I think it's an interesting question.    The trick would be a sustained fission reaction, faster than half life, but slower than a chain reaction.   A chain reaction could hardly be considered "star like" - it would just explode.

Lets say you had a planet sized object, maybe 100 parts Iron, basically inert, to 1 part Uranium  , which would generate heat through half life decay and some neutrons would spread from uranium to uranium - many wouldn't.  And as the object got hotter, more and more uranium would drift towards the center.  You'd have gradual heating, and the object, in time, would probobly begin to glow.     I'm guessing it's not possible using anything close to pure uranium, as the neutrons released in a uranium rich ball of matter would do 1 of 2 things - either blow the thing apart, or increase the rate of reaction and cause a chain reaction.   But i think, with the proper mix of uranium and inert elements, and the uranium spread out, not all in the center, you probobly could get something sort of star-like.   It would have much less available energy so it wouldn't burn nearly as long as a star.    At least, that's my thought experiment answer to this question.

The problem with nothing but heavy, neutron rich elements is that, I don't see any way that wouldn't cascade and burn or blow up quickly.   The nuclear reaction is driven by neutrons hitting the nucleus with potential energy and as the reaction happens, more free Neutrons are produced which speeds up the reaction significantly.   A ball of of Uranium many miles across would undergo fission and explode.   A ball heavy enough to not explode - say, jupiter sized, it would heat up and burn quickly and then settle down and cool off slowly. - perhaps similarish to a star going Nova then settling into a white dwarf.

If you have a material inside that slows the reaction, absorbs and re-admits Neutrons you could probobly generate a sustained burn - similar to what happens inside a nuclear plant while maintaining enough gravity to keep the object in one piece and over time, That would heat up and glow like a star in time and last for a while.   Planet sized, even moon sized would probobly be more than sufficient, but you'd need the material to slow the reaction and keep the Uranium spread out.   That's my take at least.


Maybe I'm reading too much into the question, but there's a few ways to measure a day.   I think 1912 or 1970 is an accounting answer more than the actual length of the day, similar to how 1751 in England was the shortest year, 282 days, because of a calendar adjustment.

Earth takes about 23 hours, 56 minutes and 4 seconds to complete a full rotation and that number is very consistent but gradually increasing because of the Moon and tidal effects.   It's increasing about 23 milliseconds per century or about a second ever 43,000 years or a minute every 2.6 million years.   It also works out to about 230 microseconds per year, less than 1 microsecond per day.

So, in a sense, if you're talking sidereal day, each day becomes the new longest on record, by a tiny bit, unless there's a large Earthquake which can speed up the Earth's rotation a little.  The 8.9 Richter scale Earthquake in Japan compacted Earth slightly and sped up Earth's rotation by about 1.8 microseconds, or slightly more than 2 days worth of tidal slowing.

The longest solar day would be perihelion day, when the Earth moves fastest around the Sun and a single rotation has to move that much further to get the sun back to where it started.   According to this article, obliquity plays a role too, which I'm having a hard time picturing, so I can't explain that.

This raises an interesting problem because Earth's eccentricity is currently decreasing and has been for about 15,000 years.  Earth's eccentricity was considerably higher at times in the past, perhaps a hundred or two hundred thousand years ago, perhaps as high as .04 or .05, much higher than the current .0167, which could have added perhaps 15-20 seconds to a solar day.   The longest day would probably be during a recent peak eccentricity during perihelion.   Unless I'm overlooking something.   Corrections welcome.

Eccentricity Wikipedia article 1

Eccentricity Wikipedia article 2


How does interstellar matter density vary?

The interstellar medium is a multiphase medium, and you can find (some references in this lecture and in this thesis manuscript (this one is in French, but numbers are international)):

the hot ionized medium (HIM) with density as low as 10^-3 cc (particles per cubic centimeter);
the warm ionized medium (WIM), with density of the order of 0.03 cc;
the warm neutral medium (WNM), with density of the order of 0.25 cc;
the HII regions, with densities ranging from 1 to 10^4 cc;
the cold neutral medium (CNM), with density of the order of 25 cc;
the molecular clouds, with densities above 10^3 cc, up to 10^6 cc roughly.

These different phases are due to the interplay of cooling and heating processes of its own components (Wolfire et al. 1995). The lowest density regions are hot and associated to supernova expansion bubbles. The HII regions (HII for "ionized hydrodgen") are associated with O stars (massive stars that can efficiently ionized their environment).

The highest density you can get with no evidence of forming stars are of the order of 10^4 cc (see for example this talk by Ward-Thompson on starless cores). Dense, molecular clouds that are about to collapse and form a star have a density of the order of 10^6-10^7 cc.


Could gravity waves be used to confirm the existence of a ninth planet?

No, they can't.

Gravity waves from a small, simple object moving slowly are very, very faint, to the point of being undetectable with current (or foreseeable) technology.

The waves that have been detected come from the merger (a very fast movement in the last orbits) of two black holes (two very big masses). And they were just detected over the noise level.

Do not confuse that with the possibility to detect the gravitational influence of such a planet over other bodies. That is how Neptune and Pluto were detected, indeed.


South of the tropics, where does the sun's noon shadow point?

Is it pointing south?

Yes, the shadow points south when you are south of the tropics (south of the Tropic of Capricorn).

In the tropics (between the Tropic of Cancer and the Tropic of Capricorn), the direction of the Sun's shadow depends on the location and the date.

As the angle between the sun and the earth's axis changes during the year, the apparent path of the sun in the sky changes.

Outside the tropics, this change is insufficient to ever bring the sun directly overhead.  Regardless of the date, the sun will be seen to be toward the equator at local noon.  (Same as saying the shadow points away from the equator).

Within the tropics, the changing path sometimes brings the sun directly overhead.  The position at local noon depends on the specific location and the date.

On the December solstice, the sun will be overhead for locations on the Tropic of Capricorn and will cross to the south at local noon for all locations in the tropics (and those further north).  During the June solstice, the situation is reversed for the tropics.  All locations south of the Tropic of Cancer would see the sun pass to the north.


The very first sentence of that paper makes it clear that it is concerned with observations of the 21cm line in the redshifted universe, i.e. on objects so far away from us that redshift becomes significant. And just as any other electromagnetic emission, radio waves are affected by this. So, in section 2.1, the authors describe selecting a sample of radio sources with known redshifts so that they would know how much they had to tune their receivers downwards in frequency.

This is actually quite common and relevant in today's 21 cm radio observations. Redshifted signals are observed even down to 140 MHz and below as stated on LOFAR's Redshifted 21cm hydrogen Line page.


How probably is it that galaxies will extinguish?

Galaxies are gradually being extinguished. Most star formation activity occurs near the start of a galaxy's life, or in response to merger activity with other galaxies.

The star formation rate of the universe peaked at redshifts of around 3, corresponding to a look-back time of around 9 billion years. Since then the star formation rate has declined as the universe expands; mergers are less frequent, gas is driven out of galaxies by supernovae and active galactic nuclei.

However, most of the stars that have been formed are of lower mass (K- and M-dwarfs) than the Sun and will live on for tens or hundreds of billions of years. They are however much fainter than the Sun. So although high-mass, luminous O- and B stars live their short lives and are not replaced at the same rate, the low-mass stars continue to shine. This means that galaxies will get fainter on average as the remaining stellar populations increase in average age and decrease in average mass.

It is a slow process though. High mass stars are still being formed in our galaxy after 12 billion years, and most spiral galaxies have ongoing star formation. However, star formation has more-or-less ceased in gas-poor elliptical galaxies.


Do planetary surface temperatures change in unison in a solar system?

The simple answer to your question is yes. Taking a simplified equation from Carroll & Ostlie, An Introduction to Modern Astrophysics Second Edition, the temperature of a planet can be estimated as:
$$
T_{p} = T_{\odot}(1-a)^{\frac{1}{4}}\sqrt{\frac{R_\odot}{2D}}
$$
Where $T_p$ is the predicted temperature of a planet in a circular orbit of radius $D$ with an albedo of $a$ around a star with a temperature of $T_\odot$ and a radius of $R_\odot$. If the energy output of the star were to increase, raising $T_\odot$, then there would be a corresponding increase in the temperature of all planets orbiting said star.

In practice there are factors which can make this correlation difficult to measure. The albedo  of a planet during the course of a day can vary greatly and the distance of a planet from the host star changes throughout the year. This equation also assumes that the planet is a perfect black body which most are not which can also change a planets temperature and obscure any changes caused by the host star.


Without a lot of expensive equipment the best you can do is make a rough estimate. By incorrectly assuming that the light that you see describes the black body temperature of the object, by assuming that the system is in thermal equilibrium with itself and that the temperature within the system is spatially and temporally uniform, you can use Planck's law to calculate the spectrum.

Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$.

The spectral radiance of a body, $B_ν$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body for frequency $ν$ at absolute temperature $T$ is given by:

$$ B_{\nu }(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{k_{\mathrm {B} }T}}-1}}$$

where $k_B$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light in the medium, whether material or vacuum. The spectral radiance can also be measured per unit wavelength $\lambda$ instead of per unit frequency. In this case, it is given by:

$$B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}.$$

The law may also be expressed in other terms, such as the number of photons emitted at a certain wavelength, or the energy density in a volume of radiation. The SI units of $B_ν$ are $W·sr^{−1}·m^{−2}·Hz^{−1}$, while those of $B_\lambda$ are $W·sr^{−1}·m^{−3}$.

The "Planck Calculator for Infrared Remote Sensing" can be used to make some calculations:

Temperature T(K)                 10,000
Center Wavelength μm             11.0
Spectral Radiance (W/m^2-sr-μm)  5292.388800699331

With a temperature of 10K the center frequency is 11 $\mu m$.

That's the oversimplified (inaccurate) layperson's calculation. More accurate calculations are made by allowing for everything from: relative velocity, the composition of the object's atmosphere (and our own), the effect of distance, even other elements, upon the spectral distribution (short list).

Consequently, after correcting for light travel time effects the response of the Balmer and optical helium lines should generally be strongest during low continuum luminosity states. Responsivity that depends upon photon flux and continuum state may explain a number of outstanding problems currently under investigation in broad line variability studies of these and other emission lines. [Abridged]

Once calculations are made for the blackbody curve in accordance with Planck's law it is necessary to transpose that to Rayleigh–Jeans law which is applicable at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies) where Planck's law is more useful.

See Wikipedia's "Comparison [of Rayleigh–Jeans law] to Planck's law".

Further reading:

"Introduction to Millimeter/Sub-Millimeter Astronomy" 3 Mar 2009, by Wilson
"Free-Free Spectral Energy Distributions of Hierarchically Clumped HII Regions" 30 Mar 2004, by Ignace and Churchwell
"What the Optical Recombination Lines Can Tell Us About the Broad Line Regions of Active Galactic Nuclei" 17 Jun 2005, by Korista and Goad.

If you want to include the effect of quantum gravity in your calculations a recent paper is: "Black body radiation in a model universe with large
extra dimensions and quantum gravity effects" 8 Jan 2018, by Nozari, Anvari, and Sefiedgar

  Abstract: "We analyze the problem of black body radiation in the presence of quantum gravity effects encoded in modified dispersion relations and in a model universe with large extra dimensions. In this context, modified form of Planck distribution, Jeans number, equipartition theorem, spectral energy density, Stefan-Boltzmann law and Wien's law are found and the corresponding results are interpreted. Then, entropy and specific heat of black body radiation are obtained in this setup. Finally, the modified form of Debye law and Dulong-Petit law are investigated in this framework.".

This article from the PrimaLuce Lab's website: "Radio astronomy at school with SPIDER radio telescopes" explains backyard radio astronomy.

Their 3M diameter SPIDER 500A's H142-One receiver has a 1.42 GHz superheterodyne type radiometer/ spectrometer, double conversion (type UP/DOWN) with 50 MHz bandwidth (RF=1.395MHz-1.445MHz) and 14-bit analog to digital converter. It has a spectrometer with 1024 channels (each 61 KHz) and is able to record many radio sources in the Universe, with a theoretical flow of at least 5 Jy. Price available upon request.

Their SPIDER 230C Compact Radio Telescope measures 2.3 meters and is priced at U$16K, from what I can determine from reading the literature it features the same receiver and software; dish diameter and mount capacity being the only difference, along with beam width and sensitivity.

The mesh on their dishes is good up to 5 GHz but the feedhorn normally supplied is for 1.42 GHz, I imagine that they are willing to do some amount of customization.

An explanation of the atmospheric window can be found here: "The Effects of Earth's Upper Atmosphere on Radio Signals", this image is from NASA's website:

  Question: From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?

3cm is 10 GHz and can be received with a 10 GHz receiver. The difference between synchrotron and free-free is explained on NRAO's webpages  Free–Free Radiation and Synchrotron Radiation.

Acceleration by a magnetic field produces magnetobremsstrahlung, the German word for “magnetic braking radiation.” The character of magnetobremsstrahlung depends on the speeds of the electrons, so these somewhat different types of radiation are given specific names. Gyro radiation comes from electrons whose velocities are much smaller than the speed of light: $v≪c$. Mildly relativistic electrons whose kinetic energies are comparable with their rest mass $m_ec^2$ emit cyclotron radiation, and ultrarelativistic electrons (kinetic energies $≫m_ec^2$) produce synchrotron radiation.

Acceleration by an electric field accounts for free–free radiation, the resulting emission is called free–free radiation because the electron is free both before and after the interaction; it is not captured by the ion. If the ionized interstellar cloud is reasonably dense, the electrons and ions interact often enough that they quickly come into local thermodynamic equilibrium (LTE) at some common temperature, so free–free radiation is usually thermal emission.


The usual recipe used in the population synthesis literature is that triple stars comprise ~10% of all stellar systems and the mutual inclination is uniform in the cosine.  Most of these systems have a large enough semi-major axis ratio that the inner binary is dynamically decoupled from the outer binary, so retrograde orbits will be just as likely as prograde orbits.  So, to an order of magnitude, ~5% of stellar systems should be triples with retrograde orbits.

Things get more complicated if you are interested in more details.  More massive stars have higher multiplicities, so the fraction is probably somewhat higher for them.  For more compact systems, dynamical effects like the Kozai-Lidov mechanism will become important and can lead to prograde orbits becoming retrograde and vice versa.  Moreover, in compact systems, retrograde orbits are slightly more stable than prograde orbits, so they are likely to be a little more overrepresented, though I'm not aware of any observational evidence that there is any great difference between the two.

If you want more details on the observations of multiple stellar systems you should read Raghavan et al. (2010).  However, I don't think that they measured mutual inclinations.  In general those measurements are very difficult to do.  Andrei Tokovinin has done a lot of great observational work on triple stars as well, so you might want to look at Tokovinin (2014) and Tokovinin (2008) among other papers.  There's also a good review by Duchene & Kraus (2013).


what degree in the sky does the sun change color?

Here's a partial answer:
As mentioned in the comments, the transmission of the atmosphere depends quite a lot on local factors. But given a transmission $T_0(\lambda)$ at zenith, the transmission at an angle $\theta$ from zenith can be written
$$
T(\lambda,\theta) = T_0(\lambda)\times X(\theta),
$$
where $X$ is the air mass.
The air mass can be approximated as $X=\sec\theta$ up to $\theta\simeq70^\circ$, but as you approach the horizon, the calculation becomes increasingly model-dependent. A relatively simple expression is given by Pickering (2002):
$$
X(\theta) = \csc\left(90^\circ-\theta + \frac{244}{165+(90^\circ\!\!-\!\theta)^{1.1}} \right),
$$
where $\theta$ is measured in degrees, and $\csc x\equiv 1/\!\sin x$. This model doesn't take into account refraction (which requires ray-tracing), so may be off by half a degree or so when the Sun is at the horizon.
If $S_0(\lambda)$ is the un-transmitted Solar spectrum, the observed spectrum is then
$$
S_\mathrm{obs}(\lambda,\theta) = S_0(\lambda)\times T_0(\lambda)\times X(\theta).
$$
Calculating the color that the human eye perceives, given a spectrum, is not trivial, but I recently outlined an approch in this answer.


Technique for folding sparse event data in order to detect an underlying periodicity?

While this is not exactly how folding works, one-dimensional data encourages trying some simple statistical tallying first.

What very easily reveals those periodicities is just tallying up the deltas, with an appropriate fuzziness for what periodicities are relevant:

Deltas

[0, 9308, 28857, 36709, 28857, 9308, 28858, 36708, 28858, 9308, 28857, 36708, 9309, 19549, 9309, 28857, 36707, 9310]
Already here, some patterns should be apparent. But we have some extra information that could be used to filter out the short period noise.

Rounding at an appropriate sampling frequency

Earth transits aren't going to happen every day. If there was an Earth transit yesterday, it can't happen again today, or next day, or even the next month. In fact, we know they can only happen once for every synodic period of Mars-Earth (780 days).
To not miss out in any "legitimate" part of the signal, we should sample at at least twice of that rate. "one year" happens to be pretty close to half of 780 days.
The periodicity is then readily visible:
[26, 79, 101, 79, 26, 79, 101, 79, 26, 79, 101, 26, 54, 26, 79, 101, 26]
Which also reveals an even higher extra periodicity of about 206 years.
With a strong 206 year periodicity found in a set of events only covering a little over 400 years, there's not enough data to with any certainty detect any higher order periodicities that that.


What exactly are infrared dark clouds and what is the significance of the orientation of their magnetic fields?

Molecular clouds are the birthing grounds from which stars form, which occurs when said clouds undergo a collapse.  They contain a significant fraction of molecular hydrogen ($H_2$), and are generally good places for forming other stable molecules.  Molecules and metallicity generally make star formation easier by making it easier to get the non-molecular hydrogen and helium to lose enough energy to collapse into a common core by acting as (kinetic) energy absorbers.  Imagine tossing a ping pong ball at a tennis ball: the ping pong ball bounces off at a slightly lower velocity, and the tennis ball barely moves at all from an even smaller change in its velocity.  And if the cloud is already "cold", then that's even better: there's less energy your particles have to shed.  The presence of molecular hydrogen seems to be particularly important in the collapse of low metallicity clouds, such as would be the case in the early universe. So cold molecular clouds, like infrared dark clouds, should be a great nursery for high mass stars, as you don't have to shed that much energy (relatively speaking) to get things collapsing into a core, and it's got molecules to help it shed what it still needs to.

Exactly how the collapse is predicted to happen depends on the model, and the models depend on the magnetic field of the collapsing cloud.  There are two major models.

Quoting liberally from Magnetic Fields
in Molecular Clouds by Richard M. Crutcher (c. 2012) (all quotes will be from this article, until otherwise stated):

  In the strong magnetic field theory, clouds are formed with subcritical masses, $M < M_\Phi = \Phi/(2\pi \sqrt{G})$ (Nakano & Nakamura 1978), where $M_\Phi$ is the critical mass, $\Phi$ is the magnetic flux, and $G$ is the gravitational constant. Hence, the magnetic pressure is sufficiently strong to counteract gravity and prevent gravitational collapse. Because the magnetic field is frozen only into the ionized gas and dust, neutral gas and dust contract gravitationally through the field and the ions, increasing
  mass in the cloud cores. The magnetic field strength also increases, but more slowly than does mass.

  This process is known as (gravity-driven) ambipolar diffusion. [Ambipolar diffusion, or breaking of flux freezing, may also be driven on small scales by turbulence (e.g., Zweibel 1988.)] When the core mass reaches and exceeds $M_\Phi$, the core becomes supercritical ($M > M_\Phi$), collapses, and forms stars. During the collapse, the magnetic field is dragged inward but cannot become strong enough to halt the collapse.

  The weak-field theory of star formation says molecular clouds are intermittent phenomena, with short (∼$10^6$ years) lifetimes. Magnetic fields are sufficiently weak that the low-density ISM is supercritical ($M > M_\Phi$). Clouds form at the intersection of turbulent supersonic flows. Generally, clouds do not become gravitationally bound, and they dissipate; those that are self-gravitating form stars in essentially a free-fall time (Elmegreen 2000). Supersonic turbulence will dissipate on roughly the free-fall timescale as collapse of gravitationally bound clouds proceeds (MacLow et al.
  1998). Although magnetic pressure cannot stop the collapse, it can dominate turbulent pressure during the late stages of core collapse. An extreme version of weak-field models has the initial field so weak that the medium is super-Alfvenic as well as supercritical (Padoan et al. 2004).

So, the short of the matter is that molecular clouds are important because they are connected to the formation of (high mass) stars. And the magnetic fields of those clouds is important because the exact dynamics of that collapse are dependent upon them.  And all of that is important because it gives us a better idea of how high mass stars form, which is expected to have happened much more frequently in the early universe.

It hasn't, however, yet mentioned why the direction of that field may be interesting (it only mentions the flux).

  Interstellar dust produces thermal emission and extinction of light from background stars. Linear polarization of this radiation provides a probe of the magnetic field morphology in the ISM, including molecular clouds.

So they're curious about the structure of the interstellar medium (ISM) of the galaxy. The radiation of the dust clouds, including the molecular clouds, provides a way to probe the magnetic structure of the ISM.  The paper then recalls how this polarization can arise via radiative torques creating preferential alignments in dust grains, and what this then means.

  [T]he grain extinction cross-section is greatest perpendicular to $\mathbf{B}$, so the maximum of the polarized light from background stars is parallel to $B_{POS}$, the magnetic field direction in the plane of the sky. Conversely, maximum polarized emission is perpendicular to $B_{POS}$. The predicted degree of polarization depends very weakly on magnetic field strength, so dust polarization does not directly give the magnitude of the magnetic field.

This describes how they can measure the direction of the magnetic field of a molecular cloud, a topic from your clip.  It notes that it's a poor measure of the actual field strength, and we need the flux $\Phi$ to check how a given star formation model matches up with observations.

The paper continues on and discusses some measurements of the direction of the magnetic field of molecular clouds.  The results it cites indicated that the field has been observed to be either parallel or perpendicular, and that the direction can change between different parts of the same cloud.  So I'm not sure why your clip mentions a perpendicular measurement as surprising; perhaps it's the "more-or-less" that is key, as if there's enough data to suggest a preferred (perpendicular) alignment then that would be interesting and probably unexpected.

The paper also discusses how it is nevertheless possible to obtain some estimates of $\Phi$, as well as the difficulties in doing so accurately.  This lets us tie everything together: the polarization of the radiation from the molecular cloud allows us to measure the direction of its magnetic field relatively easily; and with that and some various techniques we can then measure the actual flux $\Phi$ of the magnetic field; which we can then use to determine if the cloud is sub-critical ($M<M_\Phi$) or not.  And when we know that, we can now more readily probe how accurately star formation models, such as those mentioned above, reflect the actual reality.


Yes it can reach both Saturn and Jupiter. But it's not powerful enough to remove their rings. Even Earth's radiation belts don't get blown away.

Edit: Jupiter's magnetic field is $\approx 770 \mu T$ (at the surface of its equator) which is way more powerful than Earth's ($\approx 40 \mu T$ at the equator). Saturn's magnetic field on the other hand is a tad less powerful than that of the Earth ($\approx 20 \mu T$ at its equator). So Earth's magnetosphere shields it's particles from CMEs. Jupiter is further away and its magnetosphere is way stronger than Earth's. So it shields everything effectively. Saturn is even further away and although its magnetic field is comparable to that of the Earth, the distance from the Sun assures us that a CME wouldn't have a significant impact of any sort there.


Can one build a distributed radio telescope?

Yes, we can do radio astronomy with heterogeneous, geographically distributed antennas.  The VLBI is an excellent example.

Of course, the antennas have to be similar in some ways for it to work:

If the antennas are ground based, they need to be in the same hemisphere, or the Earth will prevent them from looking at the same target at the same time.
They must have extremely stable timing sources, since error or drift in the time tagging will result in dB losses for a detected signal and inaccuracies in angular resolution.
They must have accurate enough timing sources to allow synchronized collections.  The collection times need to be offset in time to account for path delays to the various antennas.
They must overlap in collection RF.  Any differences in center frequency and bandwidth will mean that some of the antennas will be picking up different signals than the other(s).  Practically this usually degrades the collection quality.
It's desirable to have similar antenna beamwidths or, again, the antennas won't all be picking up the same signals.
You need to synchronize the collectors and have accurate estimates for the distances between them.  The VLBI systems use methods described here: How are the atomic clocks synchronised between worldwide VLBI telescopes?,


What is the difference between asteroids, comets and meteors?

The objects you are refering to are actually two different objects: asteroids and comets. Meteor and meteorite are other names for an asteroid, at a given time of its interaction with our planet. We'll get to that.
So first, what is the difference between an asteroid and a comet?
A comet is a small solar system body that display a "coma" (an atmosphere of a sort) and sometimes a tail passing close to the Sun. They are mostly made of ice and dust, as well as some small rocky particles. We distinguish two kind of comets, with short or long orbital period. The short orbital period ones originated from the Kuiper Belt, a region composed of small bodies beyond the orbit of Neptune. The long orbital period ones originated from the Oort cloud, a scattered disk of icy planetesimals and small bodies laying around our solar system.
An asteroid is a small body, composed mostly of rocks and metals. In our solar system, they can be originated from the asteroid belt, laying between Mars and Jupiter, or from the orbit of Jupiter (the Jupiter Trojans), or from actually almost everywhere in the Solar System.
A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a meteor, what we also call a "shooting star".
Eventually, a meteor that was massive enough not to be completely distroyed entering the Earth's atmosphere and hitting the ground is a meteorite.


Could anything consume a small black hole?

Black holes don't "devour".  They can't "eat" things.  But things can "fall into" a black hole.  It really is just gravity, but really really intense gravity.

A really small black hole would have a mass of about 3 times the mass of the sun. All neutron stars are smaller than that (or they would turn into black holes).  A really large star could be more massive and would be much much larger in volume.  A black hole could orbit with such a star, but if it got close enough it would pull gas off the star which would then fall into the black hole, causing it to get larger.

There is no way that a large star can in any way "pull apart" or otherwise "consume" a black hole.

There is a theoretical notion of the quasi-star. These are similar to a Thorne–Żytkow objects.  In normal stars, the outer layers are prevented from collaping by energy from nuclear fusion in the core.  In a quasi-star, the core collapses into a black hole and the release of energy from matter falling into the black hole prevents the collapse of the star. A steady state can be achieved, as if more matter starts to fall in, more energy is released causing the star to grow, and reducesing the amount of matter falling in. Nevertheless, ultimately all the matter from the star would fall in.

Smaller black holes can't form from stars. We don't know if they exist at all, but if they do, very very small black holes could be too small to interact much with regular matter. A very small black hole could be smaller than a proton, and even if one fell into the Earth, it could pass through the gaps between and inside atoms. Such a black hole would be very hot, due to Hawking radiation.


How much solid matter is in the observable Universe?

Any non-star is negligible, their volume is practically zero.
Most matter in the Universe is gravitationally bound plasma, resulting that they do not have a well-specifiable volume, because their density continuously changes from the interstellar vacuum to many times of lead. But choosing a surface at any arbitrary point (for example, at 1 bar pressure) does not modify the result significantly (this is why we can say a radius for the Sun).
This source says that the estimated count of the stars in the visible Universe is $10^{21}$. This is obviously not very exact, there are probably better estimations which you substitute into this answer.
The majority of the stars in the Universe is a red dwarf. Their radius varies between 9% and 50% of the Sun, we get their geometric average, this is $\approx$ 21% of the Sun. Their volume is $\frac{4 \pi}{3}(0.21 \cdot 700000km)^3 \approx 1.37 \cdot 10^{16} km^3$.
Thus, the volume of all stars in the Universe is $\approx 1.37 \cdot 10^{37} km^3$.
The observable Universe is a sphere with the radius of 46billion light years.1 A light year is $\approx 9.5 \cdot 10^{12} km$. Thus, the volume of the observable Universe is $\frac{4 \pi}{3} (23 \cdot 10^{12} \cdot 9.5 \cdot 10^{12})^3 \approx 4.37 \cdot 10^{79} km^3$.
After a division, we get that
$$\frac{1.37 \cdot 10^{37}}{4.37 \cdot 10^{79}} \approx \underline{\underline{\frac{1}{3.2 \cdot 10^{42}}}}$$
part of the volume of the visible Universe is not empty space.
About 1 digit difference from the reality is possible due to the lack of not enough exact approximations.
1This is not the well-known and false 13.7 billion light years, which is the way what light could do since the Big Bang. The difference is coming from that meanwhile the Universe expanded, thus the stars we can see (as if they would be 13.7 billion ly far away) are now 46billion ly away. More details are on the wiki link.


Could not slow neutrinos orbit galaxies and clusters, thus comprising a large component of even COLD dark matter?

These neutrinos would have to be really cold. The cosmic neutrino background is at 1.9K, and they are considered hot dark matter, because they would have been highly relativistic at the epoch of structure formation. To be considered cold dark matter, and also to be captured in orbits in galaxies, the neutrinos would have to be much colder - totally non-relativistic now.
Let's assume an average energy then of about 0.1 eV for each neutrino (similar to their likely rest mass energy). In order to account for $\Omega_{\rm CDM}\sim 0.3$ there would need to be $5\times 10^{10}$ per cubic metre, or about $10^{10}$ per cubic metre per flavour, on average over the universe.
To account for dark matter in galaxies (e.g. the Milky Way) we need $\sim 10^{12} M_\odot$ within about 100 kpc, meaning a number density of neutrinos of $10^{14}$ per cubic metre.
These neutrinos are spin 1/2 fermions and thus would have a Fermi energy of about $5\times 10^{-3}$ eV. That would mean if they were any colder than 6 K, they would be degenerate and exert a degeneracy pressure. This would be sufficient to prevent the halos forming - Tremaine & Gunn (1979) showed that cold dark matter halos cannot be made from low mass leptons like standard neutrinos.
Edits:
A back of the envelope Tremaine-Gunn limit (see also Boyarsky et al. 2009) is to assume that the escape speed of the galaxy is $v$, it's dark matter halos has radius $r$, the total mass is $M$ and the neutrino mass is $m$.
The number of available quantum states to spin half fermions in this volume, up to a momentum $mv$ is
$$ N = \left(\frac{4\pi r^3}{3}\right) \left(\frac{8\pi}{3}\right) \left(\frac{mv}{h}\right)^3$$
We can write
$$ v = \left(\frac{2GM}{r}\right)^{1/2}.$$
The mass contained in these particles cannot be greater than if every quantum state is filled by one fermion of mass $m$ and if this is to explain dark matter, then this mass must be $\sim M$. Thus
$$M < m\left(\frac{4\pi r^3}{3}\right) \left(\frac{8\pi}{3}\right) \left(\frac{m\sqrt{2GM/r}}{h}\right)^3$$
and
$$mc^2 > 8.9\left(\frac{r}{\rm 100 kpc}\right)^{-3/8} \left(\frac{M}{10^{12}M_\odot}\right)^{-1/8}\ {\rm eV}\ .$$
Thus there are not enough quantum states to accommodate a halo of fermions unless their rest mass energies exceed about 10 eV. For neutrinos, there are 3 flavours and anti particles, which reduces this number by $6^{1/4}$, but conversely it must be increased because particles in the halo cannot be uniformly distributed in velocity between 0 and $v$.
10 ev is about two orders of magnitude larger than the likely rest masses of the known neutrinos.
The idea that neutrinos from stars can make any contribution to dark matter halos is untenable. The vast majority of solar neutrinos have energies above 0.1 MeV, and so for an assumed neutrino rest mass energy of $\sim 0.1$ eV, they have Lorentz factors that exceed $10^6$ - i.e. they travel exceedingly close to the speed of light and are not confined to galaxies. The neutrinos emitted during supernova explosions are even more energetic. Secondly, even if you invent a magical process that can produce neutrinos with kinetic energies less than 0.1 eV, you still have the Tremaine-Gunn limit to overcome and even if all the rest mass of all the stars in a galaxy was converted into neutrinos, that mass would still fall short by more than an order of magnitude of that required to explain dark matter in galaxies. You cannot imagine that the neutrinos would "pile up" over time because the vast majority of the mass that has ever been turned into stars is still in the form of stars today and the stars that have already lived and died are a tiny percentage of the required dark matter mass.


Can we compress any object to create black Holes?

Do you mean, "Can we in principle", or just, "Can we..." ?

The answer to the latter is no. We have no technology to create black holes. It was a remote possibility that was considered when creating the high energy densities in the LHC, but in practice the energies aren't high enough.

Can we in principle do it? The answer is yes. If you can squash enough mass (or energy) into a very small volume then the inevitable consequence is a black hole.

Most would say that the volume in question is governed by the Schwarzschild radius, $r_s$  which is roughly 3km multiplied by the mass in solar units. Thus for an Earth mass $r_s$ is about 1 cm.

Actually the radius within which collapse to a black hole is inevitable is a little bit larger because even the "hardest" equation of state, where the speed of sound equals the speed of light, cannot resist collapse at radii below about $1.5 r_s$.

The problem is that to compress materials this much would be incredibly difficult. For the example of an Earth mass, we are talking densities of order $10^{30}$ kg/m$^3$. But (long) before you got to that stage, yes you would have the problem of overcoming the nuclear fusion reactions would be initiated that would oppose your compression (ie you would end up making a mini-star before you could get to a black hole). Because the Schwarzschild radius depends linearly on mass, then less massive black holes need to be even denser. However, we are again talking about "in principle". So assuming compression could overcome the various opposing forces, then a black hole could be formed.

It really doesn't matter what the black hole is made of. In a star, the sequence of nuclear fusion events is a consequence of the physical conditions in the stellar interior. It is not anything that is required for black hole formation.


How could we tell if the Universe is infinite?

We don't know for sure, but it certainly fits into our theories. There is, of course, no way to actually test if the Universe is infinite, but right now we think it is likely.

Also, if you read my updated answer on your other post, the Universe has always been infinite in size. I explain over there how it actually works: space is created in between everything, and thus one could say the Universe is expanding.

These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.

Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.

Really, there is no limit (as far as I know) to how fast we can drift away. Farther objects will keep drifting faster and faster away, since our gravity has a much weaker effect on them.


Can SETI certify whether or not Proxima b is inhabited by beings using electromagnetic communication?

Using current technology (and by that I mean experiments and telescopes that are available now) we would probably be unable to detect life on Earth even if observed from a distance of 4 light years, which is the distance to Proxima Centauri.

A "blind" search could look for radio signatures and of course this is what SETI has been doing for lots of different stars, including, I imagine, Proxima Centauri. If we are talking about detecting "Earth-like" signals, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation.

As I have remarked in the linked question: The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches.

It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio "chatter" out to distances of 50 pc ($\sim 150$ light years) - see Loeb & Zaldarriaga (2007) - and so Proxima Centauri would be well within this range. This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals.


In Chromoscope, what is the X-ray source between Ophiuchus, Libra and Scorpius?

The source is Sco X-1. It is a point source of X-rays.

I'll have get back to you on why it looks that way in chromoscope, but it looks incorrect to me.


It's well-explained on NASA'a web site ... http://curator.jsc.nasa.gov/Lunar/index.cfm

About 380 kg of rocks and soil were brought back.

Note that ...

  "The lunar sample building at Johnson Space Center is the chief repository for the Apollo samples."

http://curator.jsc.nasa.gov/Lunar/laboratory_tour.cfm

About three hundred moon rocks were given as gifts by the US to cities and countries around the world, and you can easily find that list, and hence go see them all. (More fun than PokemonGo!)

List showing where to see moon rocks

You can usually very easily see one at the local "national museum" in your country.  I've seen maybe five all over the world at the places listed there.

It seems that at least 100 have gone missing!

There are a few moon rocks in the wild and there were cases of theft - they would be worth millions of bucks ... TV show about that.


If 50 tons or more of debris falls to earth everyday, is Earth getting heavier?

The Earth loses mass because hydrogen and helium (plus other elements in trace amounts compared to hydrogen and helium) escapes the Earth's atmosphere. The Earth gains mass because incoming asteroids (most of them very small) impact the Earth's atmosphere; a few make it all the way to the surface of the Earth. Whether those incoming asteroids burn up in the atmosphere or make it all the way to the surface is irrelevant; the Earth gains mass.

Whether the net result is a mass gain or mass loss is a bit up in the air; the uncertainties on both are rather large, and they overlap. The Earth might be gaining or losing a tiny bit of mass every year. That said, most research leans toward atmospheric losses being greater than mass gain from impacting asteroids, comets, and dust.

Whichever is the case, it's a bit irrelevant. Even the most extreme upper estimates on mass loss or mass accumulation are incredibly tiny compared to the mass of the Earth itself.


Magnification using a CMOS sensor

The concept of magnification is most easily defined when we are comparing apples to apples, not apples to oranges.
When we use a telescope (and assuming normal or corrected vision) the situation is as follows.
Light entering a telescope from a point in space, it is essentially parallel, and when it is correctly focused light exits the eyepiece parallel as well. In this case the term magnification really applies to angles. If we have a 900 mm telescope and 7.5 mm eyepiece giving 120x magnification, then a 30 arc seconds wide Jupiter will appear  1 deg apart to our eye.
So here "apples to apples" means angles to angles.
There is no easy  way to talk about magnification when a recording device like photographic emulsion or a CMOS sensor is placed at the focus, because in this case angles have been converted to positions.
Here "apples to oranges" means angles to positions.
Instead we can talk about scale.
In [this comment] the OP correctly obtained 1.65 arc seconds per pixel using pixel size divided by focal length which gives the angle in radians and then the multiplier
$$\frac{180}{\pi} \frac{3600}{1000} = 206.2648$$
which converts radians to degrees, then to arc seconds, then divides by 1000 because that equation specified micrometers for pixels and millimeters for focal length.
If you wanted an apparent magnification based on viewing your image on a screen, you can use the following which is a ratio of angles
\frac{\text{camera pixel}}{\text{focal length}}
$$M_{apparent} = \frac{\text{angle}}{\text{angle}} = \frac{\frac{\text{screen pixel}}{\text{viewing distance}}}{\frac{\text{camera pixel}}{\text{focal length}}} = \frac{\text{screen pixel}}{\text{camera pixel}} \times \frac{\text{focal length}}{\text{viewing distance}}$$
Which means of course that the farther we hold the screen away from us, the smaller Jupiter looks, or the larger the pixel size on our screen, the bigger it looks.
But of course that only works if our device displays with a 1:1 correspondence between image pixels and screen pixels. If the device scales the pixels (most do) then you have a bigger challenge.
That's why recorded images through telescopes rarely if ever display "magnification". For microscope images, if they show a bacteria and say "magnified 5000 X" then they are comparing the size of the bacteria to the size of the printed image in the book.


relation between azimuth angle of the sun and angle of the shodow

Your diagram is correct, your maths isn't.

The bearing of the object's shadow is, as your diagram shows, away from the sun. If the bearing of the shadow is 75°, then the sun must have an azimuth of of 75°+180°, or 255°.

Similarly if the bearing of the shadow was 300° (pointing towards the northwest) the sun would be at an azimuth of 300 - 180°, or 120° (in the southeast)


Although it's a little tricky to say what "bigger" means in this context, the answer is, in most senses, no.

A supernova puts out about ten to a hundred times as much energy in the form of light, and hundred or more times as much matter is ejected. (A core-collapse supernova undoubtedly puts out much more energy in the form of neutrinos as well.) What matter is ejected by a kilonova does go out faster (30-60,000 km/s, versus about 10,000 km/s for supernova ejecta).

On the other hand, a kilonova puts out much more energy in the form of gravitational waves, so they're bigger in that sense.


Actually to convert from Jy/beam to Jy/pixel you need to divide by the beam size.

Let's say you have a quantity of 1 Jy/beam, then

$\frac{Jy}{beam} \frac{beam}{\Omega}$, then to go from Jy/beam to Jy/pixel you would need to divide by $\Omega$.

The values of the beam major and minor axis must be in pixels.

Source: NRAO


Beta Andromedae (Mirach) and distances mentioned in original Cosmos series from 1980

Looking at the SIMBAD data page for Beta Andromeda shows the source of both the parallax (distance) and the magnitudes. In this case, as it will be for many bright stars, the source of the parallax is the reprocessed data from the Hipparcos satellite, described in this paper. Prior to the launch of the Hipparcos satellite by ESA in 1989, parallaxes were very difficult to obtain and only low precision values were available for the closest stars.

The parallax given in SIMBAD for Beta Andromeda is $16.52\pm0.56$ milliarcseconds which translates to a distance of $\frac{1}{0.01652"} = 60.5\pm2.1$ parsecs or $60.5\times3.26=197$ light years. This small parallax would have been extremely challenging or impossible to measure accurately with the pre-CCD technology before 1980. Errors of several hundred percent were not uncommon. The prior parallax measurement which probably was an improvement on what was available in 1980 at the time of Cosmos is from van Altena et al. 1995. This lists a parallax of $47.7\pm7.9$ millarcsec, a nearly 5x larger error and which gives a distance of 68 light years.

Similarly we can see that the $V$-band magnitude comes from this collection and the difference between Beta Andromeda ($V=2.05$) and Alpha Andromeda ($V=2.06$), the in-theory brightest star in the constellation is only 0.01 magnitude. Measuring a star that bright to that precision was, and in fact still is, quite difficult as most detectors will saturate. So it's not particularly surprising given how close Alpha and Beta Andromeda are in brightness, that the early measurements got them reversed when Beta is in fact (very slightly) brighter.

This high brightness probably means that we will not see a more accurate from the Gaia satellite, the successor to Hipparcos. The Gaia DR2 data release lists a brightness limit of $G\sim3$ which may be improved slightly in later data releases with more sophisticated data processing and treatment of saturated stars.


Is there a reason why the IAU differenciates objects in the SS orbiting the sun into a particular classification but it doesnt for satellites?

History, and recent history at that:

At one time (1980) things seemed clear enough: There were planets (Mercury-Pluto), Asteroids (mostly between Mars and Jupiter) and Satellites.  No one care much about whether a body was spherical or not, since it was pretty clear that this was all there was.

.
.
.

Then the Kuiper belt happened, and in particular it became clear that Eris was big enough to be rounded... Now why should not Eris be a planet? Could we have tens of rounded icy objects orbiting beyond Neptune?  How many would we expect grade-schoolers to memorise?

The decision was taken to classify solar system bodies into:

Planets (round and not shareing their orbit with anything comparible in size)
Dwarf planets (round, but have not cleared their orbit)
Minor planets (not round)
Satellites (not orbiting the sun)

This is a matter of convenience for humans. It does not reflect any fundamental reality.

For convenience, there is no need to distinguish between round moons and non-round ones, so no distinction is needed (for humans)


Is it easier to see the young moon when it's closer to the sun?

The further the moon is from the sun the easier it is to see.  There are two reasons.  Firstly, when it is close to the sun it gets lost in the glare from the sun.  The sky close to the sun is very bright, and there isn't enough contrast to see the moon against the bright sky.  Secondly, as the moon gets further from the sun, more of the illuminated side of the moon is visible and so the moon becomes brighter. Brighter moon → easier to see.
When I say "close to the sun" remember this only means "appears close in the sky" the actual distance to the sun only changes a little over the year.
The moon is easier to see against a darker sky, so if the sun has set, it is more likely to be visible.  There was a new moon at 2:30 GMT on 12th April 2021 and the moon will become visible in the next few days (which is astrologically significant to Muslims)


Is a supernova powerful enough to destroy the other star in a binary system?

The gravitational binding energy of a solar mass star of uniform density is $3GM^2/5R = 2.2774\times 10^{41}$ Joule. The energy of a typical supernova is $10^{44}$ J. So it looks at first glance like the explosion could disperse the other star...

...except that this assumes more than a thousandth of the supernova energy is absorbed and does work on the star. If the other star is more than 16 stellar radii away then it will definitely not be hit by this much energy. And even if it is close enough, converting radiation to mass lift is not going to be efficient. The nova will mainly heat up the surface layer down a few optical depths, meaning a few thousand kilometers get very hot but it is doubtful this does more than push the star away a bit and induces some wobble.


What is the color of Venus if it has no atmosphere?

According to the webpage from the Planetary Habitability Laboratory of the University of Puerto Rico: The Real Colors of Venus without the Clouds, did the following:

  We combined radar topography and emissivity to generate the surface texture assuming a generalized basalt composition with gray to light brown tones, depending on the iron and sulfur content. In the images we also included the effect of the 'orange sky' of Venus thus given its surface a more dark brown appearance

As in the image below.
Additionally, according to the model the scientists made, they state that

  Under white light the surface should look grayish and more similar to the Moon.

As an aside, the website also shows a fictional map of Venus with 70% surface water.


Why do proto stars on the Hayashi track get dimmer as they contract?

The Hayashi track has an almost constant effective temperature. The pre main sequence star (not protostar) is powered by the release of gravitational potential energy caused by its contraction.

If the star is getting smaller at constant effective temperature, then of course it's luminosity ($\propto R^2 T^4$) decreases.

So perhaps your question is why do PMS stars contract at constant temperature? In other words, why is the Hayashi track almost vertical in the HR diagram? The answer to this is derived in some detail on the relevant Wikipedia page and arises from the fact that low mass PMS stars are fully convective and have atmospheric opacity dominated by H- ions.


Why is the Moon receding from the Earth due to tides? Is this typical for other moons?

It's pretty simple, actually.

The Moon creates tides. Due to tides, the water bulges out towards the Moon (and also on the opposite side).

But the Earth also rotates pretty fast (once a day), faster than the Moon orbits the Earth (once a month). There's friction between the rotating Earth, and the watery bulge created by tides. The rotation of the Earth "wants" to rotate the bulge faster.

In effect, the rotation of the Earth drags the tidal bulge forward - the bulge is always a bit ahead of the Moon. When the Moon is at meridian, the tide is already decreasing.

So there's a bit of extra watery mass on Earth, a little bit ahead of the Moon. This watery bulge interacts gravitationally with the Moon.

This has two effects:

it slows down the rotation of the Earth, gradually sucking energy from it (the Moon pulls the bulge, and therefore the Earth, "back")
that energy is dumped into the Moon's orbital motion, effectively "pulling" it forward

When you dump energy of motion into an orbiting body, it settles into a higher orbit - higher orbit means more energy. Therefore, the transfer of energy from Earth' spin to Moon's orbit gradually makes Moon's orbit larger and larger.

This only happens because the Earth is spinning faster than the Moon orbits it. If the Earth was tidally locked to the Moon (spinning exactly as fast as the Moon orbits it), then no transfer would happen. If the Earth was spinning slower than the Moon's orbit, then the transfer would be opposite (from Moon's orbital motion to Earth's spin).

Note: Counterintuitively, a satellite with more energy actually moves slower, but in a higher orbit. The extra energy goes into raising the orbit, not into making its speed faster. Why this happens exactly is a whole 'nother discussion.


In zero gravity - I'm floating inside a torus as they spin it up, do I float do the "ground"?

Yes...

In the scenario you describe, you (and the air) will remain at rest, while the torus speeds up.  One hopes there are no radial partitions, or, if there are, one comes along to pick you up fairly early in the session.

Unfortunately, the rotating torus will impart its velocity to the air in contact with all the irregularities in the wall/floor/ceiling.  This would then be transferred to the rest of the air.  Floating there, you would experience a very slight breeze blowing you in the direction of rotation of the station.  This in turn would cause you to slowly drift radially outward, until you hit the floor and rolled along until you were at rest with respect to the floor.

Even if the station is not running up from zero rotation rate, you can experience some strange effects.

If the station is big enough, you can travel with low energy.  Simply run against the rotation of the station.  When you are running at the rotation speed, lift your feet, and drift in place until your destination comes along.  Put down you feet, start running, and slow to a walk.

Be careful, though.  If you face in the wrong direction and run at that same speed, your apparent weight will quadruple as you crash to the floor...


Where can I look up assigned origins of gamma ray bursts events? (GRBs)

I found a database that contains all GRBs from April 1991-August 1991. However, it has 1637 registered GRBs so it would probably be enough for your purposes. Here is the link-Goddard Space Center GRB Archive. Also, just a fair warning, that website is incredibly old and very difficult to use, it took me nearly fifteen minutes to navigate to the actual database. Hope this helps!


How to calculate observer's latitude from declination and culmination of a star?

Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.
Neglecting atmospheric refraction, the altitude at culmination is
$$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta| $$

Here are some special cases:

A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).
A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.

For the example in the question, if we solve for $\phi$,
$$\phi = \delta \pm (90^\circ - \mathsf{alt_{max}}) $$

Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N.
GCSE Astronomy's answer is incorrect.


How is interstellar gas density mapped from GAIA data?

Two techniques immediately spring to mind.

For the stars you detect, you can compare their colours and luminosities (Gaia provides photometric colours and distances) with what you expect for a star of that type at that distance. The difference between what you expect and what you observe tells you the reddening and extinction caused by interstellar dust, integrated in a column towards that star. By going through this process for lots of stars in different directions you build up a 3D picture of the distribution of dust.

The measurements are mostly insensitive to gas - simple extrapolation between dust and gas is often assumed.

A second technique would be just to compare how many stars you can see and with what brightness distribution, with a model for the Galactic stellar population and dust distribution. This is an old-fashioned way of estimating dust extinction, works well locally, but I suspect not what is used in that map, for the simple reasons that Gaia also provides the distances to stars, and one of the purposes of the mission is to improve the Galactic model, not assume it.


What celestial mechanics parameters define a lunar occultation of Venus visible from Earth?

A lunar occultation of Venus happens when light from Venus can not reach the observer because the Moon blocks it.
The simplest model for this would be to draw a line in 3D from Venus to the observer and detect when that line passes through some part of the Moon. However that doesn't take into account the time it takes for the light to travel from Venus to the observer for example.
Angular separation
If you have apparent positions in RA and Dec as seen from the observer, then those take care of several corrections.
In that case, the Moon will occult Venus from the viewpoint of the observer when the angular separation between their centers is smaller than the angular radius of the Moon, which is very roughly 1/4 of a degree but changes a lot during the month because the Moon's orbit is elliptical.
Above horizon
You need to make sure they are far enough above the horizon to be visible to the observer. If the location has a flat, dark horizon, perhaps only a handful of degrees is sufficient. If in an urban setting then it may need to be 10 or 20 degrees to clear the lights and buildings, or more.
Visibility (when above horizon)
To be visible by eye is a tough call. Venus is always roughly magnitude -4 (-3.8 to -4.8) and so if it were daytime and passing behind the dark edge of the Moon, it might be visible in the daytime to a careful observer with excellent eyesight since they know exactly where to look for it. Looking through a cardboard tube to block other light might help, and certainly as you mention the farther the angular separation from the Sun at the time the better! But this is pretty rare and many people would not see it easily.
With binoculars or a telescope it will be a lot easier to see in the daytime, since magnification reduces the surface brightness of extended objects (like the Moon and the sky) relative to unresolved objects as Venus would be in a relatively low magnification view.
If Venus were passing behind the bright edge of the Moon, this might also be difficult to see by eye. The surface brightness of the moon is about -6 magnitude per square arc minute (the apparent size of Venus for less-than-perfect vision) which is way brighter than what Venus looks like, so it's going to be a real challenge to see this situation by eye.
But again with binoculars or a telescope this will be a lot easier.


How to calculate the limiting magnitude of Hubble?

The table you linked to gives limiting magnitudes for direct observations through a telescope with the human eye, so it's definitely not what you want to use.

The quoted number for HST is an empirical one, determined from the actual "Extreme Deep Field" data (total exposure time ~ 2 million seconds) after the fact; the Illingworth et al. PDF you linked to explains how it was done. Briefly, they decided the limit would be defined by a signal-to-noise ratio of 5, and then figured out what the noise level in the final combined images was by integrating the total flux in many circular apertures of diameter 0.35 arc seconds placed in regions of blank sky background, and then computing the standard deviation ($\sigma$) of these measurements. Five times $\sigma$ was then the limiting flux. Knowing the photometric calibration (the conversion between observed counts and apparent magnitude; see the paper for specific details), they converted this $S/N = 5$ limit into an apparent magnitude.

If, instead, you want try to estimate in advance what the limiting magnitude should be, then you have to consider several factors. Simplifying a bit, the equation for long-exposure, background-limited $S/N$ is
$$
S/N \approx \frac{s t}{ \sqrt{ s t + n_{\rm pix}(b t)} }
$$
where $s$ = detected flux from the source (star, galaxy, etc.) in electrons (i.e., detected photons) per second, $t$ is the integration time in seconds , $b$ is the flux from the background in electrons per pixel, and $n_{\rm pix}$ is the number of pixels you integrate over for the measurement. $s$ can be computed from the combination of the object's apparent magnitude, the diameter of the telescope's mirror and the overall efficiency of the telescope + detector system. $b$ can be determined in a similar fashion, using the estimated background brightness (which, for a telescope like HST, is mostly sunlight reflected from dust in the inner solar system). $n_{\rm pix}$ is a number chosen as a compromise between maximizing the amount of light from the source (higher $n_{\rm pix}$) and minimizing the noise from the background (lower $n_{\rm pix}$). This depends on the telescope's resolution (better resolution = smaller images of stars and other point-like sources = fewer pixels needed); the empirical calculations by Illingworth et al. used an aperture size amounting to about 100 pixels.


Where are the two recently confirmed unusual brown dwarfs?

The relevant research paper is Schneider et al. (2020) "WISEA J041451.67-585456.7 and WISEA J181006.18-101000.5: The First Extreme T-type Subdwarfs?".
This gives the full designations, of which WISE 0414 and WISE 1810 are abbreviated forms. The designations contain the J2000 coordinates (J041451.67-585456.7 refers to right ascension 04h14m51.67s, declination -58°54′56.7″). It turns out that WISE 0414 is in Reticulum and WISE 1810 is in Serpens (Cauda).
The distances are highly uncertain, from the paper:

If we proceed with absolute magnitude–spectral type relations for field brown
dwarfs (Dupuy, & Liu 2012) and designate the spectral
type of WISEA 1810−1010 to be T0, wildly discrepant
distances are found for different passbands. We find a
distance of ∼14 pc using WISEA 1810−1010’s W2 magnitude, but a distance of ∼67 pc using the K−band
magnitude. Combined with our measured proper motion (Table 1), these distance estimates lead to a $V_\mathrm{tan}$
range of 77–360 km s−1. For WISEA 0414−5854, a similarly large distance range is found (52–94 pc), corresponding to a $V_\mathrm{tan}$ range of 170–307 km s−1. Parallax
measurements for these objects will be necessary to determine accurate $V_\mathrm{tan}$ values. Nevertheless, it is clear
that both sources are high velocity objects consistent
with membership in the metal-poor Galactic thick disk
or halo.

(emphasis mine)


Distance of extra-galactic Classical Cepheids

Since it seems you don't know the equations, I will try to keep it simple.
However, just keep in mind that, in principle, there is no difference between Galactic and extragalactic Cepheids, in this context.

Now, we know that a Period-Luminosity relation holds for the Cepheids:

$P\sim L$

Where $P$ is the pulsation period observed from the Cepheid, and $L$ is the observed luminosity.

We also know that:

$L\sim d$

where $d$ is the distance of the source.

Then it is like to write $P\sim d$.

This means that, if we observe the pulsation period of a Cepheid, we know its distance, and this allows to know the distance of the host galaxy as well.

Some basic sources:

Cepheid variable

Classical Cepheid variable with some numbers

And some diagrams too


"No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning. However, it IS possible that you might see lightning and not hear the thunder because it was too far away. Sometimes this is called “heat lightning” because it occurs most often in the summer. "

here is more information regarding lightning:
http://www.nssl.noaa.gov/education/svrwx101/lightning/faq/


When did the Scutum constellation receive its current name?

IAU's official website mentions that the names of the 88 current constellations (Scutum included) were standardised during the organizations inaugural meeting in 1922 and apparrently haven't been changed ever since.

  Celestial nomenclature has long been a controversial topic. At its inaugural meeting in 1922 in Rome, the IAU standardized the constellation names and abbreviations. More recently IAU Committees or Working Groups have certified the names of astronomical objects and features. In the following links you can find further information on how different objects and features are named.


Fate of a planet orbiting a Lagrange point behind a star?

The stability of this system depends the ratio of masses of the two stars.  If the larger star is more than 25 times more massive than the smaller star, then L5 is potentially stable, and this remains the case even if the planet does not have negligible mass
The calculation of the value is done in detail on physics.stackexchange and there you can establish that the key factor in the stability of L5 is the masses of the major and minor bodies, and not the mass of the Trojan.
If the mass of the central star is more than about 25 times1 the orbiting star, then the planet will be stable. If it is less then L5 is not a stable point, and the planet will tend to move away from L5, and ultimately may collide or be ejected: the behaviour will be chaotic.
1 note $ \frac{1}{2}\left(25+ \sqrt{23 \times 27} \right) \approx 25$ cf. section 5.3 (page 14) here


How do kepler orbits account for planetary migration?

Kepler orbits apply only to 2-body systems.

They apply only, if the masses aren't too large and the orbits aren't too close. Otherwise relativistic effects occur, as for the Sun - Mercury system.

Kepler orbits only apply, if the bodies are sufficiently spherical.

As soon as a third body comes into play, the system can become chaotic.

Too many conditions, especially for our early solar system.

Strange orbits like those around Lagrangian points, or horseshoe orbits can occur. Orbital resonance can stabilize as well as destablize orbits.


How to prove to my friend that aliens have never visited earth

I don't think you can prove that aliens have never visited Earth. It's theoretically provable if one had a time machine, but without one, we can't really prove that aliens didn't visit Earth (say) 2 billion years ago. The physical evidence would be long gone (especially with plate tectonics).

What you can say is that the burden of proof is on the person making the assertion. The null hypothesis is that aliens have never visited Earth. If your friend asserts that aliens have visited then the onus is on him to provide proof. If you can refute all the proofs he brings up, you can argue the null hypothesis remains the one to believe.


What is the YORP effect exactly? Is it just the non-central component of the Yarkovsky effect?

From What is the difference between the Yarkovsky effect and YORP effect?, both the Yarkovsky and YORP effects describe the changes in momentum to a body in orbit due to the re-radiation of photons.  The Yarkovsky effect only describes the changes to the orbital parameters of the body:

The Yarkovsky effect describes a small but significant force that
affects the orbital motion of meteoroids and asteroids

The YORP effect only describes the changes to the rotational momentum of the body:

This rotational variant has been coined the
Yarkovsky-O’Keefe-Radzievskii-Paddack (YORP) effect

Consider a perfect uniform sphere in orbit around the Sun.  For any point on the sphere, the average direction of re-radiated photons will be normal to the local tangent plane (or directly opposite the center of gravity). In this case, we expect a Yarkovsky effect (change in orbital parameters), but no YORP effect (change in rotational momentum). Graphic from this source.

In contrast, consider an irregular body with a huge flat face offset from the center of gravity.  When photons are re-radiated from the flat face, on average they will impart some angular momentum change (YORP effect) and some orbital parameter change (Yarkovsky effect).

Yarkovsky only considers the center of mass recoil from thermal
radiation of a rotating body in sunlight, is YORP just the tangential
component of this recoil?

The forces causing the YORP and Yarkovsky effects are not tangential in general. For a recoil for some impulse on a sphere, the rotational and linear forces imparted are tangential.  However, for an irregular shape like a rod, we wouldn't expect the two forces to be tangential. Here is a little drawing (extremely exaggerated for effect):

In the above drawing, the Yarkovsky vector represents purely translational momentum change and the YORP vector represents purely rotational momentum change. The sum of the vectors that represent forces that cause the YORP and Yarkovsky effects should sum to the mass recoil vector (which is the negative of the impulse):

They YORP and Yarkovsky forces don't necessarily need to be at right angles.  In three dimensions, sorting out the forces can get extremely complicated.    It can be quite complicated for a rod even if the reaction mass vector is at the tip. It's hard for me to imagine the recoil from a "crescent moon" shaped asteroid that has a center of mass exterior to the asteroid body.

Or does it require the body to be non-uniform in order to spin-up the
body?

The body must be irregular in shape to spin it up.  A perfect sphere with a non-uniform surface would not be subject to the YORP effect.  This is as mentioned above, since on average, photon emission at a specific point will be 90 degrees from the tangent plane, effectively canceling all YORP effects from individual emissions.

Would a uniform sphere in fact spin-down?

Again, a sphere would not have any YORP effects because the average re-radiation vector would be perfectly opposite the center of mass.
Note:
When an asteroid strikes a planetary sized body in an inelastic collision, momentum is also conserved.  Some of the asteroid's momentum goes into rotational momentum of the body and some of the momentum goes into the momentum of the center of mass (equivalent to the orbital parameters).  I like this analogy for thinking about the Yarkovsky and YORP effects.


Why is the observable Universe larger than its age would suggest?

The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static.

Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time.

A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is currently made of:

$$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$
where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a constant regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured today.

The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment.

Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:

$$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$

At around:

$$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$
$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}}  $$
we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an acceleration of the coordinates of universe - something which has been confirmed (see: 2011 Nobel Prize in Physics).

So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:

$$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$

where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the Friedmann–Lemaître–Robertson–Walker metric, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.


What is the temperature of an accretion disc surrounding a supermassive black hole?

It depends on the distance from the central body. This gives the temperature $T$ at a given point as a function of the distance from that point to the center ($R$):
$$T(R)=\left[\frac{3GM \dot{M}}{8 \pi \sigma R^3} \left(1-\sqrt{\frac{R_{\text{inner}}}{R}} \right) \right]^{\frac{1}{4}}$$
where $G$, $\pi$, and $\sigma$ are the familiar constants, $M$ is the mass of the central body (and $\dot{M}$ is the rate of accretion onto the body), and $R_{\text{inner}}$ is the inner radius of the disk - possibly (if the object is a black hole) the Schwarzschild radius $R_s$, in which case we can simplify this a little more. So the temperature in the accretion disk is far from constant.

Whether or not there is plasma depends on the exact nature of the disk, the central object and the region around it. For example, a supermassive black hole may have different matter in its disk than that of a stellar-mass black hole. I should think, though, that black holes in binary systems accreting mass from a companion should have plasma in their accretion disks, and supermassive black holes might also have plasma from nearby stars.


Neutron star size in different reference frames

If we ignore the rotation issue (as Pela pointed out, the correction is <1%), then we can approximate the spacetime curvature inside the neutron star using the interior Schwarzschild metric and use the exterior one for the outside. The $r$ coordinate is not the radial distance as one would expect, but defined in terms of $r=$ constant circles having circumference $2\pi r$.

A neutron star of radius $R$ in these metrics has a circumference measured on its surface as $2\pi R$ as expected. But the distance to the core (that is, the length of a hole from the surface to the core) is: $$d(R) = \int_0^{R} \frac{dr}{1-Kr^2} =  \frac{1}{\sqrt{K}}\tanh^{-1}\left(\sqrt{K}R\right).$$
where $K=r_s/R^3$ and $r_s$ is the Schwarzschild radius of the star (all of this assumes constant density in the interior). If we plot this for stars of different $R$ (but same mass, say one solar mass, producing a constant $r_s$) we get the following:

As the neutron star approaches becoming a black hole it gets "deeper": there is more volume than one would expect. Less dense neutron stars have depths that scale linearly with their radius... except that it is $r_s/3$ larger! This odd result comes from the assumption that we look at stars of the same mass even though we make them much larger. A big constant mass object will curve spacetime too, and this produces this effect. (Yes, this means that the sun is $r_s/3 = 984.73$ meters deeper than it looks!)

If we instead decide that we use a constant density (say nuclear density) and plot the depth, the collapse to a black hole instead happens to the right as the neutronium sphere becomes too large. Here I still use the solar Schwarzschild radius as a length scale to keep things comparable. Now, for small spheres the coordinate and measured depths converge:

In actual neutron stars things are complicated by rotation, the core pressure diverging as one approaches $(9/8)r_s$, and of course that it is hard to dig a hole through superfluid neutrons.


Is intrastellar commonly used by astronomers to refer to objects within our solar system?

No. (I'll note that this usage is arguably erroneous, since the nominal meaning of "intrastellar" is "within or inside a star".)

First, I personally have never heard anyone use that term in my twenty-five-odd years as an astronomer.

Second, a full-text search of the astronomical literature on the NASA Astrophysics Data System turns up a grand total of 51 sources that use "intrastellar" (as opposed to, say, 325,000 hits for "intracluster"). None of these has anything to do with the Solar System. Quick checks of a few of them show a mixture of simple typos (i.e., "intrastellar" used when "interstellar" is clearly meant), awkward/failed attempts at synonyms for "intracluster stellar light", and actually correct uses meaning "inside a star".

  would extrasolar be used by astronomers to refer to objects outside of
  our solar system?

You mean like "extrasolar planets", the discovery of which shared the Nobel prize in physics this year? In practice, "extrasolar" seems to be used mainly to refer to other solar systems and the components thereof, rather than just anything at all outside the solar system. (Though I have seen "extrasolar dust" used to refer to dust grains found in the Solar System but thought to have originated outside it.)


How to calculate how much a galaxy moves from its coordinate at redshift 0 to redshift 1?

The galaxies don't "move" (unless you have given them a peculiar velocity); space expands, such that the meaning of $x,y,z$ will change.

The Hubble parameter is simply defined as the rate of change of the scale parameter divided by the scale parameter $\dot{a}/a$. But $z$ is also related to the Hubble parameter, so if your epochs are defined by $z$, the value of the Hubble parameter at any particular epoch doesn't matter.

I am assuming that you have a coordinate system based on our Galaxy being at the "origin". In which case, the "initial" position of a galaxy (at $z=1$) is related to where it is "now" at $z=0$.

Redshift and scale factor are related by $ a = (1+z)^{-1}$. So taking your example of a galaxy at $x_0, y_0, z_0$ at $z=0$, then at $z=1$ all galaxies were closer together by a factor of two, and (ignoring any peculiar velocity) $x_1 = x_0/2$, $y_1 = y_0/2$ and $z_1 = z_0/2$.

I don't know what you mean by "halo catalogue". The simple relationship above completely breaks down on scales smaller than tens of Mpc, because galaxies then become influenced by their local gravitational potential. The relationship between galaxy separation and time then has little to do with cosmology and more to do with the dynamics of their local groups and clusters and no general answer can be given, other than to point out that $z=1$ corresponds to more than half the age of the universe - trying to predict the position of a galaxy within a cluster or group based on a snapshot 7 billion years ago is not possible!


How to choose initial state vectors for a resonant orbit numerical simulation?

I don't know if there are rules of thumb for initializing resonant orbital simulations.  I might suggest:

Setting initial state vectors' Z-Components to zero.  Keeping the whole simulation in the X-Y plane can help with debugging and visualization.

Choosing initial state vectors that are equivalent to orbits with non-zero eccentricity. Orbital resonances tend to amplify or dampen eccentricities, but the effects can be multiplicative. So if you choose circular orbits, you may see less of an effect.

Choosing initial state vectors that line up apogees and perigees with closest contacts.

Running multiple simulations with slight perturbations to initial state vectors to test stability.

Some interesting initial experiments might be to see if you could replicate the instability of the L1 and L2 points of an orbit.  Or perhaps put some little masses in the Kirkwood Gaps and see if they get expelled.

Notes:

I tried writing some of my own numerical simulations to simulate natural planetary flybys. I used an algorithm outlined by Voesenek, but I was getting round-off errors that shot my simulation. Instead of investing more time to fix my simulation, I ended up using Universe Sandbox 2, that I paid for and downloaded from Steam.

Universe Sandbox probably doesn't have the fidelity you need for resonance stability simulations.  Gallardo wrote a pretty slick long term orbital integration software package called ORBE, that is free to the public for educational purposes.  I wish I had read Tabaré Gallardo's 2017 Exploring the orbital evolution of planetary systems before writing my own simulator, since he uses a neat trick by Encke to account for the Sun's gravity analytically via 2-body problem equations while separately treating the inter-planetary gravitational accelerations numerically. This allows larger time steps and prevents some kinds of round-off error.  Gallardo's software is designed to test long term stellar system stability.  Also, it uses orbital elements rather than state vectors, which are a bit less intuitive to me. Gallardo's used of ORBE to run Kirkwood Gap stability simulations is totally AWESOME!!  Here is a capture of one of the figures from Gallardo's 2017:


If a white dwarf collides with a giant star, could it create a TZO?

Thorne and Żytkow's original paper on TŻOs actually opens with a comparison of TŻOs and the type of object you mention, with a  white dwarf degenerate core instead of a neutron star degenerate core. They note that the equilibrium states - essentially, stable configurations - of such combinations lie near the Hayashi track (actually acting a bit like AGB stars, in some cases), indicating high metallicity, as is the case with TŻOs.
These objects generate energy the same way TŻOs do: matter is accreted by the core, releasing gravitational potential energy, and the red giant envelope continues some fusion, although, of course, core fusion has been substantially disrupted by the arrival of the new degenerate core. The main difference in energy production are the ratios between nuclear contributions to luminosity and gravitational contributions to luminosity:
$$L_{\text{nuc}}/L\approx0.99,\quad L_{\text{grav}}/L\approx0.01\quad\text{for white dwarf core}$$
$$L_{\text{nuc}}/L\approx0.04,\quad L_{\text{grav}}/L\approx0.96\quad\text{for neutron core}$$
Why the difference? $L_{\text{grav}}$ is proportional to
$$\frac{GM_c}{R_cc^2}$$
where $_c$ refers to values for the core. The masses and radii of neutron stars differ drastically from those of white dwarfs. This becomes less important in the case of supergiant TŻOs (i.e. $M>10 M_{\odot}$), because convection cycles "burned" nuclear fuel back outwards into the envelope, and so energy ratios become more like those found in the case of a white dwarf core.
This difference in energy production ratios also means that the objects will remain in roughly stable states for different amounts of time; red giants with white dwarf cores can survive in equilibrium for at least an order of magnitude or more as long as TŻOs.
One interesting thing to note is that TŻOs and red giants with white dwarf cores may share some of the same problems when it comes to stability. The envelopes are expected to be composed similarly and act similarly, with the potential difference in nuclear fusion rates, and so the same dynamical instabilities are possible in both cases. However, Thorne and Żytkow state that they find this possibility unlikely.


If sun steals comets from other stars, then what is the primary source of comets?

The primary source of comets for our solar system comes from the Oort Cloud, a smaller amount coming from the Kuiper belt. The Oort cloud is thought to have originated from the remnants of a proto-planetary disk. This paragraph explains this better:

  The Oort cloud is thought to be a remnant of the original
  protoplanetary disc that formed around the Sun approximately 4.6
  billion years ago. The most widely accepted hypothesis is that the
  Oort cloud's objects initially coalesced much closer to the Sun as
  part of the same process that formed the planets and minor planets,
  but that gravitational interaction with young gas giants such as
  Jupiter ejected the objects into extremely long elliptic or parabolic
  orbits. Recent research has been cited by NASA hypothesizing
  that a large number of Oort cloud objects are the product of an
  exchange of materials between the Sun and its sibling stars as they
  formed and drifted apart, and it is suggested that many — possibly the
  majority of — Oort cloud objects were not formed in close proximity to
  the Sun Simulations of the evolution of the Oort cloud from the
  beginnings of the Solar System to the present suggest that the cloud's
  mass peaked around 800 million years after formation, as the pace of
  accretion and collision slowed and depletion began to overtake supply.


What's the brightness of Alpha Centauri from Proxima Centauri?

Not close to being able to read by.

Proxima Centauri is about 13,000 AU from the two binary Centari stars.   Together they have about twice the luminosity of the sun but at 13,000 AU, that's roughly 2/169,000,000 the visible light that the Earth gets from the Sun.

The brightness variation of the full moon to the Sun is about 1 to 440,000, so, some rough math, the two stars would shine about 1/190th as bright as the full moon.   That would make those 2 stars by far the brightest stars in the Proxima Centauri sky, but far short of reading light.   Together, the two stars would be about 10 times as bright as Venus at Venus' peak.  (Venus is about 1/2000th as bright as the moon, -4.4 apparent magnitude to -12.6 for the moon).

That would, I think, be bright enough to be visible during the day under an earth like sky a lot of the time.

The two stars would be quite close to each other too, maximum visible distance between them would be roughly 1/4 the diameter of the full moon, where they would still be visible separate stars when far apart, but they would pass close to each other too, perhaps appearing to touch to the naked eye.

The two stars complete a full orbit around each other in 79.9 years, so the variation would be noticeable over a human lifetime.


Both $B-V$ and $U-B$ are a reflection of how hot the surface of a star is. To first order (inhomogeneous surface temperatures do exist) this has a single value for a given stellar photosphere and so there ought to be a direct relationship between $B-V$ and $U-B$. Here it is, shown below as a graph appropriate for main sequence stars with a composition similar to the Sun. You can reproduce this yourself using data found (for example) in the Handbook of space Astronomy and Astrophysics (pp68-69) compiled by M.V. Zombeck.

You can see that the relationship is complicated; this is because stellar surfaces are not blackbodies and the absorption by various chemical species cuts out chunks of the spectrum, particularly in the blue and ultraviolet. However, given a $B-V$ one could look up the appropriate $U-B$. Note that this is not straightforward the other way around, where there may be two values of $B-V$ possible for a given $U-B$.

In practice things are much more complicated. The plot above assumes the stars are on the main sequence. Stellar atmospheres change a bit as stars evolve and get bigger, with lower surface gravity. Thus changes the relationship a little. The relationship also changes if the composition is not similar to the Sun, especially around the bump at $B-V \simeq 0.4$. Finally, this relationship is intrinsic, it assumes you are not viewing the star through extinction caused by dust. If that is not the case then the whole relationship slides redward in both colours (the vector direction of which is shown  in the plot above for about 2 magnitudes of visible extinction) but also changes shape slightly, depending on how much extinction there is.


Why is solar corona in visible light filamentary?

The coronal plasma is not spatially homogeneous. It preferentially exists inside heated magnetic structures, which take the form of filamentary loops.
It is better to imagine these structures as spiralling ropes of magnetic flux rather than individual field lines. They then act as tubes that the heated coronal plasma can fill.


How to calculate the ground track of the Moon's position on the Earth's surface?

Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows:

latitude = declination of the Moon
longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ascension. (LMST dependents on the date, time, and longitude.) Everything is known except for the longitude.

Accurate Method (using GMST)

Calculate the Local Mean Sidereal Time (LMST) from $LMST=GMST+long_{east}$, where GMST is the Greenwich Mean Sidereal Time and $long_{east}$ is the longitude with positive values in the eastern hemisphere. From the post Local Sidereal Time,
$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H
$ where D is the number of days (including the fraction of days) from J2000 (Jan 1, 2000 at 12 h UT = Julian Day 2,451,545.0) and H is the Universal Time (UT) in hours.

For example, at H=17 hour UT on Nov 1, 2000, I calculate the following values:
$$
RA_{Moon}=18h\;49m\;35s = 282.400° \\
Julian Day = 2,451,850.208 \\
\therefore D=2,451,850.208 - 2,451,545=305.209 \\
\therefore GMST = 296.288° \\
$$
Then
$$
RA_{Moon}=LMST=GMST+long_{East} \\
282.400=296.288+long_{East} \\
\therefore long_{East}=-13.888°
$$

Approximate Method (using Sun's position)

Your method of using the Sun's right ascension assumes that the Sun is over 0° longitude at 12 h UT. Written as a formula and allowing for different times and longitudes, it would be $LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east}$. Setting this LMST to the right ascension of the Moon gives the following:

$$
RA_{Moon}=LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east} \\
\therefore long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h)
$$

The reason this is an approximation is because of the following:

The Sun is not at 0° longitude at 12:00:00 UT everyday because of the Equation of Time. The Sun can be off by up to 16 minutes of time (which is equivalent to 4° of longitude).
The term 15(Time in UT - 12h) should be 15*1.002738*(Time in UT - 12 h), but this difference is small compared to the first approximation.

Continuing the example of Nov 1, 2000, I calculate the following:
$$
RA_{sun}=14h\;28m\;46s = 217.19° \\
long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h) \\
long_{East}=282.400-217.19-15(17-12) \\
\therefore long_{East}=-9.79°
$$

This is different by 4° because I chose Nov 1 since it is close to the date of the largest Equation of Time.

Likewise, if you wanted to calculate the sub-point of the Sun, it would be better to use the GMST and the Sun's right ascension by solving for the longitude from $RA_{sun}=LMST=GMST+long_{East}$.