How much magnetic attraction is there between the Sun and Earth?

Warning: I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.

First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts:

  There are three requisites for a dynamo to operate:

  1) An electrically conductive fluid medium

  2) Kinetic energy provided by planetary rotation

  3) An internal energy source to drive convective motions within the fluid.

  In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.

The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other "space weather" events.

To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by uhoh, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather "pushing" stream of charged that will "push" away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they "drape" themselves onto the magnetic field-less object and form the "magnetosphere" of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.

If you are looking for numbers, a hefty equation to calculate the electrical
magnetism oh an object is the Boit-Savart Law.

  If you have a
  current $I$ flowing through an infinitesimal length
  $\mathrm{d}\vec{l}$ then the field at a point $\vec{r}$ is given by
  \begin{equation} \mathrm{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{I
> \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation} The
  field does still drop of with the inverse square of the distance, but
  the vector nature of the current makes the formula rather more
  complicated. Generally this expression has to be integrated over the
  length of a wire carrying the current. This can get quite complicated
  and so this method is generally only used for simple situations such
  as the field around a magnetic dipole.  Having found the field, the
  force on a infinitesimal length of current can be found from.
  \begin{equation} \vec{F} =
> I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation} This is again
  analogous to the electrostatic case but with added complication due to
  current being a vector.

    https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction

This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation:

$$F=\frac{\mu q_{m1} q_{m2}}{4\pi r^4}$$
where

$F$ is force attraction in newtons.
$q_{m1}$ and $q_{m2}$ is the magnitudes of magnetic poles in ampere-meters.
$\mu$ is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.
$r$ is the separation in meters.

Thus, the attraction between the Earth and the Sun can be calculated in the following means:
$$\implies \operatorname{B}(\text{Earth})=q_{m1}=7.84027\times 10^{22}$$
$$\implies \operatorname{B}(\text{Sun})=q_{m2}=3.48526\times 10^{29}$$
$$\implies \mu\approx 10^{-6}$$
$$\implies r=\operatorname{AU}=149597870700$$
$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$
$$\implies F=\boxed{4.34165\text{ newtons}}$$
If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.

I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.


Problem with Dobsonian reflector telescope

Next to the eyepiece there is a screw that you can turn loose and if
  you make it as tight as possible the focus knob won't focus the image
  but when it is a little loose it starts to focus it

Sounds like the lock screw. If that's what it is (your description is a bit vague), it's designed to lock the focuser in place, like when you're doing prime focus photography with a heavy camera, and you don't want the focuser to move around and lose focus.

If it's the lock screw, don't mess with it. Don't tighten it, you don't need that for regular observations. Keep it comfortably loose. If it's a bit tight, it could be rubbing against the drawtube (or whatever moving part it's designed to stop), and then it would make that noise - and you don't want that.

I have a very heavy 30mm 82deg Explore Scientific eyepiece, and even with that thing (bigger than a soda can, and weighing about 1 kilogram) I don't need the lock screw - and this is on a regular mass produced two-speed focuser, not some fancy precision hand-crafted MoonLite. Visual observations are all about moving the focuser around, hunting for the best focus, and changing eyepieces for the best view. Let the focuser move easily.

Unless it's moving too easily, but then you have a problem with the focuser.


What is the fastest a galaxy or other body has been measured to (or can) travel?

Hubble velocities

The Universe expands and carries galaxies away from each other with a relative velocity proportional to the distance between them. This is Hubble's law, and if the Universe is infinitely large, there is no limit to how fast two galaxies may recede from each other. In our observable Universe, the most distant galaxy observed (GN-z11; Oesch et al. 2016) recedes from the Milky Way at more than twice the speed of light ($c\simeq300\,000\,\mathrm{km}\,\mathrm{s}^{-1}$).

Peculiar velocities

However, this is not a motion through space, which seems to be what you are asking about. Nothing is able to travel through space faster than the speed of light.

Galaxies generally move through space with velocities of the order of a few $100\,\mathrm{km}\,\mathrm{s}^{-1}$; in massive clusters, they may acquire velocities of a few $1000\,\mathrm{km}\,\mathrm{s}^{-1}$ (e.g. Karachentsev et al. 2006). These so-called peculiar velocities are wrt. to the "global" frame, i.e. the frame in which an observer that follows the cosmic expansion lies still. This is also the frame in which the cosmic microwave background is (statistically) the same in all directions.$^\dagger$

Black holes, stars, planets, and gas cloud move around in the reference frame of their galaxies with characteristic velocities of the order of $100\,\mathrm{km}\,\mathrm{s}^{-1}$, so this will typically be smaller than that of their host galaxy, if the galaxy lies in a large cluster.

When massive celestial bodies such as neutron stars or black holes merge, they reach velocities of the order of half the speed of light, as discussed further in Rob Jeffries' answer. This is as measured in the frame of the center of mass.

Velocities of particles

EDIT: The following was written before the question was changed to emphasize celestial bodies:
Much higher velocities are found for small particles such as cosmic rays; these are massive particles accelerated to extreme energies, e.g. by supernova explosions. The record holder of such particles is still (I think) the so-called Oh-My-God particle which traveled at $0.9999999999999999999999951c$ (in Earth's frame), much faster than the velocities reached in the Large Hadron Collider.

$^\dagger$Of course you can always define a frame that moves with almost the speed of light, such that all object acquire that speed "artificially", but then the Universe isn't isotropic anymore.


what happens when we keep going radially away from earth?

Yes. There are no edges of the universe that have been observed.

The standard cosmological models favoured by observations (open or flat universes) allow you to move outwards indefinitely. Closed universes are like the surface of a sphere, and you would eventually return to Earth unless the expansion was too fast; the same is true for the models with nontrivial topologies - but these have no evidence at present.


Is the M81 group bound to the Local group?

Interestingly enough, although we tend to group galaxy clusters together into superclusters, superclusters are actually not gravitationally bound. This suggests that, due to Hubble flow, galaxies in separate clusters will never "collide". In fact, it may be impossible to ever leave one's galaxy cluster.

EDIT:

In a previous version of this post, I stated the the Great Attractor may be reason to suspect a gravitational binding of Laniakea. In reality, the Great Attractor only reduces the expected relative velocities of these galaxies away from one another by between $200-400$ $\mathrm{km}\cdot\mathrm{s}^{-1}$—I would like to thank eshaya for pointing out my error here. Due to the current expansion rate of the universe of $67.6^{+0.7}_{-0.6}$ $\mathrm{km}\cdot\mathrm{s}^{-1}\cdot\mathrm{Mpc}^{-1}$, some galaxies in Laniakea have recession velocities of up to $30000$ $\mathrm{km}\cdot\mathrm{s}^{-1}$, as given by Tully et al. (2014)—far greater than could ever hope to be counteracted by the Great Attractor.

The question now becomes one of whether the M81 Group is gravitationally bound to the Local Group. Given the M81 Group's status as being one of the closest to the Local Group, at only $3.6$ $\mathrm{Mpc}$ distant, I would not be entirely surprised if through some chance it managed to be. Given the current estimate for Hubble's constant, due to Hubble flow the M81 Group would be expected to recede at $\sim243.4$ $\mathrm{km}\cdot\mathrm{s}^{-1}$. This is what would need to be countered through gravitational effects for the two groups to be bound, and thus eventually coalesce.


Neutron stars are generally not compact enough to have an ergosphere. Typically, the radius of a neutron star is several times the Schwarzschild radius corresponding to its mass. The ergosphere however starts at the Schwarzschild radius.


Are there any known objects in a "dual" tidally locked orbit?

Pluto and its largest moon Charon are tidally locked to each other.

  Charon and Pluto revolve about each other every 6.387 days. The two objects are both gravitationally locked to the other, so each keeps the same face towards the other. This is a mutual case of tidal locking . . .

Because of Charon's large size compared to Pluto, and because its barycenter lies above the primary's surface, some astronomers have referred to the Pluto/Charon system as a dwarf double planet.

  The system is also unusual among planetary systems in that each is tidally locked to the other: Charon always presents the same face to Pluto, and Pluto always presents the same face to Charon: from any position on either body, the other is always at the same position in the sky, or always obscured.(110)


How to find out if a galaxy is in a specific cluster

You want to use NED (NASA Extragalactic Database) rather than SIMBAD, since the former has more galaxy- and cluster-specific information.

I'm going to assume you have a reasonably accurate position for your galaxy as well as its redshift (in km/s). The basic idea is to identify candidate galaxy clusters which are close enough to your galaxy on the sky and in redshift so that the galaxy could potentially be a member of one of them.

Determine what angular radius $R$ corresponds to 1 or 2 megaparsecs at the redshift of your galaxy (e.g., using Ned Wright's Cosmology Calculator)
Do a search in NED for Galaxy Clusters within R of the coordinates of your galaxy. If your galaxy has a name, you can skip the coordinate lookup and use the "Near name" search form; if you only have coordinates, there's a "Near position" search form.

A. Put the estimated angular radius $R$ (converted to arc minutes) in the "Search Radius" box;

B. Set the "Selection in Redshift" form to "Between" and give it lower and upper limits of $V_{\rm gal} - 2000$ and $V_{\rm gal} + 2000$ in km/s, where $V_{\rm gal}$ is the redshift (in km/s) of your galaxy;

C. Under "Selection by Object Type", select "Galaxy Clusters" in the "Classified Extragalactic Objects" list.

That should give you a list of galaxy clusters within the specified radius of your galaxy which are at a similar redshift. Note that the same cluster may come up multiple times under different names (i.e., from different cluster catalogs), but it should be clear from the coordinates and the redshifts which are the same clusters.

Now the question is whether your galaxy is plausibly a member of a given candidate cluster or not. Ideally you should find some measurement of each cluster's velocity dispersion $\sigma$; if the galaxy's redshift isn't within $\pm 3 \sigma$ of the cluster's redshift $V_{\rm cluster}$, then you can ignore that cluster, since it's very unlikely that a galaxy with that relative velocity with respect to the cluster is actually gravitationally bound to the cluster.

You can't really do better than a probabilistic assessment: roughly how likely it is that the galaxy is a member of a given cluster. Galaxies which are members of a cluster are more likely to be close to the center (on the sky); member galaxies closer to the center can have a higher range of relative velocities, while member galaxies further away should have velocities closer to the cluster redshift.

A quick example: Is NGC 4889 ($V_{\rm gal} = 6495$ km/s) a member of a cluster? At its redshift, 1 Mpc = 33 arc minutes; if we search on NED for galaxy clusters within 33 arcmin of NGC 4889 with redshift limits of 4500-8500 km/s, we get 6 results, 4 of which are just the Coma Cluster in different catalogs, with $V_{\rm cluster} \sim 6900$ km/s. Since the various distances ("Separ. arcmin") translate to $\sim$ 100-300 kpc from the cluster center, and since the Coma cluster has a velocity dispersion of $\sim$ 1000 km/s and the difference between NGC 4889 and the cluster is only $\sim$ 400 km/s, it's a very good bet that this galaxy is a member of the Coma Cluster. (It's actually one of the two cD galaxies in the cluster center.)


Is there a place in the known universe that there would be no visible light from other stars/galaxies?

If you don't want to look at the stars and galaxies, you can do two things: You can travel so far that they're too far away to see, or you can block your vision.

Go far away

On large scales, i.e. above a billion lightyears or so, the Universe is observed to be roughly homogeneous. On smaller scales, however, matter is distributed in the so-called cosmic web, where galaxies lie in filaments and sheet, meeting each other in knots (where the largest clusters are found). In between these overdensities, there are underdensities known as voids. These voids are virtually free of any luminous matter, and since they're up to several million lightyears across, you could place yourself in the middle of such a void and only see darkness.

As an example, consider a void of 100 Mpc (~300 million lightyears) in radius, corresponding to a distance modulus1 of $\mu = 35$. If a Milky Way-like galaxy (absolute magnitude2 $M=-20.5$) were at the border of the void, its apparent magnitude3 would be $m = M+\mu = 14.5$, invisible to the naked eye.

A small telescope, however, would enable you to see galaxies from here.
The limiting magnitude4 of the naked eye is roughly 6–7 (though some claim to be able to see $m=8$ objects), so the Milky Way-like galaxy would be roughly 8 magnitudes too faint to see. The gain in limiting magnitude when using a telescope is roughly $g = 5\log(D_\mathrm{tel}/D_\mathrm{pupil})$, so with your 6 mm pupil, a telescope with a diameter of
$$
D_\mathrm{tel} = D_\mathrm{pupil} 10^{g/5} \\
= 6\,\mathrm{mm}\times10^{8/5} \\
\simeq 200\mathrm{-}250\,\mathrm{mm},
$$
would enable you to see Milky Way-like galaxies from the center of the void.

Note though that no matter how far you are from galaxies, there will always be some radiation around, if nothing else then at least the CMB. Of course, being microwaves, this is not visible to the naked eye.

Block your vision

Apart from going down in your basement and turn off the light, could there be "astronomical" places where you would be unable to see anything? As David Hammen comments, at the surface of a cloud-enshrouded planet or moon, you could face total darkness (at least on the night-side of the planet). But perhaps you could also, as Wayfaring Stranger comments, go inside a dense, interstellar cloud.

Bok globules are small ($R\sim10^4\,\mathrm{AU}$), dense ($n\sim10^{4\mathrm{-}6}\,\mathrm{cm}^{-3}$) nebulae of gas and dust. A sightline from the center of such a cloud would result in the apparent magnitude being extincted (i.e. increased) by up to several tens in visual light. The largest extinction through the center of a Bok globule that I've been able to find — but I'm not expert on this — is "FeSt 1-457" with a visual extinction of $A_V = 41$ (Kandori et al. 2005), so from the center and out is roughly $A_V\simeq20$. That means that the fraction of light from outside sources that make it into the center is
$$
f = 10^{-A_V/2.5} \sim 10^{-8},
$$
which is not a lot. However, Bok globules are found in the vicinity of young stars which tend to be bright. An O star has an absolute magnitude of $M\simeq-4$. Such a star located right outside the cloud (say, at a distance of $d=2\times10^4\,\mathrm{AU}=0.1\,\mathrm{pc}$) would have an apparent magnitude of $m=M+\mu+A_V\simeq6$, and thus be just visible to the naked eye, albeit very faint. And if it were just a little farther away, you wouldn't be able to see it.

I assume that you're human, but if you are in fact a cichlid, you should be able to see infrared light (Meuthen et al. 2012). In the IR, the extinction is much smaller than in the visible, so hiding in a Bok globule won't help you.

1The distance modulus $\mu$ is a logarithmic way of expressing distances: $\mu \equiv 5\log(d/10\mathrm{pc})$.

2The absolute magnitude $M$ is a (logarithmic) measure of the luminosity $L$ of an object: $M=-2.5\log(L)+\mathrm{constant}$. Note the minus sign; the brighter an object is, the smaller $M$ is.

3The apparent magnitude $m$ is a logarithmic measure of how bright an object looks to an observer, and so depends on the distance to the object: $m = M + \mu$. Stars visible to the naked have roughly $m=0$–$6$. Again, the smaller the number, the brighter the object is. An extremely distant galaxy may have $m=30$, while the Sun has $m=-27$.

4The limiting magnitude is the largest magnitude (i.e. faintest object) visible.


What exactly is the "paradox" in Olber's Paradox?

Olber's Paradox was created at a time before the idea of a finite universe was accepted.  (It was thought of in the 1600's).  In order to resolve Olber's Paradox, you have to introduce the idea that either the universe had a beginning or it is of finite size.  (Note: the solution does not require an expanding universe).  So, at the time, it was a paradox.  Pretty much all astronomers considered the universe to be static and infinite.  Therefore, the fact that their observations didn't fit with what they expected made it a paradox.


What exactly are "H30α and He30α images"? (ALMA)

These are the names of recombination lines. In terms of the Bohr model of the atom:

The electron in a Bohr atom can fall from the level ($n+\Delta n$) to $n$, where $\Delta n$ and $n$ are any natural numbers (1, 2, 3, …) by emitting a photon whose energy equals the energy difference $\Delta E$ between the initial and final levels.  Such spectral lines are called recombination lines because formerly free electrons recombining with ions quickly cascade to the ground state by emitting photons. Astronomers label each recombination line by the name of the element, the final level number $n$, and successive letters in the Greek alphabet to denote the level change $\Delta n$: α for $\Delta n=1$, β for $\Delta n=2$, γ for $\Delta n=3$, etc.  For example, the recombination line produced by the transition between the $n=92$ and $n=91$ levels of a hydrogen atom is called the H91α line.

So H30α indicates the transition from levels 31 to 30 of the hydrogen atom.
As noted by Peter Erwin in the comments, the series with low $n$ have traditional names. In particular, the Balmer series (transitions to $n=2$) are traditionally known as Hα, Hβ, etc. The first four of these are in the visible part of the spectrum.


2 Earth-like planets in co-orbital configuration

Because this has gone unanswered, and using this very nice answer as a reference, I can answer this part way.  Source article here.

  it was found that if the mass of second body becomes larger than
  1/1200 of the first body, horseshoe orbits become unstable.

So Earth mass orbiting Jupiter (1/318) won't work according to the research above.  You could try Mars mass objects orbiting Jupiter (1/2970) and you might get that to work for a fairly extended number of orbits.   Horseshoe orbits are never stable over the long term the way trojan orbits can be, but they can be stable over numerous orbits if they are properly set up.

I tried to work out the specifics for higher mass horseshoe objects, but my attempts at calculations have just sent me in circles so far.   If I make any progress I'll post an update.  You'd have to move the Mars-mass moons quite far from the planet and you'd need to keep the variation in their semi-major axes quite small.   Janus and Epimethius have only 50 km variation in their orbits, which is 1 part in 3000 of their orbital radius.

I'd explore variations in that range, and I'd start at the outer edge of stable region of Jupiter's hill sphere, and if you can't get it to work, decrease the mass some more, maybe lunar mass.


How do astronomers determine the texture of an exoplanet?

This is known as spectroscopy.  Every molecule and atom in the universe emits and absorbs light at specific frequencies.  This is a result of the quantization of the energy levels (for electrons) in an atom.  Although there are lots of complicating factors, such as redshift, to account for, the patterns are usually so distinctive that the complications can be accounted for, and do not prevent scientists from figuring out chemical compositions (and even, to some extent, how abundant each chemical is).

For example, here are the emission spectra of several common atoms at rest (with respect to the observer):

Absorption spectra, on the other hand, appear as darker regions in the band.  Here's an example of an absorption spectra, which identities several elements by knowing that it is distinguished by absorbing light at specific wavelengths that others do not:

For exoplanets with significant atmospheres, we can only expect to see the spectra for its upper atmosphere.  All other signals will be muted out by the rest of the atmosphere.  We will not be able to tell what the surface looks like, or what it is made of.


When exactly star Sirius shines into the queen's chamber of Great Pyramid?

The premise of this question is wrong: the "Sirius" shaft was not designed to point towards Sirius. Most of the shafts bend (sometimes several times), making it impossible for the light of a star to shine through. Furthermore, the angle of the shaft is not constant, making it possible for the shaft to be pointed towards tens of different stars, depending on which portion of the shaft you want to look at.

The Sirius shaft that leads to the queen's chamber faces south. It is at a 39.6° angle and the width of the window that it provides is 0.18°.

Due to precession, the maximal elevation of stars changes over time. Based on this document, Sirius would have been visible through the shaft between 2355 BC and 2295 BC. This is after the construction period of the pyramid (around 2550 BC), putting the idea that the shaft was "built to resonate with Sirius" even more in question.


Since etymologically sidereal references the stars (Latin sidus), one might expect a sidereal day to correspond to the rotation of the Earth with respect to distant "fixed" stars. But this does not seem to be the case, and because the the difference is not significant, the two definitions are sometimes used interchangeably.

I don't know of an explicit IAU definition of sidereal day. However, since sidereal time is defined as the hour angle vernal equinox (which is a local definition, although Greenwich is conventional), defining the (mean) sidereal day in terms in reference to the equinox is practically the only sensible choice.

Effective 1985, UT1 is now computed using very long baseline interferometry of distant quasars, and so can be taken to be authoritative regarding the "fixed stars". The coordinated universal time (UTC) time approximates UT1 with atomic clocks. Anyway, UT1 derived the rotational period of the Earth as $p = 86164.09890369732\,\mathrm{s}$ of UT1 time and the mean sidereal day (in 2000) would be $86164.090530833\,\mathrm{s}$ of UT1 time, following the explanation:

  The length of one sidereal day is defined by two successive transits of the mean equinox; while the Earth is rotating eastward, the mean equinox is moving westward due to precession. Therefore, one sidereal day is shorter than the Earth's rotational period by about $0.008\,\mathrm{s}$, the amount of precession in the right ascension in one day.

See: Aoki, S., et al., The new definition of universal time, Astron. Astrophys. 105, 359-361 (1982).


Corresponding BV Value to a Hertzsprung–Russell

The HR diagram is a plot of luminosity (or a proxy) versus temperature (or a proxy).

$B-V$ is a proxy related to temperature, therefore to plot your point on a 2-d plot you obviously need the other axis information - the luminosity, or its proxy the absolute magnitude. This in turn needs an apparent magnitude and a distance to the star.

It is also possible that the star has suffered some extinction and reddening by the interstellar medium, which would need to be estimated and corrected for before comparing with other stars or stellar models.

Note however that 33 Psc is a binary star, so it isn't clear what you learn by plotting its combined colour and absolute magnitude on the HR diagram.

Edit: I note that the catalogue you show in your edited question doesn't just give $B-V$, it gives the visual (i.e. $V$) magnitude. This is one component of what you need for the y-axis. The other is a distance, so that you can convert your apparent $V$ magnitude into an absolute magnitude.


How long will it take to reach the closest earth like planet

HD 219134b is apparently closest "super earth"; 21 light years away.
New Horizons, our speediest yet, is currently running at about 15 km/sec relative to the sun.
At 9.46e12 km/lightyear HD 219134b is 2e14 km away
At 3.16e7 seconds per year, New Horizons will travel about 4.74e8 km/year.
2e14/4.74e8 = 422000 years.

HD 219134b isn't even a very good earth; gravity is too high.


Where can I find a visible star dataset?

You can probably extract the data from Stellarium. Also, does it need to be all sky, or not? What Wavelength? If you want an all sky visible catalogue, the United States Navy produced the NOMAD catalog by combining 2MASS, UCAC, and the USNO-B catalog. You could also use the APASS all sky catalog, meant for helping astronomers with photometric calibration.

If you don't care about whether the wavelength is in the visible, CalTech's IRSA has a bunch of infrared source catalogs (the smallest/shallowest one being the IRAS catalog). The deepest all sky infrared catalogs are the WISE sets.

Other good databases to dig in to include the Hipparcos set of parallaxes, or some part of the Gaia catalog.


The Moon is the easiest target to start with.
After that, try whichever bright planets are currently in the sky: Jupiter, Saturn, Mars, Venus.
Then get a star map and try to find some of the brighter star clusters, nebulas, and galaxies.

I recommend the monthly sky maps and object lists at Skymaps.com.
Latitude 19°N falls midway between their northern and equatorial editions, so try the equatorial one and remember that your southern horizon hides the southernmost ~10% of the map.
Most of the "Easily Seen with Binoculars" list are also good targets for a 50mm telescope.
Objects labeled e.g. "M42" are in the Messier catalogue.
Some objects are easier to find with a more detailed star atlas.

A telescope calculator indicates that in ideal conditions, 50mm aperture can theoretically show stars as faint as magnitude 11 or resolve double stars 2.5 arcseconds apart.
In practice it's reasonable to try objects as faint as magnitude 9 or as small as 4 arcseconds.

If the telescope has a finder or other aiming device, check its alignment periodically.
No other maintenance is required if you store the scope in a clean, dry place and use the dust caps to protect the lenses from dust and fingerprints.
A few specks or spots will not significantly degrade performance.
Only clean it if it gets very dirty, and then take care not to scratch or strip any lens coatings.

As you learn your way around the constellations and acquire more challenging targets, you can sharpen your observing skill by keeping a journal in which you describe or even sketch what you see.
A local astronomy club can give you ideas for observing projects and opportunities to try more advanced equipment.

Solar safety note: filters installed near the eyepiece can overheat and break; filters which securely cover the whole objective and make everything else invisible are usually OK.
With a 50mm aperture, the eyepiece projection method is a good alternative.


What is the significance of using baseline pairs in radio interferometry?

The basic idea behind interferometry is that of interference, the combination of two waves (in this case the electromagnetic waves from distant sources).  Interference inherently implies two signals to interfere with each other, and that is why the pairs of telescopes (also referred to as baselines) are important.  The interference pattern between the signals received at two different telescopes carries information about the spatial structure of the source.  Widely-separated pairs are sensitive to small-scale structure, and close pairs are sensitive to larger-scale structure (which is why you need both to reconstruct a good image).  North-south pairs are only sensitive to structure in the north-south direction, and likewise for other orientations of pairs, so again you need pairs/baselines oriented in different directions.  (The rotation of the Earth helps somewhat by changing the orientation of baselines over time, though modern interferometers with many telescopes, such as ALMA, have pretty good coverage in a single snapshot without waiting for the Earth to rotate.)
The signal from a source is recorded at each telescope, and then those signals are combined (or interfered) with each other for each pair of telescopes, and all of those per-pair interference signals constitute the overall data collected by the interferometer.  In a radio interferometer this interfering/combining of signals is done in a computer called a correlator. For radio, the signal at each telescope is converted to a voltage and these voltages are correlated; for an optical or infrared interferometer, the light from each individual telescope is split into multiple paths so that each pair of light beams can  be interfered directory in a beam combiner before the interference pattern is recorded.


Easy interactive visualization of orbital elements

If all you need is a quick and free tool to visualize orbits around earth, websites such as the following might be enough:
First, the very simple orbitalmechanics.info, where you can just ignore the "add launch" option:

And then, the more artsy Harmony-of-the-Spheres, which can also visualize some other systems.

Maybe that already helps.


The Earth-Moon System is a system where the moon is tidally locked to the planet, but the planet is not tidally locked to the moon. Earth has a 24 hour day, whereas the Moon has a 29.5 Earth day long day. It takes the Moon 29.5 Earth days to orbit around the Earth once as well, which means the Moon is tidally locked in a 1:1 ratio. This means that if you were to stand on a particular point on the Moon, the Earth would not appear to move because the Moon is spinning on its axis at a rate such that it matches how much the Earth should appear to move from orbiting around it. Basically, the same side of the moon is always facing the Earth, which is why there is a concept of a "dark" side of the moon. The "dark" side is not always actually dark, it's just that no one had ever seen it before until we sent some satellites up there to look at it.

Here is a gif showing tidal locking. The tidal locked moon is on the left.

Now imagine you are back on Earth. Throughout the course of the day you can see the Moon move across the sky. That's because 24 hours is drastically shorter than 29.5 days. However, if you were standing on the Moon, the Earth would appear to stand still. Now, it would actually appear to wobble a bit because of the Moon's orbit's eccentricity, but it more or less stays in the same place.

Here is an artistic yet scientifically accurate rendition of what it would look like. It's from Space Engine. Notice how the gas giant (the planet) wobbles a bit but stays in roughly the same place. It would look similar if you were on the Moon, but without the atmosphere, and the Earth would obviously be smaller and further away.

Yes, the Earth would go through the same phases as the Moon. It would complete a cycle every 29.5 days because that is how long it takes for the Moon to orbit around the Earth. However, something different is that you would be able to see the Earth rotate, since it takes only 24 hours to rotate. When we look up at the Moon from Earth, as I said before, we cannot see it rotate because it is tidally locked, and thus only one side gets shown.

Brightness depends on the phase, but generally the Earth is brighter from the Moon than the Moon is from the Earth because the Earth has a higher albedo (reflects more light) than the Moon, and the Earth appears larger from the Moon than the Moon appears from the Earth. The Moon takes up roughly 30 arcminutes of the sky, whereas the Earth takes up roughly 1 degree and 50 arcminutes.

At full moon, the Moon has an apparent magnitude of roughly -12.75. At full earth, the Earth has an apparent magnitude of -16.75. That's about 40x brighter! If we were to qualify phases as a % from 0% being a new moon to 100% being a full moon, then every 10% you go below 100%, the apparent magnitude of each phase decreases by about the same amount. For example: 90% Earth would be roughly -16.5 (+.25 magnitude), and 90% Moon would be roughly 15.5 (also +.25 magnitude). The differences in magnitude only get larger when you go below 10% phase, where each little bit of the celestial body that shows adds a lot more to the magnitude.

Here is a rough chart:

If you have more questions feel free to ask.


Other than liquid water what factors could make way for life to be on Europa?

Well, Europa is one of the closest moons to Jupiter. Jupiter is an extremely massive planet (317.8 the mass of Earth) which makes strong tides on Europa and other close moons, so Europa is torn a bit which results in so-called tidal heating underwater. Through this heating, the temperature underwater makes the subsurface ocean habitable despite the Jovian system not being in the habitable zone. There are so-called hot spots underwater for which plumes erupting from under the surface are responsible. Life could thrive around those plumes.

So in other words, there is ideal temperature for life under Europa's surface due to tidal heating by Jupiter. On Ganymede and Callisto there may be a subsurface ocean too (though that's not confirmed) but they are farther away from Jupiter so there maybe isn't enough tidal heating, contrary to Europa.


What you see are clouds, big clouds, and these are as opaque as the cumulus clouds on Earth. The pale bands are ammonia clouds and the darker bands are ammonium hydrosulphide clouds. The ammonia clouds form above the hydrosulphide clouds In the polar regions you can see deeper, down to the water clouds that lie beneath the ammonia clouds.
There is a layer of atmospheric haze that is above the cloud layers, it is easiest to see in the polar regions.
The difference between the layers is on the order of 50-100km.  If you were in a cloud, your visibility would be pretty limited and as you descended through 50km of cloud, it would get pretty dark. But like on Earth, there would be gaps and clear bands, between the layers of cloud.


What is length of the universe on lights perspective

Sort of.  The Lorentz factor is

$$
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
$$

whereupon a stationary object in the stationary reference frame of length $L$ has a length of $L' = \frac{L}{\gamma} = L\sqrt{1-\frac{v^2}{c^2}}$ in the moving reference frame.  As the velocity $v$ increases toward $c$, we get

$$
\lim_{v \to c} L' = L \lim_{v \to c} \sqrt{1-\frac{v^2}{c^2}} = 0
$$

That is to say, if a spaceship were to move very rapidly from here to the center of the galaxy and back again, a clock aboard that spaceship would advance much less than an initially synchronized clock here at home.  From our point of view, the reason is that the clock on board the spaceship slowed down (time dilation).  From the point of view of someone on the spaceship, the reason is that the distance to the center of the galaxy was shortened (length contraction).  And in the limit (which cannot actually be reached by anything with mass), the clock on the spaceship becomes arbitrarily slow from our point of view, while the distance to the center of the galaxy becomes arbitrarily short from their point of view.  Of course, we can't actually ask the photon how it experiences the trip (even in principle), since the trip is instantaneous for it.  So the way to express it is to say what the limiting behavior is, as $v$ approaches $c$.

Wikipedia has a useful article on special relativity.


Why does Io cast a hard shadow on Jupiter, but the Moon casts a soft shadow on Earth?

It's due to the larger relative apparent size of the Sun. When the source of light is a point source the shadow is harder, and when it is extended it is softer.

Jupiter is approximately 5 times more distant from the Sun than the Earth, so the Sun is approximately 5 times smaller in the sky.

*Source: University of North Carolina CS


Sky view from Stellarium software vs. Sky view with naked eye

From personal experience - yes your Stellarium picture looks like a good dark sky in the Canary Islands. Certainly, the Milky Way is plainly visible, with lots of detail available to the dark-adapted eye. I more frequently view from the top of La Palma, which suffers a little less light pollution than Teide to the East.

The Canarian sky law was largely setup to protect the night sky quality at the observatory on La Palma, but covers all lighting on Tenerife in a straightline view to La Palma. So I would be guessing that the western side of Teide will be the darkest.

I found this picture which illustrates the controlled lighting areas (in green). It is not to scale, La Palma is further away from Tenerife (on the right) than illustrated.


Near-future astronomic events that could threaten Earth

Essentially, anything could happen! We can't really predict events such as you've described.
The only event we can really be certain of is that the Sun will inevitably die in about 5 billion years. When it's almost run out of hydrogen to fuse it will expand into a red giant, becoming so large it will take out the Earth. There's a good video demonstrating this here: http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm

You could add geomagnetic reversal and gamma ray bursts to your list. Also see: http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm


Was there ever a hypothesis about a planet called Biga?

If another planet in our solar system was to be named, I think that Biga wouldn't be the first name to come up, all of the known planets in our solar system are named after Roman deities and Biga is not a deity but rather a cart on which mostly Luna is depicted. I have never heard of this proposal and if the 'rules' are used for naming, it wouldn't be the first candidate.


Why is the magnetic reconnection(of sun) effect many orders of magnitude faster than predicted by standard models?

We don't know for sure.

If we knew the answer, then we'd have a new standard model and then the reconnection speed would be nicely comparable to what's predicted. (zibadawa timmy)

However, the typical models of magnetic reconnection do not account for the extra resistivity that results from small-scale electromagnetic waves, which can be upwards of ~12-15 orders of magnitude larger than that caused by Coulomb collisions. (honeste vivere)


Welcome to Stack Exchange: Astronomy!

When scientists say the universe is "flat," they don't mean that it's shaped like a dinner plate.  It's no easier to travel "up" than "forward."  In fact, off-planet, those terms don't even have any meaning.

What you're referring to has to do with the shape of the universe:

  When physicists describe the universe as being flat or nearly flat, they're talking geometry: how space and time are warped according to general relativity.

Think of it like this: if you draw a triangle on a balloon, is that triangle flat?  In one sense yes, because it never rises above the surface of the balloon.  Relative to the balloon, it is one-dimensional; you wouldn't be wrong to say that the image of the triangle is "flat."  But relative to the rest of the room, it is curved, because it follows the curvature of the balloon's shape.

When most people talk about the universe, they're talking about the triangle.  The discussion you're referring to, of the "flatness" of the universe, is discussing the shape of the balloon.

As for flying "out" of the universe and potentially into another, we're like an ant walking on the surface of that balloon: no matter what direction the ant walks, it'll just keep walking around the balloon.  The surface of his universe may be curved, but there's no direction he can go that takes him off of that curve.  For all he cares, it's essentially a flat surface, because he can only move in two dimensions along it.  Similarly, the geometry of our universe may be one thing or another, flat or curved, but either way that doesn't mean we can point a ship in any direction and "leave."


How can I measure the mass of eclipsing binary stars?

It depends.

For many categories of stars, it is possible to determine the approximate mass of the star from its non-redshifted spectrum. Stars on the main sequence: this table of spectral classifications lists the approximate mass for the various classifications. A more precise mass estimation should be possible with a more precise classification. Sol, for example, is a G2, near the high end of G, putting it near the top end of the range (0.80 - 1.04 stellar masses).


What is the largest hydrogen-burning star?

I assume by largest, you mean largest radius.

Well it won't be VV Cep B since this is merely a B-type main sequence star.

O-type main sequence stars are known and these have both larger masses and larger radii on the main sequence (when they are burning hydrogen in their cores).

A selection of the most massive objects can be found in the R136 star forming region in the Large Magellanic Clouds. If you look at this list (though I recommend having a look at the primary literature), you will see that O3V stars are listed. Such objects are also present in our Galaxy, for instance in the supercluster NGC 3603 (Crowther & Dessart 1998).

Such stars have masses of maybe $100 M_{\odot}$, luminosities of $2\times 10^{6} L_{\odot}$ and temperatures of 50,000 K. Using Stefan's law, we can deduce radii of $\sim 20 R_{\odot}$.

There are suggestions that even more massive main sequence stars have existed in R136 and NGC 3603 (see Crowther et al. 2010), which are now seen as evolved Wolf-Rayet objects, possibly up to $300 M_{\odot}$ on the main sequence (though this is a model-dependent extrapolation), and these would have had radii $>20 R_{\odot}$.

In the very early universe, population III main sequence stars without metals could have been much more massive and larger.


How far does the moon move relative to the sun during a solar eclipse

Total and annular solar eclipses occur where the Sun-Earth-Moon angle is 0$^\circ$.
Where that angle is smaller than 0.5$^\circ$ but not 0$^\circ$, there is a partial solar eclipse.
The Sun-Moon-Earth angle at those times is 180$^\circ$.
Due to the Moon's orbital inclination, most new moons miss eclipse alignment, some by as much as 5$^\circ$.


The anomaly of spiral galaxies and the need for dark matter

Supermassive black holes are really small, compared to the mass of the galaxy.

We know how big the black hole is, because we can see the stars in the centre of galaxies orbiting it. So in a typical spiral galaxy the black hole has a mass of a few million suns.  But the galaxy has a mass of a few trillion suns.

The black hole dominates the gravity in the cluster at the very centre of a galaxy, but out by us, it is entirely negligible

You don't need to use general relativity for this.  Relativity is significant when you are close to a black hole, but if you are far from it, then ordinary Newtonian gravity is an excellent approximation.  The force of gravity drops in inverse proportion to the square of the distance, which means that something as small as a supermassive black hole is much too tiny to have any effect on the the orbital speeds of most of the galaxy.

To model the orbital speeds there has to be something with a mass of trillions of suns, which entirely surrounds the disk of the galaxy, which the stars can pass through without friction and is completely invisible.  Finding that is the challenge of dark matter.


The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$.  The radius of a star is a generally a very complicated function of a star's other properties.  When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we define as the surface. So no simple formula in general.

That said, one can derive the approximate functional dependence for stars of various evolutionary states through the principle of homology. i.e. assuming that stars of a certain type are just rescaled versions of each other. Glancing at my old course notes, on the upper main sequence, where stars burn hydrogen principally through the CNO cycle and have radiative envelopes dominated by electron-scattering opacity, we derived $R\propto M^{15/19}$. The same principle (but with different assumptions about the star) is used to determine the location of the Hayashi track for pre-main-sequence stars, along which $R\propto M^{-7}T^{49}$.  Particular formulae can be found for different types of star but the relationships between $M$ and $R$ vary wildly.

Neither the two stars you mentioned are typical main-sequence stars. R136a1 is a Wolf-Rayet star, which is basically a star that has blasted away most of its hydrogen envelope.  Mass-radius relations are usually strongly dependent on mean molecular weight, which is higher without hydrogen, so the relations break down (or, rather, would have to be derived separately). But usually higher mean molecular weight gives a more compact star. UY Scuti has probably finished burning hydrogen in its core and has moved off the main sequence. So again, it'll follow a different relation.


All pulsars are observed to slow down gradually,albeit with a wide range of rates.The following diagram,known as the $\dot{P}-P$ diagram for pulsars illustrates this.The crab pulsar yes indeed would be a bad choice as a clock due to it's high $\dot{P}$.

But there are pulsars to the bottom left of the diagram which have periods of the order of milliseconds(hence called millisecond) pulsars and their $\dot{P}$ is also very low.These are the pulsars which can be used as atomic clocks,for example:MSP J0437− 4715.

  We can see that the both the period and the period derivative of this
  particular pulsar are very low providing great precision to the period
  of the system.So it can be used as a precise clock.

References:

1.http://www.cv.nrao.edu/course/astr534/PulsarTiming.html

2.http://www.cv.nrao.edu/course/astr534/Pulsars.html

3.http://en.wikipedia.org/wiki/PSR_J0437-4715


If a black hole does not emit light, how can one take a picture of the black hole itself?

The picture is of the central region of M87, taken at a wavelength where the gas is "optically thin".

The ring of bright light is pretty much exactly where it is expected to be for the synchrotron radiation emitted by the hot gas to have been gravitationally lensed by a black hole with the same mass as deduced previously from the motion of stars close to the centre of M87.

It is a "picture of a black hole", though all you see is the way it warps the light around it.

Your analogies are poor. It isn't like taking a picture of someone in a dark room. The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly. That, by the way is the only method by which you can see someone in an illuminated room, unless you have eyes that detect the mid-infrared light they emit. Of course that option is not available for a black hole, since they emit no light at all.

Your other analogy is also curious. A cursory glance at the sky reveals it is full of air molecules. That is why it is blue. Your view though lacks the angular resolution to see individual air molecules, so you would not be able to see them.

Here, the resolution of the "telescope" is just good enough to reveal the essential structure and size of the photon ring.

Finally, why isn't it a picture of an accretion disk? As I said, it is expected that the disk is optically thin and actually much bigger than the imaged area. Emission from the disk would not be expected to show anything but a haze with a vague central concentration. Given the spatial resolution of the image, what has actually been seen is a sharp, bright ring.

The accretion disk around a black hole should be terminated at or slightly beyond the innermost stable circular orbit. This is at 3 times the Schwarzschild radius, but gravitational lensing would make it appear to have an inner radius of 3.7 times the Schwarzschild radius. The bright ring has a radius of 2.6 Schwarzschild radii; correct for a lensed photon sphere explanation, incorrect for the inner wall of the accretion disk (unless the black hole is 70% of the mass previously determined from stellar motions).

The orientation might be just about ok for an accretion disk explanation. If the rotation axis is aligned with the jet of M87, the disk should be about 17 degrees from face on. However, in order to see the inner parts having such a circular shape and well defined structure would require a geometrically thin and optically thick disk, the opposite of what is deduced from the physical conditions in the plasma.

To see more context, here is a wider view of M87 taken by Chandra, and from the Chandra team's blog post:

  Chandra was used to observe M87 and other targets during the EHT campaign. While Chandra can’t see the shadow itself, its field of view is much larger than the EHT’s, so Chandra can view the full length of the jet of high-energy particles launched by the intense gravitational and magnetic fields around the black hole. This jet extends more than 1,000 light years from the center of the galaxy.


Why does the distance between Sun and Earth stay the same?

While the Sun and Earth attract each other, they cannot fall into each other because of angular momentum conservation. In a central field (where the force is acts in the direction of the distance vector and depends on distance only), the specific angular momentum vector $\boldsymbol{L}=\boldsymbol{r}\times\boldsymbol{v}$ is conserved ($\boldsymbol{r}$ is position and $\boldsymbol{v}$ the velocity). In particular $L=|\boldsymbol{L}|=rv_t$ (with $v_t$ the tangential velocity: the component of velocity perpendicular to the direction Earth-Sun). The specific orbital energy $E=\tfrac{1}{2}|\boldsymbol{v}|^2-GM/r$ is also conserved and must be negative for a bound orbit (such as Earth's). Combining these two we have
$$
E = \tfrac{1}{2}v_r^2 + \frac{L^2}{2r^2} - \frac{GM}{r}
$$
with $v_r$ the radial component of velocity. Since $v_r^2\ge0$, but $E<0$, not all radii are reachable. In particular there are two radii at which $v_r=0$: the apo- and peri-apse of the orbit.

In fact, for the Earth, the peri- and apo- apse are quite similar, and the orbit is nearly circular. The gravitational pull of the Sun is very nearly balanced by the centrifugal force due to the rotating orbit.

In GR, the picture doesn't really change much. Non-Newtonian (GR) effects are tiny in the Solar system and completely unimportant for Earth's orbit.

Note that the picture of pushing by ripples in space time (vs. pull from Sun's gravity) doesn't make a difference either. The point is that just because you're pushed some way doesn't imply that you're falling that way. All bodies also have something called inertia which makes them want to follow their path rather being pushed around. In case of a circular orbit (similar to Earth's around the Sun), gravity and inertia balance in such a way as to create a circular orbit: inertia want's to go straight, gravity want's to pull/push you into the Sun, but the result is something between those.


It looks like XEphem's format of databases as described in the manual's file format description. Does the filename end in .edb ? That is a common extension for these sort of files. Assuming it is, then the format breaks down as (taking the first line for example):

A name (Akamar)
A "type designation" for the sort of object, in this case f for fixed (as opposed to orbital elements or a bound satellite)
A subfield for the fixed type indicating it's a Star
The spectral type of the star (A3)
Right ascension (RA) and proper motion in RA (02:58:15.696 and -44.6 milliarcsec/year)
Declination and proper motion in Dec (-40:18:16.97 and 19.0 milliarcsec/year)
Magnitude (brightness) of the object (3.2)
Reference epoch for the proper motions (2000.0)


When will all other galaxies become not visible from Earth/Milky way?

That will never happen. We will always be able to see the galaxies in the Local Group. We won't get separated from them by the expansion of space because the group is bound together gravitationally. But the rest of the galaxies will be beyond the cosmic light horizon in 150 billion years or so.
However, because of the mutual gravitational attraction it's likely that the Local Group will merge into a single giant galaxy. It appears that the Milky Way & the Andromeda galaxy are already on a collision course, and will merge in around 4.5 billion years, about half a billion to a billion years before the Sun starts its red giant phase. The full merger of the group will take a long time, Wikipedia says the median time is around 450 billion years.
Due to the Sun's gradually increasing luminosity, the Earth will be uninhabitable in a billion years or so, and it's likely to get swallowed by the red giant Sun at some stage. But I guess humans may survive elsewhere in the Solar System (or outside it).
For further details, please see Wikipedia's Timeline of the far future.


Do stars twinkle when seen from Mars' surface?

Following howstuffworks and my own metereological intuition, the key effect for stars to twinkle seems less the pressure or size of the atmosphere but rather if different layers (usually of different temperature) exist, and if the gradiant between the layers is steep enough.
Light passing through zones of differently dense air on its way to the observer lead to small (time-dependent) distortions occuring for instance over warm asphalt of a road. Here, the light travels horizontally through the atmosphere, rather than vertically like in the case of twinkling stars.
I like to imagine the parcles of hot, less dense air rising from the ground as the bubbles in a lava lamp.
In the case of a hot road, a few dozends of meters (on Earth's surface) is enough to cause the flickering of e.g. cars at a distance.
The visible image then becomes somewhat blurred, and since the air is constantly (and irregularly) moving upward over the hot street, it looks as if the air is flickering.
Applied to your question: I know that dust devils are likely occuring on Mars, which need atmospheric instabilities in order to exist, so I would assume that there are indeed atmospheric conditions where one could observe twinkling stars on Mars.


The main reason is that the red mirror produces a smaller image in its image plane, which requires a smaller pixel spacing in the detector (or a stronger eyepiece) to get the same resolution.
In a Newtonian telescope the focal length cannot be too small:

(Image Krishnavedala)
Also, a more ideal shape is a paraboloid (not a sphere). Spherical mirrors are relatively easy to manufacture, and "nearly flat" ones are very close to paraboloids.


How can the temperature of stars can be determined by their color?

I think you are talking about color index and temperature. So, I will briefly explain under that circumstance.

Color index, e.g., B-V is the difference in brightness (magnitude) of the light observed by filters B and V. (The following discussion can be generalized to other color indeces, e.g., U-B or g-i). Filters B and V have effective wavelengths 4450 (bluer) and V 5510 (redder) A (see Wikipedia), respectively, therefore B-V increases when the object becomes red. (Recall that magnitude is more negative for being brighter). The temperature which is derived from the color index is called "color temperature;" this is different to blackbody temperature, or kinetic temperature, or other types of inferred temperatures (see [this 2).

For the color temperature, this is very observational inference. To map from the color index to the color temperature, it is inferred from other well-known object. See Wikipedia for this.

Another way to work around is to transform the color index into the flux ratio, then applying the blackbody function as the inference. For example, from B-V >>> flux(4450A)/flux(5510A) >>> blackbody temperature corresponding to the flux ratio.


How to differentiate between images of a gravitationally lensed object

In the case of multiple images of a background, distant object, the answer is relatively simple. You take a spectrum of the multiple images or parts of an extended lensed image and you see whether the spectrum looks the same, and in particular whether the redshift of the multiple images are the same.

Gravitational lensing affects light of all wavelengths equally, so regardless of the path taken, the spectrum should be unaltered, except that the different paths take different amounts of time to travel from the original object to us. So, if the background source, or its spectrum, are time-varying, then the lensed images could also appear different. However, distant galaxies do not change their redshift on such short timescales and so this should still be the same regardless.


Why do pulsars turn "off" from rotation?

The first thing you need to recall is that electromagnetic waves do carry momentum as well as energy.  This shows up in effects like light pressure. Specifically a photon of wavelength $\lambda$ carries momentum $h/\lambda$. In and of itself, that doesn't answer your question though, since you are asking about the rotation of the pulsar, and changed to its linear momentum don't affect that. However, there are two ways this can happen. First imagine a rotating sphere which is emitting radiation from its equator. Consider a point on that equator. The radiation it is emitting "forwards" (in the direction towards which it is rotating) will be blue shifted (as seen by a stationary distant observer) while the radiation emitted "backwards" will be correspondingly red-shifted. Since the blue-shifted radiation has shorter wavelength, it has more momentum, the net effect is a force on the source which tends to slow the rotation. (interestingly, the same effect is used at much lower energy levels to trap ultra-cold atoms in what is called "optical molasses").

The second effect is a bit more technical. The radiation which is emitted towards the poles of the pulsar has a property called "circular polarization" and a beam of circularly polarized light carries away angular momentum.


Can we observe changes in the CMB (surface of last scattering) over time?

It isn't all that hard to determine the age of the sun, provided you are content with an accuracy of 100 million years or so either way. The radiometric dating of zircon crystals, which are only a little younger than the sun, gives a date of about 4.6 billion years, and it is generally agreed that the sun is about 5 billion years old. This is compatible with its position in the main sequence of a Hertzsprung-Russell diagram. You certainly wouldn't notice any change on the CMB map in a mere ten years, but a billion years hence it will have a noticeably greater red shift, except for the fact that there will be nobody here to notice or record it. Our descendants, every last one of them, will all have gone to meet their maker in far less than a billion years.


How to convert theoretical template spectrum from luminosity density to flux density units?

Use this equation:
$$F_\nu = \frac{L_\nu}{4\pi D_L^2}.$$
This is the relationship that defines the luminosity distance, $D_L$, in a static Euclidean universe (i.e. not ours). Hogg has a good review on the arXiv of how to handle the relationship in an expanding universe, including when you need to perform something called a "$K$-correction" for spectral densities like the question asks.


Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits?

The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits.

There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time.  These give quite different results - qualitatively different.

Average distance on geometric basis.

In a way this is the most obvious one.  The mathematics is :

$$R_g = \frac a {2\pi} \int_0^{2\pi}r\,d\theta$$

And we use for an ellipse :

$$r = \frac \rho {1+\epsilon\,cos\theta}$$

The result is (eventually :-) ) :

$$R_g = a\,\sqrt{1-\epsilon^2} \tag{1}$$

Average over time.

This is a tricker calculation, but in fact it's the most "human" average.  We want :

$$R_t = \frac 1 T \int_0^T r\,dt$$

The maths on this is tedious and not at all useful, but the result ( with a little help from Wolframalpha.com ) is :

$$R_t = a\,\frac 2 {\pi} \frac{2+\epsilon^2}{(1-\epsilon^2)^{\frac 5 2}}tan^{-1}\left( \sqrt{\frac {1-\epsilon}{1+\epsilon}} \right) \tag{2}$$

Now by any definition that's a messy equation, but the main point is to note the key difference between the time averaged expression (2) and the geometric one (1).

The geometric expression gets smaller as eccentricity gets larger.

The time averaged one gets larger as eccentricity gets larger (a slight simplification).

In fact for both $\frac R a$ starts off at $1.000$ but when $\epsilon\to 1$ then  $R_g\to 0$ and $R_t\to \infty$ !

Why ?

The geometric expression doesn't allow for the orbital speed of the object.  As you increase eccentricity an orbits ends up with having a faster close approach, so less time close up to it's primary.  As it moves away from the primary, the orbital speed also decreases, so it spends much more time at large distances than at short distances.


On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept?

For planets of constant mean density you have:
$$
M=\rho \times 4\pi r^3
$$
and the surface value of $g$ is:
$$
g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r
$$
So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ as $r \to 0$

In terms of mass:
$$
g(M)={G(4\pi \rho)^{2/3}M^{1/3}}
$$
So as $M\to 0$ we have $g(M)\to 0$ and again $\log(g(M))\to -\infty$ and the intercept of the $\log-\log$ plot as $M$ goes to zero alows you to calculate the mean density.


To what extent are structure formations unexplained?

The info you reported are not opposite.
They refers to different scales on structure formation, and are both true.
As you said, ΛCDM model is very useful to explain large scale structures, as galaxies, clusters, and larger.
This means that the model fits really well to the observables, first of all the CMB map and its anisotropy.

What the model can NOT explain is the original formation of the structures, at the observed rate (take a look at this).
We know the age of universe (within this model), and we know the age of the observed galaxies. Then it results that galaxies have formed much faster then in a "simple" ΛCDM scenario. We can not explain how the primordial structures started to collapse and to grow in what now we see as galaxies.
To account for this observation we need to invoke dark matter primordial fluctuations. But this is an assumption, and does not give us the original spectrum of the fluctuations, neither it explain how it is formed. And this is a parameter that the ΛCDM model can not account for, at least at this stage.


What observational constraints are there in detecting the presence of volcanism on exoplanets?

Exoplanets are too far away to send satellites or to image them directly. So there is no way to go there and say: there it is a volcano. My guess is that we have to guess from what we know working in the solar system. And we might just get a statistical probability that the planet is active.

I would say that there are two cases:

If the planet is rocky and near enough to the star or a giant planet, you might expect volcanism caused by tidal effects, similarly to what happen on Jupiter's Io moon. In few words: the gravitational force of the nearby big planet/star deform the planet and this deformations are converted to heat due to friction. If the gravitational forces are large enough, you can end up having volcanism
If you are able to measure the spectral lines in the planet atmosphere you could try to identify signatures of gas or dust typically associated with volcanic activity. I don't know what you would want to look for, but my guess is that you want to search mostly in the infrared.


Traversing a proton within the Planck time

If you don't like math: A fraction of a second.
For math lovers:
One proton is $8.414 \cdot 10^{-16}$ meters wide, and one Planck time is $5.391247 \cdot 10^{-44}$ seconds. This means that we are traveling at $\dfrac{8.414 \cdot 10^{28}}{5.391247} = 1.5606779 \cdot 10^{28}$ meters per second. Dividing by $299792458$ m/s, or $c$, we get $5.205861 \cdot 10^{19}c$. The observable universe is 93 billion light years in size, so it would take $\dfrac{93 \cdot 10^9 \cdot 86400 \cdot 365c}{5.205861 \cdot 10^{19}c} \approx 0.0564$ seconds to traverse the observable universe. This obviously is impossible, as faster-than-light travel (username checks out) violates the rules of general relativity (as we know it).


Is the point of this paper to say that Pluto's status as a planet should have been handled scientifically, rather than by a vote?

Part of the problem here is semantics.  The meaning of "planet" was really poorly defined prior to 2006 (and some argue that it still is).  In the early 2000's, this started to become an issue. You can blame Mike Brown.

In January 2005, Mike Brown's team at CalTech in Pasadena, California, discovered an object we now know as Eris (they nicknamed it Xena early on).  The early observations of Eris suggested it was larger than Pluto (which was later conclusively proven incorrect with the New Horizons flyby of Pluto in 2015). But larger or not, it was close in size to Pluto.  Did this make it the tenth planet?  If so, then what about the other bodies like Sedna? Haumea? MakeMake? For that mater, what about Ceres, or Chiron?  By some counts, you could say we had upwards of 23 planets.  Or do we only have 9? Or 8?

For that matter, what makes a planet a planet?

The word planet comes to us from the ancient Greek "planetes asteres" which meant "wandering stars."  To the ancients, even well before the Greeks, a few of the lights in the night sky were definitely different from the others.  The stars remained in fixed patterns.  Sure, they rose and set, and they rose and set at different times through the year, but their relative positions were always the same - as far as the ancients could tell.  But five of them were different.  Their positions changed.  Two of them would rise before the sun in the morning for a while, then disappear and set after the sun in the evening for a while.  These are Mercury and Venus.  The other three, Mars, Jupiter, and Saturn, all seemed to move in the same direction, interestingly through the constellations they knew as the zodiac, but now and then they'd back-track a little bit, then move forward again.  Very odd.

The ancients had no idea that they were, in fact, other worlds, completely different in nature from the burning stars.  Of course, for that matter, they had no idea what the stars really were either.  They just knew the planets, though they looked pretty much the same as stars, moved differently. They were wandering stars.  It wasn't until Galileo turned his scope on the heavens that we learned that these aren't just bright points of light like the other stars - but whole different worlds.  The ancients barely could see Uranus, and gave it little heed, and couldn't see Neptune at all.  Other, smaller bodies like Pluto, Ceres, and others were far to small to see without a telescope.

When Herschel discovered Uranus in 1781, he first thought it was a comet.  In 1783, observations and calculations had lead to the discovery that Uranus' orbit was nearly circular, and it behaved like the other planets, and so it was accepted as one (and Herschel wanted to name it George).  When the predicted movements of Uranus didn't match the observed motions, astronomers and mathematicians suggested the reason was due to another planet, further out, pulling on Uranus and affecting its orbit.  In 1846 it was finally discovered, largely based on mathematical predictions.  At the time, no telescope could show either Uranus or Neptune as more than a very small disk.  But they were in nearly circular orbits, and that seemed to be good enough.

Ceres was discovered in 1801, and for several years was considered a planet. German Astronomer Johann Elerte Bode had previously observed what he thought was a mathematical relationship between the planetary orbital radii.  According to his calculations, there should be a planet between Jupiter and Mars, right where Ceres was discovered by Guiseppe Piazzi.  Bode's observation was actually erroneous - there was no such mathematical realationship, but it was convenient that an object was found there.

As it turned out, Neptune's orbit didn't quite fit the mathematical predictions either.  This lead to a belief that there might be a tenth planet causing the orbital discrepancies.  In the late 1800's and into the 1900's, the hunt was on.  Wealthy eccentric and astronomer Percival Lowell founded the Lowell Observatory in Fagstaff, Arizona, and set out to find "Planet X."  In 1929, the director of the Lowell Observatory hired a 23 year old farmer from Kansas who'd been constructing his own telescopes, mostly out of salvaged farm equipment, and doing his own observations. Clyde Tombaugh was brought in to work on the hunt for Planet X, a rather arduous task involving taking pictures of the same patch of sky a few days apart and then comparing them using a device called a "blink comparator" and looking or something different between the two images.  On February 18, 1930, Tombaugh noticed a tiny dot moving between images taken a few weeks before. Looking back at a previous image a few days before, he added to the evidence.  After capturing a few more images still and comparing, he finally reported the discovery on March 13, 1930.  He had discovered Pluto - the ninth planet!

Well, at least, that's what the hype said.  It didn't take too long to notice that Pluto was a bit different.  While no two planets orbit in the same precise plane, the previously-discovered eight are very close to co-planar.  They also have orbits that are very close to circular.  Pluto's orbit is tilted dramatically with respect to the other planets, and its orbit is significantly more elliptical.  In fact - Pluto's orbit crosses inside that of Neptune (though they'll never hit each other).

Pluto was also tiny. Our own moon is larger, as are several other moons of other planets. In 1978, Pluto's moon Charon was discovered. With further research, it was found that Charon was so large, that the point of orbit between them, what is known as the barycenter of the orbit, is actually well outside Pluto's radius.  Many astronomers were starting to talk of Pluto-Charon as a "co-planet."

Then, in the late 1990's and into the 2000's, a bunch of other objects started turning up out past Pluto.  Some of them were spherical, like a planet.  So, were they planets?

If you take a look back at Pluto's discovery, you can make an argument that the only reason Pluto was called a planet from the start was because Tombaugh was trying to find a planet.  He found something, there's no doubt there.  But would it have been called a planet if they weren't specifically looking for a planet?  As it turns out, Pluto's mass (including the whole Pluto/Charon system which includes four other "moons") isn't nearly enough to explain the discrepancies between the observed and calculated positions of Neptune (which, as it happens, were later explained away by calculations based on Einstein's discovery of special relativity).

When Mike Brown discovered something that was potentially larger than Pluto, but much further out, and just as weird, people started questioning what it meant.  If Eris was a planet, then what made it a planet and not Sedna, or Orcus, Or MakeMake, or the others.  And then, if those were planets, why not Ceres, which had been downgraded to an asteroid.  What does "planet" mean?

So here come the semantic police. The problem is, who decides what a word means.  We run into this in language all the time.  But for science, we need precise definitions so we can make and categorize observations.  If I simply define a star as a bright light in the sky, that includes planets, the moon, and even airplanes.  For science to work, we have to have agreed upon definitions - operational definitions.

Of course, in many cases, there's no single body to define things.  There is no "official" body to make these decisions.  The purpose of the International Astronomical Union is to provide an agreed-upon group to set standards like this.  When they met in 2006, they had a working group to formulate a definition of what should define a planet and differentiate it from another body. To some greater or lesser extent, it IS arbitrary - as is all language.

When I say fruit, you might think "banana" or "apple." But not everyone would think of "tomato" as a fruit, even though it is, by definition.  How about a cucumber?  Both a tomato and a cucumber are the fruiting bodies of their plants.  They are, by definition, fruits.  Yet how many people do you know that talk of a cucumber as a fruit?  Most people consider fruits to be sweet while vegetables are savory.  So, to an extent, it will depend on your point of view.

But for science, we need a standardized set of definitions.  And this is what the IAU did in 2006.  Their working group formulated a set of criteria to define a planet, and put it to a vote of the general assembly, which passed it.

The authors of the cited article state that "taxonomical terms like planet are determined by the scientific process, not voting."  However, when you are arguing semantics, then you are talking about consensus.  Looking at another taxonomic system, scientists agree that a "cat" is a cat because it has certain characteristics.  More recently, those characteristics include genetics and assumed evolution, but historically taxonomy was more based on observed characteristics.  In defining a planet, we have the same process at work - only the defining characteristics were put to a vote.

Regardless of what we call Pluto, no one argues that it is not a definite body of some substantial size orbiting the star Sol.  The argument does not deny Pluto's existence, only its classification.  Pluto, along with the eight other fully-accepted planets, along with the thousands of asteroids, along with the myriad other trans-Neptunian objects, Plutinos, Kuiper Belt objects, and everything else orbiting the sun, are all planet-like in some ways, and not in others.  At some point, a definition must be agreed upon, and this is exactly what the IAU did in 2006.  Not everyone agreed, but that wasn't necessary.  We weren't stating a scientific fact, we were assigning a label.  It's not a FACT that anything is a planet, it is simply a label of convenience.  In fact, if you wanted to be nit-picky, you could argue that Jupiter and Earth are not at all the same type of object, and that they shouldn't both be called planets but should have different classifications.

By reclassifying Pluto, the IAU basically played a game of "one of these things is not like the others" and drew a line.  And whether or not you agree with that line, the majority of the assembled members of the convention did, and adopted the definition.  Sure, it could be changed, but why bother? It really makes no difference other than in the minds of people who don't understand otherwise.  It's PR, and nothing more.


Distinction between metric expansion and objects just moving apart from each other?

Time will show

...although that is a very long time.

If the grid of the universe was static, objects moving away from us should continue to move away from us with a continues velocity. That is not the case from observed data: Things farther away from us seems to move away faster. This is known as Hubble's law, and is one of the basic concepts of modern cosmology. The model of an expanding metrics fit that data, in that when there is more space fabric between us and the far object, there are more space that stretches, and therefore the velocity is higher.

That things move away from us, and have an accelerating velocity, should suggest a universe that expands faster and faster. There are however other forces limiting this growth, most notably gravity. The fate of the growth of the universe is determined by that unknown balance, that must also consider dark matter and energy.

The magnitude, or even the validity of those models is the current frontier of astronomy, and cosmology in general. Nothing is settled yet.


Observing cepheid variables in other galaxies

It is correct that we can't observation stars in other galaxies with the unaided eye. But with telescopes, we can see much more, including individual stars in other galaxies. We certainly have plenty of examples including several galaxies that don't orbit the Milky Way (and some outside the Local Group!)
Indeed, it was because we can detect individual cepheids in other galaxies that we can measure the distances to galaxies. Hubble found cepheids in the Andromeda galaxy in 1924. However, if your telescope does not have the light-collecting power and resolving power it would be nigh on impossible to detect a cepheid against the background of a whole galaxy. We build bigger telescopes to get the light grasp and resolution to do such measurements.
(converted from comments)
We can resolve individual cepheids  (we can't resolve them as a "disc", but we can't resolve many stars even in our own galaxy as a disc) We can resolve them as points of light, distinct from its neighbours, and we can measure the brightness of individual stars.


Is it possible for a moon to orbit a planet floating free in the galaxy rather than orbiting a star

The answer is Yes.. Planets that don't orbit around a star are known as Rogue Planets. There is nothing preventing a rogue planet from having one or many moons.

Not so long ago, the first candidate for a free-floating exoplanet-exomoon system was presented in this paper. It looks like a gas giant several times larger than Jupiter with a sub-Earth mass moon.

Another study calculated and simulated scenarios where planets where ejected from their orbits around a star and concluded that around five percent of Earth-sized planets who are accompanied by Moon-sized natural satellites would retain them after the event.


I've been reading phys.org for years.
It's right up there with physics.aps
and physicsworld.

All good sources that provide decent refs.
Comments are often where people who like to argue, argue. Sometimes they are useful, often not.


Why is $H_\delta$ prominent in type A stars?

H$\delta$ absorption is formed when hydrogen in the level $n=2$ is excited to $n=6$.

To get strong H$\delta$ absorption lines you need large amounts of hydrogen in the first excited state $n=2$ and a radiation field that contains large numbers of photons with an energy equal to the difference between the $n=6$ and $n=2$ states.

These requirements are satisfied in stars with photospheres between about 8,000K and 15,000K, and we call these early-F and A-type stars.

At cooler temperatures most of the hydrogen is in the ground state, there are few photons with sufficient energy (3eV) to be absorbed and the H$\delta$ line weakens. At much higher temperatures (in O-stars for example) then none of the hydrogen is in the $n=2$ level, and in fact most of it is ionised.


Confirm: We can only notice planets if their orbit passes in front of their star and our view

The answer by choeger provides a correct outlook, but I wanted to provide a bit more detail on some exoplanet detection methods that have actually been used to find planets we couldn't see. Now I interpret "see" in the manner posed by the question in that "see" references observing a transit of the star by the planet. The questioner is correct that we cannot "see" the planet in this manner if the alignment is not perfectly correct. So how else might we detect the planet?

Radial Velocity Method - We can try to detect its gravitational effects on the star through a method known as the Radial Velocity method. Both the star and planet(s) will be orbiting their common center of mass. That means if we watch the star, we can see it orbit that center of mass as well. If we see the star "wobbling", we can infer what mass must exist in that system to cause the wobble. I should point out that we are only able to see radial wobbling (that is, wobbling of the star towards and away from us) which means that while we can still detect planets which are not directly aligned with the star as they would be for the transit method, they are much harder to detect. I also want to point out that the answer by choeger mentioned tidal forces, but that is actually not something we can use to detect planets or measure their effects on their host stars.
Direct Observation - This is exactly what it sounds. With some clever techniques and the right conditions, we can take a picture of the planet itself. One pretty famous example was the direct imaging of Fomalhaut b. Shown below is an image over a few years as it travels through its orbit.

Microlensing - This uses the concept of gravitational mass acting as a lens to background, luminous objects by bending spacetime, and consequently, the path of light through space. If a planet passes in front of a background object (not its host star), it will cause a slight magnification of that background object. I should say that this requires a chance alignment and so it is not a repeatable measurement, meaning there is often little chance that we can actually confirm that we saw a planet. Below is a light curve of a background object and you can see it grow brighter as the main star passes in front of the background object. Pay attention to the little blip on the right which indicates the planet also passes in front of the background object.


Are multiple geostationary satellites required to handle arbitrary locations?

Earth's center of mass must be at one of the two focus points of a satellite's elliptical orbit, or at the center of a circular orbit such as a geostationary orbit. One cannot orbit a certain latitude, except for the equator.

But there are clever alternatives for different purposes. Geosynchronous (as opposed to geostationary) orbits mean that the satellite will always be within sight from the same location on Earth, but it would move around and probably need antennas which continuously aim towards them. Japan is building its own regional GPS-system in this way.

There are also semi-synchronous orbits which put the satellite at the same place on the sky at the same time every day. This is used for some Earth observation purposes because the shadows on the ground remain the same for the same location on Earth everytime an image is taken.


Does our sun/solar system orbit around any other celestial objects?

The orbit of the Sun around the Galaxy is quite complicated, because unlike the solar system, the mass is not completely concentrated at the centre. So, in addition to the roughly circular 230 million year orbit in the plane of the Galaxy, there are superimposed motions in and out of the plane and towards and away from the Galactic centre. These roughly sinusoidal additional motions, called epicycles, do not have large amplitudes - a few hundred light years - and take about 70 million and 160 million years respectively.

The Sun does not systematically orbit any other Galactic structures or stars (see the flagged potential duplicates for more details - there is no evidence for any binary companion to the Sun larger than a few Jupiter masses)  and is unlikely to do so in the forseeable future. The space between the stars in our Galaxy is large enough that they are essentially non-interacting.

The Galaxy itself is in motion with respect to the galaxies around it. The nearest tens of galaxies form the local group and probably have complicated orbits within their summed gravitational potential. These orbits cannot be precisely determined, because although we can measure line of sight velocities using the Doppler effect, the tangential motions require extraordinarily accurate position data over many years. These are now becoming available - for instance we know our Galaxy and M31 will collide in around 4 billion years.

Further afield, the local group is part of a larger galaxy aggregate, called the Virgo supercluster, but here, determining an orbit is impossible.

Even further, large galaxy clusters are arranged into a web of larger superclusters, voids and filaments that are all in motion with respect to each other.


Where do the photons come from at the edge of an Event Horizon?

From the distant observer's perspective the person falling in becomes more and more time dilated and redshifted. As a result you see fewer and fewer photons from them and eventually none at all (although you never have any way of knowing that one more, very redshifted photon isn't going to arrive some time). So there are no extra photons.


Will the detection of colliding neutron stars by LIGO help answer the question of where heavy elements came form?

I just found out the answer to my question from a live press release on You Tube that has been covered by blogs like this: http://news.nationalgeographic.com/2017/10/gravitational-waves-discovered-neutron-stars-pictures-science/

What LIGO does is tell you what is causing the gamma ray burst that can be studied different ways. The highly anticipated announcement I just watched explained that LIGO observed the neutron star merger (they're calling the event a kilonova). With the help of VIRGO, the area of the sky could be narrowed down better than previous LIGO detections. LIGO told us the masses of the bodies, the distance to the objects, how much mass was ejected and thus what the end product of the collision is expected to be.

We have observed many gamma ray bursts and this one was not particularly bright, but LIGO told astronomers to check this one out and so it was being observed within a few hours.

The electromagnetic spectrum of this event was collected and very broad lines were detected which are believed to be caused by heavy elements being ejected at near light speeds.

By our knowledge of how often these "kilonovae" should happen and the volume of elements heavier than iron thought now to have been emitted (16 Earth masses was the number mentioned in the press release), it can be said that a majority of many of the heavy elements we see around us may have come from these events.


Is S2 still the fastest known star in the galaxy?

S0-2 (I assume you mean the star orbiting the Milky Way central black hole) actually reaches speeds of 7650 km/s at periapsis.
However, if you read further down the wikipedia article you will see that there are stars S0-102 and S62 which have shorter orbital periods than S0-2 (11.5 and 9.2 years vs 16.05 years).
Star S62 is also in a highly elliptical orbit ($e=0.976$, Peißker et al. 2020) and has a periapsis distance that is even closer to the black hole than S0-2, where it has a speed clocking in at about 30,000 km/s$^{1}$. No actual value or error bar is given in the paper.
The plot shows its orbit (in red) compared with other stars in the region.
STOP PRESS
A telegram issued by Peißker et al. 2020 claims detection of a new winner - S4714. They appear to have revised downwards their estimate of the peri-centre speed of S62 to 20,000 km/s and say that this new, very faint object, has a peri-centre speed of 24,000 km/s.

$^1$ Of course that is the speed as seen by a distant observer. If you were to go there and put a speed gun on it as it went past, then it would be a touch larger.


Observation of Gravitional waves by LIGO from binary black hole merger

Neutron stars are less massive but larger than black holes. The frequency of the gravitational waves is due to the orbital period of the objects as they spiral together. A neutron star merger would have a different orbital speed, so the frequency would be different and lower. It would rise at a different rate. The signal from neutron stars is also much weaker, so couldn't be detected from the same range.

Prior to the detection of the signal, numerical simulations of merging black holes of various sizes were done, so the theoretical shape of the waveform expected from the merger of black holes (and neutron stars) was known. It was possible from this theoretical work to estimate the mass of the two black holes taking part in the merger by matching the observed signal to a theoretical one. You can see the theoretical signal for black holes with 36 and 29 in the diagrams labelled as "numerical relativity"


Is there a relation with the noun planet and to be in a plane?

There is no close connection. Planet is from Greek asteres planetai meaning wandering stars (via Latin and French). Plane is from Latin planus meaning "level or flat".

It is possible that both words are derived from the proto-indo-European root *pele-, (flat) However the evidence linking *pele to planetai is weak. Thus there is no more connection between planet and plane than between planet and field (field is also derived from *pele, via the sound change in German from p to f)

http://www.etymonline.com/index.php?allowed_in_frame=0&search=planet

http://www.etymonline.com/index.php?allowed_in_frame=0&search=plane

http://www.etymonline.com/index.php?term=*pele-%20(2)&allowed_in_frame=0


Within that galaxy cluster, things become chaotic as these galaxies are flung around but kept in orbit by their strong gravity. And as Steven Linton mentioned,

  the galaxy's organized motion is turned primarily into disorganized
  motion within the material that is stuck together, which is what we
  know as heat. So the atoms of gas in these clouds collided and bounced
  off in all directions, where they hit other atoms, until the two
  clouds had formed one, with huge random internal motions (i.e. very
  hot).


When was the last volcanic eruption on Mars?

Berman and Hartman (2002) dated some lava flows of the Athabasca–Marte Valles system at < 20 Ma. Later, Vaucher et al. (2009) dated lava flows of the same region, finding ages even younger in some cases (around 2.5 Ma). In both studies the age estimates are derived from surface morphology (crater counts).
In Volcanoes: Global Perspectives, Lockwood & Hazlett go even further (emphasis mine):

Unlike Earth’s Moon, volcanic activity may be far from over on Mars. Detailed photographic studies by NASA’s Global Surveyor starting in 1997, and the Mars Express High Resolution Stereo Camera, which began orbiting Mars in December of 2003, provide evidence of effusive volcanism perhaps as young as 2–2.5 million years, with flows mantling the flanks of Olympus Mons erupted mostly within the past 200 million years. It cannot be assumed that Mars is now internally cold and volcanically dead, and in fact, volatile efflux may still be in progress.


What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic?

There are plenty of misleading statements in Wikipedia and elsewhere on the internet about nucleosynthesis (I am busy searching to see if I have said something similar in the past!)
The reason that the alpha chain does not proceed significantly beyond $^{56}$Ni is that in order to overcome the Coulomb barrier the temperatures need to be so high that the iron-peak nuclei are disintegrated by photons at this temperatures.
I suppose the sense in which the endothermic statement is true is when considering a core made of nickel. In order to produce alpha particles you need to disintegrate some Ni nuclei. This process is highly endothermic and can't be balanced by subsequent fusion.
e.g (And this is a bit simplistic) Photodisintegration of a Ni nucleus into 14 alpha particles requires 88.62 MeV. Then 14 fusion reactions with Ni nuclei, producing Zinc, would give back just 37.9 MeV. In contrast disintegrating $^{52}$Fe into 13 alpha particles needs 80.5 MeV, but 13 fusion reactions of $^{52}$Fe to Ni yields $8.1\times 13= 105.3$ MeV.