How did Astronomers mostly(?) agree to publish arXiv preprints along with peer-reviewed Journals? Was there pushback?

This is an interesting question, and I think it can productively be split into a couple of distinct parts, separating the publisher side from the author side:

Was/is there any pushback from journals about posting pre-publication versions of papers on the web?

Why do research article authors in astronomy post papers on arxiv at a relatively high rate?

Part 1 is easier to answer, since there are fewer individual actors to consider.  As far as I can tell, the answer to this is a flat "no".  As @ProfRob notes in a comment, there was a long tradition of dissemination of preprints on paper before the web existed, so it likely would have been difficult for publishers to forbid that practice once electronic dissemination became a possibility.  Many journal publishers are directly connected with professional societies, and even for-profit journals need to show support for wide dissemination of ideas. I don't know of any journal that publishes astronomy research that doesn't allow posting of preprints.
Contrary to what one often hears, the higher-profile journals like Science and Nature are on board with this idea as well; both of them are very explicit in stating policies that allow both pre-publication posting on sites like arxiv, and free discussion of results at meetings. (current Nature policy, current Science policy).  What they don't allow is authors discussing results with the press before publication. Nature in particular seems to emphasize the importance of preprints; they published their first editorial on it in 1997, and they seem to publish articles every few years that say, in effect, "no really, it's ok."
Part 2 is harder to answer, as is any question about why a practice evolves in some particular setting that involves individual decisions by many different people to adhere to some unstated norm.
First I thought I'd check to see whether it's really true that astronomy is different from other physical sciences in its use of preprints.  The article linked in the answer from @Thomas suggests that astro is on the high side (at least as of 2011) in use of arxiv.  I did my own spot check with a few ADS queries.  To keep it simple, I just picked a few journals. In the years 2015-2019, this ADS query shows that there were a combined 24,634 refereed articles published in The Astrophysical Journal and Astronomy & Astrophysics (to pick a major American journal and a major European journal). Of these, 21,568 have preprints available on arxiv. (You get this by adding the qualification "property:eprint_openaccess" to the query.)  So that's almost 88% of papers, which seems high indeed.  To compare to another sub-field, I chose the journal Physical Review B (which publishes papers on condensed matter physics, since I know there's a cond-mat section of arxiv), yielding 25,952 articles, of which 16,936 have arxiv preprints.  So that's 65%, still pretty high, but a fair bit lower than astronomy.
So, why might this be?  My guess is that part of the answer might lie in the creation of ADS, and in particular the ADS project to scan all of the previous astronomical research literature and make it freely available online. That made it possible to sit at your desk and do all of your bibliographic research, rather than going to the library.  Once people got used to doing that, it was natural to want to do it for the newer literature, too.  Of course those electronically published articles were likely available to many researchers directly from the publisher, through their home institutions, but that early digital access was pretty heterogeneous by journal and sometimes not so smooth.  That might have contributed to people wanting to put papers on arxiv to make them more available.
It's certainly more complicated than this, but I do think the existence of ADS (including its indexing of arxiv) may have made a difference in earlier / more complete adoption of electronic dissemination in astronomy compared to some other fields.
A related thing that I think is interesting is that the use of arxiv in astronomy has now become so ubiquitous that some researchers consider posting on arxiv to be important not just for early visibility, but for visibility at all.  By that I mean that the daily arxiv postings have become, for many, the way that they learn about new literature, supplanting the role of journal table-of-contents services.  I don't know how widespread the practice is (and I'm sure there are some generational differences), but I have certainly heard people say that arxiv is the only resource they use for regularly keeping up on new papers that are published.  In other words, there's a sense in which simply publishing your paper in a refereed journal is not enough to make it fully visible to the community, at least not in the short term (though people will likely find it in literature searches later).
To check this idea, I went back to the ADS astronomy searches referenced above and looked at citation numbers.  The 21,568 papers with arxiv preprints have 532,493 citations total, while the 24,634 papers overall have 561,738 citations.  Subtracting, that means that the 3,066 papers without arxiv preprints have 29,245 citations. Looking at average citations per paper, that means that the arxiv preprint papers have 24.6 citations per paper on average, while the non-arxiv-preprint papers have 9.5 citations per paper on average.  That is such a huge difference that it actually makes me wonder if there is some confounding factor I'm not considering - I encourage others to check the numbers and/or to try other searches.  But if that's true, it's a huge effect, and bears out the idea that arxiv plays an important role in the community in research visibility overall, not just in early access.
Finally, I'll note that the importance of arxiv is now so complete that you even see an open letter from the AAS journal editors to arxiv, asking them to include the Research Notes of the AAS (a new publication of short, unrefereed papers) in the arxiv.  There has been an ongoing discussion of this, since arxiv moderators started (mostly) rejecting arxiv postings of RNAAS papers.


How variable does a star have to be, to be a variable star?

There is no lower limit, and as you say, all stars are somewhat variable.

However catalogues of variable stars exist, and they can record a wide range of levels of variability.  For example, the general catalogue of variable stars lists stars like Alpha Triangulum, with a variability of 0.01 magnitudes.

Ultimately a variable star is a star which has had its variability measured and studied and recorded in a catalogue.


Well this is not a question that can be answered in a few sentences!

In short

We still do not know, how exactly the galaxies formed and have the shapes they possess. The large-scale structure (LSS) in the Universe as we see them today are a consequence of tiny primordial density fluctuations that arose right after the
Big Bang.
The reason for these tiny density fluctuations is believed to be quantum in nature.

So, these tiny fluctuations in the early past lead to agglomeration of gas and dust clouds, leading to certain areas becoming denser. These denser areas slowed down the expansion of the Universe, allowing the gas to accumalate into small protogalactic clouds. Gravity in these clouds casued the gas and dust to collapse, and in turn form stars. These stars burned quickly and became globular clusters (while gravity was still collapsing the dust and gas).

Also, according to the $\Lambda$CDM model, the structures form in a "bottom-up" fashion, i.e. small structures forming first (stars and galaxies) followed by large structures (galaxy clusters). This is exactly what we are observing today thanks to surveys that are probing high redshift ranges.


Will I be able to see both Jupiter and Saturn at the same time in my Astromaster 114?

This can easily be tested using software such as Stellarium, where you can visualize the field of view with given focal length.
If you have the software installed, click on "ocular view" (the most left button in the upper right corner). The following view is what you culd expect on December 21 with a 10mm eyepiece and a focal length of 1000mm:

And this is the view when attaching a Crop-sensor (APS-C) Camera to a telescope of focal length 1000mm:

So you should be able to see both planets and some moons in your telescope.
Some notes:

These simulated views do not necessarily represent the actual view to 100% accuracy. I did leave the "atmosphere" setting turned on, but actual visibility and brightness will depend on your location (light pollution and other atmospheric effects such as haze might reduce visibility)

Because I was lazy, I used the default telescopes and eyepieces. This ignored aperture (or rather, it was different than on your telescope), but it won't have an effect on magnification. If you wanted, you could add your own telescope in the software's settings.


can we calculate the velocity of galaxies at the edge of the observable universe?

I think the main reason the answer doesn't come out correct is that the simple Hubble's law equation:

$v = dH_{0}$

Is only applicable in the local universe (Hence $H_{0}$, the value of the Hubble constant today). The rate of expansion of the Universe (the Hubble parameter) is determined by the composition of its energy density. Today, the composition is about 30% matter and 70% dark energy. In the recent past the energy density was dominated by matter. In the more distant past the energy density was dominated by relativistic particles (photons and neutrinos). The rate of expansion of the universe is different in each of these "epochs". In general the universe used to expand more rapidly (expansion has slowed thanks to the effects of matter), but distances also used to be smaller (the universe itself was smaller). Combine all of this together and you have a very complicated relationship between distance and the recessional velocity we observe today. This relationship does not have a simple functional form.

I think the easiest question to ask is: "How far has a photon traveled such that if it were emitted at the beginning of the universe it would reach me today?". The answer to this question is the "cosmological horizon", which we commonly talk about as being the size of the observable universe. To answer this question you need to use the spacetime metric which applies for our universe (Called the Friedmann-Robertson-Walker-Lemaitre metric). This is basically the relationship between speed, time, and distance in the realm of general relativity. If our universe is flat (which we think it must be) this equation is actually quite simple, but unfortunately it requires that you know how the universe has been expanding over its lifetime. As I mentioned this relationship has no closed form equation or number, which is why you will never see a simple equation for the solution you are looking for. You need to integrate the solution numerically. Wikipedia has a good little article on the horizon.

If you're interested, you'd write down the FLRW metric for a photon travelling in a "straight path":

$ cdt = \frac{a(t) dr}{ \sqrt{1-kr^2} } $

Where $a(t)$ is the scale factor of the universe, which tells you about its expansion history. This is the complicated function. $k$ is the curvature constant, which for a flat universe is 0. You then get:

$ \int^{t_{0}}_{0} \frac{c dt}{a(t)} = \int^{r_{H}}_{0} dr = r_{H} $

Here $t_{0}$ is the age of the universe and $r_{H}$ is the particle horizon you're interested in. To figure out $a(t)$ you need the Friedmann equations, which are functions of the different energy densities (dark energy, matter, relativistic particles, curvature), and the rates at which they change in time. I used astropy to do the integration and got $\approx 3.35 c t_{0}$ which gives about 14 Mpc, so close enough to the answer that you could attribute any differences to a difference in measured parameters that go into solving the integral.

I'd recommend Introduction to Cosmology by Ryden as a textbook if you really want to get into it.


Not sure what you are expecting to see from the two datasets. Both datasets are examples of light curves, flux against time with different arbitrary origins for the time axis; SuperWASP TMID is integer seconds from Julian Date 2453005.5, Kepler uses BKJD = Barycentric Kepler Julian Date, but offset by 2454833.0. i.e., BKJD = BJD - 2454833.0. The SuperWASP datasets typically span 4 years (it's a little difficult to read the plot scales of the example) and are from a ground-based set of small (11.1 cm diameter) lenses with a CCD and so will have day gaps and weather. Kepler/K2 data are from a space-based 95cm telescope which was able to stare at the same patch of sky continuously for about 90 days before a break (and of course had no weather). Consequently there will be a considerable difference in appearance and photometric precision of the lightcurves.

This page at the NASA Exoplanet Archive gives details on the SuperWASP public data and this Python Jupyter notebook has info on plotting K2 light curves.

(Disclaimer: I worked on the SuperWASP project for several years)


If a black hole has a mass of a universe what will be the volume of it?

According to measurements of the cosmic microwave background the universe is geometrically flat - which means that the mass/energy density of the universe is close to the "critical value" of $\sim 10^{-26}$ kg/m$^{3}$.

The radius of the observable universe is 46.6 billion light years, so the mass/energy contained within it is equivalent to $3.6\times 10^{54}$ kg.

The Schwarzschild radius of a black hole is $2GM/c^2$. If the mass/energy of the universe is spherically symmetric then its Schwarzschild radius is 560 billion light years and thus larger than the observable universe.

Note though that the Schwarzschild solution in General Relativity is static. The universe is definitely not static.


There are several compounds referred to as 'ices' in the solar system, these are mostly hydrogen-based compounds such as water, ammonia and methane (and others) accreted around what is believed to be a rocky core, these compounds are important, as they comprise a great proportion of the two 'Ice Giants' observed in our Solar System (Uranus and Neptune), as seen in the diagram below, compared to the Gas Giants - Jupiter and Saturn:

Image Source: Wikipedia

This answer will just mainly focus on the Ice Giants. as discussion of these answer both your questions regarding the ice planets and what is sometimes described as ices existing at high temperatures.

In terms of planets with proportionally large amounts of ice, as mentioned in the comments, there is a dividing line when the solar system formed known as the Frost Line, illustrated in the diagram below:

Image Source

Inside the Frost Line, the hydrogen compounds remain vapourised, beyond the line, hydrogen compound ices can condense and solidify, as seen for Uranus and Neptune (and an array of dwarf planets, moons and other celestial objects), these 'ices' comprise the mantle of both planets. An important distinction about what these "ices" are is explained in the paper "The Interior Structure, Composition, and Evolution of Giant Planets" (Fortney and Nettelmann, 2009) as being:

  The label ice refers to a mixture of $H_2O$, $CH_4$, and $NH_3$ that are supposed to have been in an ice phase during protoplanetary core formation.

The compounds in the mantle in the Ice Giants are under immense pressure (in excess of 10 GPa), far in excess of those existing on Earth. Greater pressures change the properties of the hydrogen-based ices - so it is one of the reasons that it is sometimes referred to as 'hot ice', according to the AstroTech Gallery page 'Ice Giant', describe a model of the Ice Giant planets as being:

  Much of this mass (perhaps about 65%) is made up of hot ices, which despite being very hot are probably
  kept at least partially solid by the enormous pressures within these massive planets.

The effect of pressure on water (as an example) can be seen in the example phase diagram below (of course there would be differences depending on the actual nature of the mixtures and compounds present):

Image Source: Duke University

However, according to the article The gas (and ice) giant Neptune, the present icy (or 'hot ice') mantle of Neptune is described as:

  The mantle is equivalent to 10 – 15 Earth masses and is rich in water, ammonia and methane. This mixture is referred to as icy even though it is a hot, dense fluid, and is sometimes called a "water-ammonia ocean".

A similar situation is hypothesised to exist for Uranus.

It is known that ice exist on Mercury, Earth, Moon and Mars, but these comprise a small proportion compared to the planet itself, so not included in this answer.


Close to a black hole, but still outside the event horizon, orbiting arbitrarily fast, wouldn't result in a circular orbit, but in a trajectory towards the black hole.
In this extreme scenario, centrifugal force could be interpreted as attractive.

In more usual settings, centrifugal forces point outward.
Since the centrifugal force (in usual settings) $m \omega^2r$ is proportional to square of the angular velocity $\omega$, you would need an imgaginary angular velocity, or alternatively a negative mass to get an inward pointing centrifugal force.

I'm not aware of a physical realization of imaginary angular velocities, at least not for circular orbits,
although this paper introduces imaginary velocities for hyperbolic orbits. This may be derived from imaginary circular angles.
Negative masses aren't entirely impossible, but those potentially feasible are too tiny to be of any practical value in your sense.


Ionization Fraction of Helium in alternate, He dominated early universe

My answer is incomplete (but still possibly helpful, only posting because I don't have enough "points" to comment).

I did a bit of the math involved but did not substitute terms with n's and eta's. However, from the result at the bottom of that page, it looks to me like a Binomial Expansion (truncated Taylor Expansion). The link shows the highest order for your terms (which is okay since higher order terms can usually be regarded as negligible) with the same +/- pattern you expect in your answer.


Is it possible to calculate a star's apparent magnitude in another photometric band knowing its visual magnitude?

No there isn't if you know nothing else about the star. Knowing the distance doesn't help. Knowing the mass and size of the star (how would you know these without other sources of information?) allows you to estimate what sort of a star it is and therefore you have some idea of the spectrum of that kind of star. Given that as a template, then it is possible to estimate what colour the star is an hence how bright it would be in any other photometric band.
e.g. If you have the temperature then you should find a table of temperature vs whatever colour you are interested in - so B-V if you want the B-magnitude. You interpolate on the table, find the B-V and add that to your known V to get B.
This assumes there is no reddening/extinction involved. If there is then your estimate will be wrong (to small for bands bluer than V, to large for bands redder than V).
A suitable interpolation table for main sequence stars can be found at
https://www.pas.rochester.edu/~emamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt


In galaxy optical spectra, why are Hbeta and Hdelta sometimes absorption lines?

In the absence of emission lines from ionized gas, the (optical) spectrum of a galaxy will be the sum of the spectra of individual stars. These spectra will have absorption lines for hydrogen (H-alpha, H-beta, H-gamma, H-delta, etc.); the absorption lines are strongest for A stars and weaker for both earlier (O, B) and later (F, G, K, M) stars.

If ionized gas is present (as is typical for spiral and irregular galaxies), then there will be hydrogen lines in emission. (This is normally due to O and B stars, which ionize gas with their strong UV emission; you've misread what the Wikipedia article actually says about A stars.) The emission is strongest for H-alpha, weaker for H-beta, and weaker still for the subsequent Balmer lines. This means that emission from H-alpha and (usually) H-beta overwhelms the stellar H-alpha and H-beta absorption, but H-gamma, H-delta, etc. can often still be seen in absorption because the emission from those lines is too weak to "fill in" the absorption lines.

Elliptical galaxies usually have only F or G and later stars present; they usually lack both the gas and the UV-emitting O and B stars necessary to ionize the gas, so you see only the stellar absorption lines.

The Ca II H and K lines can be seen in emission from individual M stars, due to emission from their chromospheres being strong enough to overwhelm the absorption in the photospheric spectrum (e.g., https://arxiv.org/abs/astro-ph/0602293).


When did people discover that the Earth is not an ideal sphere, but a little flattened?

It wasn't one person. René Descartes had theorized that planets must be in the shape of prolate spheroids. Christiaan Huygens and then Isaac Newton theorized that a rotating, self-gravitating fluid body would relax itself into an oblate spheroid. So which view was correct? Giovanni Cassini argued strongly that Descartes' view was the correct one, and he argued with measurements.

The French Academy of Sciences was split on this important question in the 1730s. The question was of interest in the world of commerce because knowing the shape of the Earth became of increasing import with regard to accurate ship navigation. To this end, the king of France financed two expeditions by the Academy, one to Peru and the other to Lapland.

The group that went to Lapland included Pierre Louis Maupertuis (the head of the Academy), Alexis Clairaut, Charles Camus, Anders Celsius, and others. This group left after but returned before the group that went to Peru. Their measurements agreed with the Newton's concept of an oblate spheroidal shape and disagreed with Descartes' concept of a prolate spheroidal shape. While their observations tilted the discussion in the favor of the Newtonian view, some of those who argued for a prolate spheroidal shape remained unconvinced. Jacques Cassini (Giovani's son) strongly disagreed, arguing that the measurements the group had made in Lapland were flawed. Cassini de Thury (Jacques' son) continued the family argument.

The group that went to Peru took ten years to return. (The voyage to Peru took over a year, all by itself.) The group was plagued by bad luck, lack of supplies, and inept and corrupt leadership. The leader of this expedition wanted no part in the diplomacy and management tasks required of an expedition leader. Even worse, he spent a good part of the expedition's funds on a hottie, and then bailed on the team entirely.

The reluctant second in charge, Pierre Bouguer, took over and made the expedition work. He and his team of over a dozen other scientists made astonishingly precise measurements of the distance between two Andean mountaintops. On returning, it was those measurements that convinced the remaining oblate spheroid skeptics that the Earth was indeed more or less an oblate spheroid.

  And if you know, who was the first person to discover it?

From the above, it should be obvious that it wasn't one person. It most certainly was not Descartes or the Cassini family. (But all four are remembered for their positive contributions to mathematics and science.) But it's very hard to say who was the first to discover that the Earth is an oblate spheroid (more or less) is given that so many people contributed to this discovery.

This, by the way, is how modern science works. Most modern scientific discoveries are team efforts. In the case of physics at the Large Hadron Collider, the list of authors on key scientific articles takes an entire page. It's been argued that Newton was the first modern scientist. It's also arguable that he wasn't. He worked alone. The discovery of the figure of the Earth was very much a collaborative effort that spanned over half of a century.

References:

W.W. Payne, "Attraction and figure of the Earth." Popular Astronomy 9 (1901): 117-123.
A rather old but very nice article on this old subject.

P. Lynch, "That’s Maths: Earth’s shape and spin won’t make you thin."
A lay article on the expedition to Lapland, which was led by Maupertuis.

D. Main, "The Remarkable Story of the First Accurate Measure of the Earth"
A lay article on the expedition to Peru, which was ultimately led by Bouguer.

The Wikipedia article on Clairaut's theorem
Clairaut was one of the key members of the expedition to Lapland. His equation that describes gravitational acceleration as a function of latitude was supplanted by a slightly better expression, the Somigliana equation, almost two hundred years later.

Wikipedia articles on Descartes, Huygens, Newton, the three Cassinis (three links), Maupertuis, Clairaut, and Bouguer.
Take these with the grain of salt one should always take when reading Wikipedia articles.


There is a beautiful composite picture demonstrating exactly this over at http://www.twanight.org/

This is a composite of 90 photos across the four days, so you can easily see how far the moon moves each day (about 13 degrees)

From https://cseligman.com/

  on average the Moon crosses the sky once every 24 hours and 49 minutes (53 minutes longer than the stellar "day"). As a result, it rises (and sets) later and later every day, until after about 27 days, when it has gone once around the sky relative to the stars, it is back in its original position, rising and setting at its original time(s).


Can stars be seen in the background when looking directly at our solar system?

Stars are much less bright than the earth in full sunlight as per your picture. So with the exposure required to give that picture of earth, no, you won't see stars.

But move the camera a bit to the side, so earth is not in shot and you can see stars.

The same goes for the sun - make sure sunlight is not entering the lens, and you can easily see stars.

Sunlight doesn't block any light out (unless you want to see a star that is actually behind the sun) it's just so bright that you need to make sure it isn't in shot.


The main reason we don't see the bright center of our galaxy, which is composed of millions of stars, is dust. Visible light is absorbed and scattered by interstellar dust, but that doesn't mean we can't see it on other waves of the spectrum, for example, infrared light doesn't suffer as much because of the dust.

Notice on this image how bright the galaxy center looks on the infrared and the near-infrared pictures!:

Image taken from the MultiwaveLength Milky Way website of NASA's Goddard Space Flight Center.


How far does the Earth travel around the Sun during each month of the year?

By using the Approximate Solar Coordinates formulas on the U.S. Naval Observatory website, you can calculate the ecliptic longitude of the Sun (L) and the distance between the Earth and Sun (R) for any date. Then, The distance travelled between two dates is approximately s=R*(180/pi)*(L2-L1), where R is the average distance during the interval, and L2 and L1 are the ecliptic longitudes. Since the Earth's orbit is almost circular, this approach is reasonably accurate even when performing the calculation over an entire month. These are the results that I get, where the distance s is in Astronomical Units (AU, the average distance between the Earth and Sun).

Month     Distance Travelled,  days ave. daily  ave. change  accl.
Jan 2019      0.542 AU         31   .00175 AU       0 miles 0 f/ss
Feb           0.488            28   .00174 AU   - 475 miles 0.000672
Mar           0.536            31   .00173 AU  - 1300 miles 0.001840
Apr           0.514            30   .00171 AU  - 1490 miles 0.002104
May           0.527            31   .00170 AU  - 1210 miles 0.001709
Jun           0.508            30   .00169 AU   - 651 miles 0.000921
Jul           0.525            31   .00169 AU       0 miles 0
Aug           0.526            31   .00170 AU     279 miles 0.000395
Sep           0.513            30   .00171 AU    1300 miles 0.001840
Oct           0.535            31   .00173 AU    1490 miles 0.002104
Nov           0.522            30   .00174 AU    1300 miles 0.001840
Dec           0.542            31   .00175 AU     744 miles 0.001053

Of course, the variation from month to month is heavily influenced by the number of days in each month, where as the variation from January to July is due to the distance from the Sun.

As a check, I calculated the distance travelled for each day in January 2019 and totaled the distance. That result was s=0.5419 AU compared to s=0.5422 AU when using the calculation for the first and end of the month.


Which will destroy the Earth first: the Sun or Jupiter?

I found an article by Ian O'Neill posted on May 2, 2008 at universetoday.com with the title Could Jupiter Wreck the Solar System? which says

But here’s the kicker: There is only a 1% chance that these gravitational instabilities of the inner Solar System are likely to cause any kind of chaos before the Sun turns into a Red Giant and swallows Mercury, Venus, Earth and Mars in 7 billion years time.

This is clearly after the time when the Sun is becoming a red giant in 4.5 – 5.5 billion years - this is the number given in an article Will Earth survive when the sun becomes a red giant? by
Matt Williams from 2016.
References
The first post is refering to a scientific article On the Dynamical Stability of the Solar System by Konstantin Batygin and Gregory Laughlin.


Since the moon isn't visible all the time from any single camera location on Earth, such a website would have to be an aggregate of many different cameras around the world, displaying a single camera based on current time and weather conditions. Even in this scenario, there's a chance all the cameras will be obscured by clouds and there will be no view of the moon.

Such a website doesn't exist that I know of. Perhaps it would be an interesting project if you're interested in such a thing.


Calibrating raw photometric measurements

If I understand the question correctly, you have raw data from a CCD sensor in arbitray units (which correlate with brightness) and your challenge is to callibrate this intensity for a certain frequency range.
Not knowing much about the SMEI sensor and where on the sky it has been looking at, I struggle to give you a full recipe. If you have for an object whose intensity (time series) from other observational instrument in the desired frequency band, you could use that non-SMEI data to correlate (if not even fit e.g. linearly) with the raw data time series for the same object as recorded by the SMEI sensor. This gives a calibration function which delivers intensity from raw data in arbitraty units.
References which sound possibly helpful

P. Hick, A. Buffington, B. V. Jackson: The SMEI real-time data pipeline: from raw CCD frames to photometrically accurate full-sky maps (2005)
A. Buffington et al.: Photometric Calibration for the Solar Mass Ejection Imager (SMEI)
Martin Clayton and Anthony Holloway: Simple Spectroscopy Reductions


Metallicity of Celestial Objects: Why "Metal = Non-metal"?

To first order, the relative abundances of the heavier elements to iron (for instance) are constant. So the metal content of a star is shorthand for the content of any element heavier than He. (NB: we now know this is not true in many circumstances and elements can be grouped by synthesis process - for example we can talk about "alpha elements" - O, Mg, Si etc. produced by alpha particle capture; or s-process elements - Ba, Sr etc. produced by the s-process. We know that the ratio of say O/Fe gets larger in more "metal-poor" stars, but Ba/Fe gets smaller. So talking about a single "metallicity" parameter only gets you so far, and the truth is more complex (and interesting).
The next point is why they are referred to as "metals" rather than another term like "heavies" or something. I would guess this is down to a bit of history and the fact that initial abundance analyses in stars were (and still are on the whole) done in the visible part of the spectrum (e.g. in the early part of the 19th century by Hyde Woolaston and Fraunhofer). The most abundant elements heavier than He are in fact not metals; they are Oxygen, Carbon, Nitrogen and Neon. However, the signatures of these elements are not at all obvious in the visible spectra of (most) stars. Whereas the signatures (absorption lines) of elements like Fe, Na, Mg, Ni etc., which decidedly are metals, are often very prominent.
Thus, there is a reason and some history behind the name "metals". It is that apart from hydrogen and helium, the metallic elements have the most prominent features in the optical spectra of stars, whereas in most stars the signatures of the more abundant non-metals are hard to see.


How is the distance of a star from Earth calculated in the HYG star database

The transformation to $x,y,z$ coordinates, starting from the RA, Dec and a distance (as given in the HYG catalogue) is a simple trigonometric exercise, since the $x,y,z$ (in this case) is referred to a coordinate system aligned with the equatorial coordinate system.

$$x = d \cos({\rm RA}) \cos({\rm DEC}),$$
$$y = d \sin({\rm RA}) \cos({\rm DEC}),$$
$$z = d \sin({\rm DEC}),$$
where $d$ is the distance. How that distance has been obtained is not clear from the README file accompanying the database. It may have been by (incorrectly) inverting Hipparcos parallaxes.

The dso.csv appears to be a catalogue of deep sky objects. There is no information in the catalogue itself which allows you to calculate or estimate a distance.

You could try cross-referencing the catalogue against the  NASA/IPAC Extragalactic Database (NED). https://ned.ipac.caltech.edu/
If the recession velocity there is bigger than about 1000 km/s, then you will get a rough distance by using Hubble's law
$$ d = v/H_0$$
 with $H_0 \sim 70$ km/s per Mpc (or whatever your preferred value is). If the recesson velocity is smaller than this then the object is too close for Hubble's law to be valid.

e.g. The first line inthe catalogue appears to be a galaxy known as IC 5370, that has a recession velocity of 10372.00 km/s (no error bar given and I doubt any more than 4 significant figures is warranted and in any case $H_0$ is not known accurately to more than 2 significant figures), which gives an approximate distance of 150 Mpc.


This is a very complex answer. It has to do a lot with gravity and electromagnetism. Nebulae are made of particles that act according to those forces and give you the illusion of viscosity. Those particles are actually moving. Take a look at this simulation it might give you an idea: Simulation


What you are asking is about the existence of binary star.

There are many binary stars in the universe. They are in fact more common  than single stars. Here are interesting explanations for the latest.


What you're really asking about is less to do with Astronomy and more to do with mathematics. You're basically asking if, given a system of differential equations, will a unique solution exist for all time? For an answer, you should check out the Existence and Uniqueness theorems of differential equations. You'd be better to ask questions like this on the Mathematics stack exchange.

However, to discuss the particular, astronomical case you've asked about, the answer is yes, you can run that system both forwards and backwards if you know some initial state. Newtonian mechanics is completely deterministic in that if you know all the equations of motion involved, as well as the entire state of the system at a given time, you can figure out the state of that system at any other time, both in the past and the future.

To speak to your particular 3-body orbiting problem though, I'll say that the system of equations is not solvable in closed form - that is, you can't write down an analytic solution to the equations like you could for the 2-body case. As userLTK states, you can write down approximate solutions in the restricted 3-body problem, where one mass is significantly less than the other two and orbits under specific conditions.

To get a solution at any time $\hat{t}$, you need to use numerical methods. Of course numerical methods are inherently flawed. Numerical errors build up the longer you simulate due to time steps which are not infinitesimal, from errors within the numerical algorithms used, and general floating point errors. In theory, if you had a computer with infinite precision and infinite computing power, you could solve a 3-body (or n-body) system perfectly, but we live in the real world where such things are impossible.

To prove though, that you can figure out the state at any time in the past or present for a three body body, I've written a basic simulation in Python 3. It can run both forwards and backwards from a given start condition and start time. Essentially it puts three nearly identical masses in contrived starting positions and velocities. Below the code are plots of results.

import numpy as np
from numpy.linalg import norm
from matplotlib.pyplot import *
from time import time

# Define physical constants
G = 6.67408E-11          # Gravitational Constant, m^3 kg^-1 s^-2

# Define body 1 parameters
m1 = 2.2E30                # Mass, kg
x1 = np.array([0,1E11])  # Position, m
v1 = np.array([-3.5E4,0])  # Velocity, m/s

# Define body 2 parameters
m2 = 1.9E30
x2 = np.array([1E11*np.cos(210*np.pi/180),
               1E11*np.sin(210*np.pi/180)])
v2 = np.array([3E4*np.cos(300*np.pi/180),
               3E4*np.sin(300*np.pi/180)])

# Define body 3 parameters
m3 = 2E30
x3 = np.array([1E11*np.cos(330*np.pi/180),
               1E11*np.sin(330*np.pi/180)])
v3 = np.array([3E4*np.cos(60*np.pi/180),
               3E4*np.sin(60*np.pi/180)])

# Define simulation parameters
n = 3                    # Number of bodies, unitless
t = 0                    # Simulation time, s
dt = 1E4                 # Simulation time step, s
tEnd = 1E8               # Simulation end time, s
m = np.array((m1,m2,m3)) # All masses
x = np.vstack((x1,x2,x3))# All positions
v = np.vstack((v1,v2,v3))# All velocities
xHist = [[list(x1)],[list(x2)],[list(x3)]]
vHist = [[list(v1)],[list(v2)],[list(v3)]]

# Simulate until end time is reached
start = time()
while True:
    # Calculate acceleration
    a = []
    for i in range(n):
        a.append(0)
        for j in range(n):
            if i == j: continue
            a[-1] += - G * m[j] / norm(x[i]-x[j])**3 * (x[i] - x[j])

    # Update velocities
    for i,vi,ai in zip(range(n),v,a):
        vi += ai * dt
        vHist[i].append(list(vi))

    # Update positions
    for i,xi,vi in zip(range(n),x,v):
        xi += vi * dt
        xHist[i].append(list(xi))

    # Update time and end simulation if past tEnd
    t += dt
    if dt > 0 and t > tEnd: break
    if dt < 0 and t < tEnd: break

end = time()
print('Simulation finished in {:.4f} seconds.'.format(end-start))

# Convert xHist and vHist to np arrays
xHist = np.array(xHist)
vHist = np.array(vHist)

# Plot everything up
figure()
for i,c in enumerate(['or','ob','oc']):
    # Plot starting positions
    plot(xHist[i,0,0], xHist[i,0,1], c)
for i,c in enumerate(['-r','-b','-c']):
    # Plot path of star
    plot(xHist[i,:,0], xHist[i,:,1], c)
gca().set_aspect(1)
gca().set_xticks([])
gca().set_yticks([])
show(block = False)

Note, the plots show the initial positions of the stars as the points and then trace out their paths over time.

Path of three masses for $t<t_0$

In this scenario, the three masses end up in the contrived scenario. I ran the simulation backwards by setting the timestep dt and the end time tEnd to be negative.

Path of three masses for $t_0<t$

From here, the simulation is run forwards with positive dt and tEnd, starting from the same contrived scenario as above.

Note how chaotic and unstable this system is. The entire system only remains a 3-body system for less than 6 years. Before that the three masses are separate and doing their own thing. They "coincidentally" meet (because I set it up so they should), orbit around each other for a little less than 6 years, and one gets ejected, resulting in the other two continuing to orbit one another.

Contrived Scenario with $m_1=m_2=m_3$, $|v_1|=|v_2|=|v_3|$, and all positions $120^o$ from each other

Just for fun, a "stable" 3-body problem with all stars of equal mass and orbits. This is really an unstable equilibrium orbit and any perturbations will screw it up and you'll see what you saw in the above two graphs. In fact, if you run this long enough, the numerical instabilities of my code will result in the orbits breaking down. This numerical instability, as I said above, is inherent in any numerical solution and cannot be overcome, only minimized. I find that, using the numerical method I did, my system is resistant to numerical instabilities for about 7 years. If I want to run this any longer (and continue to be accurate), I need more robust numerical methods.


How bright could Gliese 710 appear in 1.35 million years?

WolframAlpha says (to the query "apparent magnitude of star with absolute magnitude = 8.23 and distance = 1.1 ly") that the apparent magnitude of Gliese 170 would be about 0.87 on average. I took the value of the absolute magnitude (8.23) from Wikipedia. You can plug in different values for the distance in the formula and see what comes out. The formula used by WolframAlpha is

$$D_l = 10^{{m-M \over 5} +1} pc$$

where $D_l$ = luminosity distance, $m$ = apparent magnitude, $M$ = absolute magnitude, $pc$ = parsec.


Why aren't corrector plates aligned with the center of curvature in an SCT?

In a schmidt camera, the corrector is placed at the centre of curvature, which is at a distance of twice the focal length from the primary mirror, and minimizes aberrations such as coma and astigmatism.  This makes for relatively long telescopes, particularly with large primary mirrors.

The sct design is similar, in that it uses a corrector plate, but this (along with the secondary mirror) is placed slightly closer to the primary mirror than its focal length.  This design allows much more compact telescopes, but at the expense of more aberrations.  As with most if not all optical designs, it is a compromise.


Can the Sun become a big ball of gold atoms?

Gold is heavier (further down the periodic table of elements, not necessarily more dense, although this is also the case) than iron. Iron's elemental number is 26. Gold is 79.

Any star that attempts to fuse iron will go into a supernova phase.

At that point, gold and all of the elements heavier than iron will be created, but only in the brief moments while the star is tearing itself apart.

No star can ever create gold in any circumstances other than during a supernova.  Our own sun is not nearly large enough to go supernova.

Any supernova will create gold, not just the type of supernova described above.

More reading: https://en.wikipedia.org/wiki/Type_II_supernova


This is what Wikipedia says about it:

When the MKK classification scheme was first described in 1943, the only subtypes of class O used were O5 to O9.5. The MKK scheme was extended to O9.7 in 1971 and O4 in 1978, and new classification schemes that add types O2, O3, and O3.5 have subsequently been introduced.

It references the paper A New Spectral Classification System for the Earliest O Stars: Definition of Type O2

Thus, a new earliest spectral type has been defined.

So unless that information is outdated, there's no such thing as O1 or O0.


Ejected planets during the early stages of the formation of the Solar System

We can identify two distinct realms of planetesimal formation - the inner Solar System and the outer Solar System. The initial group of small bodies1 in the inner section of the protoplanetary disk was rather quickly accreted into planets via various interactions; some bodies grew bigger and thus dominated the disk around them. In the outer parts of the disk, many small bodies were accreted into the cores of the giant planets. However, not all were accreted, and there may in fact have been a second generation of planetesimals that formed via gravitational collapse due to disk instabilities.
We can compute timescales for this "cleanup" phase, the period during which small bodies were either accreted onto protoplanets or ejected to larger orbits or out of the Solar System entirely. In the inner section of the disk, among the terrestrial planets, it was on the order of $\sim10^8$ years, while in the outer section, it was on the order of $\sim10^9$ years. However, the giant planets may actually have finished accreting matter before the terrestrial planets.
This means that the small bodies that were ejected or boosted to larger orbits were largely from the outer Solar System, terrestrial planetary migration effects aside2. Various N-body simulations agree that a total of $\sim300M_{\oplus}$ of small bodies was ejected from the outer disk, due to interactions with Jupiter and the formation and evolution of Uranus and Neptune. We know that the Oort Cloud's mass is on the order of $\sim1$ to $\sim10M_{\oplus}$, and the Kuiper Belt and scattered disk have masses an order of magnitude or two lower. Therefore, only a small fraction - certainly less than 10% - of the small bodies could have been ejected into the Oort Cloud.
If you want to visualize some of this, check out this video from a simulation of the Nice Model3. It shows the orbits of the four giant planets (Jupiter in red, Saturn in orange, Uranus in purple, and Neptune in blue). It shows Neptune and Uranus swap orbits, and also shows the dispersal of a large group of small bodies outside the original orbit of Uranus (which began further out than Neptune). The video shows the simulation over the course of a bit more than one billion years, and you can see a drastic clearing out of the extreme outer disk over this timescale.
The above results came before the development of the Nice Model and its variants, as well as the possible discovery of Planet Nine and the discoveries of many small objects in the outer Solar System, beyond Neptune. They should therefore not be taken as totally accurate descriptions of the formation of the Solar System - the video in particular is likely incorrect - and I'm guessing I'll get some flak for using them as the basis of this answer.
Obviously, many N-body simulations of the Solar System have been done, and over the last decade, a substantial number use some variant of the Nice Model. Some use planetesimal disks - excluding the masses of the eight planets - on the order of $\sim10$ to $\sim10^2M_{\oplus}$; Nesvorny & Morbidelli (2012) (two of the architects of the model) found a sweet spot, so to speak, of $20$ to $50M_{\oplus}$, depending on the exact result. As I understand it, masses of the surplus disk around these values fit the Nice Model well, and so perhaps we should drop the total mass of the ejected small bodies by a factor or about ten.
Footnotes
1 I used this term quite a lot here; it refers to bodies with masses in the planetesimal regime and lower. The reason for this is that a substantial amount of the ejection came after the inner planets had formed, and the remaining bodies were small.
2 This should not be taken to mean that no or very little matter from the inner Solar System was ejected, especially if we assume some movement among the orbits of the inner planets (see Meech et al. (2016)).
3 Originally brought to my attention through a question on Physics Stack Exchange by Kyle Oman.
References

Goldreich et al. (2004) (early N-body simulations)
Krivova & Solanki et al. (2003) (early N-body simulations)
Meech et al. (2016)
Nesvorny & Morbidelli (2012) (more recent results)


Does Triton trigger the active atmosphere of Neptune?

Converting answer in comment to community wiki

Unlikely. Neptune is warmer than Uranus, more internal heat means more turbulent atmosphere. The reason why this is so isn't known, but it's possible that Uranus had a more turbulent past and it emitted heat from its formation more quickly. It's also possible that Neptune had a collision more recently that may have added to its internal heat. Triton is too small to have the effect you describe and the retrograde motion isn't a factor.


It's too dim to be seen during a normal survey during the majority of its
orbit.

Update:  Scientists at the University of Bern have modeled a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.

The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit.  They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun.

It should however be within reach of the LSST once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty.

My original attempt to handwave an estimate of detectability is below.
The paper gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion.

To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed.  That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a radar return is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $\sim 6x$ dimmer than Uranus despite being only $50\%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$.

Using that gives a dimming of 2400x at $210~\textrm{AU}\;.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude.  $500~\textrm{AU}$ gets us to $20$th magnitude, while a $2800~\textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the faintest stars visible from an 8 meter telescope; making its non-discovery much less surprising.

This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two.

Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations.

For reference, data from NASA's WISE experiment has ruled out a Saturn-sized body within $10,000~\textrm{AU}$ of the sun.

It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion.

Orbital eccentricity can be calculated as:

$$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$

Plugging in the numbers gives:

$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$

Plugging $200~\textrm{AU}$ and $e = 0.867$ into a cometary orbit calculator gives a $58,000$ year orbit.

While that gives an average proper motion of $ 22~\textrm{arc-seconds/year}\;,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum.

Kepler's laws tell us that the velocity at aphelion is given by:

$$v_a^2 = \frac{ 8.871 \times 10^8 }{ a }  \frac{ 1 - e }{ 1 + e }$$

where $v_a$ is the aphelion velocity in $\mathrm{m/s}\;,$ $a$ is the semi-major axis in $\mathrm{AU},$ and $e$ is orbital eccentricity.

$$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$

To calculate the proper motion we first need to convert the velocity into units of $\textrm{AU/year}:$

$$205 \mathrm{\frac{m}{s}}\;  \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$

To get proper motion from this, create a triangle with a hypotenuse of $2800~\textrm{AU}$ and a short side of $0.043~\textrm{AU}$ and then use trigonometry to get the narrow angle.

$$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ   = 3.17~\textrm{arc seconds}\;.$$

This is well within Hubble's angular resolution of $0.05~\textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $\sim 500~\textrm{AU},$ it would be bright enough to be seen by the ESA's GAIA spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly.

Its parallax movement would be much larger; however the challenge of actually seeing it in the first place would remain.


Was Milky Way formed after Sagittarius A Explosion?

This is ruled out because that's not what we see when we look back in the past.

If this was the case at the beginning, we would see these huge stars, with probably different characteristics from the ones formed only recently. But we don't see anything like that.

From what we see when we look very far away - that's few hundreds of millions of years after the beginning - we see signatures of first generation stars forming, very big stars indeed, but nothing close to what you're talking about.


Will the conditions that allow life to evolve still be around in a trillion years?

According to the 1997 paper "A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects" (arXiv version) by Fred C Adams and Gregory Laughlin, which was the basis for their book "The Five Ages of the Universe", the stelliferous era (the time when star formation is ongoing) is likely to last until ~1014 years after the Big Bang. Their estimated gas depletion timescale for the galaxy is 1012–1014 years. They do note that for the stars born after ~1012 years, metallicity effects are likely to result in significantly shorter stellar lifetimes.

So it seems likely that there will still be conditions to allow life to survive in a trillion years time, unless you happen to be using the long-scale trillion.


Is the universe simply a mass of atoms in 'void'; what does it look like?

It is not known. The universe has a finite age so we cannot see beyond the speed of light times that age (and taking the Hubble expansion into account as well). This means there is a 'visible universe' which defines a boundary of what we can observe. Information does not travel faster than light.

The whole universe is bigger than the visible universe, but we don't know how much bigger (it could even be infinite). However, as the universe contains all space ánd time, the stuff outside the universe does not contain both (try to wrap your head around that!)

As I mentioned it could be infinite, we could live on the three dimensional surface of a four dimensional ball (like an ant on a balloon) or our universe could be embedded in a higher dimensional multiverse (like string theory tells us).

We probably will never know because we cannot see past the visible universe.


Are there a lot of collisions between stars in the core of the galaxy?

The answer is yes. Stellar collisions are events that occur mainly in Globular clusters due to their high stellar density. The likelihood of these events is highly dependent on the stellar density. As we know the galactic centers concentrate large amounts of matter and, of course, thousads of stars which are mainly in the old red main sequence. However, recent observations hane shown as the existence of young stellar clusters located in the Galactic center of Milky in contrast with our models and predictions about the galactic structures. It is currently believed that in these clusters occur a lot of stellar collisions. For more information there is a related paper here.


You're talking about the construction of the 'Thirty Meter Telescope'.
Currently the construction has been halted by the State Supreme Court of Hawaii as a reaction to the violation of native sacred land.

Negotiations are going on about moving the construction site to the canary Islands, which has been deemed an acceptable alternative in the northern hemisphere, even if only barely.


It might be orion as in this season it is pretty clear in India. Orion have three stars in a somewhat straight line.

From in-the-sky.org's planetarium at 19:00 Indian time near the center of the subcontinent, Orion's Belt is high in the sky and extends from the East-Southeast vertically towards the zenith.


Why are there optical differences between sunrise and sunset?

You are not the only one whom sees a difference but remember that the statistics are skewed. Fewer people are awake early to see, or photograph, the sunrise while more people are awake to see the sunset.
People and heating of the Earth are factors explaining the difference in the atmosphere through which the sunlight is filtered, that affects the appearance of the light.
It's well explained at our Photo.SE site:

Most of the physical reasons (I can think of) probably do favor sunsets as well. During the day, the sun is heating the air and the ground. This leads to air movement that tends to pick up particulates (e.g., dust, smoke, etc.) and keep it in the air. It also leads to water evaporating (relatively) quickly from anything like lakes and rivers in the area. In addition, quite a few things that people do (e.g., driving) tends to happen primarily during the day, so it "pushes" more pollution into the air during the day. Thus, at the end of the day, the light is shining through air that carries more particles, which almost inevitably does more to diffuse the light.
During the night, there's a lot less pollution being put into the air, and the lack of heating means the air tends to be calmer as well, so by morning many of those particulates will have simply fallen out of the air, and the humidity either formed into clouds or precipitated. The clearer air does less to diffuse the light, so the sunrise isn't as spectacular.

A more scientific, but not complicated, explanation is offered at Wikipedia's "Sky" webpage. The heat affects the Rayleigh scattering and the Lorenz–Mie–Debye solution or Mie scattering, that explains the differences between sunrise and the sunset - a difference in the number and movement of particles in the sky.
A somewhat related factor is explained in a Q&A over at Space.SE: "Where does space begin on planets without atmospheres?". The height of the atmosphere and its layers along with the distribution of its component gases is altered by the heating of the planet and atmospheric circulation.


How could one use the transit of Mercury to try and find the Astronomical Unit?

Thanks to the Mercury transit, you can measure the parallax from the Earth. That happens due to TRACE , which tracks the transit of Mercury along the polar diameter of the Earth.

During that tracking, the transit of Mercury goes like that:

[

Now notice that, if TRACE remained stationary, the transit would be a straight line. So, if you calculate the maximum separation vertically of the center of Mercury and divide by two (how it's another different story), you got the parallax angle $\theta$:

You know $\theta$, you know Earth radius, so you know D, the distance from Earth to Mercury (in km) because

$$\tan\theta = R/D$$

Now, if you wanna know how many km a AU is, just realise that Mercury is 0.56 AU from us, so divide D/0.56 and you got it!

All images credited to NASA/IMAGE and/or Lockheed Martin/TRACE


A highly authoritative source is Explanatory Supplement to the Astronomical Almanac 3rd ed., Urban & Seidelmann, 2013.

Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that

local sidreal time = Greenwich sidereal time + east longitude

(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)

If accuracy of 0.1 second is sufficient, the formulas given by the US Naval Observatory should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.


Given a star's ra, dec how can I compute one pair of lat, long and time where the star would be at zenith?

Latitude is the easiest; it's the same as the star's declination: -13° 25' 2.64".

Longitude and time depend on each other. A star with right ascension 7h 23m 23.97s will be at the zenith point at 7h 23m 23.97s local sidereal time (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described here. On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.


How long does it take a star to transition from main sequence to red giant?

It takes a star like the Sun about a billion years to go from the end of core hydrogen burning to the beginning of helium core burning.  One might not call that entire phase a red giant phase, however, because the puffing out process takes quite a while to get going.  In the mean time, the star is what is called a "subgiant," rather than a red giant, and physically the difference is that in a full-blown red giant, the electrons in the core have lost so much heat that they are approaching close to their quantum mechanical ground state.  This implies the core is very small, about the size of the Earth, so a lot of heat must be lost before the star gets to that state.

After that, we can say the star is a red giant, but it won't be all that large until the core builds up more mass.  A ball of electrons near their ground state will shrink as more mass is added to it, and as the core shrinks, the electrons gain kinetic energy, and importantly, so do the helium ions.  When the core mass gets large enough, the helium gets enough kinetic energy to start fusion, and that's the end of the red giant phase.  The star ceases to expand and instead contracts to the "horizontal branch."  Most of the huge expansion occurs near to the end of that process, so the Sun will look like a red giant for less than perhaps 100 million years, depending on how big and red you want it to be in order to count.  A reasonable source is https://www.space.com/22471-red-giant-stars.html.


Would a 1000 km telescope be able to resolve microscopic detail?

Ignoring the effects of the atmosphere and the engineering issues you would need to overcome to build such a monstrosity, yes you could resolve details that well.  There is no physical reason you couldn't.

There are many optical interferometers, but their baselines are currently only tens to hundreds of meters wide. See this List of astronomical interferometers at visible and infrared wavelengths.

In fact, radio astronomers regularly achieve such spatial resolution by combining multiple far apart radio telescopes together to act as one giant telescope. This is a technique called very long baseline interferometry, and it was used recently to create the first image of a black hole by the Event Horizon Telescope using sites distributed across the Earth.


How did we first determine the Earth's departure from sphericity?

Isaac Newton was the first to put it together and realized the Earth had an oblate shape in 1687.    See 1672 - Wikipedia.

In 1666 Robert Hooke reasoned that pendulums could measure variation in gravity, but this remained unobserved until 1671, when Jean Richter observed that pendulums kept different time near the equator (French Guiana compared to Paris).

Around that time, 1673, Christian Huygens observed that the wide swing of a pendulum would move in a very slight oval, so to make pendulum clocks more accurate, the swings were kept to a low angle and the swing kept perpenducular   That, however, has to do with the Earth's rotation, but not your question specifically.

Newton was the first to put it all together that the observed gravitational fluctuation was due to the equatorial bulge and faster speed of rotation near the equator.    Article here also credits him with being the first.

Also, related:   https://en.wikipedia.org/wiki/History_of_geodesy


Lunar phases if Earth was tidally locked to the Moon

tl;dr: An observer on the "Moon side" would see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. They would also see the Moon during the day (early morning, late afternoon), as we currently can, but they would never see the New Moon or crescent moons at night since these would be only visible during the day.

Background
While the Earth being tidally locked to the Moon would indeed mean that the side of the Earth facing the Moon always sees the Moon, the phases of the Moon result from the angle between the Earth, the Moon and the Sun, and probably wouldn’t change significantly.

The Moon orbits around the Earth-Moon barycentre with an orbital period of 27.3 days and a synodic period (i.e. from Full Moon to Full Moon) of 29.5 days. An observer on the Moon-facing side of the Earth would certainly see the Moon go through its different phases, but it would still take 29.5 days. [For the purposes of this question I'm ignoring the increasing lunar period due to the Moon receding from the Earth.]

As the question notes, one obvious outcome of Earth becoming tidally locked to the Moon is that Earth’s “day” becomes the same as the Moon’s synodic period: 29.5 days. Daylight will last for about 15 days, baking everything other than at the poles. The 360-hour night will be freezing, even in the tropics. Local weather will be utterly different.

Another notable outcome is there will only be really tiny “solar” tides spaced about 15 days apart (due to the Earth’s slow rotation in relation to the Sun). The loss of significant tidal flows would probably have a dramatic negative effect on marine life.

The Earth tidally locked to the Moon is a nightmare scenario!

The core question

  So standing on the "lucky" side of the Earth you should be able to see
  all the lunar phases during a single night. The Moon become gradually
  more illuminated, peak as a full Moon and then darken as the night
  ends.

No: you would only see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. You would also see the Moon during the day (early morning, late afternoon), as we currently can, but you would never see the New Moon or crescent moons at night since these would be only visible during the day.

We currently see the New Moon because when it's roughly overhead for some at midday, it's not far above the horizon just before sunrise or just after sunset for others. Since the New Moon phase currently lasts about 3-5 days (depending on how you define "new"), but the Earth rotates in just one 24-hour day, most habitable places on Earth get to see this phase every lunar month. However, in the tidally-locked-Earth scenario, the phase still lasts 3-5 days but the Earth rotates once every 29.5 days. Those viewing the Moon just after sunset will only ever see the Moon in that position above the horizon – it more or less "sits" there, apart from libration and nodal period effects – and they will miss some of the other phases because the Sun will be too bright at, say, the waning half-moon.

EDIT - Two additional things to note:

(1) Earth's phases
While the Earth becoming tidally locked to the Moon is hypothetical, the Moon is already tidally locked to the Earth, which means the above descriptions can be applied in reverse for a real observer on the Moon:

a lunar "day" (sunrise to sunrise) is 29.5 Earth days;
the Earth's position in the lunar sky is relatively constant over the lunar day, within about 15º (due to a slow but noticeable libration);
the Moon's orbit is elliptical, so over much longer periods the Earth would alternately recede and grow larger (a "Super Earth");
unlike the Earth-locked scenario, all of the Earth's phases – from Full Earth at lunar midnight to New Earth at lunar noon - would be visible, because there's no atmosphere and hence no "bright daytime sky" to block out the Earthlight.

Note also that the famous Earthrise experience of the Apollo 8 astronauts would not be available to an observer on the Moon's surface. The Earth "rose" above the limb of the Moon because the astronauts were in orbit. Nonetheless, someone located on the border of the "dark side of the Moon" would see the Earth on the lunar horizon, and libration could result in the Earth alternately setting below the horizon and rising again over a lunar day. For that observer, however, earthrise and earthset occur at roughly the same point on the horizon!

(2) Will the Earth really end up tidally locked to the Moon?
Tidal braking is slowing the Earth's rotation by 23 milliseconds (and increasing the Moon's distance from us by 38 metres) per 1,000 years. At that rate, the Andromeda Galaxy will have collided with the Milky Way long before the Earth's rotation has slowed down enough to approach tidal locking.

In fact, this AstronomySE answer suggests that as the Earth's oceans "are responsible for almost all of the tidal deceleration" (David Hammen's comment), the evaporation of the oceans in a billion years will eliminate the main drag on the Earth's spin and tidal locking may therefore never occur.


How to project galaxy data in x y z coordinates?

You could try the Hyperleda database. You will have to learn to use the database, which possibly means learning a little bit of SQL. For starters you could select all items with redshift $cz < H_0\times 150$ km/s, using redshift as a distance indicator. This should be roughly ok at 150 Mpc (though I don't see any viable alternative).

The returned catalogue should have (at least) RA, Dec, redshift, possibly a size and morphology and, for a small subset of the galaxies, a redshift-independent distance estimate. Given that very nearby galaxies (in the local group or a $d < 50 $Mpc) are not fully part of the "Hubble flow" you might choose to use these in preference to a redshift-distance where available.

X,Y,Z coordinates (based in the equatorial system) can be calculated using this prescription or in Galactic coordinates (with respect to the centre and orientation of our Galaxy can then be found using the operations defined in Johnson & Soderblom (1987).

I have no idea how complete this database will be. Certainly there will be an issue with extinction near the Galactic plane.


How likely are planets to form after neutron star collisions?

There do appear to have been some studies on the properties of potential fallback discs formed after neutron star mergers, for example:

Rosswog (2007) "Fallback accretion in the aftermath of a compact binary merger
"
Cannizzo et al. (2011) "Fall-back Disks in Long and Short Gamma-Ray Bursts
"
Lyutikov (2013) "The Electromagnetic Model of Short GRBs, the Nature of Prompt Tails, Supernova-less Long GRBs, and Highly Efficient Episodic Accretion
"

These studies focus on explaining X-ray flaring in the aftermath of gamma-ray bursts rather than the potential to form exotic planets in these environments. It does seem fairly likely that some kind of disc does form around the remnant of a neutron star merger, but it's going to be extremely hot and likely so close to the remnant that it will be unable to form planets.
As noted in Menou et al. (2001) "Stability and Evolution of Supernova Fallback Disks", planet formation in fallback discs depends on the timescales for the disc cooling and how long it takes to spread beyond the Roche limit: if the disc becomes neutral before it spreads beyond the Roche limit, spreading becomes reliant on interactions within the remaining disc of rocks. While they consider the case of merging white dwarfs (noting that this scenario leads to a more favourable environment for planets than post-supernova fallback discs around black holes or neutron stars), they do not study the case of merging neutron stars.


If you browse further down the page, there are details for each of the top table data.

So RA and Dec give you coordinates (Right Ascension and Declination, like longitude and latitude). There is a designation J2000.0 which tells you the zero point to which the coordinates relate (remember, the Earth wobbles, so any coordinate system tied to Earth changes over time, hence a zero point), in this case the year 2000. The Equ. in front of that means equatorial, which ties is directly to the earth axis (poles & equator; the latter explains the name).

Redshift (z is the symbol for redshift) for a distance estimate.

Qual probably stands for qualification, for example the type of galaxy. It's likely not related to the filter. The type of galaxy would be the classification, which is further down the page.

Magnitude is the brightness, and Filter the filter it is measured in (most astronomy is done in a filter, to obtain a brightness in a specific wavelength range. Eg, red, blue, green, uv, infrared, H-alpha, Oxygen filters).

Separ arc min: the separation (or more likely, the diameter) in minutes of arc (so in 1/60 of a degree). You can't really measure the actual size (unless you trust the distance estimate), but you can measure the size as a angle projected on the sky. That's what it says here (e.g., the sun and the full moon are about 0.5 degree, or 30 arcminutes, "wide").

As mentioned at the start, browse down to get more in depth information, which may shed a bit more light on what the top table abbreviations actually mean and indicate.


Where was this photo of Nancy Grace Roman taken, what is the display shown?

Her Wikipedia page also includes this picture (public domain), where she wears the same dress, and where a similar display can be seen in the background:

Roman sits at the control console for the Orbiting Astronomical Observtory satellite, launched in 1972 and nicknamed Copernicus. This is a publicity picture; she never actually worked in the Goddard control room.

This page from the University of Chicago also includes a picture of OAO-3 (Copernicus) control room, where a similar display can be seen (I won't paste the image here because of copyright). However, after careful examination, they look similar, but the layout is actually not the same.
This page from the University of Wisconsin has a (poor) picture of OAO-2 (Stargazer) control room (again, not pasted here), where a similar display can be seen. Despite the poor quality, this one seems to fit the one where Nancy Grace Roman is featured.
From these evidence, I'd say:

It is definitely a display from an Orbiting Astronomical Observatory, either OAO-2 (Stargazer) or OAO-3 (Copernicus).
At least one of the three sites is wrong in its picture caption—I'd guess Wikipedia rather than the university sites, which would make OAO-2 the "winner".


What's the origin of Clyde Tombaugh's 9-inch Newtonian telescope?

In the first chapter of Out of the Darkness: The Planet Pluto, Tombaugh gives a colorful, personal account of his youth in Illinois and Kansas, including five telescopes:

his uncle's uncorrected 3-inch refractor
a 2.25-inch achromatic Sears-Roebuck refractor his father and uncle shared
a mediocre 8-inch reflector he made in 1926
a better 7-inch reflector he made for his uncle
an excellent 9-inch reflector he made in 1928

The last two benefited from optical testing in a storm cellar built extra long for that purpose.
Tombaugh's planetary sketches using the 9-inch reflector got him the job at Lowell Observatory.

The same photo appears on page 23 with this caption:

  Clyde W. Tombaugh, at the age of twenty-two, with his homemade 9-inch (22.9-centimeter) reflecting telescope on the family farm near Burdett, Kansas.
  This picture was taken in the summer of 1928, eighteen months before the discovery of Pluto.

APOD 1997-02-04 (© 1996 K. Beatty) shows him at age 89 with the same telescope in New Mexico.


What changes are being made to VIRGO and LIGO (if any)?

Currently, LIGO and Virgo are in what's known as a commissioning phase, a scheduled stage where the teams work on improving the sensitivity of the detectors. A big part of this is noise reduction, but there are other technical updates (an evolving plan also mentions increasing the power of the lasers and adding a "squeezed vacuum source"). Virgo in particular is replacing the steel fibers that hold up their mirrors with silica ones. The original silica fibers failed and were replaced with steel ones before last year's detection of GW170817, but now new silica ones are being added.

As of October, the Livingston site has reached low noise levels, while Hanford is still a bit away. Virgo isn't quite there yet. Soon, though, the three detectors should begin a couple of joint engineering runs, testing the hardware updates before O3 (the third, year-long, observing session) begins in February.


NOTE: Information on question and comments has been changing a lot, so please beware that some parts of the answer may be related to previous versions of the question.

Short answer:

This is a question to check if you know the sky. Once you can identify constellations and planets on the sky, the answer on the map is quite obvious.

Long answer:

To answer this question you need to be familiar with the northern hemisphere sky. If you don't know it, you would need a properly labelled map to answer the question.

Suggested steps:

Assuming that the map just represents the visible portion of sky and the horizon is printed in it, first step would be to look for Polaris. The point of the horizon which is closest to Polaris is the North. The South is in the opposite edge of the map. To locate East and West, place the map with South lower and then East is at left and West is at right - just opposite as they would be in a map of Earth, but this is a map of Sky.
I think it would be easier to identify constellations before planets. The easier constellations are likely to be the circumpolar ones (those that we can always see in the northern hemisphere).
To find the planets, you should find the Ecliptic by looking at zodiacal constellations and then check for points that don't belong to those constellations. Of course, to do that you need to know the constellations. A helpful piece of information is that at 00:47 local time, the Sun, Mercury and Venus can't be seen. If the map shows the planets visible with naked eye, Mars, Jupiter and Saturn may be there in addition to the Moon.

Edit: addition about 1st step (finding north)

Your map is drawn inside of a circle. That circle represents the horizon, and Polaris is not in the centre of the map. Therefore, there is a point in the horizon that is the closest point to Polaris. That point is the north.

All this just comes from the fact that Polaris is always in the north. If you look at Polaris in the sky, you are facing north - just as if you look at the Sun at noon, you are facing south.

To better understand this, I suggest taking a sky map and a compass, going outside and trying to find Polaris tonight in the actual sky - tonight or the next available starry night.

Edit: finding Polaris

To find a constellation or a given star, you need to know the constellations or compare your map of the actual sky with a labelled star map. The question is not very different to (for example) "find Ireland in a blank world map": either you know the world and where Ireland is, or you have a labelled map to compare. Star maps aren't that different from geographical maps. In Earth maps you know where you are by looking at geographical features, and in sky you know where you are by looking at stars positions. To find Polaris, look for Ursa Minor; finding Ursa Major or Casiopea may help, too.

Addition after the map was included in the question:

North is in the right side of the map. Polaris is the dot halfway from the center and the right side.

The easiest clue to find Polaris is the conspicuous Ursa Major in the center of the map. As others have suggested, two stars of the Ursa Major point to Polaris. Furthermore, some stars from Ursa Minor are easy to spot over Polaris.


According to Herczeg & Hillenbrand (2014) the intrinsic bolometric luminosity of T Tau is about $7L_\odot$ (for an assumed distance of 479 light years).
The main difference between their estimate and yours is that they have: (i) Taken account of extinction towards the object, which considerably dims its visible flux by about 1.25 magnitudes (a factor of 3.2). (ii) Taken account of the spectral energy distribution, which differs from the Sun.
You cannot in general calculate a ratio of luminosities from a difference in visual magnitudes, unless the spectral energy distributions of the two sources are the same. T Tau has a greater proportion of its flux emerging at near infrared wavelengths than the Sun.
T Tau is in a multiple system with a much lower mass companion (which contributes little to the luminosity), but this has led to a quite accurate mass determination of $(2.12 \pm 0.10)\ M_\odot$ (Kohler et al. 2015).
Given a main-sequence mass-luminosity trend of around $L \sim M^3$, then it is apparent that T Tau has around the luminosity it would be expected to have on the main sequence.
This would suggest that it has finished descending its Hayashi track in the HR diagram and is currently traversing, roughly horizontally, towards hotter temperatures and slightly rising luminosity, in excellent accordance with stellar evolutionary models (see below) if its age is $\sim 2$ Myr.
T Tau is only unsusual compared with most "T-Tauri stars" in that it has quite a high mass. Most "T-Tauri stars" are a solar mass or lower, and if found in young stellar clusters and associations with ages $<10$ Myr, you can see from the plot below that they would have higher luminosities than they end up with on the main sequence.

Pre main sequence evolutionary tracks and isochrones from Steven W. Stahler and Francesco Palla, "The Formation of Stars"; doi:10.1002/9783527618675


Life planets orbiting black-holes. Can/Do they really exist?

Well, first things first.   It's not likely to have a planet orbiting near a black hole and in significant time dilation because the tidal effects would likely tear anything that close apart.   Certainly a planet orbiting a stellar mass black hole would need to be quite far away so as to not be torn apart, so any time dilation would be pretty small.

Around a super-massive black hole, the tidal effects are smaller and a nearish orbit with some measure of time dilation is possible.  (see link below for more specifics),

https://physics.stackexchange.com/questions/110044/time-dilation-factor-for-the-circular-orbit-at-3-2-schwarzschild-radius

But a stable planetary orbit, you probably max out at about 20% time dilation and only around a super-massive black hole, where there's only 1 per galaxy.    The idea of 1 day to 100 years isn't practical, it's 80 days to 100 days if you're talking about a stable planetary orbit.

and, I'm not sure you'd want to be that close to the black hole in the center of the galaxy, not because the orbit isn't stable, but because stars are in that orbit too.  It might not be a safe place to be.

So, in reality, you'd want a stellar mass black hole and a distant orbit, where the time dilation would be quite small, and in that scenario, yes, life is possible because of tidal energy, so a planet could have liquid surface water and an atmosphere, even if the black hole gave off very little light and heat.

Such a planet in a tidal-energy orbit would probably be tidally locked which would protect the far side from any gamma rays the black hole spits out when it eats, so it would in theory be a good place for life.   No significant light source, unless it was a binary system, so, plants would have it harder, but there would be heat.

There's another problem.   The creation of black holes tends to blow everything appart in a huge explosion.  It's unclear that a planet would survive a black hole's birth, so you might need a captured planet.

Finally, intelligent life . . . we really don't know enough about how common intelligent life is on other planets.   Life might be fairly enough, but intelligent life is far less clear and there's other factors than just time.

Today, we simply don't know enough to predict if there's intelligent life out there or not.   There probably is life elsewhere in the universe, though even that isn't 100% certain, but regarding intelligent life, there's far too many unknowns in that equation.   I think a black hole might not be optimal for the formation intelligent life because of the lack of light so, much less photosynthesis, so, slower oxygen formation (if it follows the same pattern the earth does) and the unlikeliness that a planet would survive the black hole's creation.


How was calculating the celestial latitude using an astrolable of any help to sailors?

Measuring the angle to a known star allows the navigator to calculate the latitude of the ship in the absence of any visible (known) landmarks.

Since they would know (or refer to an almanac) the celestial latitude (declination) or altitude of the star at their home port or some other known port, they can measure the position of the star and take the difference between the measurement and the reference to calculate their latitude at sea.

For example, if their home port were located at 54°N then the angle from the horizon to Polaris would be 54°. If they take a measurement of Polaris while at sea and find it to be 30° above the horizon then they know that they are 24° farther south, i.e. latitude 30°.

Similarly, if there were a star that is directly overhead (90° from the horizon) when it transits the celestial meridian at home, then they know that that star has a declination of 54°. If they take a measurement of it while at sea and find it to be 6° north of zenith (84° from the horizon) then they know that they are 6° south of home.

From all of this, knowing the date lets them calculate the sunrise and sunset times, and if they have an accurate clock they can work out their longitude. In the past, longitude was a bit more guesswork as clocks were not all that accurate at sea.


Why do massive stars not undergo a helium flash

After H burning has finished, the he mass of the He core gradually increases, as does its density and temperature.
Low-mass stars have denser cores when they reach a temperature at which He is ignited. The density is high enough that the electrons in the core are degenerate. In such conditions the heat from the nuclear reactions goes almost exclusively into raising the temperature of the He ions, but almost none goes to the degenerate electrons which have a very low heat capacity, yet dominate the pressure. This leads to a runaway increase in the nuclear fusion rate.
In higher mass stars the core is less dense when it reaches the He ignition temperature. This temperature is almost exactly the same as the ignition temperature for lower mass stars because of the very strong temperature dependence of the triple alpha reaction. The level of degeneracy depends on the ratio of density to temperature (it does not directly depend on the pressure).
At the lower densities in the higher mass stars, the electrons are not degenerate when He ignites and the heat capacity of the electron gas is higher than the ions (because there are more electrons than ions). In this case the heat from the nuclear reactions can effectively raise the pressure, doing work which expands the core and regulates the nuclear reaction rate.
See also https://physics.stackexchange.com/questions/174801/why-is-the-release-of-energy-during-the-he-flash-in-stars-almost-explosive


Short Answer:
I can't make any sense out of what Planetary Geology says.
Long Answer:
I can't make any sense out of what Planetary Geology says.  But here is some data about the changing number of planets believed to exist in our solar system during the 19th century.
Here is a link to the Wikipedia article Planet, specifically to the section "Objects Formerly Considered to be Planets".
https://en.wikipedia.org/wiki/Planet#Objects_formerly_considered_planets 1
It has a long list of 27 objects considered to be planets at one time but later reclassified as the sun, or natural satellites (moons), or asteroids. By 1854 all of those objects - including Ceres - had been reclassified as not being planets. So after 1854 or so the number of accepted planets was 8 until 1930, 9 from 1930 to 2006, and 8 from 2006 to the present.
I note that if someone numbered the objects still considered to be planets in order of their discovery, the list of planets would included 6 planets including Earth up to 1781 when Uranus was discovered, and then 8 planets when Ceres was discovered in 1801.  Then 9 planets when 2 Pallas was discovered in 1802, 10 planets when 3 Juno was discovered in 1804, and 11 planets when 4 Vesta was discovered in 1807.  When Neptune was discovered in 1846 it could be classed as the 8th planet by those who discounted the asteroids or as the 12th planet by those who counted Ceres, Pallas, Juno, & Vesta.
If the planets are numbered in order of increasing size, Ceres would have been number 1 when it was discovered in 1801, become 2 in 1802, and eventually become number 9 when 9 Metis was discovered in 1848, assuming that astronomers of that era could tell that Ceres was larger than other asteroids.  Ceres would become number 10 when 10 Hygiea was discovered in 1849, 11 when 11 Parthenope was discovered in 1850, and eventually number 15 when 15 Eunomia was discovered in 1851.
If the planets are numbered in order of decreasing size, Ceres would have become number 8 when it was discovered in 1801 and would have remained so until Neptune was discovered in 1846, making Ceres number 9 until being reclassified as an asteroid in the 1850s.  Unless one of the other largest asteroids was ever believed to be larger than Ceres.
If planets were numbered inwards from the outermost known one, the list in 1801 would have been 1 Uranus, 2 Saturn, 3  Jupiter, 4 Ceres.  in 1802 2 Pallas would become 4 and Ceres 5.
In 1846 Neptune would be discovered and Ceres would become # 6.  In 1849 10 Hygiea was discovered and Ceres would become # 7.  In 1852 16 Psyche was discovered making Ceres # 8. In 1852 22 Kalliope was discovered making Ceres # 9. In 1853 24 Themis was discovered making Ceres # 10.   In 1854 28 Bellona was discovered making Ceres number 11. And in 1854 31 Euphrosyne was discovered, making Ceres number 12.
But I should note that none of the asteroids after 15 Eunomia, discovered in 1851, were ever considered to be planets.
If planets were numbered outwards from the sun, as is normal, Mercury would be 1, Venus 2, Earth 3, Mars 4, and 1 Ceres would become 5 in 1801.  Ceres would become 6 in 1804 when 3 Juno was discovered, 7 in 1807 when 4 Vesta was discovered, 8 in 1845 when 5 Astrea was discovered, 9 in 1847 when 6 Hebe was discovered, 10 in 1847 when 7 Iris was discovered, 11 in 1847 when 8 Flora was discovered, 12 in when 9 Metis was discovered, 13 in 1850 when 11 Parthenope was discovered, 14 in 1850 when 12 Victoria was discovered, 15 in 1850 when 13 Egeria was discovered, 16 in 1851 when 14 Irene was discovered, and 17 in 1851 when 15 Eunomia was discovered.
So if planets and "planets" were numbered in order of increasing size, Ceres would be ninth between the discovery of 9 Metis on 25 April 1848 and the discovery of 10 Hygiea on 12 April 1849, and the tenth until the discovery of 11 Parthenope on 11 May 1850.
So if planets and "planets" were numbered in order of decreasing size, Ceres would be ninth between the discovery of Neptune on 24 September 1846 and being reclassified as an asteroid in the 1850s,and would never have been considered the tenth planet in size.  If Ceres was still considered to be a planet, it would have been considered the tenth largest planet after Pluto was discovered in 1830, but Ceres was no longer considered a planet by then.
So if planets are numbered inwards from the outermost known planet, Ceres would have been considered the ninth planet between the discovery of 22 Kallipe on November 16, 1852 and the discovery of 24 Themis on April 5 1853, and the tenth planet between April 5, 1853 and the discovery of 28 Bellona on March 1, 1854.
Except that as far as I know no asteroids after 15 Eunomia discovered in 1851 were ever listed as planets.
So if planets are numbered outwards from the innermost known planet, Ceres would have been considered the ninth planet between July 1, 1847 when 6 Hebe was discovered and August 13 1847 when 7 Iris was discovered, and the tenth planet between August 13 1847 and October 18 1847 when 8 flora was discovered.
After Urban Le Verrier used discrepancies in the orbit of Uranus to predict the existance of a planet beyond Uranus, and Neptune was discovered in 1846, Le Verrier studied the orbits of other planets.  In 1859 Le Verrier deduced that there should be a planet orbiting inside the orbit of Mercury, which he named Vulcan.
Observations of objects believed to be Vulcan were often made but never confirmed.  In 1915 Einstein's general theory of relativity provided an alternate theory of gravity which explains the motions of mercury without the planet Vulcan.
When Vulcan was was believed in it was probably considered to be larger than Ceres, which would have made Ceres the tenth largest planet, except that Ceres was no longer considered a planet.
All things considered, I believe that the only possibly ninth planet that some astronomers might have believed to exist in 1899 was Vulcan.  So if Ceres was discovered in 1899 it might have been considered the tenth planet to be discovered.  Except that Ceres was discovered in 1801, 98 years before 1899.
And by 1899 some astronomers had already attempted to calculate the orbit of a planet believed to be beyond Neptune, a planet X, "X" for unknown, a search which eventually led to the more or less accidental discovery of Pluto.
But as far as I know nobody announced any discovery of a new planet in 1899 which turned out to be false but would have been the tenth planet to be discovered if both it and either Vulcan or Planet X was real.
So I have no idea what Planetary Geology meant by stating:

When Ceres [...] was discovered in 1899, it was considered as the lost tenth planet.


What other galaxies can be seen with the naked eye in the night sky?

We can see the Magellenic clouds easily, and we can see Andromeda on very dark, clear nights, but only when we are far from the city.

Some people claim to have seen the Triangulum galaxy (the third largest member of the local group, after Andromeda and the Milky way). Exceptional conditions, and very well trained eyes would be required to see it.

No other galaxies are visible, even under the best conditions.


Eclipse comparison: May 20 2012 vs Aug 21 2017 as observed from San Francisco, CA

This is going to be the eclipse of the century for the USA. To make the most of it, you need to be on the path of totality (see map) to see a total eclipse. If you stay in San Francisco, you will only see a partial eclipse. You need to make arrangements now if you need somewhere to stay.

The gray band indicates the path of totality. Inside it, you will see a total eclipse, and will be able to see the Sun's corona in all its glory. If you are outside the path of totality, you will only see a partial eclipse (and no corona). The closer you are to the blue centerline, the longer totality will last. The black dot near Nashville indicates the location where totality will last the longest (2 minutes and 40 seconds).

During totality (but only totality), it is safe to observe the Sun's corona with the naked eye. At all other times, you will need to use a protective (solar) filter if you want to look at the Sun directly, or project an image of the Sun onto a screen.

www.eclipse2017.org


What fraction of galaxies in the observable universe have we actually observed?

There are around 2 trillion galaxies in the currently observable universe according to the latest estimates, obtained by integrating theoretical galaxy stellar mass functions above $10^{6} M_{\odot}$ between $0 \leqslant z \leqslant 8$.

It's difficult to get a precise number for the total observed galaxies as the results from new surveys are being released all the time, and updated analysis of legacy data is revealing more galaxies. For example, the latest SDSS data release identified close to 200 million galaxies, whilst the ongoing Dark Energy Survey seeks to identify around 300 million galaxies. LSST, in Chile, will observe close to 20 billion galaxies, an order of magnitude greater than anything before it. However, all of these surveys operate up to relatively low redshifts. Probing to higher redshifts requires bigger, preferably space based telescopes, and much longer exposure times. As such, the high redshift universe has been poorly documented. Our understanding of galaxy formation at high redshift is not as complete as at low redshift, and so we may be under- or over-estimating the number of galaxies at this epoch.

A vague upper limit: we've observed less than 0.01% of all currently observable galaxies.