How do different formulas for calculating the mean anomaly of an elliptic orbit correlate?

Kepler's Third Law including the constants of proportionality is:
$$GM T^2 = 4\pi^2 a^3$$
Substituting $\mu = GM$, this can be rearranged to give:
$$\frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$
Which lets you rewrite your first formula as your second one and vice-versa.
Note that in the first formula the reference epoch $t_0$ is assumed to be the time of periapsis, which corresponds to the case $M_0 = 0$. Your definition of $\Delta t$ should also read $\Delta t = t - t_0$, i.e. time since the reference epoch.


How can I find the depth of a portion of the Valles Marineris?

You can find MOLA laser altimeter data here:https://pds-geosciences.wustl.edu/missions/mgs/megdr.html.  The files are simple rectangular grids of differences between Martian "sea-level" and measure height at 1/128 degree resolution.  If you have the latitude and longitude, it is quite straightforward to find the height.

To view these files, you probably need the software found here: https://pds.jpl.nasa.gov/tools/about/pds3-tools/nasa-view.shtml

Looks like the software might be a little out-of-date, but you can give it a try.


What would our night sky look like if the solar system was 100 light years from the centre of the Milky Way

I have tried to collate some info from various other answers.
Density of stars where we are:
About 0.15 per cubic parsec
Density of stars "at the middle of the galaxy".  Unfortunately I don't know how "100 ly" affects this. It's probably about this figure.
About 50-100 stars per cubic parsec
So in this QA it's "about 500x as dense".
It would seem that this is in fact:
Not very bright. About the same amount of light overall as when there's a full moon.
However the overwhelming factor is
If one particular star was "very close" to us?
How likely would it be, in the region under discussion, that another star is "very" close to us - close enough that that one particular star is extremely bright?
The answer seems to be
No, the stars there are not so close that you'd typically be close enough to a really bright star that it would be bright.
That seems to be the bottom line.
(There's a completely pointless, if pretty, image here showing what it's like "inside a star cluster" {where? At what depth?} here  Unfortunately that image does not simply state the star density they are simulating - so it is of no help at all to this excellent QA!)
the bottom line then:
"If our solar system was only 100 light years from the centre of the Galaxy, would we still have a dark night sky?"  Answer, yes, there would just be "more stars". It would be hardly any brighter.
To improve on this answer, the next thing we need to quantify is one piece of information:
What actually is the star density at "100 ly from the center"?
Is it 50, 100, 25, 300, 72.275 stars per cubic parsec?
I don't know.

A fun point, would it be different (slightly? a lot?) if inside the galactic bar?  IDK.


Why are perihelion and the shortest day in the northern hemisphere different?

We have "days" because the Earth is rotating. The shape of the orbit has little effect on the daily cycle of light and dark.

The length of the day is determined by the tilt of Earth to the orbit of the Earth around the sun.  When the Northern hemisphere is pointed away from the sun, then the sun is low in the sky and is below the horizon for longer. So the shortest day happens when the Earth is at the point in its orbit when the angle that the Earth's axis makes with the line from the Earth to the Sun is greatest. This is on about the 21 December.

The longest and shortest days of the year occur because of the tilt of the planet, which also controls when the Equinoxes and Solstices occur. The tilt of the planet has absolutely nothing to do with its location in orbit, and it's only pure coincidence that they're so close together. Also, perihelion moves over time (apsidal precession) and in 56,000 years it would occur on the longest day in the Northern Hemisphere. Except for the fact that the Earth's axis will have also precessed (with a period of around 26,000 years) so the shortest day will occur somewhere else on in the Earth's orbit

For the sake of clarity: The equinox does not occur on the minor axis. The solstice does not occur on the major axis. The tilt of the Earth to then orbital plane does not change over the course of the year, and the Earth's rotational axis is no aligned with the major axis of the Earth's orbit.


No mass blob of stellar mass is transparent at any wavelength of interest. Opacities $\kappa_{\nu}$(inverse transparency) as function of wavelength becomes really high and broad band at pressures above > 0.1 bars, for all wavelengths.
This leads to the optical depths $\tau_{\nu}$ being enormous and as transmission is $T=1-\exp(-\tau)$, you won't be able to see through any star, ever.


What does the term alien planet actually refer to?

Very interesting article!. The term alien planet is not a scientific one, so I think you already answered your own question by saying that it was used in the article to generate more buzz.

In the English language the word is used to describe something that's not familiar, so you could argue that this is in fact an alien planet, but because of the subject in question being an exoplanet I would bet they did this as a technique to get more attention.


Is there a standard mapping of symbols to terms for celestial and orbital mechanics

There are a lot of instances (such as those you pointed out) in all of science and mathematics where a symbol can mean two different things. In mathematics, there's actually a good deal of this (and the article says that the list is incomplete!). Many are based on letters, although they may have deviated from the original shape in order to make a distinction regarding just what the heck the thing actually means.

Eventually, you run out of letters. You can opt to use symbols from another language, the first choice for many being Greek. That roughly doubles the amount of letter-based symbols available to you. But there are still a lot of duplicates, as you can see by browsing through the lists here and here.

To my knowledge, this is the most comprehensive list easily available. And yes, there are loads of duplicates. But the great thing is that you almost certainly won't have to use the same variable to mean multiple things in one equation, or set of equations. For example, I doubt you'll need to use $V$ for voltage, volume and shear force all in the same equation!


I have seen so many articles but how can finding exoplanet density determine composition

You can't get an exact composition just from density, but it is a clue.
In our solar system there are "rocky" planets with a density of about 5 to 6 g/cm³
And there are gas planets with a density of between 0.5 and 2 g/cm³
So if an exoplanet is found to have a density of 5.3 g/cm³  and a fairly small mass we can guess that the composition is rock (probably siliates with an iron core just like the rocky planets in our solar system)  And if the density is 1.1 g/cm³ we can guess that the planet is made of gasses like hydrogen and helium, perhaps with "ices" like methane or H₂O. If the mass is much greater than that  of Jupiter, then the planet is likely be be a (sub)-brown-dwarf.
You need both the mass and the density to make this judgement, as the density of gas giants depends on their mass, and larger gas giants have a greater density.


Is it possible to observe Polaris and the Southern Cross from the equator during an equinox

The March equinox would be a good time to see both; the September equinox would not.
Crux
lies roughly between RA 12h to 13h, Dec -55$^\circ$ to -65$^\circ$.
For best results, be at latitude 10-15$^\circ$ N so that both Crux and Polaris are well above the horizon, and look between 11:00 and 14:00 local
sidereal time
so that Crux is near the meridian.
That would be around midnight in March-April, late evening in May, wee hours in February, or daytime in July-December.


The dominance of Gravity vs Strength and its dependency on crater size

I guess the point is that there are two ways to calculate how big a crater a given impact will create. One is based on calculating how much rock the impact energy can shatter so the main constraint is the strength of the rock) the other on calculating how much it can lift (so the main constraint is the weight of the rock). For small impacts the first is the main consideration ("dominates") for large impact's its the second and the formula identifies the point around which it changes over.


Can you see something active in the sky apart from satellites? Can there be amateur time-domain astronomy?

If it moves or flashes it isn't astronomy, it is meteorology or technology.
There are only a few exceptions to this:  Meteors are an atmospheric phenomenon, and a meteor will appear to move rapidly across the sky. But because they "come from space" and occur well above the clouds they are often considered to be part of astronomy.
As you note, eclipses and occultations happen quickly enough for the changes to be visible. The moons of Jupiter do move notably during the night, and when one moves into or out of shadow, it can appear or disappear over the course of a couple of minutes: easily noticeable.
A supernova could brighten quickly enough for the variation to be visible over a night's observation:  You wouldn't see it suddenly appear, but over the course of a night it could appear brighter at the end of the night than at the beginning.
Similarly, Algol will fade from magnitude 2 to 3.5 over a few hours. It isn't quick enough to notice the change in the moment, but it is quite clear if you are observing over a few hours.
Pulsars are too dim to be seen with normal equipment even the bright nearby crab pulsar has a magnitude of 16.5 (and at one flash in 33 milliseconds, it is too quick to see by eye)
GRB 080319B was an exceptional object. It was a gamma-ray burst. It gave a flash of gamma rays that lasted a little over a minute.  If you had happened to know exactly where to look you could, (marginally with the naked eye, but easily with binoculars) have seen the optical counterpart to the gamma-ray burst.  This was created by the formation as a massive early star collapsed to a black hole producing a jet of energy that happened to point our way.
The sun is changing and on a small scale it does change on the scale of minutes, but it is hard to see any movement at that scale with basic equipment.  If you can get your hands on a Hydrogen-alpha solar telescope you will, be able to see prominances that change over a period of hours.
Rarely an asteroid will pass so close so as to be visible.  Apopsis will have a close approach in 2029 and will appear as a slowly moving star.
Jupiter is active at radio frequencies.  You can tune in to Jupiter with the right equipment and hear it changing at rates from a few fractions of a second to a few seconds, see e.g. radiosky.com > Jupiter Central
These are exceptions.  In general, it is rare for anything so big and powerful that it can be seen over a distance of many light-years to be able to change fast enough that those changes can be noticed by our eyes.


Do other planets with a magnetic field also have two Van Allen belts?

Yes, Jupiter and Saturn have their own radiation belts.  Jupiter's is extremely powerful, which is why all of Juno's sensitive electronics had to be radiation hardened and shielded inside a vault.  It's also why Juno is going to be on a polar orbit doing flybys of Jupiter, instead of orbiting around it closely.

These magnetic bands are indeed separated into inner and outer portions.  From Wikipedia:

  Jupiter's magnetosphere is traditionally divided into three parts: the inner, middle and outer magnetosphere. The inner magnetosphere is located at distances closer than 10 RJ from the planet. The magnetic field within it remains approximately dipole, because contributions from the currents flowing in the magnetospheric equatorial plasma sheet are small. In the middle (between 10 and 40 RJ) and outer (further than 40 RJ) magnetospheres, the magnetic field is not a dipole, and is seriously disturbed by its interaction with the plasma sheet (see magnetodisk below).
  Jupiter's inner magnetic belt was mapped by Galileo, showing it contains high-energy electrons.  Eruptions from Io result in the middle portion of the magnetosphere containing huge amounts of ions of various elements.

Saturn's magnetosphere is likewise divided into separate parts.  However, there are 4 regions to Saturn's magnetic field, and they are separated by the kinds of particles they contain.  These particles result from interactions with Saturn's various moons, as well as its ring system.

The Galileo probe discovered that Europa does produce a weak magnetic field, as a result of its subsurface ocean.  This ocean has an induced current created from Europa moving through Jupiter's magnetosphere, which in turn creates a magnetic field.  This magnetic field is actually one of the things which caused many scientists to theorize a subsurface ocean.

However, these belts are not "Van Allen" belts.  That name is reserved strictly for Earth's radiation belts.


When will a day on Earth & Mars be the same length?

Since the rate of lengthening of the day on Mars is three orders of magnitude less than that of the Earth for a first go at calculating how long it will be before the day lengths are equal we can ignore the change in day length on Mars. The difference in day lengths is $\approx 40$ minutes. The Earths day lengthens by $\approx 1.7$ ms/100 years. Therefore the day lengths will be equal in: $\frac{40\times 60}{1.7\times 10^{-3}}\times 100 \approx 141 \times 10^6$ years


What conditions would lead to this event around the black hole in the Pictor A galaxy?

This is a clear example of an astrophysical jet, in this case, most likely a relativistic jet. In short, an accretion disk forms around a black hole (supermassive or otherwise). Matter is pulled towards the black hole and further energized, before being accelerated into a jet emanating from the black hole's poles. Two different mechanisms have been proposed for the formation of jets:

The Blandford-Znajek process requires that a magnetic field forms (from the accretion disk) that is centered around the black hole. Charged particles then move along the field lines, into jets. I recently wrote an answer about the details (see How does an accreting black hole acquire magnetic fields?). For this process to work, you need an accretion disk. It is generally considered the most likely explanation for black hole jets.
The Penrose process takes rotational kinetic energy from the ergosphere outside the event horizon and gives it to particles moving in jets. Note that this does not rely as heavily on the accretion disk as the Blandford-Znajek process does. For this process to work, you need a rotating black hole surrounded by some matter, likely in a disk.

The hotspot is, to me, much more interesting. It reminds me of structures seen around young stars: bipolar outflows (streams of gas that can form shock waves) and Herbig-Haro objects (the results of shock waves from relativistic jets. Obviously, the mechanisms are different, so no clear analogy can be drawn. But what is interesting about bipolar outflows and Herbig-Haro objects is that the shock waves produced therein result from collisions with the interstellar medium.

If a similar mechanism were to cause the shock waves by the hotspot, then we could conclude that the jets have hit the intergalactic medium. But I don't think this is necessarily the case, in part because of just how long these jets are prior to the formation of the hotspot. One would think that if the hotspot and shock waves are because of collisions with the intergalactic medium, the jets would be much shorter, because they would likely have reached higher density regions of it sooner. So that's why I find it interesting, and why I can't give you a good reason as to why the hotspot formed where it did, or the precise reason for it being there at all.


Does the gravity oppose to the Universe expansion?

Expansion depends on the amount of mass

Yes, space expands more in regions with less matter — in fact this has been proposed as an alternative explanation to dark energy as the cause of the observed accelerated expansion: If by chance we happen to be located near the center of a "cosmic void", then nearby space expands faster than distant space, and since nearby space corresponds to recent times, this would be interpreted as recent accelerated expansion. This idea was popular some ten years ago (e.g. Conley et al. 2007; Wiltshire 2008), but isn't to my knowledge so hot anymore.

Similarly, in regions with more matter, space expands less. In fact, on scales as small as galaxies, and even galaxy groups, space doesn't expand at all. Gravity prevents galaxies from expanding, and keeps galaxies near each other from receding. On even smaller scales, electromagnetic (and nuclear) forces keeps objects like stars, planets, and bicycles from expanding.

Expansion is physical

For this reason, space between two stars doesn't expand. So, to address your question, let's consider instead two galaxies, separated by millions of light-years. However, fundamentally we cannot place a ruler between them and watch it expand. Electromagnetic forces would try to hold it together, fail, and the ruler would break.

Why is this? Imagine a rigid ruler, one megaparsec long (i.e. 3.26 million light-years). Since the Hubble constant is $H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, the region at the other end of the ruler recedes at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. That is, space "drags" the two ends of the ruler away from each other at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. If the ruler doesn't break, a local observer will see the end of the ruler whizzing by at high speeds.

Perhaps you might be able to construct a rod able to withstand this stress. If I were a solid state physicist I might be able to calculate the maximum possible length of a rod, given an optimal material, but I'm not. But the point is that, no matter what material you come up with it is not even practically impossible to make it arbitrarily long, but "theoretically" impossible. There is a fundamental limit to the length of a physical ruler, namely the length corresponding to the distance at which two regions of space recedes at the speed of light, i.e. $d = c/H_0 = 4400\,\mathrm{Mpc}$, or $14.4\,\mathrm{Glyr}$. If the ruler were longer than this and didn't break, then a local observer would see the end move by faster than the speed of light, which is not possible.

Luckily, there are other ways to measure distances, e.g. by "standard candles". And yes, if you measure the physical distance between the two galaxies at two different times, it will have increased in the meantime, physically.

(You could also imagine "freezing" space by magic, laying out 1-meter rods. In that case, second time you do it you will need more rods.)


How to convert from Cartesian J2000 coordinates to Cartesian Galactic coordinates?

To transform an equatorial $(x, y, z)$
into a galactic $(x_G, y_G, z_G)$, the
Gaia data release 1 documentation, section 3.1.7, uses

$$
\begin{bmatrix} x_G \\ y_G \\ z_G\end{bmatrix}
= \mathbf{A}'_G
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
$$

where

$$
\mathbf{A}'_G = \begin{bmatrix}
{−0.054876} ~ {−0.873437} ~ {−0.483835} \\
{+0.494109} ~ {−0.444830} ~ {+0.746982} \\
{−0.867666} ~ {−0.198076} ~ {+0.455984}
\end{bmatrix}
$$
is composed of three coordinate frame rotations
$$
\mathbf{A}'_G = \mathbf{R}_Z(-l_\Omega)
~ \mathbf{R}_X(90^\circ - \delta_G)
~ \mathbf{R}_Z(\alpha_G + 90^\circ)
$$
with the north galactic pole at equatorial coordinates
$(\alpha_G = 192.859^\circ, \delta_G = +27.128^\circ)$
and the galactic equator crossing the celestial equator at galactic longitude
$l_\Omega = 32.932^\circ$.

This is equivalent to the transformation defined in the
Hipparcos catalogue, volume 1, section 1.5.3.

Murray 1989 and
Liu 2011
give similar matrices whose elements differ after 6 decimal places, which matters only if you require sub-arcsecond precision.


You could confirm with taking an image of the star trails.  They would form a circle with the apparent center at the zenith of the location.  You do not need a pole star at all.  Just a night of viewing.

You would also be able to tell based on the height above the horizon that the sun is.  On the equinox, the sun would be on the horizon at noon (when it is at its maximum height).  On the summer solstice, its angle will be above the horizon equal to the angle of precession.  And below by the same amount on the winter solstice.

Measuring latitude using the sun

For any other day in the year, you will need to determine the max angle based where you are between the equinox and solstice.

Compensating for the Earth's tilt

Basically, you would determine if you were at a pole the same way you determine any other latitude.  You are just looking for a result of 90 degrees.


If you were using a telescope with a central obstruction (like a newtonian reflector (including dobsonian mounted ones) or an SCT or MCT, then what you saw is most likely an out of focus bright object (such as a star).

When things are in focus, a star shows as a point (actually, at very high magnifications, a tiny disc with concentric light and dark rings around it - this is called an Airey disc - caused by diffraction effects - but usually the magnification is low enough that it looks like a point).

When things are out of focus, you seen the shape of the telescopes aperture that gets bigger the more out of focus you are. For a refractor, with no central obstruction, this is a circle. For things with a central obstruction, you see a ring (donut) shape. This is what causes donut shaped out of focus highlights on pictures taken with mirror lenses. (and why normal out of focus highlights with normal lenses are round or close to it (the camera iris blades often produce a polygon shape when the lens is stopped down)).


Why is the discovery of merging neutron stars important?

Reasons why this is important:

It is the first simultaneous detection of a gravitational wave and electromagnetic signal, and the strongest GW signal yet in terms of signal to noise (Abbott et al. 2017a). It spectacularly corroborates the reality of the GW detection technology and analysis. The progenitor has been unambiguously located in a (relatively) nearby galaxy (Soares-Santos et al. 2017), allowing a host of other telescopes to obtain detailed measurements.

It shows that GWs travel at the speed of light, a further verification of Einstein's General Relativity (Abbott et al. 2017b).

It shows that most of the very heavy elements such as gold, platinum, osmium etc. are plausibly produced by merging neutron stars and constrains the rate of such mergers in the local universe (e.g. Chornock et al. 2017; Tanvir et al. 2017).

It shows that short gamma ray bursts — some of the most energetic explosions in the universe — can be caused by neutron star mergers (e.g. Savchenko et al. 2017; Goldstein et al. 2017).

It is the closest detected short gamma ray burst (with a known distance). That the progenitor has also been characterised allows a closer investigation of the interesting physics underlying the ejection and jet mechanisms thought to be responsible for the gamma rays and later X-ray and radio emission (e.g. Margutti et al. 2017; Alexander et al. 2017).

It provides observational constraints on how matter behaves at extremely high densities, testing our understanding of fundamental physics to its limits — for example, the details of the gravitational wave signal moments before merger are diagnostic of the interior conditions of neutron stars at densities of $\sim 10^{18}$ kg/m$^3$ (Hinderer et al. 2010; Postnikov et al. 2010).

It provides an independent way of measuring the expansion of the universe. Merging binary gravitational wave sources are known as "standard sirens", because the distance to the GW source pops straight out of the analysis and can be compared with the redshift of the identified host galaxy (Abbott et al. 2017c). The result agrees with measurements made using the cosmic microwave background and the distance-redshift relation calibrated by other means, verifying our estimation of distances, at least in the local universe.

Finally, this event will turn out to be important because it was lucky; in the sense that the source was detected well-inside the sensitivity horizon of LIGO (Abbott et al. 2017a). The detection itself, was not unexpected given the rates predicted based on studying the neutron star binary systems in our own Galaxy (e.g. Kim et al. 2015), but the fact that it was so close —
 within the closest 5% of the sensitive survey volume where it could have been detected — is fortunate.

In the end, if someone thinks none of the above is interesting or important, then nothing I can write will convince them otherwise. The vast majority of people I speak to are curious and fascinated to find out about our cosmic origins and how the universe works.


Objects entering or leaving the observable universe

I don't know if there have been any observations of objects leaving the observable universe, but I'll admit that I have a hard time keeping up with the latest discoveries in observational astronomy. I'll try to approach your question from a theoretical viewpoint.

As you pointed out, the radius of the observable universe is $4.6 \times 10^{49}$ light-years. To figure out how fast an object is moving away from Earth (or any observer), we can use Hubble's law:

$$v = H_0D$$

where $v$ is the recessional velocity, $H_0$ is Hubble's constant, and $D$ is the proper distance from the observer to the object (see Wikipedia for the difference between proper distance and commoving distance). The trouble here is that the exact value of Hubble's constant isn't exactly well-known. There have been many different observations since Hubble proposed his law that attempted to find values for the constant, but they range widely. Here I'll use $82$ $km$ $s^{-1} /Mpc$ (where $Mpc$ is a megaparsec - 1 million parsecs).

Let's say that the object in question is $13$ gigaparsecs away (so $13,000$ megaparsecs away). The light we see from the galaxy would be from when it was very young. One of the most distant observed galaxies is this far away, and only $2,000$ light-years across (note: do not confuse the "light-travel distance" with the "proper distance"), so let's say that this galaxy is smaller in size - negligibly small, in fact. We can then use Hubble's law to calculate a recessional velocity $v$ of

$$(82)(13,000) = 1.066 \times 10^6$$ kilometers per second, or $3.36 \times 10^{13}$ kilometers per year. That's pretty fast! An object $13$ gigaparsecs away would be $42.38$ billion light-years away, not far from the edge of the observable universe. Yet if the edge of the observable universe is $46$ billion light-years away, the galaxy would still be $3.62 \times 10^{21}$ kilometers away, and it would take a long time to get there. A galaxy this young would be maybe $1,000$ light-years across, so, moving at its current speed, it would take it many years to cross a distance its own length - that is, if the furthest edge of it were at the edge of the observable universe, it would still take many years to fully disappear. And that's even factoring in that it would be moving away faster!

The reason I don't have the object being further away is that any objects at the edge of the observable universe would be so young when they released the light we see today. That means that galaxies would be small, and wouldn't emit a lot of light. One of the furthest objects we have detected is only $30$ billion light-years away, and $2,000$ light-years across. So it would be terribly hard to observe an object at the edge of the observable universe (ironic, isn't it?). But here I have shown that it would take years for the object to completely disappear.

I hope this helps.

My sources for the data and conversion factors (i.e. light-years to kilometers):

Hubble's Law

Wolfram Science World

Most distant objects

Spacetelescope.org

Light-year


Help identify a bright and round object photographed through a telescope

It was a brightest object in the sky.

This already makes one think "Venus?"

At about 6 PM on 25-Jan-2019 from Riga Latvia the planet Venus was about 17 degrees above the horizon in the Southwest.

Venus is always strikingly bright when it's visible, it's apparent magnitude stays between about -3.8 and -4.8.

From in-the-sky.org's planetarium mode

This matches nicely to your photo out the window:

But what about the donut-shaped image?

Your image is way, way out of focus. The the outside of the annulus represents the diameter of the telescope's primary and the inside represents the diameter of the blockage by the secondary mirror holder. The blockage is about 45% of the radius or about 20% of the area, which seems reasonable for some telescopes. Here's a random example of a telescope with a similar blockage.

Here are some other celestial donuts to enjoy! In this case they are caused by a tiny spot of dust which is out of focus, and one is dark rather than bright, but they are other examples of how point-like objects that are out of focus show you your aperture.

above: from this answer, below: from this answer


There is an observational bias and it is taken into account when you see inferences about planet frequency.

The methods to find planets are inherently biased towards finding large, close-in planets. Both the transit and doppler-based methods suffer from this bias.

The paper you reference takes into account this bias to arrive at the statistics you highlight in your question. Note however that the figure for Earth-sized planets is that 16.5% of stars have at least one such planet in an orbit with periods less than 85 days. There are no planets of any kind at all in our solar system that have periods less than 85 days. The figure quoted for Neptune-sized planets is 31% with orbital periods less than 245 days and there are no such objects in the solar system. Finally, the figure quoted for giant planets is that 5.2% of stars have a giant planet in an orbit with a period less than 418 days. Again, no such planet exists in the solar system.

You are correct that it should be unlikely for giant planets to form close to their parent star and this is one of the reasons that they are rarely found there. It is widely believed that the close-in gas giant planets that are seen, have migrated inwards from a formation site further out.

That said, it is perhaps true that our solar system is a little unusual.
Whilst we do not yet know what fraction of stars have gas giants in Saturn-like orbits (because observations have not been going on long enough), some work (e.g. Rowan et al. 2015) suggests that only $\sim 3$% of Sun-like stars have an analogue of Jupiter (periods of 5-15 years and masses of 0.3-3 times that of Jupiter).


Why the habitable zone of a binary star is a disc and not a sphere?

If you go to the associated paper Müller & Haghighipour (2014), it shows that what they're plotting are the points which satisfy:
$$\frac{L_\odot}{l_{\rm In-Sun}^2} \le \sum_{i=1}^{N} W_i (T_{\rm star}) \frac{L_i / L_\odot}{d_i^2} \le \frac{L_\odot}{l_{\rm Out-Sun}^2}$$
where:

$L_\odot$ is the luminosity of the Sun
$L_i$ is the luminosity of the $i$th star (of a total $N$ stars)
$l_{\rm In-Sun}$ and $l_{\rm Out-Sun}$ are the inner and outer boundaries of the Sun's habitable zone (two sets of distances are used for the conservative and extended habitable zones, plotted as dark and light green respectively)
$d_i$ is the distance from the point to the $i$th star
$W_i(T_{\rm star})$ is a weight factor that accounts for the different spectral energy distribution at different temperatures $T_{\rm star}$. Details of the calculation of this weight factor are given in the paper.

This formula does not rely on the objects being co-planar. If the binary were inclined relative to the plane of the plot, the $d_i$ values would be altered at each plotted point to account for the component of the distance out of the plane, but essentially this would just be a different slice of a 3-dimensional volume.


Can a star eject a liquid or a gas that has not been ionized?

There are a variety of scenarios in which a star can emit gases, although that's probably not the best way to think about it. A better way to visualize it is in terms of mass loss rather than gas emission. Inevitably, most or all of a stars mass must eventually go "back to the galaxy in which they are in."

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

In a star the size of our Sun or thereabouts, towards the end of its life it will form a red giant. After this stage, its outer layers will "puff out" creating a planetary nebula. This is essentially the star emitting most of its material out to its surroundings, contributing its mass to the formation of new stars later on.

The other route, the supernova, happens when a star much more massive than the sun starts fusing iron and suddenly collapses, rebounding and creating a large shockwave which expels most of its material out.

Lastly, certain kinds of stars like Wolf-Rayet stars or (possibly) R Coronae Borealis variables can expel matter due to the fact that they are unstable and can suddenly eject part of their outer layers.


Rotation of the Earth about its own axis

Origin of Earth rotation
There are two main phenomena that explain Earth rotation. First one is linked to the history of the solar system: there is angular momentum in the interstellar medium (ISM), so the collapsing cloud that formed our solar system had an initial angular momentum (rotation of the galaxy and turbulence in the ISM are typically a good candidate for that; see for example Mac Low & Klessen 2004). By conservation of the latter, particles in the collapsing cloud accelerate (the norm of the angular momentum $I$ is $mrv$, with $m$ the mass of the system, $r$ the radius and $v$ the velocity; as the cloud contracts, $r$ decreases and then $v$ has to increase) and due to centrifugal force, a disk will tend to form. By conservation of the angular momentum in the protoplanetary disk, you will also tend to form bodies that rotates.
The second phenomena is linked to the formation history of the Earth. Simply put, Earth formation history is due to collisions of planetesimals (let say they are big chunks of rock, formed by a collision history of smaller chunks of rock, formed by...); these planetesimals are orbiting the forming star in the protoplanetary disk, their rotation being linked to the disk formation history (see above). When planetesimals collide, two things can basically happen: either the collision is perfectly central (meaning that the momentum of one of the two colliding objects is pointed toward the center of mass of the second object) and there is no transfer of angular momentum, either the collision is lateral and there is a transfer of angular momentum. Depending on the configuration, you can either transfer momentum that will make the other body rotating on way or the other. That being said, you can then object that there should be a equirepartition of collisions pushing the proto-Earth to rotate in both directions. Which is true, but external planetesimals are rotating faster than internal ones. Therefore, from the proto-Earth perspective (meaning, in a physical sense, in its reference frame), internal planetesimals are seen as coming backward, from front, and external planetesimals are seen as coming forward, from behind. Long story short, there are both transfering momentum in the same "direction", forcing the proto-Earth to rotate in the same direction as the rotation of the protoplanetary disk (see for example Artem'ev & Radzievskii 1965).
Rotation around the Sun
This one's easy: gravity.
The Earth is constantly falling towards the Sun because of gravity, and the centripetal acceleration counteracts perfectly gravity, mainting the Earth on its orbit.


Why do certain massive stars leave no remnants?

The gap appears because of pair instability supernovae. In short, as one looks at such massive stellar cores at increasing temperatures, an ever-larger fraction of the photons are sufficiently energetic to spontaneously form electron-positron pairs. True, they soon recombine, but there is nevertheless a loss in (radiation) pressure, which causes contraction, and thus a further increase of temperature. If unstable, as appears to be the case in such massive stars, this leads to more high-energy photons, more pairs, and a further loss of pressure.

Models of such objects suggest that the collapse leads to a sudden, rapid ignition of oxygen- and silicon-burning reactions. In a mass range between about 150 and 250 solar masses (corresponding to core masses of very roughly half of that), the thermonuclear explosion is enough to rip the core apart, and there's nothing left to collapse. This is a bit like how the sudden onset of carbon burning destroys white dwarfs in type Ia supernovae. More massive cores are more strongly bound by gravity, and models suggest that the nuclear explosion isn't enough to unbind the core.

So below about 150 solar masses (and at higher metallicities, I think), the core doesn't reach pair-unstable conditions. Above about 250 solar masses, the induced nuclear explosion isn't powerful enough to destroy the core.


Imagine an hour which goes counter-clockwise. On the left, there are the numbers from 0 to 12, and on the right, there are the same, but the negative numbers.

Now imagine yourself standing in the middle of such dial. If you have HA = 0, you know that the star is on the meridian. If HA = 6 or HA = -6, then the projection of star onto the celestial equator is on the horizon. If you have HA = 12, then the star lies on the opposite side of the meridian. The same is valid for decimal values.

Look also at this simulation for demonstration of the hour angle, and at the Wikipedia page about this.

But beware! The altitude isn't the same as hour angle. HA is based on equatorial system, but the altitude is based on the horizontal system. The altitude tels you the height of the star, but the HA tells you the position of the star in the fictional dial.

The right ascension is useful because the star is (less or more) standing still in it. RA is for the distant star constant. But HA with the combination of declination tells you more presentable coordinate system. If you know RA and if you want to know, where the star is, you have to know date and hour, and you can't imagine the position easily. But if you know HA, you know it easily, you just look at the upper dial. (Note that we are looking toward the North in the upper image.) For example, do you know where the RA = 0 is located now? Of course not. And do you know where the HA = 0 is located now? Yes, you can just look at the dial. HA = 0 line is also called the meridian. Also, you can easily convert from HA to RA and conversely using current sidereal or astronomer time. (Look at the Wikipedia article.)

Short answer: You can easily imagine hour angle (HA), meanwhile, you cannot imagine the right ascension (RA) so easily.


Is there any way a meteor can hit at less than escape velocity?

Edited.
No if you talk about the escape velocity form Earth. This follows simply from the fact that energy $E$ is conserved. An object that is not gravitationally bound to Earth has $E>0$ and hence $v>v_{\mathrm{esc}}$ when hitting ground.

Yes, if you meant the escape velocity from the Solar system, because the Earth moves with $v_{\rm Earth}=v_{\rm escape}/\sqrt{2}$ relative to (but not towards) the Sun. Here $v_{\rm escape}=\sqrt{2GM_{\odot}/1{\rm AU}}$ is the local escape speed from the Sun, while $v_{\rm Earth}=\sqrt{GM_{\odot}/1{\rm AU}}$ is the speed of the local circular orbit. An object at 1AU form the Sun and bound to the Sun cannot have speed greater than $v_{\rm escape}$.

Now, the impact speed of an object that moves at $v_{\rm escape}$ can be as low $v_{\rm escape}-v_{\rm Earth}=v_{\rm escape}(1-1/\sqrt{2})$ if it hits Earth "from behind", i.e. moving in the same direction as Earth at the time of impact.

Note also that meterors typically move not faster than $v_{\rm escape}$, for they don't come from outer space, but from the Solar system.


How does the surface gravity on Mars vary between the equator and its poles?

The following is excerpted from from this answer to the Space Exploration SE question "Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects?".

It essentially answers my question. I would appreciate if someone gave the info about the gravimeter on the Phoenix lander too.

So the surface gravity on Mars is:

3.70703 m/s² (0.378 g) on the equator
3.71683 m/s² (0.379 g) on the midlatitudes
3.73493 m/s² (0.381 g) on the poles

  So here's a complete tally using

  $$a_G = -GM \frac{\mathbf{r}}{r^3}$$

  From Geopotential_model; The_deviations of Earth's  gravitational field from that of a homogeneous sphere:

  $$a_{J2x} = J2 \frac{\mathbf{x}}{r^7}(6z^2 - 1.5(x^2+y^2))$$
  $$a_{J2y} = J2 \frac{\mathbf{y}}{r^7}(6z^2 - 1.5(x^2+y^2))$$
  $$a_{J2z} = J2 \frac{\mathbf{z}}{r^7}(3z^2 - 4.5(x^2+y^2))$$

  $$a_C = \mathbf{r_{xy}} \omega^2 $$

            magnitudes shown only (sign indicates generally "up" or "down")
                  at the equator        at the pole
              h = 0     h = -4500 m        h = 0
GM          -3.71317     -3.72303        -3.75729
J2          -0.01092     -0.01088        +0.02236
centri      +0.01706     +0.01697         0.0

vector sum  -3.70703     -3.71683        -3.73493

  Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects?

  Yes they do! At 5.2 degrees south latitude and -4500 meters altitude (bottom of Gale Crater) the acceleration is -3.7168 m/s taking into account GM, J2, and centrifugal effects.

  So as Cheap Trick tells us and Meatloaf reiterated more elegantly in Roadie:

  Everything works if you let it!

  Here's some Python for double-checking my math:

def accelerations(rr):

    x,   y,   z   = rr
    xsq, ysq, zsq = rr**2

    rsq   = (rr**2).sum()
    rabs  = np.sqrt(rsq)
    nr    = rr / rabs
    rxy   = np.sqrt(xsq + ysq)
    rrxy  = rr * np.array([1.0, 1.0, 0.0])
    nxy   = rrxy/rxy
    rm3   = rsq**-1.5
    rm7   = rsq**-3.5

    acc0  = -GM_mars * rr * rm3

    # https://en.wikipedia.org/wiki/Geopotential_model#The_deviations_of_Earth.27s_gravitational_field_from_that_of_a_homogeneous_sphere
    acc2x = x * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2y = y * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2z = z * rm7 * (3*zsq - 4.5*(xsq + ysq))

    acc2  = J2_mars * np.hstack((acc2x, acc2y, acc2z))

    accc = nxy * omega**2 * rxy

    return acc0, acc2, accc

import numpy as np

halfpi, pi, twopi = [f*np.pi for f in [0.5, 1, 2]]
degs, rads        = 180./pi, pi/180.

R_mars  = 3396200.0
GM_mars = 4.282837E+13   #  m^3/s^2  https://en.wikipedia.org/wiki/Standard_gravitational_parameter
J2_mars = GM_mars * R_mars**2 * 1960.45E-06     # https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

Req = 3396.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars
R   = Req - 4500.    # https://en.wikipedia.org/wiki/Gale_(crater)
Rpo = 3376.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars

sidereal_day  = 1.025957 # https://en.wikipedia.org/wiki/Mars
T             = sidereal_day * 24 * 3600.
omega         = twopi/T
print "omega:   ", omega
print ''

aaacs = accelerations(np.array([Req, 0, 0]))
print "Req: ", Req
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''
lat = rads * -5.4
aaacs = accelerations(np.array([R*np.cos(lat), 0, R*np.sin(lat)]))
print "R: ", R
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''

aaacs = accelerations(np.array([0.0001, 0, Rpo]))  # avoid divide-by-zero (lazy!)
print "Rpo: ", Rpo
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())


Time according to the gravity of Sagittarius A*?

Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, time passes "normally".

But to answer your question, we must specify what is meant by "the black holes's time" (let's call the black hole $\mathrm{BH}_\mathrm{Sgr\,A^*}$; see note below on the nomenclature), since it depends on how far from Sgr A* we are talking. The time pace at a distance $r$ from the center of a BH is given by
$$t = t_\infty \sqrt{1 - \frac{r_\mathrm{S}}{r}},$$
where $t_\infty$ is the time "at infinity", i.e. far from the BH, and
$$r_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 3\,\mathrm{km}\,\times \left( \frac{M}{M_\odot}\right)$$
is the so-called Schwarzschild radius (the "surface" of the BH), which is where not even light can escape. Here, $G$ is the gravitational constant, $M$ is the mass of the BH, $c$ is the speed of light, and $M_\odot$ is the mass of the Sun.

The last equality shows that a BH with the mass of the Sun would have a radius of 3 km. The mass of $\mathrm{BH}_\mathrm{Sgr\,A^*}$ is some 4.1 million Solar masses, so its radius is $r_\mathrm{S} = 12.1$ million km.

Plugging in the other numbers, we can see that at a distance from $\mathrm{BH}_\mathrm{Sgr\,A^*}$ of

1 lightyear, time runs slower by a factor of 1.00000064, i.e. unnoticeably.
1 astronomical unit (the distance from Earth to the Sun), time runs 4% slower.
1 million km from the surface, time runs slower by a factor of 3.6.
1000 km from the surface, time runs slower by a factor of 110.
1 km from the surface, time runs slower by a factor of ~3500.
1 m from the surface, time runs more than a 100,000 times slower.
At the surface, time stops.

Note that this time dilation is what a distant observer (i.e. the guy with the $t_\infty$ time) would measure for an observer at the distance $r$. The person at $r$ would just measure his/her own time as usual. For instance, according to point 5 above, if you were hovering 1 km from the surface, waving your hand every second, then I, choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see you wave approximately once every hour. And when you run out of fuel and plummet into the BH, then when you cross the surface you wouldn't notice anything particular, but I would see you frozen in time. This is the concept of relativity.

Finally, let me use this chance to clarify something that people, including myself, often have gotten wrong: Sagittarius A (without an asterisk) is a radio source in the center of the Milky Way. It consists of three parts: Sagittarius A East (a supernova remnant), Sagittarius A West (dust and gas clouds), and Sagittarius A*, or Sgr A*, which is a very bright and compact radio source believed to be formed by a supermassive BH. Sgr A* isn't actually the BH itself. I think the BH doesn't really have a name, so I'll call it $\mathrm{BH}_\mathrm{Sgr\,A^*}$. Maybe that's a bad name…


Polynesian wayfinding techniques - how does this work?

This is from an article in the Conversation, by Duane W. Hamacher, and Carla Bento Guedes:

"Voyagers measure the angles between stars and the horizon using their hands. The width of your pinkie finger at arm’s length is roughly one degree, or double the angular diameter of the Sun or Moon.

Hold your hand with the palm facing outward and thumb fully extended, touching the horizon. Each part of your hand is used to measure a particular altitude.

The hand method used by Nainoa Thompson to find the altitude of the Polaris. Journal of the Polynesian Society
In Hawai'i, the “North Star”, Polaris, is Hokupa'a, meaning “fixed star”. It lies close to the north celestial pole. The altitude of Hokupa'a indicates your northerly latitude.

In the film, we see Moana Waialiki using this technique to measure the altitude of a group of stars. Look closely and you can see that she’s measuring the stars in Orion’s Belt. The position of Moana’s hand indicates the star above her index finger has an altitude of 21º. Given that the movie takes place about 2,000 years ago near Samoa, the position of Orion indicates they are travelling exactly due East.

Moana measures altitude of Orion’s belt stars. Walt Disney Studios Motion Pictures
Later in the film, we see Moana navigating by following Maui’s fish hook. In the various Polynesian traditions, the hook was used to pull islands from the sea. It is represented by the constellation Scorpius, which rises at dusk in mid-May. This indicates southeasterly travel.

Looking at Scorpius - Maui’s Hook - in the same orientation as shown in the film. Stellarium
However, the positions of the stars are not fixed in time. Over the 3,500 years that Polynesians have been exploring the Pacific, the stars have gradually shifted due to precession of the equinoxes.

From the latitude of Samoa, the Southern Cross has lowered from 60º altitude in 1500 BCE to 41º today. Those navigating by the stars must gradually adjust their measurements as the positions of stars slowly shift over time."


If Jupiter was at 1 AU from the Sun, its orbit about the Sun would be about 4 hours and 10 minutes shorter than a sidereal year. That's an effect, but it's not very much of one, due to Jupiter's mass being about 1/1000th of the Sun's mass. For a more significant effect, if our Moon was much smaller than it is, it would orbit the Earth about 4 hours and 1 minute longer than a sidereal month. The Moon's mass is about 0.0123 Earth masses.

The period of the Keplerian orbit of an object of negligible mass about some massive object is $$P = 2\pi\sqrt{\frac{a^3}{GM}}$$ In the above, $a$ is the semimajor axis length, $G$ is the Newtonian gravitational constant, and $M$ is the mass of the central object. When the orbiting object has non-negligible mass $m$, the above expression needs to be modified to $$P = 2\pi\sqrt{\frac{a^3}{G(M+m)}}$$


Can a stellar mass black hole swallow Earth?

There's a fairly common misconception in this question that black holes "suck everything in".   They do - at distances close to their event horizon but as noted, the event horizon's diameter is about the distance across rhode island (different masses have different sizes), but many stellar mass black holes would fit inside the border of rhode island.  Virtually all of them would fit inside pennsylvania for example.   They're tiny.    I'd rate my chances of survival as much higher in a space-ship 1 million miles from a stellar mass black hole than in a space-ship 1 million miles from our Sun.

At normal planetary distances, a planet would just orbit a black hole and likely maintain a stable orbit for billions of years.   Gravitationally, they're no different than a star at those distances.  The orbital mechanics is based on mass and a stellar mass black hole has stellar mass.   What makes it different is it's shrunk inside it's schwarzchild radius, so compared to stars, it's incredibly small and very dense.   The "weird stuff" only happens at very close distances - basically at distances you can't reach around a star because you'd be inside the star.

So . . . what would happen if you put the Earth next to a black hole.

I'm not sure why you'd want to do that, but the Earth would begin to break apart form tidal forces before it got that close.  The gravity would be stronger on the near side than the far side and the Earth would stretch.  This is called spagettification.

If you theoretically made a black hole just appear near the Earth then it would begin to eat the Earth.   Think of a black hole as a drain, a drain some 20-30 miles across, and think of the Earth as a liquid trying to get down that drain.   It's a lot of material squeezing into a comparatively tiny space and in the process the Earth material would get very hot, and likely form into an accretion disk trying to get inside the black hole and the formation of the accretion disk would eject two cosmic jets along the axes.   Black holes are sloppy eaters.

I couldn't tell you the precise amount of Earth that would be absorbed if they were placed right next to each other, but that's also an impossible scenario.   Everything in space is moving and gravity guarantees that if the Earth did meet or closely orbit a stellar mass black hole, that it would be moving enormously fast.

Earth could in theory, but extremely unlikely, either orbit a black hole or collide with one, or pass very close to one, close enough to break apart.  Each scenario would have a different outcome with a different percentage of Earth falling into the black hole.

I don't feel like going through each scenario individually, but the short answer is that the Earth would get quickly ripped apart and some of it, but not all, would fall into the black hole.  Some would be ejected as high speed charged particles.   The matter spiralling in would create a very strong magnetic field and because protons and electrons are charged, much of the matter wouldn't entirely follow the gravitational forces.  Some would get shot out as jets of charged particles.

That's a rough answer anyway.


Are there bodies in the Universe that are not spinning?

Many red and brown dwarfs must have planets that are in full tidal lock with their host star, that means that the same side of the planet is always facing the star so there is always day while on the other side there is always night. Our Moon and many other moons are also in such a tidal lock with their planet but not with the Sun so that the often-used term "Dark side of the Moon" is verbally incorrect since on the far side there is a day too (when on Earth it is a New Moon).

If you mean non-spinning in the sense that on the same side of the planet it is day for half a year and night the other half I'm quite sure such planets exist too. On SpaceEngine I've seen similar planets. Nothing speaks against the existence of such planets I believe. There also might exist moons whose same side faces their planet for half a year and the other half faces away from it.


How does the Helioseismic and Magnetic Imager (HMI) compute the Magnetic Activity Index (MAI)?

I found a quote from K. Jain et al.  which defines a Magnetic Activity Index might be pointing you in the correct direction - for me it still sounds rather abstract:

To obtain the MAI, we convert magnetogram data to absolute values, average over the length of a ring day and apodise them into circular areas to match the size of the Dopplergram patches used in the ring-diagram analysis.

However, I also found http://hmi.stanford.edu/teams/rings/modules/maicalc.html which contains a link a source code repository. The man-page of maicalc states:

maicalc takes as its input a dataset of records representing solar magnetograms during a selected interval, and produces as its output the values of the Magnetic Activity Index (MAI) appropriate for a selected set of target locations as they would be mapped and tracked by mtrack for the duration of the input data interval and apodized by pspec3.


Advice on first Telescope (refractor vs reflector), Barlow lens

Further would a Barlow lens solve the chromatic aberration issues with
  a f/5 refractor.

I suppose in theory they could, but they would have to be designed to do so. In practice, they vary widely in quality, and either don't have an effect on chromatic aberration, or actually make it worse.

  Which would be a better option?

There is no objective way of answering this question. Based simply on price per square unit of aperture, a reflector is more economical, and therefore better. But if you're looking for a telescope that doesn't require collimation, a refractor might be best. You have to consider the portability of the specific models you're considering as well. Finally, what would you like to see with your telescope? A high-end refractor with correcting optics is, in theory, better than a similar reflector for high-quality imaging of planets. Ultimately, the best telescope is the one you're comfortable using regularly without feeling overwhelmed.


Seeing Azha (Eta Eridani) in Northern Hemisphere

Eta Eridani has a declination of about -8 degrees, so it is visible from most of the Northern Hemisphere. Eridanus is a constellation that is next to Orion, and like Orion, it is easiest to see during winter.

The Eta Eridanids are a meteor shower that have a peak on about 11th of August. The radiant on that date is just south of Eta Eridani, at about -11 degrees. On 11th of August, Eta Eridani rises at about 3am in London, About an hour before twilight.

Tokyo, which is much further south, and doesn't use Daylight saving, sees Eta Eridani rise at about 1am (local time).

When you change location in Stellarium the time doesn't change. So if you change from London at 3am in August (UTC+1), the time in Tokyo will be 11 am (UTC+9).

The Eta Eridanids are a minor shower with a ZHR of about 6 meteors per hour They will be visible whenever the radiant is above the horizon, but from more northern locations it is only visible for a short time. It also clashes with the Perseids. One may only see a couple of Eta Eridanids during the viewing window, which is why it was only identified in 2005.


A black hole (BH) never sucks anything in. 'Sucking' requires gas pressure, but a black hole only acts via its gravity. A non-rotating (=Schwarzschild) black hole attracts all massive objects, very much like the Sun attracts the planets.
Yet, the planets are not falling into the Sun. This is because the planets move on near-circular orbits when the centrifugal force balances the gravitational attraction. The planets orbital angular momemtum is conserved (because the Solar force field is purely radial [to good approximation]) and these orbits are stable.

The difference to a black hole is that very close to the event horizon no such stable circular orbits exist anymore. At those distances, nothing can orbit the BH without falling in.

However, an object orbiting a BH on a stable orbit (including passing trajectories) is still subjected to the BH's tidal forces. These distort the object, very much like the Moon's gravity distorts the Earth, generating tides.
In case, the trajectory passes sufficiently close to the BH and the object is rather big and fluffy, like (some) stars, the tidal forces may not merely distort the object, but rip it apart.

Such tidal disruption of stars must happen if a star passes close to a supermassive BH. This can hardly be directly observed, but the it is thought that some of the stellar matter forms an accretion disc around the BH and produces a characteristic light curve. An observed light curve in agreement with this model is often interpreted as circumstantial evidence for a stellar disruption event.


Time-frame regarding cosmic expansion and the bound state of the Virgo Cluster

I don't know how to calculate but Wikipedia has a page about it:
Timeline of the Universe

The galaxies in the Local Group, the cluster of galaxies which includes the Milky Way and the Andromeda Galaxy, are gravitationally bound to each other. It is expected that between 10^11 (100 billion) and 10^12 (1 trillion) years from now, their orbits will decay and the entire Local Group will merge into one large galaxy.[4]
Assuming that dark energy continues to make the universe expand at an accelerating rate, in about 150 billion years all galaxies outside the Local Supercluster will pass behind the cosmological horizon. It will then be impossible for events in the Local Group to affect other galaxies. Similarly it will be impossible for events after 150 billion years, as seen by observers in distant galaxies, to affect events in the Local Group.[3] However, an observer in the Local Supercluster will continue to see distant galaxies, but events they observe will become exponentially more red shifted as the galaxy approaches the horizon until time in the distant galaxy seems to stop. The observer in the Local Supercluster never observes events after 150 billion years in their local time, and eventually all light and background radiation lying outside the local supercluster will appear to blink out as light becomes so redshifted that its wavelength has become longer than the physical diameter of the horizon.
Technically, it will take an infinitely long time for all casual interaction between our local supercluster and this light; however, due to a the redshifting explained above, the light will not necessarily be observed fit an infinite amount of time, and after 150 billion years, no new causal interaction will be observed.
Therefore, after 150 billion years intergalactic transportation and communication beyond the Local Supercluster becomes causally impossible.

This wikipedia data, comes from here, which is used in the text as reference (4)


What is the Fundamental plane for Elliptical Galaxies?

I am not an expert, but it seems that the fundamental plane is a relation among characteristic quantities of the galaxy, showing a correlation that is the analogous of the Tully-Fisher relation for spiral galaxies.

You can start from the virial theorem:
$\frac{M}{R}\sim v^2$

($M$ mass contained in the radius $R$, $v$ velocity dispersion of the stars, also indicated with $\sigma$ in these cases)
meaning that the stars behave like an isothermal sphere.

Assuming that all galaxies have the same mass-to-light ratio
$L/M$, and that all galaxies have the same surface brightness

$\Sigma=L/R^2$

(with $L$ luminosity), you obtain the Tully-Fisher relation:

$L\sim v^4$

Of course, all galaxies do not have the same surface brightness.
So if you take $\Sigma=L/R^2$ and substitute that into the virial theorem (keeping the mass-to-light ratio assumption), then a relation between luminosity, surface brightness, and velocity dispersion is obtained:

$L \sim \sigma^4 \Sigma^{-1}$

You can plot the stars distributions in elliptical galaxy according to this relation, and will note that the stars do not distribute randomly, but are more concentrated along the fundamental plane (more or less the same happens for the HR diagram):

(Here $R_e$ is the so called effective radius).

For completeness, another illuminating picture from here:

References here and here.


How do scientists determine if a star's color is due to Doppler shift vs the star's composition vs temperature?

The colour of a star is due mostly to its temperature. Except at extreme velocities, the red-shift won't affect the light enough to change the visible colour. To detect a red or blue shift you need some marker in the spectrum that can be measured accurately.

The spectrum of light from a star will contain dark lines, called Fraunhofer lines, which are due to the absorption of light at particular wavelengths by gases in the star's atmosphere. The wavelengths at which these lines appear are known very finely, for example sodium causes a line at 589.592nm . If the star is moving, the light, along with the Fraunhofer lines is red- or blueshifted.

The presence of these line shows the composition of the star's atmosphere.
Measurement of the position of the lines allows for accurate measurement of the Doppler shift of a star.

A cool, red star which is stationary will have the Fraunhofer lines in the same position as a hot stationary blue star. The motion of the star is measured by the position of the Fraunhofer lines, not the colour of the star.


Differences between alt/az in Stellarium and Astropy

Perhaps the difference stems from the conversion algorithm?

Almost certainly this is the answer.  It's not clear to me exactly what Stellarium assumes for is refraction calculation, but I know Astropy's algorithm.  Given that the altitude you've listed here is below the horizon (i.e., alt is negative), it's not really clear what the correct interpretation is: in particular, atmospheric refraction is not really well-defined below the horizon (since it should be invisible anyway!).

Additionally, even well above the horizon where refraction isn't that important, there are several finicky details that you have to account for to properly do the Az/Alt conversions.  In addition to the precession that the other answers mention, if you want to get exactly the same answer on two tools, you have to worry about things like how much the axis of the Earth shifts due to, say, major earthquakes ("true polar wander"), the effects of General Relativity on the apparent direction of the incoming light ,etc.  Most of these are pretty small, but together they mean it's very hard to ensure all of the software gives exactly the same answer unless they carefully follow the definitions set by the International Astronomical Union.

That said, the differences you see are only ~20 arcsec.  The human eye isn't even capable of telling that small of a difference, and almost all telescopes that people would use Stellarium with probably wouldn't either (at least in an absolute sense).  So for this particular application they essentially do match.


(This is somewhat simplified but I hope it gets the idea across.)

The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there.

Why doesn't it start fusing helium into carbon right away? That's because the core isn't hot or dense enough yet. Different reactions broadly rely on the presence of different resonant states in the nuclei and, in the case of helium, such a state cannot be reached often enough until the core temperature is about $10^8$ kelvin.

In order to get that hot, the core has to contract and heat up. It eventually does (if the star is massive enough) but it doesn't happen instantaneously. Remember that the gas is still hot and at high pressure, which it exerts on itself and its surroundings.

Meanwhile, at the edge of the core, the star (partly as a result of said contraction) is hot enough to turn hydrogen into helium, so it does so. This is exactly the nuclear-burning shell that distinguishes the internal structure of a red giant.

So maybe think of it this way. Imagine a star at the end of the main sequence. Where is it hot enough to fuse hydrogen into helium? Everywhere up to the edge of the core! Does it fuse in the core? No, because it's out of fuel. So where does it fuse? At the edge of the core, which we recognize as the shell.


What kind of process could create such stellar radiating structures, if real?

When an embedded cluster forms and later blows away gas to become a young open cluster the stars inside interact gravitationally. They have essentially random velocities relative to each other, performing a chaotic dance. This typically makes the heavier (and hence brighter) stars end up near the core with low velocities and the lighter stars get into higher orbits. The phenomenon is called mass segregation and happens because all stars end up with roughly similar kinetic energy.

The effect is more extreme in globular clusters since they can persist for a long time, while open clusters tend to dissolve by interacting with the rest of the galaxy. In a way, as the cluster dissolves the lighter stars will be moving away ("radiating") from the core, although at this stage they are all mostly doing the kinetic energy exchange dance.


Why are radio telescopes shaped so differently than optical telescopes?

They're not different. Same principles do apply. You could have secondary, tertiary, quaternary, and so on, mirrors with instruments at any wavelength, either optical, or radio, or infrared, etc. You could also have instrumentation placed directly in prime focus (so no mirrors other than the primary) with any kind of instrument - radio or infrared or visible or whatever.

See this image of the 5 meter Hale telescope on Mt. Palomar - there is no secondary mirror in this case, the observer is sitting in a little cage at prime focus, using the primary mirror directly:

Of course, for other scenarios, the Hale telescope employs secondary and tertiary mirrors - it depends on the particulars of the telescope, the instrumentation, the experiment or research you're doing, etc.

One reason many of the large optical telescopes very often have at least a secondary mirror is that the architecture preferred in most of these cases is the Ritchey–Chrétien - chosen often for the largest professional telescopes because it eliminates coma, an aberration that is detrimental to astrometry (with coma, images of stars are not round, so it's hard to measure angular distances between them). You could use the primary mirror of such a telescope directly, sure, but being a concave hyperbolic mirror, it has strong aberrations of its own, and so requires the convex hyperbolic secondary (often a strong hyperbola, with a large eccentricity) to correct the aberrations.

The Hale telescope pictured above has a parabolic primary, so using it directly is not a problem.

Again, all of the above are not strict rules, just statistical observations.

Some radio telescopes have instrumentation at prime focus simply because it's convenient for that particular case. Other radio telescopes have secondary mirrors. Again, it all depends on what you try to achieve. E.g., the Arecibo radio telescope could be used either in prime focus, or with a secondary mirror in a Gregorian configuration - here's the image with the prime focus instrumentation and the Gregorian mirror to the left:

In the case of the Arecibo scope, the N-ary mirrors are sometimes used to correct the aberrationa of the spherical primary reflector, but that's not the only reason why they're used.

Here's a discussion comparing various architectures (classic Cassegrain versus Ritchey-Chrétien versus anastigmatic aplanat) for a large radio telescope, highlighting various design, performance and operational issues for each. TLDR: classic Cassegrain is traditional for radio telescopes, but the R-C architecture performs better and is not significantly more difficult to build; OTOH, with R-C you must always use the secondary.


Uranus' axis of rotation-when discovered?

The high inclination of the Uranus system to its orbit around the Sun has been known since the satellites were discovered.
Lassell 1851 shows Ariel and Umbriel moving at least as far north and south of the planet as they do east and west of it.
William Herschel probably noticed the same thing about Titania and Oberon in 1787.

When Uranus is at an equinox, its equator appears edge-on to the Sun, making good opportunities to observe the planet's oblateness from Earth.
Mädler 1842
and Young 1883 measured significant polar flattening.
Barnard 1896 thought the satellite orbits might be inclined 20°-30° to the equator.
It took longer to determine that the equator was within 1° of the four largest satellites' orbital planes.


Gravastar Star With no Horizon with ultracompact and very density possible?

Of course unknown "strange Quantum Effects" could do anything, but this letter reports calculations which suggest that at least one of the LIGO gravitational wave events is consistent with black holes but not with gravastars. That doesn't rule out the possibility of gravastars and black holes existing, but it speaks quite strongly against the idea that all the phenomena attributed to black holes are actually due to gravastars.


Do solstices and equinoxes shift over time?

The Gregorian Calendar was created so that annual astronomical events, specifically the vernal equinox (used to determine when Easter is), would on average keep their places in the calendar year over time.  It is the best official approximation to the definition of the tropical year, which is defined as "the length of time that the Sun takes to return to the same position in the cycle of seasons".  Because this calendar describes 97 leap years out of every 400 years, it defines the average year as exactly 365.2425 solar days, or exactly 365 days, 5 hours, 49 minutes, and 12 seconds.

However, the mean tropical year is in reality about 365 days, 5 hours, 48 minutes, and 45 seconds, or 27 seconds shorter.

Because the Gregorian Calendar is based on the tropical year, the calendar dates of the year will keep up with the solstices and equinoxes, and thus the seasons.  If this calendar were exactly the length of the tropical year, then the calendar would keep the vernal (northward) equinox around March 20th for all time.

But because of the slight inaccuracy, it will take about 3,200 years (60 s/min * 60 min/hr * 24 hr/day / 27 s/year) for these 27 seconds to add up every year to be 1 full day, and that will result in the solstices and equinoxes marching backwards in the calendar by 1 day every 3,200 years or so, depending on the accuracy of the 27 seconds difference.  This very slow shift is due to the slight inaccuracy in the Gregorian calendar in, on average, matching the tropical year, not because of the precession of the equinoxes.

3,200 years from now, if the Gregorian Calendar is still used, the date of the vernal (northward) equinox will be on average one day earlier in March.  The precession of the equinoxes will still occur, so the Earth's axis tilt will be significantly different from today.  The Earth will be at a noticeably different position with respect to the Sun on the vernal (northward) equinox from where it is today, in 2014, on the vernal (northward) equinox, but it will still be in March.

This inaccuracy may very slowly increase over time, because according to the same Wikipedia page for the tropical year, the tropical year is very slowly getting shorter, and the mean solar day is even more slowly getting longer.  But for 10,000 years to come, the Gregorian Calendar will keep the vernal (northward) equinox in March, even if it slowly shifts earlier in the month.

This is in contrast to the scenario that you imply, where the calendar date would correspond to the relative position of the Earth in its orbit around the Sun.  That describes the sidereal year, the time taken for the Sun to reach the same spot in the sky relative to the stars, which is 365 days, 6 hours, 9 minutes, and 10 seconds.  A sidereal calendar would explain why you might think that precession would cause the dates of equinoxes and solstices to change in the calendar year.  That would result in a shift in the calendar of one full month in 1/12th the cycle length of the precession of the equinoxes, or about 1 full month about every 2,000 years.


Could we verify the structure of a black hole by observing an orbiting object?

The comments from @userLTK, and @Lacklub are correct.

Lets assume there is an object of radius $R$ and mass $M$, from a Newtonian point of view, if you are at another radius $r$, such that $r > R$, then there is no difference in the gravitational field  experience by an object at $r$ if the mass spread across a shell of radius $R$ or if its concentrated anywhere between $r=0$ and $r=R$. GR doesn't do much to change this, and in fact if $r >>R$, then the result is of course exactly the same.

Now to black holes, all taken from from Wiki https://en.wikipedia.org/wiki/Black_hole

Physical Properties
"The simplest static black holes have mass but neither electric charge nor angular momentum...This means that there is no observable difference between the gravitational field of such a black hole and that of any other spherical object of the same mass."

So basically if you're outside the event horizon (not that'd you know where it was) you're experience with the black hole is the same as with a planet or star of that mass.

Again from same wiki article:

Singularity
"At the center of a black hole, as described by general relativity, lies a gravitational singularity, a region where the spacetime curvature becomes infinite... It can also be shown that the singular region contains all the mass of the black hole solution".

The "it can be shown" references page 204 of Carroll, Sean M. (2004). Spacetime and Geometry. I don't have my copy with me right now so I can't look it up, but I would say I remember reading at in Carroll back in the day.

Finally, let me just add that, once something passes into the event horizon of a black hole, there's no getting back. So we really have to ask ourselves how any information about the "stuff" inside the event horizon would come to us? I like your idea of getting some indirect evidence and perhaps gravitational waves will shed some light on this topic but I don't suppose there is any known way to getting direct access to anything beyond the event horizon.


Examples of absorption lines for main sequence stars

There are several online sources of spectra (eg SDSS), but they usually just provide a FITS file with data, or a plot vs wavelength without any annotation. Of course, you could use the  NIST atomic spectra database to identify lines, but that quickly becomes tedious.

There are many figures containing spectra for various classes of stars in the book by Gray and Corbally, Stellar Spectral Classification. Many of these figures were previously available on Gray's web site, but the links no longer work. I scanned one that may be of use, Figure 7.1 containing spectra of G through K stars. In particular, one of these spectra is for a reference standard G0V star, $\beta$ CVn, with the stronger Ca, Fe and H-Balmer lines identified.

Another good resource for annotated spectra of many star classes is Richard Walker's, Spectroscopic Atlas for Amateur Astronomers. I have the printed version, but pdf files of older versions are available online. This atlas contains spectra of a G0IV star, and the Sun, G2V, that have similar absorption lines.

There are also collections of standard spectra, such as the Miles library. These are often included in software used for processing spectra, along with tables of common lines for various elements that can be shown on the plot. Programs I have used for amateur spectroscopy that display these include Rspec (commercial), Vspec and ISIS (Christian Buil).


Is it possible to revive the Mars geodynamo?

Let's do a bit of math: magnetic energy density is at least $u = \frac{1}{2}\frac{B^2}{\mu_0}$. The magnetic field at the surface of Earth is of the order of 30µT which gives us a magnetic energy density in the order of $10^{-3} J/m^3$
Now we need to fill the entire volume of the planet and some surrounding space with this field in order to get some protection from the solar wind; let's assume a cube of $(20000km)^3 = 8\cdot 10^{21}m^3$. So that means that we need an energy in order of $10^{18}W$ to sustain a comparable field which equates to roughly 1 million state-of-the art GW nuclear power plants. We currently have significantly less than 1000 in the whole world combined and I have not considered losses of any kind (thus 100% conversion of power into magnetic field)

My rough calculation also does not include any additional power needed to sustain the field extent into space when subject to the magnetic pressure of the solar wind which will be needed in order to keep the atmosphere inside the magnetic field, especially in direction of the sun.

So please go and figure whether these scenarios are really realistic. Actually reviving any dynamo on Mars is an altogether different story and would need to understand how exactly it works/worked; it might entail to heat up the core which requires probably more energy and will pose even more severe technological challenges than a giant dipole magnet of planet size.


High quality Space-related blog/website?

Your question for "space" news sites is somewhat vague as space can cover a wide variety of topics.

I have found Innerspace.net  to be a nice blog/news source primarily concerning itself with the space exploration and technology aspects of space. It often reports on recent launches (such as those by Space X), advancements in space technologies, and space ventures by other countries outside the US such as China and Russia.

If you're looking for more sciencey topics of a higher caliber, try something like Science Daily. While this covers a variety of scientific fields, it does also cover space news. In general, this site is a curation of articles by various sources. Usually each article is linked to some specific scientific research or published paper.


How do we get to know the total mass of an atmosphere?

There is a simple$^*$ way to know the total mass of the atmosphere: measuring the pressure it exerts on the surface, which necessarily integrate all of the atmosphere above ground level.
If you take an atmospheric pressure of $1\cdot10^5$ Pa, it is equivalent to a force of $1\cdot10^5$ newton over one square meter. Multiply by the area of the planet in square meters, you get the total weight of the atmosphere: $1\cdot10^5 \times 5.1\cdot10^{14} = 5.1\cdot10^{19}$ N. Divide by the acceleration of gravity to convert this weight to a mass: $\frac{5.1\cdot10^{19}}{9.8} = 5.2\cdot10^{18}$ kg. There you go!
$^*$Well, I guess it is simple on Earth, but could be more challenging on other planets...


When the Sun becomes a red supergiant (more accurately, an asymptotic giant branch star), it will have less mass than it does now. That is because mass is lost by an increasingly massive stellar wind as the Sun approaches the end of its life. As a result, the orbital radii of the planets would increase. The surface temperature of this future AGB Sun would be lower than the temperature of the Sun, but its luminosity would be much larger.

However, even if the Earth manged to avoid being engulfed in the outer envelope of the evolved Sun (which is still uncertain), the amount of energy received per square metre at the Earth's surface would be vastly increased over the 1.4 kW per square metre it receives from the Sun now. So your conclusion that life (as we know it) would be fried is quite correct.

The Goldilocks zone for the possible occurrence of life, uses the criterion of liquid water as the basis for allowing life to evolve. As a star becomes more luminous (and an AGB Sun would probably be about 100 times as luminous as the present Sun), the Goldilocks zone moves outwards roughly proportional to $\sqrt{L}$, so about a factor of 10 further away.

The problem then with allowing life to evolve on another planet (e.g. Mars after its orbit has expanded a bit, or Jupiter) is that (i) you need the ingredients to be there already, or some way for them to be delivered, but more importantly (ii) there is little time for life to evolve, because the Goldilocks zone position evolves really rapidly towards the end of the Sun's life as the AGB Sun's luminosity increases - the timescales would be measured in millions rather than billions of years. Furthermore, the luminosity of AGB stars can be highly variable as they undergo big pulsations, which may also be a problem. (iii) After the (comparatively brief) AGB phase has ended, you would end up with a very hot ($T \simeq 10^{5}$ K) white dwarf remnant that might effectively sterilise the whole system with ultraviolet light.

I don't think we know enough about how life gets started to say that these problems rule out your scenario; but they perhaps make it less likely.


How is the path of an asteroid calculated?

How do scientists and astronomers recognize the perfect path or orbit (if it's orbiting an object) of an asteroid or a comet?

They don't, and they can't. The best they can do is estimate the object's trajectory based on observations and based on techniques used to propagate the object (and other objects in the solar system) over time. These estimates are always imperfect:

The measurements upon which the estimates are based are limited in number. This is particular true for comets and asteroids. The historical record on observations of the Sun and the planets go back thousands of years. The number of observations for a newly observed comet or asteroid might go back a few months.
The measurements upon which the estimates are based inherently have some error associated with them. These measurement errors limit the ability to determine the object's true trajectory.
The physical models used to propagate trajectories are simplified and incomplete versions of reality.

General relativity, our best model of how gravitation works, is exceeding complex and highly non-linear. Even the very best precision orbit determination programs use a linear parameterized post-Newtonian expansion of general relativity. The non-linear effects? They're ignored.
Nearby stars and the galaxy as a whole perturb the trajectories of the Sun, the planets, etc. These effects are ignored.
Comets vent gas when they approach the Sun. When you blow up a balloon and let it go, the balloon veers this way and that. There is no predicting where the balloon will go. Comets do the same thing. After time, there is no predicting where a comet will be or where it is going.
Asteroids (and comets) are subject to radiation pressure, the Yarkovsky effect, and the YORP effect. These are very hard to estimate. One of the key purposes of the upcoming Osiris-REx mission is to get a better handle on the Yarkovsky effect.

The numerical techniques used to propagate objects' trajectories are inevitably flawed to some extent. They introduce errors that grow with time into the propagated trajectories. There's always a tradeoff between precision, accuracy, fidelity to the physics, and computational expense.
The solar system is ultimately chaotic. This is particularly the case for asteroids and comets on highly elliptical trajectories. Consider a comet whose orbit takes it from beyond Neptune's orbit to inside Venus's orbit. Over the centuries, that comet has ample opportunities to make a close pass with a planet. There are but slight differences between the planet making a minor change in the comet's orbit, the comet colliding into the planet, the planet sending the comet toward the Sun or another planet, or the planet ejecting the comet from the solar system. This is chaos at its worst.


Why is the main sequence often called 'zero age' main sequence?

The zero age main sequence (ZAMS) is the theoretical locus of points in the Hertzsprung-Russell diagram where the luminosities of young stars (at a range of masses) are mostly supplied by the nuclear reactions that have begun in their cores.

Prior to this, the objects are known as pre-main sequence stars and most of their luminosity arises from the release of gravitational potential energy as they contract.

The exact definition (whether it is 99% or 90% of the luminosity) does vary, but since the nuclear reactions ramp up quite quickly this doesn't materially affect any quantitative analysis.

In any case, the age of a star is not measured from the ZAMS. The reason for this is that what is of physical interest is the time since the star formed and that must include the pre-main sequence lifetime. Whilst this is very short (maybe 1 million years) for stars more massive than the Sun, and certainly shorter than the total main sequence lifetime, it can be much longer (100 million years) for a low-mass M dwarf. This is also the reason that the ZAMS is a theoretical construct. Stars born at the same time will never be all on the ZAMS at the same time. By the time low-mass stars reach the ZAMS, more massive stars will have evolved away from the main sequence!

As to the origin of the term, I suspect it comes from the early computer/analytic models of stellar evolution. There you would likely miss out the pre main-sequence phase and start your model with "zero age" at that point where it achieves equilibrium with the luminosity supplied by fusion and with a pristine He/H ratio in its core. Thus in reality, stars of different mass would have different ages at this point, but from the point of view of a theoretical modeller, it would be their starting condition and hence a "zero age".


Can radio telescopes such as arecibo image the subsurface of asteroids or planets?

It is important to understand how this process works. The method described in your article is known as Bistatic Radar. In effect, a transmitter sends out a signal (generally a radio telescope in the microwave region) which hits the surface of some body and bounces off to be received back on Earth by a second, separate radio telescope. Now, because a microwave's wavelength is so long ($\sim 0.1-100\:cm$) the microwaves don't bounce off from the exact surface, but instead are able to penetrate slightly into the subsurface before being reflected. This means the receiver gets a reflection of the subsurface of the body.

In the case you linked, they used the Arecibo Telescope to provide the transmitted signal to the Moon and the Green Bank Telescope to receive the signal. A similar process could be done to other bodies besides our Moon. However, you run into the problem that the farther away an object is, the harder it will be to detect the returning signal.

I believe that most objects within our solar system (and certainly all objects outside it) are outside the distance where this method will work. I was able to find an instance where someone used this method for an asteroid that passed by the Earth at a distance 11 times that of the Moon. I'm not sure what the limit is to how far away something must be before this method no longer works, but I imagine it wouldn't work even for Mars unless we seriously upgrade our technology and power.

Of course your question seems to presuppose the transmitter and receiver are both on Earth. If your transmitter is much closer, say a satellite orbiting that planet, then it is certainly feasible. Here is a source which discusses the concept of using Bistatic Radar probing via a satellite, which also goes into the history a bit on what has been done before.


What is the last day a waning moon is visible in daytime (the afternoon)?

At third quarter the moon rises at about midnight and sets at about midday The exact rising times depends on the position of the moon relative to the ecliptic and the season.

So in spring, at third quarter, the moon is low and sets quite early (around 10:00 in my location)  In Autumn the moon is high at third quarter, and sets at about 15:00 (though daylight saving time affects this)

As the month progresses the moon rises later and sets later. The moon is above the horizon in the afternoon, but visibility depends on many factors, and a very thin cresent, close to the sun, may not be visible with the naked eye. There is some analysis of this with respect to the appearance of the new moon, (which has calendrical significance to some religions) There are too many uncertainties in determining visibility to give a simple answer, much depends on the observer.


What are the correct positions or alignment of planets in relation to each other and the Sun, is it vertical center, baseline or completely random

It is impossible to have the planets line up like in your image. This is done for illustration purposes only. If such an image was to scale, either the bodies would be extremely small and impossible to see, or the image would be extremely large and impractical to use.
For example, the Sun is about 1,392,000 km in diameter. The Earth is about 12,756 km in diameter (so roughly 110x smaller), and orbits at an average distance of roughly 149,600,000 km. If the Sun was a 40 cm ball, the Earth would be about 3.6 mm in diameter and located roughly 43 m away from the Sun-ball. At that same scale, Neptune is about 1,3 km away. (Before COVID, I regularly hosted “Solar System Walks” for the city of Montréal, and that was the size of my scale model.)
Now to get back to “lining up” planets… Let’s define a reference plane: the ECLIPTIC is the plane of Earth’s orbit around the Sun. All the other planets have orbits tilted with respect to that plane. For example, Mercury’s orbit is tilted 7° to the ecliptic, and Venus’ orbit 3.4°.
So there IS a point on a planet’s orbit where it lies on the same plane as the Earth—it’s called a NODE, and there’s an ASCENDING NODE, where the planet crosses from “below” (south of) the ecliptic to “above” (north of) it as well as a DESCENDING NODE where the planet goes from above to below.
HOWEVER…
1-The Earth is not necessarily in line with the Sun and the planet at that moment (explaining, for example, how the passage of Venus in front of the Sun is such a rare event [last ones were in 2004 and 2012; next ones will be in 2117 and 2125]);
2-Planets don’t [necessarily] cross the ecliptic plane as the same time as other planets; and
3-The imaginary line between a planet’s two nodes is not [necessarily] coincident with that of another planet. For example, Mercury’s ascending node is at an ecliptic longitude of 48.33° while Venus’ ascending node is at 76.68°.
4-Finally, depending on other specifics of each planet’s orbits (its so-called “orbital elements”), a planet’s orbit is not necessarily divided exactly between “above” the ecliptic and “below” it.
So… To answer your question specifically… The planets are indeed in “weird” positions and not in line, but their specific positions “above” or “below” the Earth–Sun line change with time, and it’s basically impossible for them to be in the plane of the ecliptic at the same time, even roughly speaking.
(Addendum: It applies to planetary alignments “seen from above,” but Jean Meeus has an excellent explanation of the impossibility of aligning more than 2 planets at the same time in his book Mathematical Astronomy Morsels [chapter 31].)


Relationship between the Casimir force and dark energy

There are two problems that arise here. One obvious problem is that the Casimir effect is attractive1, while dark energy is repulsive. The other problem is one of scale.

Casimir (1948) shows that, between two particles (instead of the oft-cited case of two plates)
$$\delta E \propto R^{-7}$$
That's an enormous drop-off. On large scales, this should be negligible. Dark energy, however, is clearly non-negligible on these scales, while its effects are unobservable on small scales.

The Casimir effect is certainly one manifestation of vacuum energy, which may also be the cause of dark energy (in some theories), but that doesn't mean that dark energy is an example of the Casimir effect, or vice versa.

1 There are exceptions, as pointed out by Stan Liou, which arise under certain conditions, but they wouldn't play a part here.


Have parameters been determined for the Fourier components of a Milky Way spiral density-potential model?

Yes.
Antoja et al. (2011) created two models: The Tight-Winding Approximation model (TWA) and the PERLAS model. They decided to use just the $m=2$ term for their TWA, based on data from the Spitzer Space Telescope.
The amplitude $A(R)$ was
$$A(R)=A_{sp}Re^{-R/R_{\Sigma}}$$
Their $\phi_2(R)$ (denoted $g(R)$ in the paper) was of the form
$$g(R)=\left(\frac{2}{N\tan i}\right)\ln\left(1+\left(\frac{R}{R_{sp}}\right)^N\right)$$
where $N$ measures, in their words,

how sharply the change from a bar-like to spiral-like occurs in the inner regions.

They approximated using $N=100$, although $g(R)$ should properly use the limit $N\to\infty$. $i$ is the pitch angle of the arms, which they measured to be $15.5$ to $12.8^{\circ}$. $R_{sp}$, the beginning of the spiral, was found to be $2.6$ to $3.6\text{ kpc}$. For the other constants, $A_{sp}$ was taken to be $650$-$1100$ or $850$-$1300[\text{km s}^{-1}]^2\text{ kpc}^{-1}$, depending on the exact arm shape, and the scale length $R_{\Sigma}$ was set to $2.5\text{ kpc}$.
The pattern speed $\Omega_p$ was set in the range $15$-$30\text{ km/s}$, which fits nicely with other values I've seen. $\theta_0$ (denoted $\phi_0$ in the paper) was set to $88$ to $60^{\circ}$ - a little bigger than the spiral arm pitch angle. putting all this together, we get a potential of the form
$$\Phi_\text{spiral}(R,\theta,t)=A_{sp}Re^{-R/(2.5\text{ kpc})}\times\cos\left(2(\theta-88\text{ to }60)+\left(\frac{2}{100\tan (15.5\text{ to }12.8^{\circ})}\right)\ln\left(1+\left(\frac{R}{2.6\text{ to }3.6\text{ kpc}}\right)^{100}\right)-\Omega_{p}t\right)$$
A $z$-component could be multiplied to get a three-dimensional potential, likely exponentially decaying with height.
Griv et al. (2013) also used density-wave theory. Their potential was
$$\Phi(R,\theta,t)=\Phi_{\text{mean}}(R)+\tilde\Phi\sqrt{\frac{R_0}{R}}\cos\phi$$
where
$$\phi=\phi_0+m\left[\Omega_pt-\varphi+(1/\tan p)\ln(R/R_0)\right]$$
which is essentially the same potential as discussed in the question, with
$$A(R)=\tilde{\Phi}\sqrt{\frac{R_0}{R}}$$
The authors chose $m=1,2,3$ components, with different parameters for each. $p$ (in degrees) ranged from $-6$ to $-20$, while $\phi_0$ (also in degrees) ranged from $239$ to $247$. $\Omega_p$ was from $32.4$ to $37.6$ ($\text{ km/s}$), also consistent with other models. $\tilde{\Phi}$ saw the most variation, ranging from $-65$ to $-390$ (in $\text{kpc}^{-2}\text{s}^{-2}$). $R_0$ was set to $8\text{ kpc}$, I believe. Again, an exponentially decaying function should account for the $z$-dependence.


Formation and capture of Mars' moons, Phobos and Deimos

You're right, Mars' moons likely did not form with from the same dust mass as it; their spectra, albedo, and density are suspiciously similar to those of C- or D-type asteroids. Thus, Mars presumably captured Phobos and Deimos from the asteroid belt. However, the origin of the Martian moons is quite controversial.

Burns (1992) notes that capture requires the loss of energy (and momentum, as you pointed out). So obviously, something would need to dissipate that energy. The immediate candidate is aerobraking, but we run into a problem: Mars' atmosphere is too thin to capture a Phobos-sized object through aerobraking. In fact, even if aerobraking could capture Phobos and Deimos, the densities of both moons indicate that they may not survive structurally.

Several hypotheses have been proposed to answer this. Landis (2001) suggests tidal forces are to credit:

  The existence of asteroid moons provides a mechanism for the capture of the Martian moons (and the small moons of the outer planets). When a binary asteroid makes a close approach to a planet, tidal forces can strip the moon from the asteroid. Depending on the phasing, the asteroid can then be captured. Clearly, the same process can be used to explain the origin of any of the small moons in the solar system.

According to Geoffrey Landis, Deimos and Phobos could have each been moons of parent asteroids, and they were separated due to tidal forces:

This would explain how Phobos and Deimos could survive as moons. I could mention the other hypotheses, but Landis' would answer your question quite well.

Right now, there isn't any clear consensus as to how Phobos and Deimos became moons. More research is still necessary, and we may need to modify our models for the evolution of Mars and the asteroid belt.