How does neutron star collapse into black hole?

A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place.
See Chandrasekhar limit on wikipedia for details.

A neutron star is formed during a supernova, an explosion of a star that is at least 8 solar masses.

The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star, and then into a black hole.

We know that 1 electron + 1 proton = 1 neutron;

1 neutron = 3 quarks = up quark + down quark + down quark;

1 proton = 3 quarks = up quark + up quark + down quark;

A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star.

During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back.

During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons.

In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons.

If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.

The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole.

When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged, non-rotating-charged, non-rotating-uncharged.

What we would see visually during the transformation would be a hard radiation flash.
This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs).

We know that atoms break up into protons, neutrons, electrons under pressure.

Under more pressure, protons and electrons combine into neutrons.

Under even more pressure, neutrons break down into quarks.

Under still more pressure, perhaps quarks break down into still smaller particles.

Ultimately the smallest particle is a string: open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings.

Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene.

A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity.

We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate, giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa.

So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses).

This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole.

A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star, thus gaining mass.

Here is a nice drawing showing exactly that.

Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies.


How to calculate a meteor's entrance altitude to Earth's atmosphere?

There is a example here of the process required. But basically you need observations of the track from two locations and then just perform some relatively simple trigonometric triangulation. The Nemetode network regularly does this using images from a network of cctv cameras.


how to calculate the half-mass radius and tidal radius of a (simulated) globular cluster

Your first question is about computing, not astronomy. You find a cumulative histogram of distance from the centre of mass and where that reaches 50% of the stars.
As to your second point, there is no tidal radius; you need two bodies in order to have a defined tidal radius. The tidal radius of a cluster is directly related to the gravitational potential external to it.
Stars can escape from a cluster because sometimes they can accumulate enough kinetic energy in weak interactions with each other, that their speed exceeds the escape speed of the cluster. It is a classic result that the "evaporation" timescale is of order 100 times the cluster relaxation timescale, even in the absence of an external gravitational field (e.g. Gerhard 2000).


There are a bunch of different measurements for celestial coordinates, but in regards to the solar system - or specifically the Sun as its "master" - there's the ecliptic celestial coordinate system.

In that system, centred around the sun and our solar system by extension, the North pole is in the constellation Draco. According to Wikipedia, the coordinates from Earth are right ascension 18h 0m 0.0s (exact), declination +66° 33′ 38.55″.

Source


Is there an upper limit to the mass of terrestrial planets?

I'll shamelessly reference an answer I wrote on Worldbuilding to an almost identical question. Lammer et al. (2014) suggested that "super-Earths" with masses of $2$-$5M_\oplus$1 could retain massive hydrogen/helium envelopes, up to $\sim10^{25}$ kilograms. Above this, up to about $10M_\odot$ or more, "mini-Neptunes" exist, possibly composed of volatiles and having significant gas envelopes2. In other words, there's a transition region between terrestrial planets and gaseous planets (gas dwarfs included) that depends significantly on whether the planet can hold onto an envelope for a significant amount of time.

Part of this depends on the environment. For instance, you cite Kepler-10c as a possibly upper limit to terrestrial planets. The reason Kepler-10c can't be a gas planet, as Dumusque et al. (2014) explain, is that it's way too close to its parent star to retain a hydrogen envelope. It would likely lose a lot of such an atmosphere via Jeans escape, the same mechanism that leads to chthonian planets. It's possible, then, that Kepler-10c would have been an ice giant if it had formed farther away from Kepler-10. It has a similar mass and potentially similar composition; it just happens to be much closer to its parent star than Neptune.

This is something of an unsatisfying answer, but there simply isn't an exact line that separates terrestrial planets and gas planets, just like there isn't a good line between massive gas giants and, say, sub-brown dwarfs.

1 There are, of course, exceptions, like Kepler-138a, which has a mass similar to Earth yet is a good gas dwarf candidate.

2 For more information on the transition region, see e.g. Lopez & Fortney (2013).


What is meant by sun's path through zodiac

The stars are not visible during the day, but they are still there (we just can't see them because of the brightness of the sun). We can't see the stars, but we can calculate where they would be if we could see them.

Over the course of the year the sun appears to move relative to the stars. This is actually due to the orbit of the Earth around the sun. The apparent motion of the sun is quite simple: it moves around the sky in a great circle, always following the same path, and taking exactly one year to complete. The route it takes is called the ecliptic. And most of the constellations that lie on the ecliptic are called the zodiac (they are very important to astrologers, but not especially important to astronomers).

The sun's route is not visible to the naked eye from Earth, but it can be calculated with great precision.


What happens to the 99.9 % of the sun-rays that do not fall on any planets or any other celestial body?

The light from the Sun spreads, at least initially, in an isotropic fashion into the universe.

As it gets further from the Sun, some of that light will interact with the interstellar medium (ISM) and therefore some of the energy emitted by the Sun will be used to excite atoms and molecules or even ionise some atoms. This will be the fate of almost all the light which is emitted from the Sun towards the plane of our Galaxy, which contains sufficient molecular gas and dust to block starlight travelling through it for any distance. We know this happens because we can "see" dark clouds in the Milky Way, that can be penetrated by longer wavelength radiation to reveal all the billions of Sun-like stars that lie behind them. Roughly speaking, about half the visible light from the Sun will be absorbed every 1000 light years when travelling in the Galactic plane, so it is essentially all absorbed within a few thousand light years.

But most of the Sun's light is not travelling in the direction of the Galactic plane, and interstellar and intergalactic space has a very low density of gas and dust. The equivalent extinction number for the intergalactic medium is that light travels many billions of light years with almost no chance of being absorbed (see Zu et al. 2010). This means that most of the light from the Sun will travel to cosmological distances (billions of light years) over the course of the next billions of years. Indeed light emitted from the Sun shortly after its birth has already travelled 4.5 billion light years. We know this has happened and will happen, because we can observe galaxies (the light from which is nothing more than the summation of light from many stars like the Sun) that are 4.5 billion (and more) light years away.

As the Sunlight travels towards cosmological distances, its wavelength is "stretched" by the expansion of the universe, becoming redder and redder. We know this happens because distant galaxies have redshifted spectra. If the universe keeps expanding, then its density will continue to decrease and there is little to stop the radiation from the Sun travelling on forever, with a wavelength that scales as the scale factor, $a$, of the universe.

If we follow a co-moving and co-expanding cube containing the Sun's radiation as the universe expands. The total radiative energy inside that cube diminishes as $a^{-1}$ - that is, the energy content of the universe in the form of radiation from stars (and other sources) becomes energetically less important as the universe expands and appears to be being superseded by the energy contained in the vacuum itself (a.k.a. dark energy).

In conclusion, most of the energy emitted by the Sun is not "used" for anything; it propagates into space, becoming more and more dilute.


No. The precise details of collapse of a supernova are not perfectly understood, in the way we understand (for example) the orbits of the planets. However the energy comes from the release of gravitational potential energy. The "relaxation of spacetime" is not significant. We don't understand supernovae perfectly. But we have a good model of them.

You should re-read Rob's answer to your previous question, not least because he got the order of magnitude of the energy release in a supernova correct. He also points out your error in thinking of the gravitational field changing "in an instant"

Concerning the big bang. No, as has been often pointed out, the big bang was not a explosion. There is no analogy between a supernova and the big bang.


How Long Will Earth's Year be When Our Sun Goes Red?

The year length depends on the distance between the planet's centre & the Sun's centre, not the Sun's surface. So if the Sun merely expands, Earth at 1 AU will still take a standard year to perform 1 orbit.

However, when the Sun becomes a red giant, it won't just expand. As Wikipedia mentions, red giants shed a considerable amount of mass in the form of gas and dust. It's estimated that the Sun will lose around a third of its mass over the first billion years of its red giant phase, and eventually have a mass around half of its current mass; see here for details.

It's still not clear exactly what will happen to the Earth while all this is going on. It may become engulfed, like Mercury & Venus certainly will be. But even if it's not engulfed, interacting with all that gas & dust is likely to affect its orbit.

If we ignore that complication, we can easily calculate the period of a 1 AU orbit around a Sun of reduced mass. The Newtonian form of Kepler's 3rd law says that
$$T^2= \left(\frac{4\pi^2}{GM}\right)r^3$$
where $T$ is the period, $G$ is the universal gravitational constant, $M$ is the Sun's mass, and $r$ is the mean orbital radius.

If we keep $r$ at 1 AU but halve the Sun's mass, then the period is multiplied by $\sqrt{2}$, giving us a year length around 516 days. That's using the current day length, by then the rotational period of Earth will be considerably longer.

I seriously doubt that the Earth will still be at 1 AU from the Sun, though, so please don't take that figure of 516 days too seriously!


What is the apparent magnitude of the Sun at that distance?

Fortunately, there's a formula known as the Distance Modulus which is defined to calculate precisely this information. The equation is given by

$$m-M=5\log_{10}(d_{pc}) - 5$$

where $m$ is the apparent magnitude of your star, $M$ is the absolute magnitude, and $d$ is the distance from the star, in units of $\mathrm{parsecs}$. The absolute magnitude of the Sun is easily looked up and found to be $M_\odot = 4.83$. The distance, you provide as $2.5\:\mathrm{AU} = 1.2120342\times10^{-5}\:\mathrm{pc}$. Now its simple to solve for $m$.

$$m = 5\log_{10}(1.2120342\times10^{-5}) - 5 + 4.83$$

$$m = -24.75$$

As far as apparent magnitudes go, this is still pretty bright. You can see a table of typical apparent magnitudes for comparison right here. Interestingly, this table lists $-25$ as the "minimum brightness that causes the typical eye slight pain to look at". At $m=-24.75$ (remember higher value = less bright) you'd be just above this threshold and so the Sun would be safe to look at without causing eye strain or pain.

  Is that bright enough to blot out any of the inner planets?

In order to determine the the brightness of the inner planets, we have to do some more math. Unfortunately, people don't really calculate the absolute magnitudes of planets so the form of the distance modulus listed above isn't entirely useful. Instead, we'll use a different form.

$$m_1 - m_2 = -5\log_{10}\left(\frac{d_2}{d_1}\right)$$

This equation relates an object with an apparent magnitude of $m_1$ at a distance $d_1$ to its apparent magnitude $m_2$ at a distance $d_2$. If we look up the apparent magnitude of the various inner systems from Earth, and figure out their distance from Earth at those times, we can figure out the apparent magnitude at the distance of your asteroid.

I'll work this out for Venus since Venus will likely be one of the brightest of the inner planets (given its high albedo). Venus as viewed from Earth, at its brightest, has an apparent magnitude of $m_1 = −4.89$. This occurs when it is an the point of greatest elongation. As determined from this solar system orrery, the distance is approximately $d_1 = 0.75\:\mathrm{AU} = 0.36\times10^{-5}\:\mathrm{pc}$. On your asteroid, you'll be about $d_2 = 2.17\:\mathrm{AU} = 1.05\times10^{-5}\:\mathrm{pc}$ from Venus at the same point of maximum brightness (based on a little trig). From this we can calculate $m_2$ to be

$$m_2 = −4.89 + 5\log_{10}\left(\frac{1.05\times10^{-5}\:\mathrm{pc}}{0.36\times10^{-5}\:\mathrm{pc}}\right) = -2.57$$

From the comparison table listed above, this would make Venus, at its brightest, about as dim as the "minimum brightness of the Moon". This is a rough calculation, but I don't expect the actual answer to vary too much from this value.

Keep in mind, that's the maximum brightness of one of the brightest inner planets. Any other time Venus will be dimmer and all other planets will never be brighter than this. So all in all, the inner planets will likely be outshined by the much brighter Sun. There's a chance, with the right conditions, if you managed to properly block the light from the Sun using some sort of occulting disk or flower starshade, that you could see the inner planets with the naked eye as they still would be pretty bright in their own right. You just need to get rid of the sunlight outshining them.


What is behind the Great Red Spot's longevity?

Standard answers, as found  here, (partial quote)

Unlike Jupiter, the Earth has land masses that cause major storms to
lose energy due to friction with a solid surface. Without this
feature, Jupiter’s storms are more long-lasting. However, the Great
Red Spot is long-lived, even by Jupiter standards. Researchers don’t
quite understand why, but we do know that Jupiter’s storms that are
located in cloud bands with the same direction of rotation tend to be
longer lasting.

So, we know some of the probable reasons, and don't know the full story yet.  Keep in mind that the current RedSpot might be a 3-sigma event, even for Jupiter.


How did they estimate the mass of Sgr A*'s companion G2 without knowing what it was?

To put it simply, you can't.

Both cases are explained in these two papers:

Binary: https://arxiv.org/pdf/1410.1884.pdf
Just a Cloud: https://arxiv.org/pdf/1112.3264.pdf

In both cases you have to assume some kind of model to account for the detected fluxes in different near-infrared bands.

In the "Just-a-cloud" case you would need a cloud with 3 solar masses and a temperature of about 550 K.
In the "binary-merger" case you would need a 2 solar mass main sequence star covered in gas and dust (possibly generated by a recent merger with a companion, which, however, would somewhat distort the 2 solar mass estimate).


It does not have a retrograde motion. It has a prograde rotation, and a prograde orbit of the Sun.

If you were to look down on the solar system from far above the north pole you would see the planets orbiting the sun counter-clockwise. And the planets, including the earth also spinning counter clockwise. The Moon also orbits the Earth counter-clockwise.  All this motion is in the same direction, and called prograde.

There are a couple of exceptions: Venus spins clockwise. Venus spins in a retrograde motion. There is a moon of Neptune, Triton, that orbits retrograde. Uranus spins on its side.

The reason is that most objects rotate the same way is that the planets and the sun formed out of the same cloud of gas and so have the same direction of spin.

There is another sense of "retrograde". As the Earth overtakes the other planets they seem to move backwards in the sky for a while.

From the inner planets, Earth would appear to move rather like mars, with periods of retrograde motion. From the outer planets Earth would move like Venus, moving from being a "morning star" to an "Evening star".


Is the surface of TrES-2b actually dark?

I don't know, how it works there, but perhaps it is easier to imagine, how it could look.

TrES-2b is a gas giant, so it doesn't have a well defined surface. When you are somewhere in the atmosphere then:

-When looking up, you will see the light, as deep sea creatures do. The amount of light and its color will depend on how deep you are.

-When looking down, it should be dark (black), comparing to how it looks up. Otherwise, the atmosphere would be reflecting light.

-When looking to the sides, it should also be dark, for if the light were scattered at some angles, it would also be scattered upwards.

So, one simplest possible model is: like in deep sea.

Note, that at some other wavelengths, particularly in infrared, it all can look very differently.


How is magnetic field is generated over the sun?

The Sun's interior is partly or completely ionised and therefore supports the flow of electric currents. The solar magnetic field is explained by a solar dynamo that converts the kinetic energy of electrically charged fluid flows in the Sun (due to differential rotation and convection) into electromagnetic energy.


The expression "speed of time" doesn't make much sense, since speed is the rate of change of position with time, and "time" doesn't have a position. Hence it doesn't have a speed.

If you want to know the amount of time dilation at a given speed, it is answered by special relativity. $t' = t\gamma $ where $t$ and $t'$ are the amount of time that passes in the two frames, and
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} $$

At low speeds, $\gamma$ is approximately 1, and at high speeds it approaches $\infty$.

To fit with your notion of the speed of time as being the relative rate of passing of time in the two frames, an expression for the speed of time would be $t/t' = 1/\gamma$, which is close to zero at high speeds, and close to 1 for low speeds.

So the speed of time at low speed is equal to 1.


Presenting / printing Right Ascension and Declination values

There are no standards for representing right ascension and declination apart from the abbreviations RA and Dec, and the symbols $\alpha$ and $\delta$.

The use of decimal degrees, dms, °'", hours and decimals, hms etc. depend on the [author's] style and intent. For example, determining transit times from local sidereal time and right ascension is easier when RA is expressed in hours, whereas doing calculations of angular separation is easier to calculate in software when decimal degrees are used.


Can't see any detail on planets with my new Celestron NexStar 130 SLT Computerized Telescope

Mars is a small planet and currently far from Earth (just over 1 Astronomical Unit). Therefore, it currently appears to be very small -- smaller than 9 arcseconds. At that size, it will look like a small disk.

Based on this one observation, I think that no conclusion can be made about any possible problem.

Also, a 130 mm f/5 scope is rather short for planetary observing where you want high magnification. This scope is better for wide fields of view, such as the Pleiades or Double Cluster.


What observations will be done during the New Horizons flyby of Ultima Thule?

Emily Lakdawalla in her Planetary Society blog post of 17 December describes fairly well what we can expect:

A few precious data downlinks in the days before and after the flyby will contain some small photos from New Horizons’ Long Range Reconnaissance Imager (LORRI), which takes grayscale snapshots. Lower-resolution color comes from the Multicolor Visible Imaging Camera (MVIC), a component of the Ralph instrument. The first images that we get back from New Horizons will not be the best images that New Horizons has taken.

Because of the uncertainty in the position of MU69...

... the real position of the body may be completely outside the field of view of a photo focused on the predicted position. In order to make sure it doesn’t miss the target, the spacecraft will have to scan its instruments across a region of space in which it is most likely to find the tiny world, taking a lot of data on empty space to make sure to get data on the object.

She shows what to expect in the days around closest approach, using comet 67P images as an example:

The Core phase will last 10 days, from 24 December to 2 January. Nearly all of the science will happen in just a 2-day period around closest approach, with intense imaging and spectroscopy to map MU69, study its composition, figure out its rotation rate and pole, and search for dust, rings, and satellites
Several things will make the MU69 flyby significantly more difficult than the Pluto flyby. MU69 is small, dark, and faint. New Horizons won’t see it as more than a dot until 3 days before closest approach. MU69 is darker, and the sunlight dimmer, than the LORRI camera was designed for. Also, because MU69 is so small, New Horizons has to pass much closer to it than it did at Pluto in order to see interesting detail on the surface, resulting in a rapidly changing viewpoint near closest approach.
As a result of all of these factors, LORRI’s images may be a little smeared and will be noisier than we saw at Pluto. MVIC is less affected by these challenges; the way that MVIC works, it’s possible for New Horizons to compensate for the low light and high speed by sweeping more slowly across MU69 to build up a stronger signal.

Other measurements to be made are:

Toward the end of 29 December, science data gathering will really begin. A dense set of observations will gather a lightcurve on MU69, a measure of how its surface appears to change with rotation, over a period of about 6 hours.
30 December will see another 6-hour block of science, a mix of color imaging, deep satellite searches, and lightcurve data. The same happens early on 31 December, after which New Horizons enters the most intense, Inner Core phase of the flyby.

You should read the blog post for all information, because she gives much more detail than can reasonably be quoted here.
She also quotes more resources.
Here is another simulated image of Ultima Thule, showing how the primary camera aboard NASA’s New Horizons spacecraft will “see” the Kuiper Belt object as it approaches for a 1 January flyby:
Source: Johns Hopkins University Applied Physics Laboratory
Here is the operations time line overview for the approach:

(source)
That flyby distance decision has now been made: the spacecraft will remain on course for a 3500 km flyby.


What mechanism causes oscillations of the solar system's orbit about the galactic plane?

The cause for the oscillations perpendicular to the galactic plane is the gravity of the non-spherical mass distribution (needed for a plane Kepler ellipse) in the Milky Way. Simplified, there is a dense galactic plane. The density is not exactly known; therefore there is some uncertainty (a few million years) about the precise oscillation period. Details see this article, subsection 3.3.

The idea of a correlation of mass extinctions with this oscillation isn't new, it originates back probably to about 1970, or earlier.

"The outer solar system probably does not contain a large gas giant planet, or a small, companion star", see this press release.

We are probably not orbiting around a spiral arm.

The dark matter disk is a hypotheses, an idea to investigate. Usually only a small fraction of this kind of hypotheses can be confirmed definitively later, most of them can be ruled out after some time, some remain unsolved, some can be refined to match observations.


Which 16 antenna locations were used this MeerKAT radio image?

Lorenzo Raynard, spokesperson for SKA Africa was gracious enough to send me the map with the sixteen sites below, and I've duplicated the dots on a larger format map.

It's been quite a while, probably the array has grown since this "first-light" image.

Next step (strictly for fun) will be to try to make a UV map for this configuration.


Does a merging massive binary black hole ‘emits’ more than one gravitational wave?

We can currently only detect gravitational radiation when it is extremely intense: in the last fraction of a second. For example the first gravitational wave detection lasted less 0.15 seconds. The black holes are releasing gravitational radiation with every orbit, but that radiation is too weak for us to detect. It takes a colossal amount of energy being released for it to be detected by current technology.  The inspiral of GW150914 released 3 solar masses of energy, nearly all in the final 0.15 seconds before merger.  Even so this only distorted space by less than 1/10000 of the diameter of a proton.  It is this that makes it so difficult.


Is there a theoretical minimum mass main sequence star that can become a red giant?

Perhaps you are interested in the lower mass limit (given in the comments as 0.3 solar masses), and why that lower limit exists.  This is similar to the question of why stars become red giants in the first place.  You are right that given the age of the universe stars less massive than 0.8 solar masses won't have had time, so the lower mass limit is theoretical.  But what sets it?

The answer is, red giants happen because stars can exhaust all the hydrogen in their core, turning it all into helium.  Then there is no nuclear fuel to keep the star in equilibrium, so no balance between the light that leaks out (which sets the luminosity) and the fusion rate (which mostly just responds to that luminosity).  As a result, the core shrinks and heats, and fusion initiates in a shell around the core-- a shell that used to be too cool to fuse.

The temperature in that shell, and key aspects of its fusion rate, are determined by how small and massive the core becomes (the core gets smaller with time because it is losing heat, and it gets more massive because helium "ash" keeps being added to it from the shell).  Eventually the free electrons in the core become what is called "degenerate", meaning they approach their quantum mechanical ground state, and that implies the core gets about as small as the Earth. This also means the shell is extremely hot, and the fusion rate goes essentially berserk.  This requires the envelope to puff out into a red giant.

Hence, to have a red giant, you must have a degenerate core, and you must have shell fusion surrounding it.  But stars less massive than 0.3 solar masses have an internal structure that is purely convective.  Thus they don't run out of hydrogen in their cores until they run out of it everywhere  (and like you say, none have yet done this in our universe, it's a theoretical expectation).  That is why they cannot have shell fusion, and cannot become red giants.


How would rainbows appear on other planets?

Rainbows occur when sunlight shines through rain. This is rare in the solar system. Rain (of sulphuric acid) might be common enough under Venus's clouds, but there is no sun. Conversely, there is plenty of sun in Mars, but no rain, and only very rare clouds.

It rains on Titan: methane rain. Methane has a lower refractive index than water (1.27 instead of 1.33), which would make the rainbows slightly larger (though not by much 42->52). However the atmosphere of Titan is hazy, and while there is some light on the surface, the sun's disc is not visible.

There is rain in some layers of the gas giants, but again not on the outer layers where the sun is visible.

It is likely that the Earth is the only place in the solar system where rainbows are a common phenomenon.


Is dust extinction/reddening caused by Rayleigh scattering or some other physics phenomenon (or both)?

In an astronomical context, Rayleigh scattering does not take a huge place.
The extinction is mostly due to absorption and scattering, from gas and dust. Its effect is to diminish the flux from the source (that is, a change in magnitude). Please, take care this depends on the observed wavelength!

For simplicity, we can assume the extinction factor $Q_{ext}$ as:

$Q_{ext} = Q_{scattering} + Q_{absorption}$

Where $Q_{scattering}$ is in turn composed of many elements.

As the OP above said, blue light generally suffers of higher extinction than other frequencies, so causing reddened observed light. This is due to the cross-section dependence on the frequency: if the light has a wavelength that is comparable to the size of the grain dusts (high frequency), it is more effectively absorbed, while in the opposite case it can pass through undisturbed.
This is the reason why obscured fields are observed in radio (in our galaxy, especially) or in infrared (extragalactic).

So, the answer to you question, in space is: Rayleigh scattering is not the most responsible of the extinction/reddening, you have to invoke absorption/scattering by dust and gas.
For comparison, here some explanatory numbers about the extinction efficiency in different contests:

$Extinction \sim\lambda^{-1}$

$Thomson \sim\lambda^0$

$Rayleigh \sim\lambda^{-4}$

You basically would need impressive amount of gas to account for the extinction just by Rayleigh scattering.
Instead, Mie scattering is more important in this context. Here a figure to quickly grab the concept (from here):

And here a very popular plot about extinction versus frequency from here:

Of course, if you observe from ground, you should take care of Rayleigh scattering as well.

Other references:
Lecture 1
Lecture 2


While not absolutely impossible this is not a likely situation. The reason is that for a planet to be a "giant", then it will have enough gravity to hold on to its hydrogen and helium. Conversely, if you remove the H an He from a planet, it won't be "giant". There are only two gasses that are common enough in the universe to form a massive gas mantle, and those are Hydrogen and Helium.


Does lack of atmosphere on the moon make earth-shine more, or less, bright?

The lack of a lunar atmosphere doesn't affect Earth's albedo at all, since albedo is only a measure of the diffuse reflection of a body.
However, the apparent brightness of the Earth as viewed from the surface of the Moon would be affected if the Moon had an atmosphere.  This, of course, would depend on the atmospheric composition of the Moon.
If the atmospheric composition of the Moon was such that it absorbed visible light, then the Earth would appear dimmer, or perhaps not even be directly visible (similar to how we can't directly see the Moon when the weather is cloudy here on Earth).
If the atmospheric composition of the Moon absorbed almost no visible light, and was highly refractive, the Earth-Shine might be overall brighter than if the Moon had no atmosphere at all. The refraction would bend some of the Earth-Shine into the surface of the Moon which otherwise would have passed by. This picture shows the analogous situation for Earth and the Sun:


There is a python package called Skyfield that loads, reads and interpolates the binary forms of the JPL Development Ephemerides or DEs for you, and does everything else you need to get the absolute best results possible from them. If you can use even a tiny bit of python then this would be the way to go rather than trying to figure out how to interpolate JPL's "special flavor" of Chebyshev polynomials.

See these in Space Exploration SE:

What is the exact format of the JPL ephemeris files?
How is the Chebyshev method used by JPL?
Is there a way to extract the Chebyshev coefficients for a body from a SPICE kernel?
How to obtain Chebyshev coefficients directly from the JPL Horizon's interface?
I've almost learned to spell Chebyshev, why has JPL switched to Hermite interpolation for DE438?

and also see how easy it is to do things like this with Skyfield in this answer to Sunset on [Mountain] from [Viewing Point]

Also this answer to How to calculate the planets and moons beyond Newtons's gravitational force? shows how I used JPL's Horizons interface output (which also interpolates the DEs using Spice) to compare my numerical orbit integrator results to the ephemerides.

And just for fun see How to pronounce “Ephemerides”? in English SE.


Questions related to the reflection of the moon

The moon is already a pretty dark grey. It only reflects about 12% of the light that hits it. It is about the colour of (worn) tarmac on the road. You can see how grey it is in NASA's gif from a million miles away.

"Black" paint reflects less light. Common household black paint reflects only about 5% of light. So the moon would be rather less than half as bright as it is now. But you could still see it.  There are some specialised black substances that reflect as little as 0.035%. A moon coated in these would actually be hard to see.

White paint and silver paint reflect about 80% of light, so if the moon were white it would be about 7 times brighter, but much less bright than the sun which is about 400000 times brighter than the full moon.


Is extra-planetary phosphine actually a new discovery?

You're correct; extraterrestrial phosphine is not actually a new discovery. As you said, we've known for decades (see e.g. Ridgway et al. 1976) that phosphine can and does exist in some gas giant atmospheres, including Jupiter's. However, detailed modeling has been done of the associated photochemistry beginning at around the same time, and as far as I'm aware, the important thing is that the observations aren't in conflict with the models.
The recent claim of phosphine detection (Greaves et al. 2020) went beyond all of that because as it indicated that phosphine appeared to exist in higher concentrations in the Venusian atmosphere than could be produced by known processes. From what I can tell, the group behind the detection was fairly exhaustive in their attempts to account for the concentrations by other means - lightning, known photochemical processes, meteorites, etc. All of them fall short by quite a lot, by orders of magnitude in most cases.
That's the key: It's not that phosphine was claimed to be detected, but that too much phosphine appeared to be detected. All that said, doubt has been cast on the results. Recent work (Snellen et al. 2020) found that the method used by Greaves et al. involving a 12th-order polynomial can also lead to numerous false positives when applied to other spectral features near 267 GHz, the frequency of the claimed detection. The reanalysis of the data implies a feature detected at a signal-to-noise ratio of 2, which is certainly too low to imply a statistically significant result.

(As an aside: the paper you cite (Pasek et al. 2011) isn't an observational paper - that is, it doesn't report an actual discovery of phosphine on Titan. Instead, it describes theoretical modeling and some laboratory work involving phosphorus chemistry. It's one thing to say that "Hey, here's how phosphine might participate in chemical cycles in this environment" and another to actually confirm the existence of that compound.)


Do Electric Charges in the Van Allen Radiation belt move in Opposite Directions?

Yes and no. The question @JohnRennie links to in his comment  Analogy between magnetic bottle and Van Allen's radiation belt contains a lot of relevant information, but packaged as an answer to this question.

The question correctly concludes that, at least in the plane perpendicular to the magnetic field, particles will travel in circles, and those circles will be in opposite directions, but they do not, generally simply circle around the Earth as you suggest. There are two reasons for this: firstly the magnetic fields are too strong, or the particles too light, so the circles you get are much smaller and secondly because they are also moving in the third direction. So you should think of them travelling in fairly tight spirals around the magnetic field lines, which, roughly speaking run between the North and South magnetic poles.

As they approach the poles, the field gets stronger (the lines bunch up) and it turns out that this repels the particles, so they end up bouncing back and forth between the two poles (all in spirals around the magnetic field lines).  The protons and electrons do spiral in opposite directions.


I don't know whether this  picture is a fake, but you could get this sort of image without cheating, if it were taken a long way away from Rio, and the magnification cranked up.  In other words, get far enough away from Rio that its angular diameter is about 1 minute, and take the picture from there.


Possible to make a cheap reflective objective lens with a mirror?

That won't work because the shape will be too distorted to be useful. For astronomical purposes, a good quality mirror is needed since the objects you look at are dim and small. If you want a usable astronomical telescope then the optics need to be high quality

If you want to illustrate the properties of telescopes, a simple refracting telescope can be built using the lens from (low power) reading glasses as the objective and a (high power) hand lens as the eyepiece. I've also tried with a shaving mirror, but found the distortion to be too great to form a useful image.


Is "Zero Shadow Day" necessary to perform Eratosthenes' experiment?

It's not necessary that there be zero shadow.  The altitude $\alpha$ of an object when it crosses the local meridian on the celestial sphere is $\alpha = \delta + 90^\circ - \lambda$, where $\delta$ is the object's declination and $\lambda$ is the latitude of the observation point.  This means that if you carefully measure the altitude of the same object at meridian crossing from two separate locations, and find altitudes $\alpha_1$ and $\alpha_2$, then the difference between these altitudes is
$$
\alpha_2 - \alpha_1 = \delta + 90^\circ - \lambda_2 - (\delta + 90^\circ - \lambda_1) = \lambda_1 - \lambda_2.
$$
Thus, the difference in the measured altitudes is equal to the difference in latitude between the two locations.  By measuring the distance between these locations on the Earth's surface, one can then infer the circumference of the Earth.
The condition that there be "zero shadow" simply means that one of $\alpha_1$ or $\alpha_2$ is equal to 90° (i.e, the sun is at the zenith.)  Effectively, it allows you to know what one of the angles is without physically measuring the height of your gnomon and the length of its shadow.  But so long as you're willing to measure the solar altitude in two different locations, it is not necessary that the sun be at the zenith.
Note that this method can be used to find the difference in latitude between any two points on the Earth, so long as the object's declination doesn't shift significantly between the meridian crossings at the two points.  The reason it's helpful to have the observations at two points on the same longitude line is that it leads to a simple relation between the great-circle distance between the points, the difference in latitude, and the radius of the Earth.  If the observation points aren't on the same line of longitude, then you need to know the difference in longitude as well to perform this experiment;  and Eratosthenes didn't have Harrison's chronometer at hand.


What is the temperature about 10 or 15 feet below Mars' surface?

This is one of key questions that NASA's Mars InSight probe is trying to answer but it has only recently arrived and starting taking data and it will be a while before the data is analyzed and published in peer reviewed journals. The "mole" which would burrow below the surface and make the temperature measurements encountered some initial issues and NASA and the instrument's builders were running further tests and looking at those data before proceeding according to this mission press release.


At what distance wouldn't the Sun cast a visible shadow anymore?

This depends on eye biology, so there's no purely astronomical answer.  I'll note that Venus is just about capable of casting a shadow (at greatest elongation, with ideal conditions) so let's say that a magnitude -5 object is at about the biological limit of casting a shadow.
The sun has a magnitude -27, so that is 22 magnitudes brighter than Venus.  5 magnitudes is a factor of 100 in brightness, so 20 magnitudes is a factor of 100000000 (100 million)  and 22 magnitudes is about 600 million.
But light follows an inverse square law, The square root of 600 million is about 25000.  So at about 25000 times the distance of Earth, or 0.4 light-years, the sun would be as dim as Venus at its brightest and perhaps marginally able to cast a shadow for someone with acute vision, on a good matt white surface. This puts our astronaut in the Oort cloud.
There's a good deal of uncertainty in this estimate, based on eye acuity. So let's say "something between 0.1 and 1 light-years"


It would be more accurate to say that a black hole eventually turns whatever falls in it into equal quantities of matter and antimatter.

The difference between matter and antimatter is defined by some conserved (or approximately conserved) quantum numbers. For example, matter baryons and quarks have positive baryon number, while antibaryons and antiquarks have negative baryon number. Leptons (e.g., electrons, neutrinos) and their antiparticles have a corresponding lepton number instead.

Semiclassically, a black hole evaporates through Hawking radiation. For astrophysically large black holes, this will be overwhelmingly be in terms of photons, which are their own antiparticles, but once the black hole evaporates to a small enough size, it will also radiate massive particles as well. Because there are no long-range forces that couple to baryon number or lepton number, a black hole is under no obligation to conserve them, and thus should radiate equal amounts of matter and antimatter, regardless of what fell into it.

This is unlike, say, energy-momentum, electric charge, and angular momentum, the presence of which makes a difference in the long-range gravitational and/or electromagnetic fields. Thus the total amount of charge the black hole will radiate should match the total amount that fell into it, etc.

Unless, of course, quantum gravity gives us more surprises, which is definitely possible.


Is it possible to create an artificial Black Hole?

There has been a lot of speculation, whether it's possible with particle accelerators to create micro black holes. This is only feasible, if space is higher dimensional on very short distances. This kind of black holes would be very short-lived due to decay by Hawking radiation. This kind of black holes could be detected by their decay products.

There seems to be no evidence, that black holes have formed in a particle accelerator.

In standard 3d space the smallest black hole needs the energy-equivalent of more than $10^{19}\mbox{GeV}$ of the Planck mass of about $20\mu\mbox{g}$ to form. That's far beyond the energy a particle collider can focus onto a single particle collision.

Even the most energetic cosmic rays are several orders of magnitude below the needed energy.

LHC provided up to 8 TeV (8,000 GeV) for proton-proton collisions, scheduled are 13 TeV for the upgrade. That's 15 orders of magnitude below a smallest 3d black hole, but might allow short-lived black holes for higher-dimensional space. This way LHC can constrain the structure of space on short distances.


Are rogue planets ever born in isolation?

The minimum mass of a "planet" forming from a gas cloud (definitions of what a planet is are rather slippery, and some would say this is not a planet at all) is not determined by the time available. The collapse process is rapid - less than a million years. There is a minimum mass though, and what you are referring to is something known as the fragmentation limit.

A cloud becomes unstable and collapses if its mass exceeds the Jeans mass. The Jeans mass depends on temperature to the power of 3/2 and inversely on the square root of the cloud density.
$$ M_J \propto \frac{T^{3/2}}{\rho^{1/2}}$$
When a cloud of gas collapses, its density increases. If it is able to radiate away heat efficiently, then its temperature can remain more-or-less constant and so the Jeans mass decreases. This allows the cloud to fragment into smaller pieces.

The minimum mass that can collapse in isolation will therefore be set by the smallest value that the Jeans mass can attain as the collapse proceeds. This fragmentation limit is in turn set by the cloud becoming opaque to its own radiation, which occurs when the density becomes large enough. At this point, the cloud can no longer efficiently get rid of all the heat that is generated in its interior by the work done by gravity in squashing it. The temperature rises and the Jeans mass stops decreasing. Now, the cloud may still collapse, but it won't break into smaller chunks.

The fragmentation limit is difficult to calculate with any accuracy, because it depends on the 3D turbulent dynamics of a collapsing cloud and also on whether the cloud is spinning. It is generally thought to be in the range one-to-a-few times the mass of Jupiter (e.g. Whitworth & Stamatellos 2006). This is far below the minimum mass for hydrogen fusion of about 75 Jupiter masses or deuterium fusion of about 13 Jupiter masses.

Evidence that such objects may exist can be found in answers to the related questions How are rogue planets discovered? and Is there any hard evidence that rogue planets exist?


What is the minimum mass required so that objects become spherical due to its own gravity?

Icy objects, such as most in the Kuiper belt can reach an equilibrium if they are about 400km across, whereas the rocky asteroid Pallas, at 572km clearly has an irregular, non spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical.

Rock tends to be stronger than ice. Rocky objects are able to withstand their own gravity for longer than icy ones. Pallas is a reasonable cut of point. The next smaller asteroids (Vesta, Hygiea etc) are round-ish, but not in hydrostatic equilibrium. On the other hand Small, icy moons such as Miranda and Mimas are in, or close to equilibrium. Mimas has a diameter of just under 400km.


How far apart can two people watch the moon simultaneously?

If we take 6371.0 km as the mean radius of Earth, an apogee of Moon of 405503 km, and a perigee of 363295 km, we get ratios of 6371.0 km / 405503 km = 0.01571 = sin 0.9002° resp. 6371.0 km / 363295 km = 0.01754 = sin 1.005°. So on both sides of Earth between $$40030\mbox{ km}\cdot 0.9002°/360°=100\mbox{ km}$$ and $$40030\mbox{ km}\cdot 1.005°/360° =112\mbox{ km},$$ with an average circumference of $2\pi\cdot 6371\mbox{ km}=40030\mbox{ km},~$ get lost of the 20015 km of the half circumference of Earth as zone of visibility.

The calculation is simplified to the simultaneous visibility of a point at the distance of the center of moon. More precisely one part of the moon can be visible from one observer, while an other part of the moon is visible from a second observer. This adds between 29.3 and 34.1 arc minutes or between 54.3 (apogee) and 63.2 km (perigee) to the diameter of the zone of visibility.

The visibility of the Moon is influenced by additional factors like air temperature, height of the observer above sea level or the geographic position by Earth's flattening. But this applies also for the visibility of the sun.

The corresponding calculations for the sun: With aphelion of Earth of 152098232 km and perihelion of 147098290 km, we get ratios of 6371.0 km / 152098232 km = 0.000041887 = sin 0.0024000° resp. 6371.0 km / 147098290 km = 0.000043311 = sin 0.0024815°.
So on both sides of Earth between $$40030\mbox{ km}\cdot 0.0024000°/360°=0.26686\mbox{ km}=266.86\mbox{ m}$$
and
$$40030\mbox{ km}\cdot 0.0024815/360°=0.27780\mbox{ km}=277.80\mbox{ m}$$
get lost of the 20015 km of the half circumference of Earth as zone of visibility.

(For a rough estimate you may take the 100 km loss of the moon visibility, and multiply it by the quotient of the distances Earth-Moon and Earth-Sun roughly equal to 400,000 km / 150,000,000 km = 0.002667, to get 266.7 m visibility loss of the sun.)

This is again simplified, and applies to the center of the sun. The appearent diameter of the sun varies between 31.6 and 32.7 arc minutes, adding between 58.6 (aphelion) and 60.6 km (perihelion) to the diameter of the zone of visibility for observers looking at different parts of the sun.

Depending on which scenarios to be compared, the lost area varies. As an example, a loss of a radius of the visible zone of the moon in comparison to the sun of 100 km corresponds to a roughly cylindrical area of $$100\mbox{ km}\cdot \pi\cdot 40030\mbox{ km}=12.6 \mbox{ million square kilometers}.$$

How far can two observers be apart at maximum, to see two differnt parts of the moon at the same time? $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} + 54.3 \mbox{ km} = 19869.3 \mbox{ km}$$ (apogee) for the average Earth radius. For the same part of the moon it's $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} = 19815\mbox{ km}.$$


Why do brighter stars/planets appear physically larger?

It is an optical illusion. Our brains associate brighter with bigger.  This is how software like stellarium are able to make convincing views of the night sky: they play this in reverse and use larger "stars" for brighter ones.  Your brain sees them the same.
Your brain is not tuned as a scientific device, there are lots of tricks that we can play on our visual system.


How do I reduce planetary photometry to standard viewing geometry using phase angle?

Phase functions are nonlinear in general, but Pluto presents such a small range of phases to Earth that a linear approximation will do:
about 0.03 mag/deg for Pluto and 0.08 mag/deg for Charon, or 0.04 mag/deg for the two together.
The Stern and Tholen book summarizes the estimates up to 1997.
A search for Pluto phase coefficient
yields some of the papers behind those and more recent estimates.


Could a Trappist-like planetary system exist around a sun like ours?

There are a couple of systems around solar-mass stars that follow the pattern of tightly-packed planets orbiting close to the star. These include:

Kepler-11 (0.961 solar masses), a system of six low-density sub-Neptune planets orbiting within 0.5 au of the star
HD 10180 (1.06 solar masses), a system of six Neptune-size planets of which four are within 0.5 au of the star.
Kepler-90 (KOI-351) (1.2 solar masses), a system with eight planets within 1 au of the star ranging in size from Earth-sized to Jupiter-sized.

In the case of HD 10180, planet g is located in the liquid water zone, though its mass indicates that it is an ice or gas giant planet and thus not habitable. Kepler-90's outermost planet is located at 1 au, but the star is more luminous than our Sun so it might be too hot to support liquid water and is in any case a Jupiter-sized planet. So systems like TRAPPIST-1 with packed inner planetary systems clearly exist around stars like the Sun, although the difficulty of finding small planets at sufficient distance from the star means that so far we haven't found such systems with terrestrial planets in the liquid water zone.
The question of how many planets you could pack into the habitable zone has been investigated by Kane et al. (2020), "Dynamical Packing in the Habitable Zone: The Case of Beta CVn", where they investigate the maximum number of planets that can dynamically fit in the habitable zone by spectral type, and also consider in detail the specific case of the G-dwarf Beta Canum Venaticorum. From the paper:

Our analysis
presented here demonstrates that in fact the dynamical limitations to the packing of HZ terrestrial planets
is ∼5 planets for most spectral types, and ∼6 planets
for stellar masses $\gtrsim 0.7 M_\odot$. Packing 7 planets in the
HZ is possible within certain specific stellar mass and
architecture regimes, but becomes vulnerable to MMR
perturbations that compromise the dynamical stability
of such configurations.

(MMR = mean motion resonance)
Whether real systems actually reach such a dense packing in the habitable zone is unknown, and they note some caveats to their analysis:

There are some caveats to the analysis presented here
that are worth repeating. First, Section 2.1 describes
the use of 3 Gyr isochrones, which are adopted in order
to include stars at the higher mass end of the stellar
mass range. However, the star will evolve with time,
as will the HZ (Gallet et al. 2017), resulting in an increase in both the width and orbital distance of the
HZ. Subsequently, planets located at the inner edge of
the HZ at 3 Gyr will transition into the Venus Zone
(Kane et al. 2014) where runaway greenhouse scenarios may become dominant. Second, our simulations account for Earth-mass planets and circular orbits. Higher
masses and non-zero eccentricities will reduce orbital
stability within the HZ, but lower mass planets may allow for additional planets, with the exception of strong
resonance regions.

(HZ = habitable zone)


What are the characteristics of solid carbon dioxide on the Martian poles?

On Earth we know of frozen carbon dioxide as "dry ice." If shaved, it could exist as "dry ice snow." On Mars the dry ice snow could precipitate, then become compressed into dry ice. It would probably be a stiffly packed granular solid.
I had only heard of "water ice" on Mars. Check the article on kottke.org  Link:  https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars


What is the difference between aurorae and electroglow?

Aurorae are caused by charged particles (usually electrons) being drawn into the polar atmosphere by the planet's magnetic field.
Airglow is caused by just photons interacting with the atmosphere.
Electroglow requires both electrons and photons.
Source: The Free Library  (article is under express copyright, see the paragraph that starts "Voyager 2")


MeerKAT's Epic Image of the Galactic Centre, what are we seeing?

LaRosa et al 2000 present a 90 cm VLA image and catalog of sources near the galactic center.
Here is their schematic diagram with the 2°x1°, 18-33 cm band MeerKAT image region highlighted.

The compact source Sgr A* is a small dot within the bright, pear-shaped blob Sgr A right of center.
Recent millimeter VLBI images (Issaoun et al 2019) indicate a width on the order of 200 μas, or 1/7000 of a pixel in the image you uploaded.

The bright pair of blobs left of center, Sgr B1 and Sgr B2, are massive molecular clouds.

Most of the bubble shapes are either supernova remnants (SNRs) or H II regions ionized by UV-bright young stars.
The far left pair Sgr D has one of each.

The various non-thermal filaments (NTFs) align with the local magnetic field and emit synchrotron radiation but are not well understood otherwise.
The bundle of filaments near the center of the image are known as the galactic center radio arc (GCRA) or just the Arc.


There is a relationship between $E(B-V)$ and the reddening in any other colour. The exact value depends on the type of dust, the extinction value itself and the intrinsic spectrum of the star (as does the 3.0 coefficient mentioned in your question).

However, for the purposes of estimation, the canonical relationship is $E(U-B) = 0.72 E(B-V)$ (e.g. Pandey et al. 2003).

Thus you can calculate $E(U-B)$, then $E(B-V)$, then $A_V$, then $V_0$ and use this with $M_V$ to get the distance.


Yes you can find them anywhere on Earth. The thing is that by the time they reach the Earth's surface, these meteorites are very small, usually ranging from microscopic to dust sized particles. Searching How big does a meteor have to be to make it to the ground?

  In this sense, most all meteoroids that enter the atmosphere make it
  to the ground, in the form of microscopic dust.

As you said it is easier to find meteorites in snow because it is white and contrasts with the usual colour of meteorites. It would be very hard to try and distinguish those small particles with the normal soil on Earth. As for finding one randomly in a concrete building... the chances are very low and it possibly would have done some noticeable damage to the building.

A rather simple experiment to prove that yes you can find it anywhere on earth, follow this example. This is to search for micrometeorites. I have done this before and I found it to be particularly inefficient as it is rather wind in my area, but hey you might find it different.

You can attempt to search for meteorites somewhere else but just don't expect to see large chunks of it lying around.


Why aren't other kinds of energies considered in the mass-energy of the universe?

Various missions such as WMAP and the Planck Satellite have measured the mass-energy content of the universe. You tend to see images like the one below produced by these scientific ventures.

Credit: NASA/WMAP Science Team

This breaks the entire mass-energy content of the universe into three buckets: Dark Energy, Dark Matter, and (the mass of) atoms. But what about the rest of the universe. Surely there's other mass-energy out there such as the energy in the curved space-time metric, or kinetic energy of atoms, or radiation energy in photons, or the mass-energy of neutrinos! Have the astronomers and physicists forgotten about those things? Well the answer is no, they're just not telling you about it because it doesn't matter.

Equation for the universe

The universe's content and evolution is defined by something known as the Friedmann equation (technically there's a set of equations, but that's not relevant here). One important and useful form of this equation is given below.

$$\frac{H(a)^2}{H_0^2} = \Omega_{0,R}a^{-4} + (\Omega_{0,b} + \Omega_{0,c})a^{-3} + \Omega_{0,k}a^{-2} + \Omega_{0,\Lambda} +\ ...$$

This equation may seem complicated, but let me explain the parts. The left hand side more or less is the expansion rate of the universe in the past (or future) compared to the present day expansion rate. The right hand side is the sum of all the various contributions to the mass-energy of the universe. The important parameters here are the various "energy density parameters", denoted by $\Omega$. I've included the parameters for radiation ($\Omega_{0,R}$), baryonic matter ($\Omega_{0,b}$), cold dark matter ($\Omega_{0,c}$), space-time curvature ($\Omega_{0,k}$), and dark energy ($\Omega_{0,\Lambda}$), but you could add more energy density parameters as well. The ultimate point I'm going to show is that there are only a few energy density parameters that are actually important or significant and the rest can be ignored completely.

Thermal/Kinetic Energy

You mention things like thermal and kinetic energies (which are effectively the same thing since thermal energy just comes from the total kinetic energy of a collection of particles). We can't and don't count these because there's no way to absolutely define the thermal or kinetic energy of something. To define kinetic energy of a given mass requirements specifying it's velocity, but that velocity must be defined with respect to a particular reference frame. Special relativity tells us that there is no absolute reference frame though, which means I could define your velocity with respect to me, or the Sun, or the center of the Galaxy and get a different kinetic energy for you with each definition.

Radiation Energy

Radiation energy is just the energy in light. The universe is flooded with photons (a very large portion of which are CMB photons) and these photons have non-zero energy. But that energy is almost completely negligible in the current universe. From WMAP results, it is estimated that $\Omega_{0,R} = 5\times10^{-5}$ which equates to contributing about $0.005\%$ of the Universe's mass-energy budget. It would be a tiny, tiny sliver in the above graph. Interestingly, this wasn't always the case. If you run the universe backwards, you find that very early on, $\Omega_{R}$ was the dominant source of mass-energy in the Universe.

Neutrino Energy

Similar to radiation energy, neutrino energy is negligibly small. In part, this is because the neutrino mass is so tiny. I don't have an estimate for $\Omega_{0,\nu}$ but I can say it's probably as small or smaller than $\Omega_{0,R}$ and equally insignificant.

Gravitational and Chemical Energy

These types of energies are what's known as potential energies. They're a weird sort of energy because they're not actually real. They're potential. It's the energy you would get if you performed some action (e.g., moving a body through a field). If you hold a ball up high, you might say that ball has gravitational potential energy, but the ball doesn't actually have that energy, it just has the potential to gain energy when you drop it. We simply say it does have that energy to make balancing energy more tidy and simple. When you do drop the ball, it gains energy by falling through the gravitational field and taking energy from it. So really, your question boils down to talking about the energy inherent in background gravitational fields. The same is true for chemical energy. Chemical processes such as extreme oxidation, aka fire, release energy, but that energy comes from breaking atomic bonds and that breaking of bonds involves converting electric potential energy to something else.

Ultimately you just want to add in energy from the quantum fields that permeate the universe, $\Omega_{QF}$. I can't estimate this number, nor have I ever even seen someone try, but I can almost assure you that this number would be extremely tiny, even more so that the previous considered energies. Again, we find it's negligible and would by the tiniest mote of a sliver on the above pie chart.

Space-Time Curvature Energy

There's energy in space-time itself. This is known from General Relativity. This space-time energy of the Universe is denoted by $\Omega_{k}$. The interesting thing about this is that it appears the universe is flat! This implies, to decent precision, that $\Omega_{0,k} = 0$. This energy is more than negligible, it is non-existent (to within observational precision).

Conclusion

I'm sure you could think up a lot of different types of other energies you might want to include in this, but hopefully you see the point I'm driving at. They're all completely negligible. The arguably largest contribution comes from radiation and that's been shown to be $0.005\%$ of the Universe's mass-energy budget. What we see is that the Universe is dominated by Dark Energy, has a bit of Dark Matter, a small amount of atoms, and negligible amounts of the rest of everything else. And overtime, the Dark Energy portion of that pie chart will get bigger and bigger until we'll be able to throw out atoms and dark matter from the plot as well since they'll be just as insignificant as neutrinos and radiation are now.


Formation of Cosmic Microwave Background

We observe the temperature of the CMB as a ~2.7 K blackbody, but that's the redshifted version we observe. The CMB is also know as the "surface of last scattering" at the point of  recombination when nuclei and electrons combined to form neutral atoms the universe went from opaque to transparent.

This happens at a temperature of ~3000 K. From this we can estimate the redshift (z~1100) of the CMB which corresponds to an age given our cosmology.


If you're buying the instrument, as opposed to making it from scratch, my advice is to worry about the overall quality, as opposed to a single parameter such as mirror thickness. It boils down to whether you trust the vendor or not.

In any case, there is no simple formula. You could run the PLOP simulator and try a few different scenarios, and see what the overall RMS error is in each case. But the values will vary very significantly. The main factors are:

the diameter / thickness ratio
the number of support points in the mirror cell
the relative positions of the support points

All of the above assumes a traditional mirror with a flat back, and a classic mirror cell with multiple-of-three support points. But the analysis gets even more complex if it's a cell mirror (the back is not flat), or a honeycomb structure (the back is flat but there are empty spaces in the volume of the mirror). And then there are conical mirrors, etc.

It's difficult to second-guess the vendor, especially if you've never built an instrument. Just read the reviews, ask other owners, try and gauge the overall build quality.


Is there a database of all known star names/identifiers?

The closest service to what you are describing is the SIMBAD Astronomical Database from the Université de Strasbourg/CNRS. At the time I write this post, it contains 10.8M objects and 35.5M identifiers. It does not have a single CSV you can download with this information (to the best of my knowledge, and I've asked), but there is an API and TAP service available for programmatic access.


Can Earth host another natural satellite than the Moon?

Short answer:   It might be impossible "long term".  It's problematic because of the size of the Moon and the proximity to the Sun.

Long answer:

The Earth-Moon system in empty space could hold several more moons (like the Pluto-Charon system).    The problem is proximity to the sun in addition to the size of the Moon.    Planets closer to their star have a harder time holding onto moons in part cause the Hill sphere is smaller, in part cause the tidal forces are that much bigger (and those 2 factors may go hand in hand, not in addition to).   But it's not just coincidence that Mercury and Venus don't have moons, they're also less likely to hold onto Moons.

Mars has a smaller Hill Sphere than Earth due to it's lower mass but it's 2 moons are very tiny.  Also, one of Mars 2 moons might not be there that much longer.  It may crash into Mars in as little as 100 million years.

Earth's proximity to the Sun gives the Earth a Hill Sphere of about 1.5 million km, and the true region of stability (long term) is about 1/3 to 1/2 of that, so about 500-750,000 km which isn't all that much further out than the Moon is currently.

Our Moon's average distance is 384,000 km, but our moon has a slightly eccentric orbit so it's distance varies between about 363,100 and 405,700 km.  Source.

The 2nd factor is the relative mass between the Moon and the planet.   Mars has 2 moons but it's moons are very small and they have very little gravitational effect on each other.    Our Moon is 1/81st the mass of the Earth which is the largest Moon to planet ratio in our solar-system.

Using the simplified hill sphere formula, the cube root of (the ratio of the masses/3).   Basically the cube root of (1/243) gives Earth's moon's a hill sphere of about 16% it's distance to the Earth.   Any Earth orbiting object that passes within the Moon's hill sphere is obviously not going to be in a stable orbit.   In fact the region of instability probably extends well past that 16%.   How far past, I'm not sure, but I can safely say that anything even remotely close to the Moon's orbit couldn't orbit the Earth for any period of time.

Using 16%, the clearly unstable region is between 300,000 km and 470,000 KM.   It would be impossible for an object to orbit the Earth at a greater distance than the Moon.   Solar and Lunar perturbations would be too big.

If it's possible for Earth to have a 2nd moon at all, it would need to orbit quite close to the Earth.   Probably well inside 300,000 KM.  The closer to the Earth the more the Earth's gravitation would be dominant, so if I was to guess, I think the Moon would need to be quite close, maybe in the 50,000 km range - but that's just a guess.

For comparison, Jupiter's 3 moons in orbital resonance; Ganymede, the largest of the 3 has a hill sphere is about 3% it's distance from Jupiter.   The 1:2:4 orbital resonance works out to the 1.5th root of 2 or about a 1.59 to 1 orbital distance ratio.  Those 3 moons also have very nearly circular orbits.    If the Moon's orbit was nearly circular, there might be a stable resonance 1.59 times closer to the Earth, but because of the Moon's orbit is measurably eccentric, I don't believe the 2:1 orbital period ratio would be long term stable because there's too many wobbles in it.   I think your best bet for a stable orbit for a 2nd moon would be very close to the Earth, probably in the 50-100,000 KM range (but that's a rough guess).

It's worth noting that "long term stable" isn't a precise term cause there's no clear cut-off.  A thousand orbits?  A million?   A billion?   Lifespan of the Sun?

I realize that "maybe, maybe not" isn't an answer, but I wanted to touch on some of why it's difficult.  A captured satellite couldn't get captured into a relatively circular near earth orbit because the incoming velocity is too great.  In theory, if there was enough debris, a satellite could form close to the Earth and perhaps be stable for relatively long term, but the Moon would to some extent prevent such a formation in a similar way that Jupiter prevents the asteroid belt from coalescing into a tiny planet.   (the low mass of the asteroid belt plays a role in that too, perhaps a bigger role), but Jupiter is a factor. Article here


Constant interval between pulses from all pulsars

The variation comes from the Earth's movement around the Sun. When the Earth is moving away from the pulsar in question, each subsequent pulse has slightly further to travel (up to 30km for a 1 second pulsar on the ecliptic down to zero for a pulsar located due North or South) and so is slightly delayed compared to the one before. Six months later, each subsequent pulse has slightly less far to travel, and so arrives a little earlier compared to the one before. The variation ranges from zero at the North or South poles to 1 part in 10000 on the ecliptic (this just comes from the ration between the Earth's orbital velocity and the speed of light).


Does the Milky Way orbit around anything?

The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull.

Actually, both the heavier and the lighter object orbit around their common center of mass. It's just that the heavier object doesn't move much (has a tiny orbit), while the lighter object moves a lot (has a wide orbit).

E.g. our Sun actually orbits the center of mass of the whole solar system, but that motion is tiny, it barely budges.

In the case of a double star, where both partners have about the same mass, you can clearly see how both are making similar orbits around their common mass center.

  Sun Orbits Around Sagittarius A* which us center of Milky Way.

With galaxies, including ours, it's a little different.

There is no super-heavy thing at the center, around which everything else is orbiting. Not even the very large black hole at the center of our galaxy is heavy enough for that.

Rather, galaxies are clumps of matter that create a common gravitational field. Stars, and everything else, are trapped in this common field and orbit around the common center of mass.

  So the question is that is that is Milky Way is orbiting around some object or perhaps Black Hole.

Same idea. There is no single point-object nearby massive enough for our galaxy to "orbit" around it.

Our galaxy, along with Andromeda, and a handful of other galaxies, are bound together in what is known as the Local Group. Each galaxy is moving within the common gravitational field of the whole group. The Local Group has a diameter of about 10 million light-years.

The Local Group is part of a larger structure, the Virgo Supercluster, which is about 100 million light-years in diameter and has at least 100 galaxies. However, the Virgo Supercluster is more "loose" - it is not gravitationally bound together.


The size of cosmic dust grains is in general given not by some size, but by a size distribution. The only direct measurements of such a distribution are made on dust collected on plates of satellites, which is of course a very local measurement. When we think that distributions look similar — though not exactly alike — in other locations of the Universe, so far away that we will never be able to go there, it is because we know how a given distribution and composition affects light traveling through an ensemble of dust grains.

A particle has a given probability of absorbing a photon with a given wavelength, and in general this probability peaks around wavelengths of the order of the size of the particle (for small particles). Thus, if we know how the spectrum of some light source looks if there were no dust around (and we do for many sources, e.g. stars), then the difference between the known intrinsic spectrum and the observed spectrum can be modeled assuming some distribution and composition. Often the composition needs not be assumed, but can be constrained from the emission of the dust at infrared wavelengths.

Usually, the model that fits best the observations is a steep power law of the form $P(r)\propto r^{-a}$ with an index of roughly $a\sim3.5$; that is, the probability of finding a small grain is much larger than the probability of finding a large grain (more precisely, for $a=3.5$, grains of size $r=x$ is $10^{3.5}\simeq3\,000$ times more common than grains of size $r=10x$, and $10^7\!$ more common than grains of size $r=100x$).

For very large grains, the probability of absorbing a photon becomes independent of the wavelength of the photon. Whereas the small grains as described above have a "color preference", large grains are said to be "gray". This is the case for boulders, rocks, pebbles, and even sand-grain-sized particles. Thus, if a cloud consisted of such particles, the spectrum of a background source would simply be diminished by a constant factor at all wavelengths. This is very rarely observed — rather the sources are diminished much more at the short wavelengths than at the long wavelengths, as expected if there are more small grains than large grains.


Stellarium 0.10.4: planet orbits change over time?

Probably this is due to the fact that orbits of the Earth and Mars are inclined to the ecliptic differently so our viewing angle of the orbit of Mars changes slightly over the year.  The program is probably showing the orbit of Mars projected onto the Celestial Plane as seen by us on the one specific date. Note, this is different from the ephemeris orbital data which gives the apparent position as seen by us at each date/time.   Just a guess though.


You can use one of the digital sky surveys. Examples include:

Sloan digital sky survey SDSS
ESO online digitized sky survey

Their images contain visible (red or blue) wavelengths as well as infrared. All images are monochromatic, as are almost all professional astronomy photographs. You could build your own composite color image from the different channels.

Here is an example image from the ESO DSS2 archive in red wavelength:


Why does gravity make time go slower and warp light?

Sometimes things we think of as different are actually just different forms of the same thing. Electric fields and magnetic fields are forms of each other. That is why we now call it the electromagnetic force.

Likewise, energy and mass are also just different forms of the same thing. We can even convert it too.
E = mc^2 ...

And from what I understand, space and time are connected together as well. So it gets called spacetime. If you want to think about it that way, gravity works by warping space, like a heavy marble sitting on a fluffy blanket. The fluffy blanket is spacetime. The heavy marble is a black hole or other super massive object. So now light is still going "straight" along space (the fluffy blanket) but it is bent because the blanket is bent. (Caveat: this is where i am conceptualizing light as a wave, not photon packets. It is actually both.) And if gravity bends space, of course it bends time too, because space-time is connected somehow as the same substance.


Is the boulder on the peak of Tycho Crater the core of the impactor, or is it a random rock?

It is just a rock.
A complex crater like Tycho is formed in several stages as the rock behaves like a fluid. The initial impact completely destroys the impactor and excavates a large cavity in the moon, and ejecta is shot out to all sides.
Powerful shock waves push the rock to the side building large walls. As these push down on the edges of the crater, the centre is pushed up forming a central peak. Further faulting and cracking will continue to modify the crater and the peak.
It is possible that this is part of the breccia that falls back to the surface having been shot upwards in the initial impact, but that seems very unlikely. It is more probable that this is part of the ridges of the central peak that cracked and fell into the shallow valley on the top of the central peak, which formed during the faulting as the peak was pushed up from the crater floor. In all likelyhood, this is a piece of moon rock. As it hasn't been exposed to solar radiation for billions of years, it is comparatively light. It can't be a piece of the impactor, as that would have been completely destroyed


Can equatorial mounts be used at lower latitudes?

Let me answer myself: Yes, they can. I got a 150mm x 1400mm Newtonian with a EQ mount and I am able to align its axis... even if it had no scale, it can be done with a cellphone (I have already done it... both for pointing north and adjusting latitude).
WARNING: Need to be careful that the weight doesn't go off-balance.


By how much does Haumea's fast rotation affect its surface gravity?

The numbers are going to be approximate because Haumea's size is uncertain.  It may have a ring.  If it does, then it's smaller, if it doesn't, then it's larger.
The formula is simple enough.   For force:    $F= \frac {m v^2} {r} $ where v is the rotational velocity and r is the same r in the gravitational formula.   or $\frac {v^2} {r}$ if you want the reverse acceleration.
Haumea has a rotational period of 3.915 hours or 14,090 seconds (I'm rounding) and an average radius of 780,000 meters (assuming it has a ring, I'm going to go with that assumption), but an estimated equatorial radius of about 1,050,000 meters and that's the number we need.  Divide the larger radius by number of seconds, multiply by 2 Pi and the velocity is 468.2 m/s.   A tiny bit faster than Earth's equatorial velocity.
Some Maths, the lifting force on the equator is $ \frac {468.2^2} {1,050,000} $ = $ 0.208 \frac {m}{s^2} $, slightly over half it's listed equatorial surface gravity of $ 0.401 \frac {m}{s^2} $.
That's a pretty high ratio.   Possibly the highest in our solarsystem for objects that are massive enough to be spheroid shaped by their mass and gravity.   There are asteroids that are thought to effectively have negative gravity at their equator, held together by the stickiness of the material, but those are a lot smaller.


is there any theory or observational evidence that our universe is electrically neutral or not?

By observation, gravity dominates the universe on large scales.  If there were a significant disparity in positive and negative charges then we would expect electromagnetism—which is approximately $10^{39}$ times more powerful than gravity—to dominate.  So by this observation we can conclude that the universe is approximiately electrically neutral.  Exactly why this should be so is not well-understood, and ultimately ties into the matter-antimatter problem, aka the baryon asymmetry.

Protons and electrons are produced through distinct processes.  That we have many protons is not particularly remarkable, as this would come about by conservation of baryon numbers as other particles decayed.  Why we have any electrons at all is a fair bit more mysterious.

In the early universe, we understand the energy was well above the electron-positron pair production threshold, and so the early universe had large amounts of both electrons and positrons.  When the universe cooled below this threshold, the production would have stopped and we would expect the electrons and positrons to then collide and annihilate each other, leaving essentially none of either.  But we obviously see large amounts of electrons: approximately as many electrons as protons by the aforementioned observational evidence.  And we also do not see large amounts of antimatter—positrons in particular.  If there were large amounts of positrons out there, we would expect to see tell-tale signatures in space as they annihilate with regular matter (or in any weak force interactions, such as would be evident in supernovae), and we have no such observations.

So somehow the early universe must have produced significantly more electrons than positrons.  In other words, we have strong observational evidence that there must be an asymmetry in physics between matter and antimatter production.  There are ways to work this into the theory, but the Standard Model by default does not support it, and experts have not really settled on any one particular modification.


How do I understand a brown dwarf with a M-type spectrum?

The spectral type of an object is almost entirely determined by the temperature of its photosphere. ie Saying something is type M3.5 is just a measure of its surface temperature. An M3.5 brown dwarf is at a very similar temperature to an M3.5 star.

Brown dwarfs begin their lives as hot balls of gas and gradually cool with time. They start off as M-type objects and then cool to become L-type and then eventually, T-type objects. Stars on the other hand, start their lives by cooling, but stabilise their temperatures at a mass-dependent value that corresponds to an M-type classification or even as cool as type L2 for the very lowest mass stars.

Given that stars and brown dwarfs appear to have been born throughout Galactic history, then a range of masses are possible for any spectral type. In order to use the surface temperature (or spectral type) as an indicator of whether something is a brown dwarf, then you also need to know its age. In general that is not known, unless the object can be demonstrated to be part of a cluster of stars (of known age).

In order to be a brown dwarf the mass should be below about $0.075 M_{\odot}$, but unless the object is in a binary system, this cannot be measured. So what people do is trust the models(!) and compare the luminosity (or estimated temperature) with model evolutionary tracks at a given age. This gives an age- and model-dependent mass that can be used to claim that an object is a brown dwarf.

The fact that brown dwarfs can have an M-type spectral class when they are young, but then move through L- to T-dwarfs as they get older is not a problem, it is a confirmation that the evolutionary models are (roughly) correct!

A plot might help (from the work of Burrows et al. (1997)). This shows the evolution of temperature with time. Each track represents a brown dwarf of a given mass (the tracks from top to bottom) are masses of 0.2 - 0.08$M_{\odot}$, representing stars (in blue). Then come masses of 0.07 to 0.015$M_{\odot}$ representing brown dwarfs (in green, that will never ignite hydrogen in their cores to any extent). Then below this are the "planetary mass objects" from 14 Jupiter masses to 0.3 Jupiter masses at the very bottom (in blue - the definition here is that these objects don't even ignite deuterium in their cores).

The horizontal lines mark the boundaries of the spectral type classifications (M at the top, L in the middle, T at the bottom).
Notice how the stars level-off in temperature (when they begin H-burning) but can have M-type or even L-type classifications, but the brown dwarfs cool throughout their lives. Even objects as low as $0.01M_{\odot}$ begin their lives as "M-type" objects.

To answer some of your more specific queries:
The warmest spectral type below which you can almost guarantee that an object is a brown dwarf is $\sim$ L2. Any object warmer than this could be a star.

A brown dwarf is always cooler than a low-mass star of the same age.

Current observational estimates are that there is about 1 brown dwarf for every 4 low-mass stars - the weighted mean ratio for stars between 0.08 and 1 solar mass to brown dwarfs from observations of several clusters is $4.3 \pm 1.9$, according to Andersen et al. (2008); the commonly adopted/assumed "Chabrier Initial Mass Function" has 4.9 stars for every brown dwarf.

The spectrum of a brown dwarf and an M-dwarf star at the same temperature are very difficult to distinguish. The M-dwarf is only slightly bigger and this might lead to small gravity-dependent differences. Lithium is not burned in brown dwarfs less than $0.06 M_{\odot}$, whereas it is burned completely in low-mass stars on a mass-dependent timescale. Thus an older (than say 1 billion years) M-dwarf star would certainly have burned its lithum, whereas a younger M-type brown dwarf of lower mass would still have lithium.


Why telescope mirror glass shipped as randomly broken chunks of a constant size in cardboard boxes?

The material used for the mirror is indeed E6 Borosilicate manufactured by Ohara corporation as pointed out by @notovny. The fabrication process of a borosilicate glass honeycomb  mirror has remained almost unchanged since the first one, an 8.4m LBT mirror, was cast in 1997 (here is the article of the casting). Quoting the article:

E6 is a good match to our process because it is manufactured in appropriate volumes with good quality control, and because the broken surfaces of the chunks melt together smoothly without discontinuities.

This somewhat more modern article (2019) goes deeper into detail:

A significant advantage for our process is that Ohara delivers the glass as blocks broken out of a set of one-ton melts. All surfaces are pristine fracture surfaces which melt together seamlessly in the honeycomb mold, leaving no trace of the original blocks.

And this work from the Arizona University explains further:

...E6 borosilicate glass... is melted in clay pots,..., with vigorous strirring that yields high homogeneity and low striae. It is supplied to us in the form of roughly cubical blocks with smooth, uncontaminated fracture surfaces and weighing typically 4 kg each.

So using pristine chunks of E6 borosilicate seems to greatly reduce the possible contamination (in comparison to using powder or not pristine chunks), seems to melt rather very smoothly and is probably an easy size to transport. The reason of why they use this material is explained in depth in all 3 articles mentioned.


Your scenario isn't stable.   A simple way to explain this is to imagine that the planet's orbit each other at the same rate they orbit the star (your scenario has them orbiting even slower).

At the same rate of orbit, the synodic period essentially approaches infinity.  (see diagram of the Moon's synodic orbit).

When this happens, the inner planet is at L1 to the outer planet and the outer planet is at L2 to the inner planet.   That's ofcourse, not quite right as the different masses would have different Hill Spheres, and the mutual gravitation of two objects massive to have a barrycenter outside the more massive planet would combine into a slightly faster orbit, but it's close enough to demonstrate the instability.

The L1 and L2 points are on the border of the Hill Sphere and well outside the true region of stability.

No system can be stable where the Moon (or binary planet system) orbits around each other slower than they orbit around the central star.   They need to be at least 50% to 67% closer in than that would allow.

Using the Orbital Period or "T" squared = distance ("a" semi-major axis) cubed, per Kepler's laws, the upper limit for a stable orbital period of a moon around a planet or a binary planet would be about 19%-35% the period of the planet's year.    Any longer the Moon or binary system would risk instability.

You might be able to set up a synchronous system where the planets orbited each other 4 times for every orbit around the sun, where they'd end up in the same (basically eclipse) position every perihelion with the planet experiencing greater tides (smaller and/or more fluid planet) closer to the sun at perihelion.  I suspect a 4-1 orbital period ratio is about as close as you're likely to get for any length of time for a stable binary planet-star, tidal locked system and that would be odd, but I see no reason why it wouldn't be stable.   It would only be possible around a small star where the planets were fairly close so that all 3 could have significant tidal effects on each other.   Without strong tides you don't get tidal locking.

You also can't have tidal locking with both objects when they have different orbital periods.   Tidal locking means that the rotation period is equal to the (sidereal) orbital period.


Does the redshifting of photons from the Universe's expansion violate conservation of momentum?

In relativity you can think of a single conservation law that unites conservation of energy and momentum -- conservation of four-momentum. Energy and momentum are the zeroth and the first to third components of the four-momentum respectively. Such conservation laws arise from invariance of the Lagrangian with respect to a translation in space-time coordinates.

In General Relativity these conservation laws are local concepts that (most people think) can only be applied in local, inertial (flat) frames of reference. In particular, they cannot be applied in changing space-times and so cannot be applied to situations involving the expansion of the universe.


Apparently two objects in Solar Eclipse image

The people who say it is an internal reflection are right.  This is an internal reflection inside the lens of the camera. There is no actual object there. In particular, Venus is not visible this close to the sun even during a partial solar eclipse. Moreover, Venus is currently a very very thin sliver, no more than a scrape of light at the edge of the planet, and much smaller than this image shows.
It's actually quite nice, because the sun itself is overexposed and the eclipse is not visible,  but the internal reflection is exposed quite nicely, and shows the progress of the eclipse well.


I know they will be orbiting about a common center of mass, i.e. the
  barycenter. But, do the velocities have to be equal in magnitude and
  opposite in direction (normal to R when R is their distance from each
  other) for the orbit to be stable?

The orbital velocities do not have to be and in general are not equal in magnitude. What is equal is the angular velocity, that is the angular rate (e.g., in $rad/sec$) that two bodies will be orbiting their common barycenter. The orbital radius, $r$, orbital velocity, $v$, and angular velocity, $\omega$, are related by the equation

$$\omega = v/r$$

Note that these are all scalar quantities and $v$ can be thought of as the component of the velocity vector which is perpendicular to $r$. Since conservation of angular momentum implies that $\omega$ remain constant for each body, individually, then we know that $v/r$ must also be constant which implies that bodies which orbit farther away from the barycenter must necessarily be orbiting faster and vice versa.

Now of course that simplifying explanation works for two bodies. Once you throw in more than two bodies, things become more complicated and the barycenter can move around, causing more complex motions.

  I think, if the velocity of one mass were to vary with respect to the
  other it would create a moving barycenter in which the two masses
  would collide or throw on another out of the orbit

This is a subtly different question. If one of the two masses had a varying orbital velocity, that would imply it was gaining or losing energy by some mechanism. This can occur through things like tidal interactions as it does for our Earth and Moon. As stated above, the angular velocity must remain constant which implies that the body whose velocity is changing must also migrate towards or away from the barycenter, potentially resulting in a collision or escape. Since the Moon's orbital velocity is being sped up through tidal interactions, it is moving farther away as a result. Another example might be two black holes emitting gravitational waves as they orbit which propagates away energy and thus they orbital velocities, causing them to get closer together until they collide.


How can a supermassive black hole cause so much energy to enlighten its matter when its massive gravity prevents light to escape?

It is quite correct that a black hole has so much mass that light cannot escape from a region around the black hole. The edge of this region is called the event horizon. If you cross an event horizon you are never coming back. That applies equally to light, and matter.

Around the black hole there may be matter in orbit. Since the Black hole has such strong gravity, the speed of the orbiting matter will be very fast. In fact it will be close to the speed of light. This high speed gives it lots of energy. The matter will form a disc, called an accretion disc, around the black hole, and collisions in this disc will cause the matter to heat up, to millions of degrees. At these temperatures, the disc will glow with X-rays.

On the part of the disc closest to the black hole, matter will be falling in from the disc, but before it reaches the black hole it can get enough energy to be ejected out, at very high speed, close to the speed of light. It gets ejected at right angles to the disc, at the poles of the black hole. These are the "jets". Intense radiation is produced along these jets. Blazars are distant supermassive black holes with jets that are pointed right at us.

So the black hole itself is "black", but the matter orbiting around it may be very bright.


Is there any star recognition framework/library available?

As I do not want to reinvent the wheel, is there any star recognition
  framework that you've heard of? Preferably open source, the language
  is of marginal importance as long as I can get the relevant data in
  and out. I'm thinking of something that I can feed with an image file
  and get back an array with the recognized objects with their x/y
  locations within the image.

I'd take a look at the astrometry service. You can upload a picture to their site and receive an astrometric labeled overlay. They do have an API. I'm not sure if an array output option is builtin, but I have a feeling it's doable within what they're offering.