Convert Ecliptic to Galactic Coordinates

Probably it's easiest to use some package for this, for example, if you use Python, you can use astropy:
http://docs.astropy.org/en/stable/generated/examples/coordinates/plot_galactocentric-frame.html.
Or if you are interested, here is a description on how to do it by hand, including the propagation the uncertainties:
http://adsabs.harvard.edu/abs/1987AJ.....93..864J

Of course, you also need radial (line-of-sight) velocities, but I don't know if that is in the ATNF catalogue...

Edit:
About the radial velocities: we only see the pulsars move across the sky and know the distance, so we get two velocity components perpendicular to our line of sight, a $v_\alpha$ and a $v_\delta$, from the proper motion in Right Ascension ($v_\alpha=d\cdot \mu_\alpha$) and the propermotion in Declination ($v_\delta=d\cdot\mu_\delta$).

Probably, it is useful for you to convert these proper motions from $\mu_\alpha, \mu_\delta$ to galactic longitude ($l$) and latitude ($b$) $\mu_l,\mu_b$, so you can get the velocities in cartesian coordinates: $v_y=d\cdot\mu_l$ and $v_z=d\cdot\mu_b$. This way, you have at least the two velocity components perpendicular to our LOS. Oh, and you maybe also want to subtract the rotation of the sun around the MW center.

There are ways to estimate the radial velocity sometimes: if the pulsar lives in a binary system, a spectrum of the companion maybe. Or a very crude way would be maybe something like http://adsabs.harvard.edu/full/1977ApJ...216..842H... But I don't know much about it.


How does the gravitation of Jupiter or other big planets affect Kepler's third law?

Short Answer: The gravitation of Jupiter and all the other planets makes Kepler’s third law a bit less accurate than it would be if their gravity were zero.  The gravitational interaction between the planets, however, doesn’t have much effect on either the period or semi-major axis of any of the planets, especially in the short term.
Long Answer:  You might be thinking of this pendulum motion expansion used to describe the orbital oscillations (including the big one between Jupiter and Saturn) here https://history.nasa.gov/SP-345/ch8.htm:

Where $\psi$ is the angle from the vertical, $\omega$ is the angular velocity, and $A^2$ is $g/l$, where $g$ is acceleration and $l$ is presumably the length of the pendulum.
Interplanetary resonances mostly result in precession of planet’s orbits, which are period and semi-major axis invariant.  They mostly change just the longitude of the ascending node.

Other orbital changing effects between planets are Milankovitch cycles which are also period and semi-major axis invariant, according to Wikipedia:

Eccentricity varies primarily due to the gravitational pull of Jupiter
and Saturn. However, the semi-major axis of the orbital ellipse
remains unchanged; according to perturbation theory, which computes
the evolution of the orbit, the semi-major axis is invariant. The
orbital period (the length of a sidereal year) is also invariant,
because according to Kepler's third law, it is determined by the
semi-major axis.

The more significant effects that planetary mass have on adding inaccuracy to Kepler’s third law are through their gravitational effects on the sun when treated as individual two-body orbits.
Kepler’s “laws” describe the motion of the planets.  Kepler arrived at these laws by examining observational data from the solar system in the late 1500s and the early 1600s.  On page 48 of Epitome of Copernican Astronomy Kepler asked himself the question:

How is the ratio of the periodic times, which you have assigned to the
mobile bodies related to the aforesaid ratio of the spheres wherein
those bodies are borned?

And then answered:

The ratio of the times is not equal to the ratio of the spheres, but
greater than it, and in the primary planets exactly the ratio of the
3/2 powers.

A bit of translation is necessary here. The periodic times Kepler references are the orbital periods $p$ of the planets, and the spheres are the radii of the spheres which circumscribe the ellipses of the orbits.  These are the same as the semi-major axes $a$ of the orbits.
So Kepler’s third law as stated was that, for every pair of planets 1 and 2, $(a_1/a_2)^{3/2}=p_1/p_2$.  Our typical modern restatement of Kepler’s third law is that the ratio $a^3/p^2$ is the same for every planet.
Kepler thought that the sun was the center of the solar system.  He didn’t realize that when a less massive object is in orbit with a more massive object, both objects actually rotate around their mutual center of gravity (called the barycenter).  As Newton himself said (thanks UserLTK): “......Kepler knew ye Orb to be not circular but oval & guest it to be elliptical......”: from: arxiv.org/ftp/physics/papers/0107/0107009.pdf source of quote: H. W. Turnbull (ed.): The correspondence of Isaac Newton, II, Cambridge, 1960, pp. 436-437.
Kepler’s original values for the ratios were:

Kepler thought that “a more accurate number will be produced, if you take the times more accurately.”   Kepler attributed all the inaccuracies in his data set to measurement error.  However, Newton proved Kepler wasn’t quite correct.
Newton generalized Kepler’s third law as $$\frac{a^3}{p^2}=\frac{G(m_1+m_2)}{4\pi^2}$$, where G is the gravitational constant, $m_1$ is the mass of the sun, and $m_2$ is the mass of the planet, respectively.  This generalization implies that $a^3/p^2$ will be different for each of the planets, since they all have different masses. The largest differences manifest for the planets with the highest masses, but even for Jupiter, these differences are along the third significant digit, consistent with our current best estimates here:

These values correspond well to my calculated predictions bases on Newton's generalization of Kepler's third law:

To repeat my short answer above, the gravitation of Jupiter and all the other planets makes Kepler’s third law a bit less accurate than it would be if their gravity were zero.  The gravitational interaction between the planets, however, doesn’t have much effect on either the period or semi-major axis of any of the planets, especially in the short term.


Earth's magnetic field vs. refrigerator magnet

Can any magnet (or even one much stronger) of the Earth divert the cosmic rays as Earth's?

No, because even if the strength of the field may be higher, the size of magnetic field is too small.  Although the force that your local fridge magnet exerts may be larger than the one of the Earth, it does so only for a much smaller area.  Even if you brought your magnet outside of the magnetosphere, charged particles would pass straight by it and only have a tiny deflection, before the particles are outside of the fridge magnet's reach of influence again.  The deflection would be the integral of force over path length of the particle inside the field.

  Do these magnets somehow "interfere" with Earth's?

In the immediate surrounding of the fridge magnet, the fridge magnet magnetic field is stronger than the one of the Earth, but at any larger distance, the Earth magnetic field is stronger.  The larger Earth magnetic field is not significantly affected.

  Maybe because the size (not the power)?

Yes, size does matter.

In the magnetosphere, the only significant magnetic field is that of the Earth.  The field of your fridge magnet is non-existent there, as its range is very small.  Charged particles are diverted by the magnetic field in the magnetosphere.  When they do enter the Earth atmosphere, many of the remaining one are absorbed by the atmosphere (forming the northern lights).


Why are main sequence stars more massive than the Sun less dense? e.g. Vega, Spica etc

Main sequence stars are defined by being hot enough in the core to fuse hydrogen, so their core is at least about 10 million K, and can get up to 20 million K for the more massive ones (because they are more luminous, so their nuclear fusion has to crank itself up a bit more).

The way they get their cores hot enough to fuse hydrogen is via gravitational contraction, but a more massive star does not need to contract as much to reach the necessary temperature.   Since it does not need to contract as much, it is at lower density.

To see the scaling approximately, use the virial theorem, which says that the energy per particle in the core (roughly set by the need for fusion) is about $GMm/R$, where $M$ is the mass of the star, and $m$ is the mass of a proton.  Expressed in terms of density $\rho$ instead of radius $R$, that's proportional to $M^{\frac{2}{3}}\rho^{\frac{1}{3}}$.  Ergo, $\rho$ drops when $M$ rises.


What is needed to break through the surface of Europa?

There is no feasible way for a human-made probe to crush through Europa's ice layer.

"It is predicted that the outer crust of solid ice is approximately 10–30 km (6–19 mi) thick." (Wikipedia)

It's not even known, whether there is a liqud ocean at all under Europa's ice layer.

Instead of drilling one could try to slowly melt trough the ice.


How far do we have to go to leave the galaxy?

The sun is within a few parsec (15-25 pc) of the galactic plane, slightly above. The thin disk of the milky way (containing ~85% of the stars and gas) has a density going roughly like $\rho_0 \exp(-|z|/300 pc)$, while the thick disk (older stars, a few percent) has a scale height of 1000 pc instead. So if we want to move up (galactic north) above 90% of the star density we would have to move ~680 parsec, and going down 720 parsec. At this point at least the view would be splendid, but there would still be a fair number of thick disk and halo stars above and below.

The disk has a scale length of 2.5–4.5 kiloparsec (kpc) and the sun is about 8 kpc away from the core, so we are already outside nearly 90% of the stars (not even counting the bulge). The radius is commonly given as 15.5-27.5 kpc, so we are 7.5-19.5 kpc from the nearest edge and 15.5-27.5 kpc from the far edge.

If we count the dark matter halo the distances get even fuzzier. The virial radius (a rough measure of where most mass is inside most of the time) is 200 kpc, so we are basically 200 kpc away from that distance in most directions.


Why do most meteors vanish just before hitting the ground?

Only very few meteors actually make it anywhere near the surface of Earth; most burn up 75–100 km above the surface. From your point of view, however, the curvature of Earth's surface may make it look as if they get much closer, and even fall below the horizon. But depending on where you live, the horizon often has quite a lot more background light (e.g. from cities far away). That means that when they get near the horizon, which you may interpret as "several feet above the ground", the seemingly disappear.


Coordinate transformations between reference frames in spherical astronomy

There is kind of an answer over at Math. All you can do in spherical coordinates is to change the position of your "pole", i.e. you have $(1,0,0)$ in your first coordinate system, which is mapped to some $(r', \vartheta', \varphi')$ in the second coordinate system. The two angles represent a rotation, and the $r$ represents a scaling.

I think since we still deal with a vector space (for the angular part), you can simply add the origin of coordinate system two to vectors of the first coordinate system. The radius is a scaling, and needs to be multiplied. Hence for a point $p=(r, \vartheta, \varphi)$ in coordinate system 1 you get:

$$
p'=(r \cdot r', \vartheta+\vartheta', \varphi+\varphi')
$$


Does the Moon Have Enough Water for Robert Zubrin's Mars settlement Plan To Work?

Self-sufficiency is an incredibly broad term. We could argue that yes, there is water on the Moon, and that yes, there are viable ways to produce required electricity in self-sustainable ways, but the real question is, are there areas on the Moon that would be viable for both at the same time.

You see, the most likely place where surface or near subsurface water could exist on the Moon and be suitable for mass extraction are its polar, permanently dark regions. Indeed, the ISRO (Indian Space Research Organisation) Chandrayaan-1 spacecraft has detected evidence for water locked in surface lunar regolith minerals in lunar south polar region, water that likely originates from asteroid and comet impacts embedding it deep within the lunar core and released as magmatic water closer to the surface. Any free-form water in other regions of the Moon that are exposed to sunlight and Solar radiation would sublimate to its gas form directly and with ionisation lose hydrogen atoms, so while hydrogen and oxygen atoms might still be present to some extent embedded into the surface layer minerals, extraction would likely be too elaborate there.

But, wherever you'd find your water source, you would still require a great deal of electricity to power your extraction plant, later use electrolysis to separate molecular water into its constituent atoms, and compress it in cryogenic conditions to their diatomic liquids that are suitable as propellant components, diatomic liquid oxygen (or LOX) as your oxidizer, and double as much in molecular quantity of diatomic liquid hydrogen (or LH2) as your rocket fuel. Problem with electricity is, unless you brought your own and a great deal of it on you to power your plants, you will likely want to use be it solar power, or tap into the in lunar regolith embedded helium-3 (or 3He) and power your third-generation Helium-3 fusion reactor. See for example this answer of mine on Space Exploration on how that could be done.

So the main conundrum to exploiting Lunar resources, for the time being, remains finding sufficient and viably mineable resources of water where there is also self-sustainable ways of generating required electricity. One option that I can think of is staying on the most exposed to the Sun lunar equator and extracting deuterium and tritium hydrogen isotopes, as well as helium-3 from lunar regolith, all of them embedded there from Coronal Mass Ejections (CME). Required oxygen could be produced by crushing oxidized minerals and letting them sweat with the presence of hydrogen isotopes into ionized water, and helium-3 could be used as previously mentioned to sustain a fusion reaction producing required electricity to later break water molecules into its constituent atoms of hydrogen and oxygen by electrolysis.

How much of these hydrogen and helium isotopes are actually embedded in the lunar regolith, and how long these deposits persist in it, possibly staying there for at least some time due to the static charge of the regolith as it is bombarded by the Solar radiation, this is however a whole different question and one we can't currently yet answer. The study of the Lunar exosphere and dust environment is the sole purpose of the LADEE (Lunar Atmosphere and Dust Environment Explorer), that we barely just launched there. We will know in roughly one year, if it will be able to provide conclusive scientific evidence for these theories I've just mentioned.


Number of planets in a habitable zone with large moons

It's not quite the drake equation in terms of your inputs determining your estimate but it's not too far off.   Our telescopes can't get a good look at existing or forming solar systems.  They've spotted several thousand planets but only a tiny number of possible moons and I don't think any of them are confirmed.    It's also not always possible to tell is some of the "earth-like" planets that have been observed are more like sub-neptunes or perhaps water worlds.   Any kind of gaseous/water or ice world planet would have a hard time making an impact moon.  Two rocky bodies colliding are best for that.

Young solar systems are full of collisions during their planet formation period so there are probably enough collisions with the approximate size/ratio of Theia-Earth.  At least, it's probably not uncommon, but size isn't the only factor.   The impact has to be at the right angle to generate sufficient angular momentum, because the main body ideally rotates faster than its Moon orbits.  This faster rotation and the tidal force of the moon on the planet pushes the moon outward over time and stabilizes the Moon's orbit (provided the hill sphere is larger than the Moon's distance over time).   If the main body rotates slower than the moon orbits, the moon is drawn towards the planet and would likely destabilize before long.

An impact too straight on is bad for moon formation and too glancing is also bad.  The impact needs to be within a proper range in size, ratio between the two objects and angle of impact and without a surface with lots of ice and/or water.   The requirements are pretty precise, so stable large collision moon formation should at least be somewhat rare because of all the necessary conditions.   It's also not known with accuracy how many of the "habitable" planets started out rocky.

Migrating jupiters and other planets are another problem.   It's fairly common for Jupiters to migrate which should destabilize a percentage of inner planets with stable collision moons.

The biggest issue is the size of the star.   The habitable zone of red dwarf stars is very close to the star and moons wouldn't be stable around planets in those systems.    Those planets would likely be tidally locked, which means they'd rotate slower than their moon orbits and that would likely lead to destabilizing the moons orbit and the moon crashing into the planet, that is, if the planet had a moon at all, it would need to form in a very tight orbit.

So there's a few estimates to make a guess at an input.   I don't think an accurate estimate is possible but if I was to guess, I think large collision moons will be found.  I don't think they're super rare, and I base that guess on a reasonable abundance of rocky material in solar system formation and a very large number of collisions, so even if the Earth-Theia collision was one in a million, there's enough collisions happening that other two body systems like ours probably create long term stable moons from time to time.  So, rare but I think, not super-rare.

We probably won't have an answer for this until we get a better look with the James Webb Telescope.


If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.

For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at
$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$
where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called tidal gravitational field. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.

Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.

To be relativistically correct, the tidal forces on an free-falling object are described by the equation of geodesic deviation, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:
$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$
In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called spaghettification.

Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:

$M\sim 10\,\mathrm{M_\odot}$: $\sim 10^6$ Earth gravities;
$M\sim 10^4\,\mathrm{M_\odot}$: $\sim 1$ Earth gravity;
$M\sim 10^6\,\mathrm{M_\odot}$: $\sim 10^{-4}$ Earth gravities.

The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.


On some research, most likely what I saw was StarLink train 12. Maybe due to some clouds on that part of the sky, I was unable to see the entire train. Thanks.
Also, I do not know if I am right, this is what I thought, when I read the launches page on this website.


Image sets for testing stacking algorithms?

It's pretty easy to create a dataset for yourself using remotely controlled telescopes. I've used iTelescope in the past and had a good experience. Their "starter" account costs $20 and should be good to get you going.


Let's start by making some points clear:
1. We don't know what the Big Bang was.
Rather, we know that the Universe is expanding. If you extrapolate backwards, you'd expect the Universe to be denser and denser. More specifically, we talk about this as a change in the scale factor $a$, and this gets smaller and smaller as we look further back in time. According to general relativity (our modern theory of gravity), 13.8 billion years ago, $a$ should have been $0$; however, you can't have a metric with $a = 0$.
Thus, we know that general relativity is necessarily incomplete. It breaks down at the conditions of the early universe, so we currently have no physical model to explain that time. Rather, we know that the early universe expanded, and the Big Bang is the time that perplexes cosmologists. Some theories, like quantum gravity, have emerged in an effort to explain the Big Bang; however, we currently have little understanding of what it actually was.
So no, we can't tell you what the energy output of the event was, since we don't know what actually happened.
2. The temperature of the early Universe was high
Our theories break down at the Planck epoch of the Universe. The Planck epoch was the earliest epoch of the Universe and lasted until $10^{-42}$ seconds after the Big Bang — that's 200 Planck times, which are the shortest meaningful measurement of time.
During this epoch, the entire Universe was at $1.417×10^{32} \; \mathrm{K}$, which is the Planck temperature. This is the hottest possible temperature; an object at this temperature will emit photons with wavelengths of a Planck length (you can read more about this in my answer here). The point is that there is no meaningful distance smaller than a Planck length, so the Universe couldn't be hotter than the Planck temperature.


Is Ceres a terrestrial-type (dwarf) planet?

I'm inclined to say no (and footnote, I realize Wikipedia isn't a good source for scientific proof as it's not always right, but I'm using it more to demonstrate a point than than use it as an authoritative definition).     Wikipedia:

  A terrestrial planet, telluric planet or rocky planet is a planet that
  is composed primarily of silicate rocks or metals.

and

  All terrestrial planets have approximately the same type of structure:
  a central metallic core, mostly iron, with a surrounding silicate
  mantle. The Moon is similar, but has a much smaller iron core. Io and
  Europa are also satellites that have internal structures similar to
  that of terrestrial planets.

and

  Dwarf planets, such as Ceres and Pluto, and large small Solar System
  bodies are similar to terrestrial planets in the fact that they do
  have a solid surface, but are, on average, composed of more icy
  materials (Ceres and Pluto have densities 2.17 and 1.87 g cm−3,
  respectively, and Haumea's density is similar to Pallas's 2.8 g cm−3).

So, in my opinion, half ice/half rock if a different category and not terrestrial.   There probably are planets that formed outside the frost line that are half rock-half ice in other solar systems, but I don't think I'd call them terrestrial.   We probably need a new word for them.

If Ceres was to drift closer to an Earth's orbit, it's ice would melt and it would have oceans (and an H20 rich atmosphere), at least, for a little while anyway until it lost it's atmosphere.   Planets with Ceres composition would likely be water worlds if they were warm enough.

Now, I can't speak for any the official answer, but that's my opinion.  If Ceres is Terrestrial and Haumea is more rocky than Ceres, then Ceres probably shouldn't be the only terrestrial dwarf planet in the solar-system.

edit.

Link provided that says Ceres is a terrestrial dwarf planet.

reasoning:

  Terrestrial planets have numerous similarities to plutoids (objects
  like Pluto), which also have a solid surface, but are composed of more
  icy materials.

Some of this is just semantics, but lets look at density, which is a pretty good measure of Water-Ice and other ices to Rock content.

Ceres 2.08 g/cm^3

and the other dwarfs, by size (not all of them are well measured so density isn't certain).

Eris 2.3
Pluto 1.88
Makemake ~2
Haumea  2.6-3.3
Quaoar ~2.2
Sedna (2.0?)

and I could go on.

Lets do a few moons, just for fun.

Io:  3.55  (extreme volcanism)
Europa:  3.0
Ganymede:  1.93
Calisto:  1.83
Titan:  1.88
Triton:  2.06
Enceladus:   1.61

Moons are easy, we don't call them dwarf planets even if they are larger than dwarf planets cause we call them moons. - definition averted.   :-)

But I have a hard time seeing why Ceres gets a different classification than other very similar objects which just happen to be further away.   Now, granted, Ceres, while it formed outside the frost line, it's now inside the frost line so surface ice on Ceres doesn't last.  It's got a rocky surface while most Kuiper belt objects have an icy surface.    I can see that, defining an object based on it's surface, but I still think a half rock/half ice object (of significant size, large enough to be round and meet dwarf planet criteria), should be called half rock-half ice or whatever that definition is.   I don't think it should be rocky/terrestrial if it's Ceres and icy if it's Pluto, even if that's how they look the surface, but that's just my opinion.

/// hope that wasn't too much of a rant.  :-)


Why does the number of extrasolar planets vary in different sources?

As of April 6, 2020:

exoplanet.eu shows 4241 planets;
the Open Exoplanet Catalog shows 3852 planets of which 3504 are confirmed;
the Nasa Exoplanet Archive shows 4144 confirmed planets;
exoplanets.org indicates a total of 5747 planets of which 3262 are confirmed planets;
...

The number of confirmed planets are different from one website to another simply because what exactly is being counted is different. Each one of these databases has a different criteria for deciding whether an exoplanet is going to be counted or not. To understand the differences between these numbers, you have to ask

How is an exoplanet defined?
How solid must the evidence be for the exoplanet to be counted?
What detection methods are included?

The answers to these questions are usually in the methodology section of the website, or in a dedicated article.

Here are a few examples of differences:

exoplanets.org uses a cutoff of 24 Jupiter masses for what constitutes a planet or a brown dwarf, whereas the NASA Exoplanet archive uses 30 Jupiter masses. I haven't found a methodology section for exoplanet.eu, so I emailed the people maintaining the database, and they indicated they use a 60 Jupiter mass cutoff.
exoplanets.org only shows planets that appear in peer-reviewed publications with well-measured orbital parameters, rather than giving a comprehensive list of all planet announcements.
the Open Exoplanet Catalog is decentralized and community driven, so on different instances of the database, you might have different numbers (which the authors consider to be a feature, not a bug).

In addition to this, each of these websites might also be updated at different times. For instance, the NASA Expolanet Archive is updated once a week, but exoplanet.eu is updated daily. But other websites might be updated at different times and with different frequencies. Because of this, two websites with the same criteria might still show different numbers.


Tables of aphelion/perihelion dates for other planets?

(too long for a comment, this answers why barycenter data is available for a much longer time span than planet center data)

When NASA computes planetary positions, they treat each planet and its moons as a single point mass (the barycenter of the planet and its moons). They compute these positions +-15000 years from now.

For most planet systems, the barycenter is very close to the planet center (Mars' planet barycenter is only a few inches from Mars' system barycenter because Phobos and Deimos have such little mass-- for the moonless Mercury and Venus, the system barycenters and the planet barycenters are the same).

They then compute the locations of the moons and the planet's center itself. These calculations are done for a much shorter period of time.

Therefore, you can compute the location of the barycenter of Jupiter and its moons for +-15000 years, but the actual barycenter of Jupiter itself for a much shorter period of time, even though Jupiter is massive enough that its own barycenter is very close to the barycenter of Jupiter and its moons.


Is a black hole heavier than the star from which it was created?

The mass of a black hole is always much less than the star from which it formed.

A very large star such as Eta Carinae (which is expected to form a black-hole some day) is about 100 times that of the sun, Having already lost as much as 50 solar masses, blown off the star by powerful outbursts in the past.

The future of the star is not fully understood. It is possible that it will collapse in a supernova, in which most of the rest of the star will be blasted into space, forming a supernova remnant, and the core of the star will collapse. The black hole formed may have a mass of more than 3 times the sun, but much less than the mass of the original star.

So while the black hole is not heavier, it is much denser. A 3 solar mass black hole would have a diameter of about 20km.


Why are planetary orbits not all circular?

For a given amount of angular momentum, a circular orbit is the orbit with lowest energy.

Some orbits are elliptical (actually, all orbits approximate to ellipses) with a non-zero eccentricity, because they have yet to reach their lowest energy state, or because they have been excited from their minimum energy state circular orbit.


What increases or decreases the mass and density of a fixed radius star?

Stars are broadly divided into 3 types. These 3 types are divided on the basis of their composition of gases and elements other than H and He. So even if these stars are of same radius, their composition is different.
I am not sure but I think that as different stars have different composition and their rate of fusion is different, their density is different and so is their mass. Also while some stars are just forming and are at that radius, their gas composition is different than what a dying star of that radius is and so their masses is different.
I don't know if my version of the answer is right, but I hope this helped.


Likelihood of a stable system with a dwarf planet's orbit inside that of a gas giant

This is just a repository of useful links to software and papers along the way.

NEMO
On Toolboxes and Telescopes , by Hut and Sussman, (1986) in: The Use of Supercomputers in Stellar Dynamics, Springer Verlag, p 193-198.


How to determine the eccentricity, knowing the orbital speed and distance

How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?

We are given three things:

$v$ "instantaneous orbital speed" (a scalar)
$r$ "distance from the Sun to the planet at that moment" (a scalar)
$GM$ since the OP specifies "the Sun"

Let's see how far we can get with the vis-viva equation and look at the problem purely from an energy standpoint.

$$v^2 = GM(2/r-1/a)$$

becomes

$$a = 1/(2/r-v^2/GM).$$

Let's look at a numerical example. Say $r = 1AU \approx 1.5 \times 10^{11}$ meters, and $v = 25,000$ m/s.

We know then that the semimajor axis of the orbit must be 0.773 AU. If the orbit reaches 1AU, the velocity will be 25,000 m/s. If it reaches 1.1 AU (which it might or might not, depending on eccentricity) it's velocity will be 21,500 m/s.

We're almost there but not quite. We need one more piece of information.

The periapsis and apoapsis $r_p$ and $r_a$ are related to the semimajor axis and eccentricity by

$$r_p = (1-e)a$$
$$r_a = (1+e)a$$

So for example if you had mentioned that the apoapsis was 1.1 AU, then we could calculate eccentricity from

$$e = r_a/a -1$$

and get $e = 0.4228$ AU and a periapsis $r_p = 0.387$ AU.


Does the position of a star on a HR diagram depend on the magnitude system/color index used?

Yes, of course. But note that HR diagram refers to a luminosity vs. effective temperature plot, while you're dealing with a color-magnitude diagram (or CMD).

  Why is this happening?

First of all, you should check that it's not your fault. If you see unphysical things, like the horizontal branch going beyond the red giant branch, you know you've messed up. This can happen with bad calibrations. But I think you're safe, it's not your case.

Then, I assume that the dependence of magnitude on filters is trivial.

Now, how colors change with filters?
Let's start in first approximation. Start imagining the Planck's function (here you go), and assume your filters are monochromatic. Recall that color is the difference of two magnitudes, i.e. \begin{equation*}m_1-m_2= 2.5\log_{10} I(\lambda_1)/I(\lambda_2)\end{equation*} where $I$ is the intensity and $\lambda_1>\lambda_2$. Now take the points on the Planckian corresponding to your two wavelengths, i.e.  $I(\lambda_1)$ and $I(\lambda_2)$, and draw a line between these two points. If the slope of the line is negative (resp. positive), then the ratio of intensities is less (resp. more) than 1, and the color is negative (resp. positive). If the ratio is 1, then slope and color are 0. So color is somehow proportional to the Planckian slope between the two $\lambda$s. Here comes the point: color is most sensitive when the peak of the Planckian is near the wavelength of the filters. If the peak is too far from the filter wavelengths, then the Planckian slope becomes almost constant, and the color loses its sensitivy to temperature. That's the reason why in some CMD you see the bluer part of the horizontal branch suddenly going down: the color saturates and doesn't catch the temperature variations. First take-home message: different colors probe different ranges of the Planckian, and hence are sensitive to different ranges of effective temperature.

But I think we didn't hit the point. Going to the second-order approximation, we need to consider the whole width of the filter. This fact brings in a lot of effects. For instance, color excess starts depending on the star's spectral type. That is, hotter stars are more reddened, because reddening is more effective at shorter $\lambda$s (e.g. the famous Whitford's law), so early spectral-type stars emit most of their flux at short $\lambda$s. Another effect is that broad filters are highly affected by absorption lines and molecular bands, and I think this is your case. You have photometry in two broad UV filters, i.e. $F275W$ and $F438W$, of a putative blue straggler, so a source with a lot of blue flux. Moreover, the OH absorption band falls right in the $F275W$ band, while several CH and CN bands fall into the $F438W$ band. So your colors are not only proxies for effective temperature, but also for chemical abundances. Notoriously, the UV filters have been used to separate multiple stellar populations in the CMDs, exploiting their different abundances. So, second take-home message: reddening depends on spectral type, but most importantly colors can depend on chemical abundances.

Finally, as other answers have pointed out, the position of a star on the CMD can be changed by other effects, such as variability.


How large part of the sky has Hubble observed?

25 years of say 16 orbits/day is a (very over-estimated) maximum of 146,000 pointings.

The largest instrument is the various incarnations of the optical imager with a FOV of 6 (WFPC2) to 11 (ACS) square arcminutes. Thus the absolute maximum fraction of the sky (40,000 square degrees) that can have been covered is 1 per cent. And this is a big overestimate, since many parts of the sky have received many repeated pointings (e.g. the Hubble deep field).


Are objects in the universe moving away from each other at the same acceleration?

I assume you mean is $\frac{dv}{dt}$ the same for all objects and/or is it constant for all objects, where $v$ is recessional velocity (and $t$ is cosmic time)? The answer is no it is not the same for all objects and it is not constant for all objects.

$$v=\frac{dD}{dt}=H(t)D$$

Where $D$ is proper distance and $H(t)$ is the Hubble parameter. If we take the Hubble parameter as a constant, then:

$$D = D_0e^{Ht}$$

Where $D_0$ is the proper distance at the present time. So,

$$D''(t) = \frac{dv}{dt} = D_0H^2e^{Ht}$$

So the "recessional acceleration" of an object depends on its present proper distance and is exponentially increasing with time.

Now the Hubble parameter isn't a constant and in LCDM cosmology is currently asymptotically decreasing to a constant. In the current epoch though it is fair to say that the "recessional acceleration" of an object depends on its  proper distance and is increasing with time.


Are there any simulations of the Grand tack?

Kevin Walsh, lead author of the original Grand Tack paper has a page on his website discussing the Grand Tack model and subsequent work. At the bottom of this page there is a movie of the evolution taken from the numerical simulation that was the basis of the paper and the Grand Tack model. Sean Raymond (one of the other authors on the Walsh et al. 2011 paper) also have some movies on his research webpage on the Grand Tack model. There is also a presentation from the 2019 Sagan Workshop which discusses terrestrial planet formation in more detail and presenting the 2 other possible solutions, along with the Grand Tack model, to the early Solar System formation problems such as the so-called "small Mars" problem.


When the set of official dwarf planets is expected to increase?

As I noted in a previous answer, the official body for naming and classification of astronomical objects is the International Astronomical Union, or IAU. They are the ones responsible for designating bodies as "planets," "dwarf planets," etc. The scientific community recognizes them and their decisions as "official."

The IAU meets periodically for General Assemblies to discuss general issue; the next such meeting is scheduled for 2015. The topics discussed may range widely, covering all sub-fields of astronomy. Unfortunately, I don't know the exact topics for discussion, and while I imagine there will be a wide range, issues relating to dwarf planets may not be covered. However, the IAU does meet in between these meetings to discuss a variet of topics.

That might answer your question, but if you're talking about the possible discovery of dwarf-planet-like bodies, I can't help you. There are probably many on-going efforts to search for small bodies within our solar system, but I don't know of any recent discoveries of possible dwarf planet candidates.

I hope this helps.


If you're interested in a low cost, but usable telescope that provides decent performance, look into the Galileoscope.

http://galileoscope.org/

It provides 50 mm (2 in) of aperture, at a focal length of 500 mm - so it's an f/10 instrument.

The kit is very easy to assemble, and instructions are provided on the Internet in great detail. Even kids can assemble it, with some supervision.

The objective lens (the primary optics) is an achromat doublet - it's made of two actual lenses combined to reduce chromatic aberration. I was impressed by the quality of that lens, at least when compared to its price - it seems to work as well as optics in instruments that cost several times more. This is one of the main reasons why I recommend this instrument to beginners.

It uses standard 1.25" eyepieces. That means you can either use the eyepieces that come with the kit, or you could borrow any 1.25" eyepiece from another telescope and just use it. This is rare in an instrument this size.

The kit gives you the option to build either a galilean eyepiece, or a 2x barlow. My advice: don't bother with the galilean eyepiece; its field is far too narrow. Instead, build the 2x barlow.

There's another eyepiece in the kit, use it either alone or combined with the barlow.

In theory, the maximum magnification is 100x, but in practice things will get a bit difficult when you push magnification much over 50% of the max value. This is typical for many telescopes.

Important: Like any telescope used for astronomy, this scope will not work well if you just hold it with your hands. You must use some kind of mount. The scope comes with a standard mounting nut that fits any tripod used for photography or astronomy. Use the sturdiest, most rigid tripod you can find. Even a very little amount of wobble will make it very difficult for you to use the scope.

Alternatively, make and use a DIY mount such as this one (results will vary depending on skill and materials used):

https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/

At the very least, support one end of the telescope on a fixed, tall object such as a fence; this will make things a little better, but it's not a replacement for a true mount.


What is the official classification and naming system of objects in the universe(, or what would my URI look like)?

Objects aren't named like that.

Either objects are given "names" that have no structure. For example, an asteroid may be named "Vesta".

Or they are given essentially meaningless catalogue numbers. For example, a star might be catalogued as HD 138987. It is number 138987 in the Henry Draper catalogue. Some major stars do have their constellation as part of the name. Alpha Centauri is the brightest star in the Centaurus constellation.

Comets are often just named according to the year of discovery

Other objects are named by their position in the sky, using polar coordinates.

There is no URI-like system in use.

The reason for this is that there is no natural hierarchy. When naming species of animal, they naturally group into genus, family, order, class as a result of evolution. No such structure is found among astronomical objects, so a hierarchical approach is not appropriate.


How gravity behaves for an Earth body and for a celestial body?

Gravity is the same qualitatively.  It is the same type of thing: It pulls you down.  It can be described by Einstein's equations.  It can be very closely approximated my Newton's equations.  Force of gravity is proportional to mass and and inversely proportional to distance²

Newton proposed a law of Universal gravity.  The same law of gravity applies everywhere.  Although the details of what the law actually is have been tweaked (by Einstein) the principle of "the same law everywhere and at all times" seems to be true.

Gravity is different quantitatively. It is stronger in some places and weaker in others. The moon has a much smaller mass, and so your weight on the moon (ie the force of gravity on you) would be much less.


Is the expansion of the universe greater than the speed of light?

After inflation, the expansion of the universe did indeed slow down. During the inflationary epoch (lasting roughly $1 \times 10^{-33}$ seconds), the universe expanded by a factor of $10^{26}$. That's incredible! However, the inflationary epoch didn't last long, and that incredible expansion ended pretty soon after it started.

The universe is, as you said, still expanding. In fact (like you also said), that expansion is increasing. However, we have to be careful when we talk about the rate of this expansion. Currently, the rate of expansion between two objects depends on the distance between them, which is encoded in Hubble's law:
$$v=H_0D$$
where $v$ is the recessional velocity, $H_0$ is Hubble's constant, and $D$ is the proper distance between the objects. This relation proves that objects that are farther away are moving away at a greater speed. We can extrapolate from that that, at a sufficiently far distance, two objects would be moving away from each other at the speed of light or greater! However, this is only true for objects very far from each other. But yes, eventually, any two objects sufficiently far from each other will move apart at the speed of light - and then greater.

Source (for length of inflationary epoch): http://www.universe-galaxies-stars.com/Cosmic_inflation.html

Source (for Hubble's law): http://map.gsfc.nasa.gov/universe/uni_expansion.html


What Are Some Good Southern Hemisphere 7x50 Binocular Targets?

For anyone else who might be asking a similiar question, I found a great resource for this. The PDFs found at the URL below contain a good list of binocular, naked-eye and telescope targets for each month of the year in both hemispheres.

http://www.skymaps.com/downloads.html


Would "layered" radio interferometry work?

Absolutely this can work as can be seen in  The Murchison Widefield Array or MWA in Western Australia. This is essentially exactly how it works. You definitely do loose information (the first stage of combining antennas significantly reduces the field of view, and the number of interferometric baselines you get is reduced).
From this page:
Frequency Range            70-300 MHz
Number of Receptors        2048 dual-polarization dipoles
Number of antenna tiles    128
Number of baselines        8128

There are 32 receptors per tile (2 polarizations for $4x4=16$ elements), and $N = 128$ tiles. The number of baselines is $N(N-1)/2 = 8128$.
The advantages is a reduction in the data rates, so reduced digital processing. Note the first stage would be treated as beam forming - you cannot do interferometry between the first stage and somehow do interferometry between the results. Ie the first stage is summing the voltages (with appropriate phase corrections) and the second stage would be interferometry.

One of the tiles making up the 32T, a prototype instrument for the Murchison Widefield Array (source)

An MWA antenna consists of a four by four regular grid of dual-polarisation dipole elements arranged on a 4m x 4m steel mesh ground plane. Each antenna (with its 16 dipoles) is known as a "tile". Signals from each dipole pass through a low noise amplifier (LNA) and are combined in an analogue beamformer to produce tile beams on the sky. Beamformers sit next to the tiles in the field. The radio frequency (RF) signals from the tile-beams are transmitted to a receiver, each receiver being able to process the signals from a group of eight tiles. Receivers therefore sit in the field, close to groups of eight tiles; cables between receivers and beamformers carry data, power and control signals. Power for the receivers is provided from a central generator.


Is the galaxy made of a nebula or the solar system?

No, those are two different nebulae.  There are stars in our galaxy much older than 5.6 billion years.  The problem is that "nebula" is a very generic term, so it can be a whole galaxy or even the remains of a single star.  It's not even clear what is meant by the nebula from which our Sun formed, it could be the giant molecular cloud if our Sun formed in a huge cluster, or it could be something called a "proplyd" (which is much smaller) if the Sun formed in relative isolation.  There are even proplyds that break off from giant molecular clouds in a kind of combined scenario where the GMC forms the proplyd and then the proplyd forms the star.  In any event, it is important to realize that the galaxy formed from a huge nebula some 10 billion years ago, and the Sun formed from a much smaller nebula much later.


Why does east and west appear to be mirrored in the celestial sphere?

You're viewing the celestial sphere from the inside looking outwards, as opposed to maps of the Earth's surface which are viewed from the outside looking inwards.

Imagine drawing a map on a sheet of glass. When you view the map from the other side, it will appear mirrored.


How could the polarizer in early coronagraphs make the Sun's corona more visible against the daytime sky?

When you look at the Sun in white (visible) light, what you see is actually sunlight that has scattered from free electrons in the solar corona. This is known as "Thomson scattering".

If we view the corona around the Sun (from our perspective), then the scattered light that reaches us has scattered through a large angle (roughly 90 degrees). The sunlight incident upon the electrons in the corona is unpolarised (that is, the electric field vector of the light is randomly oriented but at right angles to the direction of travel of the light). Such light accelerates the free electrons which re-emit light (this is what scattering is). But the scattered light is polarised when it is scattered through a large angle, because we effectively can only "see" the electrons accelerating along a line that forms a tangent to the limb of the Sun. The component of the acceleration along the line of sight does not cause any radiation to be emitted towards us.

The light we see from the daylight sky is also produced by scattering. In this case it is "Rayleigh scattering" from electrons that are bound into molecules and atoms in the Earth's atmosphere. The polarisation effects of this scattering are almost identical in nature to that caused by Thomson scattering. However, when we look at the sky right next to the Sun, then this will have been scattered through a very small angle. In these circumstances the scattered light will be unpolarised.

Thus we have a means of seeing a contrast between (polarised) light coming from the corona and (unpolarised) light coming from the sky in the same line of sight. By viewing the coronal area using a linear polariser, and then presumably rotating the plane of polarisation and taking a series of pictures, and comparing this with images taken without a polarising filter, we can isolate the small fraction of light from around the Sun that is due to Thomson scattering in its corona.

You made the comment that you don't see the sky darken when looking through a polarising filter. Well, either your filter isn't working or you were looking at the sky roughly 90 degrees away from the Sun and had the polarising filter in just the right orientation, or your sky is very polluted (the scattering caused by particulates does not cause the same degree of polarisation). Daytime sky is most definitely polarised depending on angular separation from the Sun. This map shows how the degree of linear polarisation changes with position as the Sun moves across the sky. The Sun is where the black circle is (black indicating unpolarised light). See also this answer.


How are microlensing events used to constrain the size of innermost stable circular orbits around spinning black holes?

I think most of the paper is devoted to describing how they do it.  (This may be more of a comment than an answer, because I'm an amateur and not prepared to go through the paper in detatil).

Note that microlensing here is a little different than microlensing as described in the Wikipedia.  In the Wikipedia page and the Planetary Society page you reference, microlensing refers to observing the change in brightness over time of a small, relatively nearby object when a closer but non-luminous object passes in front of it from our line-of-sight.  In this paper, microlensing refers to the extra magnification that can be provided to tiny sub-sections of a "macrolensed" (gravitationally-lensed cosmologically distant object) by the chance placement of a star at the right position in the lensing geometry.

The authors of the paper figure out how this sort of microlensing might apply to the innermost parts of a quasar, where it might highlight small parts of the accretion disk, and so the contribution of those parts to the spectrum (in this case, one particular X-ray emission line characteristic of active galactic nuclei accretion disks) of the "macrolensed" object as seen from Earth.  They apply a statistical model of the likelihood of such microlensing and how it would affect the measured spectrum of a range of possible active-galactic nuclei (resolving a particular part of the accretion disk with microlensing would enhance that particular parts contribution to the whole, and each part of the accretion disk would have a different relativistic shift of the emission line as it whips around the central black hole), and then see how their model matches the actual measured spectrum.  It sounds like they are operating at the limits of statistical significance-- at least for most of the objects in their study-- but got more robust data for one particular object.


Gaia: What is the difference between CCDs used for astrometry, photometry, and spectroscopy?

CCDs are optimized for a certain wavelength range, and for a certain expected signal level. In astronomy, we tend to be short of light, so here we almost always want them to be as sensitive as possible (an exception may be observations of the Sun, which I don't know much about). But for instance, the Nordic Optical Telescope has a CCD which is optimized for blue wavelengths, but has quite a lot fringing in the near-infrared. And further out in the IR, CCDs aren't even used, instead using something which are just called "detectors".

However, whether the CCD is used for imaging (photometry and astrometry) or spectroscopy does not have anything to do with the CCD; it's just a matter of inserting a grism or not. I'm not really into the instruments of Gaia, but I assume that differences in the CCDs are due to different wavelength regions being probed. There may be a difference in how its sub-parts (it's actually an array of CCDs) are positioned (for instance, for spectroscopy in principle you don't need a large field of view, but can do with a long array rather than a more square one), but the design of the individual CCDs are the same.


If Venus and Mars changed places, would we then have 3 habitable planets?

This is a very interesting question. Of course, as you noted, you have simplified things quite a bit; there are other factors besides temperature that affect habitability.

Regarding Venus, you probably know that Venus is extremely hot at its surface not just because it is closer to the Sun, but because it has a thick CO2 atmosphere and is warmed by the greenhouse effect. There are, in fact, two things about Venus that would prevent it from being habitable wherever you put it. One is the lack of a magnetosphere, which is necessary to prevent ionizing radiation (particularly from the solar wind) from reaching the planet's surface. The lack of a magnetosphere appears to be due to the lack of a geodynamo on Venus, which has to do with the structure of its core. Second, Venus appears to lack tectonic plates, which you may know are responsible for earthquakes here on Earth. Interestingly, tectonic plates play a major role in controlling the buildup of CO2 in the atmosphere (see here for details). Lacking tectonic plates, Venus is doomed to have a large CO2 atmosphere wherever you put it, which would not make it a nice place to live.

Mars, on the other hand, is a very different matter. It has both a magnetosphere (albeit it is very weak) and it likely has tectonic plates (although last I heard it is thought to only have two). One of the reasons that NASA has sent so many probes to Mars is that it was thought to be habitable at one point. It is thought that Mars's growth was stunted because of gravitation effects from Jupiter and Saturn. So, in another universe, Mars could have ended up very much like another Earth.


Cases of gravitational lensing resulting in a recognizable image of an extended object?

There are actually quite a few examples of this. A particularly nice one, where you can see star-forming substructure within the distant, lensed galaxy (the extended, vertical "snake" structure just right of center), is this HST image:

This is another HST-based example, in which the same background galaxy is lensed multiple times (indicated by white ovals) by an intervening galaxy cluster, with a reconstruction of the background galaxy in undistorted form shown in the lower-left:

https://www.space.com/14481-hubble-photo-brightest-galaxy-gravitational-lens.html

People have even obtained spatial resolved spectroscopy within the lensed galaxies, as discussed in this paper (which includes spectroscopy of the first lensed galaxy pictured above).


Are there home-brew photometers for amateur observing of variable stars and are they widely used?

It should be doable with a standard photographic setup, a star catalog with distances and measurements for a some standard stars and a bit of software.

With the usual "ritual" for obtaining images from the sky with a given setup, chip and temperature (obtain dark frames, flat field frames and ground noise frames for division and subtraction) one can calibrate the brightness of an area (subframe around a star) in a photography by comparing it to known stars. The capture software or one of the many post processing packages for amateur astronomy can calculate the brightness of a region, if it is not totally saturated, i would guess if it is well away from any saturation.

A more detailed description: http://www.bellatrixobservatory.org/cvaaI/17/

Edit: An add-on script for one of the amateur software packages for doing photometry: https://pixinsight.com/doc/scripts/AperturePhotometry/AperturePhotometry.html.


People have demonstrated high-pressure diamond formation using laser-driven shocks, much in line with the original idea. The diamond rain idea seems to be doing well (popular science article about it, with references).

However, it is quite likely that the diamonds could have different properties because of formation mechanisms. There are a lot of unusual allotropes of carbon, and local conditions of formation (not to mention impurities) may cause the carbon to turn into these forms rather than the standard diamond lattice.

Simulations show that beyond 1000 GPa diamond has a BC8 structure. However, this is a pressure unlikely to occur inside Neptune's atmosphere: it occurs deep in the mantle, so sinking diamonds may shift crystal form.


Is the Moon not in hydrostatic equilibrium?

The paragraph notes previously that Mimas is not in hydrostatic equilibrium for its current rotation. A quick search for the moon and hydrostatic equilibrium turned up M. Burša (1984) "Secular Love Numbers and Hydrostatic Equilibrium of Planets". According to this paper the Moon, Mercury and Venus are all far from hydrostatic equilibrium. The discrepancy is much smaller for the Earth. The paper goes on to note that the rotation periods required for the flattenings of these objects to be explained as hydrostatic are 3.7 days, 4.7 days and 17 days respectively, all substantially faster than the current rotation periods.

According to the abstract of C. Qin's presentation "Formation of the lunar fossil bulge and its implication for the dynamics of the early Earth and Moon" the usual hypothesis in the case of the Moon is that the shape is a "fossil bulge", a relic from when the Moon was spinning faster early in its history. Possibly this explanation may also apply to Mercury and Venus, which also have been spun down by tidal forces.


Is there a possibility of finding a system where a star is orbiting a planet?

Technically stars and planets orbit a common barycenter. In case of single star systems the barycenter is usually within the star. However, the barycenter between the Sun and Jupiter is outside the Sun (though very close to it). Otherwise, colloquially, when we say something orbits another body, the body that's "being orbited" is usually the most massive one. Since stars are more massive than planets, it's the planets that orbit them.
A gas giant needs more than 12.5 Jupiter masses to start deuterium fusion, thus becoming a brown dwarf star. That doesn't mean it will. From about 15 Jovian masses on I guess it definitely will, so there's a grey zone between 12.5 and 15 Jovian masses where a more massive gas giant might not fuse deuterium (therefore being a sub-brown dwarf planet) and a less massive could (thus being a brown dwarf star). This is the only case where you might say a star orbits a planet though in fact they'd orbit a barycenter that would be outside either body.
And since black holes count as stars, you may want to consider primordial black holes stars which are less massive than gas giant planets, so if they orbit a gas giant that would be another case of a star orbiting a planet. But we're unsure whether such PBHs exist.
There might also be planets close enough to a stellar black hole that are big enough so that the barycenter is closer to the planet but I'm unsure whether the black hole's tidal forces wouldn't tear the planet apart in such case.


What other definitions for a planet were proposed?

The IAU website has this draft definition of "planet" and "plutons" page, published on 16 August 2006 in Prague. It says (emphasis mine):

The world's astronomers, under the auspices of the International
Astronomical Union (IAU), have concluded two years of work defining
the difference between "planets" and the smaller "solar system bodies"
such as comets and asteroids. If the definition is approved by the
astronomers gathered 14-25 August 2006 at the IAU General Assembly in
Prague, our Solar System will include 12 planets, with more to come:
eight classical planets that dominate the system, three planets in a
new and growing category of "plutons" - Pluto-like objects - and
Ceres. Pluto remains a planet and is the prototype for the new
category of "plutons."

The proposed 12 planets Solar System looked like this (credit: The International Astronomical Union/Martin Kornmesser):

The final press release, published on 24 August 2006, says:

The IAU members gathered at the 2006 General Assembly agreed that a
"planet" is defined as a celestial body that (a) is in orbit around
the Sun, (b) has sufficient mass for its self-gravity to overcome
rigid body forces so that it assumes a hydrostatic equilibrium (nearly
round) shape, and (c) has cleared the neighbourhood around its orbit.
This means that the Solar System consists of eight "planets" Mercury,
Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. A new
distinct class of objects called "dwarf planets" was also decided. It
was agreed that "planets" and "dwarf planets" are two distinct classes
of objects. The first members of the "dwarf planet" category are
Ceres, Pluto and 2003 UB313 (temporary name).

Which looked like this (credit: The International Astronomical Union/Martin Kornmesser):

So in 8 days, the IAU switched from a 12 planets system, with a planet subcategory called "plutons", to a 8 planets system, with an dwarf planet category distinct from the planet category. How did that happen? It's probably something only a General Assembly attendant who witnessed the discussion would know. It must have been quite a debate!


Is it still the case that the transit of Mercury can't be measured with sufficient accuracy to determine the Sun's distance?

When transits of Mercury and Venus occur, amateur groups do try to observe those transits so as to measure the astronomical unit. To do so, observations of a transit must be made at widely separated points on the Earth. The separated observers will see slightly different versions of a transit, with the differences eventually yielding an estimate of the astronomical unit.
Observations of transits inevitably yield less accurate estimates than do modern estimates based on radar observations of the distances to other planets and on even more accurate distance measurements based on probes sent to other planets. The radar and communications measurements are extremely accurate, twelve places of accuracy or more. Modern transit-based estimates using accurate clocks and GPS are lucky to get better than four places of accuracy.
Transit-based estimates are less accurate than radar-based and communications-based estimates because of the very small parallax involved in Earth-based observations of a transit. Observations of transits of Mercury yield even less accurate results than do observations of transits of Venus because Mercury is closer to the Sun / further from Earth than is Venus during its transits.


Could ancient tectonic activity on Mars be due to it having oceans of sulphuric acid?

First, tectonic activity have not ceased on Mars, although the planet is in a "stagnant lid" stage, which represents the last stage before a body becomes tectonically dead. See the answer to Is there convection in Mars' mantle?
Second, tectonic activity is caused by internal heat. As a planet gradually loses its internal heat, tectonic activity slows down, until it eventually stops. There is no need for more convoluted explanations.


Stargazing (Tenerife) - different quality today and yesterday

If you want to see the milky way, you really need the moon to have set. The moon sets about 50 minutes later each day, so you will need to go later.  On the 28th of July, the moon sets at about 2 am and on the 29th it sets at about nearly 3am.  If the moon is up that will significantly affect the background sky brightness.  But when the moon is down, Tenerife has dark enough skies to see not only the milky way, but also the zodiacal light.
The other parameter is high cloud.  Although you are above low lying stratus, High cirrus or cirrostratus cloud can still obscure your view. Sand might also have an effect, but the reason that you can't see the milky way is primarily because the moon is up. So go back at 2am or later, until dawn at about 6am.


What accounts into calculating the Hubble constant?

I think this might answer your question, I have quoted the important paragraph below.

  If peculiar velocities could have any value, then this would make Hubble’s law useless.
  However, peculiar velocities are typically only about 300 km/sec, and they very rarely exceed
  1000 km/sec.
  Hubble’s law therefore becomes accurate for galaxies that are far away, when H0d is much larger
  than 1000 km/sec.
  Furthermore, we can often estimate what a galaxy’s peculiar velocity will be by looking at the
  nearby structures that will be pulling on it.

So it seems that in answer to your question, if the galaxy is far enough away then the peculiar motion can be neglected, or if its close by, a reasonable approximation found using estimates from nearby galaxies (in a cluster for example).


Could something temporarily break the Moon's tidal lock?

Yes, theoretically it is possible to impact the Moon with enough moment and low enough energy to have it change its rotation periode, which actually is the same as its orbital periode (this is the tidal lock). Best candidate will be a high-mass, low-speed asteriod impacting almost tangentially and on the rotation plane.

Is this really possible? I would bet that it will never happen.

Supossing that it happens, will Moon get tidal locked again? Yes. As long as there are water on Earth suffering tides from Moon, these same water will attract one side of the Moon differentially from the other, so Moon will get locked again with the same face towards us.


When will the next 'Super Blue Blood Moon' occur?

If you are looking for the perfect 'Super Blue Blood Moon', then, according to this article (assuming you live in the US):

  The next Super Blue Blood Moon won't happen until Jan. 31, 2037, and it will be visible in North America, Europe, Asia, Australia and parts of Africa.

However, if a Super Blood Moon is enough for you, you can see the next one on Jan. 21, 2019.


Can you create an orbit in a space station using balls?

As PM 2Ring mentioned, you seem to have a misunderstanding that spin is involved in creating an orbit. What matters for gravitational attraction is the mass of the bodies and their distance, the effects of spin are more subtle and only an issue when bodies are non-spherical. Also, one body doesn't need to be smaller than the other, they can be the same size.

A lead sphere 10 cm across would mass 6 kg. If two such spheres were 10 cm apart (20 cm center-to-center), they would have a gravitational pull on each other of 0.06 micronewtons...not nearly enough force to feel. They would technically have an orbital period of about 5.5 hours, with an orbital velocity of a few millimeters per minute, but pretty much any random air current would blow them out of orbit and no human could place them precisely enough in the first place, and nearby masses (like the humans occupying the station) would throw them off.


The Earth's heliocentric ecliptic longitude is opposite to the Sun's geocentric ecliptic longitude. If the Earth's perihelion is at 103 degrees heliocentric longitude, then the geocentric longitude of the Sun at that time is 283 degrees, or 13 degrees past the December solstice.


Is the Drake Equation an accurate way of finding the probability of life on planets?

The Drake equation is little more than a randomly assembled guess of relevant factors for intelligent life, and any use of it requires a number of hard to justify guesses for the various probabilities.

It is otherwise an attempt to give a foundation for making Fermi estimations on the number of civilizations, and the associated Fermi paradox. A Fermi estimation is an order of magnitude estimate. It doesn't try to get a particularly accurate estimate, it only tries to tell you how many digits you can expect the answer to have. And the Fermi paradox is simply stated as "There is an absurd number of stars. So absurd that finding extraterrestrial life should be easy. But in fact it is hard. Why?"

The resolution to the paradox is the second sentence: it assumes we have reasonable basis to suppose life is easy enough for the sheer number of stars to give us lots of observable life. We do not. The Drake equation is in one sense a way of singling out as many factors as possible so we have a concrete goal to figure those factors out, and so the entire problem.

But mostly it is used by pop science and sensationalists to rile us up with lines like "there should be thousands of civilizations more advanced than ours in our own galaxy!" and "there could be life in Jupiter's atmosphere." These things have no solid science behind them, and are just attention grabbers dressed in pseudoscience.


wavelength-based IFU data cubes VS velocity-based IFU data cubes

Your logic seems entirely correct. Whatever you do is an approximation, the difference between a Gaussian profile in cube B and what it "should" look like will be small for any sensible resolution. By sensible, I mean if your wavelength/velocity axis covers many resolution elements.

However, there is a "correct" way. Generate a Gaussian in wavelength space $f(\lambda)$ (if you think that's what your LSD is, where $\lambda$ here is short hand for $\lambda - \lambda_0$, the separation from the central wavelength) and then convert to velocity space in a flux-conserving manner:
$$f(\lambda)\ d\lambda = g(\log \lambda)\ d\log \lambda$$
$$ g(\log \lambda) = \lambda f(\lambda)$$

Maybe a better way to do this is to get some data, like an arc calibration and see how unresolved lines look in your observational space?


How do I calculate the inclination of an object with an amateur telescope?

Any telescope can be made to give you the information that you are looking for.  The first thing that you will need to know is the location of the ecliptic which varies throughout the year.  Or are you looking to find in relation to the the celestial equator?

https://en.wikipedia.org/wiki/Celestial_equator

Either way, you would start of the same by find the Declination (D) and Right Ascension (R.A) of the object in question.  You do not need a telescope to find this, unless you are unable to see the object with the naked eye.  If the telescope has a polar mount and is properly set up.  You can read the R.A. and D off of the mount.

Otherwise it will require some trigonometry and knowing where you are (at least your latitude)

You will need to determine the Altitude and Azimuth of the object.  Which is really just the direction from North and the angle from horizontal that your telescope is pointing.  You could do this with a compass and protractor or even an astrolabe.  Knowing this you can then convert to R.A and D with the following formulas:

$$RA = \arctan(\frac{- sin(Az) \times \cos(Alt)}{\cos(Lat) \times \sin(Alt) - \sin(Lat) \times \cos(Az) \times \cos(Alt)})$$

$$Dec = \arcsin(\sin(Lat) \times \sin(Alt) + \cos(Lat) \times \cos(Alt) \times \cos(Az))$$

You are looking for the inclination to the ecliptic, so you are mostly concerned with the declination.  The ecliptic changes in declination throughout the year from 0 at the equinox to +/- 23.5 at the solstice.  So your inclination from the ecliptic would be the Declination of your object +/- the Declination of the ecliptic.


The supermassive black hole (SMBH) in the center of the Milky Way (MW) — called Sgr A* [Sagittarius A-star] — has no direct impact on our galaxy. Its mass is only a few million Solar masses, and if you remove it$^\dagger$, it will only affect the most central stars, which would suddenly continue in straight paths out through the MW. These stars would almost surely not hit any other stars or something like that (since stars are really, really far apart), but some of them have velocities high enough that they may escape the MW.

If Sgr A* weren't there to begin with, things might look a little different. There seems to be a relation between the mass of a galaxy's SMBH and the velocity dispersion of the stars in its central bulge; the so-called M-sigma relation. so MW without Sgr A* would mean a more ordered center. Our Solar system is located in the disk, far from the center, and their is evidence that SMBHs have little impact on the disk (Gebhardt et al. 2001). However, in their early phase (as an active galactic nucleus), their extreme luminosities cause galactic superwinds which blow out gas and may quench star formation (Tombesi et al. 2015).

$^{^\dagger}$Removing Milky Way's SMBH is left as an exercise for the reader.


Naked-eye limiting magnitude: mismatch between Wikipedia, online converter and Shaefer's formula

Being by no means an expert of light pollution, I still find my intermediate results worth sharing.
On the one hand, there is the Bortle Scale, and I find the flow chart from darksky.org the most accesible way (and probably it is also a reliable source):

This flow chart refers to the original publication by Bortle  from 2001 which was mentioned in the question. For all objects mentioned in a flow chart, one could determine their magnitude and obtain a magnitude-Bortle-scale conversion table (I might do that in a later edit).
On the other hand, there is a conversion site directly by a manufacturer of SQMs which converts ${\rm visual\,mags}/{\rm arcsecond}^2$ to ${\rm cd/m^2}$ according to the following formula:
$$[{\rm value\,in\,cd/m^2}]=    10.8\cdot 10^4 \cdot 10^{-0.4 \cdot [{\rm value\,in\, mag/arcsec^2}])}$$
The impression I got from my research is that there is no single, golden source publication linking NELM with physical properties such as ${\rm cd/m^2}$.
Last but not least, I stumbled up Roy Bishop's visibility factor (see German wikipedia) as more advanced (aka quantitiative) approach.


Angular distance/apparent separation between AE Aurigae and Orion's belt

Given the cords of AE aurigae:

RA 05h 16m 18.1s DEC +34° 18' 49"

And Alnitak in Orion's belt:

RA 05h 40m 45.52666s DEC  −01° 56′ 34.2649″

I get an angular separation of 34.83°

By this formula:

$\cos(A) = \sin(\mathrm{DEC}_1)\cdot \sin(\mathrm{DEC}_2) + \cos(\mathrm{DEC}_1)\cdot \cos(\mathrm{DEC}_2)\cdot \cos(\mathrm{RA}_1 - \mathrm{RA}_2)$


Where does the CO2 in Mars atmosphere come from?

The early atmospheres of Venus, Earth and Mars all had a lot of CO2, Carbon and oxygen are some of the most common elements in molecular clouds, and so it is no surprise that their most common and stable compound should be common in atmospheres.

On the Earth, life happened.

The carbon in the atmosphere was fixed by autotrophs (particularly blue-green algae and the later plants) into the bodies of living things, and then into rocks formed from those bodies: Chalk, Limestone, Coal and Oil. This left an atmosphere dominated by Nitrogen, and later by the toxic waste products from those algae: oxygen.

Mars was too small to hold onto lighter gases like Nitrogen, and these escaped into space, there is just a trace of the heavier gases: CO2, but at a very low pressure.

Venus retains a dense atmosphere dominated by CO2


Naive star filter visible at night in certain country

In the answer https://physics.stackexchange.com/a/26787 you can find an example how to convert celestial coordinates to horizon coordinates an than check, if the altitude is above the horizon


Spotting the International Space Station (ISS)

The official site you point to has 3 pieces of location information:
1) the max height angle
2) the point it appears
3) the point it disappears
The only one of these that is especially accurate is the max height angle, and that isn't the number you want to use to try and see it. However, passes with a bigger number for the max height angle will be easier to see.

To find the ISS, get the reference to the point it "Appears" - that is the "11 above N" you have described. Now, notice that the reference to the compass direction is simply N or NNE or SW or whatever. This is not a precise directional reference, but it doesn't need to be, as if you're looking towards that direction, you shouldn't have any problem seeing ISS appear. It isn't clear to me if the website is using True North or Magnetic North for the direction; Magnetic North might be easier for people with a simple compass to find, but usually these directions will be relative to True North, which is not where a compass points. From a quick check using my own location, it appears that they have used True North, which is maybe a bit unfortunate. If you don't know where True North is for your location, you can look it up, and then you have to remember to adjust the reading off your compass by the offset for your location.

Notice that the description is "11 above N". This means 11 degrees above the N direction. The reason for it appearing some distance above the horizon is because it is coming up out of the shadow of the Earth, which is being cast some distance above the horizon.

To quickly and easily estimate angular distances, use the trick of observing your hand at arms length. Different parts give different angular dimensions. It is a pretty fair estimate, because no matter how big or small you are, the relative size of your hands and the length of your arms is probably about the same as everyone else... not good enough to use as a survey instrument, but good enough to guesstimate locations on the sky.

The second position description is the "Disappears" location. That is where the ISS will again drop into the shadow of the Earth. Similarly, it is described as degrees above a True North compass direction.

You can figure out (approx) the path the ISS will take by drawing an imaginary line between these two points. However, don't draw that imaginary line so it goes over your head, note the "Max Height" angle again: this is the highest point on the line that you draw between the "Appears" and "Disappears" spots.

Just give it a go, especially one with a pass that starts around 30mins or so after sunset, and that has a high "Max Height" angle. It won't be hard to see.


Maximum and minimum gas giant & ice giant densities

Here is a plot I generated in 5 minutes at the site exoplanets.org

To construct this I took planets discovered by the transit method and which had a $M \sin i$ measured using radial velocities. I divided the $M \sin i$ by the sine of the measured inclination angle (this is required to avoid using masses that have been estimated using an assumed mass-radius relation). The y-axis is density, which comes directly (and quite precisely) from transit measurements. Of course transiting planets are the only ones with radii and density measurements.

You have to do this to avoid some very uncertain values that are given for planetary masses that have been simply assumed from a theoretical mass-radius relationship.

As you can see, there is a wide spread (factor of three) in density at a given mass for hot Jupiter's (most of the transiting giant planets are hot Jupiters), but there is a strong correlation. The density is at a minimum for a few tenths of Jupiter mass, but then smaller planets (presumably rocky and icy, rather than gas giants) appear to show higher densities.

Remember, these are all transiting exoplanets, and therefore predominantly orbiting close to their parent stars. There could be biases and selection effects at work! For example although the cores of gas giants are governed by degeneracy pressure and this makes the theoretical mass-radius relationship quite flat, there is the perturbing effect of radiation from the parent star ("insolation") that can make some objects larger. Even beyond this there appears to be scatter that is difficult to understand.

EDIT: For an easy-to-use empirical formulation you could try the relationships proposed by the Planetary Habitability Laboratory.


What is the difference between $E_\gamma$ and $E_{\gamma,iso}$ in gamma ray bursts?

I will answer with what I now believe to be the correct answer. We see some gamma-ray burst emission hit our satellite which carries an amount of energy $E_{obs}$. If we say the burst was emitted equally in all directions (isotropic) then we derive a quantity $E_{iso}$. If we say that the burst was actually emitted in a tight beam then we derive $E_\theta$ where $\theta$ is the opening angle of the burst and we will find $E_{iso} > E_{\theta}$. $E_{iso}$ is used to compare burst energies in the literature without needing to find the opening angle.


How to isolate H$_\alpha$ from Nitrogen doublet emission lines

I would've posted this as a comment, but I lack sufficient reputation.
I like your idea of fitting three Gaussians, but I think you should try fitting with one large positive Gaussian and two small negative Gaussians.
My physical justification for this is that I would expect a lot of self-absorption from the H-alpha line, which should be well-modeled using the larger negative Gaussian. I'm not sure if the other notch is absorption from another line, but it also looks like a negative Gaussian would fit well there.


What does this "web on the surface of the Sun" image reflect? What does 789 nm show us?

Well let me take a stab at it. The line in question is said to be a probe of an Fe XI line, that is iron atoms with 10 electrons removed.

You do not get such ions in the solar photosphere, it is far too cool; the radiation from the photosphere is probably a pseudo-continuum at that wavelength.

However, much hotter material in the chromosphere and corona may contain Fe XI ions. Plasma in these structures could absorb light from the underlying photosphere, if it were dense enough, or more likely, if you point your camera above the limb of the photosphere, you could see optically thin coronal structures emitting light at this wavelength.

Why is this important? Normally you would need to use EUV or X-ray emission to probe the coronal plasma, but the imaging quality is not so good. You can do much better at optical wavelengths, but there are precious few diagnostic lines that can be used.

Edit: Indeed this presentation on DKIST coronal diagnostics appears to confirm this hunch and also mentions the additional possibilities that polarimetry of optical/IR spectral lines offer in investigating coronal magnetic structures. The Fe XI line in question is sensitive to Zeeman splitting, offering the opportunity to prove the strength and direction of magnetic fields where the line is formed. The Zeeman effect is proportional to the square of the wavelength, so the more usual X-ray and EUV diagnostics just can't do that.

On p.2 of that presentation it clearly states that these kinds of measurements are limited to coronagraphic images taken off the limb of the Sun. Pointing at the photosphere will not yield useful information for these diagnostics, because the weak light from the chromosphere and corona is swamped by the normal photospheric emission.

Sextus Empiricus has pointed to a press release, which although unclear in its exact meaning, implies that the dark features around the granulation may offer sufficient contrast to see the much thinner and hotter chromospheric gas directly above, producing bright points in the dark lanes. Observing these through different polarising filters could then reveal details about the structure and strength of the magnetic field.

However, on further research, these photospheric bright points are nothing of the sort. They are concentrated magnetic flux tubes that allow a deeper (and therefore hotter and brighter) view of the Sun. The typical temperatures of the deeper material is still  only around $10^4$ K (e.g. Shelyag et al. 2010) and nowhere near enough to excite Fe XI.

The picture below, which comes from this site, referred to by Sextus Empiricus, shows the situation. A bundle of magnetic flux "hollows out" a passage further into the solar interior and the light we see comes from deeper, hotter, brighter regions. Nothing to do with coronal emission.

I arrive at the conclusion that this image (which was taken for science verification purposes) was just using the Fe XI filter as a narrowband filter. Everything we are seeing in the image is essentially continuum from the photosphere at temperatures between 4000K and $\sim 10^4$K. The contrast therefore just arises from the differing monochromatic intensity of material at different temperatures.


Has the Hubble ever been pointed at earth?

xkcd already did the math on this one!
Like Gauti's answer, xkcd also links to this question on Hubble's website. Hubble moves too fast for its minimum exposure time to be able to focus on anything on the surface. He goes one step further, linking to this article on Bad Astronomy that points out that Hubble frequently points at the sunlight side of Earth to help calibrate its WFPC2 camera.

Still, you may be surprised to find out that Hubble routinely points at the Earth! It uses the bright, daylit Earth to help calibrate one of the cameras on board. The Wide Field Planetary Camera 2 (WFPC2, pronounced ``whiff pick 2'') is an electronic device which detects light, and is similar in principle to a normal digital camera. However, it is far more sensitive, and astronomers are very particular about how well they understand such devices. They want to make sure that every part of the detector is very well calibrated.
One way to calibrate the camera is to look at some bright, evenly lit object. If one part of the detector is more sensitive than another, for example, then part of that object will look brighter, even if in reality it isn't. This can be used to correct any later images.
In space there aren't any flat fields you can use, but we do have one right here: the Earth. To calibrate WFPC2, sometimes Hubble is pointed straight down at the Earth. As the terrain (or water or whatever) streaks across the field of view it forms a very crude flat field image. The image is called a ``streak flat'', and looks really weird. Trees, houses, all sorts of objects blur across the image. It takes a lot of sophisticated computer processing to turn this into a real flat field for Hubble to use, but luckily there are a lot of smart people working on the problem.

xkcd also shows an approximation of what it would look like if you tried to take a picture of an average desk from orbit. It is just a smear of colors, devoid of any discernible detail. I tried attaching the pictures here, but SE doesn't like the png files for some reason and I don't have time to do the file conversion. Just check out the first link to see what it looks like, plus a more detailed explanation of why.


Is the colour of a wave from a far galaxy the same for us as for a galaxy which lies between?

If I understand you right, you're asking whether or not the redshift of the photons emitted from a far-away galaxy happens the instant it leaves the galaxy.

Redshift is gradual…

If so, the answer is no. The redshifting of photons happen gradually as they travel through the expanding Universe. You can find the derivation here where you'll see that every infinitesimally small increase $da$ of the scale factor $a$ of the Universe (its "size") increases the photon's redshift by an amount $dz$, or, in terms of wavelength, by an amount $d\lambda$.

If galaxy $B$ lies at redshift $z_\mathrm{B}$, then an observer in galaxy $A$ at redshift $z_\mathrm{A}$ lying between us and $B$ (so that $z_\mathrm{A}<z_\mathrm{B}$) would measure $B$'s redshift to be
$$
z_\mathrm{B\,seen\,from\,A} = \frac{1+z_\mathrm{B}}{1+z_\mathrm{A}} - 1,
$$
which is less than $z_\mathrm{B}$.

…at least in our Universe

The redshifting is not due to the source moving away from us. If the expansion hadn't been gradual, but we instead lived in a crazy universe that were static when the distant galaxy emitted the light, and static when we observe, but somehow expanded suddenly by some factor in the meantime, then we would still observe a redshift, even though the galaxy were static both when it emitted the light and when we observed it.


Can "rogue" supermassive black holes be made this way?

I assume you're asking about central supermassive black holes (SMBHs, one per galaxy), not stellar-mass black holes.

The answer is yes, but what actually happens is the two SMBHs have to merge first, and then the resulting combined SMBH can sometimes be ejected from the combined (merged) galaxy.

[Edited to add: Since you've updated the question with a series of diagrams, I should state explicitly that the scenario suggested by the diagrams -- stars in smaller galaxy merge into big galaxy, but SMBH continues on almost unaffected -- is not physically possible. Most of the stars from the smaller galaxy will not end up in the center of the big galaxy, but because of dynamical friction, the SMBH will.]

This NASA press release from 2017 describes the discovery of a quasar apparently ejected from a recently merged galaxy. I'll go ahead and quote their description of the suggested mechanism (this possibility has been suggested by theoretical studies going back at least ten or fifteen years):

  According to their theory, two galaxies merge, and their black holes
  settle into the center of the newly formed elliptical galaxy. As the
  black holes whirl around each other, gravity waves are flung out like
  water from a lawn sprinkler. The hefty objects move closer to each
  other over time as they radiate away gravitational energy. If the two
  black holes do not have the same mass and rotation rate, they emit
  gravitational waves more strongly along one direction. When the two
  black holes collide, they stop producing gravitational waves. The
  newly merged black hole then recoils in the opposite direction of the
  strongest gravitational waves and shoots off like a rocket.

Since most massive galaxies -- including those that have undergone major mergers in the past -- have a SMBH in their center, the gravitational recoil usually isn't strong enough to eject the SMBH; instead, the SMBH loses energy to the stars in the inner part of the merged galaxy via dynamical friction, and settles back into the center. But it appears that sometimes there's enough of a kick to allow the SMBH to escape.

Another possibility is that if two galaxies merge and their SMBHs form a binary, and then another galaxy (with its own SMBH) merges before the previous two SMBHs have actually merged, then you can have a three-body interaction between the late-arrival SMBH and the binary SMBH, which could result in one of the SMBHs being ejected. But this requires the right timing, and probably doesn't happen very often.


Interfacing FORTRAN programs for model atmospheres with IDL or Python

After looking into the various wrappers and ways Python and IDL work with FORTRAN I came to the conclusion that it is more of a preference which language you use. However as people continuously add astronomy support to Python this may change, but as of now, it seems to be a question of preference.


How to read 5 degree data of green line intensity

The format of the files seems to be described in this file, the content of which I've included below.
The annual files (1939-1993) contain the green coronal emission line
530.3 nm observations.  The coronal intensities are given in millionths
of intensity of the solar disk (coronal units) and converted to the
photometrical scale of Lomnicky Stit Station at a height of 40" above
the solar limb.

Several stations were used in this database with Lomnicky Peak being the
primary station since 1965.

Reference:  Rybansky, et al, 1994, Solar Physics 152, p 153-159.

Data format:

 COLUMN  FORMAT   DESCRITPION
  1-  2    I2     Year, last two digits
  3-  4    I2     Month
  5-  6    I2     Day
  7-  8    2X     Blank
      9    I1     Coronal station code:
                  0 - Interpolated data
                  1 - Lomnicky Stit
                  2 - Sacramento Peak
                  3 - Norikura
                  4 - Kislovodsk
                  5 - Pic Du Midi
                  6 - Wenselstein
                  7 - Arosa
                  8 - Kanzelhohe
 10- 13    I4     Time of the observation in hours (or zero)
 14- 17    I4     Time of the observation in minutes (or zero)
 18-305  72(I4)   Coronal intensities (72 values) from 0 degrees to 355
                  degrees of position angle in increments of 5 degrees.

Using the first row data as an example...
DATE        SC  H   M   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69  70  71  72
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
19390101    0   0   0   0   0   0   0   0   0   0   0   3   3   3   8   15  22  34  41  52  46  40  52  68  59  49  35  17  8   3   3   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   3   3   6   8   12  19  31  40  50  47  43  38  53  73  71  65  56  47  38  24  17  12  6   0   0   0   0   0   0

19390101 is the date (1939-01-01).
The first zero (Marked as SC) is the station code, the second the recorded hour, and the third the recorded minute (H/M). Most of this is 0 due to being interpolated data.
After this information, there are 72 remaining columns (numbered 1-72 above), which represents collected data consolidated in groups of 5 degrees, presumably starting at 0.
Given the amount of interpolation of old data the value may be questionable, but newer data appear to be more complete.
The data format might be a little overwhelming to work with, but you can organize it in excel quite easily - paste it in and each line break will form a row, then use the text-to-column function to break each data into its own column.


Why don't we have a browsable spacetime augmented virtual solar system?

There is NASA/JPL's Solar System Simulator. It doesn't cover all points, but does include many of the interesting ones.


The expansion of space is something that happens on the largest scales.  At small scales, such as distances between nearby galaxies, other forces, such as gravity, dominate.  Galaxy clusters are held together by the attractive force of gravity between these galaxies.  Space in these regions is still expanding, but the gravity pulls on these galaxies much more than space's expansion moves them apart.

Galaxies that are close enough together, such as the Milky Way and Andromeda, are pulled together by the force of gravity between them.


Calculate the shadow on earth of a large orbital disk

It might be easier to use similar triangles.

$$\frac{Rd}{b} = \frac{Rs}{a}$$

$Rs$ is 696,000 km and $Rd$ is 5 km. $a$ is about 150,000,000 km and c is 80 km.

The the smallest triangle is used next:

$$\frac{r}{b-c} = \frac{Rs}{a}$$

Solve for r and you get

$$r = Rd - Rs \frac{c}{a} = \text{4.63 km}.$$

Extra digits are not helpful because the exact diameter of the Sun depends on how you define it and the distance from the Sun to the Earth varies by almost +/-2 percent.

$Rs/a$ is about 0.00464 and that's also the half-angle of the Sun in radians. Convert it to degrees my multiplying by 180/pi and you get 0.266 degrees, or a quarter of a degree. The full diameter of the Sun is double that, or about a half degree.