Why do the upper-atmosphere clouds of Venus appear to have that V shape?

This is supplementary to antispinward's excellent answer and provides additional sources and a visualization from the JAXA spacecraft Atasuki. It has been shamelessly borrowed from Would it be possible to "ride the wave" on Venus?

The recently published paper in Nature Geoscience Atmospheric mountain wave generation on Venus and its influence on the solid planet’s rotation rate has open-access links in Science News, Motherboard, and Science.

I haven't found a preprint in arXiv yet.

There is a stationary gravity wave in Venus's very dense atmosphere. It is intermittent, but has been detected several times. See the NYTimes article Venus Smiled, With a Mysterious Wave Across Its Atmosphere for a discussion of recent observations by JAXA. See Akatsuki and Happy Birthday, Akatsuki!, celebrating it's first Venusian year at Venus.

above: "A sequence of images showing the stationary nature of the bow-shape wave above Venus when it was observed in December 2015. Planet-C" from NYTimes. credit: Planet-C/JAXA

above: "An illustration of how gravity waves travel up mountains and into Venus’s atmosphere. Credit ESA" From How Mountains Obscured by Venus’s Clouds Reveal Themselves.


As the commenter states, $e$ is indeed called the orbital eccentricity. If you add a radial scale length (e.g., semi-major axis) to both of those values the $1-e$ describes the periapsis (closest approach of orbit) and the $1+e$ apoapsis (furthest) of an elliptical orbit. They don't have a specific special name, as they are dimensionless measures, but can be quite useful in determining the orbit of planets and other Keplerian systems.

Periapsis:
$$
r_{p}=a(1-e)
$$
Apoapsis:
$$
r_{a}=a(1+e)
$$

Perhaps the could be named the maximum radial eccentricity and minimum radial eccentricity, if you needed to describe these parameters in a report or class homework etc.


Slow moving star like objects in the night sky

Unlike meteors, which burn out in a second or two, satellites can be visible for a few minutes at a time.
In twilight they may be visible all the way across the sky; at darker times of night the Earth's shadow cuts their visibility short unless they're in high orbits.

One good resource for such things is Heavens Above.
From Hereford, UK, their Daily predictions for brighter satellites list a few matching your description on the morning of 2019-08-07.
For example, the polar-orbiting Russian intelligence satellite Cosmos 1975 passed from WSW to N between 02:00 and 02:04.


Enceladus; why use the words "geysers", "jets", and "plumes" interchangeably?

Question:
I would know why the words "geysers" and "jets" would be used interchangeably. Is this because it hasn't been confirmed yet what they are, and the authors don't wish to show favoritism by choosing one or the other, or are they synonymous and can always be used interchangeably, or is there some other reason ?

A jet is a generic term to describe the ejection of material.  You'll see it used all over the place in physics (and astronomy).  Jet's can be any form of matter ( solid, liquid, gas, plasma ).
A geyser is a specific type of geological process and, as I understand it, means (at the least) a liquid is being vented (typically water).  The exact process is, of course, not known for certain for non-terrestrial bodies.  I don't think using the term "geyser" is a statement of certainty, but would be used to indicate the general type of event.
I think you're more likely to use "jets" if it's not properly understood and "geysers" when you're either proposing a mechanism that's geyser-like or it's generally accepted that they're geyser-like already.  But I don't think that's written in stone.


Shooting stars are indeed meteors falling down, but consider this: meteors which are part of a meteor shower actually appear to come from a point in the sky called the radiant point:
google image

Since you're looking upwards, you may be looking straight at the radiant point, and therefore the meteors are 'radiating' away from that point in all directions; not only 'downwards' towards the horizon, but 'upwards' from your point of view.

You can see that, even for a meteor which is not part of a shower, it's just as likely for a meteor to be going 'upwards' across the sky, as it is to be going 'downwards'.


Is there any ephemeris generator that gives you accurate distances between Earth and other bodies?

JPL HORIZONS gives the distance from the observer to the target as delta.
In Table Settings this option is 20. Observer range & range-rate.
For asteroids and comets, it can also give distance uncertainty (39. Range & range-rate 3-sigmas) as RNG_3sigma.
You can choose units of au or km.
For example, with Table Settings: QUANTITIES=1,20,39; range units=KM, the geocentric ephemeris for asteroid (153201) 2000 WO107 around its 2020-11-29 close approach to Earth looks like:
 Date__(UT)__HR:MN     R.A._____(ICRF)_____DEC            delta      deldot RNG_3sigma RNGRT_3sig

 2020-Nov-29 03:00     08 48 22.83 +14 26 33.8 4.3065970451E+06  -1.1137937   143.5915  0.0000166
 2020-Nov-29 04:00     08 43 44.11 +14 51 56.9 4.3035317776E+06  -0.5890828   144.6349  0.0000171
 2020-Nov-29 05:00     08 39 04.01 +15 17 00.3 4.3023570178E+06  -0.0635364   145.6159  0.0000175
 2020-Nov-29 06:00     08 34 22.78 +15 41 42.1 4.3030745286E+06   0.4621549   146.5331  0.0000180
 2020-Nov-29 07:00     08 29 40.62 +16 06 00.2 4.3056835832E+06   0.9872992   147.3852  0.0000185

In other words, at 05:00 UT, the asteroid is 4,302,357 ±49 km from the center of Earth.
Though its orbit is determined well enough for a permanent number designation, even 1 km precision would be unreliable.
Its perihelion distance q is only known to ±8.7 km and perihelion time tp to ±1.7 s.
New observations may improve this.
Update: new observations of (153201) 2000 WO107 improve its q to ±6.9 km, tp to ±0.3 s, and distance at 2020-11-29 05:00 UT to 4,302,522.2 ±1.1 km.
That precision is largely due to radar data, which are not always available.


Taking Mars' average distance from the sun of 1.52 AU, the sun would be 43% as bright.

Phobos has an albedo of about 0.071 which is pretty dark.  Darker than the Moon with an average albedo of about 0.12, so it reflects about 59% as well as the Moon does.   That gives it a brightness to area of just over 25% the brightness of the moon.

Phobos is an odd shape, about 27x22x18 km.   Averaging that out to 22, then Phobos is 1/158 the diameter of the Moon, but it's closer to Mars' surface.  about 5,986 km, that's about 63.5 times closer than the Moon is to the surface of the Earth.

Working out the numbers, the size of phobos in the Martian sky would be on average 63.5/158 or 40% the diameter, or 16% the size of Earth's Moon from Earth.    Multiply that by 25% the brightness to area and that's just 4% the brightness of a full moon.  That would be peak brightness  (and if you want to fuss about Mars' Perihelion and say 5% the brightness of a full moon, fair enough, but lets not get ahead of ourselves).

Brightness of our Moon varies depending how good a reflector it's surface is, and how much of the surface is in shadow.  The Moon is particularly cratered so a half moon is significantly less than half as bright as a full moon.  Phobos is so close to Mars that from the surface of Phobos, Mars takes up about 40 degrees of arc in the sky.  That casts a huge shadow, so there's no such thing, from Mars, as a full Phobos.  As Phobos approaches full, it enters Mars' shadow and it's brightness drops significantly  (whether it turns red like a blood moon or just vanishes - I'm not sure, perhaps that's best for another question).   But Phobos never becomes full, it gets a bit over 3/4s, then it enters Mars' shadow.

"Enough to see" is a tough question, because sight depends on not just the light source but the reflectivity of your surroundings and there's some variation in how well individuals can see in the dark and, obviously, on how clear the night sky is on Mars on that particular night.  I think 3%-4% of the brightness of a 3/4 moon would be pushing the limit though and probably not bright enough to see for the most people.    BUT reading the comments, and reading the very cool Venus shadow article in the questions comments, my assumption might be wrong.   Limited sight might/(should?) be possible with some recognition of shadows and shapes.


Is there a difference between the terms 'elliptical' and 'elongated' for galaxies?

tl;dr: "Elliptical" refers to a special kind of galaxies. "Elongated" is a loose term meaning that a galaxy is stretched out in one direction.

Etymologically, "elliptical" means having a shape that can be described by a certain mathematical function — the ellipse. For a three-dimensional object, the proper term is "ellipsoidal", but still it refers to being described by a certain equation. In contrast, "elongated" is a more loose term, meaning something like "significantly deviating from a spherical shape in one direction".

In the context of galaxies, the term "elliptical" refers to a certain type of galaxies that tends to be not actively star-forming, have no or very little structure (as opposed to the beautiful spiral arms of spiral galaxies), be quite massive, as well as other characteristica. Having no ongoing star formation, the massive, short-lived stars, which have blue colors, have died long ago, leaving behind the less massive, red stars. In addition, elliptical are often quite dusty, further reddening the light. Thus, ellipticals appear red in color.

The shape of an elliptical galaxy is described by the relative size of its three axes, $a$, $b$, and $c$. If two axes, say $a$ and $b$, are roughly the same size and are larger than $c$, the galaxy is said to be "oblate", whereas if they are less than $c$, it is said to be "prolate".

As in the etymological meaning, an elongated galaxy is not a specific type, but rather refers to a galaxy departing from spherical shape in one direction. This could for instance be a prolate elliptical, but it could also by any other galaxy that has been distorted by merging with another galaxy.

Usually, though (I think), it will refer to ellipticals. The "elongation" of an elliptical is defined as $10\times(1 - b/a)$, where $a$ and $b$ is now the observed axes, i.e. of the two-dimensional projection on the sky. Ellipticals are classified according to this number as E0 (being spherical, such that $a=b$) to E7 (being very elongated). In principle, you could have even higher E-numbers, but that's not observed.


The other question has some very good detailed answers, worth reading, but briefly and more specific to your question:

  Would there not be interactions between their Oort cloud objects and
  the planets of the Solar system? For example, the outer edge of the
  Sun's Oort cloud is approximately 100,000 AU away. So, if another star
  comes at a distance of 1 light year from the Sun, its Oort cloud
  should similarly go through the inner solar system.

While very little is actually known, it's not crazy to think that smaller stars might have smaller oort clouds and larger stars, larger ones, though there's a great deal of uncertainty on what an average oort cloud is, how far an average one extends, and how densely packed they are.

First part first, how often does a star pass within one light year of our sun?

Not very often.   Ballpark, maybe every million to couple million years or so.  The last time was 70,000 years ago, a small star, 15% the mass of the sun.  See here.

In the fly-by section of that same wiki-article, it says a star passes that close (0.82 light years or less) about once every 9 million years.

Scholz's star is quite small, only about 15% the mass of our sun, and it's unclear if it had an Oort cloud that extended .82 light years at all, so while it passed through our Sun's Oort cloud, it's unclear if the inner solar-system passed through it's Oort cloud, assuming it has one.  It's also possible that the sun stole much of Scholz's star's oort cloud during the fly-by, but that's just speculation.

Did Scholz's star send Oort cloud objects heading towards Earth?  That's a good question and the answer is, probably, but we don't know, because if it did, it would take longer than 70,000 years for any re-directed Oort cloud comets to reach the inner solar-system.   By the same article, an estimated 2 million years for re-directed Oort cloud objects to reach the inner solar system, so, in about 1.93 million years (give or take), we'll start to see what Scholz's sent our way.    :-)

In 1.36 million years, Gliese 710 could pass within 1 light year of Earth.  The estimate is 1.1 light years plus/minus .577 light years.  Gliese 710 is larger than Scholz's, about 60% the mass of the sun but we know nothing about how large or how dense it's Oort cloud might be.   But, very generally, it seems likely that the Sun has passed through another stars Oort cloud at some point, perhaps every 1-10 million years, perhaps every hundred million years or so, but it seems reasonable to me that has happened and will continue to happen, on rare occasion.

2nd question, what would happen if the inner planets passed through another star's Oort cloud.

If we use our Oort cloud as a model, and, warning, these estimates are enormously ballpark, but, a trillion Oort cloud objects 1 KM or larger and the Oort cloud extends between 100,000 and 200,000 AU (lets say 150,000 as a middle estimate), that works out to one 1 KM or larger object per cubic region of space about 15 AU across.   That's 3/4ths the distance between the Sun and Neptune, so if you figure just one, 1 KM or larger comet or object in a 3D space over such a vast distance, most would pass by the inner planets unaffected.   The occasional one would get it's orbit measurably altered with a fly-by of one of the gas giant planets and rarely, there would be an impact.  Now presumably there's more smaller stuff in Oort clouds too, so there should be smaller impacts more frequently, though there's almost nothing that's actually known about how dense Oort clouds are, so anything beyond "infrequent", is impossible to say.

Passing through another Stars Oort cloud should take tens of thousands, perhaps 100,000 years years or so, so there probably would be the occasional impact on a planet here and there, but on a human time scale, significant impacts should be very rare.


How often are there five planet apparently aligned within 25°?

It is not super-unlikely. Let's look at alignments of the three interior planets and two superior ones.

From Mars, the three interior planets (Earth Venus and Mercury) all appear to remain quite close to the sun, and so quite close to each other.  Moreover they will seem to move slowly as they reach greatest elongation. The chance that two interior planets will be on the same side of the sun as each other is 50%, and they will be "close" quite often. I've not done careful calculations, but let's say there is a 1/3 chance that two interior planets are close.

So to get the third planet close is another 1/3 chance, it would have to be close to both other two planets but the probability of all three interior planets aligning somewhat is around 1/3^2 or roughly 10%.  (This is a Fermi estimate so I'm taking the liberty of rounding)

Now the other planets can appear anywhere on the ecliptic. If you want them to be within 25 degrees of the others, that is a 25/360 chance (ie roughly 7%) and to get two planets both aligned is 7%×7% or about 0.5%. But there are four such planets, so six possible choices of two planets. That means that there is roughly a 3% chance that two superior planets will align with the inferior ones.

In conclusion, there is roughly a 0.3% chance of an alignment in the morning (or evening) sky on Mars. It is notable, but not super-unlikely.  Your particular alignment includes Uranus (which is almost to faint to see) and Mercury (which is all but lost against the sun.)  Close alignments of Jupiter Saturn Earth and Venus would look more spectacular.  A similar four-planet alignment was visible on Earth a few years back. How to watch at best this week's planet alignement?


Why does Gaia use only calcium NIR lines for stellar radial velocity measurements?

The Ca triplet in the near infrared are extremely strong resonance absorption lines. They are by far the strongest features in the near infrared spectra of cool G,K,M type dwarfs and giants, which will be the majority of the stars observed by the Gaia RVS.
The Ca triplet lines are so strong that even in low metallicity halo stars, that have little Ca in their photospheres, these lines are still strong enough to measure radial velocities.

The lines are much weaker and much broader for hotter O, B and A stars, and measuring radial velocities for these will be difficult and much less precise.

You can have a look at an atlas of the Gaia Ca triplet region for stars of  different spectral types in Figure 2 of Munari et al. (2001). http://cds.cern.ch/record/531022/files/0109057.pdf

I should also add that these three lines are not the only features used to determine the velocities, they are just the strongest features in the spectra of most stars.


Does anybody recognize this kind of prism?

My guess is that it's not a prism (or at least doesn't function as one optically).  It looks like it is a right triangle in one cross section.  The diagonal face would be silvered to act as a star diagonal. If aligned correctly, the black plate it's glued to that attaches to the support arm would be perpendicular to the tube axis.

It looks like it's rough beveled on other sides, maybe to get an ellipsoidal cross section?


Is there any update on the Availability of Event Horizon Telescope Images?

Today, April 10th 2019, there was a press conference where finally an image of M87 was released:

Scientists have obtained the first image of a black hole, using Event Horizon Telescope observations of the center of the galaxy M87. The image shows a bright ring formed as light bends in the intense gravity around a black hole that is 6.5 billion times more massive than the Sun. This long-sought image provides the strongest evidence to date for the existence of supermassive black holes and opens a new window onto the study of black holes, their event horizons, and gravity. Credit: Event Horizon Telescope Collaboration


Can there be an infinity of stars in the Universe?

Theoretically yes, there can be infinitely many stars. Since this space you're talking about is better described by general relativity and Einstein's field equations. One of its solutions describes the universe as being spatially infinite.


Can the interferometer called "Gravity" measure "a few centimeters on the Moon"?

A 130m baseline operating at 2 microns gives a theoretical resolution of $2\times 10^{-6}/130$ radians. At a distance of 400,000 km this translates to 6m.

My guess is that Genzel is referring to the accuracy with which the position of a point source of light can be measured. This not the same thing as the smallest thing you can image. The centroid of a Gaussian can be measured much more precisely than its FWHM.

The moon landing sites have been photographed by the lunar reconnaissance orbiter. If the 'foil hat brigade don't believe that, they aren't going to believe a reconstructed interferometric image in any case.


Centerless expansion or a matter of perspective

The first principle of cosmology is called the "cosmological principle", which asserts that the universe is more or less the same everywhere spatially, but changes with time.  This is a fundamentally new way to think of the history of the universe, and has revolutionized our ability to understand long-range observations.  We do not, however, know that this principle is true, we only know it is both useful and consistent with all observations.  So we adopt it as a working hypothesis, and all of cosmology follows.  Some day we may discover it is inadequate, but until then, we will continue to use it.  The same could be said for other principles, like conservation of energy-- we don't know that's true either, but it's useful and consistent with observations.  The cosmological principle rules out that the universe has a unique center.

Your question is whether the apparent lack of a unique center stems from our location not being special enough to be able to tell that it does have such a center (say, by encountering a boundary).  Ironically, the cosmological principle turns that question on its ear, and points to the possibility that one's perspective always puts themselves at the apparent center.  So we would say the illusion of having a center is all about perspective-- not the illusion of not having a center!


What are some applicable problems with the correct usage of G?

I think one of the reasons for using different units is how you take the measurements for the other variables in the equation.

For example, let's say you have two marbles, and you want to calculate the gravitational force between them (not a lot!). You place them a certain distance apart, and measure it with your meter stick. You find the distance between them to be $0.146 \text{ meters}$. Now you weigh them, and find their masses to be $0.00156 \text{ kilograms}$ apiece. You decide to use the value of $G$ with units including meters and kilograms.

Now let's say you have two stars far, far away. You take measurements based on parallax and find that they're roughly $35,400 \text{ light-years}$ away from each other. Your instruments can't measure this to any greater accuracy. You measure their masses by calculating how much they deflect light, and come up with a figure of $10 \text { solar masses}$ apiece. You therefore decide to use some form of $G$ that, for some reason, has solar masses and light-years as some of its units.

Basically, you choose different forms of a constant depending on the units you're able to measure with your instruments. You won't be able to figure out exactly how many meters your stars are apart from each other, but you can figure out just about how many light-years apart they are, with less than a light-year of uncertainty.


What allows WFIRST to have similar resolution to Hubble over a 100x larger solid angle?

The ultimate resolution of a telescope depends upon its diameter. The larger the telescope aperture, the finer the resolution.
Hubble and WFIRST are both the same diameter 2.4m so they both have the same theoretical resolution.
However the image quality on any telescope degrades the further you are away from the optical axis. In WFIRST there is a 3rd mirror that corrects for these inherent distortions for a much wider field of view.
The last factor is the improvement in detector technology. The individual pixels are smaller, smaller gaps, and there chips are much larger. Also WFIRST has mosaic of these  CCD chips to cover the larger corrected field of view.
The paper Optical Design of the WFIRST-AFTA Wide-Field Instrument (Pasquale et al. 2014, Proc. SPIE-OSA Vol. 9293, 929305) discusses the design of the optical path (cycle #4). In addition to a small figure adjustment to the primary, the figures of the secondary and a tertiary mirror are optimized to provide high resolution over a much wider field than the HST. This design also provides a pupil at which filters can be located as well as an intermediate focus (1:1 with the final focus) as shown below.

The basic optical design is decidedly simple for fabrication, integration and test purposes. This approach begins with the base optical design; all three powered mirrors are optically co-axial and simple conics. We considered more complicated designs, including the use of tilted and de-centered mirrors and Zernike and/or anamorphic surface figures.[2,3] TMA designs are widely preferred for high A*Omega telescopes such as JWST and ATLAST.
The primary mirror, referred to here as Telescope mirror 1 (T1), is a fast f/1.2 primary with a linear central obscuration of 31%. It is a light-weighted mirror using a hollow honeycomb core. T1 and Telescope mirror 2 (T2) together form an ~f/8, intermediate focus. This focus is uncorrected, with a large caustic image and strongly curved image surface. You will note the off-axis field bias (seen in Figure 1); this is typical in all TMA systems.
Within the instrument, Mirror 3 (M3) is almost a 1:1 magnification relay of the intermediate image to the focal plane. Working in concert with T1 & T2, these 3 powered mirrors form the large corrected field of the TMA. A pupil is formed between M3 and the focal plane. With a diameter of approximately 100mm, the pupil allows for the insertion of bandpass filters and spectral dispersion elements via an element wheel. Finally, in order to package the system into the volume constraints, two fold mirrors were used.


The orbits are not all in exactly the same plane, that is why we do not have eclipses every month. With the orbits at a slight skew it is possible to have the geometry very near to that in your diagram without the Earth blocking the Sun.

Lunar eclipses, the Moon passing through the Earth's shadow which is what your post is about, do occur relatively frequently compared to Solar eclipses because the Earth is bigger than the Moon.

At New Moon the lit side of the Moon is facing away from the Earth and so we do not see it.

You will find a fairly good explanation, and a better diagram, of all this here


What does black hole evaporation correspond to in the accelerating universe / black hole analogy?

An analogy in the settings of the question would be a deceleration down to an expansion rate, such that eventually light from anywhere in the universe can reach us, meaning the cosmic horizon can vanish. A constant expansion rate of the universe above the speed of light would be the analogy of a black hole of fixed mass.

In an infinite universe this would mean, that its expansion needs to slow down to an asymptotic speed for infinite distance below the speed of light. For a finite 3-spherical universe for a fixed cosmic time a slow-down to below the speed of light would be sufficient. This would correspond to a massive follow-up object of the black hole without event horizon.

For an analogy to a completely evaporating black hole the expansion rate needs to slow down to zero. The slow down between speed of light and zero corresponds to the very last phase of black hole evaporation, when its mass drops below the Planck mass, then without event horizon.

The analogy to Hawking radiation would be radiation or matter emerging from behind the former cosmic horizon.

Thinking in stereographic projections may help for a better understanding: Project infinity of a plane expanding universe to the singularity of a black hole at the north pole of an expanding sphere, as an initial point of further reasoning.


Is it true that we see the center of the milky-way for only half of the year?

No, although there are times when it can't be seen, it isn't true that it is visible for 183 days of the year.
The general question could be "If I take an arbitrary location on the sky (say a randomly chosen star) will it be visible for half the year?  The answer is "it depends on the star!" Polaris and many other stars are circumpolar for Northern hemisphere observers and are visible every night,  contrariwise, Polaris is never visible for Southern Hemisphere observers.
Now, the galactic centre happens to be located in Sagittarius, so co-incidentally is quite close to the plane of the solar system.  This means that for about two months of the year, the sun is too close to make observations easy, certainly in visible light from the Earth's surface.  At other times it will be visible, perhaps briefly.  Sagittarius and the galactic centre are also quite a few degrees South of the equator, so it is more easily visible from the southern hemisphere.
At 50 degrees North, The galactic centre sets at twilight at the start of October and doesn't rise until dawn at the middle of February,  between February and October (about 7½ months) the galactic centre is above the horizon at some point during the night. (The exact number of days depends on your definition of "dawn" and "above the horizon") More southerly observers will see the galactic centre for longer. 50 degrees South, the galactic centre is above the horizon at some time during the night between January and November.  In Antarctica, the galactic centre is circumpolar, and so is above the horizon every night. But as the summer sun in Antarctica doesn't set, there are days when you can't "see" it. The optimum location would be 30 degrees South (or in space)


How to check what telescopes were looking at a certain portion of sky at a particular time?

The areas of the sky covered by the major Near Earth Object (NEO) surveys are reported to the Minor Planet Center. You can plot visualizations of that sky coverage data using the sky coverage form where you can filter by depth, date and survey. The raw data is available, after the surveys give permission for it to be released, from the raw data page. This provides a download link to a tarball of all the pointing data so far, broken down by survey site and then a file per night. The description of the pointing fields is also shown on that page; hopefully it shouldn't be too hard to use a combination of e.g. astroPy's Table and 'SkyCoord` to read and parse the data.


Telescopes for beginner astrophotography

Ah, welcome to the world of AP... check your sanity at the door, and that loud sucking sound you hear is your bank account being emptied.

My first and best piece of advice for you is to NOT buy anything at all right off.  Find your local astronomy club(s) and join.  Most clubs have at least a few people doing imaging, and you would be well-advised to meet these people and learn from them.

AP is not like conventional photography at all, for the most part. It's closer to digital signal processing.  It is a blend of science, engineering, and art.  What you know from conventional photography is not useless information, but you have a whole slew of new things to learn.

When we talk about telescopes and equipment for AP, we don't talk about "tripods" all that often.  A tripod is one kind of support structure for a mount head, which does the actual work.  The mount's role is to carry the telescope, camera, and associated equipment and point it at the object you wish to capture.  While the concept is quite simple, the execution is extremely complex.

There are two fundamental problems that we must overcome to capture good images of astronomical bodies.  The first is the fact that the majority of these objects are very faint.  While people often think the purpose of a telescope is to magnify, this is a misnomer: an astronomical telescope's primary purpose is to collect more light and concentrate it into a smaller area (the eye or the image sensor).  Magnification, while not irrelevant, is a secondary concern (and comes with its own issues).  Because these objects are so dim, we must resort to long-exposure photography.  While typical daylight photography uses exposure times measured in tenths and hundredths (even thousandths) of a second, exposure times for astrophotography often are measured in whole minutes.  Actually, total exposure is often measured in hours, with individual sub-exposures done in minutes.

And this is where the second key problem emerges.  While our shutter is open to capture an exposure of 1 or more minutes, we have to contend with the fact that the objects we are capturing are moving relative to our instruments.  Due to the rotation of the earth, a given star, planet, galaxy, nebula, etc... in the sky above moves across the sky from east to west in an arc with a radius depending on the object's position relative to the celestial north or south pole.  In a full day, an object moves 365 degrees around the pole (well, we rotate, but for practical purposes, the object moves).  In a single hour, it moves 15 degrees.  In one minute, 15 arcminutes.  And in one second, 15 arcseconds.  This may seem like a small amount of motion, and when you look up at the stars in the sky, they do seem to be fixed in place.  But, when you start to reduce your field of view by using a telescope or camera lens, the motion starts to become more and more obvious.

As a conventional photographer, you are probably familiar with the standard alt-az style mount motion.  Alt-az is short for altitude-azimuth, and a typical camera tripod works this way.  The mount angles up/down (altitude) and left/right (azimuth) to point at an object.  In order to follow an object across the night sky, then, you have to move the mount in both axes.  The rates of motion between the axes will vary from one another as well as depending on the location of the object in the sky above.  To automate this process requires a great deal of computerization, and even then it is difficult to do so accurately enough to handle the demands of long exposure imaging.  And even if you do overcome those issues, you will find that the object you are following appears to rotate in the field of view as it crosses the night sky, a phenomenon we call frame-rotation.  This is because the frame of reference of an alt-az mount is the horizon, not the sky itself.

The solution to this issue is the Equatorial mount.  An equatorial mount has four axes.  The first two, altitude and azimuth, are only used for alignment purposes, not for motion.  They are used to align the third axis, known as Right Ascension (RA), with the celestial pole.  Once that is accomplished, motion in the RA axis shifts the telescope's aiming point east and west.  The fourth axis, Declination (or Dec), handles north and south.  When properly aligned, the mount then needs only to move in the RA axis and at a single rate of speed to follow an object across the sky.  This doesn't even require a computer, and thousands of mounts have been manufactured with clock-drive mechanisms to handle this motion.  Computerization only enhances the process and makes it more accurate.

But not all equatorial mounts are created equally.  Many just do not have the accuracy needed to handle long exposure imaging, and the level of accuracy needed increases as you increase focal length, focal ratio, and exposure time.  A mount that could handle a 300mm camera lens or 80 mm short-tube refractor and produce 60 to 120 second exposures might not be able to handle an 8" f/4 Newtonian telescope.

Weight is a common issue.  Mounts like this are all rated for a specific payload weight amount.  Payload here includes not only the telescope, but also any cameras, adapters, finders, guide scopes/cameras, and mounting hardware - everything attached to the top of the mount.  But these payload ratings are typically given for total payload maximums. When doing AP, the general wisdom is that you should keep your imaging payload to 1/2 or less the mount's rated maximum.  So if your mount can handle, say, 40 lbs. of weight, for imaging you shouldn't load it with more than about 20 lbs.  The rule isn't a hard limit, and not all manufacturers rate their mounts the same way, but it is definitely a strong suggestion.

As I said, the tolerances become tighter as you increase focal length and ratio.  First, let me discuss ratio. The focal ratio of a telescope is calculated by dividing the focal length by the aperture.  Thus, if you have a telescope with a 200 mm (roughly 8") aperture and a focal length of 1,000 mm, you have a focal ratio of f/5.  If you have a focal length of 2,000 mm with the same aperture, it's f/10.  In general, the lower the focal ratio, the better for imaging.  The reason is the concentration of light.  Imagine you have the two telescopes I just mentioned, both 200 mm apertures, one f/10 the other f/5.  Now imagine you have identical cameras in each and are tandem-mounted on a GEM that can handle the payload without issue.  Now, let's say you are imaging an object that appears roughly circular and evenly illuminated.  Now, let's say in the shorter scope, f/5, the object you are imaging covers a spot on your camera's image sensor about 100 pixels across.  The total area of pixels covered is then around 7,850 pixels.  Let's say each pixel receives about 20 photons of light per second and you run a 1 minute exposure.  Thus each pixel gets a total of 1,200 photons of light in the exposure, for a total of about 9.4 million photons collected.  Now, if we look at the f/10 telescope, we find the image is larger.  A longer focal length increases magnification.  Double the focal length, you double the magnification.  So now the object we are imaging has a diameter of about 200 pixels on the image sensor.  This means that it has an area of about 31,400 pixels - double the diameter means 4 times the area.  Now, because the aperture is the same, we do NOT get more light, we get the same number of photons, they are just spread out over the larger area.  Instead of 20 photons per second, each pixel gets only 5.  In one minute, each of them gets only 300 photons.  The total number is the same, about 9.4 million, but they are less concentrated.  This means the image is only 1/4 as bright as that in the f/5 scope.  To get the same level of exposure, you would need to increase your exposure time by a factor of 4 to 240 seconds.  Yes, you get a larger, more detailed image, but it takes you four times as long, and this can be a problem.

No mount is perfect and all will show drift over time, even with autoguiding technology.  Let's say that the mount you are using has an error that is equal to about 1 pixel per minute in the f/5 telescope.  In that 1 minute exposure, everything is stretched by 1 pixel.  This is hardly noticeable and not likely to be a problem.  Let's say you had a star near the center of the image that has a diameter of 5 pixels.  In the exposure, you would see it as 5 pixels wide by 6 long.  Hardly noticeable.  But now let's look at the f/10 telescope.  Because we double magnification, we now have 2 pixels of stretching.  Now the star is stretched 2 pixels.  Still, this isn't a huge problem, but we also need to increase exposure, so we quadruple the stretching to 8 pixels.  Now we have significant stretching in our image and blurring of our object (and that's assuming the mount is THAT accurate, which is not a given).  This is why astrophotographers refer to shorter focal ratios as "faster" telescopes.

Aperture is typically the critical measure for a telescope.  But for AP, this isn't always the case.  A lot of very high quality imaging is done using short-tube refractors with fairly small apertures.  This also gives you a much wider field of view.  And this, in turn, reduces the exacting tolerances necessary for the mount.

However, this isn't ideal for all targets.  And this is another key problem with AP: there is no one-size-fits-all solution.  What is ideal for planetary imaging is not ideal for imaging large nebulae, which is not ideal for imaging most galaxies... you need to consider multiple factors in equipment selection. Personally, I use an 8" f/10 SCT (with a Barlow lens giving me f/20) for planetary imaging, an 8" f/3.9 Newtonian for globular clusters, medium-sized nebulae, and most galaxies, and an 80mm f/3.75 refractor for wide-field nebulae and the Andromeda Galaxy (which is 3° across, so very large).

But the mount is the key to all of it.  The conventional wisdom in AP is to focus on the mount first.  You can have a million dollar top-of-the-line camera and a million dollar top-of-the-line telescope, but if you don't have them on a mount that can provide accurate motion, you won't get good images.  Alternatively, you could put a mediocre camera and telescope on a good mount and get pretty decent images - the mount is THAT critical to the process.

Prices on these kinds of mounts varies, and you can easily spend into the tens of thousands.  The cheapest mounts in this range are designed to handle DSLR Cameras and lenses, usually not more than 300 mm in length (perhaps a little shorter).   The price on a mount like this is around \$350 to \$600 USD, depending on quality and capability.  And, again, these are limited in what they can handle.

Stepping up from that are basic German Equatorial mounts with powered tracking ability.  While computerization (What we call GoTo) is not required, it is usually found on these kinds of mounts and allows for the addition of autoguiding technology.  Autoguiding is the use of a second camera and, usually, telescope that is harnessed to software to lock onto a guide star and make minor adjustments to the mount's motion to keep it on-target. Autoguiding will not make a bad mount into a good one, it will simply allow an already good mount to capture longer exposures.  The lowest-price mounts in this range usually have a payload weight maximum of 30 lbs. or less and a price tag between \$800 and \$1,200.  However, I wouldn't actually recommend ANY of them.  The two most popular here are the Celestron AVX, running about \$900, and the SkyWatcher HEQ5, at about \$1,100.  The SkyWatcher might actually be usable, but the 30lb limit means only 15lbs for imaging, and that's very low indeed.  You can still probably fit a short-tube refractor on here with a DSLR and lightweight guiding package.  The Celestron AVX promised to provide entry-level AP capability, but the reviews of the mount show that it hasn't lived up to its promises.  This mostly seems to be due to poor quality control in the production line, resulting in some that work just fine, and a lot that do not provide the accuracy needed for imaging.  They are still usually fine for visual observing, just not for imaging.

The next step up finds mounts capable of handling between 40 and 45 lbs. of payload with prices between \$1,500 and \$2,000 USD or so.  This is where I STRONGLY recommend people start.  While they are not cheap (though you can often find them used for about 30% or so less), they often have the quality and accuracy necessary.  There are four models in this range I recommend: The Celestron CGEM II, Orion Atlas EQ-G, SkyWatcher EQ6-R, and the iOptron iEQ45 Pro.  My personal preference is the iOptron (I have the early version of this mount and love it), but all should be competent.  My mount is capable of 45lbs and I run it at around 26 (yes, over the recommendation, but hot hugely so, and it handles it fairly well) with an 8" f/3.9 Imaging Newtonian and CCD Imaging camera.

You CAN go with a cheaper mount.  But you rapidly lose capability and accuracy.  The lower mounts aren't really intended for imaging use, the heavier ones are generally engineered to stricter tolerances needed for imaging.

One nice thing about these four mounts, and most others in this range, is that they all use a common interchangeable modular mounting system for attaching the payload.  I personally have three sets of payloads and can switch them out in a matter of minutes. It also means that if you buy the mount now with a low-end telescope option, you can later upgrade the telescope itself without having to get a totally different mount.  It also means you can have different payloads for imaging and for visual observing.  I've never heard of anyone having too much mount for imaging.

Now, as for cameras....

DSLRs are not the best tool for the job, but they are widely used all the same.  There's a few considerations here.

First, DSLRs tend to have a great deal of signal noise - noise which must be dealt with.  The preference for AP is generally thermoelectrically cooled CCD imaging cameras.  These use Pelletier cooling systems, heat sinks, and fans to cool the image sensor and dramatically reduce signal noise.  Remember, we're dealing with faint objects, and the noise can often drown out the image signal in the exposure.  These kinds of systems also typically better isolate noisy components and reduce their effect on the image. Also, the nature of CCD technology makes them inherently less noisy than CMOS sensor technology prevalent in DSLR cameras. However, CMOS cameras are improving and many manufacturers are now offering cooled CMOS cameras.

With or without a cooled system, you still will need to capture calibration frames. There are three common types: Dark frames, Bias frames, and Flat frames. Dark frames are images taken with the shutter closed.  They are done with the same exposure time/gain/ISO settings and, preferably, the same temperature range.  These are then digitally subtracted from the "light" frame image you capture.  Some cameras actually have built-in technology for this (my Nikon D5000 does).  But for AP, it is usually done through capturing separate dark frames and processing them in bulk after the fact.

The Bias frame is also done with a closed shutter, but for the shortest exposure time possible. This maps the noise created when the image data is read off the image sensor itself.  It is also digitally subtracted in processing.  I don't believe any DSLRs have technology to do this automatically, though I could be wrong.

The flat frames are a different beast altogether.  They are taken with the camera still attached to the telescope or camera lens.  To produce flat frames, you want even illumination entering the optical train (e.g. the front of the lens/scope) and an exposure length that produces an average of 50% illumination.  These images are used to map the optical system's light and dark spots and even out the illumination (compensating for problems like vignetting).

As I said, noise is a major problem in DSLR cameras. In well-lit images such as daylight, well lit interior, or flash photography, the noise is nearly irrelevant.  But in low-light long-exposure imaging, it can be dramatic.  This is especially true if you use higher ISO levels to increase sensitivity.  CCD imaging systems are not immune, but they are designed specifically to handle it.

CCD and CMOS systems designed for AP are also typically more sensitive.  The term Quantum Efficiency (Qe) describes the sensitivity level of an image sensor to various wavelengths of light.  Essentially, it indicates what percentage of photons that hit the image sensor are detected.  A typical DSLR has a peak Qe around 30-40%, with some newer ones moving up closer to 50% and a handful of newer ones in the 70% range.  CCD and CMOS systems for AP typically have peak Qe starting around 50% and some have peak Qe around 95%.  This means much more efficient imaging and shorter exposure times.  The actual Qe varies across the spectrum, with the peak being the highest sensitivity.

These issues, so far, are fairly simple to deal with, though often require longer exposure and/or more frames in the stack (I'll come back to stacking).  There's one problem with DSLR cameras, however, that isn't as easy to deal with.  With the exception of the Canon 20Da, Canon 60Da, and Nikon D810A, all DSLRs, and, in fact, nearly all digital cameras in general including cell phone cams, webcams, etc..., have a built-in filter for color-balance improvement.  In most cases, it's mounted directly over the imaging sensor "chip" and also provides some degree of protection from dust and damage.  This filter is designed to reduce the amount of deep red and near IR energy that reaches the sensor.  Human eyesight is not very sensitive to these longer wavelengths, and the filter helps produce a more "life-like" color balance to match human vision. Without the filter, images tend to be shifted toward the red.  Software color balance profiles can reduce this significantly, but for conventional imaging, the filter is the best option.

However, for astronomy, this is a problem.  There is a lot of astronomically interesting light in the deeper red part of the spectrum.  When an electron in a Hydrogen atom falls from its second-highest to third-highest energy state, it emits a photon at a wavelength of 656.28 nanometers.  We call this Hydrogen Alpha, and it's a major component of emission nebulae and star-forming regions in galaxies.  When you see red in a photo of a nebula, it's often H-Alpha.  Some nebulae, such as the Rosette Nebula, are extremely heavy in H-alpha.  Unfortunately, the filter in a DSLR blocks the majority of this light from reaching the sensor - something like 90% if I remember correctly.

The solution for this problem is to not have the filter in place.  To date, only three factory-issued DSLR models have been released with this filter removed (actually replaced with one that is transparent to these wavelengths).   For other DSLR cameras, you would need to have the camera modified. There are several services you can find on the internet that will do this for you, but there's a cost.  It is not something I recommend people do themselves unless they have experience with camera repair and/or can afford to run the (very large) risk of ruining the camera.  Doing so also makes the camera ill-suited for continued conventional imaging, unless you get a clip-in or screw-on replacement filter (more cost).  You can do AP without a modified camera, but you will be significantly handicapped.

Of course, for best results, a monochrome CCD and filter wheel is the best option of all, but this significantly increases the budget.

You mentioned one of your cameras has more/smaller pixels.  In most cases, this is not actually desirable.  Larger pixels tend to be more sensitive (more collection area to gather photons), and more pixels may provide higher resolution, but also means larger file sizes and MUCH slower processing.   The Hubble Space Telescope's primary cameras have a resolution around 2048 x 2048 pixels (about 4 megapixels). The issue is not resolution as much as sensitivity.  I've worked with some very high end astronomical imaging CCD's with resolutions of only around 1 megapixel: but they produce FAR better images than any DSLR on the market.

When it comes to brands, Canon and Nikon are far and away the leaders, with Sony gaining some ground.  But overall, Canon is really the brand of preference among astrophotographers using DSLRs.  Canon has been very welcoming and accommodating to the AP community. They produced the first two pre-modified cameras well before Nikon produced their first.  Canon has made a great deal of information available, including detailed specifications and even firmware source code. Nikon is slowly coming around, but has been a LOT slower to do so and still has not offered anywhere near as much support as Canon.  Because of this, you will find a lot more resources and, especially, software for camera control of Canon cameras than Nikon or anyone else.  For people just getting started in AP and on a budget, I generally recommend looking for a used pre-modified Canon DSLR as an imaging camera.  Sites like CloudyNights and Astromart frequently have these in their classified ad sections.

And now... software...

A major difference between conventional photography and AP is the processing.  I already mentioned calibration images.  But the big deal here is image stacking.

Most Digital AP, whether DSLR or CCD based, uses stacking processes to produce better images.  Stacking is, essentially, a signal processing method using statistical analysis.  You capture a set of images of the same subject and exposure time.  You then run each through a calibration routine which removes the dark and bias signals and then applies the flat frame to even out the illumination.  The images are then aligned (nearly always through lining up stars in each frame).  At the end of the alignment phase, any pixel all the way through the stack represents the same point in space.  Then, each pixel in the stack is statistically analyzed.  For example, if you have 100 images in a a stack, and a particular pixel in the stack has brightness values between, say, 100 and 105, you average them out to come to what the most likely "true" value is, say 103.   In many cases, a weighted average is used in which pixels outside of 1 or 2 standard deviations above or below the average are discarded entirely, and only values in a consistent range are finally averaged together to produce the most likely "true" value.  This has the effect of reducing noise and, using drizzle techniques, can actually increase resolution.  Even without drizzle, you end up with a much better overall image, with greater clarity and less noise.  The larger the sample size, the better.   My most recent image of M51, the Whirlpool Galaxy, uses 3.5 hours of 120 second (2 minute) exposures (so over 100 individual exposures). The more exposures I use, the better the final result and the more faint detail I can gather.

Of course, this processing takes time.  The larger your image files, the longer it takes.  My main imaging camera produces image files in FITS format (the most common format for AP) that are about 50mb in size each.  100 of those comes to about 5 gigabytes, not including the flat, dark, and bias files (usually 10-20 of each).  This ends up taking a lot of storage space and a lot of processing time.  There are other file format options, but it is critical that you use a lossless format (like FITS or TIFF), and the only way to reduce those files is through compression, which works against you in processing (you have to decompress each frame in processing, so you save space, but dramatically increase processing time).

There is good processing software available, some for free.  I do most of my calibrating and stacking currently using DeepSky Stacker, which is a free product.  I do most of my image capture in Nebulosity, which is not free, but works well for what I need and I got before the last price increase.  While live view on a camera is helpful, it's not necessary with a program like Nebulosity which will loop exposures and provide metrics to help achieve accurate focus.  You can get MUCH better focus this way than through the live-view on a camera.  It will then also allow you to automate the capture, setting the number of exposures, length of exposures, what to name the files, where to store them on your hard drive (typically these store directly to hard drive, not on the camera), and in what format (usually FITS or TIFF, not the camera's proprietary RAW/NEF/etc... or JPG - and you really do NOT want to use JPG, it's FAR too lossy).

After you've captured, calibrated, and stacked, you will then want to do post-processing.  Some people use GIMP, which is an open-source option.  I mostly use Photoshop, which has a lot of capability and there's a wealth of information and tutorial advice.  I am also trying to learn to use PixInsight, which is probably the best astrophotography processing software on the market, but is neither cheap nor easy to use.

At this point, you probably are starting to see how very difficult Astrophotography can be.  Doing "pretty decent" AP is neither cheap nor easy.  I'm about 10 years into it and have spent about \$10,000 total (that includes several bad choices in equipment and also includes a backyard roll-off-roof observatory), and I'm just now feeling relatively competent.  My main issue now is time and weather.

For someone just starting, my recommended starting budget is at least \$2,000 and preferably closer to \$4,000 or more.  At \$2,000 you can get a competent mount and low-end telescope to get you started (and, depending on deals and what used equipment is available, a pre-modified DSLR or lower-end CCD camera).  You CAN get started cheaper than that... but you rapidly start losing capability and flexibility.  A \$1,500 or so mount can handle a camera/lens combo for very wide field imaging, a smaller scope for normal wide field, a medium sized scope for narrower field deep-sky (e.g. galaxies and smaller nebulae) and longer focal length scope like an SCT for planetary imaging.  If you go with a lesser mount, you're not likely to have as much flexibility there.  And, again, most of these mounts allow for interchangeable payloads.  So you can start with something simple like a short-tube refractor, and add to your equipment options as you have the ability to do so (my annual tax refunds have often ended up going to the hobby).

In the end, I return to my original advice: joint a club.  Most clubs have people doing this you can learn from.  A lot of clubs have people regularly selling well-maintained used equipment to finance their next upgrade.  Some clubs have loaner equipment to borrow.  Most either have a dedicated observing site or have scouted out good locations away from light pollution. And many also have their own small observatories with higher-end equipment, and, sometimes, even imaging equipment.  Most clubs in the US have annual dues less than \$50 per year.  It becomes a very cost-effective way to pursue the hobby.

I know I've given you a lot here.  And I've honestly barely scratched the surface.  AP is not for the weak!

Good luck and clear skies!


What discoveries have come from the DASCH project (Digital Access to a Sky Century @ Harvard)

The most notable one I'm aware of is the detection of the secular dimming of Boyajian's Star (KIC 8462852), as described in Schaefer (2016). Follow-up by Schaefer et al. (2018) suggest that the ratios of the dimming across the BVRI bands are consistent with dust, though the reason for the lack of detectable infrared excess remains unknown.

DASCH data has also been used to constrain the orbital period of the ringed object transiting J1407, see Mentel et al. (2018). The DASCH plates did not show any previous transit in the data, ruling out a large fraction of orbital periods and opening the possibility that the ringed object J1407b may not be gravitationally bound to J1407.


What's an order-of-magnitude main sequence star look like?

To get an order of magnitude estimate you can just use the total mass $M$ and luminosity $L$ of the star and an assumption of your fusion process. Main sequence stars fuse Hydrogen in to Helium through the proton-proton chain, which converts 0.7% of mass into energy. So the estimated lifetime of the star would just be: $0.007\frac{Mc^2}{L}$ ($c$ is the speed of light). For example, the sun would have a lifetime ~$10^{11}$ years, but that's assuming that all Hydrogen is being converted to Helium via the proton-proton chain.

Given the scaling from observations of the Sun, you can use the mass-luminosity relation to estimate lifetimes for other main sequence stars.


Where do the symbols $\lambda$ and $\beta$ come from for celestial latitude and longitude?

German: $\lambda$ (or l) like Länge (longitude) und $\beta$ (or b) like Breite (lattitude). A century ago it used to be a language many science papers were published in and English was far from dominant.
Mind though that it is just symbols of notation and you can define them as you see fit - if you define them. They are just one choice among many possible and one among a few commonly used, depending a bit on context where you come from.


How many stars are between Milky way and Andromeda?

The stars between galaxies are called "intergalactic stars", and one study (2012) claims to have identified about 675 of them between the Milky Way and Andromeda:

  Now, Vanderbilt astronomers report in the May issue of the Astronomical Journal that they have identified a group of more than 675 stars on the outskirts of the Milky Way that they argue are hypervelocity stars that have been ejected from the galactic core. They selected these stars based on their location in intergalactic space between the Milky Way and the nearby Andromeda galaxy and by their peculiar red coloration.

  -"Rogue stars ejected from the galaxy found in intergalactic space", Research News @Vanderbilt (2012-04-30)


Relationship between photometric colour and redshift

The main reason is that the intrinsic spectra of galaxies are complex and therefore a redshift of their spectrum, whilst leading to a redder spectrum overall, does not necessarily lead to reddening in all colours.

For instance if there is apeak in the intrinsic spectrum, then as that peak moves redward, then colours formed from bands on the same side or straddling the peak in wavelength would behave differently.

The dramatic redward turn up at high redshifts (note, a large colour is redder) is caused by the "Lyman break" moving through the colour bands. Basically, the intrinsic spectrum is self-absorbed at wavelengths shortward of 91.2 nm by neutral hydrogen. This absorption edge is redshifts into the visible region at redshifts bigger than 3, causing the bluer bands to essentially disappear and any colour formed with them to become very red.


Is there evidence for galactic "seasons"?

Mass extinctions have been linked with the Solar System's oscillations up and down through the galactic plane (the Galactic Cycle). You could also take a look at this question.


Am I using my Celestron 8 Schmidt-Cassegrain telescope correctly?

Keep in mind that many of the photos you see with colors are composite images.  To differentiate colors try to use some various color filters.

To see bands on Saturn, try using a yellow.  This will bring out different details in what you are seeing.

This link provides some more details for using various filters.
http://www.telescopes.com/telescopes/howdifferentfilterscanbetteryourviewarticle.cfm


How many solar masses is the central black hole in the Whirlpool Galaxy?

It is not well known, but a paper by M. Brightman et al gives a value of $10^{6.3\pm0.4}$ or between 8 hundred thousand and 5 million solar masses, while noting that this estimate is lower than previous estimates which had $10^{6.95}$ or about 9 million masses.
It seems that, although the galaxy is face-on to us, its black hole is viewed from the edge.  This means that the black hole itself is hidden behind a torus of relatively cool dust and gas. [ESA] which is visible as a dark streak in the image of the central part of the galaxy
There is much that we don't understand about the M51a black hole.  It is active, but not as active as would be expected, given the interaction between the two galaxies in M51. There is a second dark streak making an "X" the cause of which is unknown, and, as suggested by the wide error margins above, the mass of the black hole is not well constrained.


Galactic winds/outflows: why and how are they detected via blueshifted absorption lines in spectra?

Yes, a blueshift here would mean relative to the systemic velocity of the galaxy - which means that, relative to the galaxy, the material is moving at least partly in the direction of Earth.
In many cases the outflow may be a pair of jets being ejected in opposite directions out of the nucleus of the galaxy, but we are less likely to see absorption lines from the jet on the far side of the galaxy (relative to Earth) because the disk of the galaxy may be blocking our view of it, and in any case it won't produce absorption lines because it isn't between us and the rest of the galaxy.


Well, I'm not sure exactly what you mean by "collapse" (if it takes $10^9$ years to die, the last few hours satisfy what your asking about). However, maybe some explanation would help:

You're asking about the interaction between two timescales - the Dynamical Timescale, which is the time it takes for a star to free fall into a nova, and the Nuclear time scale (I think), which is the rate at which the sun exhausts it's fuel. The dynamical timescale for the Sun is like 1000 s, and the nuclear timescale is like a Gyr (usually what we think of when we think "lifetime of the Sun").

Any nuclear effects would totally dominate over dynamical effects (since they take much, much longer), and without some specific collusion of behavior (angular momentum, metalicity, what have you), collapse would occur with the burning of fuel and not with the dynamical timescale until the Sun exhausted all it's fuel, and would collapse in a matter of minutes.


Is lithium considered a metal in astronomy?

The latter.
To astronomers, a metal is any element that is not hydrogen or helium, because these elements together constitute most of the elements in the Universe, by far.
This means that, in many circumstances all other elements can be neglected, at least to first order.
By mass, H and He account for some 74% and 24% in the present-day Universe, respectively, while the next-most abundant elements are on the <1% scale. And because these elements are also heavier, if you consider the abundance by number, H and He account for roughly 99.9% of all atoms.
Lithium, on the other hand, only account for a few billionths of all atoms. Here's an overview of the abundances in the Solar System, in terms of number densities, i.e. not mass densities:

Credit: Wikimedia Commons user 28bytes, under C.C.-by-S.A.-3.0.
The reason for the zig-zag pattern is that, in general, elements with an even number of protons are more stable than those with an odd number, as discussed in this post.
Note that the $y$ axis is logarithmic, i.e. each tick mark denotes a ten-fold increase in abundance. The axis is arbitrarily scaled to $10^6$ silicon (Si) atoms, so the Si dot is at $\log A \equiv 6$. Oxygen, for instance, is at $y \simeq 7.15$, so it is $10^{7.15-6}\simeq14$ times more abundant than Si.
Are we wrong?
The terminology used by astronomers may be considered wrong by other physicists and chemists. But why do we use the term at all? To distinguish some of the elements from other elements, based on their properties, I'd say. In astronomy, it very often makes sense to consider the elements of the Universe as divided in three:

hydrogen, which is most of the mass and most of the atoms, and which provides most of the energy in stars,
helium, which adds significantly to the mass, lowering the mass fraction of hydrogen, and
all other elements, which together help cooling the gas (so it can condense), form dust grains (altering observed spectra), have a very roughly equal amount of protons and neutron (so $A\simeq2Z$), are formed in other processes than the Big Bang (except for a small part of the lithium), etc.

In physics, you may be more concerned with the material's ability to conduct electricity and hence consider sodium a metal until the pressure becomes too high, after which you call it a non-metal.
In chemistry, you may be more interested in the materials ability to form bonds, and hence do not consider arsenic and antimony metals.
So, I think you can say that there isn't a formal definition of the term "metal" — it depends on the context. From Wikipedia:

Around 95 of the 118 elements in the periodic table are metals (or are
likely to be such). The number is inexact as the boundaries between
metals, nonmetals, and metalloids fluctuate slightly due to a lack of
universally accepted definitions of the categories involved

Apart from these practical reasons, there are also historical reasons, as discussed in this answer by Rob Jeffries.


How to calculate the temperature of a star

Empirically (I fit a regression on log(mass) vs log(surface temp)), using the table of values in the article on Main Sequence stars, I get a fairly well-fitting formula:  $\mathrm{estTemp} = 5740*\mathrm{mass}^{0.54}$, where estTemp is in C and mass is in multiples of the sun's mass.  Seems to work very well for all but the largest and smallest main sequence stars (and not TOO bad for those).


Calculating Field of View of a Telescope Given Time of Another Object

This technique for determining the field of view of a telescope and eyepiece is called drift timing. Since the sidereal day is about 4 minutes shorter than the usual solar day, a fixed point on the celestial equator rotates at about 15.04 arc seconds per second of time. Thus the true field of view in arc seconds is given by
$$TFOV=15.04 \hspace{2pt} t \cos(\delta)$$
where $t$ is the time in seconds for the star to drift across the field, and $\delta$ is the declination of the star. Divide by 60 to convert to arc minutes, or by 3600 to degrees.


Are galaxies moving away from us faster than before?

The recessional velocity $v$ of an object depends on two things: firstly it depends on how faraway an object is in terms of proper distance $D$, and secondly on the rate of the Universe's expansion as a function of cosmological time $t$, which is best expressed as the Hubble parameter $H(t)$. Specifically:

$$v = D \times H(t)$$

It's worth noting that this equation is slightly vacuous as it is merely the definition of recessional velocity, which is not something that can be directly measured.

As recessional velocity depends not just on a function of $t$, but also on $D$, the question as to whether objects are receding from us faster than ever before could be answered in several ways. Before I look at the different ways we could answer your question, I will note a few things. Firstly the definition of the Hubble parameter is:

$$H^2(t) = \bigg( \frac{\dot{a}(t)}{a(t)}\bigg)^2$$

where $a(t)$ is the scale factor which describes how the scale of the Universe changes with $t$, and $\dot{a}(t)$ is the first derivative of the scale factor with respect to $t$. Due to cosmological observations, the Universe is said to contain dark energy which causes the Universe's expansion to accelerate. What is meant by this is that at the current time $\ddot{a}(t) > 0$ where $\ddot{a}(t)$ is the second derivative of the scale factor with respect to time.

The first way we could look at your question is we could ask whether galaxies currently at a distance $D_0$ are receding faster than other galaxies that were previously at distance $D_0$.

From the definition of accelerating expansion and the Hubble parameter we can see that accelerating expansion does not imply that the answer to this question is "yes" and in fact if we assume dark energy takes the form of a cosmological constant (ignoring cosmic inflation), and we delve into the dynamics of the Universe, we find that galaxies currently at $D_0$ must be receding from us slower than the galaxies that were previously at $D_0$ were receding when they were at $D_0$. So in this particular sense the Universe's expansion is slowing down, even though we usually describe it as accelerated.

The second way we could answer your question is to ask whether the recessional velocity of any given galaxy is larger now than it has ever been in the past.

The answer to this question is more difficult as accelerated expansion does imply that the recessional velocity of a given galaxy increase with time, but the Universe's rate of expansion in previous epochs was decelerating. However, again taking dark energy as taking the form of a cosmological constant, we see that the answer is that galaxies achieve their highest recessional velocities twice: firstly at the Big Bang, and secondly in the infinite future. So the answer to this question is that galaxies are not currently receding from us faster than they have been at all previous times.

Recession velocity is different from peculiar velocity (i.e. the local velocity with respect to the CMB). We could add the two to find the 'real velocity', but as I've noted recession velocity doesn't have a direct physical meaning so what this 'real velocity' actually means is not straightforward.

Expansion is homogeneous, whereas the vacuum around a black hole is not homogeneous, so in this sense the 'sucking' of a black hole is not the opposite of expansion.


Do photons slow down as they approach the event horizon of a black hole?

Photons travel on null geodesics through space time. In the curved space-time near a non-spinning black hole, a distant observer could infer that the speed of light (travelling on a radial path) is given by
$$ v_{\rm light} = c \left(1 - \frac{r_s}{r}\right),$$
where $r_s$ is the Schwarzschild radius.

As $r \rightarrow r_s$ then we see that $v_{\rm light} \rightarrow 0$. So yes, according to an observer far from a black hole, the speed of light approaching a black hole appears to slow down and asymptotically approaches zero at the event horizon.

However, the locally measured speed of light is always $c$.


Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?

That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force pushing them up, away from the center of the Earth. The gravitational force on astronauts in the International Space Station is about 90% of what they experience on the Earth's surface, but they feel none of that.

Another relevant factor is that the Earth as a whole, along with objects on the surface of the Earth, accelerate toward the Moon (and the Sun, and Jupiter, and Venus, and ...) gravitationally. The gravitational acceleration of those surface-bound objects toward those other bodies is not exactly the same as is that of the Earth as a whole.

The difference between these accelerations results in a force that can be measured. This is the tidal acceleration. An extremely sensitive scale will show that you weigh slightly more when the Moon is on the horizon than when it is directly overhead. For a 61 kg person, this difference in weight between the Moon being on the horizon vs directly overhead is about 10-4 newtons.

Compared to the ~600 newton weight of that 61 kg person, this is a very small effect. This very small effect, along with an even smaller effect from the Sun (roughly half), are however responsible for the tides in the oceans.


A [very basic?] question about light scattering and colors

There may be a particular effect. Smoke from wildfires or cities, fine dust in the atmosphere etc. can have the effect of making the moon appear redder. However it is most likely an example of observation bias.  You noted that the moon appeared orange, and on subsequent nights you were looking for this effect. In fact it happens each moon rise. You just didn't notice it before (or if you did, it didn't make enough of an impression on you.)
There are two forms of scattering that are applicable. Rayleigh scattering, and Mie scattering. There is also absorption of some colours. Rayleigh scattering occurs as the electromagnetic wave that is light causes the molecules of the air to act like little electic dipoles. The light then bounces of these electrically active molecules. Shorter wavelengths are affected more than longer ones.
The light emitted by the sun is very slightly yellow. It is slightly reddened by scattering through the atmosphere. However it still contains a mixture of all visible colours, and so a prism will still show a complete rainbow. The reddening is not apparent to the eye, but can be measured by instruments.


Ptolemy’s understanding of the planet’s whereabouts

I suppose you mean in three dimensions, but anyhow…
Yes, Ptolemy did know that planets move also in latitude, not only in longitude, and thus the Almagest includes sections for how to predict the latitude of a planet for a given moment. In fact, Book XIII of the Almagest is entirely written about this specific subject of the latitude of the planets.


The amount of insolation (the solar irradiation a planet receives) follows the inverse-square law: half the distance to the Sun results in four times the insolation for a given surface area. However, Earth's orbital eccentricity is very low – currently just 0.0167 – which means our orbit is nearly circular, so the change in insolation is never so dramatic as a quadrupling: our perihelion is 147.1 million km and our aphelion 152.1 million km – a difference of just 5 million km – which means the Earth receives $\left(\frac {152.1}{147.1}\right)^2$ = 1.069 times (i.e. 6.9%) more insolation at perihelion.

Another effect of orbital eccentricity is that under Kepler's second law, Earth's orbital velocity is greater at perihelion than at aphelion: Earth spends about 4.7 fewer days in the seasonal quadrant at perihelion than in the opposite quadrant at aphelion. In other words, the northern winter and southern summer are slightly shorter, and the northern summer and southern winter slightly longer.

These variations do indeed affect the seasonal weather: the southern hemisphere currently has slightly hotter, shorter summers and colder, longer winters, and the northern hemisphere has slightly milder, longer summers and milder, shorter winters. 13,000 years ago, when the Earth's axis was pointing in the opposite direction, it was the reverse: slightly hotter but shorter northern summer, etc.

However, there are many other things that affect insolation and its impact on Earth, including variations in the Sun's radiation (for just one example, see the Maunder Minimum in sunspot activity), variations in albedo, scattering and absorption (e.g. greater northern hemisphere land mass means more heat absorbed; loss of Arctic ice in summer reduces surface albedo so more heat absorbed), variations in Earth's axial tilt, etc. It's very complicated!

Some of these variations are cyclic in nature. The combined impact on Earth's insolation from cyclic variations in orbital eccentricity, axial tilt (obliquity of the ecliptic) and axial direction (precession of the equinoxes) are described by the Milankovitch cycles.


Why is it so difficult to discover the earth Trojan?

2010 TK7 has a diameter of about 300 meters (1,000 ft). Its movements do not bring it any closer to Earth than 20 million kilometers (12.4 million miles), which is more than 50 times the distance to the Moon. (from the wiki entry for it)

That should make it pretty clear? Can you see a fly from 200 km away?


what is the current explanation for the formation of cosmic voids?

Void formation is due to baryon acoustic oscillations. These can be thought of as fluctuations in the dense matter in the early stages of the universe; they became anisotropies, which can be seen in the cosmic microwave background. The fluctuations grew, until they hit the 150 Mpc scale, which correlates nicely to the sizes of voids in the universe today.

There's an interesting section (4.2) in this paper by Robert Brandenberger, which gives a mathematical treatment of how fluctuations of all kinds grew in the early universe. There is a better version of the explanation in Zeldovich (1972), which in fact puts forth the theory. The number density (if you'll pardon the pun) is lesser in that latter, but the former has a good overview, as it is meant to be a broad summary.

The simple explanation is that tiny fluctuations grew over time to become enormous structures.


Could non-supernova carbon, oxygen, or silicon flashes be observed?

There is also something called a "giant eruption", like eta Carinae, which is a type of "failed supernova" because it didn't destroy the star.  It is not known what causes giant eruptions, but perhaps one thing to think about is a flash of the type you mention.  Giant eruptions certainly are visible, eta Car got quite bright when it blew off much of its envelope.


Why do some artificial satellites decrease and increase in brightness multiple times?

I do not know for sure, but you are viewing the satellites by the reflection of sunlight. As the satellite moves across the sky, the Sun-satellite-you angle changes, therefore there is no reason to expect that the brightness should stay the same.

As for why it should fade and brighten again - if you imagine the satellite to consist of a number of surfaces, these surfaces may essentially cast a beam of light onto the Earth, which move across the Earth's surface. The satellite will brighten as one of these beams crosses you. I suppose this could happen several times for the same satellite.

The extreme examples of this are the Iridium satellite that "flare up" to magnitude -8, before fading in a matter of seconds.


Could the accelerating expansion of universe inidicate we are surrounded by "denser" space?

It's a pretty clever idea and a solid question that unseen mass might cause gravity outside the observable that tugs on the universe and might be the cause of dark energy as opposed to some unknown energy that seems to permeate all space.  It's not a new idea, but it's worth answering.

There's 2, possibly 3 (or 4) pretty big problems to that approach.   The first is that Dark Energy expansion is very uniform.  For it to be explained by gravity, that gravity would need to be close to uniform around the observable universe, with Earth close to the center.    A giant "edge" as you called it, or maybe we're in a low mass bubble surrounded by higher mass.    Not impossible, but somewhat improbable that the Earth is in the center of this theoretical gravity field pulling the observable universe apart.

The 2nd problem is time.  Gravity travels at the speed of light, so this gravity field would need to have existed prior to the big bang or be very close to the edge of the observable universe.  Still not impossible, but it's a difficult theory to work with as it requires a few leaps of faith.

A possible 3rd problem is that measurements show that the universe is very flat.  I don't know if a gravity field of the sort you suggest would be flat or not, I suspect not, unless it was enormously large.

Given enough time, maybe they can find a test to detect gravitons (if they exist) or, find dark energy and answer this one with better certainty.  The question is asked here on Quora if interested.

I'll add a 4th problem.   The velocity of observed young galaxies at the edge of the observable universe is close to the speed of light away from us in all directions.  Relative velocity that high is difficult to explain by high mass outside of the observable universe cause that mass would need to be enormous, measurably greater than the mass of the observable universe.   All the factors taken into account, it's an improbable explanation

If I've missed anything obvious or broken any laws of physics with the answer above, corrections are always welcome.


I want a galaxy count database, also how do I use SDSS skyserver to fetch galaxies?

You might find the NED database and Astrometry.net useful. NED is NASA's extragalactic database. You can search by a position in the sky with a radius and it will give you a list of known objects in that field.

Astrometry.net lets you upload images and will resolve what you're looking at, also giving you a list of known objects in your image.

I like to use CDS for a quick look up on objects. This will let you browse images from different surveys easily and links you to other sources on that object as well as publications about your object.


Is it possible that some stars are already black holes yet we see light emitted from before becoming a black hole?

It depends what you mean by "see".  I assume that you mean with the naked eye.  Let's do some rough calculations to estimate the odds.

The farthest single star visible to the naked eye is Deneb, estimated distance around 1550 light years. Most stars visible to the naked eye are much closer.  Since even the sort of star that forms a black hole lives a few million years, we would expect perhaps as many as one sufficiently large star in a thousand to have formed a black hole during the time their light was on the way to us.

There are about 5000 stars visible to the naked eye under good conditions, but only a hundred or so massive enough to form black holes.  So on that basis, the odds are that there is at most one such star, and probably none.

To estimate this another way, consider that there is on average one supernova in a galaxy per century.  Comparing the distance to the farthest visible star with the diameter of the galaxy I estimate we can see at most 1/2500th of the Milky Way, so we should expect only one visible star to go supernova every 250,000 years.  Dividing that by the distance in light years to the farthest visible stars, we get only about one chance in a hundred that any of the visible stars have gone supernova during the time their light was on the way to us.

One caveat is that I've gotten these figures from disparate sources, based on Google searches.  However, it seems clear that the odds are not good.  Hopefully one of the experts can provide better data.


Why doesn’t the Sun fill the sky on Mercury?

When looking at the first image, you need to take into account perspective distortion in images with a narrow field-of-view. This is the same effect that makes people look closer together than they actually are when photographed through zoom lenses (which is useful to bear in mind when reading the news these days!). Mercury is not as close to the Sun as the first image makes it appear.

You also need to bear in mind that objects in space tend to lack "depth cues": in the atmosphere, objects further away will become blurrier and more "washed out" due to scattering. In space, this does not apply.


It is a question of where, but also to better understand, you must know when it happened. Because earls structure was different from today's.
From wiki on Universe's chronology,
By NASA/WMAP Science Team - Original version: NASA; modified by Ryan Kaldari, Public Domain, Link
you will learn that first stars are thought to appear when the Universe is about 3% of its current age, i.e. 400 million years old. This is the time it took for first atoms to cool down sufficiently and gather together in the clouds in question.
To see where first star formation happened, we have the possibility to look back in time.

we can see the star being born, and for this we need to use radio frequencies, because infrared was the only thing coming trough the surrounding gas clouds, from that time is so red shifted...
if we are very lucky, we can see these stars exploding - since they are so big, they will quickly come to an end and explode - but because of surrounding gas clouds it is regarded as a rare event.
there are simulation tentative like this one where we use currently know physic laws to calculate the evolution of matter since it first emitted light (CMB - the bluish-greenish thing on above image); the very small differences in this primordial light we are able to see allow the simulation to explain densities differences and the apparition of filaments, planes and clusters (nodes at the intersection of planes and filaments). These things can be artistic, as you can see here filaments and a Protogalaxy simulated:

from Sky & Telescope
Clusters where the denser places of the universe, and due to this highest matter density, is where first stars where created.

Here is an images where it is thought a filament is actually seen, lightened by a catastrophic event it hosts:

The Lyman-alpha blob (blue fuzz) surrounding the quasar UM 287 (white dot in center) extends 1.5 million light-years long, far too big to be contained within the quasar's host galaxy or the host galaxy's halo. The blue fuzz is likely to be part of a cosmic web filament.
S. Cantalupo / UCSC

I invite you to get a better idea of what we can see of the distribution of stars from our planet here, where you can zoom in and out and get an understanding of scales.


Retrogradation movement of planet Mars relatively from Earth by Copernic

You are right.  If Mars orbited in exactly the same plane as the Earth, instead of an S or a loop, we would see Mars moving prograde relative for the stars along the ecliptic, then slowing and stopping, moving retrograde for a few months, as Earth overtakes it, still on the ecliptic, then moving prograde again.
But Mars doesn't orbit in the same plane, so it has some motion perpendicular to the ecliptic. when these motions are combined, the usual effect is a "loop" or sometimes an "S".
(note that the axial tilt of the Earth is not relevant here)


What star is directly north of the suns north pole?

The North Pole of the sun points towards the constellation Draco. There is no bright star close to the North Pole of the Sun. The closest fairly bright star is Delta Draco, "Altais". A magnitude 3 star.

The other marks on the image are the "North Pole of the planets' orbits" The Earth's rotation is inclined by 23 degrees from its orbit, so the Earth's rotational North Pole is not the same as the pole of its orbit.


"Iron stars" are a hypothetical kind of stellar remnant - similar to a neutron star or white dwarf - that are hypothesized to form in the far future of a Universe undergoing heat death, under the assumption that protons do not decay, which is something that we do not know if is or is not the case. Some of the more promising theoretical models for proton decay (a fair number of so-called "Grand Unified Theories" [GUTs] which one should note, despite confusion in the popular press, are not the same thing as unified quantum gravity theories like string theory) have already been experimentally rejected, but that doesn't mean it doesn't occur at a rate slower than we can currently detect (or may ever be able to).

In particular, the idea rests on the fact that, given infinite time, there is no such thing as a stable element except, perhaps, for iron. What we call "stable" elements are really just radioactive elements with a half-life too long to observe, or too long to matter, whichever you prefer. For example, up into maybe about 16 years (300 megaseconds) or so ago, the element bismuth (element 83) was widely believed to be entirely stable, but then its most stable isotope was discovered to be unstable to alpha decay with a half-life of around $10^{19}\ \mathrm{yr}$ ($10^{20}\ \mathrm{Ms}$).

Feynman (I believe) once said that "everything not forbidden is compulsory". What this means is that, according to the laws of quantum mechanics, there is a nonzero probability that an agent can register any process not forbidden by, say, a conservation law, as having occurred, and will register such as having occurred given infinite time. In the case of atomic nuclei, if there is a nucleus with less nuclear energy than a given one, there is a probability to register the latter as now being the former, i.e. as having undergone a decay to it. (It cannot go the other way, because that would imply a violation of the law of conservation of energy.)

All elements heavier than iron have more nuclear energy than iron. Thus, by this principle, ALL isotopes of them are radioactive: we just don't know the half-lives. E.g. even common copper is radioactive (unstable) and will likely undergo alpha decay to iron, two atomic numbers below it. It's just that the timescale for this decay is likely so inordinately long that we cannot ever have any hope of observing it directly.

Thus, given unlimited time, anything made entirely of such elements heavier than iron will decay until it becomes purely composed of iron.

However, now one might be wondering something: clearly, stars are not made of heavier elements, and white dwarfs are typically also made of light elements like carbon and oxygen, and iron cores collapse into neutron stars if not black holes, neither of which can reasonably be called "iron", so how can we have "iron stars"? Well, it turns out that something opposite to radioactive decay will also happen for at least aggregate concentrations of light elements too: they will undergo cold fusion. Helium has less nuclear energy than 4 (protonic) hydrogens - that's why the Sun and main-sequence stars are able to perform hot fusion to a net energy gain to begin with. Contrary to what you may have heard, cold fusion is not impossible: it's just really damned slow, as in, taking far, far longer than the elapsed age of the Universe, just like it is for the decays just mentioned. But in any case, in "eternity", cold fusion will likewise raise the atomic number of the nuclei in a white dwarf, to again, converge on iron. This is an "iron star".

But, as said, this all depends on two things:

that the ultimate fate of the Universe is heat death,
that there is no proton decay.

Neither of these are things we can "know" to good confidence. In particular, there are still major open questions in fundamental physics that leave the door wide open to these being things that could go differently. In the case of the ultimate fate of the Universe, it is intimately bound up with the nature and governing laws controlling the still as-yet poorly understood "dark energy". Heat death will occur if the dark energy remains constant, i.e. acts as a true cosmological constant as per Einstein. However, if the dark energy either rises or falls with time, we may end up with two other "catastrophic" scenarios respectively called the Big Rip and Big Crunch. We have no evidence to rule these out because we don't have any regarding the physics governing future evolution of dark energy. The formation of iron stars by cold fusion and radioactive decay of so-called "stable" heavy elements is estimated to take a half-life of $10^{1500}$ (years, megaseconds, pretty much doesn't matter at this scale as that exponent itself is likely approximate).

Likewise, there are potential mechanisms - such as quantum tunneling via "virtual black holes" - that could still cause the decay of protons with half-lives on the order of $10^{200}$. A detector of the type we use now to look for proton decay (basically a big tank of water) could not detect these directly, as it would require many orders of magnitude more mass than there is in the current observable Universe and, moreover, would collapse into a black hole long before one got that far.


If any object could become a black hole, could any object become a neutron star?

There can be no such thing as a "supermassive neutron star". The theoretical upper mass limit for a neutron star is somewhere between 2.2 and 3 solar masses. Any more massive and they inevitably form black holes. So I am not clear what kind of "neutron objects" you were thinking of?
Nor is it clear what you mean by "non-stellar" objects that will have the densities required to make neutron-degenerate matter? There aren't any apart from (I) the cores of massive stars at the ends of their lives. (II) Massive white dwarfs if they accrete matter over and above their Chandrasekhar limit. Furthermore, there is a minimum mass for a neutron star. Although the observed lower limit to those seen in nature is (so far) around $1.15 M_{\odot}$, there is a theoretical lower limit of about $0.1$ to $0.2M_{\odot}$ set by the stability of the neutron against decay in self-gravitating material ( See https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174 ).
It is true if you could arrange to compress any matter to densities above $\sim 10^{15}$ kg/m$^3$ it would form neutron-degenerate material. But this requires (as far as we know) the conditions I listed above.


Of course this is possible. Whether it will fill its Roche lobe depends on the masses and radii of the stars and the distance between the stars. If the distance is big enough then the second (less massive) star will not fill its Roche lobe and continue evolving as a single star would. The (large) majority of binaries will fall in this category, only close binaries will evolve into cataclysmic variables.

For instance the Alpha Centauri/Proxima Centauri system will not evolve into a system where mass transfer takes place via the Roche lobes as the distance between Alpha Cen A/B and Proxima Centauri is much too big.

Even when the stars are very close, a supernova might not occur if the masses of the stars are low. The supernova occurs when the mass of the existing white dwarf increases (because of the accretion) to a mass surpassing the Chandrasekhar limit (1.4 solar masses). If the transferred mass is too low then no supernova will happen.


Purple and green star exists. For example, a blue supergiant star can have its surface temperature to 50000 K, which peaks at purple in the blackbody spectrum. However, our eyes are more sensitive to blue than purple, and the contrast between blue and purple in the spectrum is insignificant, therefore we see blue.

Also, check this article: https://parade.com/295384/marilynvossavant/can-stars-be-green-or-purple/