How is the size of a distant planet determined?

One method of finding planets orbiting stars is to analyze the transit effect. This is when a planet crosses the face of a star between the star and observer.  This will block some of the light reaching the observer. By measuring this decrease in light from the star we can determine how much of the surface area of star is occluded.

The speed of a planet across the face of a star will let us calculate the orbital radius considering a circular orbit. The temperature of the star can be determined from its brightness and location on HR diagram and spectroscopic data. Combining this information lets us determine the size and of the planet.

Using space-based or ground-based telescopes and automated data collection and analysis methods large regions of the sky can be viewed for these transits. This method is highly accurate and will probably surpass the radial velocity method as most common method to discover new planets.

Another method for finding planets is known as the radial velocity method. When a planet orbits a star it is really the planet and star rotating about the center-of-mass of the system. Hence the star wobbles due to the bodies orbiting the star. It wobbles because usually the center-of-mass is inside the star and this results in the star rotating about an axis running thru the star but offset from the center. By doing a doppler analysis it can be determined whether the star is revolving or not and the period of revolution. The radial velocity method is the most effective method. The transit method is very limited by the viewing angle, the planet must pass between observer and star.

A good source to better learn about these methods is here.


How can I calculate how the debris of an object ripped apart at the Roche limit will spread out?

I think there are two key aspects to the answer.  1) Solid/rocky bodies should tend to collide before they reach the Roche limit.  2) When gaseous bodies reach the Roche limit (and undergo 'Roche-Lobe Overflow'), the dynamics are basically those of test-bodies and are fairly straightforward and well understood from binary stellar dynamics.  To expand on both:

1) Rocky Roche Limit.  If you are thinking about a gaseous donor (the object being disrupted), then this is an irrelevant point, but It sounds like rocky is what you had in mind.  To an order of magnitude, the Roche limit is the same as the Hill sphere, or the Tidal Radius (e.g. Rees 1988) --- which is simply the radius at which the density of the donor equals the average density of the primary in a sphere of that radius:

$$R_t^3/M \approx r^3/m \rightarrow R_t \approx r \left( M/m \right)^{1/3}$$

Rocky material varies very little in density (e.g. Iron meteorites are only about twice as dense as chrondritic ones), which means that the densities will only match very near the radius of the primary.  The tidal bulge in the secondary (donor) also makes it easier for the objects to collide before meeting this criteria.

e.g.

2) Gaseous Roche Lobe Overflow.  Basically every text on stars and star systems will have a section on mass transferring binaries which will describe the dynamics of mass transfer (this PDF is by Philipp Podsiadlowski who is a wizard of the field).  In the type of situation you are describing, like in (semi-)stable binary systems, it is a very gradual process where material is slowly syphoned off of the donor.  This material can either form an accretion disk (high angular momentum material) and gradually accrete onto the primary, or directly impact the primary (low angular momentum material).

e.g.


Why do we only have one major meteor shower from 3200 Phaethon?

Note the short straight lines that hang down (or stick up) from the orbital track of 3200 Phaethon you reference.
These indicate the distance above or below the plane of earth's orbit (the ecliptic) 3200 Phaethon is at that point.
So 3200 Phaethon passes well above Mars' orbital track and passes through the Earth's orbital track (just about)
Then it passes under Venus, under Mercury, makes a U-turn around the Sun, and passes back over Mercury, Venus Earth, and Mars.
So your supposition in the last paragraph is correct.


Found this link by Emily Lakdawalla, she wrote a detailed explanation about Pluto`s seasons and its movement thru the solar system. Here research seems to contradict the atmosphere collapse and and here 3 models and research on Pluto. Seems the most likely model gives the highest pressure in the atmosphere between 2020 ad 2040, but none a atmosphere collapse. source


The contribution of components like dark matter, stars and gas to the mass of a galaxy

This is a rather complex question, for several reasons.

First, galaxies come in many variations, regarding such diverse
properties as mass, morphology, and environment.
Second, different observational techniques, and different models,
yield different observables — you can observe the same field of the
sky with two different instruments and deduce a different
distribution of galaxies and their properties.
Third, as you mention, galaxies evolve, and you will not necessarily
obtain the same ratio between, say, gas mass and stellar mass, even for a
given galaxy type, at different redshifts.

Nevertheless, some things can be said to be true "in general":
Stellar mass – gas mass relation
The more massive a galaxy is (in terms of stellar mass, $M_*$), the more efficient it is at forming stars. Hence, the gas fraction $f_\mathrm{gas} \equiv M_\mathrm{gas}/(M_\mathrm{gas}+M_*)$ decreases with $M_*$. Moreover, although part of the gas in stars is returned to the interstellar medium (ISM), as time goes, and a galaxy forms stars, it will "deplete" the ISM, further reducing $f_\mathrm{gas}$.
This can be seen in this plot from Magdis et al. (2012), showing the gas fraction as a function of stellar mass today (open circles) and ~10 billion years ago (closed circles):

The galaxies used in this survey are "main sequence" galaxies, and other selection criteria apply as well.
Stellar mass – halo mass relation
The dark matter (DM) component of a galaxy is much more extended and diffuse than the baryons (because DM is collisionless), rather lying in a large "halo" around the gas and stars. Of course, we cannot see DM, making measurements of its mass difficult. Only in numerical simulations, we know exactly its mass.
The larger the DM halo mass ($M_\mathrm{h}$), the more stars the galaxy has. But the relation is not straightforward. In general, $M_*$ increases with $M_\mathrm{h}$ more rapidly for low-mass galaxies, while for $M_\mathrm{h} \gtrsim 10^{12}\,M_\odot$ (roughly Milky Way-sized galaxies) the relation flattens out:
This is seen in the left panel of this plot from Behroozi et al. (2013):

The different colors correspond to different epochs in the Universe. The data is from a cosmological simulation, but the simulation was calibrated to match various observations.
Another way to show this relation is seen in the right panel, where the stellar fraction $M_*/M_\mathrm{h}$ is seen to rise until around $M_\mathrm{h} \sim 10^{12}\,M_\odot$, after which it decreases again.
Why is this? In general it is thought that star formation is suppressed at low masses because gas is more easily expelled from a shallow gravitational potential, while a high masses, active galactic nuclei become very efficient at blowing out gas, thus quenching star formation.
How are masses measured?
There are several techniques to measuring these masses.
Stellar masses are measured using known relations between the amount of stars and the amount of light from some physical process — either a single emission line, or a broader band of light. For highly star-forming galaxies, where there are still many hot O and B stars that ionize the surrounding gas, nebular lines such as H$\alpha$ or Ly$\alpha$ can be used, while for non-star-forming galaxies you can use e.g. the continuum radiation from heated dust.
The conversion depends on the assumed initial mass function of the stellar population.
Likewise, gas masses and molecular masses may be measured knowing how much light a given amount of gas emits (at a given temperature, pressure, …).
Measurements of halo masses are typically done looking at the width of various spectral lines, thus deducing the velocity dispersion $\sigma_V$ of the gas and stars. Then, the total mass $M$ can be calculated from
$$
\sigma^2 = \frac{GM}{CR},
$$
where $G$ is the gravitational constant, $R$ is the radius, and $C$ is a geometrical factor (see this answer for an explanation).


What is the longest natural bound orbit chain observed?

Certainly 4 (perhaps 5):
The longest certainly-known chain is "3":  Sun-Earth-Moon.
Rhea (a moon of Saturn) has been observed to have a ring of orbiting material: That gives Sun-Saturn-Rhea-Ring for four levels (including unresolved dust)
Looking further afield:
DH Tauri and DI Tauri are a binary pair of similarly sized T-Tauri stars (both small red dwarfs); they are in a mutual orbit.
DH Tauri has a large sub-stellar companion.
The companion has an accretion disc of orbiting matter.
So we have DI-Tauri > DH-Tauri > DH-Tauri-b > matter in disc, for four levels of orbit.
Even this isn't very good, because DH Tauri is bigger than DI Tauri, so it is a bit of a stretch to say that DH Tauri orbits DI.  If anything it is the other way round.  There are very few planets found in orbit around the secondary star in a system and none of those have likely moons. Moreover the disc of matter around the sub-stellar object is not known to have actual "moons" embedded in it.
So "4" seems to be the best we can do...
But if we can add in Galaxy - Star - planet -moon - dust/ring? we get 5 levels.


How complicated is the heating of meteors?

Your analysis is correct. Compression of the air is the cause of heating.

You call this "simple" but the compression depends on the ability of the air to flow around the meteoroid, and as the modelling the hypersonic flow is far from trivial.

"Air fiction" is the resistance to motion felt by an object moving relative to the air. That includes the reaction to the force that is compressing the air in front of the meteoroid, but also results from work done in moving the air out the way, and is dependent on shape, surface effects, turbulence etc.

The visible meteor is the trail of plasma produced in the air by the meteoroid. The actual piece of rock is too small and dim to be seen. The meteor is glowing air.


Calculating azimuth from equatorial coordinates

lat2 = 0 and lmst = 0 place the vernal equinox at the zenith, so most points on the ecliptic should have azimuth either obliq degrees north of east or obliq degrees south of west. Try other values for lat2 or lmst and see if the resulting azimuths make sense.


"If earth had rings like Saturn": A few further thoughts on the consequences

This is a big one, so I'll split it into parts based on the questions you asked:

  1) Earth has roughly a distance of 1 AU to the sun, Saturn between 9 and 10. Sun's gravity would therefore have a much stronger influence on these rings. Would it already be enough to distort the rings (e.g. even leading to the sun "stealing" parts of it) or would the rings still have a ring shape?

Sun would not have too much of an influence because the rings are all individually gravitationally bound to the earth (similar to the moon, and you don't see moon being affected by the sun, gravitationally; the rings will be much closer to the earth and hence even less affected).

The moon, on the other hand, would distort a few rings due to orbital resonances (similar to Titan Ringlets: https://en.wikipedia.org/wiki/Rings_of_Saturn#Colombo_Gap_and_Titan_Ringlet and resonant gaps in asteroid belts due to Jovian planets: https://en.wikipedia.org/wiki/Kirkwood_gap), and its effect would be much more pronounced because it is closer to the earth than Titan is to Saturn, though limited to a few resonant regions. There will also be ripples if the plane of the moon is different from the plane of the rings (http://blogs.discovermagazine.com/badastronomy/2009/06/11/saturns-rings-do-the-wave/#.U7VjpvmSw08).

  2) How well would the objects in such rings be visible from earth? The
  Saturn rings are mostly dust and ice as far as I know, but there are
  surely some larger objects. Depending on material and size some object
  would probably be visible to the naked eye, right? Is there (probably
  yes) a formula to calculate what materials and sizes are required to
  achieve this?

Most matter in Saturn's ring is particles up to a few metres in size (http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q6). Resolution of the human eye is about 1' (Is the moon only 60 pixels?), which means at a distance of around 1 to 1.5 Earth radii (given by Roche limit of the earth; this is where the rings will be), the eye can resolve things above a couple of kilometres in size. However, Saturn also has Shepherd moons, which are a few tens of kilometers in size. Such moons, if present, would be visible to naked eye. Also, the gaps would be visible clearly, since they are often big enough.

  3) Would those rings not be constantly raining particles down on
  earth? Would there be a much higher risk of dangerous impacts for the
  earth's population?

There would be collisions/interactions in the rings (I'm guessing, correct me if I'm wrong) which would send particles earthwards. However, the particles which will get thrown will be the lighter, smaller ones (conservation of momentum etc.), and usually small meteoroids are fairly harmless. They will also be slower than the usual meteoroids that enter the earth at above 20 km/s, since the orbital motion of the earth doesn't add up in this case. I am a little unsure about the consequences of them being slower, but anyway, smaller meteoroids should technically be fairly harmless. (Also I couldn't find decent references other than: http://science.howstuffworks.com/question486.htm)

  4) Are the colours used in the video precise? Don't they depend on the
  angle? What influence would our atmosphere have onto them?

I believe so. This section shows true colour images of the rings from two different angles, and there doesn't seem to be a significant change in the colouration: https://en.wikipedia.org/wiki/Rings_of_Saturn#Subdivisions_and_structures_within_the_rings. The atmosphere will probably affect the rings closer to the horizon, with redder colouration similar to sunrise/sunset and moonrise/moonset.

  5) Would those rings have the same orbital inclination or would theirs
  differ from earth's to an extent where the difference becomes visible?

Depends largely on the formation history. This FAQ question here discusses this fairly well: http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14

  6) What shadow would these rings cause on earth? Would they be dense
  enough to "shade out" complete stretchs of land?

There will be shadow similar to that due to clouds, since the rings are not too thick or dense. In many of the Cassini photos, you can see through the rings, so it is likely that the sunlight would still pass through, but would be dimmed (it would be interesting to see by what factor; I cannot say for sure with my knowledge). Also, remember that the earth will still be rotating but the ring system will be dynamically static, so there will be extended periods of darkness during the day for some regions at different times of the year.

I will add more to this if something comes to mind. I hope this was helpful.


The Wikipedia article on voids is pretty good (though IMO unusually awkwardly written.) The key thing is that voids are not empty, they are just large volumes which have a lower density (typically around 10% of average) compared with the rest of the universe.

These low density areas still contain stars and galaxies, just fewer of them and the galaxies they contain tend to be smaller. The Lambda-Cold-Dark-Matter (LCDM) model, which is currently favored as our best description of the universe, predicts that the voids should not have many large galaxies in them, but should have a fair scattering of small, diffuse galaxies (SDGs).  (A small diffuse galaxy is, well, small compared with the Milky Way --
sized more like the Magellanic Clouds or one of the small galaxies which are being devoured by the Milky Way today such as the Sagittarius dwarf. They are also more spread out than the Magellanic Clouds.)

Here's an interesting senior thesis reporting on a search for small diffuse galaxies in voids. (They think they found some, but need confirmation.)

You ask what the night sky would look like from within a void. First, we'll assume you're in one of the SDGs, though this wouldn't make a huge difference.  What you'd see mostly is whatever planets share your solar system -- they'd be comparable in brightness to our own planets from Earth. You'd also see a few dim stars.  SDGs tend not to have much star formation, so there would be very few young stars and thus very few intrinsically bright stars. The visible stars would all be ones which happened to be nearby. And because SDGs are diffuse, there'd be many fewer nearby stars. At a guess there would be 20-200 naked eye stars, with only a handful of magnitude 3 or brighter. (Compare that with 6000 naked eye stars visible from Earth.)

There would be no naked eye galaxies. The SDG itself would not appear as a milky way, but more like the Zodiacal Light -- a vague glow dimly visible under exceptional conditions.

This would be a world whose astronomy would be heavily focused on the planets.

(Planetary formation might be quite different in SDGs, but I don't know even to try to speculate on how it would be different. So I assume that the planetary system would be similar to ours.  YMMV.)


No. Problems:

Hawking radiation would have very small temperature of $10^{-6} M_{sun}/M_{BH}$ Kelvins. Primordial black holes below 1kg already did evaporate, so there would be no sources from small black holes. The only hope is artificial black holes created by CERN.
Falling matter inside black hole would expirience enormous amount of acceleration leading to emittance of photons (particularly X-rays). So you would first are required to cut out this noise.
Black holes are surrounded by ergosphere consisting of hot plasma, which effectively blocks all photons below some critical temperature. For example, we have our radiowaves blocked in the same way by ionosphere of Earth.
(Reasons above were not enough?) There is other quantum-mechanical mechanisms of creating particle-antiparticle pairs which potentially lead to heavy emittance of antiparticles, mostly positrons from black holes. So you would somehow need to shield from positrons, and the photons coming after $e+e^{+}\rightarrow \gamma + \gamma$ kind of reactions.

You would be required somehow prepare black hole for investigation by removal surrounding physical layers from ergosphere, stopping its rotation (really impossible?). I don't know of any "surgical" instruments available today for such task ("surgical" in a sense that we will need to modify our black hole).

Our measuring instruments are totally capable, though. We also know where black holes are located. Tiny problem of communication still persists. But I think scientists would be perfectly fine to wait few millions of years for station to return. Also there is problem from the field of mechanics. It is enormously hard to cross ways with gravitationally strong object, because your trajectory (hyperbolic type) would intersect with a lot of other trajectories (of elliptic type). But I did not list it, counting on humankind getting smarter with this kind of things in a few centuries.


Can a star emit a larger fraction of its total light in the visual range than the Sun does

Surely this is exactly what the V-band bolometric correction assesses. The numerically larger the bolometric correction, the more of a star's flux (as a fraction) is in the V-band.

The Sun's bolometric correction is about -0.06 to -0.11 mag, depending on which sources you look at. Cool stars and hot stars have numerically smaller BCs and emit a bigger fraction of their flux outside the V band.

Some sources put the BCs of A5-G0 stars as very slightly larger than the Sun's, by up to 0.1 mag (e.g. Pecaut & Mamajek 2013).


Do great spot like features favour appear in south hemisphere?

Given that we have very few examples, we could not, statistically, draw any significance from any apparent bias.  It's also hard to see any theoretical basis for expecting a bias, so my gut feeling would be there probably is no bias.

It's also worth remembering that human observations of these things are essentially a very brief snapshot in a huge period during which we have no idea about what may have been happening.  So again the statistical significance of what happens at the moment needs to be carefully considered.


Stars of spectral class Y are dim brown dwarfs, with temperatures of about 300-400 K (For comparison, Earth's average temperature is around 288 K, give or take, the surface of the Sun is 5,800 K, and the core of the Sun is 15 million K). Their temperature range means that they don't fuse hydrogen in the same way that "true" stars do. They may fuse deuterium ("heavy hydrogen"); Y dwarfs may not fuse lithium, as other brown dwarfs do, because they are relatively low-mass.

The fact that Y dwarfs (and brown dwarfs, as a whole) don't undergo normal stellar nuclear fusion means that they'll never undergo core-collapse supernovae, which typically happen when an extremely massive star gets to the point of iron fusion and then can no longer generate energy from fusion. Brown dwarfs work completely different; this is not at all a possibility.

Instead, they will slowly radiate away their heat until they become extremely dim, via the Kelvin-Helmholtz mechanism. For more information, see the basic example evolutionary tracks on this page; if you're really feeling ambitious, you can try going through these notes to get a better feel for brown dwarf evolution.


Would you expect this planet to be hotter or colder than earth?

Yes, it is absolutely silly to predict an atmospheric/surface temperature without knowing it's composition or density. And we don't understand the formation process of atmospheres well enough to exclude any type of atmospheres around different stars.

There is however one more or less objective quantity that you can calculate: The equilibrium temperature. That is the temperature that the planet would have without an atmosphere.

This is essentially given by the planetary distance (which you have) and the stellar irradiation (which you have still to find), and the planetary albedo (a free parameter between 0 and 1).


How many horizontal branch stars are within 2 kpc of the sun?

You could try selecting them in an absolute magnitude versus colour diagram.

e.g. Use the apparent G magnitude and parallax to find absolute G magnitude and plot this versus a colour ($B_p-R_p$).

The HB stars occupy a reasonably distinct region (box?) in this diagram.

Your problem is that this distinct box is blurred by differences in composition and by extinction/reddening. You could sidestep the latter to some extent by avoiding low Galactic latitudes.


Anecdote of an astronomer looking for correlations in antipodal pairings of stars; what am I remembering?

This is a partial answer because while I'm quite sure that

Fang Lizhi is the astronomer I'm thinking of
The paper Periodicity of Redshift Distribution in a T-3 Universe (FERMILAB-Pub-90/26-A
January 1990) is a description of the research I'm remembering

I can not understand the paper well enough to connect it to how I remember the author describing his computational experience working on the small Apple computer while trapped in an embassy, and what this paper has (if anything) to do with comparing antipodal points.

Hopefully this is enough information for someone to go further and post a proper answer.

There is also the book The Most Wanted Man in China: My Journey from Scientist to Enemy of the State by the same person. On Page 294:

  What I needed was a computer. Again I was fortunate, because one of the diplomats, who originally had been a student of mathematics, was being transferred back home and was ready to part with his first generation Apple computer.


Ramifications of black hole stellar system

For those who haven't seen it:

Some human explorers land on a planet orbiting a black hole. The black hole is surrounded by a large accretion disk. The planet orbits at a distance such that going any closer to the black hole will mean that your odds of getting out are slim; it's also composed of water. Finally, time dilation from the black hole means that even though the characters spend about two hours on the planet, a decade or so passes for their colleague on board.

The basic answer is that a planet can orbit a black hole. There are stable circum-black-hole orbits, just as there are stable orbits around just about any celestial body. There's a problem: A black hole typically forms as a result of a supernova. This will eject most nearby planets out of the stellar system. Alternatively, it's unlikely that a planet could be captured by a black hole and be in a stable orbit, so the whole premise - while possible - is highly unlikely. Then again, it's improbable that a planet will be made out of water, a wormhole will open up near Jupiter, or Matthew McConaughey will star in a decent sci-fi movie, so why should anything else in the story be normal?

However, you can't just put a planet anywhere near a black hole, give it a strong enough push, and hope it orbits. The innermost orbit is at the boundary of the photon sphere. On this sphere, only photons can orbit. Inside it, nothing can orbit. However, the only stable orbit is twice as far away, at $2r_p$.
The radius of the sphere is
$$r_p=\frac{3GM}{c^2}$$
We'll assume that the object is not rotating (I don't remember exactly if it is or isn't rotating, but it's simpler in this demo to say it isn't.). The formula for gravitational time dilation is
$$t_0=t_f \sqrt{1-\frac{3r_0}{2r_f}}$$
where
$$r_0=\frac{2GM}{c^2}$$
Assuming that $t_0$ (the time for the observer inside the field) is two hours (7200 seconds) and $t_f$ is ten years (315360000 seconds),
$$\frac{t_0}{t_f}=\frac{1}{43800}=1-\frac{(3)2GM}{(2)r_fc^2}$$
Simplifying, and saying that $\frac{2GM}{c^2}=\frac{2}{3}r_p$,
$$\frac{r_p}{r_f}=\frac{43799}{43800}$$
$$r_f \approx 1.0000228315715 r_p$$
which is outside the photon sphere, but just barely. However, it's well inside $2r_p$, and so most likely instable. The planet is, in short, not going to survive for long. And so giant waves - mini-spoiler - are the least of Matthew McConaughey's problems.

Post-question-edit modifications:
It couldn't have been a supermassive black hole; these form at the center of galaxies. It could have been a stellar-mass black hole, though an intermediate-mass black hole is also likely - if not likelier, if the massiveness is emphasized.
The existence of the neutron star is interesting. If the black hole were intermediate-mass, I would expect that it would have gobbled up the neutron star by now - and the planets, too. So I'd bet the black hole is a slightly-more-massive-than-average stellar-mass black hole. I highly doubt that multiple planets could orbit a black hole - for the reason I gave above; the supernova would have destroyed them or flung them out of the system.
Would any of the planets be habitable? I doubt it. The accretion disk could heat up enough to provide some light, but there probably wouldn't be a lot. I'll write up the calculations either later today or possibly tomorrow, as I'm a bit LaTeXed-out after writing a math-heavy answer on Worldbuilding to find the luminosity, but I suspect it'll be negligible - as will Hawking radiation, in case any smart-aleck was planning on bringing that up.


Do free protons and neutrons absorb much radiation? To affect astronomers' observations? If so, at what wavelength(s)?

You cannot have free protons without electrons. Plasmas, in general, are electrically neutral.
It is usually electrons that dominate the scattering (note that a point-like charge cannot absorb a photon and conserve energy and momentum) in a plasma at low photon energies. That is basically due to their much lower masses (classically you can think of the electric field of the incoming light accelerating the particles). NB. Electrons in the electric field of ions can emit or absorb photons in a process called (inverse) bremsstrahlung.
High energy (>MeV) photons are capable of causing transitions in the energy levels of nuclei, so can be absorbed. At even higher energies the photons can interact with a nucleus and create particle/anti-particle pairs - pair production.
Very high energy (>100 MeV) photons can scatter from protons and neutrons because these particles have internal quark structure. It is even possible for photons to be absorbed, creating short-lived 3 quark resonances.


Elements of answer:

It is not an easy question, as we lack of data to constrain strongly the density and mass of Saturn's rings.

However, a first clue of their density is their optical depth (that is a measure of the transparency of a medium). The densest rings are the one of the main ring system (rings A, Cassini Division, B and C) plus ring F; they are characterized by optical depth larger than 0.1 (up to about 5 for the densest part of ring B). Rings in the faint ring system (E and G) have very small optical depth (lower than $10^{-5}$).

With this regard, B is indeed the densest ring of Saturn's rings. (edit) Looking around ([1] and [2]) you can find numbers that help to figure out the density of this ring. Knowing its mass (about $2.8\times10^9$ kg) and its extend (from 1.527 to 1.951 Saturn radii, with a typical thickness of 100 m), you can deduce a mean density of about 0.0167 g.m$^{-3}$ (about $10^5$ times less dense than the terrestrial atmosphere at sea level).

Just for fun:

This gives you an idea of what means the optical depth. In the densest parts of ring B, optical depth is between 2.5 and 5, so you'll definitely notice you're in the ring. It is probably more complicated than that (with ring particles ranging from dust size to house size), but you get the idea.

Side note:

Optical depth is a measure of transparency of one medium; it measures how radiation is neither scattered nor absorbed by the medium. It expression is given by

$$\frac{I}{I_0} = {\rm e}^{-\tau}$$

where $I_0$ is the source intensity, $I$ the observed intensity, and $\tau$ the optical depth.

Sources:

Charnoz 2009


Cosmology calculator where curvature density $\Omega_k$ can be independently set?

The density parameter of cosmology, $\Omega$, is defined as the ratio of the energy density of all forms of matter vs. the critical density.
The forms of matter may include nonrelativistic matter (dust, $\Omega_m$, itself perhaps a sum of baryonic matter, $\Omega_b$ and dark matter, $\Omega_{\rm DM}$) with equation of state, that is, the ratio of its pressure to its energy density, $w=p/\rho\sim 0$; dark energy, aka., the cosmological constant, $\Omega_\Lambda$, with $w=-1$; radiation, $\Omega_\gamma$ (or maybe $\Omega_r$) with equation of state $w=1/3$; or indeed, anything else. So rather than spelling out all possibilities, let me just denote the sum of it all as $\sum\Omega$.
As I mentioned, every one of these $\Omega$-s is a ratio of the corresponding energy density to the critical density: $\Omega_x=\rho_x/\rho_{\rm crit}$, where $\rho_{\rm crit}=3H^2/8\pi G$ with $H=\dot{a}/a$.
The "critical density" is the density at which the universe is spatially flat, i.e., has no spatial curvature, $k=0$. In other words, when $\sum\Omega=1$ exactly, we have a spatially flat universe.
If $\sum\Omega\ne 1$, the universe has (positive or negative) curvature. This is expressed using $\Omega_k$, which can be thought of as being defined by
$$\Omega_k=1-\sum\Omega.$$
To offer a little more detail:
This $\Omega_k$ behaves formally as the density parameter of a perfect fluid with negative pressure, $w=-1/3$. To see this, it is instructive to look at the Friedmann equations:
$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2+\frac{k}{a^2}&=\frac{8\pi G\rho}{3}+\frac{\Lambda}{3},\\ \frac{\ddot{a}}{a}&=-\frac{4\pi G}{3}\left(\rho+3p\right)+\frac{\Lambda}{3},\end{align}$$
where $a$ is the scale parameter, $\rho$ represents the energy density of everything that is not the cosmological constant or spatial curvature and $p$ is the corresponding pressure. However, we can also move the $k/a^2$ term to the right-hand side of the first Friedmann equation, and then re-express both the $k$ term and $\Lambda$ in the form of effective densities and pressures, namely $\rho_\Lambda=\Lambda/8\pi G$, $p_\Lambda=-\rho_\Lambda$, $\rho_k=3k/8\pi Ga^2$, $p_k=-\rho_k/3$, and get
$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2&=\frac{8\pi G(\rho+\rho_\Lambda+\rho_k)}{3},\\ \frac{\ddot{a}}{a}&=-\frac{4\pi G}{3}\left[(\rho+\rho_\Lambda+\rho_k)+3(p+p_\Lambda+p_k)\right].\end{align}$$
This form makes it explicit that the sum of all $\rho$-s is indeed the critical density and thus unavoidably, by definition, $\Omega_k=1-\sum\Omega$.


How do I calculate the sun's azimuth based on zenith, hour angle, declination, and latitude?

I've confirmed that this formula is in fact correct. I did have the wrong units in the code, which @Glorfindel suggested: sun's hour angle was in hours instead of radians.

I did need to adjust the azimuth's angle offset based on the hour angle:

     // The angle offset needs to be adjusted based on whether the hour angle
     // is in the morning or the evening.
     //   https://en.wikipedia.org/wiki/Solar_azimuth_angle#Formulas
     if (hourAngle < 0 || hourAngle > Math.PI) {
         return az;
     }

     return (Math.PI * 2) - az;


It isn't the only way, Aaron. See the Wikipedia Methods of detecting exoplanets article. The first method listed is radial velocity. That's where we measure the Doppler shift of the star's "wobble".

The second method is transit photography. And as you say it has a drawback. In fact there's more than one:

"This method has two major disadvantages. First, planetary transits are observable only when the planet's orbit happens to be perfectly aligned from the astronomers' vantage point. The probability of a planetary orbital plane being directly on the line-of-sight to a star is the ratio of the diameter of the star to the diameter of the orbit (in small stars, the radius of the planet is also an important factor). About 10% of planets with small orbits have such an alignment, and the fraction decreases for planets with larger orbits. For a planet orbiting a Sun-sized star at 1 AU, the probability of a random alignment producing a transit is 0.47%. Therefore, the method cannot guarantee that any particular star is not a host to planets...

The second disadvantage of this method is a high rate of false detections. A 2012 study found that the rate of false positives for transits observed by the Kepler mission could be as high as 40% in single-planet systems..."


Why do type Ia supernovas produce more iron than type II

Context

Iron has the highest nuclear binding energy of all the elements (not completely true, but sufficiently accurate in an astronomical context). So, fusion of light elements into iron or something lighter is an exothermic process - you gain energy doing it, allowing the star to function. This is what happens in the last stages of a type II supernova. The core of a massive star in its last moments of life is hot and dense enough to fuse silicon into iron. Just before the supernova explosion, there is an iron ball of about 1.4 solar masses at the centre.

The progenitor of a supernova type Ia is a binary system where a "normal" star loses mass to a compact stellar remnant (a white dwarf). Once the white dwarf Contexthas accreted enough mass to be above a limit of 1.4 solar massses, fusion starts again, completely disintegrating the compact object.

Explosion

A SN Ia completely destroys the white dwarf progenitor in a runaway fusion process.

In a SN II, the pressure on the central iron ball exceeds the degeneracy pressure exerted by the electrons in the iron atoms' electron shell. The Fermi principle in quantum mechanics states that no Fermion (such as an electron) may occupy the same quantum mechanical state as another. The pressure exerted here is so large that the electrons of the iron atoms can no longer obey it and are forced into the nucleus, where they react with the protons to form neutrons.

Iron abundance

Why do SN Ia enrich their environment with more iron than SN II? This is not so much a matter of iron production, but about how much of that iron ends up in interstellar space where it can be part of a new generation of stars. In a SN Ia, the progenitor is completely destroyed, scattering all its constituent atoms into its host galaxy. A SN II forms a compact remnant, either a neutron star or a black hole. A lot of the later, heavier fusion products end up not being carried outward in the supernova explosion but become part of the compact remnant.

Note that a lot of the heavy elements scattered from a supernova of the "exploding massive star" result from an abundance of neutrinos escaping the central explosion and reacting with the outer shell of lighter elements that is blasted away.


Is gravity a source of infinite energy at a cosmological scale

Gravity travels at light speed (or less, possibly), so even in an infinite non-expanding universe of finite age you'd only be gravitationally interacting with a finite mass in a finite volume. Our universe is observed to be expanding, further inhibiting us from coming into contact with new objects.

Furthermore, a common assumption of cosmological models is that the universe is homogeneous and isotropic on large scales. Meaning it is pretty much identical in all directions. Observations support this assumption. One net effect of this is that the gravitational potential energy you experience from objects outside (roughly) your supercluster is zero (and really close to it from things outside your own galaxy, actually). They just balance each other out.


Are any other planetary bodies warming up as well in the Solar System?

The short answer to your question is "decades", only for Mars, and only a two Mars decades (about 4 Earth decades)

I don't know if your question was related to climate change, as there were a number of extremely unscientific arguments made that Pluto and Mars are getting warmer, "and there are no SUV's on Mars" . . . so, maybe it's not the CO2 but those claims weren't science.

Silly articles and bad arguments aside, global temperature on other planets is a legit scientific question and there were scientific articles that said Pluto is getting warmer (2002, story below) and Mars is getting warmer (2007, also below), so there is some truth to this.   I'll give an answer as best I can.  Each planet deserves it's own bit of research, so this will be longish.

Pluto

Here's an article on Pluto, the title is iffy but the article is good.   It's from 2002, so it's long before New Horizon's fly-by, it's even before New Horizon was launched, but "pluto is getting warmer" was a real scientific story in 2002, and Mars, in 2007 was real science too, but some of the conclusions drawn, loosely based on those two stories were ridiculously bad.

The gist of the article is, we can't see Pluto's atmosphere well, even with Hubble, but they were able to measure some aspects of Pluto's atmosphere by watching what happens when a star passes behind Pluto and that was measured twice, in 1988 and again in 1997.

Pluto's Perihelion was in 1989 and most everyone figured that the little ball of ice/dwarf planet would get colder after 1989 when it passed it's closest point to the sun and stated moving further away but the readings said otherwise, and that was one of those really cool "oh, wow, things are more interesting than we thought they were" moments.  Pluto's atmosphere had apparently gotten thicker and it was about 2 degrees warmer.  Not what anyone expected.

The problem is, the sun couldn't make Pluto 2 degrees warmer while it moved further away without making the Earth 20 degrees warmer at the same time, so, clearly, something else was at play and it wasn't the sun warming Pluto.  Generally speaking, the warmest periods aren't necessarily precisely timed to closest to the sun, and there's a lag time caused (on Pluto) by sublimation of the frozen gases on it's surface, which thickens the atmosphere adds trace amounts of Methane and can warms the planet past Pluto's perihelion, just as on the Earth, June 21 isn't the warmest day on the Northern Hemisphere, nor December 21st the coldest, and 12:00 noon isn't the hottest time.  Could that seasonal "lag time", apply over a number of Earth years on Pluto?   It's plausible.

But we should remember, that's just two data points, which isn't nearly enough.  Or, 3 data points now with New Horizons, but it's still just 3 data points spanning 27 years on a planet where one year is 248 earth-years, so, there's a lot we don't know about seasons on Pluto.

New Horizons got better data on Pluto's atmosphere and frozen surface than we've ever had before, and what I remember reading about it, though I'm very far from well read on the subject, is that Pluto has a much more dynamic atmosphere and surface than anyone expected.

Gas Giants

you asked:

  Do we have a few decades worth of surface temperature data to compare?

Surface temperature isn't well defined on gas giant planets because we can't see the surface.   Also (I think all 4, I know for sure Jupiter and Neptune), they give off more heat internally than they receive from the sun.  Temperature on the visible surfaces / upper atmospheres of the gas giants has more to do with circulation than anything else.

Mars

There's a number of articles on Mars warming, but there's still a lot we don't know about Mars' climate.   It's not too difficult using modern technology and with Mars' thin atmosphere, to read surface temperature on Mars by measuring the infra red light that comes from it's surface.  The surface temperature on Mars was first measured in 1920 (see 4th paragraph from the top), and NASA says here that they've been taking accurate surface temperature readings on Mars for 30 years (article from May 2007), so it's pretty safe to say we have about 38-39 Earth years (20 or so Mars years) of reasonably good global temperature readings on Mars and a few decades more than that with less accuracy.

There's many problems with reaching any kind of conclusion on Mars based on that data though.  Mars has significantly greater eccentricity than Earth, so it may be prone to larger temperature swings than Earth and just 20 Mars years of data is too few.

Winter to Summer on Mars' enough CO2 can freeze to reduce it's atmosphere by as much as 25% and that can drive temperature changes and mars is also prone to planet wide dust storms that can last months.  These can have the effect of blocking sunlight and after settling, darkening the ice caps, which lowers albedo.  Like Pluto, Mars' atmosphere, climate, temperature and orbit are dynamic and there's still a lot we don't know about how they worth together to form a predictable climate on Mars.

As an FYI, dust storms are nothing like in the movie the Martian, but they can cause a visible haze over the planet that can obscure viewing it's surface.   A nice short article on Mars' dust storms here and here.

I think it's kind of cool that Mars forms an annual "cloud", not unlike how parts of Earth form annual snow and ice covers in winter that thaw in Summer, but we don't know nearly enough about Mars, even with 20 Mars years of observation, to make good long term predictions on Mars' climate.

Venus

Venus, as far as I know, pretty much never changes.  It's the same temperature, Winter, Summer, Pole, Equator.  There might be some slight variation of Venus temperature depending on elevation, like the gas giants, but I'm not sure.   Venus traps heat so effectively that is changes very little, in fact, I've never seen one article mentioning Venus' surface temperature changing as much as 1 degree.   Venus reflects most of the sunlight that hits it away anyway, so it would be less prone to solar variation anyway.

I won't do Mercury cause I don't know enough about it, but that's 7 out of 8 anyway.

. . .

If your question is about having a good understanding of climates on other planets, then decades of surface temperature data wouldn't be enough even if we did have it.  Centuries of accurate surface temperature data, you could begin to recognize patterns, for the inner 4 planets anyway, but decades isn't enough to differentiate between a baseline, a natural fluctuation or a transition period.

If anyone tells you there are no SUV's on mars, so how could Mars be warming, get away from them as fast as you can.  They're not interested in science.


Of the two, I would go with the Celestron - it's from a named and respected manufacturer with sensible specifications and relevant additional equipment. The eyepiece selection is thoughtful.

However, you pay the price because the Celestron's focal length at 400mm is much more limited. The best magnification you'll be able to get is 40x without a Barlow (good for the Moon but rather anaemic for the planets).

The other scope promises much higher magnifications but I worry that the manufacturer's name isn't stated, that there seems to be a number of redundant extra items and about the overall build quality.

Personally, I would consider the https://www.bhphotovideo.com/c/product/485261-REG/Celestron_21061_AstroMaster_70_AZ_2_7_70mm_Refractor.html as an alternative, though it is a little more expensive.


Because planets actually do twinkle. Most people were told that the major difference between stars and planets is that only the former twinkle - but that's an oversimplification. Given the right conditions, planets will twinkle too, it just happens more rarely.

Several factors that contribute to it:

lots of air turbulence; or, as astronomers call it, "bad seeing"
closeness to horizon; if the planets are high in the sky, the air column is shorter so there's less chance they will twinkle; but when they are low, their light goes through more air and so it is perturbed to a larger degree

The observation you've made, Venus twinkling, is not very unusual. Many stargazers are used to seeing that once in a while. I've seen Venus scintillate several times in the past, always at sunset when it was about to drop below horizon; I would presume you could see the same behavior very early in the morning as Venus has just risen.


Determine sun azimuth and elevation for shading a window throughout the year

You can use a package such as the pyephem package for python. It can calculate the position of any astronomical object.

However for the sun since it moves along the ecliptic, roughly at a uniform rate, you can approximate the position by doing coordinate transformations. The sun's longitude moves by 365.24/360 degrees a day from 0 at the vernal equinox and has an ecliptic latitude of 90 degrees. This can be converted into equatorial coordinates (right ascension and declension), and then, given the hour, the alt and az positions can be found. The formulae can all be found on https://en.wikipedia.org/wiki/Celestial_coordinate_system


Sun constantly converts mass into energy, will this cause its gravity to decrease?

If the sun is constantly converting the mass into energy, then will its gravitational field go on decreasing?

It's a very interesting question and the answer is yes!

The solar constant indicates the mean solar radiation of electromagnetic waves (mostly in visible and near infrared light and I'll answer based on that.

While the conversion of mass matter† to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun!

There is some prompt mass and energy loss via neutrinos and it's significant, perhaps several hundred keV per neutrino I simply don't know the number yet. I'll ask a separate question about it. I'm guessing that losses due to the stellar wind are small, but I'll update here as soon as the following is answered:

How much mass does the Sun lose as light, neutrinos, and solar wind?

update: the answer there is that loss via neutrinos is only about 2.3% of the radiative loss, and on average loss via solar wind and coronal mass ejections is about 4E+16 kg/year, or about another 30% relative to the radiative loss described below.

The value $I$ is about 1360 Watts per square meter at $R$ = 1 AU which is about 150 million kilometers or 150 billion meters. So the total energy lost per second $P$ is

$$P = 4 \pi R^2 I$$

Taking the time derivative of $E = m c^2$ we get

$$\frac{dE}{dt} = P = \frac{dm}{dt} c^2$$

so

$$  \frac{dm}{dt} = \frac{1}{c^2} \ 4 \pi R^2 I$$

That means that the value of the mass that we use to calculate the Sun's gravitational attraction changes by about 4.3E+09 kilograms per second, or 1.3E+17 kilograms per year.

The Sun's current mass is about 2.00E+30 kilograms, so this effect changes by a very tiny fraction per year, about 6.7E-14.  Over the age of the Earth of 4.5 billion years, that's 3E-04, or about 0.03% if the Sun's output were constant. It has probably changed over this time of course, so this is just a rough estimate.

†Thanks to @S.Melted's answer for clarifying this.

added to my answer by @Tosic:

  The Earth feels no torque from any force during this (the force from radiation is radial), which means its angular momentum is conserved. This means
  $$R_1 v_1 = R v$$
  $$R_1 \sqrt{\frac{GM_1}{R_1}} = R \sqrt{\frac{GM}{R}}$$
  $$M_1 R_1 = M R$$
  We can see the Earth's orbital radius would change by a factor of 0.03% as well (M1 and M are solar masses).


By inspection of a star charting app, I'd say the circles lie in the celestial equator, ie declination equal 0 degrees.  Stars above the circles are to the north and vice versa.


Why is right ascension measured in hours?

In order to know when a star will be above horizon, you'll need an equation with times, not "celestial latitudes".

So you end up adding RA and sidereal time.

EDIT

In order to find out the position of a star you need to find out how its Hour Angle varies with time.

The Hour Angle is how far the star is from the observer's meridian:

In this image, it is the angle between the site's celestial meridian (North Celestial P ole-Z enith-S outh horizon) and the star's celestial meridian (P-X-Y). It is the same angle as between where X culminates and X.

So it comes than $HA=ST-RA$ and since you may not have a "local sidereal time" clock at hand, you can use Greenwich's Sidereal Time: $HA=GST-Lat-RA$

$HA$ (local) Hour Angle of a star
$ST$ Local Sidereal Time
$GST$ Greenwich Sidereal Time
$RA$ Right Ascension of a star
$Lat$ (local) Geographical Latitude


Are there equations to simulate creation of a solar system?

I may have to answer this in parts, a bit at a time, so I apologize if I'm not able to answer everything at the moment.

  If I wanted to find out how to estimate, say, a planets surface temperature based on a stars temperature and distance, what is the best way to do this?

Ah, I just studied this about a week ago! What you're looking for is the effective temperature of the planet, which can be calculated as
$$T=\left(\frac{L(1-A)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$
where $L$ is the star's luminosity, $A$ is the planet's albedo, $\sigma$ is the Stefan-Boltzmann constant, and $D$ is the distance from the star.

  how does the mass of a star relate to the distance of the frost line, and is there an already existing equation to estimate this?

I wasn't able to find an explicit equation for this, but it's calculable. The frost line is a certain distance from a star in a stellar nebula such that the temperature is about 150 Kelvin.

If you know the mass of the star, you can use the mass-luminosity relation:
$$\left(\frac{L}{L_{\odot}} \right)=\left(\frac{M}{M_{\odot}} \right)^a$$
Re-arrange that to solve for the luminosity. The inverse-square law says that the intensity of power drops off as the distance from the source increases:
$$I=\frac{P}{4 \pi r^2}$$
You can solve for $r$ if you can figure out the required intensity:
$$r=\frac{1}{2}\sqrt{\frac{P}{\pi I}}$$
Here, $P \approx L$. Figure out the necessary intensity to heat the nebula to 150 K and you've found the frost line.

  To find the most likely average mass of a potential planet, what information would I need?

I can't help you much there, but I can give you a starter. The density $\rho$ of a circumstellar disk at a distance $r$ (as given by Michael Woolfson in On the Origin of Planets) is
$$\rho (r)=C \exp \left[-\frac{(r-r_{\text{peak}})^2}{2 \sigma ^2} \right]$$
where $\sigma$ is one standard deviation and $r_{\text{peak}}$ is the peak density, and $C$ is a constant.

As promised, here are some more, set forth by Woolfson.

The disk decays over time; its density at time $t$ is approximated by
$$\rho_t=\rho e^{- \gamma t}$$
where $\gamma$ is a decay constant. If you've taken certain math course, you've probably seen this equation many times before, just in with $\rho=y$ and $\gamma=k$.

Another equation - really, a couple equations - deal with accretion by a small dust particle in a circumstellar disk. The variables involved are: $m$ (mass of the particle), $t$ (time), $s$ (the radius of the particle), $\rho_s$ (the density of the particle), $\rho$ (the density of the disk), $f$ (the fraction of the medium that is dust), $T$ (the temperature), and $k$ (Boltzmann's constant, not be confused with the Stefan-Boltzmann constant).

The relevant pair of equations are:
$$\frac{dm}{dt}=4 \pi s^2 \rho_s \frac{ds}{dt}$$
$$\frac{ds}{dt}=\frac{3f \rho}{4} \left( \frac{kT}{4 \pi \rho_s^3} \right)^{1/2}s^{-3/2}$$
Using these, you can find the size and mass of a given dust particle at any given time.


Einstein General relativity and bending of space

We cannot say that it bends in a particular direction because space-time in relativity is 4-dimensional. The space-time diagram which we see frequently is just a simple way of explaining it in two dimensions because 4-dimensional space is hard to visualize.


How small can a planet be and still have Earth-like gravity?

The surface gravity of a planet is very close to
$$g=\frac{4\pi G}{3}\rho r.$$
With $g$ to be kept constant, and $\frac{4\pi G}{3}$ a constant, we need
$\rho_Pr_P=\rho_Er_E$, or
$$r_P=\frac{\rho_E}{\rho_P}r_E,$$
with $\rho_E=5.515 \mbox{ g}/\mbox{cm}^3$ the mean density of Earth, $r_E=6371.0 \mbox{ km}$ the mean radius of Earth, $\rho_P=22.59\mbox{ g}/\mbox{cm}^3$ the density of densest natural element osmium, and $r_P$ the radius of the fictive osmium planet.

Hence $$r_P=\frac{5.515}{22.59}r_E=0.2441~r_E=1555\mbox{ km}.$$

Some compression of the core of an osmium planet due to pressure is neglected.


No. The Roche limit refers to the radius at which a satellite which is held together only by its own self gravity is disrupted by tidal forces. Spacecraft are not such satellites, but are much harder to tear apart.


Will the Heat Death scenario ever occur or just the Big Freeze?

The Wikipedia article is somewhat confused on this point (at least one of the references does not support the sentence it is supposed to support).

The classic concept of the heat death of the universe was that eventually it will run out of free energy and there will hence be no possibility of any work. A more modern view of heat death is an approach to maximum entropy where the system will be in equilibrium. This does not rule out random fluctuations bringing it away from equilibrium, but they are so rare it does not matter.

Note that it is entirely possible to have a heat death at any temperature.

The big freeze would happen if expansion of the universe led to temperatures decreasing towards absolute zero. However, if there is accelerating expansion this generates a constant finite temperature horizon radiation, about $10^{-29}$ K. This means that the big freeze will never happen.

Had there not been horizon radiation it would in principle have been possible to avoid heat death by storing some background heat in a perfectly reflective box, waiting until the universe become colder, and use it to perform work. This can be repeated indefinitely. (Not really: the timescales quickly become long enough that random fluctuations destroy your device, but that it a separate issue).


How much light comes from quasars in an AGN?

Active Galactic Nuclei (AGNs) are very luminous galactic nuclei. In Quasars, a type of an AGN, the nucleus outshines the host galaxy. Therefore the main source of radiation is the nucleus, not the stars and the host.


Are there more stars in the universe than grains of sand in the Earth?

A quick google gave me these (approximate) figures:

7.5 x 1018 grains of sand in all the beaches and deserts of the world
7 x 1022 stars in the observable universe

If these are reasonable estimates, then there are approximately nine thousand stars in the observable universe for each grain of sand on Earth. (By observable universe, I mean in all the galaxies that we can detect with our most powerful telescopes.)

Robert Krulwich: Which Is Greater, The Number Of Sand Grains On Earth Or Stars In The Sky?


Actual observable horizon for satellites on an orbit line map?

It is not that satellite. As it passes over Finland, it has an altitude of 27000km and a magnitude of about 12. You won't see it. All Molniya orbits are similar.

The website Satflare to shows the region in which satellites are visible (it uses a blue ring around the satellite) Your satellite was so high that it was visible from nearly the whole hemisphere.

I won't try to identify your object, except to say that an object as described is not in orbit.


What celestial body (inside the solar system) has the highest flattening ratio?

I believe besides Telesto the dwarf planet Haumea in the Kuiper belt could also be in the mix for the highest flattening of objects in our solar system. Haumea is a triaxial ellipsoid where the 'f' value would around 1/2 between the largest and smallest major axis.


What is the current routine of modern astronomy?

What you say is not quite true: the search for exoplanets is clearly intensive, but it is far from the only things astronomers are looking at. Most of the time, in two words, the situation is: resolution & wavelength. Whatever the field (if you are interested in galaxies, interstellar medium, stars and so on) you want more resolution, to resolve smaller scales (most stars are still points even with our best telescopes, or we are still far from resolving individual stars in galaxies!), to have more informations, to understand better the underlying physics. You want more wavelength, because spectroscopy gives you much more phsyical information than a single wavelength observation, for example. And to combine both is sometimes challenging: to have high resolution observations in infrared is not that easy, and it can be crucial for some fields (if you ever want to see a star forming, you better observe it in infrared, since this baby is embedded in its gas cloud that shields very efficiently its radiation).

That being said, the routine tasks of an astronomer would be

extract information from the current data. It involves a lot of coding, with Python, IDL, or more specific astronomy-oriented languages as IRAF or MIDAS. Data reduction is an important part of the job, because it is in general challenging to extract data from the raw signal you will get.
write papers about these data and the inferred informations
read a lot of papers to stay tune with the latest discoveries of other teams
write proposals to ask for more observation time/better observations/bigger telescopes
drink a lot of coffee

The three first points probably takes an almost equal amount of time for any astronomer; point 4 takes even more time for older astronomers; point 5 is also crucial for all the good things that come out of discussions over a good ol' bowl of coffeine.

Complements:

To answer to your comment and to give you an overview of the current research, I can think of:

Hershel data in infrared. People try to understand better the interstellar medium and the star formation processes in our galaxy, the formation of early galaxies, and the chemical composition and evolution of the Universe with these data.
Planck data in longer wavelength. These data are useful to understand the first age of the Universe (searching for anistropy in the CMB), but also to have an other view of the galaxy and the interstellar medium in these wavelengths.
Very Large Telescope data. There are plenty of different kind of data out of these telescope, mostly in the visible and infrared ranges, and mostly in spectroscopy. Almost everything is studied with these data, from galaxies evolution to stars in the nearby galaxies.
ALMA data in millimeter/submillimeter ranges. The same kind of objects are studied with ALMA and Herschel: early galaxies, interstellar medium and molecular clouds. How galaxies form and evolve? How stars form? In which environment? What are the dominent processes in star formation?
HESS data, in the gamma ray range. Gamma rays offer a window on the non-thermal Universe, i.e. all the extreme events occuring in the Universe. It can give precious informations about gamma-ray bursts, supernovæ, AGN (the active galactic nuclei) etc.

That's for the big projects (with a strong European bias, sorry folks I know better what's done on this side of the ocean). You can add to that all the missions to study exoplanets (like Kepler), missions to study our solar system planets (Cassini, Huygens, Messenger, Juno, all the Mars missions etc.), plus all the other facilities around the world to study anything and everything, from stellar dynamic to planet's composition. The main problem is always to understand how such structure (from the large scale structures in the Universe to small scale structures in the galaxy), object (from galaxies to satellites), phenomenon occurs, forms, appears. To understand what are the dominant physical processes at play.

Astronomy is still craving for data; the more data you have, the better your statistic will be, and hopefully the better your understanding will also be.


No, no stars have an exactly spherical shape. The reason for this is that the centrifugal force of the star's rotation is much greater at the equator of the star than it is at the poles, for the simple reason that the rotational velocity is greater. This greater centrifugal force pushes the equator outwards, stretching the star into an oblate shape. This is called gravity darkening.

Because we have only visually resolved the surfaces of a few other stars, it's not something that is commonly directly observed (though the effect can also be observed from stellar spectra). Regulus is one star that has been observed as an oblate spheroid spectrographically. The image below shows the star Altair, directly imaged using the CHANDRA space telescope. Go here for an animation!

Some stars are also non-spherical due to the effects of a nearby star in a close binary orbit. Much like how the Moon causes tides on Earth, two stars can stretch each other's surfaces. If they're very close to one another, as in the picture below, there can even be mass transfer between them (ref common envelope).


Why is there no Proxima 'a' planet? I'm confused about the naming convention

Usually, the primary host star reserves the right to have the 'a' designation. Therefore 'a' never gets assigned to a planet.

Please check the following post for a detailed explanation and some examples.
Where did TRAPPIST-1a go?


Black Hole / Hawking Radiation: Why only capture anti-particle?

I thought that the anti-particle was annihilating with "normal" mass inside the black hole? No?

No. First, both particles and anti-particles have "normal" mass (should they have mass in the first place) and "normal" (positive) energy. The distinction between them is either a matter of convention or a question of which type is more common in the universe. Furthermore, for typical-massed blacked holes, the bulk of Hawking radiation would be made of photons, which properly speaking do not even have anti-particles, though one could also say that they are their own anti-particle.

  Shouldn't both particle and anti-particle have equal chance to be the one to fall in, or just manage to escape?

Yes, and uncharged ones do. A smaller black hole would radiate both neutrinos and anti-neutrinos, assuming all neutrinos are massive (otherwise, all black holes would do it already), and a sufficiently small (and thus sufficiently hot) one would radiate both electrons and positrons. Very roughly, a black hole will radiate non-negligible amounts of massive particles when the temperature of the black hole is on the order of the particle mass or greater, in natural units.

  It seems that there should be an equal chance of either the particle, or the anti-particle, would be captured while the other "ejected."

Correct, with a minor exception that if a hot black hole has electric charge, it is more likely to radiate particles of the same sign of charge.

  So it seems that the black hole should be somewhat steady-state as far as mass change with respect to virtual particles.

If either a particle or an anti-particle falls into a black hole, its mass will go up. It doesn't matter. Fundamentally, the "reason" for Hawking radiation is that the vacuum state in quantum field theory is a state of lowest energy, but different observers can disagree about which state is the vacuum. Thus, since particles are fluctuation on top of the vacuum, they can disagree about whether or not there are particles.

I don't think there is a good way to repair the "antiparticle falls in" story except some roundabout appeal to energy conservation: if the escaping particle is real and have positive energy, the one that fell in must have negative energy, and would therefore decrease the mass of the black hole. Unfortunately, that only shows what must happen for the situation to be consistent, not that it does actually happen.

Although with some knowledge of general relativity, one can motivate this slightly further--e.g., for the Schwarzschild black hole, there is energy conservation given by a Killing vector field, which goes from timelike to spacelike at the horizon--so what an external observer considers time/energy would be space/momentum inside the black hole, and momentum is allowed to be negative.


How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?

An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas.

You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal acceleration. It is equal to
$$a_\text{tidal} = \left|\frac{GM}{(R\pm r)^2}-\frac{GM}{r^2}\right| \approx 2 \frac{GMr}{R^3} = 2\frac{GM}{R^2}\frac{r}{R}$$
where $R$ is Mimas' semi major axis length and $r$ is Mimas' mean radius. The approximation assumes that $r\ll R$, which most certainly is the case given that Mimas's radius is about 1/1000 of the semi-major axis length of its orbit about Saturn. The result is rather small, about 0.002355 m/s2.


Estimating close approach between all asteroids themselves

Thanks folks for helping.
I knew about SPK files but from Horizon's telnet inteface there is a limit of 200 bodies per request.
Making over 5000 request might be possible but not very productive.
I also looked at the  DE421 file from ftp://ssd.jpl.nasa.gov/pub/eph/planets/bsp/ but it looks like the corrections for the 8+1 planets
here is the output from BRIEF, a SPICE utility: https://naif.jpl.nasa.gov/naif/utilities.html
BRIEF -- Version 4.0.0, September 8, 2010 -- Toolkit Version N0066
Summary for: de421.bsp
Bodies: MERCURY BARYCENTER (1)  SATURN BARYCENTER (6)   MERCURY (199)
    VENUS BARYCENTER (2)    URANUS BARYCENTER (7)   VENUS (299)
    EARTH BARYCENTER (3)    NEPTUNE BARYCENTER (8)  MOON (301)
    MARS BARYCENTER (4)     PLUTO BARYCENTER (9)    EARTH (399)
    JUPITER BARYCENTER (5)  SUN (10)                MARS (499)
    Start of Interval (ET)              End of Interval (ET)
    -----------------------------       -----------------------------
    1899 JUL 29 00:00:00.000            2053 OCT 09 00:00:00.000

I can always do a statistical estimate, any close approach being false but the overall count being close to reality.
Too bad to drop the project, I was getting nices pictures from it. The coloring is done against eccentricity, red=0


The other answer posted is correct, but let me try to explain it a bit more fully, and actually correct you on your terms.  I did an interview on this topic for a podcast literally just two days ago, so it's fresh in my mind.

Answer: The way we determine an astronomical object's mass is through Newton's form of Kepler's Third Law, which relates the distance another object orbits the object in question to the distance it orbits from that object, and some fundamental math and physics constants.  Doing this requires that there be a smaller, "test" object in orbit of the object whose mass you want to measure.
Conceptually, this means that we can only measure an object's mass via that mass's effect on another object.

What does that mean?  Kepler's Third Law states that the time it takes one object to orbit another, squared, is proportional to the semi major axis of its orbit around that object, cubed.  We often use that in our solar system in units of distance in A.U. (1 A.U. = Earth's average distance from the Sun) and Earth years.  So, before we knew how big an A.U. is in km, we still knew that Jupiter was about 5.2 A.U., and it took 11.9 Earth years to go around the sun.
Kepler's Third Law was formed in 1619, but it took Newton's Law of Gravitation in 1687 to provide the physics behind it.  That physics changed the "is proportional to" to "equals," and added 4π2/G*M into the formula.  We know what 4 means, we know what π means.  M is the mass of the big object that something is orbiting.  It's what we want to solve for.  And, assuming we can observe an orbiting object, we know its period and distance from the main object we want to solve for.
So then, all we have to do is figure out G, the gravitational constant.  Gravity is really weak, so it's really hard to measure, and it took nearly 100 years for it to be calculated kinda accurately, and the first real measurement that people like (done by Cavendish) didn't happen until 1798.
With a value for "big G" as it's often called in physics class, you can now calculate any object's mass, in space.  So long as it has another object orbiting it.  That smaller object is required so we can use Kepler's Third Law to observe that object's orbit around the big one we want to measure.  If we don't have that smaller object in orbit, we can't* measure the big one's mass.  So, our first reasonable measurements of Pluto's mass only came once its main satellite, Charon, was discovered.  Only asteroids with moons have reasonable mass estimates.
*There are slightly other ways to do this.  One is to put an artificial satellite in orbit, so once Dawn orbited Vesta, and then Ceres, we could get their masses.  Another way is a gravitational slingshot around an object, like sending a probe past Venus, we can watch how much Venus deflects the probe and get its mass.  We can also watch for gravitational perturbations in a system caused by smaller masses and figure out what those smaller masses must be to cause those perturbations, but this is much harder (but led to some early estimates of the mass of Venus).
The bottom-line is still that, conceptually, you can only do this by measuring the object's (mass's) effects on another object.

Terms:  Mass is a fundamental property of matter, it's a way to think of how much "stuff" is there.  It has the familiar units of grams (or kilograms) or pounds.  Weight is different.  Weight is how mass acts in the presence of an acceleration, like gravity.  It has the metric units of Newtons.
In the presence of a basically-the-same gravitational acceleration – Earth's surface – mass and weight are proportional because the gravitational acceleration is the same.  So my mass – which doesn't change (much) – is going to mean I weigh the same wherever I am on Earth's surface.  If my mass is 100 kg, then I will weigh about 1000 N on Earth.  If I go to the Moon, my mass is still 100 kg, but I will be in a different gravitational field and I will weigh about 160 N.
Because my weight and mass are proportional practically anywhere on Earth, we use the terms "mass" and "weight" completely interchangeably, but we still use them wrong.  In fact, when I did the podcast recording I mentioned above, we had to do up to seven takes of just one sentence because I kept using the terms wrong, even though I was explaining how to use them correctly!
This is important because your question is actually phrased wrong: The question is, "How do we determine the mass of a planet?"


What experiment does this "Tracing the History of Starlight" video try to explain?

Blazar sequence
Let's first look at which kind of sources are studied.
Blazars are special type of radio-loud AGN. One of their special properties is that their spectral energy distributions (SED) form the so-called Blazar sequence. They all have a characteristic double peak SED and the position of the peaks mainly depend on their luminosity. The higher the luminosity the lower the peak frequencies (see Fossati et al., 1998 for details)

(image source: Donato et al., 2001)
Leaving details like sub-classes aside, this means that SED of Blazars can be modelled very well, especially if looking at a huge ensemble of Blazars.
Extra-galactic background light (EBL) absorption
Now the high energetic light has to travel from the Blazar to us. The Blazars being far away, much can happen during this travel, notably it can interact with the extra-galactic background light (EBL absorption). If a high energetic photon from the Blazar encounters an EBL photon in the optical or ultraviolet range they can pair produce an electron/positron pair (this is the dominant process for the involved energies).

These charged particles will most likely be subject to magnetic fields or other interactions diverting them from their direct way towards earth.
This means not all of the high energy photons from the Blazar reach Earth and how large the absorption is depends on:

the distance between the Blazar and Earth
the energy of the high energy photons from the Blazar
the energy of the interaction partners from the EBL

Bringing both together
Fermi looked at a large sample of SEDs from Blazars of various luminosities and grouped them based on their place in the Blazar sequence. For each group they could compare the expected SED from the Blazar model with their measured SED and thus determine how large the EBL absorption is depending on the energy. This allows to calculate the density and energy distribution of the EBL.
Knowing the spectral energy distribution and density of the EBL allows to go even one step further and make assumptions on where the photos in the EBL came from. How many stars it took, how long they had to emit photons etc.
As if this alone would not already be interesting enough, Fermi even repeated this for samples of Blazars at different distances, thus probing the EBL at different ages of the universe (remember that we observe the light that reaches the Earth now, thus the light from Blazars further away was emitted earlier than the light from "nearby" Blazars)

(image source: https://fermi.gsfc.nasa.gov/science/eteu/ebl/)


Does any iron fuse in stars before they go supernova?

The "iron core" in a supernova is actually the end product of a nuclear statistical equilibrium that begins when the silicon core begins to fuse with alpha particles (helium nuclei). Exothermic reactions are possible right up to Nickel-62 (which is actually the nucleus with the highest binding energy per nucleon). In fact, successive, rapid alpha captures produce nuclei with the same number of protons and neutrons, but at the same time, the competing processes of photodisintegration and radioactive decay work in the other direction. The process is thought to mostly stop at Nickel-56 which, because heavier nuclei are more stable with $n/p>1$, then undergoes a couple of $\beta^{+}$ decays via Cobalt-56 to Iron-56. However, the core of a supernova just before it explodes is likely to contain a bit of a mixture of iron-peak isotopes.

Before all this happens it is possible for iron and nickel to undergo nuclear reactions if there is an appropriate source of free neutrons. The elements beyond iron in our universe are predominantly created by neutron-capture in either the r-process or the s-process.

The r-process is thought to occur after a core-collapse supernova (or a type Ia supernova) has been initiated. The neutron flux is created by the neutronisation of protons by a dense, degenerate electron gas in the collapsing core.

However, the s-process can occur outside the core of a massive star before it explodes. It is a secondary process because it needs iron nuclei to be present already - that is, the iron that is used for the seed nuclei is not produced inside the star, it was already present in the gas from which the star formed. The s-process in massive stars uses free neutrons produced during neon burning (so at advanced nuclear burning stages beyond helium, carbon and oxygen burning) and results in the addition of neutrons to iron nuclei. This builds up heavy isotopes, which may either be stable or undergo $\beta$ decay and/or further neutron captures to build up a chain of "s-process elements" (e.g.  Sr, Y, Ba) all the way up to lead. The overall process is endothermic, but the yields and reaction rates are so small that it has no major influence on the overall energetics of the star. The newly-minted s-process elements are easily blasted into the interstellar medium shortly afterwards when the supernova explodes.


Moving-Cluster method for determination of the distance of Hyades. A starter problem

The angle you are looking for
$$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (a_1-a_2),$$
where $(a_1, \delta_1)$ are the RA and Dec of the star and $(a_2,\delta_2)$ the RA and Dec of the convergent point.


Is there an upper limit on the mass of black hole mergers we can detect?

It is quite likely there is an astrophysical upper limit to the mass of a black hole that can be produced during the core collapse of a massive star, caused by the pair instability supernova phenomenon. There isn't an observational bias against detecting more massive black holes in the range of 100 to a few hundred $M_{\odot}$.

Details:

The frequency of the gravitational waves is twice the orbital frequency of the binary system. The general scenario is that early in their evolution, a mrging binary system will be orbiting at relatively long periods (seconds !), but as gravitational waves take energy out of the orbit, the system becomes more compact, the orbital period gets smaller and the frequency of the emitted gravitational waves goes up. This continues until the black holes get so close together that their event horizons merge.

Very roughly, we can derive (from Kepler's third law, not going into detail), using Keplerian orbits
$$ f_{\rm max} \sim \left( \frac{GM}{\pi^2 a_{\rm merge}^3} \right)^{1/2}\ ,$$
where $f_{\rm max}$ is the peak frequency at merger (when the gravitational wave signal is also maximised), $a_{\rm merge}$ is the separation of the masses at merger and $M$ is the total mass of both black holes.

If we let $a_{\rm merge} \sim 2GM/c^2$, the sum of the two Schwarzschild radii of the black holes, then
$$f_{\rm max} \sim \frac{c^3}{GM} \left( \frac{1}{8\pi^2}\right)^{1/2} \sim 2\times 10^4 \left(\frac{M}{M_{\odot}}\right)^{-1}\ {\rm Hz}$$

Now, LIGO is limited to observing frequencies above about 20 Hz. The sensitivity drops off rapidly below that because of seismic noise and other factors. If the mass of the merging black holes exceeds some critical value then the frequencies of the gravitational waves they produce never get into the sensitivity wndow of LIGO. Using the expression above, we can estimate that this happens only if the total mass exceeds $1000 M_{\odot}$. Observing the mergers of more massive black holes would require a detector that is sensitive to lower frequencies, probably beyond the surface of the Earth (e.g. LISA).

This calculation is only good to a factor of 2 or so, but we can check it. GW150914 had a total mass of around $65 M_{\odot}$ and merged at frequencies of about 120 Hz. Since $f_{\rm max}$ scales as $M^{-1}$ this suggests 360 solar mass mergers should be just about detectable, but clearly demonstrates that LIGO could detect black holes of 100-200 solar masses. What's more, at a given distance and frequency, the signals from such mergers would be more powerful than for less massive black holes -- something like $h \propto M^{5/3}$, which means the volume in which the mergers would be visible goes as $M^5$. Thus more massive black hole binaries would have to be extremely rare in order to have evaded detection.

The astrophyscal reason for an upper limit is the pehenomenon of  pair instability supernovae (e.g. Farmer et al. 2019), which blows the star apart rather than leaving a black hole (or any other kind) of remnant. This likely happens for stars with initial masses of $130+\ M_{\odot}$, and means that leaving behind black holes with $M > 50M_{\odot}$ is very difficult, with even lower mass limits for stars with a metallicity more similar to the Sun, since they lose more mass in stellar winds during their lives.

For initial masses of $250+ M_{\odot}$ it is possible that the pair instability supernova mechanism ceases and direct collapse to a black hole becomes possible. In which case their might be a population of $300+ M_{\odot}$ mergers that are just below LIGO's sensitivity window. New Earth-based gravitational wave detectors like the Einstein Telescope and Cosmic Explorer aim to push their low frequency response down to a few Hz and might be capable of detecting mergers in the 300-1000$M_{\odot}$ range.

This means you could not get a merger pair between about $100 M_{\odot}$ and $300 M_{\odot}$ (unless they themselves were the products of a merger).


Does the sky get darker faster during the winter?

The further north you go, the time between sunset and darkness becomes longer, no matter the season. The reason is due to the velocity at that latitude. If there is 10 min of twilight on the equator, then there is $10\sqrt{2}$ min at 45° latitude, 20 min at 60° latitude, ...

Added:
I used the website that @barrycarter listed above and discovered that there are 2 maxima and minima during the year. The solstices have the longest twilight times and the equinoxes have the shortest twilight times.


If an object with mass were to somehow go the speed of light, would it destroy the whole universe?

Would an object with mass traveling the speed of light destroy the whole universe because it would have infinite energy / mass?

If we understand the question as a limiting process, which is the only way it makes any sense, the answer is no. For illustrative simplicity, take a spherically symmetric isolated body, so that its exterior gravitational field is the Schwarzschild spacetime. Now boost this to ultrarelativistic speeds. In the speed of light limit, the result is a linearly polarized axisymmetric gravitational pp-wave, the Aichelburg-Sexl ultraboost.

Its effect on test particles it passes is an impulse: it instantaneously bends the worldlines, but does not destroy anything. You can think of this as infinite force acting for an infinitesimally small time, making the overall effect finite and nondestructive, somewhat analogously to a particle encountering a Dirac-delta potential in Newtonian physics.

This is similar to the ultrarelativistic limit of a moving electric charge. The electric field lines are Lorentz-contracted along the direction of travel, squeezing them together (image credit: wikipedia):

In the ultrarelativistic limit, the electric field in the transeverse direction becomes infinitely strong, but it is also Lorentz-contracted to be infinitely thin. Without ignoring the magnetic field, the result is an electromagnetic plane wave with a Dirac delta profile. The gravitational case is analogous, though quantitatively different. Some details on this can be found in arXiv:gr-qc/0110032.

  My premise behind it possibly destroying the whole universe is that an object with infinite energy should also have infinite mass (E=mc^2, and infinity / c^2 is still infinity).

The concept of 'relativistic mass' is largely deprecated in modern physics. Mass is more properly related to energy and momentum via $(mc^2)^2 = E^2 - (pc)^2$.

  One thought is that it could just form a black hole instead, ...

No, that's completely wrong. It doesn't become a black hole for the same reason it doesn't destroy the universe: the gravitating body moving relativistically past a stationary observer is physically equivalent to a relativistic observer moving past a stationary gravitating body. But the observer moving relativistically obviously does nothing to the gravitating body!


Free neutrons and stellar nucleosynthesis

I have yet to find good information regarding the s-process, so I'll just talk about the r-process here.

The key difference between the two is the conditions under which they take place. The r-process takes place during supernovae, specifically those due to gravitational collapse. In a collapsing star, many of the protons and electrons are squeezed tightly together - so tightly, in fact, that a process called neutronization begins. The protons and electrons actually combine to form neutrons (and neutrinos). This creates a large neutron flux, and so there are now lots of neutrons available for the r-process.

None of this lasts incredibly long, though, because the "explosion" doesn't last long. It soon propels much of the outer material of the star out into space, and the remains become either a neutron star or a black hole. That said, an interesting follow-up might address the relationship between the r-process, neutronization and the formation of neutrons stars in more detail.

Sources:

https://en.wikipedia.org/wiki/R-process

http://www.ucolick.org/~bolte/AY4_00/week8/SNeII.html


If the Sun disappeared, could some planets form a new orbital system?

The issue here is whether pairs of planets can become gravitationally bound to each other.
In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or remain bound. This situation (finite speed when far away) corresponds to a hyperbola, so the generic case will not form a new bound system: some of that energy needs to be shed somehow.
One way this can happen is an inelastic collision, i.e. a merger. A less extreme case is a close encounter where tidal forces dissipate some of the kinetic energy. But the kinetic energy of planets is large, so a single encounter is unlikely to dissipate enough to make them bound.
Another important possibility is a three-body encounter. This is how comets get into short-periodic orbits when falling into the solar system: an encounter with Jupiter (or some other planet) leads to a gravitational "assist" that makes it bound with the sun.
In your scenario where the rogue planets fly off along their tangents it is very unlikely that any start out bound to each other since planetary orbital velocities around the sun are much bigger than the orbital velocities of a bound planet-planet pair at a large distance. Hence they will just fly apart. In theory they might have an encounter before leaving the solar system vicinity (e.g. Mercury by sheer chance sweeping past Jupiter) and that leaves a tiny chance of a three-body interaction with moons or a merger, but the chances are truly tiny. The generic case is just scattering.
Addendum: Some estimates
For a body to be bound to another body of mass $M$ it need to move slower relative to it than escape velocity $v_{esc}=\sqrt{2GM/r}$. If we assume planets move in circles we can then take their orbital velocities for all possible positions along the orbit, calculate the relative velocity, and compare it to the escape velocity.
The result is this diagram, where I have plotted the separation on the x-axis and how many times escape velocity the relative velocity is on the y-axis. The numbers denote which pair of planets I am comparing. The blue horizontal line is $\Delta v=v_{esc}$.

Looking just at the lowest relative velocities shows that the closest to being potentially bound is Saturn to Jupiter were the sun vanish just as they are moving in parallel, but even in this case Saturn has several times Jupiter escape velocity.
One could complicate it with inclinations and elliptic orbits, but it will not change the qualitative picture: the planets are not even close to being bound to each other in the absence of the sun.


For computer software, the easiest way to take a sphere (and/or hemisphere) and flatten it into a flat shape (usually a rectangle) is the equi-rectangular projection (also known as the plate carrée), because it has the simplest formula relating pixels and coordinates:

$$x = w \times \frac\lambda{360} + \frac w2$$

$$y = -h \times \frac\phi{180} + \frac h2$$

where $x$ and $y$ is the pixel point, $w$ and $h$ are the width and height (in pixels) of your map's rectangle, $\lambda$ is the longitude in degrees and $\phi$ is the latitude in degrees

This all assumes that your map puts the 0 longitude, 0 latitude point in the center. In other words, the prime meridian runs down the map's middle vertically and the equator runs down the map's middle horizontally. It also assumes that north and east are positive, while west and south are negative.

Notice that you don't even have to take the sine or cosine of anything. That's why the equi-rectangular projection is so simple. It basically just takes longitude and latitude and makes them $x$/$y$ coordinate with only an offset. The offset comes because, usually in computer programming, the origin is at the top left instead of the center of the screen.

Notice also that I negated the latitude in the $y$ formula, because most computer systems in 2D coordinates have the positive $\bf y$ direction pointing down. This is opposite of a standard graph in math, where positive $y$ goes up.

In your case, longitude and latitude will actually be right ascension and declination respectively. They are basically stellar coordinates for the celestial sphere as opposed to ground coordinates for the geode.

There are many, many other map projections possible, but equi-rectangular is the simplest.

However, your question is a little vague because you mentioned a "virtual 3D world", yet also asked how much you need to flatten it. If you already have a 3D model of a sphere in your computer graphics, you don't have to flatten it at all. You just draw the 3D model (sphere) with a texture on it. If you want to represent it as a map, then of course you have to flatten it, and equi-rectangular projection is the simplest way.


Seeing refers to turbulence in the atmosphere that distorts wavefronts passing through and results in the blurring of distant objects.

Unfortunately, there is no foolproof way of predicting what the seeing will be from night to night, though building up a database of local seeing and meteorological measurements might allow you to find some crude correlations using principal component analysis.


A moon is held together by its own gravity, and pulled apart by the tidal action of a planet. If a moon comes too close to a planet it will be ripped apart by the planet's gravity and become a ring. The closest a moon can come to a planet is known as the Roche limit, and it is dependent on the mass and density of the planet and moon.

A large planet, such as as Saturn, has a large Roche limit, and has collected many moons over its history, some of which have been ripped apart to make rings.

The Roche limit for the Earth-moon system is at a radius of about 10000 km, which is very close to the surface of the earth (about 6400km). And unlike the gas giants, terrestrial planets don't have many moons.

So with few moons, and small Roche limits, the opportunities for ring formation are rare. That said, Phobos, one of the satellites of Mars, is now so close to Mars, that it is held together as much by its rigidity as by gravity, and is slowly getting closer to Mars. Within 50 million years it will be ripped apart and form a (small) ring around Mars


How much light would be reflected from Jupiter to Europa (in Europas night)?

It's a pretty straight forward calculation of 3 factors.   Distance from the sun, apparent size and albedo.   I'm going to compare Jupiter to our full moon, since we're all familiar with that.

Jupiter averages between 4.95 and 5.45 AU from the sun.   That puts it, in comparison with our full moon (about 1.02 AU on average), that means individual square meters on Jupiter would receive between 1/23.5 and 1/28.5 the light that hits our moon.  I'm going to round and say between 1/24 and 1/28.

Brightness is also important.  The Moon is very dark with an albedo of about 12% - about the color of worn or greyish asphalt.  Jupiter's bond albedo is about .34, so it would reflect nearly 3 times more light.   That means per square meter, Jupiter is up to about 1/8th to 1/10th as bright.

The big difference is size.   From Earth, the Moon takes up about 1/2 of 1 degree of arc.   Jupiter's diameter is about 37 times the Moon's diameter.    It's distance to Europa is about 1.73 times Earth's distance to the Moon, so it's apparent size would be about 21 times the diameter (about 10 degrees of arc), the reflective area is closer to 450 times.   If you divide that by the ratio of 8 to 10 depending on how close or far Jupiter is to the Sun in it's orbit, A full Jupiter would be the brightness of roughly between 45 and 55 average full moons.   It wouldn't glow as brightly, though some of it's lighter bands would be almost half way there.   But the entire package would be 45 to 55 times as bright (ignoring any atmospheric reflection, which would make it appear even brighter).

That's probably bright enough to read under, but not so bright that it would hurt your eyes if you stared at it.

This only answers your question for a full jupiter, but the ratio, 45 to 55 times brighter should be about right for all the phases.   And about 450 times larger than an average Moon.   The Moon's size varies fairly significantly from Earth, though we don't always notice.   Europa has a much more circular orbit around Jupiter, so it's size would stay about the same from the surface of Europa.


Why doesn't the solar azimuth vary uniformly

The Sun's hour angle, measured around the celestial equator, changes at a nearly constant rate.
The gnomon of a sundial is aligned with the polar axis to take advantage of this.
The celestial equator is oblique to the horizon except at the Earth's equator and poles.

The lines of azimuth converge at the observer's zenith and nadir, so the Sun's azimuth changes more rapidly near one of those points (e.g. summer midday, winter midnight) and more slowly near the horizon.
The amplitude of this fluctuation depends on the Sun's declination and the observer's geographic latitude.

For an observer in the tropics, on a day when the Sun passes directly overhead, its azimuth swings abruptly from 90° to 270° at noon.
For an observer near the north or south pole, the Sun remains near the same altitude all day, and its azimuth rate stays near the average 15°/hour.


How to deal with data that's billions of years old?

The redshift you're referring to is a cosmological redshift, which is fundamentally different from a Doppler redshift. At the heart of it, it has nothing to do with neither time, nor distance. It is a consequence of light traveling through an expanding space, no more and no less.

Because space expands, the emitting source is carried away to a certain distance from us, so we can relate the redshift to a distance. And because it takes time for the light to reach us, we can relate the redshift to a time. But this requires a model for how the Universe has expanded through time, and how light travels.

The model that best describes our reality is the so-called standard cosmological model, or ΛCDM ("Lambda-CDM") model, which is a mathematically based, physically motivated, and observationally constrained model. It uses the so-called Friedmann–Lemaître–Robertson–Walker metric, which is a description of space itself, the Friedmann equations, which describes how space expands, and an equation of state, which describes the relation between pressure and density. The various parameters that enter the equations are then observed and constrained in a number of different ways.

With this model, we can relate the distance to an object, and its age, to the redshift we observe.


Planetary observations: what to expect from the Powerseeker 50AZ?

Currently, Mars is very far from the Earth, so it appears to be very small. I doubt that any amateur telescope will show any details on Mars in this condition. Perhaps when it comes to opposition and is close to Earth, you might be able to see some faint markings with your scope.

Jupiter should show a sharp disk and nearly pinpoint moons with the 12.5 mm eyepiece (48 power). You should see two prominent bands on Jupiter.

The 4 mm eyepiece is probably poor quality, and since it provides 150 power, is too much power for a 50 mm scope. In ideal conditions, the general rule is 100 power for a 50 mm scope (2 times the scope size in mm). The clarity of the sky is not what makes a sharp image. There is a condition named "seeing" that indicates how sharp the viewing is. On some nights, the seeing is poor, so the images will be soft (not sharp) and perhaps shimmering, as if you are viewing the object through a stream of water. (Essentially, you are, except it is miles or kilometers of air!)