What bright body is next to the Moon in these photos of the ISS transiting the lunar disk?

Given the appearance of the moon, that must have been photographed on the weekend of the 21-22 (the moon was nearly full on the 28-29)

The object is Jupiter, and from the position I'd say the photograph was taken at about 0400 UTC on the 21st July 2018.

I think Antares can be dimmly seen towards the left of the photo, and perhaps alpha and beta Libra area visible above and below the moon, but I can't see any other stars.


No. These words are English, not Greek.  "Periapsis" means the point on the orbit when the two bodies are at their closest.  It doesn't matter if this good Greek or bad Greek, it is correct English.
Apsis actually derives from "arch" and indicates the two points where the orbital curve is most "arched": ie the points of greatest curvature.  There is the "periapsis": The close point of greatest curvature and the "apoapsis" "the distant point of great curvature".
There is no such thing as a "misconstruction".  The idea that a word does not mean what it means is obviously nonsensical


Number of lenslets in wavefront sensor array

Generally speaking as many as possible because the number of lenslets determines the lateral wavefront resolution. But in reality there are a few factors to be considered.
Wavefront sensor
Say your lenslet array is of focal length $f$ and subaperture size $D$, given a maximum desired detection angle $\alpha$ (in your case is $2'' \approx 1\times10^{-5}\, \text{rad}$), there is the following relationship:
$$
2'' = \alpha \approx \tan \alpha = \frac{\Delta x}{f} \leq \frac{D}{2f}. \tag{1}
$$

To ensure a good resolution and reduce cross-talk effect, $D$ is chosen to be at least a few image sensor pixel size. In other word, for a given image sensor to measure the seeing angle, the number of lenslets are limited to a maximum value.

Usually you do not design the lenslet arrays; those are pre-designed, e.g. https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2861. So Eq. (1) serves as a metric to select between different microlens array designs.

For AO applications, the primary goal of wavefront sensors is the measurement speed rather than the lateral resolution. The sensors need to be fast. The smaller the number of pixels, the smaller the sensor latency is, resulting in a low number of pixels, and hence low number of lenslets.

Adaptive optics (AO) system
$w \in \mathbb{R}^m$: Measurement wavefront of a lenslet array of number $m$.
$c \in \mathbb{R}^n$: Correction wavefront of the deformable mirror that has $n$ number of actuators.
An adaptive optics (AO) system works by solving a linear system to output $c$ given $w$:
$$
w = A c
$$
where interaction matrix $A \in \mathbb{R}^{m \times n}$ is pre-calibrated before usage.

To compute $c$ in a least squares sense, it requires $m \geq n$, meaning an over-determined system, mentioned also by @nflemming2004.

As for the aperture sizes of the two, it does not really matter that much because the conjugate lenses/mirrors in your optical system can scale down/up the apertures. That said, the sub-aperture of the wavefront sensor (i.e. $D$) and the actuator cover area $D_{\text{deformable_mirror}}$ do not have to be equal.

If $m \gg n$ (as in most cases), as you can imagine, the extra number of lenslets only serves to make a robust estimate, and is wasted from AO's perspective because of insufficient correction degree-of-freedom.

Example
Using WFS30-7AR as our wavefront sensor, that $f = 5.6 \,\text{mm}$ and $D  = 150\,\text{um}$, given Eq. (1) we know:
$$
2'' \leq \frac{D}{2 f} = 1.34 \times 10^{-2} \, \text{rad} \tag{satisfied}
$$
The number of lenslet is $m = 80 \times 80$, far larger than $n = 140$ the actuator number of deformable mirrors (say https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3208).
So for this design, we conclude it satisfies the requirement of this question.


Is there an element of chance/chaos in stellar evolution?

If you want to go into truly random stuff, you should look into quantum. The internal processes inside a star are not random, however they are bviously undetectable. Some factors other than mass and metallicity are temperature of the original dust cloud, or angular velocity. The temperature obviously affects the brightness, while the angular velocity will affect the gravity. I am not an expert, but I'd assume that any minor perturbation inside the star could cause a butterfly effect and result in, say, solar flares, so on average there would be stars who would be losing more mass than others with no  particular reason.
I assume there would be other factors linked with more complicated physics, too.


In how far is Hipparcos data still being used?

The Hipparcos results can be used as an earlier epoch to improve the astrometry produced by Gaia. This was the linchpin of data release 1 (DR1), but less important in DR2 and DR3, because the astrometry of Gaia alone is so much better in general.
An exception to this is bright stars where much data is still missing in DR3, but can be found in the Hipparcos catalogue.
Apart from precision-based concerns, the earlier Hipparcos data, combined with Gaia, can be used to look for changes in proper motion: i.e. accelerations (see Brandt 2018).


When graphing radial velocity measurements, whats the relationship between days and the orbital phase

If the orbital period is $P$ (in units of days) and the time of periastron passage (a specific date when the stars are closest together) is $T_0$ (generally given as a Julian Date, or JD), then the conversion from orbital phase $\phi$ to JD (days) will be:
$$ JD = T_0 + \phi P $$
In other words, $\phi$ is just the fraction of the orbital period, so multiplying it by the period gives the number of days since periastron within the orbit.
If the orbit is circular, there is no specific periastron date so $T_0$ can be set to any arbitrary date, but is generally chosen to be a date when the radial velocity is zero (or at the systemic velocity if that is non-zero) and crossing from negative to positive radial velocities.


Stellar Classification Table and Star RNG

I found a table that was based on data from 2003 for Stellar Classification, and I can't get to the given data source ...

The website that was the source of the data was last active  in 2006. I've replaced the broken URLs with working URLs. I wouldn't recommend relying on data that old, there's likely a newer source.

Original (404) Text and URLs:

  Credit:

  The table data was obtained from SpaceGear.Org's SCTable version 3.0 09/23/2003. The following data source was obtained from their SCTable web page:

Original Text and repaired URLs:

  Credit:

  The table data was obtained from SpaceGear.Org's SCTable version 3.0 09/23/2003. The following data source was obtained from their SCTable web page:

The 3rd links says that they derived the data from:

  Absolute magnitude was derived from the standard Hertzsprung-Russell (HR) diagram. The book Stars and Their Spectra by James B. Kaler was used as a resource for deriving the absolute magnitude for each possible luminosity class. Classifications N, S, WC, WN, C, and R are mapped as follows:

N and S are equivalent to M.
WC and WN are equivalent to O.
C spans G4 to M9.
R spans G4 to M0.

  Temperature was derived from the relationship between luminosity class and spectral class. The book Stars and Their Spectra by James B. Kaler and Mitchell Charity's work on star colors was used as a resource for deriving temperature for Main Sequence (V), Supergiant (Ia), Giant (II), and Dwarf (D) luminosity classes. Luminosity classes Ia0, Ib, III, IV, and VI are mapped as follows:

Temperatures of Ia0 and Ib are equivalent to Ia.
Temperatures of II is equivalent to III.
Temperatures of IV and VI are equivalent to V.

  Color was derived from Mitchell Charity's work on star colors. Each designated color is defined with at least one field (R, G, or B) at full saturation (value equals 255). Actual color values for display would be attenuated across all color components based on the star's apparent magnitude in a given situation.

  Mass was derived from many sources including Astrophysical Formulae by Kenneth R. Lang. Many rows are heavily interpolated from known data on star mass.

See how I fixed all the URLs so they work correctly. If there is more from that webpage that you wanted sources for then you can use the same method. Since I was able to get to the raw CSV data that should be easier to use in your program.

See also Wikipedia's webpage on Stellar Classification.

An example of a website providing similar data (from 2015) is: "The Mag 5 Star Catalog" or see: "The HYG Database", his latest update to GitHub is a couple of years ago.


Will the JWST be affected by dust at L2 (gegenschein?)

The Wikipedia article cites "Zdeněk (1962)" for the statement that the dust responsible for the Gegenschein has a possible concentration at L2. I haven't been able to obtain that paper, but I can't really see why that would be the case, since L2 is not dynamically stable.

However, the dust consists of millimeter-sized grain (see e.g. this APOD image), which is quite large for dust. Such large grains probably has a highly anisotropic "phase function", with a high preference for backscattering. Since when you look from Earth toward L2 you have the Sun in your back, you will thus see an increased brightness compared to other directions, even though the density of that dust is the same.

If this is the case, then placing JWST in L2 is no problem (except when it looks exactly away from the Sun, but that would be a problem anywhere in the ecliptic).


Ok, not sure where you get 200m from, the maximum baseline for the VLT interferometer is 130m. Nevertheless, your question still stands.

There are a number of reasons. Firstly, it is important to understand that with relatively limited number of baselines (4 VLTs = 6 baselines) one does not get actual images out of the interferometer, but rather "visibilities" that enable one to work out whether sources are double, extended or whatever. Second, the interferometry only works at infrared wavelengths. This is a technical issue that might be solved in the next decade or so, but a general rule of thumb is that the longer the baseline, the longer the wavelengths you will be limited to use. Thirdly, the light gathering power is limited to the light that can be gathered by the constituent telescopes. I think this also limits the kinds of instruments you can use since the required delay lines and optical components of the system make everything inefficient. Fourth, there are limitations of the abilities of optical/IR interferometers caused by turbulence in the atmosphere. This limits them to very small fields of view and the performance does not reach the ideal case. It also means that you are limited to bright objects because individual exposure times need to be very short.

The E-ELT is designed to work at optical and infrared wavelengths and will gather enormous amounts of light. Not everything in astronony needs spatial resolution; sometimes what you need is just a bigger light bucket. Examples include: deep imaging of the universe, gathering spectra from faint objects; accumulating enough photons to do transmission spectroscopy using exoplanet transits.

It's horses for courses really. At optical and IR wavelengths the future will likely be big telescopes in space and even bigger interferometers in space.


Could the mineral composition of planets explain the absence of megastructures?

In our paper we just used Mercury because it was convenient (close, airless, low binding energy). But we did not try to optimize the Dyson sphere design at all. It turns out that one can likely construct a far more efficient sphere out of very thin aluminium foil reflectors balancing light pressure and solar gravity to stay in place. That only takes the mass of the asteroid Vesta or so. Plus, were the reflectors manufatured out in the asteroid belt one could light-sail them into close orbits with modest effort.

Aluminium is just one possible material. Essentially any metal works for the reflectors (the trick is to keep the mass low enough), while energy gathering subsystems could be based on semiconductors or even heat turbines - silicates and water are not uncommon in the solar system, and presumably similarly available in others. Even if the inner region has been swept clean by a hot jupiter parts can be manufactured and spiraled in from the outer system.

So unless we expect even asteroids to be rare (a pretty implausible idea) there is likely plenty of building material. As an answer to the Fermi paradox lack of resources is not reliable enough.


Would a low albedo substance warm up Mars?

Of course! Actually this is one of the possible ways we have in order to make Mars friendlier
to humans. This technique is called Terraformation and it inludes the altering of the Mars surface's albedo. However, the feasibility of this plan is yet questionable and there are maybe better ways to achieve this increase in the temperature like CO$_2$ emissions. Here is a nice article discussing exactly this subject.


Decreasing rate of Earth's rotation: where does the power go?

You're correct in assuming the net angular momentum of the system in question here will remain constant. The Moon's orbit around Earth is responsible for the slowing of Earth's rotation. This effect is extremely small.

The decrease in Earth's angular momentum is transferred to the moon, which resultantly sees it's orbit accelerate. This acceleration also causes the moon to move further and further from the Earth. This trend will continue until they reach a common speed.


Do comet tails increase as comets move away from the Sun?

The only data I could find indicates that the brightness is similar approaching and receding:

Source
This is for the unusual almost "tailless" Comet Wirtanen .


There's a mistake in the book (see page 280 on google books): the values for $\lambda_{ecl}$ and $\phi_{ecl}$ should be swapped. Your result is correct:
$$
-6338.688^\circ\text{ mod } (360^\circ) = -218.688^\circ,
$$
which is the value that is erroneously listed under $\phi_{ecl}$.


Is it only possible to look at solar systems with stars at least as old as ours to be able to find intelligent life?

Assuming the speed it takes to development intelligent life is the same in all solar systems,

We don't know if this is true.

We don't know if:

Life has started on more than one planet in the Universe.
How long, on average, it takes for life to start.
Whether life normally evolves past the "bacterial slime" stage to form complex multicellular life, and how long this takes.
Whether multicellular life normally develops intelligence, and how long this takes.
How long intelligent life survives, on average.

So we don't know what time it takes to develop intelligent life. However, planets are the same age as their stars, so it is reasonable to suppose that older stars are more likely to harbour intelligent life.

Moreover, visiting any extrasolar planets is too hard with current technology. Stars are just too far away.


Yes, the are angles in radians. With "i=0.5" to make things clearer:

$p=0$ is the right half illuminated
$p=\pi/2=1.57$ is top half illuminated  (rotated anticlockwise by 90 degrees)
$p=\pi=3.14$ is left half illuminated  (rotated 180 degrees)

and so on.
The r parameter is the rotation of the moon, so $r=0$ is "lunar north up" $r=pi/2$ is rotated anticlockwise by 90 degrees.


Translating a zenith position to the nadir

In Astronomical Algorithms, Second Edition, Jean Meeus shows us how to convert local coordinates to equatorial and vice versa.
Formulas 13.5 and 13.6, p. 93:
$$ \mathrm{tan}\ A = \frac {\mathrm{sin}\ H}{\mathrm{cos}\ H\ \mathrm{sin}\ \phi - \mathrm{tan}\ \delta\ \mathrm{cos}\ \phi} $$
$$ \mathrm{sin}\ h = \mathrm{sin}\ \phi\ \mathrm{sin}\ \delta + \mathrm{cos}\ \phi\ \mathrm{cos}\ \delta\ \mathrm{cos} H $$
Unnumbered formulas, p. 94:
$$ \mathrm{tan}\ H = \frac {\mathrm{sin}\ A}{\mathrm{cos}\ A\ \mathrm{sin}\ \phi + \mathrm{tan}\ h\ \mathrm{cos}\ \phi } $$
$$ \mathrm{sin}\ \delta = \mathrm{sin}\ \delta\ \mathrm{sin}\ h - \mathrm{cos}\ \phi\ \mathrm{cos}\ h\ \mathrm{cos}\ A $$
where $ \alpha $ is the right ascension in degrees, $ \delta $ is the declination in degrees, $ \phi $ is the latitude (+ North, $ - $ South) in degrees, $ H $ is the hour angle measured westwards from South in degrees, $ A $ is the azimuth in degrees, and $ h $ is the height of the object in degrees. Please note that Meeus measures azimuths from the South heading East (90°) then North (180°) and West (270°).
With these formulas, you can find the required position of the point you want to know about.
In order to get the nadir, I would just put $ h = -90° $. Another way would be to find the position of the object at zenith, add/remove 180° to its right ascension, and change the sign of its declination.
Hope this helps!
Clear skies.


Can we see asteroid 1998 OR2 with unaided eye?

According to the page on The Sky Live the asteroid is currently at a visual magnitude of +11, which is also what Sky Safari says. It is not due to get much brighter so no, it will not be visible with the naked eye. According to the predicted light curve, there is a chance it will get to naked eye brightness (from a dark sky site) around April 2079.... So there is quite a wait! And these predictions are based on current models which can change with new information.

You can use theskylive to check on things like comets and asteroids, or if you prefer a mobile app, I always recommend Sky Safari


What would a moon with a six-month orbital period look like from the earth?

A moon with a sixth month period would be about 1.3 million km from Earth. That puts it close to the edge of the Earth's Hill sphere, and probably isn't stable in the long term.  So the biggest thing you might notice is "goodbye moon" (and having the moon in an Earth crossing orbit wouldn't be pleasant in the longer term, a collision would melt the crust)
Your questions:

One day.  The moon's motion is mostly due to the rotation of the Earth.
The moon would be very distant and small but moonrise would happen just as it does now.
Tides would be much reduced.
The cycle from new moon to full moon would take much longer. Nights at full moon would be much darker than they are now. The moon is less likely to be tidally locked, so might show different faces at different times.


Was is the angle needed to see a glory (a rainbow-like effect) from space off the Earth?

A glory is a rainbow-colored bullseye appearing directly opposite the Sun due to diffraction by water droplets of a certain size.
From an airplane, you can sometimes see this effect surrounding the airplane's shadow.
As the NASA article explains, the polar orbiting Terra satellite's scan pattern transformed the circular rings into elongated ovals.


What is that donut-shaped object I see in my telescope?

This is because your image is not in focus. So you’re seeing the shadow of the secondary mirror.
You should have a knob near the eyepiece, that you can turn to adjust focus. You need to turn it, one way or another, until the image is as small as possible and as bright as possible.
Good luck!


You have a clear view of the history, and it wasn't an independent confirmation, it was simply the best evidence.  Whenever one talks about the discovery of dark matter, one should start with Zwicky, and move to Rubin.  One might do something similar with the Big Bang, starting with Hubble, and moving to Penzias and Wilson.  Normally we give credit both to the initial discovery, and to the confirmation, but it is the most conclusive evidence that generally gets associated the largest credit.  After all, it was Edgar Allan Poe who first described the Big Bang model, but he gets no astronomical credit because he provided no evidence.  Zwicky certainly did better than that, so does deserve (and should receive) much credit, but there were still many doubters in his day, much as with Hubble and the Big Bang.  Perhaps Zwicky's role contributed to the reason that Rubin never received the Nobel prize, I hope it wasn't simply the unspeakable reason, but of course it's too late now for both of them.


The Wikipedia Precession article explains this.

Precession of the Earth's axis is caused mainly by the Sun's gravity acting on the oblate spheriod of the Earth. The Earth bulges at the equator and the force of the Sun's gravity on the bulge makes the axis wobble like a child's spinning top - the axis traces out a cone shape in space. It has no effect on the Earth's orbit.

Perihelion precession is caused by the gravity of Jupiter and the other outer planets.


Why is Chandrasekar Limit expressed relative to Sun?

The mass of the sun is just a unit of convenience in a astronomy. The sun's mass and luminosity, in particular, are relatively easy for us to measure precisely, and when we're talking about stars, provide a convenient scale where the numbers won't be too "astronomical" (be too high a power of $10$ to picture easily). You could derive the Chandrasekhar limit in grams, kilograms, or slugs from the relevant physics, if you wanted. The relevant equation (from the Wikipedia article) is:
$$M_{\mathrm{limit}} = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left(\frac{\hbar c}{G}\right)^{3/2} \frac{1}{(\mu_e m_\mathrm{H})^2},$$
where $\mu_e$ is the average molecular weight per electron (stellar composition dependent), $m_\mathrm{H}$ is the mass of hydrogen, and $\omega_3^0$ is a numerical constant that is approximately $2.018236\ldots$.

It should be noted that the Chandrasekhar limit is a limit on the mass of the final white dwarf, not of the object that will produce the white dwarf.


How will they know when to start taking the picture of the black hole at the center of the Milky Way?

The Milky Way's central supermassive black hole (SMBH) is feeding, albeit at a very low level. Radio emission from the accretion disk (and/or weak jets) is responsible for the long-lived "Sgr A*" radio source.
Here is a paper from 2000 (Falcke et al.) arguing that VLBI (as used by the Event Horizon Telescope) should be able to image the "black hole shadow", based on the known sub-mm and mm-wave emission. And in fact the EHT has been observing the Milky Way's SMBH.
As I understand it, the real reason we haven't seen a formal, published detection of the Milky Way's SMBH by the EHT is that its emission is highly variable on short time scales (e.g., minutes to hours). In the case of M87's SMBH, the variability of the (sub-mm and mm-wave) emission is slow (days to weeks), so they could combine observations taken over several hours and two nights in April of 2017 under the assumption that it was all of the same static configuration. Figuring out how to properly account for the short-term variability of the Milky Way's SMBH emission is much more difficult, which is why the (relatively) easier case of M87 was solved and published first.
See also Rob Jeffries' answer to this physics.stackexchange question.
Edited to add: Unfortunately, I don't think there's any validity to the idea that we can track incoming "food" and predict future accretion flares for the Sgr A* SMBH with any useful accuracy. There was some excitement a few years ago when a group reported detection of an apparent gas cloud ("G2") on an orbit that would take it in to about 2000 Schwarzschild radii from the SMBH at pericenter (in 2014), possibly allowing it to be tidally shredded and increasing the accretion rate. But as a review article published in 2013 pointed out, "The actual free-fall time scale from ∼2000 $R_s$ is roughly one month and the viscous time scale could be anywhere between months up to hundred years depending on the viscosity parameter $\alpha$."
And in fact the actual pericenter passage produced... nothing much at all. There's a discussion of the "fizzle" here: "with the majority of the simulation parameters used, only 3–21% of the material Sgr A* accreted from 0–5 years after periapsis is from the cloud".
So in the one case where potential "food" was identified and tracked, one couldn't be sure in advance whether the possible increased accretion would happen on timescales of months to years, and so far nothing significant has happened. I very much doubt the EHT team is basing their observing schedule on this kind of thing.


Accretion of in-falling material for a young main sequence star

At the end of the protostar phase a vigorous outflow from the star develops called the T-Tauri Wind and this could cut off accretion.  Eventually, it develops into a normal star and the strong wind dies down.  Material that was not totally blown away could then continue to  fall down and be accreted


We know that two celestial bodies rotate around a center of mass.

"We" do not know that. Rhetorical question: Pick up an apple and drop it at high noon. Does it fall towards the solar system barycenter? The answer is of course "No." It falls away from the barycenter, towards the Earth.

I'll steal a graph from an answer I wrote to a question at the sister site, physics.stackexhange.com, Do the planets really orbit the Sun?:

Given the above graph, it appears that it's much better to say that Venus orbits the Sun (red curve) rather than the solar system barycenter (black curve). So why do people consistently use a barycentric frame rather than a heliocentric frame? The answer is simple: The barycentric frame is the frame in which the equations of motion undertake their simplest form. This is particularly so when one wants to incorporate some of general relativity into the equations of motion.

Our solar system is well-behaved. Other star systems aren't quite as well-behaved. For an example of a hypothetical system that is extremely far from well-behaved, I suggest you read Saari and Xia, "Off to infinity in finite time," Notices of the AMS.


How has the Earth's orbit changed over hundreds of millions or billion of years?

What governs the Earth's orbital period is its orbital angular momentum and the mass of the Sun. Two events have certainly changed the Earth's orbital period (a) whatever collision formed the Moon and (b) the continuous process of mass loss from the Sun. A third possibility (c) is that tidal torques from the Sun have increased the angular momentum of the Earth.

Given that (a) probably happened sometime in the first tens of millions of years and likely did not alter the Earth's angular momentum greatly  - it depends on the speed, mass and direction of the impactor and the amount of mass lost from the Earth-Moon system - I will ignore it.

(b). It seems, from observations of younger solar analogues, that the mass loss from the early Sun was much greater thanthe modest rate at which it loses mass now via the solar wind. A review by Guedel (2007) suggests a mass loss rate over the last 4.5 billion years that increases as $t^{-2.3}$ (with considerable uncertainty on the power law index), where $t$ is time since birth, and suggests an initial solar mass between 1% and 7% larger than it is now.

Conservation of angular momentum and Kepler's third law means that $a \propto M^{-1}$ and $P \propto M^{-2}$. Therefore the Earth's orbital period was 2-14% shorter in the past due to solar mass loss, but has been close to its current value for the last 2-3 billion years.

If the solar wind power law time dependence is very steep, then most of the mass loss occurred early, but the total mass loss would have been greater. On the other hand, a lower total mass loss implies a shallower mass loss and the earth spending a longer time in a smaller orbit.

(c)
The tidal torque exerted by the Sun on the Earth-Sun orbit increases the orbital separation, because the Sun's rotation period is shorter than the Earth's orbital period. The Sun's tidal "bulge" induced by the Earth applies a torque that increases the orbital angular momentum, much like the effect of the Earth on the Moon.

Quantifying this is difficult. The tidal torque on a planet from a the Sun is
$$ T = \frac{3}{2} \frac{k_E}{Q} \frac{GM_{\odot}^{2} R_{E}^{5}}{a^6},$$
where $R_E$ is an Earth radius and $k_E/Q$ is the ratio of the tidal Love number and $Q$ a tidal dissipation factor (see Sasaki et al. (2012).

These lecture notes suggest values of $k_E/Q\sim 0.1$ for the Earth and therefore a tidal torque of $4\times 10^{16}$ Nm. Given that the orbital angular momentum of the Earth is $\sim 3\times 10^{40}$ kgm$^2$s$^{-1}$, then the timescale to change the Earth's angular momentum (and therefore $a$ and $P$) is $>10^{16}$ years and thus this effect is negligible.


What scientific evidence is there to support or refute the Iron Sun Hypothesis?

If the sun was principally composed of Iron, then this would be apparent in the spectrum of light from the sun (which is how we know what stars are made of) (but see comment by @Keith Thompson below, the spectra only tells you about the surface, rather than the core).  The fact that all solar physicists bar one seem to think it is made up of hydrogen and helium strongly suggests this is not the case.

If the sun were made of iron, then I suspect that would require a revision of our understanding of nuclear fusion, as iron is the product of the final fusion reaction that stars can support, unless of course the theory is that the sun is no longer undergoing fusion reactions and all we have been seeing for the last 4 billion years+ is the residual heat of the core.

Apparently heliosiesmology suggests there isn't that much iron in the suns core (although the physics is beyond me, so caveat lector).

As a further point, if the sun is primarily made of iron, it becomes difficult to explain why the planets (especially the gas giants) have so little of it.  It is reasonably straightforward to explain why the inner planets have lost much of their original hydrogen and helium, but rather more difficult to explain how the outer planets could have become hydrogen enriched.

Update.  Dr Manuel's website claims that "About 99% of the sun has the same composition as ordinary meteorites.".  The density of carbonaceous chondrite meteorites is about 2100-34700 kg/m3 (with porosity being an issue).  The mean density of the sun is 1410 kg/m3.  Given the temperature and pressure within the suns' core, if it had a similar composition to a meteorite its' porosity would be zero, and hence even more dense than an ordinary meteorite.  There is a problem with accounting here, if 99% of the sun had a composition similar to that of a meteorite, it would have to be full of voids to have the suns observed density.  The figures just don't add up as far as I can see (although how materials may behave at ultra high temperatures and pressures means intuition may be invalid here).

For Manuel to be right, pretty much all we know about stars must be wrong.


How far away are the events that caused the gravitational waves that have been detected?

Yes, it is possible to calculate (within an error range) the distance of observed gravitational wave events.  It is known that a variety of parameters will affect how the amplitude and frequency of the observed gravitational waves will change over time as recorded in the "chirp" event from the interferometers: the parameters include distance of the event, the mass of each of the colliding objects, the angular momentum of each of the colliding objects, the orientation of the objects' angular momentum vectors with respect to each other and with their orbital plane.  With general relativity, you can build a model that calculates the expected "chirp" given a value for all these parameters; when a chirp is observed, it is possible to determine the combination of these parameters that result in a chirp that best matches the observation.

The effect of a larger distance parameter is to decrease the amplitude of the expected waves from colliding objects of a given mass, as well as to "slow down" the entire event due to cosmological red shift.

From GWTC-1: A Gravitational-Wave Transient Catalog of Compact Binary Mergers Observed by LIGO and Virgo during the First and Second Observing Runs

  Gravitational waves from compact binaries carry information about the
  properties of the source such as the masses and spins. These can be
  extracted via Bayesian inference by using theoretical models of the GW
  signal that describe the inspiral, merger, and ringdown of the final
  object for BBH [23–30] and the inspiral (and merger) for BNS [31–33].
  Such models are built by combining post-Newtonian calculations
  [34–38], the effective-one-body formalism [39–44], and numerical
  relativity [45–50].


How the watches are showing perfect time?

Those 23hrs, 56mins, 4.1secs are actually the sidereal rotation period of Earth. It is how much time the Earth needs to perform one rotation relative to the stars. But that is not exactly the same as solar time. By the time the Earth has completed one rotation, it has also moved a little forward in its orbit around the Sun. Thus, it needs a few more minutes of rotation to catch up so the Sun is at the same location of the sky.


Why Comet C/2020 F3 Neowise returns even it's orbit is near- parabolic?

Comet naming conventions limit "periodic comets" to those comets whose orbital period is less than 200 years. Comets in elliptical orbits with a period of greater than 200 years are classified as non-periodic. Comets in parabolic or hyperbolic trajectories are also classified as non-periodic (and they will never again approach the primary).


How can "Geysers" on Europa reach heights of 100km?

Technically those aren't "geysers" on Europa, they're cryovolcanos.    Though that definition may be a bit imprecise as well, but the 100 km eruptions on Europa probably have more in common with volcanic eruptions on Earth than geysers.   Europa undergoes significant tidal flexing which, in combination to it's solid icy crust, could be compared to explosive volcanic eruptions on Earth and not the cyclic geysers, which are far smaller.

Volcanoes on Earth can shoot material high into the atmosphere on Earth but volcanic ash isn't a good example because it's not so much "shot" into the air as it rides rising hot air to altitudes 30 and 40 KM high.

Magma fountains on Earth can reach a few hundred meters and Tephra (anything larger than Ash, from pebbles to large rocks/boulders) can be found as far as a mile from a volcano and can exit the volcano at several hundred meters per second.   From the Tephra article:

  Blocks and bombs as large as 8-30 tons have fallen as far away as 1 km
  from their source (Bryant, 1991). Small blocks and bombs have been
  known to travel as far away as 20-80 km (Scott, 1989)! Some of these
  blocks and bombs can have velocities of 75-200 m/s (Bryant, 1991)

Europa's Tidal-flexing can cause significant pressures within it's icy crust.  See here.  and here.

So, if we draw an Earth-Europa comparison, and assuming my math is right, using 1/2 at^2, all you'd need for material to reach 100 km on Europa is an exit velocity from the ground of a bit over 500 meters per second.   Or, some 20%-30% faster than a 22 caliber bullet.   That's faster than material usually leaves an Earth Volcano, but not all that much faster.   When frozen, Europa's outer icy crust is probably very hard and brittle.  In pressure spots where the pressure exceeds the structural integrity of the icy crust, it's not all that surprising that the results are pretty explosive, leaving the surface at 500 meters per second.

Ice is very strong and brittle and when cold enough, where the Earth's crust, while rocks are quite brittle and strong, the crust overall can also be quite malleable, especially a few miles down in hotter temperatures under an erupting volcano.

Very cold ice is resistant to cracking, but under enough pressure, when it finally does crack it can release explosive energy.   Ice cracks in water due to differential expansion, (Silicate rock doesn't expand with temperature nearly as much as frozen ice does), and while the cracking of an ice cube dropped into water might not seem like much, a big enough ice cube dealing with enough tidal heating can have explosive results.

While this fun little video might seem unrelated (turn down your volume before watching), it demonstrates the explosive force that rapid warming can have on a hard block of ice.  This isn't steam pressure, as the steam is all escaping.  This is the inside of the block expanding faster than the outside can and the result is impressive.   https://www.youtube.com/watch?v=epkRd-w3TGw   So, with that in mind, explosive energy from a hard, cold 10 mile thick icy surface under the influence of tidal flexing shouldn't be all that surprising.


Do we get any benefit from being in a galaxy?

I've had my eye on this question for some time, but I haven't had the time to answer it. Sorry for not getting to it sooner.

As zibadawa timmy said, you'd be hard-pressed to find the materials necessary for a star system if you're outside a galaxy. For one thing, there aren't any stellar nurseries in the intergalactic void (well, none that we know of. But I'd guess it's highly unlikely that they exist plentifully). So you have no hydrogen or helium, or dust, or other elements - in fact, you have nothing that you could use. That's a wee bit of a problem.

There is a way you could get around this, though - if your star system is in a globular cluster. They're groups of stars that orbit a galaxy. We don't know just how they got out there. They could be from dwarf galaxies, which then lost them, or they could have been partially chucked out of the larger galaxy but stayed in an orbit. The trouble is, as Wikipedia says,

  However, they are not thought to be favorable locations for the survival of planetary systems. Planetary orbits are dynamically unstable within the cores of dense clusters because of the perturbations of passing stars. A planet orbiting at 1 astronomical unit around a star that is within the core of a dense cluster such as 47 Tucanae would only survive on the order of $10^8$ years.

Yep. Not a great place to live.

But I take it you didn't intend for our solar system to be in a globular cluster, so I'll go back into intergalactic space. It's not actually a vacuum - not that most parts of outer space are. There's a lot of plasma and hot gas there (the intergalactic medium), sometimes getting to a toasty $10^5$ or $10^7 \text { K}$. Could it harbor a stellar system? That depends on just what material the star brings with it. If planets haven't formed around it yet, and the star has a protplanetary disk, then perhaps that disk could dominate over the intergalactic medium. Remove the ICM and maybe planetary formation could proceed normally.

What happens then? Well, there would be some interesting effects. For example, the Oort Cloud might not have become spherical, because it was partly deformed by the galactic tide. Here, tidal forces probably wouldn't be as strong (although this depends in part on just how far away we are from the galaxy). We probably wouldn't have a lot of negative effects, because we don't have a lot of extreme benefits here in the Milky Way. Our galaxy does not appear to have an AGN, there aren't any monstrous black holes nearby, and we aren't undergoing any catastrophic galactic collisions. And the Milky Way-Andromeda collision won't be too bad.

We probably wouldn't be in too bad of a position if we were in intergalactic space. There's not a lot of benefits from being in a galaxy - well, no significant ones, at least. Well, besides star formation depositing all the materials we need for the solar system to form. That's a biggie.


Yes, the Earth does receive energy from the stars, but not much. The effective temperature of the night sky is about 3 Kelvin which is not much more that that of the cosmic microwave background at 2.7 Kelvin. The difference being due to the total energy of starts etc.(room temperature is ~295 kelvin)

The coldness of the night sky is interesting in itself see Olber's Paradox


How do scientists know if an Earth-like planet is really Earth-like?

The earth similarity index (ESI) is a weighted geometric mean of four similarities.

The formula documented on
http://phl.upr.edu/projects/earth-similarity-index-esi
(as of march 23, 2014) should be adjusted a bit, since
$n$ should be the weight sum instead of the number of planetary properties.
With the weights provided on the site we get

$$ESI =\left( \left(1-\left | \frac{r_E - r_P}{r_E + r_P} \right |\right)^{0.57}\cdot
 \left(1-\left | \frac{\rho_E - \rho_P}{\rho_E + \rho_P} \right |\right)^{1.07}\cdot\\
 \left(1-\left | \frac{v_E - v_P}{v_E + v_P} \right |\right)^{0.70}\cdot
 \left(1-\left | \frac{288\mbox{K} - \vartheta_P}{288\mbox{K} + \vartheta_P} \right |\right)^{5.58}\right)^{\frac{1}{7.92}},$$
with $r_E=6,371 \mbox{ km}~~$  Earth's radius, $r_P$  the radius of the planet,
$\rho_E=5.515\mbox{ g}/\mbox{cm}^3$ Earth's bulk density, $\rho_P$ the bulk density of the planet,
$v_E=11.2 \mbox{ km}/\mbox{s}~~$ the escape velocity on the surface of Earth, $v_P$ the escape velocity on the surface of  the planet, and
$\vartheta_P$ the surface temperature of the planet; the weight sum is $0.57+1.07+0.70+5.58=7.92$.

Mars as an example:
With $r_P=0.53 r_E$, $\rho_P=0.71 \rho_E$, $v_P=0.45v_E$, $\vartheta_P=227\mbox{ K}$ we get
$$ESI_s =\left( \left(1-\left | \frac{0.47}{1.53} \right |\right)^{0.57}\cdot
 \left(1-\left | \frac{0.29}{1.71} \right |\right)^{1.07}\cdot\\
 \left(1-\left | \frac{0.55}{1.45} \right |\right)^{0.70}\cdot
 \left(1-\left | \frac{61\mbox{ K}}{515\mbox{ K}} \right |\right)^{5.58}\right)^{\frac{1}{7.92}}=\\
(0.811241627\cdot 0.819676889\cdot 0.716163454\cdot 0.494865663)^{\frac{1}{7.92}}=
0.833189885$$ as surface similarity. (Some of the data used from here.)
Global similarity combines surface similarity with interior similarity.
Global similarity of Mars with Earth is about 0.7.

Surface gravity can be calculated from radius and bulk density of a planet.
The radius of an expoplanet can be estimated by the transit method, relating the estimated diameter of the star to the brightness change of the
star during planet transit. The mass of the planet can be estimated by the wobble of the radial velocity of the star (using Doppler shift).
By mass estimate of the star and the orbital period of the planet the distance of the planet to the star can be estimated. An estimate of the
absolute brightness of the star can then be used to estimate the surface temperture of the planet. There exist more methods.
The accuracy of these estimates vary with the quality of the observations.

ESI values between 0.8 and 1.0 are considered as Earth-like.

Details of the planet's atmosphere, amount of surface water, and other details are not considered in the formula. So it's just a very rough prioritization. With future spectroscopic data further refinement could be possible.


The Pluto-Charon system is similar to this.

It's unlikely that you'd have two planets of nearly the same mass because some variation is far more common, but a moon orbiting a binary planet system is certainly possible.   It's easier if it's not that close to the sun, for the same reason that it's difficult for Mercury and Venus to have moons.   A larger hill-sphere and less solar gravitation would make your system more stable in the long term.

And the Moon couldn't be in a Lagrange point in your system because, for starters, L1, 2 and 3 are never stable and the potentially stable Lagrange points, L4 and L5 require a large central planet and a significantly less massive object, (less than 1/26th the mass) orbiting the central planet.   The L4 and L5 points are to the orbiting object.  That doesn't work in a binary system of 2 objects of similar mass.

Any working Lagrange point scenario you'd have two moons and one planet, not binary planets and one moon.    You could, in theory have two moons at Lagrange points to each other orbiting a binary planet system but that would be adding even more improbabilities, but theoretically possible and stable for some time.


Cosmic microwave background radiation: why did it take so long to get here?

Here's a very, very simple explanation:

"why did it take so long to get here?"

There's two aircraft carriers, A and B, fifty miles apart on the ocean. You've got a helicopter that flies 50 mph.  Of course, it takes one hour to make it from A to B.

Now, imagine A and B are both moving - they're steaming away from each other at a good speed.

Of course, it now takes longer than an hour for the chopper to make it between A and B.

Regarding the many, many mysteries of the metric expansion of space, this trivial observation is just one thing to bear in mind. Note that of course this is not a full and complete analogy, and the analogy is indeed utterly wrong / totally irrelevant to many aspects of the broader discussion.

"I thought that the photons were "trapped" in an expanding ball" sure, the chopper can only move at 50 mph over the water underneath it meaning that at any given time it can only have moved so far over the water underneath it (a sort of "circle of achievablity", if you will).  But you can see with the carriers moving the hell all over the place, it's just an open question whether the chopper reaches B in much less than an hour, much more than an hour, or can never reach B at all.


Laser Collimator For Dobsonian vs Newtonian

A Dob is a Newtonian, so yes it will work.  Strictly speaking the Dob bit is the mount, but it's nearly always a newtonian that's mounted.


A well-known analytical planetary theory is VSOP.  The paper describes six versions, most of which are heliocentric.  One of them (E) is barycentric and has an extra (ninth) body: the Sun.  Hence, for VSOP you have the choice of using the Solar System's barycentre as the origin and computing the position of an extra body, or of computing heliocentric positions of the planets directly.

If you use a different theory, I would expect something similar: a heliocentric version and/or a barycentric version with equations for the Sun.


First. In about 6 billion years the Sun will become much larger than it is now, maybe a factor of 20. It will be a red giant. Then a little while later it will get really big, as large as the earth's orbit and perhaps the Earth will be engulfed.

Second. Only after this does the Sun lose about half of its mass and the remainder is exposed as a small, hot, white dwarf star, about the same size as the Earth but half the mass of the Sun. It will have a density about 1 million times that of water and it is held up by a pressure that only arises because of the quantum mechanical nature of the electrons within it.

The Sun is not currently a white dwarf. It is a G-type dwarf star, supported by "normal" gas pressure and will remain reasonably stable, though will become about 50% more luminous over the next 5 billion years or so.


How would the combined masses of the Centaurs or the scattered disk compare to the combined mass of the asteroid belt?

According to Sheppard et al. (2000) "A Wide-Field CCD Survey for Centaurs and Kuiper Belt Objects", the total mass of the centaurs is likely around 10-4 Earth masses.

Gomes et al. (2008) "The Scattered Disk: Origins, Dynamics, and End States" estimate the mass of the scattered disc as between 0.01 and 0.1 Earth masses. (Going by the abstract, I don't have a copy of the book to check the details.)

As you noted in your question, the asteroid belt is around 4% of the mass of the Moon, or about 0.0005 Earth masses. So the centaurs have a smaller or similar mass to the asteroid belt (bearing in mind the uncertainties on the values), while the scattered disc is considerably more massive.


Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.
Kerr Black Holes
A rotating black hole is known as a Kerr Black Hole (named after Roy Kerr who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole $M$. The second is the spin $a$. Really $a$ isn't the spin itself $-$ it's defined by $a=J/M$ (see footnote) where $J$ is the angular momentum of the black hole $-$ but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that
$$0 \le a/M \le 1$$
Black Hole Event Horizon
The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by
$$r_e = M+(M^2-a^2)^{1/2}$$
In the case when $a/M=0$ (and thus $a=0$), this reduces to just $r_e=2M$, or in regular units (instead of geometrized units) $r_e=2GM/c^2$. Hopefully you can see that this just reduces to the normal Schwarzchild radius for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when $a/M=1$ (and thus $a=M$). In this case, you find that the radius is $r_e=M$. When $a/M=1$, you have a maximally rotating black hole, and your radius is half the normal Schwarzchild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.
Ergosphere
As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the static limit is given by the equation
$$r_{s} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$
Notice that this is almost exactly the same as above, except for that extra $\cos^2(\theta)$. This defines a different, slightly larger and somewhat "pumpkin-shaped" horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the Ergosphere. Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, $r_e<r<r_s$) must rotate exactly with the black hole - it is physically impossible to stay still here!
Answers

They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?)

When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by $r_e$ above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by
$$J_{max} = a_{max}Mc = M^2c$$
Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra $c$ now. Remember that $a_{max}$ is achieved when $a/M=1$.
We can also define the angular momentum using the standard equation from physics 101, $J=rMv_\perp$, where of course $r$ is the radius of your object, and $v_\perp$ is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, $r_e=M$ so we also have that
$$J_{max} = r_eMv_{\perp} = M^2v_\perp$$
You can see that these two equations for $J_{max}$ only equal each other if the tangential velocity $v_\perp$ is equal to the speed of light $c$. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!
I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - "how can a singularity have a tangential velocity"? As it turns out, Kerr black holes don't have point singularities, they have ring singularities. These are "rings" of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.

They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR).

We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, $M_\odot$) is given by
$$r=\frac{GM_{\odot}}{c}=1.48\:km$$
So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that $a=M$. What happens to our event horizon radius when $a>M$ (and thus $a/M>1$)? For arguments say, lets say $a=2M$. Then our event horizon radius becomes
$$r_e=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$
Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus cannot exist. Now that we don't have an event horizon, our singularity can't hide behind it and is "naked", exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.

Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.

Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from this Nature paper which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.

Footnote: In GR, to make the math easier, often scientists adopt special units known as geometrized units. These are units chosen in just such a way that the gravitational constant, $G$, and the speed of light, $c$, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have $G$ or $c$ in them, but they're implicitly there, they're just equal to one and so not shown.


At the moment of total solar eclipse what are exact distances of Moon & Sun to the Earth

First off, your definition of "total solar eclipse" is different from that generally used. A total solar eclipse occurs when the moon is directly in front of the sun and is bigger so it blocks the whole of the photosphere. Also note that the sun and moon are both somewhat oblate (not perfect spheres, but flattened) and not exactly the same shape, but for the sake of estimation I'll use the diameter information from wikipedia.

The moon has an average diameter of 3,474 km, and the sun has an average diameter of 1.392684 million km. So the Sun is 400.9 times larger than the moon. When the sun is 149600000 km distant, then if the moon is $149600000/400.9 = 373200$km, the sun and the moon would appear equal in size.

When the moon is 384000km from the centre of the Earth, the distance to the Earth's surface is smallest at the sub-lunar point, about 378000km, but still greater than 373200, so when the Sun and Moon are both at their average distance, an annular eclipse occurs. Annular eclipses are slightly more common than total eclipses.

However if the distance from an observer to the moon is closer than 1/400.9 of the distance to the sun, then a total eclipse is possible, and this allows you to generate infinitely many solutions to the question "How far is the moon during a total solar eclipse"


What’s the closest earth like planet we have discovered in our galaxy?

Well, this really does depend on what is considered an Earth-like planet. For instance, one could conclude that Venus is an Earth-like planet based off of them being similar in mass and also in regards to the similarities in radius. As for Earth-like moon, Titan would be a great candidate if located inside the so-called Goldilocks zone.

Counting planets outside our solar system, the nearest planet I could find was about 4.243 light-years away. Astronomers found it on August 24, 2016. It is a rocky planet in the habitable zone of its parent star Proxima Centauri, the closest star to Earth. The name of the planet is Proxima b, this planet is approximately  1.3 times the mass of Earth and possesses an orbital period of roughly 11.2 Earth days.

However, since Proxima Centauri is classified as a red dwarf this casts reasonable doubts on whether Proxima b is habitable. To be habitable Proxima b would have to deal with low stellar flux, which causes a high probability of tidal locking, and high stellar variation.

List of nearest planets

List of possible habitable moons


It can't happen. The moon is quite big and can't get knocked out of orbit easily.

If it were somehow knocked out of orbit and collided with the Earth, the energy released would melt the entire crust. Huge amounts of matter would be knocked into space. After everything had settled down, you would get a lifeless world, still spherical but now slightly larger.

The moon can't get knocked out of orbit.  If it did, It can't "land". If it did "land" it would be crushed by its own weight, and all life would end in one way or another.


What was the length of the solar day 73 million years ago?

There is evidence from Silurian coral fossils that, 430 My ago, there were 420 daily rotations in each annual season, compared to 365 days per year now.  If we assume that the year (fixed by mean Earth-Sun distance and Kepler's Laws) hasn't changed much in that interval, but that instead the day has gotten longer due to the Earth-Moon tidal interaction, that means that 430 My ago there were about 20.9 hours in each day.

Linear extrapolation in between says that only about one-sixth of that change has taken place in the past 73 My, which would give a day length of about 23.5 hours.

In a comment, JohnHoltz cautiously extrapolates your currently-known slowing of 0.0016 seconds/century over that interval, which gives an ancient day length of closer to 23.7 hours.

JohnHoltz's estimate is probably better. Tidal slowing is occurring as the mean distance between the Earth and Moon increases, which makes all tidal effects (including tidal slowing) weaker. So a linear extrapolation from today's well-measured rate of rotational slowing is probably more accurate for recent times than a linear extrapolation over a much longer period that probably includes more rapid slowing in the past. It inspires confidence that the two methods come out close to each other, though.


What's the furthest distance that something could travel and eventually come back to Earth?

There's an easy way to work out the distance to the mirror at the present time, if it moves with the Hubble flow: imagine that instead of being emitted by us, the light is emitted by the mirror image of the Earth and simply passes through the mirror to us. The farthest comoving distance that the mirror Earth could be is the same as the farthest present-day distance from which light emitted in the present day can ever reach us; assuming ΛCDM is correct, that's around 16.5 billion light years. The comoving distance to the mirror (and the metric distance to it in the present day) is exactly half that.

Working out the metric distance to the mirror at the time of reflection is trickier. If (nonstandard terminology warning) $D(a)$ is the comoving distance light can travel from scale factor $a$ to $∞$, then the scale factor at which the light hits the mirror is $D^{-1}\left(\frac12 D(1)\right)$. If all energy was dark energy and the expansion was exactly exponential then the solution would be $2$ exactly. Using central parameters from here I get $a\approx 2.09$, so the mirror distance is a bit higher, around 17 billion light years.
For what it's worth there's an exact expression for this function in a flat perfect dust+dark energy universe:
$$D(a) = \frac{c}{a \sqrt{Ω_Λ} H_0} \; {}_2F_1\left(\frac13,\frac12;\frac43;-\frac{Ω_m}{a^3Ω_Λ}\right)$$
where ${}_2F_1$ is the hypergeometric function.


There's no clearly defined definition of what a stable orbit is.  Orbits can last for hundreds or thousands of years, millions, billions or even trillions of years.  There's no neat and tidy definition between stable and unstable.

Hubble for example, orbits the Earth every 97 minutes, and in a few years, primarily slowed by the very slight air resistance, it is expected to fall to the Earth.   Many satellites, because their equipment doesn't last forever, are designed to gradually slow by air resistance and fall to the Earth after a few decades of operation.   Satellites can be positioned where they stay in orbit much longer simply by moving them a bit further out, but there's little benefit to that.   It's advantageous to have satellites fall to the Earth after they've served their usefulness.

Orbits around the Moon are more unstable than orbits around the Earth for 2 reasons.   One, because the Moon is gravitationally lumpy so low orbits around the Moon don't stay in place very long.  See here.

and two, because the Earth is gravitationally dominant, so the Moon doesn't have a very big hill sphere.  Celestial mechanics doesn't like satellites to have satellites.   See here and here.  Our moon could have a satellite, but it likely wouldn't be stable for very long, at least, not long in a celestial time-scale.

Trojan points, L4 and L5 are generally stable in a System where the larger body (Earth) is more than 25 times as massive as the smaller body (moon), but our Moon's orbit is too eccentric and too chaotic for lunar Trojan points to be stable for any period of time.     Earth's L4 and L5 points around the sun are more stable and there are a few small asteroids in Earth's Trojan points.  The Moon's L4 and L5 aren't stable cause the Moon wobbles too much in it's orbit.  Eccentricity and perturbations tend to destabilize L4 and L5 points.   That said, the Moon's L4 and L5 are more stable than L1, L2 and L3.

Mars moon Phobos is in a decaying orbit that may only last 10 million years or so.

Our Moon, however, even though it's slowly moving away from Earth, is sufficiently stable that it will outlast our sun.  It's stable on the scale of many billions of years.

As to your question "prolong an orbit via moon power", The moon tends to destabilize, not stabilize orbits around the Earth.

But while L1 and L2 points aren't gravitationally stable, they are still useful places for satellites because they require less adjustment.   Satalites with orbits that move around lagrange points are often used as useful locations.

Take this bizar orbit around the Moon, where the NASA satellite transfers from L2 to L1 and could, in theory, be transferred back and forth without expending too much energy.  This isn't a "stable" orbit by any means but it's an example of an unstable but useful orbit for the life of a satellite.

Source

The James Webb Telescope, when it's launched will orbit around Earth's L2 Lagrange point, which isn't stable, but it requires relatively small adjustments to stay in that order.  It's also useful because it will be nearly entirely shaded from the sun by the Earth and it works best in cold space, away from sunlight.

Source.

Those are 2 examples of useful but unstable orbits.

Jupiter's orbit around the sun, for example, will probably be stable for trillions of years.   Jupiter's moons around Jupiter?  Hard to say, with 4 Galilean moons, it's difficult to predict their orbits over long periods of time.
.


Short answer, no, they don't.

Longer answer, it's complicated. There are, in essence, five different measures for the centre of a galaxy cluster, based on different physical properties of the cluster, and occasionally not in agreement with each other.

BCG, the brightest cluster galaxy. It should sink to the bottom of the gravitational well, but that's only the case if the cluster had enough time to evolve. Of course, galaxy clusters are, cosmologically speaking, quite young objects and still in the process of forming, so this is not a given.
Mean position of the cluster galaxies. Works for spherically symmetrical clusters with lots of galaxies. Problems: Poisson noise due to limited number of galaxies, identification problems (is this galaxy really a part of the cluster), asymmetry.
X-ray: the hot intra-cluster gas emits X-ray radiation that we can observe - more and hotter gas means more and harder X-rays and if the gas is in equilibrium, the highest intensity radiation comes from the centre. Unfortunately, the gas is typically not in equilibrium, there are cooling flows, streams, cold cores and other things that make the measurement less straight-forward.
(Weak) gravitational lensing. Measures the total mass of the cluster and doesn't care about ugly gas physics. Problems: Shape noise (measurements are based on mean ellipticity of background galaxies), resolution / smoothing issues with the mass map created from averaging over lots of background galaxies.
Sunyaev-Zel'dovich effect. Again the hot cluster gas, but this time we have inverse Compton scattering of photons in the cosmic microwave background on the electrons in the gas. The problems with gas physics should apply here, too.

So, there are other methods to find the centre of a galaxy cluster. This paper suggests that ususally the weak lensing centre is offset with respect to both X-ray and BCG. I am not aware of anyone using the mean position of the galaxies as a measure for the cluster centre.

This paper evaluates the cluster centre offset between measurements in X-ray and the SZ. This is interesting because they both trace the hot intra-cluster gas.


Do the orbits of satellite galaxies show the influence of dark matter?

The short answer is "Yes, the orbits of satellite galaxies definitely show the influence of dark matter (or something like MOND, if you prefer)." For galaxies outside the Local Group, you can only measure their radial velocities from Doppler shifts (i.e., velocities along our line of sight to the galaxy), but you can still look at the distribution of these velocities for satellites around some central galaxy and work out the approximate amount of mass required for the satellites to remain bound -- and thus you need dark matter. (The first hints of dark matter, going back in the 1930s, were from a larger-scale version of this: Zwicky and others measured the radial velocities of galaxies in galaxy clusters and found they were too high to be explained by the visible mass of the galaxies in the clusters.[*])

For satellites of our own galaxy, it has become possible, in the last twenty-five years or so, to measure proper motions (motions across the sky, perpendicular to the line of sight) as well as their radial velocities and so we can measure their actual 3D space velocities. This paper by Fritz et al. (2020) uses proper motions from Gaia observations for 45 satellite galaxies to estimate the mass of the Milky Way's dark-matter halo. They find this to be about $1.5 \times 10^{12} M_{\odot}$, consistent with other measurements.[**] (Note that unlike the case for measurements of gas clouds orbiting in a disk -- the original data indicating dark matter in spiral galaxies -- you can't assume the orbits of satellites are circular, so the calculations involved are more complicated, but they still give roughly the same answers.)

[*] Later observations with X-ray satellites demonstrated that galaxy clusters contain significant amounts of normal baryonic matter in between the galaxies, in the form of very hot, dilute, X-ray-emitting gas. However, even when you add this to the stellar mass in the galaxies, you still don't have enough mass in the cluster to explain the radial velocities.

[**] This value can be compared with a baryonic mass for the Milky Way of about $5 \times 10^{10} M_{\odot}$ in the form of stars, plus maybe another $1 \times 10^{10} M_{\odot}$ in the form of gas.


If Earth didn't rotate, would we feel heavier?

The centrifugal force at the equator is only 1/289 of normal gravity, or 0.35%. That's too small to be perceived.

http://books.google.com/?id=mE4GAQAAIAAJ&pg=PA358&lpg=PA358

It gradually diminishes as you move away from the equator, and becomes zero at the poles.

In any case, if Earth stopped rotating, the added weight at the equator would be the least of your worries. There would be major perturbations in terms of weather, etc, that would amount to global catastrophe.

  Likewise, if Earth's rotation increased, would we feel lighter as
  centrifugal force lifts us from the ground?

In theory, yes. But the rotation increase would have to be huge. Again, in that case, there would be other, much bigger effects, that would be much more important than your loss of weight. E.g. the whole planet would bulge at the equator and become more flat at the poles, with incalculable effects on plate tectonics, weather, etc.


Why are orbits elliptical instead of circular?

Assume the planet has a negligible mass compared to the star, that both are spherically symmetric (so Newton's law of gravitation holds, but this normally happens to a very good approximation anyway), and that there aren't any forces besides the gravity between them. If the first condition does not hold, then the acceleration of each is going to be towards the barycenter of the system, as if barycenter was attracting them a gravitational force with a certain reduced mass, so the problem is mathematically equivalent.

Take the star to be at the origin. By Newton's law of gravitation, the force is $\mathbf{F} = -\frac{m\mu}{r^3}\mathbf{r}$, where $\mathbf{r}$ is the vector to the planet, $m$ is its mass, and $\mu = GM$ is the standard gravitational parameter of the star.

Conservation Laws

Because the force is purely radial $(\mathbf{F}\parallel\mathbf{r})$, angular momentum $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ is conserved:
$$\dot{\mathbf{L}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{r}\times\mathbf{p}\right) = m(\dot{\mathbf{r}}\times \dot{\mathbf{r}}) + \mathbf{r}\times\mathbf{F} = \mathbf{0}\text{.}$$
If the initial velocity is nonzero and the star is at the origin, then in terms of the initial position and velocity, the orbit must be confined to the plane of all points with vectors $\mathbf{x}$ from the origin that satisify $\mathbf{L}\cdot\mathbf{x} = 0$. If the initial velocity is zero, then the motion is purely radial, and we can take any one of infinitely many planes that contain the barycenter and initial position.

The total orbital energy is given by
$$\mathcal{E} = \frac{p^2}{2m} - \frac{m\mu}{r}\text{,}$$
where the first term part is the kinetic energy and the second term is the gravitational potential energy of the planet. Its conservation, as well as the fact that it invokes the correct potential energy, can be proven by the fundamental theorem of calculus for line integrals.

Define the Laplace-Runge-Lenz vector to be
$$\mathbf{A} = \mathbf{p}\times\mathbf{L} - \frac{m^2\mu}{r}\mathbf{r}\text{.}$$
It is also conserved:
$$\begin{eqnarray*}
\dot{\mathbf{A}} &=& \mathbf{F}\times\mathbf{L} + \mathbf{p}\times\dot{\mathbf{L}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\
&=& -\frac{m\mu}{r^3}\underbrace{\left(\mathbf{r}\times(\mathbf{r}\times\mathbf{p})\right)}_{(\mathbf{r}\cdot\mathbf{p})\mathbf{r} - r^2\mathbf{p}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\
&=& \mathbf{0}\text{.}
\end{eqnarray*}$$

Finally, let's also take $\mathbf{f} = \mathbf{A}/(m\mathcal{E})$, which has the same units as $\mathbf{r}$, and since $\mathbf{L}\cdot\mathbf{f} = 0$, it lies along the orbital plane. As it's a conserved vector scaled by a conserved scalar, it's easy to show that $\mathbf{f}$ is conserved as well, as long as $\mathcal{E}\neq 0$.

Simplifying

By employing the vector triple product, we can write
$$\begin{eqnarray*}
\frac{1}{m}\mathbf{A} &=& \frac{1}{m}\left[p^2\mathbf{r}-(\mathbf{p}\cdot\mathbf{r})\mathbf{p}\right] -\frac{m\mu}{r}\mathbf{r}\\
&=& \left(\mathcal{E}+\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\\
\mathcal{E}(\mathbf{f}-\mathbf{r}) &=& \left(\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\text{,}
\end{eqnarray*}$$
the norm-squared of which is easy to crank out:
$$\mathcal{E}^2|\mathbf{f}-\mathbf{r}|^2 = \left(\mathcal{E} + \frac{m\mu}{r}\right)^2r^2\text{,}$$
where $\mathcal{E}$ was used throughout to switch between kinetic and potential terms.

Why Ellipses?

Since $\mathcal{E}$ is energy relative to infinity, to have a bound orbit we need $\mathcal{E}<0$. Thus, from the previous section, $|\mathbf{f}-\mathbf{r}| = -\mathcal{E}^{-1}\left(\mathcal{E}r + m\mu\right)$ and therefore
$$|\mathbf{f}-\mathbf{r}| + |\mathbf{r}| = -\frac{m\mu}{\mathcal{E}}\text{,}$$
which defines an ellipse with foci $\mathbf{0},\,\mathbf{f}$ and major axis $2a=-m\mu/\mathcal{E}$.

Why Not Circles?

The circle is a special case where the foci are the same point, $\mathbf{f} = \mathbf{0}$, which can be restated as
$$\mathcal{E} = -\frac{1}{2}\frac{m\mu}{r} = -\frac{p^2}{2m}\text{.}$$
In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $\mathcal{E}<0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)

Note that hyperbolic orbits have $\mathcal{E}>0$, and we can still find the foci using the above method, though being careful with the signs. For $\mathcal{E}=0$, the second focus $\mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.

Additionally, the eccentricity vector $\mathbf{e} = \mathbf{A}/(m^2\mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.


Time-density and time-temperature functions of the Universe since the Big Bang

Both the temperature and the density of a toy universe can, in a way, be linked to the age of the universe ($t$). Certain properties of the universe are dependent upon a scale factor $a(t)$ that is a function of time. The expansion of such a universe can be written in the Friedmann-Lemaitre-Robertson-Walker (FLRW) metric, generally written as
$$ds^2 = dt^2 - a(t)^2(d\Sigma^2)$$
where, in spherical coordinates, $d\Sigma$ is given by
$$d\Sigma^2 = \frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2$$
where $k$ is a constant. This shows that, in a toy universe, $r$ and $dr$ is proportional to $a(t)$, and so in a perfectly spherical toy universe, you could relate the volume to $t$. We know that, for a sphere,
$$V=\frac{4}{3}\pi r^3$$
and because $d\Sigma^2$ (and by extension $r^2$) is related to $a(t)^2$, we can relate volume to time. If you knew the mass-energy of the universe, you could then relate the change in density to $t$. Unfortunately, our universe does not appear to be perfectly spherical, and so this may not apply.

You can relate something else to the age of the universe: the temperature of the cosmic microwave background. It has cooled over time, and is presently at about 2.7 Kelvin. While the temperature of the CMB may not be governed by the scale factor $a(t)$, it is certainly governed by time, and thus one you could in fact write the temperature of the universe as function of time. Unfortunately, I haven't been able to find data on the evolution of the CMB, but I can try, and perhaps this will lead you to an answer.

To summarize: In a perfectly spherical universe, assuming you know the matter-energy content and the scale factor $a(t)$, you could write an equation for the density of the universe. In any universe, you can also write an equation for the temperature of the CMB - the "temperature of the universe".

I hope this helps.

Source for FLRW metric: http://www.universe-galaxies-stars.com/Robertson-Walker_coordinates.html


Can't find Cassiopias Schedar or Caph in Tycho-2 catalog supplement 1

Alpha Cas (Schedar) is  TYC 3663-2668-1 or HIP 3179 in the Hipparcos catalogue.

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Schedar&NbIdent=1&Radius=2&Radius.unit=arcmin&submit=submit+id

Beta Cas (Caph) is not in the Tycho-2 catalogue but is in the Hipparcos catalog as HIP 746

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Caph&NbIdent=1&Radius=2&Radius.unit=arcmin&submit=submit+id

Gamma Cas is also does not have a Tyc number but is listed as HIP 4427 in the Hipparcos catalogue.

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Gamma+Cas&NbIdent=1&Radius=2&Radius.unit=arcmin&submit=submit+id

All three of these stars are right at the bright limit of the Tycho-2 catalogue.


Difference in time between sunset and moonrise on a full moon

Technically, the Moon is Full at a specific time. Therefore you need to check the time of sunset/moonrise OR sunrise/moonset that is closest to the time of Full Moon. Then you can more accurately compare the variation. Here is the list for 2019 created using the U.S. Naval Observatory website.
Longitude E9° 8', Latitude N41° 35'
All times in Universal Time (UT)
Date of Full Moon   -------- Sun ----------     -------- Moon --------  Moon-Sun
     Date    Time              Date    Time              Date    Time     Diff
 1/21/2019    5:16   Rise  1/21/2019    6:45     Set 1/21/2019    6:54   +0:09
 2/19/2019   15:53    Set  2/19/2019   17:01    Rise 2/19/2019   16:55   -0:06
 3/21/2019    1:43   Rise  3/21/2019    5:26     Set 3/21/2019    5:58   +0:32
 4/19/2019   11:12    Set  4/19/2019   18:08    Rise 4/19/2019   18:14   +0:06
 5/18/2019   21:11    Set  5/18/2019   18:39    Rise 5/18/2019   18:14   -0:25
 6/17/2019    8:31   Rise  6/17/2019    3:49     Set 6/17/2019    3:54   +0:05
 7/16/2019   21:38    Set  7/16/2019   18:56    Rise 7/16/2019   18:49   -0:07
 8/15/2019   12:29   Rise  8/15/2019    4:32     Set 8/15/2019    4:09   -0:23
 9/14/2019    4:33   Rise  9/14/2019    5:02     Set 9/14/2019    4:59   -0:03
10/13/2019   21:08    Set 10/13/2019   16:46    Rise10/13/2019   17:02   +0:16
11/12/2019   13:34    Set 11/12/2019   16:07    Rise11/12/2019   16:25   +0:18
12/12/2019    5:12   Rise 12/12/2019    6:40     Set12/12/2019    6:42   +0:02

The difference is much smaller than your original list.
Why this happens
MCG and Carl Witthoft explained why it happens, namely because:

If the moonrise/set is not the same time as the exact Full Moon, then moonrise/set  occurs earlier or later than the sunset/rise.
Because the Moon passes above and below the ecliptic (the path of the Sun in the sky), the Full Moon is rarely directly opposite from the Sun. Therefore, it can rise earlier or later than the Sun even when exactly "Full".
Even if the Full Moon were exactly opposite from the Sun (180 degrees apart, and in a total lunar eclipse), there would still be a difference between the time of sunset/moonrise. Those times are when the limb of the sun/moon is visible, and the limbs are closer together than 180 degrees apart. Also, atmospheric refraction shrinks the separation a small amount.

Where and when are they closest?
This is more involved but can be reasoned using the following figure. The Sun is in the direction of 1, so the ring of sunset-sunrise is the circle A-A. The Full Moon is in the direction of 2, so the ring of moonrise-moonset is the circle B-B. Where these two circles cross at C is where the sunset and moonrise (or sunrise and moonset) occur at the same time. The highlighted area is where the Moon and Sun are both above the horizon at the same time. Note that this figure is highly exaggerated for illustrative purposes.


What does designation VY, NML, UU in star names stand for?

VY Canis Majoris and UY Scuti are variable star designations. The first discovered variable star in a constellation is called R, the second, S, and then unto Z. After Z comes RR and so on. For a full description see this wikipedia page. NML Cygni is also a variable star, but its variable star designation is V1489 Cygni. The letters NML come from its discoverers: Neugebauer, Martz, and Leighton.


What created the big bang's singularity?

No one knows what came before the Big Bang if, indeed, anything did.  Theories include:

The Ekpyrotic universe theory where the BB was the result of the
collision of branes.
Various oscillating universe theories where
the expansion of the universe ultimately reverses into a Big Crunch
Quantum fluctuation models where a zero-net-energy universe is a
quantum fluctuation from nothing.
Various multiverse models
where the universe we see is only an infinitesimal bubble in a much,
much, much larger universe.
The Hartle-Hawking idea where
spacetime gets a bit scrambled and there is no before. (Really:
time turns into a space dimension if you go far enough back, and
there is no time.)
Stringy stuff. Very difficult math.

I've only scratched the surface -- making models like this is an avocation to some physicists, while others have discovered that it sells books and generates publicity.  All of the theories have one thing in common: There is no experimental evidence for any of them. Additionally, many of the theories have been shown to contradict currently-understood physical law. (Especially the 2nd Law of Thermodynamics.) The authors generally just shrug and write another paper.

(On the religious side, speculation about what happened before Creation has been around as far back as written history, with roughly equal success. In the 4th century St. Augustine, after being pushed to answer that question retorted that before the Creation, God was creating a Hell for people who ask such questions.
 That didn't stop the speculation either.)


How to find the mass of a planet not knowing the gravitational constant?

How can I do this by not knowing the gravitational constant?

Sarcasm: You could do what Henry Cavendish did, which was to measure how the horizontal deflection rather heavy balls separated horizontally by a small distance and held in place by torsion rods. The concept of the Newtonian gravitational constant did not exist in Cavendish's time, and would not exist until almost a century after Cavendish's paper on Experiments to determine the Density of the Earth was published. At the time, Cavendish envisioned his experiment as a way to "weigh the Earth".
After the fact (about 80 years after the fact!), physicists saw that Cavendish had measured something much more fundamental than the average density of the Earth. We now see that Cavendish had indirectly measured what we now call the Newtonian gravitational constant. G does not come into play in Newton's formulation of his law of gravitation.
The modern version of the Newtonian law of gravitation does use the Newtonian gravitational constant, which frees you from having to redo Cavendish's experiment.


Examples of Equal Mass, Unequal Mass and Double Binary Star Systems

Equal mass
Alpha Centauri, the nearest star system to us, is composed of a binary star and Proxima Centauri (making it a triple star system). Alpha Centauri A has a mass of 1.1 M$_{\odot}$, while Alpha Centauri B has a mass of 0.9 M$_{\odot}$.
If you count brown dwarfs, Luhman 16 are a good target, with each being about 0.04 M$_{\odot}$.
Unequal mass
Eta Carinae contains a Luminous Blue Variable (LBV) with a mass of 120-160 M$_{\odot}$ and a smaller (but still very massive) star with a mass of 30-60 M$_{\odot}$.
Sirius is okay, though with two stars of much smaller masses, 2 M$_{\odot}$ and 1 M$_{\odot}$.
Double binary
Mizar, in reality part of a six-star system (Mizar/Alcor), has four stars separated into two pairs. This  ensemble then interacts with the binary star Alcor.

You can find plenty of stars here.


Why do certain moons have their rotational period equal to their orbital period?

The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in tidal locking, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.

To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being deformed such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.

Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.

That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.

In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.


Is there a way to tell the difference between earth andesite from Mars

If this is something that you have found (rather than purchased as a meteorite) the chances are very small that it is a meteorite. Even if it is a meteorite, the chances it's a Martian one are even smaller still and none have been found in the United States.

According to the Meteorites in the US page, which draws from the Meteoritical Society database, only 1821 meteorites have been found in the US over the past 200 years. Of all meteorites found in the world, less than 0.1% are from the Moon or Mars (source statement, graphs of meteorite fractions) and none of these have been found in the US, with the vast majority (99%) found in Antarctica or the African or Arabian deserts (lunar meteorites)).

There is a long "Meteorite Realities" page and a shorter, graphical "Self-Test Check-List" that it would be good to check and go through before it gets to chemical testing.

If you are determined to get testing done, the same meteorites.wustl.edu site on its page on meteorite chemical composition recommends chemical testing by Actlabs; there is more information on what they need (a 5g sample) and the type of tests to ask for on this page. Andesite is a type of basalt formed by volcanism and while it is true that most of the Martian meteorites are basalts (as discussed here), so are a lot of Earth rocks. On the chemical composition page, in plots of chemical composition such as silicon dioxide (SiO$_2$) vs total iron and magnesium oxide content  e.g. the martian meteorites (red squares) separate from most terrestrial/Earth rocks and the "meteorwrongs" (white circles) due to having higher iron+magnesium content in the form of pyroxene, olivine and ilmenite (from 'Chemistry' section of How Do We Know That It’s a Rock from the Moon?) However as noted on the basalt page:

  Unfortunately, the only way to distinguish a terrestrial (Earth) basalt from a basaltic meteorite (Moon, Mars, asteroid) is with expensive chemical and mineralogical tests. If you find a basalt, it's probably not a meteorite.

So I would guess that these additional tests for the contents may be enough to distinguish a non-Earth basalt from an Earth one when the appearance to the eye or under a microscope is very similar (due to the similar formation mechanism via lava). However additional tests needed for trace elements may also be needed (Lunar basalts are Chromium-rich but have much lower concentrations of the alkali elements of potassium, sodium, rubidium, and cesium)


Does binarity play a role in spectral fitting?

Yes you can, to some extent.

If the binary has a short period, then the chances are that the two stars have a different radial velocity. In other words, there are two sets of spectral lines in the spectrum. A cross-correlation or direct fitting technique can then pick the two stars out.

If the two stars are unresolved and have indistinguishable velocities (true for the majority of binaries), then it is more difficult. A fit to a single spectral template may be poor (and you could judge this from the chi-squared of the fit), but there needs to be a significant difference in the spectra of the two stars in order for a two-star model to adequately separate them.

There is basically a trade-off. You want the stars to be of similar brightness but to have quite different spectra. This is difficult to arrange for main sequence stars and there is normally a sweet-spot where one star is a bit cooler and a bit dimmer, but can still be seen in the combined spectra. Other combinations are easier - a hot white dwarf and an M dwarf for example have very similar luminosity but very different spectra, such that each dominates in different parts of the spectrum. Others are harder - a red giant and a main sequence star of lower mass.


As you are suspecting, the sun burns in a different sense, not by chemical reaction with oxygen.

Atoms consist of a tiny, heavy nucleus, surrounded by an almost empty space, populated by electrons. Burning by chemical reaction with oxygen doesn't change the nucleus of atoms, but takes place in the hull of atoms: Atoms may assemble to form molecules; electrons change their orbitals (the way they surround the nucleus), and release some energy as heat.

Atomic nuclei are (positively) electrically charged, and repell each other.
But if small nuclei, like those of hydrogen atoms, come close together, they can fuse and form a larger nucleus. This nuclear fusion of hydrogen to helium (in this case) releases much energy, more even than fission of uranium in a nuclear power plant. The notion "burning" is used sometimes for reactions of atomic nuclei, too, if they release energy as heat.

To overcome the electrostatic repulsion of hydrogen nuclei, high pressure and temperature are needed. These conditions occur in the core of the Sun.


The specifics of two Earth-like worlds orbiting each other?

A couple thoughts:
Tidal lock depends on the age of the system, so you can have it or not as you wish, depending on how long they came to orbit each other -- which is to say, of course, orbiting a common epicenter very nearly halfway between the pair.
Depending on your whim, you might want the orbit plane to be in or 90 degrees out of the ecliptic, the latter leading to interesting effects on tumbling.  I'd recommend staying in the ecliptic, which is more stable and allows each of the planets to rotate about an axis which allows night/day similar to Earth.  Notice that there will be a dramatic bunch of total solar eclipses of decent length, on both planets in turn.

As to their spatial separation, take a look at Newton's Laws.  Nearly any separation is allowed, though if they were very close they'd have suffered severe gravitational distortions (of shape) , messed up each others' atmospheres, and almost certainly have crashed together with horrific results.


How to connect my phone to a motorized telescope via a cable?

Group dedicated to all versions of the Celestron "C14" 14-inch Schmidt-Cassegrain Telescope.  This includes the original C14 and the newer "EdgeHD" series. All other Celestron SCT telescopes are welcomed to participate.
https://groups.io/g/C14EdgeHD


Can we tell how fast bodies are moving away by measuring their frequency?

Your understanding is correct. The doppler shift observed from a galaxy is the sum of its peculiar velocity with respect to the "Hubble flow" and the redshift due to the Hubble flow, which is caused by the expansion of the universe.

There is no direct way from a spectrum to separate these two components - they have the same qualitative result.

In principle, the expansion of the universe (or a change in the peculiar velocity) could be directly measured by looking for a change in redshift with time, which would depend on the cosmological parameters.

This is an extremely small effect and is confused by the peculiar motions of individual galaxies. Nevertheless, measuring this redshift drift is one of the prime goals of the Codex Instrument on the E-ELT (see Pasquini et al. 2010, http://esoads.eso.org/abs/2010Msngr.140...20P )
using Lyman alpha absorption systems towards distant quasars. This experiment is also planned for the Square Kilometre Array, using the 21cm line (Kloeckner et al. 2015 http://arxiv.org/abs/1501.03822 ).

In both cases, to overcome the experimental uncertainties (eg at 21 cm, it amounts to line drifts of 0.1 Hz over a decade), then observations of millions of galaxies must be combined.

There is no prospect of measuring this effect in an individual galaxy, furthermore I fear your understanding of cosmological redshift is flawed. The dependence on distance is a statistical average, not an absolute dependence. Individual galaxies are moving in individual gravitational potentials from objects around them. This gives them their peculiar velocities with respect to the flow. This velocity could increase or decrease as a galaxy got further away, but is never expected to be large enough to be detectable on human timescales for any individual galaxy. In addition, any change in peculiar velocity should average to zero when looking at millions of galaxies, leaving the redshift drift due to the expansion.


Apparent Centering Problem When Using the Perifocal System to Draw Orbital Ellipses

I solved the issue.
Everything I posted in the OP was correct; I made a simple, stupid sign error when transcribing the expressions into code.
It should also be noted that in the OP, the expression for Pk is wrong.  The correct expression is
Pk =   sin (inc) sin (aop)
Just in case anybody is trying to pull this off themselves!
Thanks for the ideas everyone!
Correct Orbits:


How much delta-v does a typical geostationary satellite carry?

There's a mountain of readily available information, of which the following is one example.  From   GOES page
edit
There is more information there about the thrust produced, so I suspect you can calculate the total delta-kinetic energy available.

A2100 has been used for over 40 missions and is available in different versions catered to specific mission needs in terms of
dimensions and power availability with satellite mass varying between
2,000 and 6,740 Kilograms.
Each GOES-R-class satellite is 6.1 by 5.6 by 3.9 meters in size and has a dry mass of 2,857 Kilograms. At liftoff, the satellite
weighs 5,192 Kilograms, loaded with propellants to support a mission
of at least 14 years.
GOES-R, S, T, U launches with a propellant and pressurant load of
2,335 Kilograms with Mixed Oxides of Nitrogen and Hydrazine acting as
main propellant components and Helium gas used for the pressurization
of tanks.


How do you define the diameter of the Sun

Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.

The base of the Photosphere is defined as the region where the optical depth is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.

Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.


Do boulders erode differently on asteroids than on the Moon?

The moon does attract more particles.  But more importantly, those particles hit the moon with velocities of the order of the Moon's escape velocity, 2.4 km/s.  That's a very powerful sandblaster...


Any new info about solar flares hitting the Moon added by the LADEE mission?

You'd probably be most interested in the results of the Lunar Dust EXperiment (LDEX).  A 2015 paper states

  LDEX data show no evidence for an electrostaticallylofted
  dust component at densities greater than a few
  per m3

I am assuming the solar flare stuff didn't pan out, otherwise it would be mentioned in the various LDEX summary papers.  In addition, the NASA DREAM group site doesn't seem to have anything relevant LADEE results, and they're the ones that brought the subject up in the first place.