Multiple theories and hypotheses have been proposed as an alternative to dark matter (DM). The most popular are, arguably,

MOND (MOdified Newtonian Dynamics):A term for various theories where the gravitational force falls off less steeply than $1/r^2$ at large distances.
TeVeS (Tensor–vector–scalar):A relativistic generalization of MOND.
f(R) gravity:A generalization of general relativity, where the so-called Ricci scalar — which describes the curvature of space(time) — may take different forms than just a number.
Entropic gravity:A string-theory-based theory describing gravity not as a fundamental interaction, but as a so-called entropic force.

Note, however, that DM is not just "a fudge factor to explain weird dynamics". What makes most (but not all) physicists believe that DM exists is, I think, that several independent observations yield consistent results, not only qualitatively, but quantitatively. In addition to rotation curves of spiral galaxies, or velocity dispersions in stellar clusters, elliptical galaxies, and galaxy clusters, you have for instance

Gravitational lensing:Massive clusters of galaxies "bend" space around them, causing the light from background sources to diverge from a straight path. Measuring this divergence tells us the mass of the foreground object, which can then be compared to the visible mass..
Cosmic microwave background:The patterns in the inhomogeneities of the background radiation is sensitive to the amounts of both dark and "normal" matter.
Structure formation:DM explains how matter may collapse and form galaxies a few 100 million years after Big Bang, before the Universe expanded too much.
Supernova observations:The way that the brightness of distant supernovae decreases with redshift depends on the expansion history of the Universe, which in turn depends on the densities of the various components of the Universe. A certain amount of matter is needed to explain the expansion, and this amount is larger than the observed amount by the same factor that the other experiments yield.

Some of the above phenomena are explained by the alternative theories as well, but in general several ad hoc hypotheses have had to be made. In my view (but I am no expert on this!), DM is simply the simplest explanation, and so, by Occam's Razor, is favored by most physicists. I think the theory, unfairly, has a reputation for being "some magic that astronomers pulled out of their hat to explain what they don't understand, and which we made sure can't be observed so you can't prove us wrong". But firstly, postulating an unknown object to explain some phenomenon is quite normal in physics — this is how e.g. Neptune and several elementary particles were found. And secondly, there are several candidates for what DM might be, and at least some of these make falsifiable predictions. If DM is a particle, then that particle might produce a signal that can be observed. There are many efforts to search for a DM annihilation signal, which is expected to be in the (tens of) GeV region (e.g. H.E.S.S Collaboration 2011 and Cui et al. 2017). There are also efforts for direct detections — see e.g. Liu et al. (2017) for a review.

One of the most compelling pieces of evidence yet for DM came, I think, with the observation of the Bullet Cluster (and subsequently many other clusters), where lensing clearly shows how unaccounted-for mass follows the collisionless galaxies, but not the collisional gas, as seen in this beautiful image:

The Bullet Cluster. Pink shows X-rays from the gas, whereas blue shows where the mass is concentrated. Credit: NASA/CHANDRA.


How Wide Would the Habitable Zone Be in a Singular Red Giant System?

Habitable zones have a width that scales as the square root of the luminosity. So a 10000 solar luminosity star will have a zone about 100 times wider.

However, the temporal width is another thing: the luminosity changes relatively fast, so planets will not be in the zone over long timescales.

(The reason for the square root dependence is that a piece of surface at semimajor axis $a$ gets $L/4\pi a^2$ Watt of incoming energy, and if a blackbody with emissivity $\epsilon$ radiates away $\sigma \epsilon T^4$. Energy balance gives $L/4\pi a^2 = \sigma \epsilon T^4$, or rearranged, $a = \sqrt{L/4\pi \sigma \epsilon T^4}$, proportional to $\sqrt{L}$. The life zone is given by the range where $T$ allows liquid water. In practice this is more complex and fuzzy since it all depends on complex atmospheric properties, but as a first approximation it feels OK.)


Could anyone recommend a good book about the basics of astrophysics?

Answers to these sort of questions tend to be opinion-based which is frowned upon, but Neil DeGrasse Tyson's recent book Astrophysics for People in a Hurry has had very positive reviews. Injecting some facts, it has apparently spent a year on the New York Times bestseller lists and has sold over a million copies. It could be a good place to get you started before starting on a e.g. introductory college/university textbook.


Because we are standing on a rotating globe.  We are changing direction as the Earth spins. This appears exaggerated in the video as the horizon is not visible, so there is no way of seeing how the camera was rotated to view the moon as the eclipse progressed.

Relative to the stars, the moon appears to enter the Earth's shadow from one side and leave it from the other,


Why isn't IMBH candidate GW190521 very "chirpy"? Which types of events will produce the most chirp-like behavior?

All the binary mergers chirp, but the overall timescale of the event depends on the total system mass (or rather the chirp mass - see below). The more massive the system, the more rapid the evolution of the amplitude and frequency and the lower the orbital frequency when it finally merges. What you observe is also governed by the response of the detector - that is, its strain sensitivity as a function of frequency.
Thus high mass stellar black hole mergers have rapidly evolving signals and end their chirps at relatively low frequencies. Lower mass black hole mergers or neutron star mergers are much more slowly evolving and end their chirps at much higher frequencies.
The key equations (assuming circular orbits and to first order) are:
$$ \frac{df}{dt} \simeq \left(\frac{96}{5}\right)\left(\frac{G\mathcal{M}_c}{c^3}\right)^{5/3}\pi^{8/3} f^{11/3}\, ,$$
where $\mathcal{M}_c$ is the "chirp mass" and is given by $(m_1m_2)^{3/5}/(m_1+m_2)^{1/5}$, where $m_1$ and $m_2$ are the component masses. The chirp mass basically gets larger with the total mass of the system for a given mass ratio, but strictly speaking, what I said in the first paragraph about high- and low-masses means high and low chirp masses.
You can see from this formula that at a given frequency $f$ (in this case in the LIGO sensitivity band), that the rate of change of frequency is higher for higher chirp masses.
The chirp ends when the objects "merge". Exactly what this means is a bit loose, but is when the separation of the components is a small multiple of their radii in the case of neutron stars or a small multiple of their Schwarzschild radii for black holes. A simple use of Kepler's third law (not strictly applicable in General Relativistic conditions, but it gives the right scaling), and recalling that the GW frequency is twice the orbital frequency yields
$$ f_{\rm max} \sim \left(\frac{GM}{\pi^2 a^3_{\rm merge}}\right)^{1/2}\ ,$$
where $ a_{\rm merge}$ is the separation at merger and $M$ is the total system mass. If we let $a_{\rm merger} \sim 4GM/c^2$ (for black holes), then we see that $f_{\rm max} \propto M^{-1}$.
$f_{\rm max}$ is around 130 Hz for a $30+30$ solar mass black hole merger (like GW150914, which had $\mathcal{M}_c \simeq 28M_{\odot}$, so would be just $\sim 50$ Hz for a merger resulting in a 150 solar mass black hole like GW190521 (which had $\mathcal{M}_c \simeq 64M_{\odot}$).
So that I think is your answer. The sensitivity of LIGO is quite poor below 30 Hz due to seismic noise. Thus in this event, the merger and chirp is barely seen at all in the sensitive LIGO frequency window of 30-2000 Hz - just the final few orbits (in fact only two for GW190521) before the merger and ringdown phase begins at a peak frequency of $\sim 50$ Hz.
To get a more chirpy signal that is detectable by LIGO you need (a) a lower mass system with a higher $f_{\rm max}$ and (b) it needs to be close enough so that LIGO can detect the lower amplitude GWs with $f<f_{\rm max}$ that are emitted prior to the merger.


What is the level of tidal heating between bodies that are already in mutual tidal lock?

There are actually several components of tidal forces which serve to distort a planet or moon - diurnal, nonsynchronous rotation, ice-shell thickening, orbit obliquity, and polar wander.  A moon could be stressed by any combination of these mechanisms, leading to frictional forces and heating.

Diurnal Stress - Since orbits are ellipses, not circles, the moon will experience a differential gravitational field.  When the moon is closer to the planet, the tidal stresses will be slightly larger than when it is closer away, since the gradient will be steeper. In addition, since Kepler's 2nd Law informs us that a body moves faster when it is closer to its primary, this means that the tidal locking is imperfect.  When the moon is close to its primary, it is moving slightly faster than it is rotating.  Likewise, when it is further away it is moving slightly slower than it is rotating.  This leads to the moon seeing its primary as oscillating slightly in the sky.  Source
Nonsynchronous Rotation Stress - If the moon's crust is decoupled from its core by a liquid layer (either liquid rock or liquid water), the crust can rotate freely over the core.  The core will stay tidally locked to the primary, but the shell can move around.  Since the core will have a tidal bulge, when the crust moves around it will become stressed.  The crust feels a torque because its thickness varies across the moon's surface.  Source
Ice-Shell Thickening - Icy moons, such as Europa, can experience stresses caused by their icy outer shells freezing and thickening.  As the moon loses heat, the water at the bottom of the shell will freeze.  This will increase its volume, creating extensional stresses.  At the surface, the cooling of the ice will contract it, causing compressional stresses.  While this is not a tidal stress, it is still a source of stressing which can act on these moons, so I thought I'd throw it in.  Source
Orbit Obliquity - Most moons do not rotate exactly perpendicular to their orbital plane.  Rather, their rotational axis has some amount of obliquity.  This obliquity changes the latitudinal orientation of the tidal bulge as the moon orbits the planet.  This creates additional stresses as the tidal bulge gets pulled on.  Source
Polar Wander - Large impacts can cause a moon's lithosphere to rotate and reorient itself relative to its rotational axis.  When this rotation changes the apparent location of the rotational poles, it is called "polar wander".  Polar wander causes stress in a similar manner to nonsynchronous rotation.  The lithosphere rotates over the core's bulge, pushing on the crust.  Source

There may be other stress mechanisms that I am not aware of, but these are the main ones.  If you want more information about the topic, or want to see some visualizations of what these various stresses look like, check out SatStressGUI, a program I helped developed which models stresses on icy moons.


When will the next series of mutual eclipses of Jupiter's moons begin?

Mutual eclipses and occulations of Jupiter's moons occur during Jovian Equinox, twice in each orbit of Jupiter, and so roughly every 6 years.

The last series of events was in 2015, so the next will start in 2021. There will be a series of events between January and August.

The site http://lnfm1.sai.msu.ru/neb/nss/nsszph517he.htm can generate ephemerides for these events.


Is there a simple, analytical formula for the lunar phase brightness curve?

The magnitude of the Moon without the opposition surge is given by a simple equation:
$m = -12.73 + 1.49 \cdot |\psi| + 0.043 \cdot \psi^4,$
where $\psi$ is the phase angle in radians (Allen 1976).  This can be converted to flux using $m \propto -2.5 \cdot ^{10}\!\log F$,
where $^{10}\!\log$ denotes the 10-base logarithm (as opposed to e.g. the natural log).
For the increase in flux due to the variable distance and opposition surge, I multiply F with the factor
$\left(\frac{\Delta_0}{\Delta}\right)^2  \cdot  \max\left(1, 1.35 - 2.865 \cdot |\psi| \right),$
where $\Delta_0$ and $\Delta$ are the mean and current distance to the Moon, respectively.  However, I can't find a reference for the last term at the moment, and I can't remember where I obtained it.  I'll post that information if I find it.


Basic explanation of baryon acoustic oscillations; what if anything is actually oscillating?

tl;dr nothing is oscillating now, but it was in the early universe. Now we just see the frozen relics
This is just an attempt at a simplified explanation of the Wikipedia article linked in the question.
There were pressure waves in the very early universe. They came from a balance of gravity and radiation pressure (the matter at the time was hot enough to be opaque). Whereever there happened to be a locally higher density (of matter, radiation and dark matter) the excess pressure would cause an expansion and then gravity would cause a contraction sending waves out at (according to the article) half the speed of light.
As the universe expanded and cooled, matter eventually became transparent (it was hydrogen and helium gas) letting the radiation pressure escape and leaving higher densities both in the location of the original concentration and at the crest of the wave a certain distance away (but the same certain distance for the first waves from each concentration). Those high densities are where the galaxies formed, so what we look for today is an excess of galaxies at a specific separation and that's what we see.


The lowest exposed point on land is the dead sea shores, at -413metres below sea level. so theres a good chance you could find a spot that lines up correctly to give you a view of a piece of the horizon well under sea level. I dont believe that elevating yourself is the key factor here though. At Around 35000 ft you can start noticing the curvature of the Earth, the horizon being 200 some odd miles away (Pythagorean theory -theorem?- can be used for this, several sources on google explain the steps to find your equation, and far better than I can. ) but the distance to the horizon grows rapidly. So you would certainly  want to find the sweet spot of height /distance, too high and you wont know when the land has ascended back to sea level, as the terrain will become more vast and harder to judge detail or distance. Too close to the shore and your only seeing a few miles but the angle of sight wouldnt be the best possible for finding the lowest horizon. You would want to work out the math and find a spot that puts the dead sea shores as your visible horizon, bc if your standing there, watching the sun rise, the sun is rising over a horizon that isnt the lowest accessible location...


How to calculate the stellar visibility?

Maybe that piece shouldn't consider necessarily individual stars rising from the East but entire constellations/parts of constellations. Trying to express the sky movement as a whole sounds really nice to me. There could be planets setting (like Mercury or Venus right after dusk in the West) or others rising (quite bright) Jupiter, Saturn Mars. Even large deep sky objects portrayed there could be amazing - like the Orion Nebula, Andromeda Galaxy, The Double Cluster, Pleiads, M44 Cluster, meteors, etc. I found a good expression of items of the night sky here.


Orbital radius depends on speed of the body. Size is largely irrelevant. Imagine a  space craft and much larger space station, able to orbit meters apart at the same speed, with engines switched off. An astronaut going on a spacewalk would be launched away from the station, being much smaller and so on wrong orbit. No. It's as relevant as "Boeing 747 shouldn't be able to fly because it's painted the wrong color." False and based on wrong premise.
For anything to ring you need atmosphere, a medium for sound to travel through. Moon, floating in vacuum, doesn't ring, nor create any other sound for that matter. Even the most powerful meteor impacts are entirely silent. False and based on wrong premise.
Countless iron meteorites hit the surface of the Moon, often shattering on impact. Specks of iron are very common in lunar soil, and I have no doubt it's entirely possible shavings-shaped pieces of lunar meteor might have found their way into the samples. Likely True and nothing odd about it.
...how does that even relate?


The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can only be supported by electron degeneracy.

Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be determined from the binary motion (with Sirius A), or gravitational redshift or from estimating the surface gravity from the width of absorption lines in its spectrum. These tell us that Sirius B has a radius of $0.0084 R_{\odot}$ and a mass of about $1 M_{\odot}$ (e.g. Holberg et al. 2012). This leads to an estimate of the average density of $2.4 \times 10^{9}$ kg/m$^3$.

Now, we know that for such an object to find any kind of equilibrium that the pressure must be greatest in the middle. Since for any plausible equation of state, the pressure increases with density, then the white dwarf must be even denser than the average at its centre.

From there it is a standard piece of bookwork to show why the white dwarf must be supported by electron degeneracy pressure and I will repeat it here.

An electron gas can only escape degeneracy if $E_F - m_e c^2 < k_{b}T$, where $E_F - m_e c^2$ is the kinetic energy associated with the Fermi energy of the electrons, $m_e$ is the electron mass and $T$ is the temperature.

The Fermi energy can be derived from the following formula (again, standard bookwork, this is not exotic physics, it is the same physics that describes how conduction takes place in metals).
$$ E_F = \left[ \left(\frac{3h^3n_e}{8\pi}\right)^{2/3}c^2 + m_e^2 c^4 \right]^{1/2},$$
where $n_e$ is the electron number density. Now it turns out that the temperatures we are going to need to avoid degeneracy (see below) are easily going to be sufficient to completely ionise the contents of the white dwarf. If we assume the white dwarf is made out of carbon or oxygen, then the relationship between mass density and $n_e$ is just $\rho = 2 n_e m_u$, where $m_u$ is an atomic mass unit (the 2 just comes from the fact that there are 2 mass units per free electron). We could mess about with the composition a little to change this relationship, but it doesn't matter very much.

Making the appropriate substitutions and using $\rho = 2.4 \times 10^{9}$ kg/m$^3$, we obtain $E_F = 0.75$ MeV (or $1.2\times10^{-13}$ J). If we now use the test
$$ T > \frac{E_F - m_e ,c^2}{k_b},$$
we find that to avoid electron degeneracy then the temperature inside the white dwarf must exceed $2.75 \times 10^{9}$ K (and be much higher in the centre of the white dwarf where the density and Fermi energy are larger). If temperatures were this high inside a white dwarf and it was made of something like carbon or oxygen then nuclear fusion would take place (rapidly) and the luminosity of the white dwarf would not be as low as it is.

But what if it were made of iron? There would be no nuclear fusion, even at these high temperatures. OK, leaving aside that if the interior really were that hot, it could not be hidden very easily by an atmosphere, then this would not be a stable situation. The star would be losing copious heat via neutrino losses. If the gas were non-degenerate, the star would rapidly contract and become even hotter and even denser, and if you are not going to allow for degeneracy (i.e. you postulate that quantum mechanics does not exist), then rather quickly the white dwarf ends up being a black hole.

A back of the envelope estimate would be to divide the gravitational potential energy by the current luminosity. Taking Sirius B as an example, the timescale is
$$ \tau \sim \frac{GM^2}{RL},$$
where $L$ is the luminosity. Taking the current luminosity from photons at the surface this timescale would be a few $10^{10}$ years (plausibly stable). But if the interior were at $\sim 10^{10}$ K, the neutrino losses would be many orders of magnitude greater and the timescale would be just millions of years.

We know that this isn't happening to white dwarfs because we know they are stable and long-lived. We see them in clusters of stars of known age and associated with stars that are much older than millions of years. For instance we know that stars like Sirius A are hundreds of millions of years old.
We can calculate using the theory of electron degenerate structures that the faintest white dwarfs we see have been cooling for 10 billion years without collapsing - precisely the age expected for the galactic disk. The cooling ages of white dwarfs in clusters match the ages of those clusters very well.

Degenerate matter is very common on earth. The conduction electrons in a metal form a degenerate gas. If you mean the kind of degenerate matter in white dwarfs, then no, at least not in any stable state. The pressures required to constrain such energy densities are too large to create in the lab.


Hypothetically, could a civilization in a solar system near a supermassive blackhole use the blackhole as a huge gravity lens telescope?

It is indeed true that we are contemplating the use of a star: namely, our Sun, as a telescope. The concept is called the Solar Gravitational Lens (SGL). See here for our very latest manuscript, which also contains references to our earlier work: https://arxiv.org/abs/2104.08442
The tremendous light amplification and angular resolution of such a gravitational lens telescope arises from the fact that its light collecting area is effectively the Einstein ring that is seen around the lensing object. The diameter of the Einstein ring determines its angular resolution; its diameter and thickness together determine the light amplification. When using the Sun as a hypothetical telescope, in conjunction with a modest (~1 m aperture) observing telescope, light amplification is of ${\cal O}(10^9)$ and angular resolution is below a nanoarcsecond, though deviations from spherical symmetry in the solar gravitational field tame these numbers somewhat.
These numbers are determined solely by the size of the Einstein ring, i.e., the radius of the Sun (since the Einstein ring has to appear around the Sun; the Sun itself is opaque.) What the mass of the Sun determines is how far away from the Sun images can form. Obviously, the more massive an object is, the more it deflects light, so the sooner light rays can meet. For the Sun, the earliest point where initially parallel light rays passing on opposite sides of the Sun intersect is about 550 astronomical units from the Sun. For a more massive object, this could happen sooner. Or conversely, at a given distance, the Einstein ring seen around a more massive object will have larger radius, corresponding to more light collected and higher angular resolution.
However, the fact that to reach, e.g., Sgr A* at the center of the Milky Way requires travel over several kiloparsecs kind of negates any practical advantage that an SMBH might offer as a practical gravitational lens telescope. Any civilization possessing the technology to travel across the Milky Way (assuming such technology even exists within the realm of physics) probably has other, more practical means at their disposal to make astronomical observations of the deep sky.


Does the radio signal decay when it travels through the intergalactic space?

All electromagnetic radiation from a point source - which a normal radio transmitter is - propagates according to the inverse square law which means that the intensity of the signal is inversely proportional to the square of the distance. This happens on earth and in deep space equally.

So this will mean that for any signal there will be a distance at which it's becomes indistinguishable from the background noise of the universe. However, that distance will depend on the initial strength of the signal.


Why does the Sun's surface contain lithium?

The Sun has lithium because, like all the elements that it has heavier than helium, it was born with it.

Many stars, particularly older metal-rich stars or cool stars with deep convective envelopes show very little Li in their atmosphere - and that is true for the Sun too. But this trace of lithium is still detectable (just).

By comparison with meteorites, we know that the Sun was born with about 100 times the abundance of Li that is seen in its photosphere now. Roughly speaking, about 20% of this Li was produced by big-bang nucleosynthesis and the rest was likely produced, before the Sun was born, in (giant) stars and (classical) novae via rather complicated synthesis and mixing routes.

Most of the Sun's Li has been destroyed in its interior. It reacts with protons to form helium at temperatures lower than required for hydrogen fusion. However, the exterior of the Sun has not been completely mixed with the interior, so a little still remains in the photosphere.

As to why stars with planets deplete their Li more efficiently, well there is still a bit of a question mark over whether that is true or not. If it is, then it probably has something to do with the history of rotation and angular momentum loss in the star. Stars with long-lived, planet-forming discs may have spun down more readily, creating an internal differential rotation that aids the mixing process. It is a clear observational result, that slower-rotating stars in young clusters have depleted more Li than their faster rotating siblings.


Is antimatter also attracted by gravitational field?

Yes, common believes are because gravitation has effect on particles with mass. Antimatter are equivalent to there counterpart, sharing the same mass but opposite charges. Therefore, antimatter is attracted by gravitation.
But, this is still not experimentally proven, therefore a more theoretical explanation of the gravitational interaction.
Some researches indeed suggest anti-gravitational interaction between matter and antimatter.

source


Which are stars and which are noise in this comet photo?

Is that right?

Yes.

  Is the fuzzy one an extended object?

That would certainly be my guess (probably a distant galaxy).

  What causes so many isolated pixels to be so much brighter than the background? Is this just the tail of a statistical distribution of
  shot noise, or are there other mechanisms that can produce single
  pixel noise many standard deviations above the statistical shot noise
  distribution?

Some of those might be "hot pixels", which are manufacturing defects that cause individual pixels to overproduce electrons. Most of them are due to so-called "cosmic rays", a generic term used in astronomical imaging to refer to the effects of energetic charged particles hitting the detector. In ground-based images, these are often not actual cosmic rays, but instead are particles from the decay of radioactive isotopes in or near the detector. In the case of space-based images like these, they more often are proper cosmic rays, though many of them are really charged particles trapped in the Earth's magnetosphere. (There is a part of Hubble's orbit which takes it within the South Atlantic Anomaly, where the Van Allen belt dips down toward the Earth's surface; observations are not scheduled during those periods, because of all the extra cosmic-ray hits the detectors receive.)

For example, here is a preview image from one of the other exposures in the sequence of observations. If you look closely, you can see that many of the cosmic-ray hits are slightly elongated, because the particle was traveling at an intermediate angle and passed through two or more adjacent pixels. The bright linear streak right next to the comet is an extreme example of this, where the particle was traveling almost parallel to the detector and so passed through an extended sequence of pixels.

The usual approach to dealing with this is to take several short exposures, one right after another, and combine them afterwards by rejecting extreme outlying pixel values from a single exposure (stars and other genuine objects will have similar pixel values from one exposure to the next, while cosmic rays are randomly located and do not repeat). This is why the HST images you see usually don't have cosmic-ray artifacts: they are the result of combining multiple exposures. (See here for a discussion and an example.) There are also statistical analysis tricks one can use to try to clean up individual exposures, though this necessarily involves interpolating data.


Where do new stars get their hydrogen from?

Stars only burn hydrogen in their core, where the temperature gets high enough for nuclear reactions to occur. They end their lives when they run out of fuel in the core, but lots of hydrogen still exists in their envelopes.

The Sun will burn 5-10% of its mass before exhausting its core. When this happens, the core will contract due to the radiation pressure disappearing. It then starts to burn hydrogen in a shell around the core. Eventually, when the Sun dies, it will have burned less than half of its hydrogen. Larger stars burn an even smaller fraction.

This means that, when stars die they still leave hydrogen behind for the next generation.

Galaxies can still run of of gas, though. Since after all, each $M_\odot$ of star formed burns of the order of $1\,M_\odot$, if a galaxy isn't fueled with new gas it will become depleted on a timescale of order $1$ over its specific star formation rate sSFR, which is its star formation rate SFR measured in Solar masses per year, divided by its stellar mass $M_*$ in Solar masses:
$$
t_\mathrm{depl} \sim \frac{1}{\mathrm{sSFR}}=\frac{M_*/M_\odot}{\mathrm{SFR}/M_\odot\,\mathrm{yr}^{-1}}.
$$
For instance, a $10^9\,M_\odot$ galaxy with a star formation rate of $10\,M_\odot\,\mathrm{yr}^{-1}$ will become depleted of gas in roughly $10^8$ years.

Gas-depleted galaxies do exist, and the more depleted they are, the lower the star formation rate is (e.g. Rose et al. 2010), but in general the timescale is longer than the above, since galaxies also accrete gas from the surrounding circumgalactic medium.


Why is a black hole's Hawking radiation temperature inversely proportional to its mass/size?

In a sense, you're right.  Vacuum energy over a surface area is proportional to the surface area so larger area means more of the particle-antiparticle occurrences and that's the source of Hawking radiation, that is, if you don't mind the virtual particle explanation, which is the only explanation I sort of understand.  I look at the equations and my brain goes blank.
But surface area isn't the only factor.  The more important factor is how quickly the gravitational field drops off, which, for stellar mass black holes or larger, the drop off over particle-anti particle distance is tiny, given a radius of at least a few miles and a very small particle-anti particle distance or high probability of a small distance.  That small drop-off in gravitational field leaves little chance for the outermost of the two virtual particles to escape.
That's why the only black holes that are thought to be hot are theoretical micro black holes, which are tiny.  The smaller the theoretical black hole, the faster it's gravitation falls with distance, so, even small distances like those between a particle and an anti-particle, make it much easier for one of the particles to escape the gravity, effectively stealing energy from the black hole and radiating outwards.


First things first ...

  Tachyons

These are entirely hypothetical and no evidence for them exists.

  How much does subatomic particles such as photons, tachyons, and electron being the biggest particles etc. occupy space?

How long is a piece of string ?

There's no simple answer to what size an elementary particle is.  You can view them, depending on context, as having no size (point like) or being spread all over space (wave distribution).

We can't say how many free particles there are (i.e. particles not bound to other particles).

This is a lot more complicated by the current mainstream thinking that dark matter and dark energy make up most of the mass/energy content of the universe.  Whatever dark matter may be, if it turns out to be a particle or particles we don't know about yet (quite likely on existing evidence) then those particles will almost certainly exist as free particles (only gravitationally bound together) as current thinking is that dark matter does not interact much at all with other matter or itself.  And we have no idea what dark energy could be.  So with most of the mass/energy is in forms we cannot identify we can't really say anything much about what you're asking.

A very, very, very rough estimate on the amount of dark matter is about 25% of the mass-energy of the universe.

  If put in a ball how big of a ball would that be?

In a ball it would be (possibly) a huge black hole comprising anywhere from 25% to 95% of the mass-energy of the universe.  I won't even bother estimating the size of the event horizon as it's just a useless number.

However "how big a ball" could equally mean it was, as it is now, spread out over the entire universe.  So a perfectly correct answer is - universe sized.


How big would the Milky Way be if all the stars were emptied into one sphere?

Obviously the answer depends on what the density of such a sphere was!
The only sensible way I can interpret this question is to ask what is the combined volume occupied by all of the galaxy's stars and then get a spherical radius from this.
A modal star in the galaxy has a mass of around 0.3 times the Sun and a radius of 0.3 times the Sun. Since the stellar mass distribution is quite peaked at that value, we could do worse than simply assume that we have about $10^{11}$ stars like this, since stars of higher and lower masses are less common.
In which case, the radius containing a similar total volume is just $0.3\times 10^{11/3}\ R_\odot= 1400R\odot$.
The problem with this calculation is that although the masses of stars are quite peaked, their radii are not. Some evolved stars individually have radii that are a significant fraction of the combined radius figure. Thus although the figure above might be good for the combined radius of all main sequence stars it certainly isn't a good number if you wanted to include all giant stars, with the densities they have now.


Is it possible make artificial gravitational waves?

Typical accelerating masses will generate gravitational waves.  but because gravity is so weak of a force; and current/plausible near future detection apparatus will be unable to detect signals from anything less dense than gravitationally stable masses of degenerate matter (ie neutron stars or white dwarfs) moving at near relativistic speeds.


How many stars can stay close to each other without collapsing?

Well, if they have an angular momentum with respect to their center of mass, or in other words , if they are in orbit, gravitational collapse is not the issue.

This is what happens when three bodies are in orbit around each other. Don't get disappointed so soon. There is indeed a way to avoid this. If the three bodies are of comparable mass, and one of the bodies is situated at L4 or L5, tuning the orbital parameters may work just fine, but then you gotta worry about tidal forces.

So, if the size of our third star exceeds the area given to it at L4/L5, it will experience tidal forces and will start ejecting mass to either of the two other stars, or both.

But as Zolan commented above, astronomy with an infinite universe has a possibility for everything. What are listed above are the most common problems which arise, that is the reasons which are mainly responsible for making such systems highly unlikely of existence.

But here's a system which works the way you described :

A binary system exists, and the third star is far enough that it experiences the gravity as that of a single star. And the nearest planet(as you want it to be a "solar system"), is so far that it experiences the gravity of the three stars as that of one. Now, since "close" is a relative term, for the planet, the three stars are appreciably close.

So, the answer to your question, YES and NO


How to convert VSOP's xyz coordinates to heliocentric spherical lbr coordinates?

Looking at the numbers and comparing to JPL Horizons, the best match is to the position of the Earth-Moon Barycenter in heliocentric coordinates. That's of course consistent with the data being labeled as:  EMB, Ecliptic Heliocentric Coordinates.

If you can live with some small uncertainties due to the details of how coordinate transforms may evolve over time, I believe you can just apply some simple trigonometry. You can confirm here.

$$r = \sqrt{x^2 + y^2 + z^2}$$

$$l = tan^{-1}(y/x)$$

$$b = sin^{-1}(z/r)$$

Data in Question: for JD 2415021.0
position (AU):     [-0.2054467990,  0.9615310525,  0.0002141233]
velocity (AU/day): [-0.0171063024, -0.0036576954, -0.0000011815]

JPL Horizions Earth-Moon barycenter J2000 Heliocentric for JD 2415021.0
position (AU):     [-0.2054465857,  0.9615310983,  0.00021393097]
velocity (AU/day): [-0.0171063032, -0.0036576916, -0.00000118253]

JPL Horizions Earth J2000 Heliocentric for JD 2415021.0
position (AU):     [-0.2054522688,  0.9615603123,  0.00021303115]
velocity (AU/day): [-0.0171136285, -0.0036594066, -0.00000182603]

JPL Horizions Earth J2000 Solar System barycenter for JD 2415021.0
position (AU):     [-0.2022720210,  0.9679285296,  0.00010939456]
velocity (AU/day): [-0.0171209826, -0.0036556277, -0.00000165111]


What is meant by "there is no refraction at the zenith"?

To predict the position of a star one has to account for atmospheric refraction. In a simple model, the atmosphere is treated as being composed of layers of gas with variation in density and temperature due to height, but no variation with latitude or longitude and no turbulence. In such a model, one can predict the position of a star to a great deal of accuracy (so it is a useful model), and in this model there is no refraction at the zenith (as the light doesn't pass at an angle through a medium of varying optical density).

In such a model there would be no twinkling, however modelling twinkling is not needed for positional astronomy. Large variation in optical density that could cause stars to appear significantly displaced at the zenith doesn't occur. The variation in optical density with height is much more significant for positional astronomy.


Your scenario is, of course, rather speculative and considers the very distant future. I don't like that article you're referencing, some stuff seems inaccurate (for example the fraction of baryons locked into stars is tiny and will always remain tiny).

If two brown dwarves collide (extremely unlikely, but if you wait long enough, anything with some tiny non-zero probability eventually will happen), they form a new star with their combined mass. Since the brown dwarves have not undergone any hydrogen burning, the new star has the same composition as a new-born star of the same mass would have had (except for lithium) and, if massive enough, will evolve just like an ordinary M dwarf (or so). In other words, it will fuse hydrogen into helium as all stars on the main sequence.


What is this "time travel" these popular blogs refer to?

My question is to know if will you "live for a thousand years" (...) or "only" 10 years in your spaceship - time "froze" during the travel.

If you were to travel to close the speed of light (not doing the math, plainly speaking), you would appear to live for a thousand years to the people that are outside of your frame of reference. Time would actually pass the same to you and you would observe the time outside your frame of reference to pass significantly faster.

  Is it a scientific fact?

Human time travel is a very real scientific fact, but the implications are too complex to be presented to the public sphere since it has no practical applications yet. If we were to sent people to another star system (Alpha Centaury A/B - at 4.3 light-years from Earth), accelerating their ship close to the speed of light would make the time in their frame of reference pass slower relative to the outside of their frame of reference, thus allowing them to reach the destination in a shorter time from their perspective. It would be useful when managing supplies on this vast journey.


How can I tell if my solar glasses are up to ISO standards?

if I had a pair of glasses, what would be the proper way of determining if they met ISO 12312-2 standards?

The process involves a lot of lab equipment, calibrated sources of infrared, visible light, and ultraviolet, and calibrated detectors. Extremely unlikely you'll be able to whip something up in your garage.

Buy from vendors with good reputation; these are online stores that sell nothing but astronomy equipment, and they know what they're dealing with. If you're an astronomer, you know who those are. If you aren't, go on various astronomy forums and see which vendors people are recommending. Or see the AAS page for a list of vendors.

Amazon is now cracking down on bad eclipse glasses, but realistically speaking they are not an online store specializing in astronomy, so their product vetting may or may not be perfect.

It's getting a little late now, so better hurry up and get your eclipse glasses already.

You can always dash to the local welding supplies store and get a #14 welding glass. It's completely safe and passes the ISO standard. Maybe get the whole mask to go with it, so your face won't get sunburned.


Must time pass more slowly, relative to our inertial reference frame, inside galaxies that are currently located at half way to the Hubble Horizon?

The point of the unification of space and time in relativity is that there's no sense in asking what's happening "right now" at a different position. It makes as much sense as asking what's happening "right $y$" at a different $x$ in Euclidean geometry. If you fix a Cartesian coordinate system then "right $y$" is mathematically well defined, but being mathematically well defined doesn't make it any more sensible to talk about. The laws of nature don't have any notion of distant simultaneity and don't care about your coordinates.
There are situations in which there's a coordinate-independent, geometric difference between two curves (worldlines) in spacetime that justifies saying that one is shorter (less elapsed proper time) than the other in some absolute sense. One such case is when the curves meet at two spacetime points and you're only interested in the length between those points (as in the twin paradox/effect). Another case is gravitational time dilation, where the curves are analogous to circles of constant latitude on the earth.
Worldlines moving with the Hubble flow are more like lines of constant longitude on the earth. The distance between them varies (as a function of latitude), and as a result if you draw rhumb lines from one of them to another (analogous to lightlike worldlines), they'll arrive at a different separation than they departed at (analogous to red/blueshift). But the situation is symmetric, and it makes no sense to say that one longitude line is longer than another.


What enhances the capture and merge rates of pairs of small black holes orbiting around supermassive black holes?

I would think that the same reason that a gas disk enhances the growth of planetesimals. Drag from the disk enforces circular, co-planar orbits, which in turn means that objects that get close to each other have small relative velocities.
Edit: So what is thought to be going on is a little bit more complex than the simple answer above. The density of massive objects embedded in the accretion disk can be enhanced at certain orbital radii. These "migration traps" (Bellovary et al. 2016) are caused by torques exerted on the orbiting objects (in this case black holes) caused by gas accretion. For certain disk conditions it is possible to find locations where at smaller radii the torque acts to increase the orbit radius, but at greater orbital radii the opposite is true. This would cause objects to accumulate in these migration traps, giving them more chance to interact as per my original answer.


Tongue-in-cheek quote on dust extinction

Well, I managed to contact the person I heard this from.

Apparently it was Walter Baade who was asked:

  If you had your life to live over, would you be an astronomer again?

To which he answered:

  Only if the ratio of total to selective absorption is everywhere the same.

I will not accept this answer as I have not been able to find a source yet. If anyone can find one, feel free to answer and I'll accept.

EDIT: Decided to accept after all, since apparently no one here knows a source. Follow up question regarding the source here.


Statistically, what would the average distance of the closest black hole be?

Let us assume that $N$ stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100$M_{\odot}$. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - $n(m) \propto m^{-2.3}$. Then assume that all stars with mass $m>25M_{\odot}$ end their lives as black holes.

So, if $n(m) = Am^{-2.3}$, then
$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$
and thus $A=0.065N$.

The number of black holes created will be
$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$
i.e 0.06% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors.

Now, I follow the other answers by scaling to the number of stars in the solar neighbourhood, which is approximately 1000 in a sphere of 15 pc radius $\simeq 0.07$ pc$^{-3}$. I assume that as stellar lifetime scales as $M^{-2.5}$ and the Sun's lifetimes is about the age of the Galaxy, that almost all the stars ever born are still alive. Thus, the black hole density is $\simeq 4.5 \times 10^{-5}$ pc$^{-3}$ and so there is one black hole within 18 pc.

OK, so why might this number be wrong? Although the number is very insensitive to the assumed upper mass limit of stars, it is very sensitive to the assumed lower mass limit. This could be higher or lower depending on the very uncertain details of the late stellar evolution and mass-loss from massive stars. This could drive our answer up or down.

Some fraction $f$ of these black holes will merge with other black holes or will escape the Galaxy due to "kicks" from a supernova explosion or interactions with other stars in their dense, clustered birth environments (though not all black holes require a supernova explosion for their creation). We don't know what this fraction is, but it increases our answer by a factor $(1-f)^{-1/3}$.

Even if they don't escape, it is highly likely that black holes will have a much higher velocity dispersion and hence spatial dispersion above and below the Galactic plane compared with "normal" stars. This is especially true considering most black holes will be very old, since most star formation (including massive star formation) occurred early in the life of the Galaxy, and black hole progenitors die very quickly. Old stars (and black holes) have their kinematics "heated" so that their velocity and spatial dispersions increase.

I conclude that black holes will therefore be under-represented in the solar neighbourhood compared with the crude calculations above and so you should treat the 18pc as a lower limit to the expectation value, although of course it is possible (though not probable) that a closer one could exist.


rotation curve in the innermost part of galaxies

So they are talking about different scales. Rubin is finding the rotation curve on kpc scales. And Sofue is talking about how hard it is to get the rotation curve in the central less than a kiloparsec. In the center you cannot use the same techniques as Rubin did.

Rotation curves of galaxies are most often measured by the gas speed. The speed of the stars can be used, but owing to dynamical heating over time, the stellar rotation speed does not indicate the true rotation speed of the galaxy as well as the gas speed.

When looking for the speed of something, astronomers will look for the redshift or blueshift of the stars or the gas which they are observing. So, they need to identify a gas, which will be present in all galaxies, which has a well known and easy to observe "line" in the spectra which will allow a velocity determination by seeing how red or blueshifted it is compared to that emission line in a laboratory.

Two such lines are H$\alpha$ and molecular CO (the one mentioned in that paper you've linked).

The H$\alpha$ line is an optical line at 656.28 nm. The CO lines are radio waves at 2.6 and 1.3 mm.

It looks like the Rubin+1978 paper took spectra of the H$\alpha$ lines, and from Figure 2, it looks as if they've collected spectra from various radii from the center. If you do it to many different points inside the galaxy at different distances from the nucleus, you can see how fast they are moving relative to each other.

That steep increase at the beginning of the curve usually occurs for around the extent of the bulge, a few kiloparsecs (as in this image from Rubin+1978).

It looks like they had a resolution of about a kpc, which would not tell you much about the center, except that the rotation curve is very high, very quickly (you assume the very center is 0 rotation):

Sofue+1999 seems to be talking about even more central than that, like the "central few hundred pc region" which will require a new technique.

A large limitation, as mentioned in the Sofue+1999 paper, is that HI is often obscured in the very centers (<0.5 kpc) of these galaxies by dust. Galactic dust, in general extincts shorter wavelength light much more effectively than long wavelength light. In this way, looking at the galaxy in longer wavelengths allows you to see through the dust. The HI line is much more easily erased by galactic extinction than the CO line at a longer wavelength. So CO makes it easier to look into the galactic central hundred parsecs than H$\alpha$. You can see this in the galactic extinction curve:

The CO line is in the radio band (mm range), a type of astronomy that was just picking up speed in the 1960s and 1970s. Having been predicted in the 1800s by Maxwell’s equations, scientists were unable to observe any radio signals from the Galaxy until a serendipitous observation of Sagittarius A* by Karl Jansky at Bell Labs in the 1930s.

However, radio astronomy comes with a problem of resolution, the resolving power of a telescope is basically:

$\Theta = 1.22 * \frac{\lambda}{D}$

Where lambda is the wavelength of the observation and D is the diameter of the telescope. And, for reference, the distances to two galaxies from that rotation curve above are:

NGC4594 = 9.45 Mpc
NGC7217 = 15 Mpc

At the distances of these galaxies, we need to be looking in the right wavelength to penetrate, and we need enough resolution to get the rotation speed at maybe 10 parsec scales. This resolution, at 15 Mpc, is maybe 0.1 arcseconds.

$3600 * arctan(\frac{10}{15e6}) * \frac{180}{\pi} = 0.1375 arcseconds$

To get that resolution observing at the 2.6 mm you need for the CO line, you need a:

$1.22 * 3600 * \frac{180}{\pi} * \frac{2.5e-3}{0.1375} = 4581.48$

4.5 km telescope… Hmmmmm. A bit of a problem considering the unfortunate tale of the 300 foot telescope.

Then, computers to the rescue. Interferometry is an idea whereby one can use two telescopes to increase their resolving power. When combining their signals, they have an effective diameter of their separation. Of course, they are not equivalent to a single dish telescope of that diameter, because they collect far less total starlight.

The technique was successfully utilized scientifically as early as the 1920s, to resolve the star Betelgeuse. But it was difficult to use this technique in it’s modern form, combining tens or hundreds of telescopes at once without the advent of computers which could handle the calculations in the 1960s and 1970s.

And now, the final piece. Interferometry grows much more difficult the smaller your wavelength. Interferometry is only possible if amplitude and phase of the incoming wave is recorded and recombined precisely. So an optical interferopeter needs: more precisely polished mirrors, more precisely measured separations, stable signal information transport from the dishes (fiberoptics), and more computing power to run more fourier transforms. While radio interferometry was performed in the 1920s, it took all the way until the 1990s to do optical interferometry, and it's still far from routine (see the Large Binocular Telescope).

So, to summarize:
1) To get a rotation curve of a galaxy, we need a molecular line that we can get a redshift or blueshift from. HI (optical) and CO (mm) are good choices.

2) The very centers of galaxies are obscured by dust, which destroys optical signals, but allows mm to pass, so CO is the molecule of choice for rotation curves of the nuclei of these galaxies, as H$\alpha$ signal is poor in the centers.

3) However, for a mm telescope to be able to spatially resolve small parts of the galaxies at the large distances of 10+ MPC in this wavelength, we need a telescope a few kilometers wide.

4) We can solve this with interferometry, but only recently has the technology to do interferometry at this wavelength been accomplished.

The first interferometer observing at this wavelength, the SubMillimeter Array (SMA; eight, 6-meter telescopes), opened to guest astronomers for the first time in 1990. The SMA was still being expanded at the time of the paper you linked.

These days, the Atacama Large Millimeter Array (ALMA) is just now nearing completion, and is focusing on the same wavelength range only with huge resolution and surface area (66 telescopes from 7-12 meters). They are collecting CO rotation curves to such precision that they claim to be able to estimate the central black hole masses of galaxies to 5%.

Stuff:

https://en.wikipedia.org/wiki/Astronomical_interferometer
https://en.wikipedia.org/wiki/Aperture_synthesis
https://ned.ipac.caltech.edu/level5/Fitzpatrick/Figures/figure1.gif
https://www.nrao.edu/whatisra/images/300ft-before.gif
https://www.nrao.edu/whatisra/images/300ft-after.gif
http://www.atnf.csiro.au/outreach//images/wallpapers/atcamorningsml.jpg
http://www.almaobservatory.org/


How can GAIA's accuracy for position, parallax, and annual proper motion for a given star be the same number?

Parallax and proper motion are determined from a series of position measurements taken over the course of (for Gaia DR2) 22 months. A "5-parameter" astrometric model is fitted to these position measurements, consisting of a sky position at some epoch, a parallax and a proper motion in each of the celestial coordinates.

The precisions of each of these parameters improves as the number of measurements increases, but parallax and mean position precision are basically unaffected by the timespan of the observations (as long as they are taken over more than a year), whereas proper motion precision is improved by a longer timespan given the same number of observations.

If a star had no proper motion, then a basic parallax measurement would be the difference between two (extreme) positions on the sky, and the parallax precision would be around $\sqrt{2}$ of an individual position precision. The mean position of the star is something like the average of the two measurements, so might have a precision that is a factor 2 better.

In practice, you have many more than 2 observations and measuring the parallax and mean position is like trying to fit a sine wave (with a DC offset), of known period (1 year), where the mean position is the DC offset and the parallax is the amplitude.  In such cases, the mean position has a variance of $\sigma^2/n$, where $\sigma$ is the uncertainty in a single position measurement and $n$ is the number of observations. It took me a while to track it down, but it can be shown that the variance of your best estimate of the amplitude is also $\sigma^2/n$, when the frequency is known and $n$ is large (e.g. https://pdfs.semanticscholar.org/df9b/0c876b488be088fd416387866cf8499fa438.pdf). Thus we expect the mean position and parallax to have similar uncertainties and for these to get smaller as $\sqrt{n}$.

Proper motion precision is different. If we imagine a set of measurements where parallax can be ignored (e.g. at the same time every year), the proper motion precision is akin to the precision in the gradient of a line fitted to two or more points (with time as the x-axis). The precision improves with both the number and precision of the position points and with the timespan of observation.

The answer to your question is therefore coincidence. You have picked on a magnitude of star for which the typical individual position measurement precision, typical number of measurements and typical timespan lead to a rough coincidence in the precision of the 5 parameters when proper motion is expressed in units of per year. For other magnitudes or for stars observed for an atypical number of times or duration, that isn't the case.

Quoting the abstract of Lindegren et al. (2018)

  For the sources with five-parameter astrometric solutions, the median uncertainty in parallax and position at the reference epoch J2015.5 is about 0.04 mas for bright (G < 14 mag) sources, 0.1 mas at G = 17 mag, and 0.7 masat G = 20 mag. In the proper motion components the corresponding uncertainties are 0.05, 0.2, and 1.2 mas yr−1, respectively.


Our own magnetic field is generated by convection currents in Earth's liquid outer core.
A useful summary from Physics.org:

Differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal as cool, dense matter sinks whilst warm, less dense matter rises.
This flow of liquid iron generates electric currents, which in turn produce magnetic fields.
The spiralling caused by the Coriolis force means that separate magnetic fields created are roughly aligned in the same direction.

Mars used to have a liquid iron core, but it was never as extensive as Earth's and has long since solidified.
Once our own core cools enough to solidify, we too will lose our magnetic field.


It is.  Tidal disruption occurs when $\rho_{Jup}(R_{Jup}/R)^3$ > $\rho_{comet}$ where R is the separation and $\rho$ is the density.  The density of Jupiter is 1.33 g/$cm^3$ and the comet was estimated to have a density of only 0.5 g/$cm^3$ based on the radius at which it disrupted.  That is, disruption took place at separation R ~ $(1.33/0.5)^{1/3} R_{Jup}$.


Are nebulae only visible from a distance?

The colorful patterns of gas are caused by excitation of gas atoms, e.g. oxygen, from radiation of a nearby star. You would see the same colors as from a distance, as the light is emitted in almost random directions. But you wouldn't see the same region of nebula shining at the same time as seen from Earth, when excitation is caused by a flash of radiation from a central star, because the radiation travels concentrical away from the star, and the distances the light has to travel from the excited part of the nebula to the observer is dependent of the location of the observer.

Light may be scattered non-uniformly by dust-particles, so looking different from differnt positions, e.g. forming a halo around stars, depending of the composition and shape of the particles. The interplanetary dust in our solar system causes zodiacal light.

Our solar system is in an interstellar dust cloud right now, but it's not very dense.

Very dense clouds can form, but would become instable and collapse to a star or a protoplanetary disk, compare this site about dark nebula.


Do far away stars that we observe today still exist?

Simple answer is yes. It will take light from an explosion that happened 1 mya at a distance of 200 Mly, about 199 million years to reach "us" - not that any of "us" will be around then. The speed of light is the fastest speed we can get information from anywhere in the Universe. There's nothing special in asking about what happens to that star after that light starts out on its journey here. The fact is, we can't know, until we are able to gather more light from that same source. So, that star would appear to explode if you were looking up at it 199 million years from now (keeping in mind that both the Solar System and the Milky Way are also moving through space, so the exact time it takes the light to get here will be different than the time it took the light we are now seeing from that star to have gotten here.)


Why can't our Sun be a binary with Jupiter as a T or Y dwarf?

But why can't Jupiter be a Y Dwarf who is in the binary relationship with the Sun?

There are two reasons: One is that Jupiter is too small to have ever undergone fusion of any sort. To qualify as a brown dwarf, an object needs to be large enough to have undergone deuterium fusion in its core. This requires a mass of at least 13 Jupiter masses. The other is that Jupiter formed by planetary formation mechanisms rather than by gravitational collapse.

There is debate on what to call objects of less than 13 Jupiter masses that formed by gravitational collapse. Are these sub brown dwarfs or free-floating gas giants, and does it even matter? There is also debate on what to call objects above 13 Jupiter masses that formed by planetary formation mechanisms as opposed to by gravitational collapse. Are these brown dwarfs or hyper planets, and once again, does it even matter?

In any case, Jupiter is not a brown dwarf.


Why is Ganymede's aurora only visible in ultraviolet while Earth's is bright green and red?

Quoting from Nat.Geo. article (which has that same UV image),

  Then, in 2010 and 2011, Hubble took a close look at the moon. More
  specifically, it looked at the auroral bands ringing Ganymede’s poles.
  Hubble studied the auroras in the ultraviolet, but Saur said the
  shimmering lights would be visible to human eyes.

  “If somebody could be standing on Ganymede looking up into the night
  sky, it would appear as red aurora, to you,” he described.


Parts of the Universe Shows Fewer Named Stars

The database used for the above was for Deep Sky Objects, not stars.
When I changed the database the correct one, then the following image is distribution of 120,000 named stars. Thanks for flagging my error.


What consequences does a positive muon anomalous magnetic moment have for astronomy?

This is unlikely to have any implications for astronomy. They may have seen an extremely small deviation from the predictions of the standard model. The result may actually have to do with technical issues in theoretical methods, not with any new physics: https://arxiv.org/abs/2002.12347 . Even if it does have implications for new physics, there is no reason to believe that it would have implications for astronomy. It could, for example, be an indication of the existence of a new unstable particle with a high mass, whereas dark matter needs to be made of stable particles.


Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe.

But dark matter is always "found" (or "inferred") either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter associated with colliding galaxy clusters / flat rotation curves of spiral galaxies / abnormal velocity dispersion in galaxy clusters) or not associated to anything visible but nevertheless forming clumps (weak lensing detection of galaxy clusters previously unseen). That is why dark matter is thought to be cold.

Additionally, there is a clear distinction between both types: there is not such thing as dark matter that is "not too cold but not too hot either" (see footnote as well). Dark matter is either made of particles with less than ~10 eV (hot dark matter, made of light particles, mostly dispersed everywhere) or particles with more than ~2 GeV (heavier, slower particles gravitationally bound to some structure). Both limits are found when imposing the maximum amount in which the candidate particles (neutrinos or something more exotic) can possibly contribute to the actual value of the density parameter due to matter in our expaning Universe.

Thus, either DM appears gravitationally bound (cold DM) or dispersed (hot DM), and both types are clearly distinct (10 ev vs 2 Gev). Observations favour the first case. However, Cold Dark Matter is not the ultimate solution, and still faces some problems.

Regarding the possibility of mixed solutions, many of them have been already ruled out. Microlensing has ruled out the possibility of unseen compact objects (brown dwarfs, stars, stellar black holes) in galactic halos, in our galactic neighbourhood as well as in the extragalactic domain. Ordinary matter (stones, bricks, dust) cannot possibly be, otherwise they would become hot and re-radiate. Any exotic mix of known particles doesn't work.

All we think we know is that DM must be made of some heavy particles yet to be discovered. In order to introduce a more complex model (e.g. different types of particles depending on the structure they appear attached to) one needs a justification (i.e. some predictions that better agree with reality) and nobody has been able to do that yet.

Remark Note that Dark Matter particles, either from the hot or the cold type, cannot possibly "slow down" and clump too much (e.g. forming planets) because they don't interact electromagnetically like ordinary matter, that is why DM is said to be collisionless. Wherever infalling ordinary matter forms any structure (e.g. protostars or accretion disks), a very important part of the process is thermalisation, i.e. the redistribution of energy of the infalling particles by means of numerous collisions. This cannot happen with Dark Matter.


Distribution of spectral types of "giant" stars

OK, I see what you want.

What I did was take the revised Hipparcos catalogue and select all stars closer than 50pc. At this distance there should be a complete sample of the types of stars you were interested in. I can plot an absolute magnitude versus colour diagram for these 7096 stars (see below) and use this to select giants and OB main sequence stars.

I do this by using data from chapter 2 of Zombeck (1992), I define O and B main sequence stars to have $B-V<0$ and $M_V<5$ (to exclude hot white dwarfs). Evolved giants are selected (a bit arbitrarily) to lie significantly above the main sequence.

Then I can plot a frequency histogram of the B-V values of those stars and label them with spectral types, again according to tables in Zombeck (1992).
The result is seen below.

Of these 235 stars: None are O stars, 29 are B stars, about 120 are G giants, about 80 are K giants and only 2-3 are M giants


Location of the milky way in August in Death Valley

The Milky Way will be in a straight line that goes from SSW, (a bearing of about 200 degrees) almost directly overhead and to the NE. The brightest part of the Milky way will be low in the sky in the SSW.

In future, you can check the visibility and direction of any astronomical object with planetarium software, such as the (free) Stellarium.


What other hints of possible life in Venusian atmosphere have we dismissed?

Now that phosphine has probably been discovered in relatively large
quantities in the Venusian atmosphere

It is not large, it's only 20 ppb.

I heard that there have been other unexplained phenomena (i.e. dark
patches) seen.

Yes, dark patches has been observed on Venus from quite a time. Scientist aren't able to figure it out and are speculating to be clouds that are absorbing UV light thus making it look dark. Some suggest uneven mixing of clouds creating dark and light spots or some unknown chemicals present or even a sign of microbial growth.
These "dark patches" are later named "unknown absorbers" because they absorb large amounts of solar radiation.

What makes the patches dark? Why is it dark?

The dark patches are made up of concentrated sulfuric acid and some unknown chemicals that is absorbing the UV radiations making it appear dark. Scientists have claimed the chemicals to be some forms of sulfur, disulfur dioxide or ferric chloride.
These dark patches seems to affect Venus's climate by affecting its albedo. The clouds absorb solar radiation, which causes temperature changes that affect wind patterns and this affects albedo (how much energy is reflected back to space). If we talk about numbers then, Venus’ albedo had been observed to diminish by about half between 2006 and 2017, before then returning to normal. This affected the upper atmosphere, including variations in the “super-rotation,” which is driven by winds exceeding 200 miles per hour (322 kph).

It is also further noted that around 30 miles (48 km) above the planet's surface is a lower cloud layer with temperatures around 60° C (140° F) and pressures similar to that of Earth and it is hypothesized that there is some sort of microbial life or algal bloom that is absorbing light and radiation and causing the dark color.
References

Sanjay S. Limaye, Rakesh Mogul, David J. Smith, Arif H. Ansari, Grzegorz P. Słowik, and Parag Vaishampayan, Astrobiology, Sep 2018, 1181-1198. DOI: 10.1089/ast.2017.1783
Lee et.al. Long-term Variations of Venus's 365 nm Albedo Observed by Venus Express, Akatsuki, MESSENGER, and the Hubble Space Telescope, The Astronomical Journal, Volume 158, Number 3, 26 August 2019. DOI: 10.3847/1538-3881/ab3120


Blood Moon Identification in NASA reports

"Blood moon" is not a term used scientifically. Therefore look for total lunar eclipses.

As some light is diffracted by earth's atmosphere, the moon is not entirely black but appears red. (Blue light is scattered, see why is the sky blue).

All total lunar eclipses will show a reddish colour. The brightness may vary according to the state of the atmosphere but the moon will never be black, or any other colour.

(James K contributed equally to this answer)


Is there a timekeeping word for the orbit of a moon?

You should use one of the precisely defined terms of specific measures of orbital period, which are described on this Wikipedia page.

You might get away with just "orbital period" is you're not using it in a context that requires precision.

For length of day there's synodic day as well as sidereal period.  There's a discussion on the difference between these here.


Can black holes be used as an energy source

What you are thinking of is called the Penrose process - a means of extracting rotational energy from a black hole.

The basic idea is that you allow a composite body to fall towards a rotating black hole. Once it reaches the ergosphere - that region, outside the event horizon, where the "frame dragging" due to the rotating spacetime is so severe that no body can remain stationary from the point of view of a distant observer - the body should split. One half falls into the black hole and the other orbits and flies out again. If things are arranged correctly then the returning part of the body can have a kinetic energy that exceeds the rest mass energy of the half that has fallen into the black hole. The energy comes at the expense of slowing the rotation of the black hole.

The Penrose Process (from Brito et al. 2015)

If the rotating black hole has an accretion disc (as many do if they are in binary systems - that's how they are identified) then another method is the Blandford-Znajek process. This may actually be the way that quasars and many of the energetic phenomena in gamma ray bursts and black hole binary systems are powered. In simple terms, a magnetic field that threads the accretion disc can be twisted up by the ergosphere. Material can then be propelled in jets along the rotation axis, but is given a spin by the field lines, which extracts angular momentum (and rotational energy) from the black hole. This energy can then be released when these jets run into other things.


Can a solar system exist where the second planet rotates fast, and the third planet is tidally locked to their star?

The tidal locking timescale depends on several factors: $$\tau_{lock} \approx \frac{0.4 \omega_0 a^5 m Q}{3 G M^2 k_2 r^3}$$ such as the initial spin rate $\omega_0$, the semimajor axis $a$, the mass $m$, the solar mass $M$, the radius $r$ and various dissipation parameters $Q$ and $k_2$.

Two planets that merely differ in $a$ will have the inner one lock much earlier than the outer one. But give them slightly different $\omega_0$, $m$ and especially $r$ and the inner might dissipate more slowly than the outer.

There is also the possibility of spin-orbit resonances. Mercury is in  3:2 resonance, and tidal forces might lock planets into 2:1 or 5:2 resonances. That way the inner planet may be in a resonant lock while the outer is tidally locked 1:1. In fact, Venus is turning very slowly (and retrogradely) so it is not too far off, although the causes of the retrograde spin are unclear.

So, yes, you can have tidally locked planet orbiting outside non-locked planets.


Are astronomers researching or trying to find signs of worm holes?

This has been considered long ago (Here's a paper talking about this).

Wormholes are not forbidden by physics, but the creation of wormholes is iffy ground. THere are two possible paths one can take to create a wormhole:

Choose a pre-existing wormhole in the quantum foam and "expand" it by feeding it exotic matter.
"Tear and sew up" space — we're not sure if this is allowed by physics, as it ventures into the area of physics that we don't have an adequate explanation for.

Whatever it is, the creation and sustenance of a wormhole requires us to have control over exotic matter (in this case, particles/waves with negative mass/energy density). Vacuum fluctuations already have regions of negative energy density, but they're an uncontrollable quantum phenomenon by current technology.

"Faster than light travel" is a different matter, though. While wormholes let one jump to another point in space, one does not attain a speed greater than c whilst in them.


Schindler et al. 1999: Morphology of the Virgo Cluster: Gas versus Galaxies has details for $\beta$ model fits for Virgo and its subclusters.


There is no reason it should be.

There are bound to be irregularities due to local geological differences.

EDIT:

The theory as to why planets tend to be spherical is because any irregularity would cause a non-uniformity in the gravitational field at the surface, and the forces on the surface rocks would ultimately cause it to be spherical. Though, ignoring local irregularities, the Earth has a smaller radius (from its center of mass) at the poles, about 42 km less, than that at equator.

This can be attributed to Earth's rotation. The distance of a point on the Earth's surface located at the equator is larger from the axis of rotation than at any other latitude. As it rotates, this causes a bulge at the equator. The technical term given to this shape is the oblate spheroid.

You can read about the mathematical formula which describes the flattening and more about the formation here.


Fresnel lenses for a makeshift Galilei telescope?

My first telescope was made from the lens and eyepiece of an old discarded pair of binoculars. I used a paper towel roll cardboard tube with the objective taped on the front and a flat, circular piece of card stock (with radial cuts near the edge for folding) taped over the other end. It had a hole big enough for the eyepiece to fit, and pieces of tape slightly covering one edge of the hole for friction.
Sliding the eyepiece in and out slightly against the friction provided focus.
It's exciting to see the world big and upside down.
I'd go ahead and try the fresnel lenses for both objective and eyepiece if you can find a pair with different focal lengths, but I think you can also find short focal length lenses in novelty, "dollar", toy, or other random-object store as pocket tools or toys, or in hobby or tool stores as cheap imitation jewler, gemologist or stamp-collector's loupes.
For a short focal length lens you can even try a drop of water stuck on a glass slide, or a transparent material that hardens transparent.
Or a spherical transparent marble or bead: Ball lens calculator, Ball lens.
If you'd like to "go small" afterwards and explore the microcosm, see foldscope.com, teamfoldscope, Foldscope for more ideas how low-cost projects can be fun and educational.


Approximate spectral type and luminosity given apparent magnitude and distance

You can calculate the absolute magnitude of a star:
$$M=m-5\log_{10}(\frac{d}{10\,\text{pc}})$$
where $M$ is absolute magnitude, $m$ is apparent magnitude and $d$ is the distance.
Then you take a look at the HR diagram.

One can easily see, that you need two data to obtain the third one, but we have only one data (absolute magnitude). That means, that you also have to know the luminosity classification of a star:

I: supergiants
II: bright giants
III: giants
IV: subgiants
V: main sequence
VI: white dwarfs

Then you just look at the intersection of the class and absolute magnitude.

If you know radius, then this is even easier. All you need is this formula:
$$T={\Big{(}\frac{10^{0.4(4.77-M)}}{4\pi R^2 \sigma}\Big{)}}^{0.25}$$
$T$ is the temperature in Kelvins, $M$ is absolute magnitude of the star, $R$ is radius of the star, $\sigma$ is Stefan-Boltzmann constant.
With these data you look at the table of temperature and spectral classes or HR diagram.

If you are given color, then you already have an answer (spectral class is defined with color). If you are given temperature, you can look at the HR diagram.


What effect does stellar granulation to have on a chemical analysis of a star's spectrum?

There is no simple answer to these questions - although I could be brief and say
(i) No it doesn't and (ii) no they won't.
If you make a simple two component atmosphere then the observed spectrum will be the flux-weighted combination of two spectra.
$$ S_{\rm obs} = \frac{A_1 T_1^4 S_1 + A_2 T_2^4 S_2}{A_1 T_1^4 + A_2 T_2^4}\ , $$
where $A_1, T_1, S_1$ are the area, temperature and spectrum of material at $T_1$ and the quantities with subscript 2 are for the regions at temperature $T_2$.
Now, depending on the response of any spectral feature to a change in temperature this could make the equivalent width of that feature become stronger or weaker in the flux-weighted average spectrum. If you then analyse that spectrum assuming that there is a single temperature then what you derive for an abundance based on that feature could get bigger or smaller.
For example most lines of neutral species (e.g. Fe I, Li I) get stronger in cooler atmospheres. If you put cool starspots on a star then the equivalent widths of these lines will become stronger. The average temperature of the star will decline very slightly. The net effect would be to estimate that the abundance of iron or lithium in that star was larger.
On the other hand if the line was very strong such that it was in the saturated part of the curve of growth, then it wouldn't get much stronger in a cooler atmosphere and it could even be that by adopting a lower average temperature, the lack of a significant increase in line equivalent wdth would lead one to infer a reduced abundance.
A more clear-cut case might be lines of ionised Fe II. These would become weaker in a cooler atmosphere, so you would end up inferring a lower iron abundance from the flux-weighted average spectrum.
Then there is the issue of the over-simplification that a simple two-temperature model imposes. Real granulation also involves the movement of plasma with consequent implications for micro- and macro-turbulence and how those affect the line formation and assumption of local thermodynamic equilibrium. These can only be addressed using expensive 3D stellar atmosphere models. Such models do exist and they offer grids of 3D NLTE corrections to abundances determined using 1D LTE atmospheric analyses. The corrections can go in either direction depending on the strengths of the lines and the intrinsic parameters of the star. As an example, Wang et al. (2021) present a grid of corrections for the Li I optical/NIR lines that I have a particular interest in.


What determines which frequency range a radio telescope has?

A parabola does indeed focus over a broad frequency range. The lower limit is determined by the dish diameter, the upper by the construction (mesh size, parabolic accuracy etc.).

The collector placed at the focal point may be a simple dipole or other fixed frequency antenna, or more commonly, a waveguide that leads the collected signals to a low noise amplifier and receiver. The physical dimensions of the antenna or waveguide set the operating frequency range of the antenna system. More than one range may be tuned by means of multiple switched collectors.

The receiver is fundamentally no different to any other receiver; a tuning element selects the signals of interest which are then detected and amplified.


If the tilt of the Earth changed, would the tropics latitude change too?

Yes, the Tropics of Cancer and Capricorn are defined as the highest and lowest latitudes where the sun is directly overhead at least one day a year. The sun's highest position appears to move up and down on the sky over the year because of the Earth's tilt, so if the Earth's tilt were greater or lesser, so would be the extent of the Tropics. (Side note: I may be wrong, but I think the Tropics were named after the Zodiac sign the sun is in when it reaches its highest or lowest point.)


What exactly is wrong with this plaque in Neil deGrasse Tyson's "back yard"? Has it been fixed?

I see 2 candidates:

the universe is 13.8 billion years old, so should be rounded off to 14
the cosmic horizon is 46 billion ly away


Calculating the Size of Earth in the Apollo 8 Earthrise Photo

What is your actual question? Are you asking others to confirm your process/formulas?
Correct, specifically whether the diagonal fov is always the correct method?

No, an incorrect approximation is never "always the correct method".
Even with the basic pinhole camera model (the basis for almost all of the better approximations) position on the focal plane is never linear with incident angle. Instead, you draw similar triangles.
Here is a simple 2D analogy, usually people use 3D, Projective Geometry and Homography and the matrix math it requires.
$$\frac{w}{d}=\frac{x}{f}$$
$$x=f\frac{w}{d}$$
where $d$ and $f$ are the distances along the optical axis from the source and image plane to the lens pupil, and $w$ and $x$ are distances from the axis to the points on those planes which are perpendicular to the axis.

Most higher order methods that incorporate lens distortions star with the pinhole camera approximation, and quantify the imaging system's deviations from it.
no arctangents were harmed in the making of the projective plane.


Historical Lunar Tables and the Associated Error Bound

The first question (how are the tables generated?) is easy to answer. The acknowledgement on the Astropixels page says:

It is based on procedures described in Astronomical Algorithms by Jean Meeus (Willmann-Bell, Inc., Richmond, 1998).

For the second question (how accurate are the calculations?), there are 3 sources of error that I can think of:

The procedure may be for a limited time span. Meeus' book does not mention what time span the formulas are good for, but the source from which the procedure was based on (Chapront's ELP-2000/82) used the years 1900 to 2000 to determine the variables in the equations. (See NASA Astrophysics Data System - The lunar ephemeris ELP 2000.) It is hard to know what the accuracy is when going backward or forward several hundred years. Mathematically, extrapolating beyond the chosen time span could be reasonably accurate, or it could be "completely wrong". (I did not read all of the Astropixels' pages, but I know that a lot of historical research was done when calculating the solar eclipse tables. So I would anticipate some cross-referencing between the Moon phases and eclipses. After all, you cannot have a solar eclipse unless you have a New Moon!)
Time is based on the rotation of the Earth, such as GMT (Greenwich Mean Time), UT (Universal Time), or the time in a local time zone. This type of time is not consistent over long time intervals. The Earth's rotation slows down and speeds up which requires adjustments to our clocks. The motion of the planets is (more) consistent and runs at a steady rate. Astronomers call this type of time Dynamical Time (formerly known as TDT Terrestrial Dynamic Time or ET Ephemeris Time). The difference between UT and Dynamical Time is explained in the Astropixels' page Uncertainty in Delta T. The uncertainty in Delta T is over an hour for the year 2000 BCE, so the time of the phase can be off by this amount just due to the uncertainty of clock time (the Earth's "clock") versus Dynamical Time (the solar system's "clock").
Computer languages can introduce inaccuracies. For example, it may not be possible to store the number of days, hours, minutes between now and the year 2000 BCE to full precision due to the limited accuracy of storing really large number. There are approximately 1.05 billion (1.05E9) minutes in 2000 years. In order to calculate a result 2000 years ago to a precision of 1 minute requires the ability to store a number such as 1,050,000,001 and 1,050,000,002. (I am not a computer expert, but these numbers can probably be stored easily and to a higher accuracy. The potential problem is that the procedure uses values of time, time squared ($T^2$), time cubed ($T^3$), and time to the fourth power ($T^4$)! Thus, numbers can quickly become larger than what the computer can store to full precision.)

When put all together, I would expect that the times 2000 years ago (or 2000 years in the future) may only be accurate to a couple of hours.


Orbital eccentricity variation of the other planets?

You may take a look at the lates parametrization file by JPL-NAIF for the precession, nutation and pole orientation of the largest known bodies.

Although, for the large time scales you are asking, I expect you will need to propagate the data and make your own wild guess, or dig into appropiate literature about solar system physics.


Can solar luminosity & activity be predicted?

All of these effects are related to the 11 year solar cycle. And while we know there are times when the sun is "active" versus "inactive" we aren't necessarily predicting exactly where sunspots or solar flares will occur, but how many we see in total.


Why are the time zones calculated as 360°/24 and not 361°/24 or 360°/23.933?

The Earth takes 23 hours 56 minutes to rotate once.  But that is not relevant to most people. Sure, the stars will be in the same position again after 23 hours 56 minutes, but the sun will not be in the same position.
It is far more important, for most people, to measure the time from noon to noon. And the average time from noon to noon is 24 hours. This is because the motion of the sun is a combination of both the spinning of the Earth and the orbit of the Earth around the sun. The orbital motion of the Earth adds four minutes. You should also teach the students

In twenty-four hours the sun advances 360 degrees. (solar day)

Time zones are based on clock time, which is based on the motion of the sun and not the motion of the stars.


What is the name of meteoroids which hit Moon, Mars, or pretty much anything that isn’t the Earth?

Oxford English Dictionary is an authoritative source, but it's aimed at the general public. I think no astronomer will frown upon you when you're using the terms meteoroid and meteorite to describe a small body impacting another body in the solar system. As an example, Wikipedia doesn't impose the limitation that it must hit Earth:

  A meteorite is a solid piece of debris from an object, such as a comet, asteroid, or meteoroid, that originates in outer space and survives its passage through the atmosphere to reach the surface of a planet or moon.


Equations for coordinates of solar system objects

It depends a bit on how precise you would want to be. A very good discussion on how to calculate the orbits of solar system objects is given in the book by Jean Meeus, Astronomical Algorithms (1999), which is at an advanced amateur level. At professional level you have the Explanatory Supplement to the Astronomical Almanac by Urban and Siedelmann.

For precise calculations Meeus uses the VSOP theory from Bretagnon (1987) (wikipedia link). You can download all files needed for these calculations from VizieR. These files contain a lot of numerical terms that are needed in the equations. The resulting positions are however very precise.

For the highest accuracy, you can download the predicted positions from the Jet Propulsion Laboratory (JPL) Horizon system.

There are also quite a few software library projects that implement the equations from Meeus. This might be the best option. Then you do not have to implement the equations yourself. For C/C++ you have for instance LibNova.

For highest precision you have professional software libraries such as for instance the NOVAS libraries from the Naval Observatory for Fortran, C, or Python. But to use that correctly you will have to have a good understanding of celestial mechanics.


Is a branch on HR diagram a line formed by stars in the same evolutionary phase, or the evolutionary track of an individual star?

The answer is: both, but for different contexts.

The HR diagram is used to display the properties of observed stars. A point on the diagram represents a star with a certain effective temperature and luminosity, which can be interpreted as being indicative of the stage of development of the star. Equivalently, the diagram can be expressed in terms of absolute magnitude and stellar classification/type. The diagram was originally constructed in 1911 by E. Hertzsprung and H. N. Russell independently, though we predominantly use Russell's version nowadays.

As one simulates the theoretical evolution of a star, the star traces an evolutionary "track" on the HR diagram, which is a line that runs across the diagram in various directions representing the evolution of the star through its various development stages, as in the second figure in the OP. This is useful for displaying the effect of various parameters on the complicated evolution of stars, see here for example.

These two efforts are related historically: as better telescopes were made, there were more and more stars observed, and more and more points were added the HR diagrams, but eventually we needed computers to handle the data. Along that route, the development of theories of stellar evolution were/are intimately tied to the development of observational astronomy, as experiment and theory always are, and the HR diagram has been a powerful tool for advancing theories of stellar astrophysics.

Looking at the first image, we might think that the subgiant branch is the approximate line of stars formed by different subgiant stars on the HR diagram. While looking at the second one, we might think that the expression branch means the evolutionary track of an individual star.

So you can think of the stars in the first image of the OP as snapshots of each of the stars at a point in their lifetimes. The stars that group in the sub-giant branch (so called for low and intermediate mass stars, but is called the Hertzsprung gap for high mass stars because it's so short lived) are in the part of their evolution where hydrogen burning in their core has ceased and hydrogen shell burning has begun. The evolutionary track in the second image in the OP does indeed show the evolution of a single star through its stages of development.


How would the Tycho Impact have appeared from Earth?

I found a partial answer in this article on the smaller and younger Giordano Bruno crater.

  The impact creating the 22-km-wide crater would have kicked up 10 million tons of debris, triggering a week-long, blizzard-like meteor storm on Earth...

So the secondary effects of the Tycho impact would have been even more spectacular. However, I don't know whether the fragments would have been large enough to devastate the Earth's surface.

Experimenting with an online impact simulator it appears that the Tycho impact was somewhere in the order of 1.00x10^23 joules, judging by the size of the crater. This is about four times that of the G fragment impact of Shoemaker-Levy 9. When you consider all the fragments it would be fair to say humans have witnessed an impact of similar magnitude to Tycho.


How come, in our lifetime, we will see the first stars which ever formed?

We will not find Galaxies that have existed before the big bang.
There's a principle in Astronomy that states: The further you look, the earlier you look. That means that the more distance the light had to travel to reach our telescopes, the more time it needed to reach us, because the speed of light is not infinite. Therefore seeing the first galaxies of the universe isn't some lucky coincidence, we are just looking the right distance away (13 billion light years). If we had lived earlier we could still see these galaxies, but they would be closer to us. And the same goes vice versa; If we had lived at an earlier time, we would still see said galaxies, but much farther away. I hope this helps.


Python controllable amateur astrophotography cameras?

I use AutoHotKey scripts to control my observatory. As for the individual devices, cameras focusers etc I use the manufacturers software and control that software with Macro scripts.

I do some minor RS-232 control with AHK.

Below is a video of the scripts starting up my remote controlled observatory.

The video shows the two screens of the computer in the house as the scripts power up the observatory and start the four computers out there. Then load, position and make all the necessary software connection between the computers and equipment.

Example. The color bars of the cameras. The color bar shows the camera video and control connections are working and the script has turned on the color bars.

I use AHK to automate TheSky, MaximDL and a number of other apps. I have a script that will use Maxim to plate solve when finding objects.

I have no idea if or how Python fits in AHK but the question is here.

https://www.youtube.com/watch?v=UN4VoOKOcXo&feature=youtu.be

As I said you are watching the action from inside the house as the four computers in the observatory are powered up all the software is loaded.

You will probably need to watch it a few times before it makes much sense. What you see mostly are the windows of individual apps loading and being reposition. Individual setting in the windows are being changed but for the most part the screen is just too small to show everything.


Gravity makes objects compress to their centers of mass, since gravity extends in all directions.

For planets, gravity continues compressing the object until the rocks cannot be compressed anymore, since the pressure will fight against gravity. At this point (called hydrostatic equilibrium), the object has become a spheroid. One of the requirements for an object to be called a "planet" is that it has reached hydrostatic equilibrium.

For black holes, the shape of their event horizons really depend on whether the black hole is spinning or not. For non-spinning black holes, gravity extends in all directions, so the event horizon will become spherical in shape. For spinning black holes, it will likely be an oblate spheroid.

Most galaxies are only spherical when they are forming. Their angular momentum usually causes them to flatten out over time. However, one part of galaxies, called the "halo", is spherical.

Wormholes are purely hypothetical, so I won't really entertain that.


Could space time curve from the vantage point of earth and out into the universe, and would d = 1 / P in that case give false estimates?

Yes, but the amount of curvature is too small in scale proportionally to affect such things, especially when one thinks of how away from us the objects are that we observe.  Even the moon is a quarter of a million miles away, and the Earth's gravitational well is quite small.


When one body has a much larger mass, so it's motion is negligable, it can be called the central body (even for eccentric orbits, where the central body is a the focus of the ellipse, and far from the centre)

This is the terminology used by Wikipedia editors

  The central body in an orbital system can be defined as the one whose mass (M) is much larger than the mass of the orbiting body (m)...

Also by Glasgow university

  The most straightforward orbit calculations occur when the central body is much more massive than the orbiting body.

and elsewhere

  The angular momentum of a satellite depends on both the mass of orbiting satellite and central body as well as the radius of the orbit.

Normally, in a specific situation "planet", or "star" can be used.


Can the order of planets change in a young planetary system?

Yes, the order of planets in a system can change over time. Indeed, in some variations of the Nice model, Uranus formed further out than Neptune and then switched places when Jupiter and Saturn underwent their resonant interactions. According to Tsiganis et al. 2005, one of the original three papers on the model, Uranus and Neptune switched positions in ~50% of all simulations (presumably meaning that in some where Uranus formed past Neptune, ended up further in, and in some where Neptune formed beyond Uranus, it traveled further inwards). Therefore, a number of models reproducing the parameters of the giant planets in the present-day Solar System involve Neptune forming closer to the Sun than Uranus. This seems to fit better with models of the Sun's protoplanetary disk.