere as defined by the International Astronomical Union. In total, there are 88 constellations.

Astronomy and Astrology are not the same thing. Astronomy is a science while Astrology is not. As such, I'll restrict myself to the historical and modern constellations of the Zodiac.

According to Encylopedia Brittanica

  Zodiac (is) a belt around the heavens extending 9° on either side of the ecliptic, the plane of the earth’s orbit and of the sun’s apparent annual path.

In historical astronomy, the zodiac is a circle of twelve 30° divisions of celestial longitude that are centered upon the ecliptic, the apparent path of the Sun across the celestial sphere over the course of the year. Historically, each of these divisions were called signs and named after a constellation: Sagittarius, Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, and Scorpius.

In 1930, the International Astronomical Union defined the boundaries between the various constellations, under Eugène Delporte , who,

  ... drew his boundaries along vertical lines of right ascension and horizontal parallels of declination. One governing principle was that all variable stars with an established designation would remain in that constellation, as requested by the IAU’s Variable Stars committee.

"Constellations ecliptic equirectangular plot" by Cmglee, Timwi, NASA - Own work, http://svs.gsfc.nasa.gov/vis/a000000/a003500/a003572. Licensed under Public Domain via Commons.

As a result, the path of the ecliptic now officially passes through thirteen constellations: the twelve traditional 'zodiac constellations' and Ophiuchus (which was one of the 48 constellations listed by the 2nd-century astronomer Ptolemy), the bottom part of which interjects between Scorpio and Sagittarius.

Source: journeytothestars.wordpress.com

So, the 13 constellations of the Zodiac are Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius and Ophiuchus.

As seen from the (first) figure, the ecliptic also touches the edge of the constellation Cetus, though it in not usually included in the Zodiacal constellations.

Note about Precession: Because the Earth in inclined (by $23.45^{\circ}$), it rotates like a top. This is called precession, which results in a shift in the position of the constellations relative to us on Earth. The result is that the 'Signs of the Zodiac' are off by about one month.


What must/do astronomers reveal beyond their academic papers?

Most catalogs have a public release somewhere on the web. Here is the best listing of astronomical catalogs that I know of:
http://vizier.u-strasbg.fr/viz-bin/VizieR

There is also a NASA Extragalactic Database (NED) that can be helpful: http://ned.ipac.caltech.edu/

In general, astronomical observations are available to the public to be analyzed. Different telescopes/space missions maintain their own archives. Some give an observer a proprietary period (6-7 months is the time frame I have heard given for most observatories) to give the person who requested the observations time to analyze the data before it is released to the general public.

In publications the author(s) will list they catalogs the used to come to their conclusions so replicating scientist can use the same dataset.


How might Thales have predicted a solar eclipse?

I recently read the relevant parts of the Cambridge Concise History of Astronomy and I'd suggest it was partly good observation, partly luck.

Good observation in that Thales would have noted that the movements of the Sun and Moon were such that eclipses were possible on certain dates, and luck in that the shadow of the eclipse he suggested was possible passed over him or those to whom he had suggested the eclipse took place.

I very much doubt he had the observational data (about the size of the Earth for instance) to have been precise about the shadow path - but would have had the data to show when the Sun and Moon were in alignment.


Align multiple fits images without WCS for future aperture photometry analysis

You can use astrometry.net to get the exact WCS of your image.

From the website:

  If you have astronomical imaging of the sky with celestial coordinates you do not know—or do not trust—then Astrometry.net is for you

There is a web application where you can just upload your image and it returns accurate calibrations.


(I can't comment)- See https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point
TLDR, The big bang happened everywhere in the universe at the same time, because it was the universe. The top answer goes more in depth.


Could there really be a preferential direction to the speed of light?

Prior to Einstein's 1905 paper, the Lorentz transformation had already been worked out by Lorentz and others. Only their interpretation of it was lacking. They still clung to the idea that there was a Newtonian absolute time, and the times in the Lorentz transform were only apparent times. Einstein was the first to realize that there doesn't need to be a Newtonian time; the Lorentz transform stands perfectly well on its own.
The guy in this video is thinking the same way as Einstein's predecessors; he's an aetherist, though he doesn't realize it. He's stuck to the idea that there's a real time with respect to which the true speed of light is defined, but various "effects" prevent any experiment from actually determining what it is. This is most obvious starting from 11:32, where he says he wants to show how differently the universe works if light is anisotropic, but then shows that it works exactly the same in every experimentally measurable way.
The reality is that only what is operationally measurable matters. What we mean when we say that the speed of light is constant is that there exist coordinates with respect to which it's constant. In a Newtonian corpuscular world, no such coordinates would exist, so the fact that they do exist in the real world is physically meaningful. It's not necessary to use these isotropic coordinates, but it's frequently convenient. That's the only reason we use them. To put it another way, the Einstein synchronization convention really is a convention; it isn't an assumption.
There also exist coordinates with respect to which the speed of light isn't constant. This is not physically meaningful, because no theory could ever avoid them; you can always do a formal substitution of variables, as long as it's invertible and you're consistent about it. The result of any experiment in these coordinates is always the transformation of the result of the experiment in inertial coordinates, because they both describe the same reality.
If $(x,t)$ are standard inertial coordinates, then with respect to coordinates $(x,t')$ where $t'=t-x$, the speed of light $|dx/dt'|$ ranges from $c/2$ to $\infty$ depending on direction. Why don't we see this as an anisotropy in the sky? Because the universe in different directions has aged by different amounts, and their ages differ by just the right amount to compensate for the different light travel times. This is similar to the way that length contraction, the relativity of simultaneity, and so forth always conspire to make things consistent in different inertial frames.


Are there good animations of the movement of proximate stars over long time periods?

This image should give you a good idea of the distances of stars relative to the Sun over the next 80,000 years (and the past 20,000 years).

Image courtesy of Wikipedia user Lambiam under the Creative Commons Attribution-Share Alike 3.0 Unported license.

I've created a modified version below. The nearest star's track in time is in red, and the four blue boxes indicate the five transitions (the second box contains two):

Proxima Centauri remains the nearest star to us for about 25 thousand years, at which point Alpha Centauri passes it. ~7.5 thousand years after that, Ross 248 swings in closer. It then moves away after about 10 thousand years, when Alpha Centauri moves just ahead of it for a short period of time. Gliese 445 becomes the nearest star for about another ~7.5 thousand years, until it wings away and Alpha Centauri again becomes the closest star. 80 thousand years in our future, Ross 128 nudges ahead.

A few interesting things to note are that for most of this time, the nearest star is closer to the Sun than Proxima Centauri is now - indeed, there are times when quite a few stars are closer simultaneously - and that this lasts for some time as a brief spike.

However, this is only a short-term pattern. In the past, many stars have come much closer - even inside the Oort Cloud.

Also, to address the mention of Barnard's Star: It will swing close to us within ~ 10,000 years, but will just as suddenly swing away.

As an aside, a cool animation shows the movement of stars that are near the Sun throughout their past trip around the galaxy.

Animation courtesy of Wikipedia user Henrykus under the Creative Commons Attribution-Share Alike 3.0 Unported license. Data and original information from the European Southern Observatory.

The above animation was the result of 1,000+ nights of observation over 15 years by astronomers working at the ESO. They observed or calculated the ages, velocities, and compositions of 14,000+ F and G stars. They then used the information to create orbital trajectories for these stars.


Would a Einstein–Rosen bridge change size and/or position in an expanding Universe?

The latter is closest to the truth, although I wouldn't use the phrasing "stretch". The "mouths" of the wormhole are (more or less) fixed in comoving coordinates (i.e. the coordinate system that expands with the Universe, and in which galaxies lie approximately still). But the bridge is sort of outside our three-dimensional space, and doesn't necessarily follow the same expansion.

If the mouths retained a fixed physical distance, they would accelerate in comoving coordinates beyond bound, eventually moving through space at superluminal velocities, which is forbidden. For example, consider the Milky Way and the galaxy GN-z11 which today lies at a distance of 32 billion lightyears from us. Roughly 13.3 billion years ago, the size of the Universe was 1/10 of today's value; that is $a=0.1$, and the distance to GN-z11 was only 3.2 billion lightyears. If at that time you created a wormhole$^\dagger$ bridging MW and GN-z11, and if the distance between the mouths were fixed in physical coordinates, then today, at least one of the mouths would have moved so far outside its host galaxy that it must have traveled superluminally.

$^\dagger$Creating a wormhole 13.3 billion years ago is left as an exercise for the reader.


About two-thirds of the Milky Way’s estimated 100 billion stars reside in the disk, and these all move in direct orbits. Most of the remaining third are in the football-shaped central bulge, and while their orbits are more varied, on average they too are direct. A few percent of our galaxy’s stars occupy a vast, spherical halo. These travel on randomly distributed orbits — accordingly, half are direct and half retrograde.

Source: Do astronomers have any idea what percentage of our galaxy’s stars move in retrograde orbits?

And while formation of disk galaxies is a complex process, which is not yet fully understood, the reason that almost all stars move in the same (direct) orbits is the conservation of angular momentum when gravitational force pulls gas clouds to come together. It is the same mechanism that is responsible for the planets to orbit the star in the same direction.


I suggest taking a look to https://en.wikipedia.org/wiki/Clearing_the_neighbourhood .

Quoting it,

  In the end stages of planet formation, a planet (as so defined) will have "cleared the neighbourhood" of its own orbital zone, meaning it has become gravitationally dominant, and there are no other bodies of comparable size other than its satellites or those otherwise under its gravitational influence.

The Trojans are textbook example of objects under the influence of a planet (Jupiter).

For another example, Pluto is said to not have cleared its neighbourhood because Neptune is in it. Neptune has cleared its neighbourhood although Pluto is in it because Pluto is a lot smaller and Pluto orbit is governed by resonance with Neptune.


How do you call it when two celestial bodies come as close to each other as they will in their current orbits?

This is a syzygy.  "A syzygy of the Sun, Jupiter and Saturn." In general a syzygy is when three bodies are in a line.  In your case the three bodies are the sun, and the two planets. You can also talk about "Saturn is at opposition as seen from Jupiter. Not every syzygy is "proximity event" but many are.  I know of no word or phrase for the "approaching phase" or the "departing phase".

I don't think that there is a specific term for "closest approach". Establishing absolute distance is generally difficult, compared with finding the direction. And the closest approach doesn't happen exactly at opposition (due to orbital tilt and eccentricity). Terms like "opposition" or "syzygy" are about the direction rather than distance.


Does the Sun have a numerical designation in any star catalogues?

It's somewhat difficult to prove it either way given the large number of catalogues out there. The majority (at least of the ones on VizieR) do not include the Sun.
The only case I'm aware of that almost meets your requirements is in the Preliminary Version of the Third Catalogue of Nearby Stars by Gliese and Jahreiß (1991). This catalogue is well-known as the source of stellar designations beginning "Gliese" and "GJ", and is queryable via VizieR.
The first record in the catalogue is the Sun, which includes its spectral type, apparent magnitude, colour indices and space velocity as for other stars. Right ascension, declination and proper motion aren't included for obvious reasons. It doesn't match your requirement for having a numerical designation because the name field is just "Sun", while the other stars in the catalogue have numbers. The "NN" stars originally didn't have numbers either but these were added in 1997: in the literature these objects are still typically referred to as "GJ ####" rather than "NN ####".


There is a limit to how quickly a black hole can accrete matter. As a star gets close to a black hole it may be tidally disrupted and pulled apart. The material from the star will then fall toward the black hole and as it does so, the gravitational potential energy will go into heating it up. Now the stellar material would not normally be able to fall directly into the black hole - angular momentum must be conserved, so first it will form an "accretion disk". Viscosity in the disk can transport angular momentum outwards, allowing material to then fall through the event horizon. The ultimate rate at which the black hole can swallow mass will be controlled by the rate at which material can reach the inner parts of the accretion disk.

If the accretion disk is very hot and luminous, there could be a limit to how fast material can be added to it. This is known as the Eddington limit. Basically what happens is that the radiation from the inner parts of the accretion disk is so intense that it can "blow away" any new material falling towards the black hole. This is then a self-regulating phenomenon because the luminosity of the disk is derived from the material falling onto it - and so an equilibrium will be set up that limits the accretion rate to that which produces only enough luminosity that allows the accretion to proceed. This limit on the accretion rate is what the article you refer to means by "choking".

An estimate of the maximum accretion rate is set by equating the accretion luminosity and the Eddington luminosity (the maximum luminosity that an object may have before radiation pressure will halt accretion). This in turn depends on the efficiency $\epsilon \sim 0.1$ with which gravitational potential energy is turned into radiation as material falls into the black hole.

The journal paper that reports this discovery - Vinko et al. (2015) - analyse the light curve of the event and suggest this is black hole of $10^{5}$ to $10^{6}$ solar masses, tidally disrupting and swallowing a star of about a solar mass.

The Eddington luminosity (for spherical accretion) would be $\leq 3\times 10^{10}$ solar luminosities , but this thing had a peak luminosity (even in X-rays) of $10^{11}$ solar luminosities indicating that it must be in the regime where radiation pressure would reduce the accretion rate. To supply this luminosity would require $\epsilon \dot{M} c^2 \sim 10^{11} L_{\odot}$.
For $\epsilon \sim 0.1$ this gives a peak mass accretion rate $\dot{M} \sim 2\times 10^{-4} M_{\odot}$ per day. Thus it would take more than 10 years to "swallow" the star even at this peak rate.


Location on (or in) the celestial sphere

Well, at least you did some thinking and proposed a couple of personal thoughts instead of just asking for the answer to a homework question. Both observations you make are pertinent.

Remember, a celestial sphere doesn't actually exist. It is an imaginary concept, a simplification based upon our perspective of the universe.

A planetarium, whether it is an old-fashioned one that depicts an earth-bound perspective of the universe, or a modern digital projection system that can "fly" you through space to view the universe from a different place, is projecting a perspective of the view of the universe from a particular place, that is, you see the "celestial sphere" as it would appear from that place. In the case of a view from elsewhere in space, it would be a different celestial sphere from our celestial sphere.

So yes, you would be at "the centre" of the celestial sphere. Maybe not the "very" centre in the case of old fashioned projectors with fixed star fields, as they wouldn't account for the 150 million kilometre imprecision caused by the movement of the earth around the sun. But, as this is an insignificant distance in astronomical terms, that doesn't matter a great deal, so "the centre" is probably a sufficient description.


Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun

You might find this paper helpful:

http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf

Look specifically at Eq. 5 for the wavelength dependence.  To get the intensity, use Equation 1.  The value of $u$ can be found in the caption to Figure 2.


Is there a relationship between stars' absolute magnitude, spectral class and relative location in a cluster?

Stars born together in clusters have more-or-less the same age. As a rule of thumb, any spread in age, measured in millions of years, is smaller than the extent of the cluster in parsecs.

For most stellar clusters, smaller than a few pc, the only chance of measuring age differences occurs in the first 10 million years of life. There is no evidence for age spreads in clusters beyond 30 million years old and therefore no chance of identifying any trend of age with position.

In young star forming regions, the picture is very complex. Estimating the ages of individual stars is difficult. Generally speaking, newborn stars are closely grouped together, whereas slightly older stars are a bit more dispersed. In gravitationally bound clusters, such differences tend to be eliminated in dynamical timescales of less than a few million years.

Mass segregation can occur in clusters. Some argue thus might be primordial, but it certainly can happen dynamically. The net effect is that more massive objects tend to be more centrally concentrated in a cluster.

There is evidence for multiple populations in some old globular clusters. The picture here is very complicated, still being researched, and not something that is easily summarised.


Do tides occur on planets like Neptune, having water?

There will be tides, but they will be not very large.  It's pretty easy to get a good estimate of their size.

There are two things that control tides on Earth and both will be present anywhere else: The tidal forces from celestial bodies, and the size of the body of liquid in which the tides are raised along with resonances int he body of liquid.  Taking them in turn.

The tidal force is a bit funny, since the size of the tides raised on a planet by a celestial body is

(1) proportional to the mass of the celestial body raising the tide,

(2) inversely proportion to the cube of the distance away,

(3) inversely proportional to the surface gravity of the planet,

(4) and (roughly) proportional to the diameter of the body of liquid in which the tides are being raised.

That inverse cube of the distance means that the ability of a body to raise a tide drops off very quickly with distance.

Comparing Earth and Neptune: The Sun is 30 times further from Neptune than Earth, so Solar tides will be roughly 27,000 times smaller on Neptune.  They can be completely ignored.

Triton (Neptune's largest moon) is the jackpot, since it's a third as massive as Earth's Moon, and about the same distance away.  Neptune is 17 times more massive than Earth and four times the diameter.  So Triton tides of Neptune will be about fifteen times smaller than lunar tides on Earth.

All the other moons are much smaller or further away or both and will raise completely negligible tides.

So from gravitational effects alone, Triton will raise small tides and nothing else will be measurable without good instruments and a lot of care.

(Note, I'm talking about tides in oceans comparable in size to the planet -- for smaller bodies of liquid, the tides will scale down.)

On Earth, tides vary all over the place, and this is due to the configuration of the body of water and resonance effects.  Basically, if a body of water has a resonance -- a natural period for sloshing back and forth -- that matches the period of the tidal force, the tide can build up like repeated pushes to a kid in a swing.

Additionally, there can be a funnel effect with two arms of land coming together, or shallowing water that can amplify the tides. (E.g., the Bay of Fundy.)  Since as far as we know, Neptune has no land, it's effectively an planetary ocean and will get the full effect of the gravitational tides, but no resonances.

So, bottom line: Small Triton tides and nothing else.


Why don't we use amateur astronomers' telescopes to create a huge interferometer?

You can't make an interferometer by combining photographs. You need to combine the light sources so that the light waves can "interfere", which means you need to have not only the intensity of the light, but also the phase of the wavelength. A CCD only detects the intensity.
You also need to position your telescopes to sub-wavelength accuracy. For Radio telescopes, with long wavelengths, it is possible to detect and then combine the radio signals from several telescopes electronically. That's not possible with light. Visible wavelength interferometers capture the light and combine it as light. It requires very high precision.
Finally, we still have the atmosphere to contend with. Amateur observers know that, beyond a certain size (about 10in), larger telescopes don't have greater resolution, because of the blurring due to the atmosphere1. Bigger mirrors mean more light, mean dimmer objects can be seen, but no more detail is visible. There are ways of getting round this, (Lucky imaging) but these are not suited to interferometery.
So since placing amateur telescopes at nanometer accurate positions, and linking their output with single strand optical fibre is beyond the budget of amateurs, and since the results would be limited by the distortion of the atmosphere, this is not normally done.
1Large observatories like the Very Large Telescope use Adaptive optics to obtain dramatically improved resolution through Earth's atmosphere but that kind of technology is sophisticated and out of reach of amateur instruments.


How do telescopes "zoom" and change angle of view?

Telescopes tend to have a fixed focal length. What changes is the size of the sensor in the instrument used. If a small sensor is used, then a smaller section of the field of view is exposed, resulting in a narrower field of view being imaged than the equipment is capable of. If a larger sensor is used, more of the field of view of the telescope is utilised.

Additional optics in the light path that are associated with the specific instrument being used on the telescope will affect the field of view also.

Telescopes do not offer the capability of 'zooming' as you understand it from your consumer digital camera.


Why are cirrus clouds more common on Mars?

Hmm... here's my guess.
Cumulus clouds form due to warm, moist air rising. With the atmosphere of Mars being so cold, there's little heat to facilitate the formation of cumulus clouds. Cirrus clouds, on the other hand, are composed of sparse ice crystals, which makes them probably more common on Mars than cumulus clouds. After all, supercooled water does become ice.


Basically, yes, it's possible.  Our moon is slowly moving away due to energy transfers from Earth related to tidal drag.  This is very slow spiral.

I do not know whether a star can experience sufficient tidal drag relative to the center of its galaxy to cause a change in orbit - and there are a few billion other objects in the galaxy producing gravitational force in other directions, making the analysis kind of messy.


Appropriate battery for overnight viewing to drive telescope and GOTO?

Before buying anything, get out your Amp-meter and measure what the motors pull in clock drive and GOTO mode, starting from a variety of scope positions.
An 18 incher will have a lot of mass, and Dobsonians are not known for their excellent bearings, so the amount of juice you need for a night of observing could be considerable.
Do you track most of the time, or GOTO...GOTO...GOTO?
How many hours is your typical observing session.
Multiply your average tracking amperage by the approximate number of hours you spend tracking, then do the same for your GOTO work.
Add the two numbers (amp hours) together, and double the result.
That should tell you about how big a battery you need.

The doubling is done to provide leeway, and because taking a standard lead acid below 50% is a sure way to kill it after only a few recharges. If you go Li, you can probably cut the multiplier.


Was Earth closer to Europa on 1983-11-25 or 1985-07-22?

tl;dr Miki Sudo

Using JPL's SPICE toolkit, I computed the positions of Earth and Europa for the times in question.

On 1983-Nov-25, Earth and Europa are 935.2 million km apart, while 1985-Jul-22, they are 612.5 million km apart. Miki Sudo wins by 323 million km, given the assumed date for her birthday. If we don't trust famousbirthdays.com for Sudo's birthday, but assume it was still sometime in 1985, she still wins.

In fact, it turns out Joey Chesnut was born at a time when Earth was close to its farthest point from Europa, so he's fairly unlikely to ever win in this tiebreaker:


What effect does time dilation have on bodies orbiting close to Black Holes?

Trajectories close to a black hole can't be approximated by Newtonian mechanics, General Relativity is needed. The black hole doesn't have a surface, but the Schwarzschild radius is point of no return. However it is not possible to orbit just above the Schwartzchild radius.

If you are less than 1.5 radii from the event horizon, there are no orbits. The orbital velocity closer to a black hole than 1.5 radii would be greater than the speed of light. Inside this radius, the faster you try to go, the less the centrifugal force you experience becomes, and so the faster you would fall towards the blackhole.

In fact you won't find a stable orbit closer than 3 radii from the black hole. If the black hole is spinning (as most younger ones are expected to do) the space around the black hole is dragged along by the spinning black hole. It is possible to gain energy by entering this region of dragged space around a black hole. This all means that a space ship can't remain in the region of ultra-intense gravity for many hours by orbiting, It can't experience (from the point of view of a distant observer) extended gravitation time dilation by orbiting a black hole only a few km above its Schwarzschild radius.

It is possible to approach closer than 1.5 radii. An outside observer would see the spaceship approach the black-hole in a spiral at very high speed, but not cross the event horizon. The orbit would not appear to be elliptical and it would appear time dilated. Outside of 1.5 radii the dilation would be less.

Trajectories close to the black hole would have to be calculated using GR. Newtonian mechanics would not be a close approximation. The calculation in GR would be computationally intensive, since the GR equations can't be integrated exactly, even for a two body problem, so a numerical technique would have to be used.

Provided you stay rather further from the black hole, then the usual Newtonian approximation is valid. There's no certain boundary at which Newtonian mechanics is a good approximation. It depends on the accuracy needed.

Remember also that the region close to a black hole is a violent region of space. It would be dangerous for a ship to be so close to a black hole due to the intense radiation produced by infalling matter, and tidal forces.


There is no observational evidence that the sun is a member of a binary (trinary, or more) star system, where "star" means an object that is at least ~80 times the mass of jupiter and emits energy/light via standard hydrogen fusion.

Some evidence that people point to is that the majority of stars in the Galaxy (perhaps 60% or so) are binary.  However, that does not mean that we are a binary system, just that we are among those which are not.

There is the Nemesis star idea which was a hypothetical binary companion to the sun to account for periodic mass extinctions roughy every 26 million years (as in, the two stars would therefore be on a 26 million year period, so the companion is a few light-years away).  There are two problems with that:  First, that a 2010 reanalysis of the data showed that it was too periodic over the past 500 million years, during which time we've gone around the galaxy twice, and so our hypothetical binary orbits should have been perturbed and not be perfect (though others say that the precision of the geologic timescale isn't good enough to say this).  But, second, we can see stars out to a few light years -- we can see small, faint stars out to thousands of light years.  We have all-sky infrared surveys that should be able to pick up the faintest even star-like objects to at least a few light-years, and yet ... nothing.

The other problem is what LDC3 pointed out in their answer: We should see some systematic motion of our own star in orbit around the common center of mass of the hypothetical binary.  This would not be a yearly wobble, but rather it would be a slow motion of the entire sky on top of the other >1 year motions that we see.  We now have very accurate astronomical records dating back at least a century of star positions, especially close stars.  Even if we were on a 26 million year orbit - and especially one much shorter as some people claim - we should see effectively that our star is making a small arc, a part of a circle as it orbits the center of mass of it and the binary.  We don't.  There does not seem to be any systematic signal in the motions of stars that require the binary star model.

So, to summarize: A binary companion simply lacks any observational data; if it existed, we should not only have been able to see it by now, but we should also observe not only the stars in our sky show a systematic motion due to our orbit around it, but we should again see THAT binary star also move very quickly, relative to other stars, as it orbits the common center of mass.


Tools for simulation of theoretical astronomical problems?

Give GADGET-2 a check if you haven't already.

GADGET is a freely available code for cosmological N-body/SPH simulations on massively parallel computers with distributed memory. GADGET uses an explicit communication model that is implemented with the standardized MPI communication interface. The code can be run on essentially all supercomputer systems presently in use, including clusters of workstations or individual PCs.
GADGET computes gravitational forces with a hierarchical tree algorithm (optionally in combination with a particle-mesh scheme for long-range gravitational forces) and represents fluids by means of smoothed particle hydrodynamics (SPH). The code can be used for studies of isolated systems, or for simulations that include the cosmological expansion of space, both with or without periodic boundary conditions. In all these types of simulations, GADGET follows the evolution of a self-gravitating collisionless N-body system, and allows gas dynamics to be optionally included. Both the force computation and the time stepping of GADGET are fully adaptive, with a dynamic range which is, in principle, unlimited.
GADGET can therefore be used to address a wide array of astrophysically interesting problems, ranging from colliding and merging galaxies, to the formation of large-scale structure in the Universe. With the inclusion of additional physical processes such as radiative cooling and heating, GADGET can also be used to study the dynamics of the gaseous intergalactic medium, or to address star formation and its regulation by feedback processes.


Questions on apparent motion of superior planets

The planets do not orbit in the same plane as the Earth, they are inclined by a few degrees. This means that as they orbit the sun, they are sometimes above the plane of the Earth's orbit and sometimes below.  And it takes one orbital period to go the full cycle from above, to below and back to the start.
The reason that they appear further from the ecliptic at opposition is a matter of perspective.  At opposition the planet is close to the Earth. And nearby things appear bigger. So the planet will appear (in general) to be further from the ecliptic at opposition.


Why does this large Newtonian telescope's front cover have two or three holes in it?

What are these?
These are used to reduce the aperture of the instrument. While you see three in this image, likely only one is removable.
The instrument appears to be a Newtonian reflector. Which means it would have a secondary mirror near the front (and center). So the thing that looks like a knob in the center of the telescope cover is probably just for grip (and not likely removable).
One of the off-axis caps is probably actually a cap. That can be removed.  The other thing that looks like a cap is typically non-removable. It is a holder for the cap on the opposite side (the one that can be removed).
Purpose
If you observe a bright object. Such as the Moon. You can use this to reduce the total amount of light.
Solar filters are large aperture telescopes will often employ the same idea. The Sun is plenty bright. Collecting "enough" light isn't the problem. By using a solid cap with an off-axis (off-axis only because it is necessary to avoid the central obstruction in the telescope) can help limit the total energy entering the telescope to avoid over-heating internal components and damaging the instrument.


I'm going to write a short answer to get you going.

Here is a random link that claims to track StarLink satellites after launch, it might be helpful to predict when a newly launched train of satellites passes over your location, I don't know.

http://www.satflare.com/track.asp?q=StarLinkLaunch#TOP

It's not secure and it asks you for personal information. There are instructions in URGENT – Time change for tonight’s Starlink Satellite Train visible over Washington County where there is also this video that can give you some idea what kind of pointing and field of view might work.

Okay on with the question:

  What is the best lense to use and do I need any additional filters?

You don't want a long focal length lens, use a normal or somewhat wide-angle lens. Your best strategy is to search the internet or other Stack Exchange sites for examples that you like and then just see what they use. Satellites move across a big chunk of sky and their visibility is limited by what fraction of their path is in sunlight.

Photography SE has over 300 questions tagged astrophotography and it seems 17 posts also contain "satellite'.

  How do I point my camera at the spot where the satellite will pass through, do I need some kind of tracker for it?

If you know what satellite you want to photo, then there are many websites and programs available to predict when they will pass and where in the sky to look. I use a python program called Skyfield but in-the-sky.org and heavens-above are two websites that are very useful. This has been covered in more detail in other answers here and in other sites.

Even if the trail goes all the way across the sky, it won't always be sunlit, so look west for a few hours after sunset or east a few hours before sunrise. During the middle of the night only satellites in very high orbits will be in sunlight and therefore visible. For more on that see answers to Why do satellites appear to move faster when overhead and slower closer to the horizon?

Satellites move very roughly 1 degree per second, but it varies a lot depending on where they are in the sky (overhead vs near the horizon) and their altitude (see below). So if you tracked them everything else would be a complete blur and your satellite would just be a dot. For more on that again see answers to Why do satellites appear to move faster when overhead and slower closer to the horizon?

If you track the motion of the stars which is only 0.004 degrees per second, then you could expose for a long time and get many stars to show up in your photos. The satellite track will then be a roughly straight line across your photo. That's what most satellite photos look like.

But if you have a short focal length lens and your exposures are only a few minutes, I don't think tracking is necessary at all, just use a tripod. Sometimes people program their cameras to take a new photo every say 10 or 30 seconds and they stack them together later with small shifts so that the stars don't appear to move or blur. In that case they have to process the background objects on the horizon separately or else they'll be blurred.

Have fun!

 from this answer


What would happen to a polished marble statue left in space for a million years?

There are three main space weathering processes that will affect the surface of the marble.

Cosmic rays, high energy particle from the sun and beyond, will hit the surface. This can change the chemistry of the surface.
Solar wind particles, hydrogen and helium, can become implanted in the surface
Micrometeoroids will impact the surface, causing small craters, melting, and the inclusion of other elements such as iron.

These processes will tend to change the surface, developing a patina on a timescale of a hundred thousand years. The surface will darken (though as marble is not a typical rock in asteroids, there isn't any direct evidence of what happens with marble.

Marble is largely CaCO3, and this is in a equilibrium with CaO and CO2. At  standard temperatures and even the very low partial pressure of CO2 in the atmosphere, this equilibium favours CaCO3. In our atmosphere one needs a temperature of 550⁰C to decompose Calcite. However in space there is no CO2, and so the Calcite would very slowly decompose to CaO. Calcium in meteorites is mostly in the form of CaO.


Why is the Sun's Luminosity not equal to the Zero Point Luminosity?

The zero point luminosity L0 is the luminosity of a star whose absolute bolometric magnitude Mbol is 0.
IAU recommends the value
L0 = 3.0128 × 1028 W.
If the Sun's Mbol is +4.74, then
L0 / Lsun = 104.74 / 2.5 = 78.7,
which is consistent with Lsun = 3.828 × 1026 W.

Your understanding that Lsun ≡ L☉ is correct.
Maybe there was an accidental substitution of L☉ for L0 at some point.
Curiously, the number 3.75 × 1028 appears in the Wikipedia:Sun sidebar as the Sun's luminous flux in lumens, a quantity more applicable to a light bulb than to a star.


Can Pluto be seen with the naked eye from Neptune when Pluto and Neptune are closest?

No, it cannot. Far from it.
The closest approach between both planets is roughly 16 AU due to the 3:2 orbit resonance. Pluto will even then be a tiny dot among many with a brightness around 14 mag.
You can try that with Stellarium yourself, placing the observer on Neptune and looking for Pluto. You just have to find the right time. One such time is approx. in the year 2877.


Black holes are not "made of matter". They are better described as structures of gravity/warped spacetime. However, they do grow when absorbing things with positive mass-energy.

Antimatter still has positive mass-energy, so the black hole cheerfully gobbles it up.

Now, exotic matter with negative energy density might indeed decrease the size of the black hole, akin to some form of enforced evporation process. (paper1, paper2) This is highly speculative since there are various theoretical reasons why exotic matter does not exist, and many kinds of hypothetical exotic matter that might work differently.


There are apps which will help guide you to more significant stars (Google skymap comes to mind if you're looking for a simple mobile system - Stellarium may also be a good option), but it's highly unlikely that any app will include the location and names of so-called 'personal' stars, as the bodies who sell such things don't have any the authority to do so and are not generally recognized in the actual world of Astronomy.

Note that the IAU (International Astronomy Union) is the body generally recognized to have the responsibility of naming and classification of stars and other bodies. They also come out strongly against such practices and services:

  As an international scientific organization, the IAU dissociates itself entirely from the commercial practice of "selling" fictitious star names or "real estate" on other planets or moons in the Solar System. Accordingly, the IAU maintains no list of the (several competing) enterprises in this business in individual countries of the world.  Readers wanting to contact such enterprises despite the explanations given below should search commercial directories in their country of origin.

Link to IAU page on buying of star names


Is energy required for gravity to have effect, or is it an energy itself?

Gravity is an effect of the stress-energy tensor. Put differently, energy and momentum (they're related concepts) are the cause of gravity in general relativity. A fully charged electric battery has a very slightly stronger gravitational field than a depleted one. You may have heard that $E = mc^2$ or in as the terms are more descriptively defined in modern usage, $E^2 = m^2c^4 + p^2c^2$ . Just thinking about these equations from special relativity, you can see that more energy means more mass, which means more gravity.

Beyond that, gravitational waves carry energy. You may remember the recent discovery of gravitational waves from a black hole collision, in which about three solar masses worth of energy were calculated to have been given off as gravitational waves.

Getting back to plain old Newtonian mechanics, when masses move around in a gravitational field energy is converted from one form to another as Dean noted in their comment on the question. For example when a comet is near the sun on its highly elliptic orbit, it has high kinetic energy and low potential energy. It loses kinetic energy as it moves outward, being slowed by gravitational energy, but gains gravitational potential energy. This potential is, quite simply, a representation of the fact that it can get its "lost" kinetic energy back when it falls in again on the inbound leg of its orbit, being helped along by gravity instead of hindered when inbound.


There is a Dantes Standardized Subjet Test (DSST) in astronomy. I have no idea if and how widely colleges give credit for these exams. Typically, you will start with physics and choose to specialize in astronomy. Therefore, most standardized exams are for physics which may have a handful of astronomy questions.


Lunar Soil: what are those "bumps" and how were they formed?

It's a bit difficult to understand what exactly you're asking about, but I will presume you meant the topological features in images like this one, taken from the video of which you provided a link to:

For the most part, they are meteorite and micrometeorite impact craters, with some regolith boulders and impact ejecta scattered along the edge of the larger impact crater at the bottom right of the image.

These impact craters appear strange to an untrained eye when they're the only visible feature on images, because there is not atmospheric scattering of light on the Moon that would somewhat illuminate the shadow side too and reduce contrast to what we're used to seeing on Earth, so the high contrast can appear as bumps or dents interchangingly and at will, similar to how you can see either either a vase / candlestick in the middle (yellow) or two faces facing each other (black) on this image:

A trained eye will have less problems recognising them for what they are, though. One nifty trick is to search the image for different looking topological features, such as already mentioned larger crater ejecta along its edge at the bottom right of the image, or perhaps prominent features within the features themselves, like some larger micrometeorite impact craters show shockwave formed uplifted centers, and so on. It also helps to understand how these features were likely made, but once you orient yourself, it will prove difficult to convince yourself into seeing what you saw initially, if you saw indentations where there are perturbations, or vice versa.


Does Tobler's First Law of Geography Apply to Star Composition?

As you say stars composition depend on the composition of the molecular cloud where they were created. You have a molecular cloud where many stars are being created at the same time, all with the same composition and age, but some are bigger than others. As time goes the bigger stars will die earlier, adding new heavier materials to the cloud which will lead to a new generation of stars with a bigger metalicity.

So the star's composition not only depends on where it was created but also on how old it is.

The best way this can be checked is by studying the stars within a star cluster. You may find this site interesting.


Which star has the highest mass loss rate?

Some very hot and massive stars have strong stellar winds, known spectroscopically as "P Cygni profiles" after the type star. P Cygni 'erupted' in the 17th century to magnitude 3 and is now visible as a mag.5-6 star near the central star of the 'Northern Cross' that is Cygnus. An ordinary pair of binoculars will show it easily. So P Cygni stars have probably the highest mass-loss rate as I believe you are thinking of.


What are other names for planetoids that aren't orbiting a solar system, but hurtling through space?

See nnnnnn's comment below.

The NASA website, describing 'oumuamuam, uses the term "interstellar object." Extrasolar asteroid would seem to be another option.  I haven't seen a unique word different from asteroid that originated outside of the solar system.

It seems a simple prefix is sufficient to describe objects beyond the solar system, i.e. exoplanet.

Scienfitic American has an article which uses the term "interstellar object."

Also, see answers to:

Is intrastellar commonly used by astronomers to refer to objects within our solar system?
Adjective for things outside our solar system


Why can't we observe a solar eclipse and the preceding/following lunar eclipse from the same place?

For a solar eclipse, the Moon has to be between the Earth and Sun. For a lunar eclipse, the Earth has to be between the Moon and Sun.

Since it takes the Moon about 4 weeks to orbit the Earth, it will take it about 2 weeks to go from a lineup of Moon-Earth-Sun to a lineup of Earth-Moon-Sun.

To make this more complicated, the Earth is rotating so that it takes about 24 hours to go from midday to midday.

So, say we have a solar eclipse at midday - there's a line between the centre of the Earth, through you, the Moon and the Sun. Two weeks later, when the Moon has moved around to the other side of the Earth this means that when you go out at midday, the line is now Moon-Earth-you-Sun. So you can't see the lunar eclipse, because the Earth is now in the way.

You can see both eclipses two weeks apart when they are around sunrise/sunset because you're no longer on the line through the centre of the Earth between the Sun and Moon.

Update:

I was going to do the math - but I think that this map of eclipses in 2019 demonstrates it better. If you compare the solar eclipses with their paired lunar eclipses, we see that there is a bit of overlap (since eclipses are not instantaneous) - but since these are at the edges of the eclipse, it's practically impossible to see a full solar eclipse two weeks after seeing a full lunar eclipse from the exact same location.


Doing a bit of reading up on this, I might have an answer, though credit where credit is due, the answer isn't really mine:

https://www.reddit.com/r/askscience/comments/3wsy99/why_is_neon_so_rare_on_earth/

When the planets coalesced, it's likely that there was very little ices/gas around the inner planets when they formed and the Earth's atmosphere and water (CH4, NH3, CO2 and H20 being the 4 most common outside the frost line ices).  These likely came from asteroids and meteors that formed outside the frost line and later crashed onto earth.

Neon is the 5th most common element in the milky-way but because all noble gases have very low freezing points, it's likely not be very common even on comets or meteors for the same reason that water or CO2 aren't common inside the frost line, Neon, and other noble gases likely stay free and don't collect on comets or meteors in high amounts.   (I looked, but couldn't find an article to verify that).

But if comets have low noble gas content, then we have to look for an alternate source.  With that in mind, and going back to the first link, Argon is produced by radioactive decay of Potassium 40 and that would explain it's relative abundance compared to the more common noble gas, Neon.   Helium (Alpha particles) is also produced inside the earth and Radon is is too in small amounts but Radon also decays - that's not related to your question though.

If Argon on planets comes primarily from Potassium 40, you should expect the amount of Argon to have a roughly similar ratio to the amount of potassium on a planet and not be relative to the percentage of atmosphere.   A 2nd factor, how much gets blown off the planet over long periods of time is a factor too.   Venus in general should be able to retain much of it's Argon based on atomic weight (40) similar to CO2 (44), but if it loses even a tiny percentage of it's Argon over time, that would be a factor too.

Now, to see if this is possible, I should run some numbers, but I warn you, my math can be a little rusty.

Potassium is the 7th most common element in the Earth's lithosphere at about 0.26% and about 0.0117% of that Potassium is Potassium 40.   Using a very rough estimate of $2.3 \times 10^{19}$ tonnes for the Earth's crust, $(2.3 \times 10^{19}) \times (2.5 \times 10^{-3}) \times (1.17 \times 10^{-4}) =$ about $6.7 \times 10^{12}$ or 6.7 trillion tons of Potassium 40 currently in the Earth's crust. (There's probably a fair bit more in the mantle, so these numbers are rough)

With a half life of about 1.248 billion years, that's sufficient time for over 3 half lives if we start after the late heavy bombardment, which suggests a bit over 7/8ths of the original Potassium 40 in the Earth's crust has decayed into Argon 40, so there should be, given the age of the Earth and abundance of Potassium 40, a bit over 7 times 6.7 trillion tons or, lets ballpark and say a bit over 50 trillion tons of Argon that formed on earth by Potassium decay.  (I'm ignoring any that might have been produced prior to the late heavy bombardment, cause I assume that could have blown some of the atmosphere off the earth or heated the atmosphere enough for the sun to blow some of it off).    Also, doing a bit of research, only 11% of the Potassium 40 decays in to Argon 40, 89% undergoes beta decay into Calcium 40, so for this to work, there would need to be a fair bit more Potassium in the earth than I estimated, but that's still likely the case.

The mass of the atmosphere is about 5,140 trillion tons, and 1.288% of that (By mass, not volume) = about 66 trillion tons, so the Argon we should expect from Potassium 40 decay and the amount of Argon in the atmosphere are pretty close.   Some Argon gas might have escaped and some should still be trapped inside the earth but the numbers are close enough to work and I think that's very likely the answer.  It also suggests that the Earth has lost relatively little Argon to space, which also fits with the Atmospheric Escape article.

A 2nd way to look at this is that Argon 40 makes up 99.6% of the Argon in the atmosphere and Stellar Nucleosis likely wouldn't account for a ratio anywhere close to that (not a typical stellar link but Wikipedia says Argon 36 is the most common isotope).   The decay of Potassium 40 does explain the 99.6% Argon40 ratio.

If we apply a similar estimate to Venus, with Venus atmosphere about 94 times the mass of Earth's, and we assume a similar amount of Argon-40 being produced in Venus' crust we could roughly expect 1.28%/60 or about 0.02% Argon by mass in Venus's atmosphere or perhaps, if the Earth lost a pretty high share of it's lighter crust elements after the giant impact, we might expect a bit more than that on Venus, perhaps 0.03% or 0.04% as a rough estimate.   Using your number of 0.007%, that's lower than I calculate it should be, but Venus could have lost a higher share of it's Argon than Earth and it also might be slower to release trapped gas inside it's crust than Earth because it doesn't have plate tectonics, so the number for Venus looks "about right" too.    It's the Potassium 40 in the crust.   I'm convinced.

Interesting question.  I learned something researching it.


Do narrow lines in the spectra of O- and B- type stars always indicate magnetic fields?

The narrow lines don't necessarily require a magnetic field, but any star that large will have one.

I'm wondering two things:

Has the presence of such comparatively narrow lines in the spectra of O- and B- type stars been observed in cases where there is no magnetic field present?

O-type main-sequence stars have high metallicity, they will have a magnetic field. Tau Scorpii and Theta$^1$ Orionis C info at their links. The Phys.Org article: "Strong magnetic fields discovered in majority of stars" says:
"An international group of astronomers led by the University of Sydney has discovered strong magnetic fields are common in stars, not rare as previously thought, which will dramatically impact our understanding of how stars evolve.".
This Wikipedia article: "Solar Dynamo" could use some citations, but here's what it says:
"The solar dynamo is the physical process that generates the Sun's magnetic field. A dynamo, essentially a naturally occurring electric generator in the Sun's interior, produces electric currents and a magnetic field, following the laws of Ampère, Faraday and Ohm, as well as the laws of hydrodynamics, which together form the laws of magnetohydrodynamics. The detailed mechanism of the solar dynamo is not known and is the subject of current research.
Mechanism
A dynamo converts kinetic energy into electric-magnetic energy. An electrically conducting fluid with shear or more complicated motion, such as turbulence, can temporarily amplify a magnetic field through Lenz's law: fluid motion relative to a magnetic field induces electric currents in the fluid that distort the initial field. If the fluid motion is sufficiently complicated, it can sustain its own magnetic field, with advective fluid amplification essentially balancing diffusive or ohmic decay. Such systems are called self-sustaining dynamos. The Sun is a self-sustaining dynamo that converts convective motion and differential rotation within the Sun to electric-magnetic energy.
...".

If so, are there other mechanisms that could be responsible for this in cases without magnetic fields? I'm grasping at straws for ideas, like low collisional broadening.

Our Q&A: "Different width of spectral lines for different groups of stars" has Rob Jefferies answer.
This seems well explained in section 3.4 of "Magnetically Confined Wind Shocks in X-rays - a Review" (22 Sept 2015), by Asif ud-Doula and Yael Nazé, on page 10:
"3.4. Structure of confined winds, as revealed by high-resolution spectra
High-resolution spectra can yield a wealth of information. With current instrumentation, line widths and shifts can be evaluated with precisions down to a few tens km s$^{−1}$ in the most favorable cases (a few hundreds km s$^{−1}$ more typically). Furthermore, the comparison of lines from H-like and He-like ions and of components of $f ir$ triplets of He-like ions constrain the temperature and location of the emitting plasma. However, such measurements are currently only possible for the brightest X-ray sources, so that few magnetic massive stars have been investigated
in this respect (τ Sco - Mewe et al. 2003; Cohen et al. 2003, θ$^1\!$ Ori C - Schulz et al. 2000; Gagné et al. 2005a, b, HD 191612 - Nazé et al. 2007, HD 148937 - Nazé et al. 2008, 2012, β Cep - Favata et al. 2009, IQ Aur - Robrade and Schmitt 2011).
Within noise limitations, the X-ray lines of magnetic massive stars were found to be symmetric, and globally unshifted. This agrees well with MHD models. In the case of θ$^1\!$ Ori C, global fitting however suggests small variations in velocity (Gagné et al., 2005a): from −75 km s$^{−1}$ when the star is seen pole-on to about 100 km s$^{−1}$ when seen edge-on. This change needs to be confirmed as the errors are large but also because one cannot exclude a stochastic variation when only a single observation per phase is available. If further observations provide evidence that velocity varies with phase, then refinement of models will be needed, as no such changes are currently predicted (Gagné et al., 2005a).
Reported widths of X-ray lines largely depend on object and ion considered. The narrowest widths, so far, were found for β Cep, whose lines are dominated by instrumental resolution, yielding only upper limit on intrinsic widths (<600 km s$^{−1}$, Favata et al., 2009). Larger widths, full width at half-maximum (FW HM) ∼ 600 − 800 km s$^{−1}$, were reported for ions with high ionization potential (Mg, Si, S) in τ Sco, θ$^1\!$ Ori C, and HD 148937, three stars with more rapid winds than β Cep. Such widths are much smaller than observed for “normal” O-type stars (FW HM ∼ v∞), where lines arise in embedded wind shocks distributed all over the wind, hence cover a larger velocity range. They indicate formation in slowly moving plasma, in agreement with the confined winds scenario. However, most MHD models predict even narrower lines (Gagné et al., 2005a).
Furthermore, lines from ions with lower ionization potential, notably oxygen, appear broader (FW HM ∼ 1800−2000 km s$^{−1}$ Gagné et al., 2005a; Nazé et al., 2007, 2008). These lines are associated with cooler plasma, which could have a different origin than hotter plasma. For example,
the dominant hot plasma in θ$^1\!$ Ori C is thought to arise in confined winds while the cooler one could arise in embedded wind shocks as in normal O-stars (Gagné et al., 2005a). This dual origin could be supported by the different temperatures derived from the different ions (Schulz et al., 2000). In Of?p stars, however, the spectra are dominated by the cooler component, i.e. confined winds emit soft X-rays in these objects (see above), but it cannot be excluded that current errors, which are large, are somewhat blurring the picture.
In He-like triplets, the forbidden line is suppressed when the density is high or the UV radiation is intense. In the case of massive stars, the latter effect is the most important one and, thanks to dilution with distance, enables us to locate the emitting region. In τ Sco, θ$^1\!$ Ori C, HD 148937, and β Cep, the start of the emitting region is found to be close to the photosphere, at radii r ∼ 1.5 − 3 R$_∗$ for the first three stars and r ∼ 4 − 6 R$_∗$ for the latter case. These values are only slightly lower than the corresponding Alfvén radii of these stars, they thus appear qualitatively compatible with MHD simulations. In IQ Aur, the forbidden line was found to be normal, suggesting a formation radius larger than 7R$_∗$ (Robrade and Schmitt, 2011) - despite a large value, this also appears compatible with the supposed location of confined winds in this star.".
Other reading:
"The surprising magnetic topology of $\tau$ Sco: fossil remnant or dynamo output?" (7 Jun 2006), by JF Donati, ID Howarth, MM Jardine, P Petit, C Catala, JD Landstreet, JC Bouret, E Alecian, JR Barnes, T Forveille, F Paletou, and N Manset
"The magnetic field and confined wind of the O star θ$^{1\!}$ Orionis C" (26 Jan 2006), by G.A. Wade, A.W. Fullerton, J.-F. Donati, J.D. Landstreet, P. Petit, S. Strasser
"X-ray hardness ratios for stars of different spectral types" (2006), by Meurs, E. J. A., Casey, P., & Norci, L.
Wikipedia - Zeeman Effect
"Magnetic Fields of the A-Type Stars" (30 Apr 1958), by H. W. Babcock


Is every star formed in an open star cluster?

The question is still an open matter of current research.

It seems to be true that the vast majority of star formation takes place in groups and aggregates of various sizes - from a few stars to millions of stars in "super" star clusters.
This is likely because collapsing clouds of gas are normally much more massive than a star and the collapse process reduces the Jeans mass and renders the cloud unstable to fragmentation into smaller cloud cores.

However, it seems that the vast majority ($>90$%) of star clusters/associations are either born in a gravitationally unbound state or become gravitationally unbound within a few million years. The gravitationally bound open clusters, who's exemplars include the Pleiades, are comparatively rare survivors (or at least partly survived) of this "infant mortality". So, in that sense we can say no, most stars are not born in open clusters, but the likelihood is that most were born in aggregates with near neighbours that went their separate ways shortly after birth.

Current thinking is that our Sun was born in a cluster of some ten thousand stars (Adams 2010). This is an argument based on the shaping of the early solar system by dynamical encounters and the early presence of radioactive nuclei that were likely injected by the explosion of a very nearby massive star (probably a cluster sibling).


How to explain the equation 'Altitude of a star at culmination = co-latitude + declination'?

Unless the observer is at a geographic pole, the celestial equator is not parallel to the horizon but crosses it at an angle equal to the observer's colatitude.
A star's path across the sky is parallel to the celestial equator, with maximum altitude midway between the eastern and western horizons, i.e. on the meridian.

Rendered by Stellarium


Most of Saturn's rings are inside it's Roche limit, which means they will never clump together.   Tidal forces prevent this from happening.

small objects that are already together can withstand the tidal forces.    A sufficiently large body inside a planet's Roche limit should break apart by tidal forces.   That may be how the rings formed in the first place, or they may have formed by collision.

Saturn's rings are thought to be relatively young, if 100 million years qualifies as young, but it is for objects inside the solar system.    Being young doesn't tell us how long they will last.   I think the 300 million year estimate may be accurate enough, originally mentioned in the comments, but until that happens, the rings will remain, just grow smaller over time and they'll never coalesce into moons.

As a footnote, Saturn's thinner outer rings or gossamer rings are outside the Roche limit, and those might, one day, form into moons, or they may be too thin for that to happen.  A certain density is probably required for moon formation.


Are there ways to determine a star's spectral type without the spectrum itself?

The color index of the star can give you that information. I can look it up for you if I know which exoplanet you're interested in, or you can find it in the NASA Exoplanet Database here.


How to identify component of velocity of a star from its red shift?

The first part of this question is already answered at How Do we know about redshift?

Regarding the second part, measuring tangential velocity is much harder, and indeed impossible for stars that are far away. Basically the method is to observe the position of the star very accurately compared to very distant stars which are assumed not to be moving and then observe it again a few years later. After adjusting for known effects like aberration and parallax, what remains is "proper motion".

The most sensitive current telescope for this kind of measurement is the Gaia mission, which is able to measure the movemnt of stars up to a few thousand light years away with reasonable accuracy.


Eta Cassiopeiae star type; G0V? F9V? Both?

$\eta$ Cassiopeiae A has an estimated mass of 0.972 M$_\odot$, an estimated temperature of 5973 K, and a B-V color index of ~0.58[1]. In addition to the spectrum of the star, we look at these and other properties when attempting to classify main sequence stars. You can see a table [here] which shows the bulk properties of each spectral type that we can use in addition to spectrum to help classify the star. The mass estimate places it into the G5V category, while the temperature and B-V place it definitively into the G0V category[2]. Hence, I would go with G0V. The mass is a little on the light side for the G0V category, but the temperature and color index and right on for the characteristic values for G0V.


How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?

ALMA produces stable, mutually coherent ~THz LO (Local Oscillators) for all the antennas by...
Using a single central LO and piping it to every antenna via fiberoptic cable!  The fiber expands and contracts due to temperature fluctuations so a laser system is used for calibration among the antenna.  If you can believe it, they manually lengthen or shorten the fiber to each antenna to adjust for the contraction/expansion. The LO for the high frequency observations for ALMA requires femtosecond accuracy.
Check out this youtube for a more thorough explanation than I could provide:

Notes:

I think this is an excellent question, since solving the LO synchronization problem for high frequency observations was one of the most difficult technical and engineering problems associated with bringing the full capabilities of ALMA online.

Building and designing interferometers decades ago, we always remembered that light travels about a foot per nanosecond.  To get femtosecond accuracy for synchronization, they have to adjust fiber lengths by the micrometer.  Picoseconds?  Ain't nobody got time for that!

The video shows and explains details of how the LO signal included in the modulated laser is split and then distributed by optical fiber to each element in the array. From the explanation on the video's page:

Complex electronics accurately stitch ALMA’s individual wave detections together into one dataset. The first step in this process is to have exact measurements of where and when the antenna picked up its waves. On each antenna is a clock that timestamps the data using a kind of atomic metronome, or rhythm-keeping device, kept near the supercomputer. The timekeeping waves from this central oscillator beam out to each of ALMA’s antennas. Onboard the antennas, a local oscillator injects this timekeeping beat into a microscopic mixer with the waves coming through the receiver, and a mixed-down signal is digitized and sent back along the fiber into the supercomputer.
The timekeeping signals generated by the central local oscillator are sent through optical fiber, which presents an additional challenge. The length of the optical fiber can vary with temperature, but in order to achieve the incredible level of accuracy that ALMA requires, the fiber length must not change by more than one millionth of a millimeter over the entire 15-kilometer-distance to the farthest antenna. Stabilizing the length of the optical fiber is another job of the central local oscillator.


Why don't we have 2 Summers and 2 Winters?

There are a few incorrect assumptions in your post, so it is difficult to answer as asked. But I can address the misconceptions.

1. The seasons are not caused by our distance from the sun
The seasons are caused by the 23.5° tilt in Earth's axis. When the Northern Hemisphere is tilted towards the sun (summer), the Southern Hemisphere is simultaneously tilted away from the sun (winter). So the seasonal temperature difference has little to do with the Earth's position in its elliptical orbit. Without this tilt, there would be no seasons and the temperature day to day across the globe would be relatively uniform.

2. Even the GLOBAL temperature is NOT consistent with our change in distance
As a matter of fact, the average temperature of the Earth globally is hottest when it is the furthest from the sun — hotter by about 2.3°C (ref). That's because there is a lot more landmass in the Northern Hemisphere facing the sun (when Earth is farthest away in its orbit). So even though there is less intensity of sunlight, the land is able to be heated up much faster than the vast oceans which have to be heated at perihelion.

This distance-temperature inconsistency isn't unique to the Earth. Look at the average temperature of the other inner planets as we move away from the sun:

Mercury (167°C)
Venus (460°C)  ← farther, but hotter than Mercury?
Earth (14.0°C)
Mars (-60°C)

Venus is actually warmer than Mercury because of the thick carbon dioxide atmosphere causing runaway global warming. So it isn't simply the distance from the sun that determines the average temperature of a planet.

3. There's only ONE aphelion/perihelion
The closest point of the Earth's orbit (perihelion) and the farthest (aphelion) only happens once per year; not twice. That is because the elliptical orbit of the Earth is such so the sun is at one of the foci, not the center (as illustrated below).

Note that the size of the bodies and the eccentricity of the orbits are greatly exagerated here.


Can a solar flare destroy every electronic item on earth?

Here's the how: very energetic charged particles interact with the Earth's magnetic field, and when they do so they emit electromagnetic radiation. If the energy of this radiation was high enough then when it reached wires/conductors at the Earth's surface the opposite effect would take place: ie the EM radiation would result in the flow of charged particles (a current). Such currents, if strong enough, could damage very many electronic and electric devices.

As to the possibility: it seems unlikely that such a high energy event would take place (ie to destroy all/nearly all electronic equipment), but it could be possible. We don't have sufficient data.


What does it mean for a star to go nova or supernova? Can I safely observe these?

It's a matter of size and stellar evolution.  There are many, many types of stellar explosions.  The University of Arizona has one page that describes these types.  generally, a Novs is not what we think of (i.e. a star exploding). That's actually a Type II Supernova.
According to that site:

Novae are frequently (perhaps always) members of binary systems where the more massive star evolves more quickly and becomes a white dwarf. The less massive star loses some of its hydrogen-rich outer atmosphere to the white dwarf where it collects.  The high gravity and intense heat of the white dwarf make conditions suitable for hydrogen fusion into helium. When enough hydrogen collects, it explodes in a fusion reaction, causing a nova. The gas ejected in the explosion moves outwards at 1000-2000 km/sec.

The Supernova has two distinct types:

The "Type I" supernovae arise when the white dwarf star member of a binary system accretes so much matter from its companion that it is tipped over the 1.4M Chandrasekar limit. The white dwarf collapses very rapidly (just a few seconds) until the infalling layers hit the very hard surface of the neutron star. The resulting shock goes back through the collapsing layers, and they explode by hydrogen fusion.

And the one we are more familiar with from popular media:

The "Type II" supernovae are the result of a massive star consuming all of its nuclear fuel and then exploding. The black line traces the evolution of a massive star schematically, until it reaches the supernova stage.

Portland Community College actually lists a lot of the variations that happen in the different types of supernovas out there.

Type I: supernovae WITHOUT hydrogen absorption lines in their spectrum
Type II: supernovae WITH hydrogen absorption lines in their spectrum.
Type II
The type II supernovae are massive stars whose iron cores collapse and then rebound, shock heating the outer layers of the star, which then explode outward.
Type I
The type I supernovae are subdivided into three subclasses, called (boringly enough), Type Ia, Type Ib, and Type Ic.
Type Ia: no hydrogen lines, no helium lines, strong silicon lines
Type Ib: no hydrogen lines, strong helium lines
Type Ic: no hydrogen lines, no helium lines, no silicon lines
Type Ib and type Ic supernovae are massive stars which lost their outer layers in a stellar wind before core collapse.
Type Ib supernovae lost their hydrogen-rich outer layer, revealing the helium-rich layer immediately below.
Type Ic supernovae suffered more mass loss as supergiants, losing both the hydrogen-rich layer and the helium-rich layer (revealing the carbon-rich layer below).
Type Ib and type Ic supernovae are essentially the same as type II supernovae.

As for observing them, since none of them are expected to be close to us (i.e. no stars near us able to go supernova), it should be safe. The problem is predicting when one will happen.  For instance, Betelgeuse is expected to go supernova "soon", but in stellar terms, that could be well over 100,000 years from now. The list of candidates in the Milky Way doesn't bode well for one in your lifetime. Although astronomers are often seeing them in other galaxies, and some are observable by amateur astronomers.  In fact, just this past August (2013) an amateur discovered a nova.


Can angular information be known more precisely than the diffraction limit?

Yes, Gaia makes extensive use of this technique.

The measurement accuracy of star positions for the brightest stars is 10 to 20 micro- arcseconds – up to 100 times better than that the precursor mission, Hipparcos; this angular resolution corresponds to the observed diameter of a Euro coin on the Moon, viewed from Earth.

The underlying principle is the standard deviation of the mean. It depends on the sample size, more precisely on the square root of the sample size. In this case the sample size is the number of the measured photons. Therefore it works better for bright stars.


Evidence that galaxies are made of billions of stars?

The individual stars can be resolved in galaxies close to us...including our own, the Milky Way Galaxy.
More distant galaxies are morphological similar to those that can be resolved into stars, so a natural induction would be that they are similar in nature to those we can resolve into stars.
Galactic spectra are consistent with them being composed of stars in the sort of relative motion we expect.
Supernovae are observed in distant galaxies, and these are similar to those seen close up which we know are stellar in origin.
... etc


Why do the planets in our solar system orbit in the same plane?

In the protostar stage of the Sun, it was surrounded by a (spinning) gas cloud. This cloud behaved like a fluid (well, a gas is a fluid), so it flattened out into an accretion disk due to conservation of angular momentum. The planets eventually formed from the dust/gas in the disk from compression of the dust in the disk. This process won't end up moving the dust out of the plane (all the vertical force of gravity is toward the disk), so the final planet is in the plane too.

Why does an accretion disk have to be flat? Well, firstly, let's imagine the protostar and gas cloud before the accretion disk formed. Usually such a setup will have particles spinning in mostly one direction. The ones spinning in retrograde orbits will end up reversing themselves due to collisions.

In this gas sphere, there will be an equal number of particles with positive and negative vertical velocities (at a given point in time; due to rotation the velocity signs will flip). From collisions, eventually these will all become zero.

A particle revolving around a planet will always revolve such that the projection on the planet is a great circle. Thus we cannot have a particle with vertical velocity zero but with vertical position nonzero (as that would imply an orbit that isn't a great circle). So as the vertical velocity decreases, the orbit inclination decreases too. Eventually leading to an accretion disk with very little vertical spread.


How does one use SPICE to calculate sunrise and sunset?

I'm no SPICE expert but here are some potential solutions (unless of course you want to try Skyfield's Almanac methods!)
Firsrt possibility, but all these may not work
this answer links to this comment links to SPICE Tutorials, Updated December 11, 2019 links to the 69 slide presentation SPICE Geometry Finder (GF) Subsystem; Searching for times when specified
geometric conditions occur dated January 2020.

Much SPICE software computes a geometry parameter at a given time, t, i.e. x = f(t). Example: on 2011 MAR 30 14:57:08, what is the spacecraft’s altitude
above Mars?
The Geometry Finder subsystem does the inverse: it finds times when specified geometric events occur

GF provides two primary types of event-finding APIs

Boolean: a geometric condition (an event) is true or false
– Numeric: a geometric quantity has a given value, is within a given range or has achieved a local or global maximum or minimum

In your case you would specify a position on a body, then look for times when elevation was 0 or 0+/- small.
Another possibility, this seems to address exactly what you are looking for

https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/spkcpo_c.html
Examples
The numerical results shown for these examples may differ across
platforms. The results depend on the SPICE kernels used as
input, the compiler and supporting libraries, and the machine
specific arithmetic implementation.

Compute apparent solar azimuth and elevation as seen from a
specified surface point on the earth.
Task Description
In this example we'll use the location of the DSN station
DSS-14 as our surface point.
We'll perform the solar azimuth and elevation computation two
ways:

Using a station frame kernel to provide the
specification of a topocentric reference frame
centered at DSS-14.

Computing inline the transformation from the earth-fixed,
earth-centered frame ITRF93 to a topocentric frame
centered at DSS-14.

Note that results of the two computations will differ
slightly. There are three sources of the differences:

The station position is time-dependent due to tectonic
plate motion, and epochs of the station positions used
to specify the axes of the topocentric frame are
different in the two cases. This gives rise to different
orientations of the frame's axes relative to the frame
ITRF93.

The two computations use different earth radii; this
results in computation of different geodetic latitudes
of the station. This difference also affects the
topocentric frame orientation relative to ITRF93.

The station movement between ET and the epoch at which
the DSS-14_TOPO frame is specified contributes a very
small offset---on the order of 10 cm---to the station-sun
position vector, expressed in the ITRF93 frame.

and there's a lot more there.
It seems you can do the same thing in Python using SpicyPy

Find the azimuth and elevation of the apparent position of the Moon as seen from the DSN station DSS-13 by the following steps:

Find the apparent position vector of the Moon relative to the DSN station DSS-13 in the topocentric reference frame DSS-13_TOPO at epoch ET. Use light time and stellar aberration corrections.

For this step, you'll need to have loaded a station SPK file providing geocentric station position vectors, as well as a frame kernel specifying topocentric reference frames centered at the respective DSN stations. (Other kernels will be needed as well; you must choose these.)

Convert the position vector to latitudinal coordinates. Use the routine spiceypy.reclat for this computation.

Compute the Moon's azimuth and elevation as follows: azimuth is the negative of topocentric longitude and lies within the range 0-360 degrees; elevation is equal to the topocentric latitude. Display the results in degrees.