How to measure mass of planets' core from orbit

I certainly don't know the details of these kinds of calculations, but as my thought is a bit too long for a comment I'll write it up as an answer.
If you measure the flattening of a planet due to rotation (e.g. by measuring its rotation period) and the gravity (measured while keeping the orbiter at constant height above the planet), then you have everything you need to calculate the moment of inertia I/(MR2) using Radau's approximation (not the most informative link, but I couldn't find a better one). After that, it's a matter of taking a model for a planet, e.g. a two-layer model with a core and a mantle, which has free parameters x, the fractional core radius, and f the difference in density between both layers of the planet. Compare your measured moment of inertia with the model and you've got x and f for that planet, which allows you to calculate the mass of the core (if you also have the total mass of the planet from gravity measurements). The figure below shows the two-layer model and the measurements for some solar system bodies. The value 0.4 on the y-axis is the value for a homogeneous planet. From that you can deduce that Ganymede is a very differentiated planet. Hope that helps a little bit.

A note on how this graph was created: first of all, this isn't my graph, I took it from my lecture notes. The moment of inertia factor for a homogeneous body is calculated as
$$I = \frac{8\pi}{3}\int_0^R \rho r_0^4 dr_0 = \frac{8\pi}{15}R^5 $$
$r_0$ is the mean radius of a shell within the planet (because planets generally aren't round).
Then you take a simple two-layer model for a planet where $\rho(r) = f\rho_0$ for $0\leq r \leq xR$ and $\rho(r) = \rho_0$ for $xR < r \leq R$,  where $R$ is the radius of the planet.
The moment of inertia factor is then
$$I/MR^2 = 2[1 + (f - 1)x^5]/5[1 + (f - 1)x^3] $$
which is the curve that you see in the figure.


No, the expansion of the universe appears to be accelerating. Hence distant objects get more and more red-shifted. Our horizon is shrinking instead of expanding. At the end, only galaxies which are sufficiently bound by gravity remain visible; this could turn out to be the Virgo supercluster.

More reading about the accelerating universe on Wikipedia, and on National Geographic.

Scientific background to the according Nobel Prize in physics 2011.


Is there any way to distinguish the doppler redshift from a distant galaxy from the probably much greater cosmological redshift?

The two effects are observationally indistinguishable for an individual galaxy. The redshift associated with universal expansion is often interpreted as a recession velocity and there is nothing in a spectrum that tells you what fraction of the redshift is contributed by which effect.
If a distant galaxy is associated with a cluster of galaxies, then you can make an attempt by assuming that the average recession velocity of the cluster represents the cosmological redshift and that the perturbations from that for individual galaxies are their peculiar velocities.
Nevertheless, the cluster will have it's own peculiar velocity as an additive term to all these individual peculiar velocities. Fortunately, that doesn't really matter when it comes to studying cluster dynamics.
Since the cosmological redshift increases with distance (Hubble's law), whereas peculiar velocities do not, then at large distances cosmological redshift can be assumed to dominate. See What does "located in the Hubble flow" mean?


What are the most valuable meteorites and asteroids in terms of precious metals?

I have found an outdated study that says that some siderites contain as much as .001% of gold. Is it genuinely a viable mining research subject?

Not yet, and not for quite a long time.

Gold is precious because humans only mine about 2.5 million kilograms of the stuff per year. Compare this to the huge amount of infrastructure (and huge cost) needed to retrieve the few kilograms of stuff that has been brought back to Earth from space.

We haven't the foggiest idea how to mine minerals in space, let alone minerals that at best might have a concentration of ten parts per million (another way to look at your 0.001% figure). On the other hand, there are many ideas regarding taking advantage of untapped resources right here at home. For example, there are manganese nodules all over the sea floor. These nodules contain not just manganese , but also iron, nickel, and precious metals. Space mining cannot compete with those untapped local resources until they run out. (And then we'll find yet another untapped local resource.)

Mining stuff in space will be profitable only if there's a need for that stuff in space. If space mining becomes reality, the first stuff to be mined will be rather mundane volatiles such as water and methane rather than metals. Those volatiles are the low hanging fruit in space. How to extract and make use of those mundane volatiles is a good problem for our current generation of young adults. How to go beyond that is a problem for their grandchildren.


What is this web on the surface of the Sun?

The dark lines are colder areas at the edge of the convection cells, where the cooled down plasma sinks towards the inside of the Sun. Now "colder" for the surface of the Sun, is still pretty hot, as explained here.

The yellow parts are where the plasma rises to the surface.
Each yellow spot (which is actually the size of a country) is called a granule, and this web-like appearance is called granulation.

In the outer part of the Sun (the convection zone in the image below), there is convection, that is hotter plasma floats towards the top, cools down at the surface, and sinks back down, like in a lavalamp.

The existence of a convective zone in the outer part of the star is determined by the mass of the star, and all stars with a convective zone in their upper layer are thought to have such granulation patterns. So stars like our sun, or smaller have these patterns.

For larger stars, though, the convective zone is in the inner part of the star, and the outer part of the star is the radiative zone, so there might not be the same patterns on the surface.


What is the first recorded reference to the Moon being a satellite of the Earth?

From before the dawn of history people naturally assumed that the sky was a solid dome above the flat earth.  The dome was assumed to rotate once a day, so the stars were assumed to be lights attached to the dome.

Anyone who assumed that the sky dome was opaque had to assume that the sun and the wandering planets were nearer than the sky dome.  I think that it was usually assumed the sky dome was opaque and colored blue, thus making the sky seem blue when it reflected the light of the sun during the day, thus making the sun closer than the sky dome.

Those who kept records of astronomical observations soon noticed that the moon occulted or passed in front of stars, planets, and the sun (solar eclipses), and thus was closer.

Everyone assumed that the sun and the moon were tiny, until traveler's reports showed that they had the same apparent diameters every place that as visited, and so were about equally far away from every place that was visited.  Thus as the size of the known world grew larger and larger, the minimum possible size and distance of the moon and the sun grew larger and larger.

The idea of a spherical earth was proposed and gradually accepted by Greek philosophers between the sixth and third centuries BC.  Thus the earth became known as Earth, a sphere instead of a flat disc, and the dome of the sky became a hollow spherical shell around it.

In Hellenistic times the diameter of the Earth was calculated with reasonable accuracy, as well as the distance to the moon and thus its diameter.  So it became known that the moon was about a quarter the diameter of the Earth and over sixty Earth radii distant.

Because of solar eclipses, it had been known since prehistoric times that the moon was closer than the sun, and keepers of astronomical records of events when the moon occulted (passed in front of) stars and planets knew that the moon was closer than planets and stars.

Anyone who assumed geocentrism, that the Earth was the enter of the universe, naturally assumed that every object revolved around the Earth, including the moon, and that the moon was thus the closest satellite of the Earth.  A few ancient Greek philosphers supported the Heliocentric theory, that the Earth and the planets revolved around the sun.  Some of them could have believed that the moon also revolved around the sun, but as far as I know all heliocentric believers also had the moon revolve around the Earth.

By the last few centuries BC there were many educated persons who believed that the Earth, the moon, and the sun were giant balls of rock (and thus somewhat similar objects), that the sun was a giant burning rock, and that the stars and planets might also be giant burning rocks far, far away from the Earth and appearing as dots as seen from Earth.  Since everybody already since prehistoric times believed that the moon revolved around the Earth, by the last few centuries BC educated people in Western civilization believed that the moon was what we now call a natural satellite of the Earth, though some of them believed and some did not that the sun and planets were also natural satellites of the Earth.


What happened to the third planet in the Kepler-47 system?

Well it looks like it has finally showed up! The Institute for Astronomy in Hawaii has put up the following press release: Scientists Fill Out A Circumbinary Planetary System. Kepler-47d turns out to be the largest of the planets, with a radius 7 times that of the Earth.

The relevant paper is Orosz et al. (2019) "Discovery of a Third Transiting Planet in the Kepler-47 Circumbinary System".


Yes, stars can form outside galaxies if the conditions are right. An impressive example is D100, a galaxy that is moving through a cluster so fast that the ram pressure from the ambient gas forces galactic gas out of it leaving a long tail. That tail is dense and cold enough to allow star formation, and there are newly formed clusters in it.

In principle star formation can happen anywhere where there is dense, cold gas that can collapse gravitationally. Stars can sometimes help by forcing gas together by radiation pressure. There are intergalactic gas clouds, but they are usually too hot.  Dark matter halos are not great at capturing hot gas, but intra-cluster gas is losing energy from Bremsstrahlung so over time some of it would accumulate there.

Space is large, so no doubt there is some star formation happening in corners where gas clouds have cooled off remotely from galaxies. After all, that was how stars and galaxies once started.


Point Spread Function size: Full Width Half Maximum (FWHM) vs Sigma

FWHM is the indicator of the width of a Gaussian that is easiest to measure, and is least error prone in terms of actual physical measurement (due to the slope of a Gaussian being the highest near half maximum - not exactly sure about this, but it does seem like that's the case - so error in y contributes least in the measurement of width, relatively. If this is not clear, look at a noisy Gaussian - something like this: http://www.astronomie-amateur.fr/Documents_Supernovae/Mesure_Vexp_gauss.PNG - and you'll know what I mean. It seems most sensible to estimate width using FWHM). And it has a simple correlation to the sigma with a factor of 2.35 if the distribution is actually Gaussian (which is often the assumption for PSFs - correct me if I'm wrong). Measuring actual standard deviation of data with errors is much more complicated than just simply measuring FWHM, and often it is enough to know the FWHM estimate. Hope this is the answer you were looking for.


On the day of total solar eclipse why was early sunrise observed in India?

If your sunrise had been earlier than expected, it would have been the same around the world, and would have been noticed very widely.

Here is a chart showing times of sunrise/set at your location.


What is the temperature of the black hole?

See Hawking radiation. The temperature of a black hole depends on its mass, and is given by:

$T=\frac{hc^3}{16\pi^2GMk_B}$

Where $h$ is the Planck constant, $c$ is the speed of light, $G$ is Newton's gravitational constant, $M$ is the mass of the black hole, and $k_B$ is the Boltzmann constant.


Why isn't the dark energy getting decreased?

AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe.

If some articles on the web suggest that, they are mistaken. Dark energy affects the acceleration of the cosmic expansion, but it is not necessary for the universe to be expanding, so it cannot be "the reason behind the expansion".

How the dark energy density changes as the universe expands depends on its current energy density and its equation of state--the relationship between its energy density and pressure. For cosmological constant provides the simplest dark energy model in cosmological models--it's the $\Lambda$ in the ΛDCM model--for which the pressure is exactly the negative of its energy density: $p = -\rho$. In general relativity, both energy density and pressure gravitate, and in three spatial dimensions expansion the acceleration is proportional to $\rho + 3p$ (cf. also Friedmann equations). As a result, a cosmological constant with positive energy density $\rho>0$ leads to accelerated expansion, because the contribution of its negative pressure means the overall effect is repulsive.

  If so, why isn't it getting used up in doing so.

Somewhat ironically the answer to your question for cosmological constant can be interpreted like so: it doesn't "get used up" because of a kind of energy conservation.

As a given small volume expands, it keeps a constant energy density $\rho$ and a constant pressure $p$. Is this situation sensible? Suppose its volume increased by $\Delta V$, so its energy content increased by $\rho\Delta V$. If the pressure is constant, this is some kind of isobaric process, so the work done is $W = p\Delta V$. But if $p = -\rho$, then everything is consistent because the dark energy does negative work as it expands to exactly balance the increase in its energy.

In reality, you shouldn't take the above reasoning too literally. There are many issues in interpreting anything like "how much energy" there is in general relativity and whether or not it's conserved, some of which I've covered in this answer, though some of which can be alleviated by the stressing of the "small volume" qualifier. Still, "it doesn't get used up because the equations of the theory say it shouldn't be getting used up", while perfectly true, would not be a very interesting answer.

Additionally, it should be clear from the above that the cosmological constant case of $p = -\rho$ is very itself special. The reason for its simplicity is that it's completely Lorentz-invariant: in different local inertial frames, the energy density and pressure of a perfect fluid get intermixed and transformed into different values... except in the case where the pressure is exactly the negative of energy density. Every observer would agree about the cosmological constant, so one can interpret it as an intrinsic energy density of the vacuum. In fact that's how it was originally introduced; other forms of dark energy are a generalization of $\Lambda$.

  And consequently, the rate of expansion should reduce!

With a slightly more general equation of state $p = w\rho$, the cosmological constant corresponds to $w = -1$. Other cases give a perfect fluid, which might indeed "get used up" in the sense of getting more dilute, and for positive energy density, $w<-1/3$ is necessary to outwardly accelerate cosmic expansion. But something curious happens if $w<-1$: we should expect from the work done argument for the energy density to accumulate instead of remaining constant. This kind of "phantom energy" would, if positive, produce a Big Rip scenario, in which the cosmic expansion accelerates to an infinite value in finite cosmological time, ending the universe by tearing everything apart.


I found :

$$B-V = -2.5 \log_{10} \left(\frac{f_B}{f_V}\right)$$

From this question. It works for my purposes.

Those are magnitudes on the left. Five magnitudes are a factor of 100, so 2.5 magnitudes are a factor of 10. So if $f_B/f_V =$ 100 (i.e. 10$^2$) then $B-V =$5.


What is driving the expansion of the Universe?

As far as I understood it, no one really knows. It is observable that the universe is expanding with increasing speed, therefore there has to be a form of energy driving it. We can meassure (or at least try) the amount of energy and find $\Omega_\lambda = 69.11\%$ (see $\lambda$CDM-Model)

To your questions:

1: I wouldn't imagine it as Big Bang momentum, for it is also changing. After the Big Bang, there was (or at least had to be, to make our comological model work) a period of inflation, meaning rapid expansion, which then slowed down again. I can't think of an explanation which includes some kind of finite momentum.

2: On small scales (speaking of galaxies), where densities are high, expansion doesn't act. Gravity is working against it.

3: It is responsible for the expansion. How exactly is not understood. (Little personal comment form my side. If a physicist calls something 'dark', it mostly likely means it is not understood)


Is there a non-trivial limit to a number of nested orbits?

There is no theoretical maximum. It would be possible for the solar system to orbit a sufficiently large mass, (while the planets' orbits remain stable). And it would be possible for that large mass to orbit another, even larger mass. The process could, in theory, be repeated without limit. The masses of the central bodies would have to be exponentially larger with each embedding, and so could not be conventional stars, beyond a certain point. (They could be black holes)

This is the absence of theoretical limit imposed by the theory of gravitation. In practice, multiple levels of orbits don't exist, since there does not exist a mechanism that can create the massive central bodies in the right orbits.


What is the movement of stars within the nucleus of a barred spiral galaxy

It's a little hard to know how to answer your question, partly because you're using "nucleus" in a confusing fashion. So let me try answering what might be your general question, which is "how do stars move in the region spanned by the bar?"

("Nucleus" is generally used to refer to the inner few hundred parsecs -- or even, these days, just the inner ten parsecs or so. Since bars extend to anywhere from several hundred parsecs to as much as ten thousand parsecs in radius, they are usually well outside the "nucleus". [I am deliberately ignoring so-called "nuclear bars" -- which are generally only a few hundred parsecs in size -- in order to keep things simple ;-)])

Stars in a disk galaxy generally all orbit in the same direction and in the same plane. Some of the orbits are close to circular, while others are more elongated. These elongated orbits precess -- that is, the axis of the apparent ellipse rotates with time, so the star traces out a rosette. In a barred galaxy, there are a lot of stars which are on elongated orbits which are aligned, and which precess at the same rate. Thus, they maintain a common oval region that is dense with stars (the "bar") which rotates as a whole at a more-or-less constant speed (even as the stars making up the bar are doing their own individual orbits at different speeds). The combined gravity of all these stars in an oval configuration is what keeps the whole thing synchronized and consistent, although there are some stars on more chaotic orbits.

There is a kind of magic radius called "corotation". This is where an imaginary star in a circular orbit would go once round the center of the galaxy in the same time that the bar as a whole make one complete turn. Stars inside corotation go once around faster than the bar turns, while stars outside this radius would go once round slower than the bar. Stars orbiting inside the corotation radius can form part of the bar (and generally have to be part of the bar if they're not on some very chaotic orbit). Stars outside corotation cannot be part of the bar; their orbits will actually tend to be elongated perpendicular to the bar, though as you get further away from the bar, its gravitational influence becomes less and less important, so that stars further out can have orbits that are pretty much circular.

There is good evidence -- both theoretical and observational -- that bars cannot extend further out than the corotation radius, and that bars in fact tend to extend about 3/4 or 4/5 of the way to corotation (with the stars further out but still inside corotation probably having very chaotic orbits).


Orphaned star systems and intergalactic travel (in fictional context)

The Canis Major dwarf galaxy is about 8 kpc from the Sun, but is only 8 degrees below the Galactic plane (and further out than the Sun). So it is about 42,000 light years from the Galactic centre and about 1150 light years below the plane.

This almost within the disc of the Galaxy itself. The Galactic disc has a density that varies pseudo-exponentially in both radial distance from the centre and with height above the plane:
$$ n \propto \exp(-r/R_0)\exp(-z/Z_0),$$
where $R_0$ and $Z_0$ are appropriate numbers for the radial scale length and the height scale length respectively, and $r$ and $z$ are the radial and vertical coordinates with respect to the Galactic centre.

In this coordinate system, the position of the Sun is about $r=25,000$ light years and $z \sim 0$ light years. The appropriate numbers for the scale-lengths/heights are $R_0 \sim 10,000$ light years and $Z_0 \sim 1000$ light years.

The density of stars in the solar neighbourhood is about 0.1 pc$^{-3}$ or about $3\times 10^{-3}$ stars per cubic light year. We can now scale this number according to the equation above, to work out the approximate density in the vicinity of the Canis Major dwarf galaxy, which will be loer than the solar neighbourhood.
$$ \frac{n_{\rm CM}}{n_{\rm Sun}} \simeq \exp[(r_{\rm Sun} - r_{\rm CM})/R_0]\exp[(z_{\rm Sun}-z_{\rm CM})/Z_0],$$
from which we get $n_{\rm CM} \simeq 0.06 n_{\rm Sun} = 2\times 10^{-4}$ stars per cubic light year.

To work out the average distance between stars we take the reciprocal of the cube root,
$$n_{\rm CM}^{-1/3} = 18\ {\rm light\ years}.$$

So, if you have a range of 500 light years per "jump" you have no problem skipping all the way to the Canis Major dwarf galaxy.


How much of the ambient radiation on Mars is arriving line of sight from the Sun?

Besides solar energetic particle (SEP) events, it's mostly (almost isotropic) galactic cosmic rays (GCR). GCRs may cause secondary radiation by interaction with the atmosphere or with rock. The smaller the solid angle of the visible sky, the smaller the mean galactic radiation dose. On Mars you get additional shielding by the atmosphere, hence dependence of the atmospheric pressure. More detail in this paper.
Shieldings from SEP events should be rich in protons, e.g. water or food, or you need to go deeper into underground, as soon as a SEP event is detected.
GCRs are partially shielded by the heliosphere. Hence solar maxima protect partially from GCRs, but SEP events get more likely during solar maxima.
Since GCRs are unpredictable, and SEP events can be detected, for solar maxima, protection strategies resultng in an overall reduced radiation dose can be elaborated.


What's the difference between grism and grating?

Transmission gratings on their own introduce chromatic aberration.  This is because they change the effective focal length and do so as a function of wavelength.  The chromatic aberration can be eliminated by introducing a prism of the correct dispersion.  Known as a grism, the grating/prism combination provides an unaberrated image at the 0th order in addition to the spectrum, although the spectrum has fairly low resolution.


Information Paradox with Hawking's Radiation

How exactly can Hawking's Radiation carry the information?

It does not, if you mean information about the interior of the black hole.

The relevant theorem is the No Hair Theorem.  Basically you can only get gross statistical properties of black holes, nothing else and no information about the interior structure (meaning any info inside stored in any way, stays there or is lost as randomized radiation).

If you accept the No Hair Theorem then Hawking radiation would be essentially random in character, giving you no more information about the black hole than the No Hair Theorem allows you to know.  This would mean "information" being lost as a black hole evaporates.

  How is it possible for information to exist at the event horizon?

I don't think it's possible for anything to exist at the event horizon for any time.  It's a surface of zero thickness (as I understand it) and no actual substance.  Nothing can stop there and no real thing could be said to be "on" the horizon, just passing through it.

One of the complication of discussing black holes in relation to things like quantum-scale elementary particles is that while we can't say exactly where a particle is (the uncertainty principle) "classical" general relativity (i.e. without quantum theory) doesn't incorporate this at all and so the idea of a particle having a precise position is perfectly reasonable in classical general relativity.  But that's unlikely to be the case in whatever theory we end up with that links general relativity and quantum field theories.

I suspect you are hearing things related to the Holographic principle.  This is pretty much beyond my real understanding in any depth, but note that it is intimately linked with trying to connect String Theory, gravity (and hence a kind of quantum gravity), Thermodynamics and a notion of "information" within those multiple contexts.  I personally don't intend to try and read too much into this as it's beyond cutting edge theory for me (beyond my "event horizon" I suppose :-) ).  However I'd suggest being wary of reading too much into this idea given the present state of the theories involved.


How is distance in light-years converted to years back in time?

Yes, for stars in our galaxy, if a star is 736 light years away, the light took 736 years to reach here.

For very very distant objects, you need to account for the expansion of space. The light might have been travelling for 10 billion years, but in that time space has expanded and so the distance that you would find if you froze time and measured would be greater (to get the proper distance at the current time) it would be larger.  This effect is not significant at distances less than about a billion light years.


I wonder if the area I am looking for is GSH 238+00+09.
In the color image quoted in my question, the galactic core lies beyond the top center. This would be 0° galactic longitude. Going counterclockwise, 238° is somewhere around Beteigeuze.
Here is a "[p]rojection of the Local Cavity onto the Galactic plane":

Figure 9 from Lallement et al. (2003)
But if you try to superimpose this image on the other one (I matched the positions of the Sun, the Ophiuchus clouds and the Pleiades bubble, the Lupus cloud is a bit off), it seems as if the area marked in green was actually part of our Local Bubble! And GSH 238+00+09 then appears to be identical to the Orion-Eridanus-Superbubble:

I'm sure I didn't match this correctly. Especially Loop III would be in totally the wrong place. But I found an article that used the same black and white image to illustrate possible locations of Loop III, and the second alternative to the one shown here was in the upper right hand corner, where it would then overlap with "Loop II" of the color image. But probably we should not worry too much over Loops II and III. The creator of the 3D Galaxy Map writes: "I haven't added Loop II, III or IV because I couldn't find any accurate information about their size and location." I'm sure he has been looking more thoroughly than I have, so I feel the mismatch of Loop III is not a problem.
But then the color image is from 1994, and understanding and cartography of our Galactic neighborhood has advanced quite a bit in the ten years between the two publications, so maybe the placement of the Local Bubble in the color map is only a rough equivalent and the mismatch between the two images is not my fault but a result of scientific progress.
Here is another image (rotated 180° in relation to the color image from my question) of local loops, bubbles, and superbubbles, where it becomes clear that GSH 238+00+09 (termed "New Superbubble" (NSB) in this image) is not identical to the Orion-Eridanus-Superbubble:

Figure 8 from Heiles et al., (1998)
If you compare the scales of this last and my first color image, you will note that the color image falls completely inside the circle denoting the Local Bubble, supporting the thought that the green area lies within it (and not within GSH 238+00+09 or the Orion-Eridanus-Bubble). But this second image also supports the insight that the color image might not be that accurate, and that maybe the area whose name I am looking for does not exist in that shape at all.
In the following image (which shows Loop I as the Scorpius-Centraurus Shells and the Orion-Eridanus-Loop as the Orion Shell) the red blot immediately rimward (lower right) of the sun is the Taurus Dark Cloud. This is not from a scientific but a popular publication (just as my color image), so we should not expect it to be totally accurate, but it still suports the notion that the green area might be part of the Local Bubble.

Unnumbered 3rd figure from Frisch (2000)
Here is a final image, apparently a different rendering of the image from Lallement et al. (2003), the first image given in this answer. In this image the location of some important stars is marked. Most notably ε CM, which is the scientific name for the Arab name Adhara, with which the same star is marked in the color image from my question and there lies far out in the green area, **lies still within the Local Bubble in this image, close to the edge.

Figure 3 from Sfreir et al. (1999)
Here I overlayed the previous image and the image from my question. I matched the sun an Adhara. Other objects are still a bit off, but not much, and we get a good idea of where the Local Bubble ends.

So it seems the local bubble extends to a bit beyond the star Adhara, halfway from the Sun to the right edge of the image in my question and does indeed include the closer part of the green area.
This thought is supported in the abstract of Heiles (1998), where it says: "[GSH 238+00+09] is at least partially responsible for the apparent extreme elongation of the Local Bubble toward l ~ 230° ..." (my emphasis).
I'll leave my first thoughts and the badly matched image in this answer, to document how I came to the final conclusion.
I'd welcome any thoughts and corrections of this attempt at an answer.

Sources:

Frisch, P. (2000). The Galactic Environment of the Sun The heliosphere appears to protect the inner solar system from the vagaries of the interstellar medium. American Scientist, 52-59. Available online at https://www.americanscientist.org/issues/issue.aspx?id=862&y=0&no=&content=true&page=5&css=print
Heiles, C. (1998). Whence the Local Bubble, Gum, Orion? GSH 238+ 00+ 09, a nearby major superbubble toward Galactic longitude 238. The Astrophysical Journal, 498(2), 689. Available online at http://iopscience.iop.org/0004-637X/498/2/689/pdf/0004-637X_498_2_689.pdf
Lallement, R., Welsh, B. Y., Vergely, J. L., Crifo, F., & Sfeir, D. (2003). 3D mapping of the dense interstellar gas around the Local Bubble. Astronomy and Astrophysics, 411(3), 447-464. Available online at http://www.aanda.org/articles/aa/full/2003/46/aa3635/aa3635.right.html (under "Article" in the sidebar)
Sfeir, D. M., Lallement, R., Crifo, F., & Welsh, B. Y. (1999). Mapping the contours of the Local bubble: preliminary results. Astronomy and Astrophysics, 346, 785-797. Available online at http://adsabs.harvard.edu/full/1999A%26A...346..785S


One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones.

Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay.

Time goes up on the diagram, and a typical infalling trajectory is in blue.

Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity.

Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually shorten your lifespan. You'll actually get to the singularity sooner if you struggle in that manner.

This particular image came from Prof. Andrew Hamilton. Note that it pictures an eternal Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no "antihorizon"). However, in all respects relevant here, it's the same situation.


How close to a host star can a tidally locked planet be and its dark side still maintain a moderate temperature?

Your general idea about this process is correct. At close semimajor axis distance, rock can evaporate and will form a Silicate-oxygen atmosphere.
For low-mass rocky planets, the condensation flow from day to night-side, as it necessarily is very hot, will have to compete with the possibility of instead escaping vertically from the nightside, instead of precipitating. For higher-mass planets, vertical escape will be too difficult, and the hot Silicate-Oxygen gas will recondense on the nightside and therefore heat via thermal radiation and condensation latent heat.
This is a problem on which current research is in progress, and not many groups have worked on this, with the exception of one fantastic article that came out this year. They glue four different 1-D models together to represent vertical and horizontal for each day and night-side of catastrophically evaporating planets together, in order to create a sort of fake-2D model of the planetary Silicate-atmosphere.
Their nightsides are very hot in general (500-1000K), but they sadly do not give silicate densities or pressures. Hence, without the density, it is difficult to estimate how much the heat transfer between the Silicate gas and a human would be, i.e. how much a human would 'feel'.
If you are very curious though, I am sure you can construct this effect, by assuming Silicate saturation pressures, which are given in this article.


What is the faintest magnitude a telescope can see?

This is a term known at the limiting magnitude. This term refers simply to the faintest apparent magnitude your instrument can detect. Wikipedia has an article describing limiting magnitude.  I pulled the info from there.

The wiki article begins off explaining the basic way of finding limiting magnitude.

$$5 \cdot \log_{10}\left(\frac{D_1}{D_0}\right) $$

Now, if we simply plug in some values, 254 mm telescope and assuming the eye is 6 mm, we get an answer of 8.13.  This can be rounded to nearest base and we can say that it increases the magnitude by 8.

The article goes into more description about the contrast of the given medium. With Earth, we have the atmosphere, lights, and pollution to worry about. The more you are able to reduce these effects, the better. That's why for telescopes at observatories, we place them in high places, usually away from populated areas to help reduce these worrisome effects. This helps increase the contrast between the object you are looking at and its medium (in astronomy the night sky).

In the article it explains this with the formula below:

$$m \cdot v = m_{nakedeye} - 2 + 2.5 \cdot \log_{10}(D \cdot P \cdot t)$$

$D$ = objective or main mirror diameter in mm
$P$ = power or magnification
$t$ = transmission factor, usually 0.85–0.9

Simply putting in our values from above and estimating a 10" telescope has a 250x magnification with a transmission factor of 0.85, we get how the wiki article states that the answer of a 10" telescope easily getting a magnitude 15 is possible.

Now for observational laboratories and something like the Hubble telescope, they use additional factors, including image processing mechanisms that help reduce noise in the image and integration time of the object you are viewing.  This means using a sensor that can capture the image at low light levels AND while it moves.

If you really want to know more, go here. This link actually explains in very technical detail the technology behind the imaging discovery efficiency.  Basically, it helps boost the limiting magnitude to detect extremely faint objects.


Why don't astronomers use meters to measure astronomical distances?

In addition to the answer provided by @HDE226868, there are historical reasons. Before the advent of using radar ranging to find distances in the solar system, we had to use other clever methods for finding the distance from the Earth to the sun; for example, measuring the transit of Venus across the surface of the sun. These methods are not as super accurate as what is available today, so it makes sense to specify distances, that are all based on measuring parallaxes, in terms of the uncertain, but fixed, Earth-Sun distance. That way, if future measurements change the conversion value from AU to meters, you don't have to change as many papers and textbooks.

Not to mention that such calibration uncertainties introduce correlated errors into an analysis that aren't defeatable using large sample sizes.

I can't speak authoritatively on the actual history, but solar system measurements were all initially done in terms of the Earth/sun distance. For example, a little geometry shows that it's pretty straightforward to back out the size of Venus's and Mercury's orbit in AU from their maximum solar elongation. I don't know how they worked out the orbital radii of Mars, etc, but they were almost certainly done in AU long before the AU was known, and all of that before the MKS system existed, let alone became standardized.

For stars, the base of what is known as the "cosmological distance ladder" (that is "all distance measures" in astronomy) rests on measuring the parallax angle:
$$\tan \pi_{\mathrm{angle}} = \frac{1 AU}{D}.$$
To measure $D$ in 'parsecs' is to setup the equation so that the angle being measured in arcseconds fits the small angle approximation. That is:
$$\frac{D}{1\, \mathrm{parsec}} = \frac{\frac{\pi}{180\times60\times60}}{\tan\left(\pi_{\mathrm{angle}} \frac{\pi\, \mathrm{radians}}{180\times60\times60 \, \mathrm{arcsec}}\right)}.$$
In other words, $1\operatorname{parsec} = \frac{180\times 3600}{\pi} \operatorname{AU}$.

Astronomers also have a marked preference for the close cousin of mks/SI units, known as cgs. As far as I can tell, this is due to the influence of spectroscopists who liked the "Gaussian units" part of it for electromagnetism because it set Coulomb's constant to 1, simplifying calculations.


Pointing direction movement of a fixed camera

As has been noted, the camera, if truly fixed, will point at spot with fixed declination. The RA it points to will be a simple function (namely so many hours and minutes ahead/behind) of the local sidereal time.

Either your camera is not truly fixed (a bit of shake) or (possibly more likely), you have not correctly determined the local sidereal time.


Seeing is typically measured as the FWHM of the seeing disk, but can also be expressed though the Fried parameter $r_0$, which measures the size, or strength, of the parcels of gas that cause the turbulence in the atmosphere. The relation between the two is (e.g. Vernin & Munoz-Tunon 1995)
$$
\text{seeing} = 0.98\frac{\lambda}{r_0},
$$
where $\lambda$ is the wavelength of the light.

$r_0$ can by calculated by integrating the "turbulence strength" along the line of sight$^\dagger\!\!$, and  scales as $\lambda^{6/5}\!$. Because of this dependence, the "usual" seeing has only a weak ($\lambda^{-1/5}$) dependence on wavelength.

From the Wikipedia article on seeing:

  At visible wavelengths, $r_0$ varies from 20 cm at the best locations to 5 cm at typical sea-level sites

$\dagger$Actually, $r_0$ refers to a distance corresponding to the line of sight towars zenith; observing at an angle $\zeta$ from zenith introduces a factor $\cos^{3/5}\!\zeta$.


Can New Horizons probe turn back and start orbiting Pluto

A Pluto orbiter would have been possible, but then the entire mission would have to be designed around that. New Horizons was designed as a small fast probe that could reach Pluto relatively quickly (almost 10 years), but not be able to stop there. An orbiter would have to take a lot more fuel with it, and approach Pluto slower. Such a mission would therefore take a lot longer to reach Pluto, i.e. it could easily take double the time or more, and would be more expensive.


If your telescope has an equatorial mount, its polar axis should point toward the south celestial pole (SCP).
The bad news is that all the stars within 5 degrees of the SCP are magnitude 5 or fainter.
The good news is that visual observing does not require precise polar alignment.
Even if you are off by 5 degrees, the mount's drive should keep most targets in view at least 10 times as long as if the drive were not running.

You don't have to aim the telescope's optical tube at the pole, only the polar axis of its mount.
Find your geographic latitude, set the polar axis to the same angle above the horizontal, and point it south.
The skymaps.com Evening Sky Map,
Southern Edition, can help you get your bearings.
The SCP makes a right triangle with Crux and Canopus,
both of which are fairly high in the southern evening sky in March, April, and May.
If you really need a precise alignment, this
Australian article
has a finder chart for
$\sigma$ Oct
and describes the drift method.
This article in Spanish may also be useful.


On a lunar map, why is it labelled East on Earth, West on Moon?

Imagine you're lying down outside, looking up at the sky, with your body aligned so your head is pointing North, and your Feet are pointing South. If you look to your Left, you'll be looking East, and see the East part of the ground and sky.  If you look to your right, you'll be looking West, and see the West part of the ground and sky.
Now imagine a friend  holds a globe above your head, centered on your home city, with the north pole pointing North, and the South pole pointing South.
When you look at your country on that globe, the parts East of your city will be on the right side of the globe, and the points west of it will be on the left side of the globe.
The same thing happens with the Moon. When you're standing on the Earth, looking up at the Moon near the meridian,  from the Northern Hemisphere (so it appears generally south of you), The parts of the Moon that are Lunar-Surface-East of the Moon's center will appear on the right side, and the parts of the Moon that are Lunar-Surface-West will appear on the Left.


What determines the Earth's orbital separation from the Sun is the total mass of the Sun and Earth, as described by Kepler's third law.

The orbital radius of the Earth is increasing because the Sun is gradually losing mass both by emitting radiation and through the solar wind at a rate of around 5 million tonnes per second (e.g. Noerdlinger 2008).

Any gain of mass through the accretion of dust or asteroids (by the Earth or Sun) is negligible by comparison.

This mass loss rate is only sufficient to widen the Earth's orbit by about 1.5 cm per year.


What is that light of the middle of the milky way galaxy?

The galactic bulge is formed by tightly packed stars and interstellar dust
Also most stars are in the direction towards the galactic center
Thus it would appear brightest if it were viewed from outside the plane of the galaxy as this image is intended to show.
However, we don't see this from Earth, and the reason why is explained in  Phys.org's Why can't we see the center of the Milky Way? explains why we don't see this at night.
A brief outline of what is described there is as follows:

When it is dark enough, and conditions are clear, the dusty ring of the Milky Way can certainly be discerned in the night sky. However, we can still only see about 6,000 light years into the disk with the naked eye, and relying on the visible spectrum. Here's a rundown on why that is.

Size and Structure
Low Surface Brightness
Dust and Gas
Limited Instrumentation

However we could see a bright spot if we were outside the Earth's atmosphere and could see in certain wavelengths of infrared. The article shows the following image from COBE

Milky Way in infrared. Credit: COBE Source

False-color image of the near-infrared sky as seen by the DIRBE. Data at 1.25, 2.2, and 3.5 µm wavelengths are represented respectively as blue, green and red colors. The image is presented in Galactic coordinates, with the plane of the Milky Way Galaxy horizontal across the middle and the Galactic center at the center. The dominant sources of light at these wavelengths are stars within our Galaxy. The image shows both the thin disk and central bulge populations of stars in our spiral galaxy. Our Sun, much closer to us than any other star, lies in the disk (which is why the disk appears edge-on to us) at a distance of about 28,000 light years from the center. The image is redder in directions where there is more dust between the stars absorbing starlight from distant stars. This absorption is so strong at visible wavelengths that the central part of the Milky Way cannot be seen. DIRBE data will facilitate studies of the content, energetics and large scale structure of the Galaxy, as well as the nature and distribution of dust within the Solar System. The data also will be studied for evidence of a faint, uniform infrared background, the residual radiation from the first stars and galaxies formed following the Big Bang.


The answer is, of course, that a singularity can't exist.  The fact that General Relativity predicts that singularities are inevitable under certain conditions is proof that GR is wrong. A singularity is an infinity and a place where the equations of GR literally break down and become meaningless. (Infinities are decidedly unphysical!) When (if!) a more complete theory of space, time and gravity is developed, the new theory will probably show finite results where GR shows a singularity forming.  (I say probably because whatever more complete theory of gravity we develop next is not guaranteed to be the final, completely correct theory -- it may just be an improvement which solves some issues with GR, but not all.)

There is good reason to think that the successor to GR will reproduce nearly all of GR's predictions -- GR is, after all, a very well verified theory.  No prediction of GR that we have been able to test has been less than 100% right, and any new theory of gravity will have to match GR in those realms while showing differences in places (like the center of Black Holes) where gravity is very, very, very..., very strong.

Now the existence and properties of a BH's event horizon is a (no pun intended) gray area.  Unlike near the singularity at the center of a GR black hole, there is no obvious signal in the mathematics of GR that it fails there.  Nor have we made any measurements of the properties of an EH to test against GR's predictions. So our confidence in those predictions is due to the believed-high quality of the theory rather than due to test successfully passed.

(In fact, there's good reason to think that any post-GR theory of gravity will make different predictions about the nature of the event horizon, since it's very difficult to  reconcile the GR's event horizons with quantum mechanics, and quantum mechanics will almost certainly will be part of the next theory of Gravity.)

Having said all that, there's a second answer to your question, that the question itself is ill-formed. As @mmeent says in a comment to another answer, you are assuming a flat space-time and stationary observers, needed in order that simultaneity between those distant observers and observers near the BH be meaningful:  It is meaningful to talk about what any observer sees, but not meaningful to speak about what a distant observer and an observer near the event horizon see at the same time.

There is no inconsistency with observers at infinity seeing things slow down asymptotically as they approach the EH so as to never quite reach it, and observers falling into the same BH passing the EH in a finite time (and noticing nothing special about space at the EH).  The whole point of the distant observer seeing nothing ever quite reach the EH is a consequence of there being no timelike or null curves (light follows null curves and matter always follows timelike curves) which reach out of the BH to outside observers.  You can't get here from there.

It's also worth remembering that nearly every statement you have ever read about BHs is about equilibrium BHs. But a BH where the mass is falling in asymmetrically isn't going to be at equilibrium. Fortunately, GR is a dynamic theory and GR's BHs can be dynamic objects, though they settle down to equilibrium very, very quickly.  We have observed this process using LIGO (the ringdown of two merging BHs is almost the poster child for "Not At Equilibrium") and the observations appear to be consistant with the predictions of GR. (This is a hot area of research, so new, surprising, results are very possible.)

I'd also note that if you fall into BH, the incoming blue-shifted light won't fry you because you, yourself will be falling along with the light.  What an observer at infinity sees as a huge flux of high-energy radiation as it falls close to the EH, you, falling with it, will see as ordinary starlight. (If you could somehow stand still, you'd see the blue-shifting, though.)  If you like, you can think of your high-speed fall red-shifting that blue-shifted light back to normal.


Neutrino interaction with dense molecular structures

Neutrinos wouldn't pass through white dwarfs or neutron stars as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.

To scatter a neutrino with a chance of > 50%, you "just" need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:

The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to this paper. A proton has a diameter of about $1.75\cdot 10^{-15}m$.
With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.


Questions about Hubble's brand new view of an interstellar comet (video)

TL;DR version, Yes, yes, distortion. In more detail:

HST's orbit is approx. 90 minutes so this will cause the breaks. Some more information is given in the Orbital Constraints section of the Cycle 27 guide for proposers.
Since HST is having to move to track the moving comet to keep it in the (small) field of view of the instruments, it will likely need to roll the spacecraft. This is to keep suitable guide stars in the Fine Guidance Sensors, to keep the Sun on the solar panels for power and to keep the pointing constraints of not looking to close to the Sun, Moon or Earth satisfied. These are discussed in more detail in the Orientation and Roll constraints section of the primer/guide for proposers mentioned earlier.
This is due to the geometric distortion of the optics in UV/Visible (UVIS) channel of the WFC3 instrument as described in more detail in this section of the WFC3 Instrument Handbook,


Where can I find historic rates of meteor impact events?

The database of the meteoritical society might be what you are looking for. You can do a search by year.
Based on this data, I drew this quick plot of number of recorded meteorites per year

Now there are a couple things to see here. First is that we find as roughly as many meteorites as we did ten or twenty years ago. So these events are not more common now than they were in recent memory. This answers your friends' question. Second is that we have many more records of meteorites than we did 100 years ago. This is simply because these numbers are not the exact number of meteorites that occurred, but rather the number of meteorites that were recorded.

The International Meteor Organization also has some databases you can browse to find more information.


How to fix a Direction in the context of universe?

I had a nice long answer all mapped out, then I came across a similar (identical?) question...
refer answer here:
How does the earth's plane of orbit (ecliptic) vary over time?

Basically it is changing about 1.4 degrees per 10000 years about now, but that will change chaotically over longer periods.


Mars night sky - how many stars visible?

I'll add to Wayfaring Stranger's comments. In fact most of the time you would be able to see fewer stars in the night sky of Mars, than in a good dark night sky on Earth, because of dust obscuration.

Even in favourable conditions, the optical depth of the Martian atmosphere is usally somewhere between 0.5 and 1 per airmass. (Petrova et al. 2012; Lemmon et al. 2014) and is nearly wavelength independent. This corresponds to a reduction in flux to between 37% and 60% of its value above the atmosphere. This compares to a typical optical (V-band) extinction of around 0.1 magnitude at a good site on Earth, which allows 90% of the flux through.  There are times when it can be much worse than this on Mars.

This means that the limiting magnitude on Mars would be somewhere between 0.44 and 0.96 magnitudes brighter than it is on a dark site on Earth and that all the stars would be fainter by these values.

As a rough guide to the effect this has I took the Hipparcos catalogue and made a frequency histogram of V-magnitude of the stars. If we define a Hipparcos magnitude of 6 as the limit you can see from a good dark site on Earth, then you would see 4559 stars (that's over the whole of the sky in both hemispheres). If that limit was reduced to 5.56 or 5.04, then this number decreases to 2745 or 1560 respectively. Thus there would be a reduction in the number of stars you could see by typically between a factor of 1.66 and 2.92.


How to tell a pulsar is rotation-powered or accretion-powered?

The spin behaviour of the two types of pulsar would usually be quite different. The $\dot{P}$ for a rotationally powered pulsar is always positive, and higher order time derivatives of $P$ are quite small. This is because the rotational kinetic energy is powering the pulsar emission and the neutron star continually spins down as it loses rotational energy.

Accretion powered pulsars can have very variable spin down or spin up characteristics, because they are powered by mass transfer and accretion in binary systems, and are influenced by a variety of factors affecting the accretion flow and how it couples to the pulsar magnetic field. The rates of spin down can be much higher than can be plausibly be accounted for by magnetic spin down as in a rotationally powered pulsar. Spin up cannot be accounted for in a rotationally powered model and neither can sign reversals or dramatic variability in $\dot{P}$.

Additionally, an accretion-powered pulsar would necessarily need to be in a short period binary system - therefore there would be a very obvious periodic modulation in the pulse period caused by the doppler shift as it travels in ts orbit.


Can the paper narrowing Solar System's barycentre to within 100m help find Planet Nine?

Short answer
Short answer is no, it can't prove or disprove the existence of Planet 9. The reason is because, even if there is a significant difference between the barycenter of the solar system with and without Planet 9 we wouldn't be able to tell without hundreds if not thousands of years of precise data. If we don't have Planet 9 as a reference and are limited to the Sun and the eight known planets for reference then without another reference in the same frame we have no way to observe the motion of our reference.
It is perhaps useful to consider the solar system (i.e. the Sun plus the eight known planets) as orbiting together around the barycenter of the solar system plus Planet 9 system. Without seeing Planet 9 as a reference how would we have any notion that this other barycenter even exists?
Of course Planet 9 would still be exerting gravitational influence and over time this would cause a deviation between where the barycenter should be and where it actually is -- that is the only way we could use the motion of the barycenter to infer anything about individual bodies. Essentially, the barycenter is a convolution of the gravitational influence of all the mass,5 its deconvolution requires analysis of temporal variation.

 Unnecessarily long answer
Note   -   When I discuss variations in the barycenter I will be referring to the barycenter of the solar system in heliocentric, not barycentric, coordinates (in barycentric coordinates of course there is no variation in the barycenter, by definition). This simplifies some of the qualitative analysis.

Rigorously answering this requires answering several separate questions (some which aren't necessarily relevant in answering the broader question but do provide useful background):

Does the location of the barycenter over time contain enough information to recover information about the contributing masses?
How precisely can we predict the location of the barycenter?
How much deviation would a $5$-$10~M$⊕ planet at $400$-$800$ AU cause to the center of mass of the solar system?
If no such deviation exists, does that rule out the existence of Planet 9? Conversely, if such a deviation does exist, does it rule in the existence of Planet 9?

 1 - Location of the barycenter over time
The short answer to question 1 is yes (sort of). There is information on the orbits of the contributing masses which can be backed out from simple analysis of the location of the barycenter over time.
The emphasis here is because at any given point in time the location of the barycenter contains no information about the location of individual bodies, only the entire mass distribution. However, since most of the mass in the solar system is discretized into large bodies (as opposed to a gas or dust cloud, protoplanetary disk, etc), the influence of each massive body on the location of the barycenter will vary with a frequency proportional to its orbital period -- and, critically, this is fairly measurable (but only if we can also know the relative positions of the bodies). Consequently, the influence of the individual masses can be retrieved (this is not necessarily the case for gas or dust clouds and protoplanetary disks).
In a simple system, retrieving information about the orbits is fairly straightforward (although I will neglect the rigorous demonstration for brevity). The issues arise when the system becomes more complicated.
One source of confound is resonant orbits. Certain resonances would defy deconvolution because the frequencies of their respective influences on the barycenter are synchronous. Furthermore, multiple resonances can't necessarily be disambiguated: the frequency pattern in the motion of the barycenter caused by 2 bodies in a resonance could be replicated by 3 bodies in a resonance. Pluto and Neptune are in a 2:3 resonance, so I don't think we can immediately dismiss the possibility of a theoretical Planet 9 being in a resonance of some kind and that this could hinder deconvolving its effect on the barycenter.
Another significant challenge is the qualification that the barycenter must be observed over time, and more specifically how much time is needed to deconvolve the various influences. As I mentioned, each body's influence on the barycenter is cyclic based on their orbital period and eccentricity. Using the Bulirsch-Stoer integrator from Vulcan and tracking an energy analog of the barycenter's motion1 as a function of time we can analyze the effect of the different orbital frequencies.
Let's consider a three body system of the Sun, Saturn, and Jupiter. In this system the influence of Saturn and Jupiter should clearly be visible in the resulting sinusoid. After integrating this system for $2$ Saturn years ($60$ Earth years) the energy-analogy over time looks like this:

It should be pretty clear that the above behavior can be created by two sinusoids with different frequencies, to any who find this unclear I recommend having a look at this tool. If wanted to we could use Fourier analysis to deconvolve those two frequencies and recover the orbital periods of Jupiter and Saturn (I will also neglect this for brevity).
Let's consider a slightly worse case: a three body system comprised of the Sun, Jupiter, and a hypothetical hot jupiter with a semi-major axis of $1.55^{}$x$10^{10}$ m on a relatively circular orbit. Our energy-analog after $3$ hot-jupiter years ($12$ Earth days) looks like this:

Where did the other sinusoid go?
Its effect is still there, but we haven't sampled sufficient time to see it. We would need to look at $1,095$ hot-jupiter years ($12$ Earth years) in order to see one orbit of our toy Jupiter2 Fortunately we don't have to integrate quite that long to see the effect of the toy Jupiter on the energy-analog. After $320$ hot-jupiter years ($3.5$ Earth years):

This becomes even more complicated with non-zero eccentricities and unstable orbits. For a similar three body, hot-jupiter system but with the toy Jupiter having an eccentricity of $0.65$ over a period of $7$ Earth years the energy-analog is:3

There are other ways to draw inferences about the orbital parameters from the relative motion of the barycenter (e.g. the shape of individual cycles), but that is for a different question.
In sum, the location of the barycenter over time could certainly be used to find a mass imbalance, but but not necessarily argue for the existence of Planet 9 -- that would depend on the specifics of the observations.

 2 - Precision of barycenter predictions
There are number of significant barriers to accurately calculating the barycenter of the solar system but the most troublesome are the uncertainties regarding the interiors of Saturn and Jupiter. Specifically, the behavior of liquid metallic hydrogen at such massive pressures and (more so for Saturn) understanding their gravitational moments (Fortney 2004).
The issue this raises is that without knowing the centers of mass for Jupiter and Saturn (which contain 92% of planetary mass in the solar system) with sufficient precision we don't know well enough what the solar system barycenter should be to determine whether or not the true barycenter differs enough to indicate the existence of Planet 9.
With information from the Cassini spacecraft combined with data from the VLBA radio telescope the Saturn ephemerides was constrained to $4$ km in 2015. This represented an improvement of about an order of magnitude. The Juno spacecraft provided a similar improvement to the Jupiter ephemerides, constraining it to within $10$ km in 2019.
What is less well constrained is how constant these parameters are, so it bears noting that an inaccuracy of $350$ km in the location of the center of mass of Saturn's core within the planet corresponds to an uncertainty of $100$ m in the position of the solar system barycenter. Jupiter lacks a dense core like Saturn's, so it is more difficult to quantify a direct relationship between inaccuracies, but suffice it to say that an inaccuracy of $100$ km in Jupiter's center of mass corresponds to an uncertainty of $100$ m in the barycenter.4
So assuming Saturn's center of mass has not drifted much since September of 2018, I would estimate the precision of the calculated solar system barycenter to be roughly $\pm15$ m.

 3 - Deviation from predicted barycenter
Calculating the effect Planet 9 would have on the barycenter is actually quite trivial thanks to superposition (see this answer for a more rigorous treatment of the limitations of this concept when it comes to gravitational interactions). The equation for calculating the distance from the primary body to the barycenter in a two-body system,
$$
r_1 = \frac{a}{1 + \frac{m_1}{m_2}}
$$
can also be applied to calculating the barycenter of two barycenters where $a$ is the distance between the two barycenters, $m_1$ is the mass which contributes to the primary barycenter, $m_2$ is the mass which contributes to the secondary barycenter, and $r_1$ is the distance from the primary barycenter to the shared barycenter. And since $99.86$% of the mass of the solar system is contained in the sun it is a rigorous approximation to use
$$
r_1 = \frac{200 \cdot 1.496 \text{x} 10^{11}}
           {1 + \frac{1.989e30}{5 \cdot 5.972 \text{x} 10^{24}}}
    = 4 \text{x} 10^8 \text{ m}
$$
as a lower bound, and
$$
r_1 = \frac{1200 \cdot 1.496 \text{x} 10^{11}}
           {1 + \frac{1.989e30}{10 \cdot 5.972 \text{x} 10^{24}}}
    = 5 \text{x} 10^9 \text{ m}
$$
as an upper bound for the distance from the barycenter of the solar system without Planet 9 to the barycenter with it.

 4 - Interpretation
Even the lower bound of $4\text{x}10^{8}$ m is a huge deviation, so did we just demonstrate that Planet 9 can't exist? Unfortunately no. This goes back to number 1 - the problem is time. There are a number of ways to think about this, but the salient point is that the influence on the barycenter of the solar system would have been persistent since well before the earliest observational records. Planet 9 would have to have an orbital period on the order of $10,000$ years. Over periods of thousands of years it would have obvious and readily measurable effects on the motion of the other planets (and that motion would be reflected in the motion of the true barycenter), but otherwise the effects are simply too small to disambiguate from noise.

1 I create this analog by calculating a hypothetical energy of the system if the center of the sun was the barycenter of the system and comparing it to the initial value. It is not tremendously physically meaningful but it does reduce the parameter-space of the temporal position of the barycenter to two dimensions, thereby making it easier to show the effect I discuss graphically.
2 This is one of the reasons simulating hot jupiters is so difficult (and one of the things the Vulcan project is intended to remedy): the limiting factor for orbital integration is always the fastest-orbiting body (although multi-step integrators mitigate this somewhat) so systems with large differences between orbital periods take proportionally longer to simulate. Using the Mercury code, a gigayear solar system simulation with a hot-jupiter on an 8 day orbit would require roughly 10 months of simulation time on a high-end desktop computer.
3 Note that the energy analog is degenerate here because the hot-jupiter will be ejected after approximately 10 Earth years.
4 These uncertainty relationships are from my own calculations and may be flawed as they have not been rigorously tested or reviewed.
5 And I do literally mean all the mass. As in all the mass in the universe, gravity has an infinite range after all....


How is the Milky Way on a collision course with Andromeda when the universe is expanding?

I'm not an expert, but I looked into this recently when toying with the idea of a scale model of the observable universe. To the best of my understanding, the following is approximately correct.

TLDR: The expansion happens on the very large scale of the universe. At the scale of the observable universe the Andromeda galaxy is a next-door neighbour, and at the local scale gravity is strong enough to override any effect of expansion.

The Milky Way is 100,000 light years across, 2.5 million light years brings us to the Andromeda galaxy.

Galaxies cluster in galaxy groups, and 10 million light years span about 50 galaxies and brings us to the edge of what's called the Local Group.

Galaxy groups cluster into superclusters. We're part of the Laniakea Supercluster, which spans about 100,000 galaxies and 500 million light years.

For any appreciable expansion (redshift 0.1) you need at least a billion light years. It's something that happens in the voids between superclusters, and doesn't become really pronounced until you're halfway across the observable universe (redshift 1, 7-8 billion light years as measured in light travel time).

For the "raisins in a loaf" analogy, I think it's not too wrong if you think of the raisins as superclusters rather than individual galaxies. Note that although the dough expands in a baking loaf, individual raisins do not, because there are other forces at work that hold them together.


Observable universe equals its Schwarzschild radius (event horizon)?

There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is not equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of $\gtrsim3$.

The Schwarzschild radius of the Universe

Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is $R \simeq 46.3\,\mathrm{Glyr}$.

The mean density of the Universe is very close to the critical$^\dagger$ density $\rho_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$.
Hence, the total mass (including "normal", baryonic matter, dark matter, and dark energy) is
$$
M = \rho_\mathrm{c}V = \rho_\mathrm{c}\frac{4\pi}{3}R^3 \simeq 3.0\times10^{57}\,\mathrm{g},
$$
and the corresponding Schwarzschild radius is
$$
R_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 475\,\mathrm{Glyr}.
$$

The Universe is not a black hole

Even worse! you might say. If our Universe is much smaller than its Schwarzschild radius, does that mean we live in a black hole?

No, it doesn't. A black hole is a region in space where some mass is squeezed inside its Schwarzschild radius, but the Universe is not "a region in space". There's no "outside the Universe". If anything, you might call it a white hole, the time-reversal of a black hole, in which case you could say that the singularity is not something that everything will fall into in the future, but rather something that everything came from in the past. You may call that singularity Big Bang.

The error in the source

You will see questions like this many places on the internet. As I said above, they all assume the Hubble radius, $R_\mathrm{H} \equiv c/H_0$ for the radius. But this radius is well within our observable Universe, and doesn't really bear any physical significance. In this case, the age in Gyr works out to be exactly equal to the radius in Glyr, by definition.

So, what does that tell us?

Nothing, really. Except that our Universe is flat, i.e. has $\rho\simeq\rho_\mathrm{c}$, which we already knew and used in the calculation.

That is, setting $R = R_\mathrm{H} \equiv c/H_0$,
$$
\begin{array}{rcl}
R_\mathrm{S} & \equiv & \frac{2GM}{c^2}\\
             & =      & \frac{2G}{c^2} \rho V \\
             & =      & \frac{2G}{c^2} \rho \frac{4\pi}{3}R^3 \\
             & =      & \frac{2G}{c^2} \rho \frac{4\pi}{3} \left(\!\frac{c}{H_0}\!\right)^3 \\
             & =      & \frac{8\pi G}{3 H_0^2} \rho \frac{c}{H_0} \\
             & =      & \frac{8\pi G}{3 H_0^2} \rho  R,
\end{array}
$$
so if $R=R_\mathrm{S}$, we have
$$
\rho = \frac{3H_0^2}{8\pi G},
$$
which is exactly the expression for the critical density you get from the Friedmann equation.

$^\dagger$The density that determines the global geometry of the Universe.


If the coordinates are precessed to epoch B1875.0, the constellation boundaries lie neatly along the equatorial coordinate grid, allowing simpler tests against boundary segments.
Roman 1987 describes a basic algorithm and provides implementations in C and F77.
Glaschke 2010 makes a binary partition of the sky for fast lookup and provides an implementation in C.
Skyfield implements another fast lookup in Python, using lists sorted by right ascension and by declination.
It builds its data structure in build_constellations.py and uses it in constellationlib.py.


How old would I be if I travelled 1000 light years in one year

The answer is sort of trivial. If you travel 1000 ly so fast that in your own reference frame it takes one year, then you will have aged by one year in your own reference frame. To do so, you will need a speed of almost the speed of light, so in the reference frame of Earth, you will have spent just a tad more that 1000 yr to travel 1000 ly.

In general, the time dilation is given by the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ so to be exact, your speed must be
$$
1000^2 = \frac{1}{1 - v^2/c^2} ⇒ \\
     v = (1-10^{-6})^{1/2}\,c = 0.9999995c
$$
so your journey will take
$$
t = \frac{d}{v} = 1000.0005\,\mathrm{yr},
$$
i.e. 1000 years, 4 hours, and 23 minutes in Earth's reference frame.

"Realistic" acceleration

As David Hammen comments below, this assumes that your spaceship accelerates instantaneously to $v$. There are infinitely many ways to achieve that speed.
The proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $d$ when traveling at a constant acceleration $a$ is
$$
\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ad}{c^2} +1 \right).
$$
Solving for $a$ yields the acceleration needed. The most pleasant way would arguably be accelarating at $a\simeq 19.2\,\mathrm{G}$ for half a year, and the decelerate at the same amount for the rest of the journey. A more pleasant way in the beginning, but less pleasant in the end, would be to accelerate at $a\simeq9.6\,\mathrm{G}$ for one year, and then crash into planet WASP-142b, which lies at a distance of roughly 1000 ly.

Your journey, as measured by people on Earth, would then take
$$
t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),
$$
which works out to roughly 18 and 36 days more, respectively, than in the instantaneous case. The reason that it doesn't differ that much is that you actually reach relativistic speeds pretty fast at this acceleration.

The largest G-forces can be with endured in the "forward direction", i.e. corresponding to lying on your back and accelerating upwards (accelerating along the direction of your spine tends to break it, and accelerating "backward" makes your eyes pop out). According to Wikipedia, "acceleration pioneer" John Stapp withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.


Why can we observe the Cosmic Microwave Background no matter the direction we look?

Until the Universe was 380,000 years old, it was filled with a gas of protons an electrons. There was also radiation, in thermal equilibrium with the matter, and because it was so hot, the protons and electrons couldn't form neutral hydrogen, since every time it "tried", an energetic photon would knock off the electron.

This gas was everywhere. And photons traveled and scattered in all directions:

Photons (purple) scatter on free electrons (green), and both are mixed with protons (red).

380,000 years after the Big Bang, the temperature had fallen sufficiently that neutral atoms could form (this is called recombination). The radiation, which until now had scattered continously on free electrons, could now stream freely between the atoms (this is called decoupling).

So they did. Still in all directions:

This free streaming is still taking place. Photons travel in all directions, and are everywhere. The photons that you are able to see, are the ones that started out at a particular distance from you, and in a particular direction, but other photons started out at smaller and larger distances, and in other directions. You just don't see them, because you happen to be right here. But a person in another place of the Universe would see the same as you.

The photons that we observe as the CMB come from a region we call the surface of last scattering, because it corresponds to the surface of a shell centered at us. But there is nothing special about this "surface", except that is consists of all points in the Universe that are so far from us, that it takes a photon roughly 13.8 billion years to travel. And because of the expansion, these points are now roughly 47 billion lightyears away from us.

In the figure below, the arrows show CMB photons. All have the same length; they start where they were emitted and end where they are today. What we observe as the CMB are all arrows that end at the Milky Way (in the center). Other arrows may be observed by other observers in other galaxies that have their own surface of last scattering around them.


Opportunities to see a near earth object via unaided eyes or binoculars?

Your best viewing opportunity will be 99942 Apophis which will pass Earth on April 13, 2029, where it will pass about 19,600 miles from Earth; look out for that. According to Wikipedia:

On that date, it will become as bright as magnitude 3.1 (visible to the naked eye from rural as well as darker suburban areas, visible with binoculars from most locations)

Other ones include:
(137108) 1999 AN10, passing Earth at a distance of 240,000 miles on August 7, 2027, peaking at apparent magnitude 7.3, visible with binoculars, but questionably visible in urban areas.
(153814) 2001 WN5, approaching at 155,000 miles on June 26, 2028, at magnitude 6.7.
I hope this helps.


Your radio information source is wrong.
The Large Hadron Collider has not discovered mini black holes.
When it comes back on line this spring, the LHC will begin looking for mini black holes: Large Hadron Collider Could Prove the Existence of Star Trek's Parallel Universe

Journalists do like their headlines, but the gist is this:

  When the Large Hadron Collider is brought back online in the spring, researchers will be looking for the existence of mini black holes. These mini black holes would lend support to string theory, which posits that different dimensions and parallel universes are possible.

No miniholes yet, maybe never.


Could the closest approach of stable orbits of two earth sized planets allow life to jump between?

From comments

no, you can't have orbits anything like that close. Nor can you have a large moon anything like as close as you are thinking. 2 earth sized planets near each other's Roche limits would cause catastrophic earthquakes. Independent orbits would need, probably a few million miles apart for stability. Orbiting each other you could work them to perhaps, earth-Moon distance or even closer assuming they were tidally locked to each other to avoid too much disruptive tidal forces.

BUT if this is for a story: don't forget the other direction. One could say it is perhaps generally agreed today that "panspermia" can happen: life - microbes - via asteroids/etc can travel large astronomical distances. If we learned today that Mars has life - and indeed it's the "same" as Earth life - we'd assume that it travelled via ejecta - the Mars-Earth distance is no problem for such a putative story

Comment converted to community wiki


Is the age of the Universe really 13.8 billion years?

The age of the universe is not calculated based on the size of the visible universe. The age of the universe is being calculated based on the fact that the laws of nature have no direction. This means that you can use the laws of nature to predict future behavior, but also assume previous behavior. Based on calculating backwards with the laws of nature, for instance, general relativity, physicists can to a pretty certain degree calculate backwards how the universe started.
The fact that the universe is expanding and the size of the universe is what it is, has no relation to the age of the universe. Please, look at another video from Don Lincoln about the size of the universe and its age. Here it is explained that the universe already had an extra large expansion making it bigger than the age.
What will happen, though, is that the visible universe, which is smaller as the actual size of the universe, will be more empty over the course of time. In the 5 billion years you mentioned, there will be more galaxies outside the visible boundaries of the universe, because they are moving away from us faster than the speed of light. The expansion of the universe, hypothesized due to dark energy, makes the universe bigger, not older.
If the laws of nature don't drastically change, and we use these laws in 5 billion years to calculate the age of the universe, we should end up with the right number, 14 billion plus 5 billion years.


Is broadcasting the location of Earth to potential extraterrestial civilization regulated?

No, except in the sense that national broadcasting and so on is regulated. Our location can be fixed, as you suggest, from simply looking at where the signals come from.

Perhaps, though, you mean is it sent out "regularly" (i.e., at specified intervals). In truth we have been sending very powerful signals continuously into space since the dawn of the television age. Those signals may now start to wane in strength as we move to digital terrestrial television but that won't bother any aliens for some time yet.


There are several theories on how the universe might end by time stopping. The problem with cosmology and some other parts of theoretical physics is that it cannot be proven, nor reproduced or tested. This makes it often more a faith/religion than a science. So, if you notice some irony in my answer, please, it's because of this.

There are two types of theories which fantasize about the end of the universe by end of times. I freely call those type the formula type and transcendental type theories.

The first theory I will only briefly describe, because they end up transcendental anyway. There are several models on the universe that will describe what the universe (or universes) looks like and how they will evolve in time. The problem with these models is the variable time. If you have an endless time or an endless amount of universes, you end up with an endless amount of possibilities, which will result in all things will eventually happen. So, somewhere in time or a universe we will all become filthy rich and marry a gorgeous, intelligent, funny, friendly, etc., woman/man that never says no (to you). Because of this anomaly scientists have started to theorize that time must end at some point. If you believe these models to be correct, the scientists might be right. The problem in this lies that you first must believe the model is right.

The transcendental theories are a bit more difficult to explain and I will briefly leap out of astronomy into philosophy. In the 18th century a great philosopher wrote a book on the critique of pure reason (Kritik der reinen Vernunft). In this book he was looking for a transcendental doctrine of method and elements. The transcendental part of his doctrine was that it had to be build up on a priori knowledge that was not based on experience. This was his pure reason, because the elements and the methods are not tainted by (personal) experience from an observer. They are true in itself.

Why is this interesting? Because Kant (that great philosopher) said that there were two transcendental elements, space and time. I will skip space, but must say something on time, because someone thinks it might end. Kant states that time is an a priori element of knowledge in pure form. Kant uses five reasons, why time is universal and not empirical. They all can be brought back to the fact that we cannot see things/phenomena to coexist together or successively when there is no concept of time. Therefore, time is an a priori element of pure form.

Now, what has this to do with the ending of the universe in respect to time? The theory you mentioned as heat death is such a transcendental theory. The theory states that the universe expands until the universe is too big to be heated by the energy provided by the matter in the universe. This theory models that the universe follow the same rules as any thermo dynamic system and therefore at a certain point in time all energy, matter and temperature are evenly distributed in this huge universe. Because all is evenly distributed no stars will be created and all processes come to a grinding halt. Since the state before, now and after are the same, the scientists now determine this as the end of time. The formula theories also end up making some theory that time will end.

The problem with these theories is that time will never end. According to Kant time is an a priori element of pure reason and will be there despite of what we observe. Because the state of the universe doesn't change, doesn't mean that there will be no time. Time has just stopped. Theories that state that time will end, do not comply with Kant's critique of pure reason, which states that time is the formal condition of all phenomena whatsoever.

If it didn't entirely answered your question, I hope it was entertaining and educational.

Kind regards,
MacUserT


Recommendation for introductory cosmology text

I don't know any specific text but I can recommend you a course I took last year at Coursera.org. It says "introduction" but I can assure you it is really complete, 12 week that include a lot of mathematics and physics.

https://www.coursera.org/course/introastro


Population density of stars within $13$ light-years of us

The criteria for sample selection are admittedly not made overly clear in the preprint itself (Wysoczańska et al.). It appears that the authors started with a list based on stars that were previously believed to have come close to the Sun at some point in the past or future. That list was then augmented based on data from Gaia's second data release, to get a total of 820 objects, and the motions of those stars were then integrated backwards and forwards in time from the present, based on a model of the galaxy's gravitational field.

Here's where that "over 600 stars" came from. The authors ended their integration for a given star when either (i) the star went more than 3 kpc from the Sun or (ii) the time reached 50 million years in the past or the future. With those cutoffs in place, the astronomers found that during that timespan (50 million years before or after the present), 647 stars were within 4 parsecs of the Sun at some point in time. That's the figure Fox is referring to.


While more of a physics than astronomy question. Entropy as a whole increase in the universe due to statistical probability. By rearranging a deck of cards looking at the back of them, no matter what order, it is the same macrostate with different arrangements of the cards (microstates). A better example would be if you decreased entropy by picking up cards from the floor and put them together. In that situation, while in your small area the cards lost entropy, you expended energy to do that task. The energy you expend is enough to make the entropy of you and the cards remain constant and arguably increase. However, if you then take into account the entire universe, the rest of it is increasing in entropy at a large rate anyway. This means no matter what you do, entropy will increase overall due to other effects.


How long should it take for us to observationally determine if Caltech's Planet Nine is really there?

Those are model calculations, which hint to the existence of a possible body of about 10-times the mass of Earth. Calling this a discovery would clearly be premature. The confidence level is just a little above the "evidence" level of 3 sigma, under the assumption, that the discoveries of the KBO objects leading to the inference aren't observationally biased.
It's a long way to direct observation, since precise orbital data haven't been inferred.
Another option is, that there has been a planet. But it might have left our Solar system. Hence predicting a date for direct observation doesn't appear reasonable at the moment.


The # of Estimated Galaxies increased 10 fold. Where does that leave our # of stars estimate?

The sensationalist headlines mask a rather more prosaic result that has absolutely no effect on the number of stars we might estimate are present in the observable universe today.

The study in question has discovered (actually it was always supposed and suggested that this was the case), that the distribution of galaxy masses was much more skewed to lower mass galaxies in the past. That is, there were many more small galaxies in the past. But this happens at the expense of the large galaxies - there were fewer large galaxies in the past. To first order, the overall mass contained within galaxies stays the same and so long as the ratio of stellar mass to luminosity (since this is actually what is measured) remains roughly constant, than any estimate of stellar numbers stays the same.

Actually, a better estimate for the number of stars in the observable universe uses the present day galactic mass function and then extrapolates that to the size of the observable universe today. This is a far more honest estimate, since we have a much better understanding of how many stars there are in our local part of the universe and can then invoke the cosmological principle to say that other places (but not times) in the universe are similar. This cannot possibly be affected by the new study since that is dealing with the numbers of galaxies in the past.


Is there an orbit that could cause regular debris showers like in the movie Gravity?

If I understand you correctly (and I haven't seen the movie ^_^), you want to know if it is possible for two different objects in orbit to come close to each other periodically, with roughly the same period as one of the object's orbital period.

Yes, that is possible, but not probable.

Obviously, if the objects are in the same orbit, they are always close to each other
If they are in slightly different orbits, one of the objects will seem to move about the other object in regular or irregular patterns, depending on orbit specifics
If they are in completely different orbits, the necessary conditions are

the orbital period of both is equal
they have at most 1 point in common (or at least very closely approach each other)
the phasing of both objects is such that the objects are in the close approach region at the same time.

In ordinary two-body Keplerian celestial mechanics, an orbit's semi-major axis is the only orbital element that determines the orbital period. Therefore, one orbit may be circular and the other elliptic and inclined in all sorts of ways, but as long as the semi major axis of both is the same, they will have the same period.

However, this whole scenario is completely unstable for any real-world celestial body. The (relatively small) asphericity of the Earth, as well as the presence of the Moon for example, causes all orbits around it to drift in some way or another. If you put two objects in orbits like you describe, they will not likely meet up very often before never seeing each other again for centuries.


How to calculate the expected surface temperature of a planet

The formula

$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi  R ^ 2  L_{\odot}(1 - a)}{4  \pi  d ^ 2}$$

is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to

$$T ^ 4 = \frac{ L_{\odot}(1 - a)}{16  \pi  d ^ 2 ơ}\;.$$

You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as

$$σ = 5.670373 × 10^{−8} \;\mathrm{W}\; \mathrm{m}^{−2}\; \mathrm{K}^{−4}.$$

The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:

$d = 149,000,000,000 \;\mathrm{m}$

$L = 3.846×10^{26} \;\mathrm{W}$

Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get

$$
T ^ 4 = \frac{ 3.846×10^{26}(1 - 0.29)}{16  \pi \times  (149,000,000,000) ^ 2 \times (5.670373 × 10^{−8})}=4,315,325,985 \;\mathrm{K}^4\;.
$$

After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.


Where is the Moon for the last five or six days?

Two things: You either have a new moon or waxing crescent moon that is difficult to see or even invisible to the eye. Or (this is more likely), you viewed the sky before moonrise.
Try viewing the night sky in a few days (it will be clearly visible), or view the sky later in the night.


At what distance does Proxima Centauri become visible to the naked eye?

"Back of an envelope" calculation:

Proxima Centauri has an apparent magnitude of about $11$. The faintest objects visible to the unaided eye have magnitudes about $6.5$. So we need to decrease Proxima Centauri's magnitude by about $5$ in round numbers, which corresponds to an increase in brightness of about $100$ times. To achieve this we would have to be about $10$ times closer to Proxima Centuari than we are now, which gives a distance of about $0.4$ ly.

At a distance of $0.75$ ly the apparent magnitude of Proxima Centauri would be about $7.5$, so I think Celestia is being a little optimistic.


Is it possible that the shadow of The Moon is a single dot during solar eclipse?

What you are calling a focal point is the end of the umbra, the point at which the umbra changes to the antumbra. In a total solar eclipse, that point is below rather than above the surface of the Earth. An annular eclipse occurs when that point is above the surface of the Earth. The Moon appears to be smaller than the Sun, leaving an unbroken ring of the Sun that appears to surround the Moon.

There are some solar eclipses that transition from annular to total, and from total to annular. A point on the surface of the Earth is exactly at the end of the umbra at those times of transition points in time. At this point in space and time the Moon will appear to be exactly the same size as and exactly in line with the Sun.