Acquirable Raw Data in Amateur Astrophotography

First off, pairing a classic dob with a DSLR is a bit like a shotgun marriage. A dobsonian is fundamentally a visual telescope. Most manufacturers don't even consider the possibility that these instruments could be used for data collection via a sensor. There are 2 issues here:

1. The dobsonian is not tracking

The sky is moving, the dob stays still. You have to push the dob to keep up with the sky. Any long-exposure photo would be smeared. To remedy this, you'll need an equatorial platform, which will move the dob in sync with the sky.

Please note that only the best platforms allow reasonably long exposure times. Then the results can be fairly good.

2. There isn't enough back-focus

The best photos are taken when you remove the lens from the camera, plug it into the telescope directly, and allow the primary mirror to focus the image directly on the sensor. This is called prime focus photography. But most dobs can't reach the sensor within the camera, because their prime focus doesn't stick out far enough. There are several remedies for this, like using a barlow, moving the primary mirror up in its cell, etc.

The bottom line is that it takes some effort to make a dob and a DSLR play nice together. Is it doable? Yes. Is it simple and immediate? No. So the literal answer to your question is that there isn't much you can do with just a dob and a DSLR.

You can take photos of the Moon and the Sun, because the short exposure there does not require tracking, but that's pretty much it. Here is an image of the Moon I took with a home-made 6" dob (with home-made optics) and a mirrorless camera (prime focus, about 1/320 sec exposure):

Makes a cute little desktop background, I guess, but it's definitely not research-grade.

Now add a tracking platform and things become more interesting, and the possibilities open up quite a lot.

In a more general sense:

There are telescopes that are specifically made for astrophotography. They have lots of back-focus, they are short and lightweight and therefore can easily be installed on tracking mounts. More importantly, there are tracking mounts made specifically for imaging - very precise, delicate mechanisms that follow the sky motion with great accuracy. In fact, the mount is more important than the scope.

A typical example would be a C8 telescope installed on a CGEM mount, or anything equivalent. Barring that, a dob with lots of back-focus sitting on a very smooth tracking platform (probably not as accurate as a GEM, but good enough for many purposes).

Make sure you don't exceed the load capacity of the mount. If the mount claims it can carry X amount of weight, it's best if the telescope weight doesn't exceed 1/2 of that amount. Close to the weight load limit, all mounts become imprecise. The exceptions are high end (the most expensive) mounts which cost many thousands of dollars and usually honor their promises in terms of load capacity 100%.

Once you have: a tracking mount, a good camera, and a telescope (listed here from most important to least important), you can start imaging various portions of the sky for research. There are 2 main classes of objects that you could image:

1. Solar system objects

They're called "solar system objects" but the class includes anything that's pretty bright, not very big, and it's high resolution. Tracking is important but not that crucial.

You need a sensitive, high speed camera that can take thousands of images quickly (a movie, basically). These are called planetary cameras. They generally have small sensors, are high sensitivity, and can operate at high frame rates (hundreds of frames per second).

As a cheap alternative in the beginning you could use a webcam, there are tutorials on the Internet about that. A DSLR in video mode in prime focus might work, but it's going to do a lot of pixel binning, so resolution would be greatly reduced unless you use a very powerful barlow (or a stack of barlows).

You'll load all those images in a software that will perform "stacking" to reduce them all to one single, much clearer image.

The scope needs to operate at a long focal length, f/20 being typical, so a barlow is usually required. The bigger the aperture, the better.

2. Deep space objects (DSO)

These are anything that's pretty faint and fuzzy, like galaxies, but some comets are also DSO-like in their appearance. You need to take extremely long exposures; usually a dozen or a few dozen images, each one between 30 sec and 20 min of exposure, sometimes even longer. Extremely precise tracking is paramount, so you need the best tracking mount you could buy. Autoguiding is also needed to correct tracking errors.

The scope needs to operate at short focal ratios, f/4 is pretty good, but as low as f/2 is also used; focal reducers (opposite of barlows) are used with some telescopes, like this or like this. Aperture doesn't mean much; small refractors are used with good results.

The camera needs to be very low noise; DSO cameras use active cooling that lowers their temperature 20 ... 40 C below ambient. Typically they have large sensors.

DSLRs can also provide decent results, but their noise is typically higher than dedicated cameras, so you need to work harder for the same results.

Specific software is used for processing, stacking, noise reduction, etc.

So what can you do with such a setup?

Comet- or asteroid-hunting works pretty well. Terry Lovejoy has discovered several comets recently using equipment and techniques as described above. Here's Terry talking about his work.

Tracking variable stars is also open to amateurs. This could also be done visually, without any camera, just a dob, meticulous note-taking, and lots of patience.

With a bit of luck, you could also be the person who discovers a new supernova in a nearby galaxy. You don't need professional instruments, you just need to happen to point the scope in the right direction at the right time and be the first to report it. This also could be done purely visually, no camera, just a dob.


How do scientists determine the age of stars?

Measuring the ages of stars is an incredibly difficult process. It is not something that can be summarised adequately in an SE answer. Please take a look at these references.

Soderblom et al. (2013); Jeffries (2014).

Methods split into -

Fundamental - e.g. radioisotope dating - only possible for the Sun

Semi-fundamental - kinematic traceback of stars to their birthplace; observing the depletion of lithium in a star.

Model-dependent - comparing the luminosity and temperature (and composition) of a star with that predicted by stellar evolutionary models. Or equivalently, its surface gravity and temperature.

Empirical - looking at how things like rotation and magnetic activity decline with age and using this to estimate the age of a star.

Statistical - using the metallicity or galactic space motions of a star to provide a (crude) estimate of its age.

There is also a lot more detail given in this SE answer.
Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status?


Lucky imaging with Celestron 14 - is this result reasonable?

I don’t think it’s any of the points you suggest.  The scope and camera are fine for what you are doing.
I’d say that the focus is slightly off. It takes a lot of effort to get it spot on


Why are there no terrestrial planets with a subsurface ocean?

Some hypothesize that the Earth did have a subsurface ocean during the Cryogenian period, which lasted from 720 to 635 million years ago. The Cryogenian saw the two greatest known ice ages in the Earth's history, the Sturtian and Marinoan glaciations. There is some evidence that the Earth was completely covered with ice and snow during those glaciations. (There is also some evidence that it was not.)
Whether even older periods in the Earth's history also succumbed to snowball Earth episodes is even more debatable. The evidence has been wiped out by a billion plus years of tectonic activity. That even older periods of the Earth's history than the Cryogenian did suffer snowball Earth episodes does however make sense.
The Sun is considerably more luminous now than it was shortly after the Earth first formed. Once the Earth cooled from its formation (and that appears to have happened fairly quickly, in a geological sense of "quickly"), that faint young Sun should have resulted in a cool Earth. That the young Earth had periods where it obviously wasn't covered with ice and snow from pole to pole is the faint young Sun paradox. The apparent paradox is almost uniformly explained away via greenhouse gases. But which ones?
That said, once plants started converting carbon dioxide into oxygen, and once the oxygen stopped combining with iron to form most of the world's iron ore deposits, the greenhouse effect that kept the young Earth from freezing over should have dropped significantly. There are some signs that this happened, some that it didn't. Puzzling out what happened well before the Cambrian has always been problematic because rocks that old are hard to find.


Does this smartphone photo show Mars just below the Sun?

That is a camera artefact caused by the bright sunlight reflecting within the lens on your phone. It’s more pronounced than on a large camera because of the small lens size. This is a secondary image of the sun, as the brightness of the source allows for the reflection to be still intense enough to be detected.

Here is a photo I took with the same artefact. It appears for me as a cyan dot with a magenta outside, but other photos I’ve taken have one with the same colour as the sun.

Notice if you zoom in on yours, there’s a halo around the artefact as well.

Also notice that the artefact is directly below the sun in my image, whereas in yours it’s slightly to the left. However, this is because in your image the sun is slightly to the right, and hitting the lens at an angle.

As uhoh says, you can see that the sun is the same distance from the centre of the image as the lens flare (or very close to):

It should be possible to take a photo with the artefact wherever you want, by re-angling the phone - including inside the sun, where it will be drowned out.

I actually took my phone into an Apple Store to find out what this was, where they explained the lens flair effect. Unfortunately I don’t have a source for this, as it was in person. Hopefully the above is convincing enough.


What irradiance does the Moon receive during a total lunar eclipse?

The following is more a series of observations than a full answer, but I hope other people can fill in the details.

The apparent magnitude of the full moon is -13. The apparent magnitude of the moon during a lunar eclipse will differ from eclipse to eclipse (correlated to the Danjon scale value), but let's suppose it is 0 for a certain eclipse. (Note that this magnitude is different from the (pen)umbral magnitudes mentioned on the NASA website; these magnitudes can be computed and do not depend on the Earth's atmosphere.)

This means the full Moon is normally reflecting $100^{13/5}$ times more light than during this eclipse. I'm not sure if the albedo of the Moon is different for red light than it is for normal sunlight, but if not, that means the answer is $\frac{1361}{100^{13/5}} \frac{W}{m^2} \approx 8.6 \frac{\text{m}W}{m^2}$.


Chances of life on other planets, how watertight is the idea that 'As soon as life could occur, it did occur’?

Yes, early life on Earth does not imply easy life.
Full disclosure: I have a paper on this, Snyder-Beattie, A. E., Sandberg, A., Drexler, K. E., & Bonsall, M. B. (2021). The Timing of Evolutionary Transitions Suggests Intelligent Life is Rare. Astrobiology, 21(3), 265-278.
The mental model most people have is that there is some constant rate $\lambda$ of life emergence on lifeless worlds, so we should expect the probability of life at time $t$ to grow like $1-\exp(-\lambda t)$, with the expected emergence time at $t=1/\lambda$. Earliest life was conservatively 3.77 Gya and oceans 4.5 Gya, so that gives estimates like $\lambda \sim 1.37$ per Gy. One can of course quibble about whether the rate is constant, how long the window of emergence is, and many factors.
But as the question asks, that we observe life early in Earth's history may not be good evidence. Suppose $\lambda = 10^{-12}$ per year, making the probability of getting life over a 10 billion year sun-like star lifespan 1%. Any intelligent observers that evolve will be on one of these rare 1% planets. If they try to guess how likely life is on other worlds they will estimate a $\lambda$ that is much bigger, typically $10^{-9}$ or so based on when life emerged on their world.
But this bias is likely to be much worse, since intelligent observers come about after a long evolutionary chain that may contain several fairly unlikely steps. For most of Earth's history life was very simple, and becoming eukaryotes, land-living or intelligent may have required very unlikely steps with their own low probabilities. Observers will always find themselves on worlds where these steps have been passed, even if they have their own super-low probabilities and the observers are extremely rare lucky cases.
When you run the math on this, you will find that the hard steps get roughly evenly distributed across the habitable time of a planet. If you need 5 hard steps (say: life emerging, photosynthesis, eukaryotes, multicellularity, intelligence) the observers will typically see life show up around 1/6th of the habitable time, and they show up 5/6th of the way to the end.

Random samples of histories where rare transitions happen but a full set of transitions happens before the end of the time interval. Colour indicates the number of the transition.

Probability density of the transitions over the habitable period of the biosphere for 5 very hard transitions.
Getting back to the questions: if we were to discover several independently evolved kinds of life on Earth (or elsewhere in the solar system, even if extinct), then we would have good reason to think that at least the life emergence step was easy and there will be lots of life in the universe. Same for discovering life that can thrive in very different environments.
It might also be that true life emergence is very rare, but then spreads by panspermia across the galaxy. In that case it would seem as if life was easy, but it is just received from elsewhere.
However, later hard steps will still tend to push appearance of life early since there is a need for a lot of time to have the other unlikely things to happen so that the very rare observers can emerge. What we did in our paper is to show how the pattern of steps can be used to constrain estimates of life emergence, but there is plenty of room to argue what steps were truly hard or even when they happened. But I would agree with the original post: early life does not have to imply that life is easy. But the way to know the difficulty of life is of course to look for it, not just do statistics.


Does Ceres have an exosphere similar to the Moon?

In general, asteroids experience the same fundamental processes as the moon: solar wind, micrometeorites  and photons from the sun are striking them and sputtering material off to form an exosphere.  So there must be some sort of exosphere, but its density and composition will vary with distance from the sun, as well as the gravity and composition of the asteroid.

A 1992 ultraviolet study of Ceres found $OH$ emission, consistent with escaping water, and suggesting a ice cap.

But a more recent study with the more sensitive VLT  did not detect $OH$, limiting the maximum amount of water produced  per second to $7 \times 10^{25}$ molecules per second at the time of the observation.


Population of excited H levels in a Strömgren Sphere

There just aren't any other mechanisms to keep the population excited in a Stromgren sphere.  In principle collisions can also lead to excitations, but if you do an order of magnitude calculation you'll find that the time between collisions is much longer than the time to receive an ionizing photon.  In a laboratory setting you might be able to coerce some particles into remaining in an excited state using the Quantum Zeno effect, but that won't happen in interstellar space.

So without something to keep the atoms in an excited state, even if you start with everything in the excited state, in time $\tau$ you will find that everything has decayed to the ground state, with the exception of some very small fraction $\tau_{\textrm{ion}} / \tau$, which may have moved up to a higher energy state.  Assuming the lifetimes of those very excited states are comparable, after a few more times $\tau$, you will find that even those will have decayed to the ground state.


Why isn't the constellation of that month's zodiac prominent or even visible during that month?

The Zodiac sign of a month is decided by the month when the Sun is "in" that sign from the point of view of Earth.
(See Mike G's comment below - because of the precession of the Earth, Zodiac signs are now 30 degrees "off" from their original markings/associations with the months of the year.)
I believe that August is actually the month of the Leo, not October - more specifically, July 23 to August 23. (see here for more date info) If you look at the diagram, you'll see that the Sun is right in front of Leo from the point of view of Earth during the month of most of August, and thus we call that the month of the Leo. Some more examples - during October, the month of Libra, the Sun is directly in front of Libra, and during March, the month of Pisces, the Sun is directly in front of Pisces.


Does a parabolic trajectory really exist in nature?

No, but nor do elliptical or hyperbolic trajectories.  These are features of a model we have for gravity, not part of nature.
The modelling process involves constructing a mathematical framework that describes a natural system and allows for predictions. It is often said that "no models are correct, but some models are useful".  For gravity, one common model is the Newtonian model.  In the Newtonian model, you can set up two particles in such a way that one follows a parabolic curve (in some inertial frame) Parabolic trajectories exist in the Newtonian model.
In nature, there are more than two particles in the universe and so there are no parabolic, elliptical nor hyperbolic trajectories.  Moreover, the Newtonian model is known to only approximate gravity (even General Relativity is probably an approximation, gravity is probably some kind of quantum interaction. But the details are not yet known)
Nevertheless.  The Newtonian model is a very useful model for solar system dynamics.  Long-period comets, falling from the distant Oort cloud, will have orbits that are very long and are well approximated by parabolas. (but only approximated, as comets, in particular, have significant non-gravitational accelerations due to reaction forces from outgassing, in addition to gravitational perturbations of the planets, and effects from the non-spherical shape of the sun)


Are Trans-Neptunian Objects (TNOs) asteroids?

The IAU does not give a definition of "asteroid" (in contrast to its defintion of "planet") The definition of "asteroid" is therefore determined by actual use, rather than a specific authoritative source.

The IAU in its guidelines on "naming" doesn't use the term "asteroid" (except qualified as "near earth asteroid". The term is somewhat ambiguous.  Instead it considers "planets", "dwarf planets" and "minor planets".

Minor planets can be dwarf planets, asteroids, trojans, centaurs, Kuiper belt objects, and other trans-Neptunian objects.

Initially solar system bodies were classed as "comet" (appears fuzzy due to a coma of dust and gas) "asteroid (appears star like, a single point of light with no fuzziness). The planets appear disc-like.

TNOs have no coma, but their surfaces are dominated by ices (and would start to evaporate if ever brought to the inner solar system).  Thus they are rather different from the traditional rocky asteroids.

So you should avoid using an unqualified "asteroid" as it may be ambiguous. You can say "minor planet" if you want to include both TNOs KBOs and classical asteroids.  Or "Main belt asteroid" for those between Mars and Jupiter.

If you read "asteroid" it would normally exclude the outer solar system bodies. But as it may be ambiguous, so read the context.


Underground sea in Titan(Saturn's moon), water or hydrocarbons?

Titan is believed to have a layer of liquid water under its surface, due to tidal heating of the icy crust.

Titan has a surface formed of ices with some hydrocarbon (methane) lakes. The crust of Titan is mostly water ice.  Below this, there is a liquid water layer and perhaps more ice and silicate rocks below that.

This is described at https://science.nasa.gov/science-news/science-at-nasa/2012/28jun_titanocean


Do comets contain any significant amounts of iridium?

I also googled "iridium content of comets", and the first result was https://news.dartmouth.edu/news/2013/04/dartmouth-researchers-say-comet-killed-dinosaurs

Now there is currently no consensus on the nature of the Chicxulub impactor. The observations of iridium and osmium suggest an asteroid. A minority opinion is that a comet may be responsible. The authors of the article in the linked news report suggest that measurements of iridium concentrations from ocean cores may be erroneously high, due to sorting in marine sediments. If these cores are excluded, the amount of iridium measured is lower, and consistent with some models of the composition of comets.

So, comets do contain some iridium. The amount of iridium is not certain. Some models of comet nuclei contain more dust and less ice, and so have more iridium. The exact amount of iridium in the KT boundary layer is also uncertain. The low end of estimates would be consistent with a cometary impact. The nature of the impactor is not yet settled science, but the balance of opinion currently favours a stony asteroid.


When did "resilvering" large telescope mirrors actually refer to aluminization, and why was it necessary?

Amateur telescope and mirror maker here. Not sure if I qualify as a "citable source" but anyway, here it is:

All metals will eventually tarnish. It may take a long time, but it will happen. The process is not entirely chemical always. Sometimes it's purely mechanical (abrasion). Other times it's in between. Surface phenomena are complex.

Even gold-coated mirrors will tarnish. You could use gold for infrared observations. But no matter how much care you take of it, eventually you will have to re-coat it. Sure, gold lasts a pretty long time, but not even gold is forever.

Aluminum is covered with a natural layer of colorless sapphire (aluminum oxide), which protects the metal from further destruction - sapphire is very hard and very inert chemically. But oxygen still diffuses through sapphire (albeit extremely slowly), and so do other corrosion agents like sulphur oxides, marine salt, etc.

So, yes, replacing silver with aluminum was an improvement. But not even aluminum is forever. In practice, you will maybe double the lifetime of the reflective layer when you migrate from silver to aluminum.

It should be noted that silver is still used. Aluminum is most popular nowadays, but you'll still find silver-coated mirrors (its reflectivity is higher than aluminum), gold-coated (for infrared), etc.

There are further coating techniques, like layering silicon oxide on top of metal, which further improve the lifetime of any kind of metal.


How would the characteristics of a habitable planet change with stars of different spectral types?

The following table gives the mass, radius, temperature, and luminosity of an average star of several selected spectral types:
$$\begin{array} {d|c|c|}
\text{Spectral Type} & \text{Mass} (\odot) & \text{Radius} (\odot) & \text{Temperature (K)} & \text{Luminosity} {(\odot)} \\ \hline
\text{M8V} & \text{0.082} & \text{0.111} & \text {2500} & {0.00043}\\
\hline
\text{M5V} & \text{0.16} & \text{0.199} & \text {3030} & {0.00299}\\
\hline
\text{M2V} & \text{0.44} & \text{0.434} & \text {3550} & {0.0268}\\
\hline
\text{K8V} & \text{0.59} & \text{0.587} & \text {4000} & {0.079}\\
\hline
\text{K5V} & \text{0.68} & \text{0.698} & \text {4410} & {0.165}\\
\hline
\text{K2V} & \text{0.78} & \text{0.763} & \text {5040} & {0.337}\\
\hline
\text{G8V} & \text{0.94} & \text{0.909} & \text {5490} & {0.673}\\
\hline
\text{G5V} & \text{0.98} & \text{0.982} & \text {5660} & {0.887}\\
\hline
\text{G2V} & \text{1.02} & \text{1.01} & \text {5770} & {1.014}\\
\hline
\text{F8V} & \text{1.18} & \text{1.25} & \text {6170} & {2.031}\\
\hline
\text{F5V} & \text{1.33} & \text{1.46} & \text {6510} & {3.434}\\
\hline
\text{F2V} & \text{1.44} & \text{1.61} & \text {6810} & {5.001}\\
\hline
\text{A8V} & \text{1.67} & \text{1.81} & \text {7500} & {9.3}\\
\hline
\text{A5V} & \text{1.85} & \text{1.94} & \text {8080} & {14.392}\\
\hline
\text{A2V} & \text{2.05} & \text{1.97} & \text {8840} & {21.263}\\
\hline
\end{array} $$
The following table gives the orbital distance, period, and velocity of an Earth-like planet receiving the same flux from its star as Earth does from the Sun, along with the radial velocity of the star caused by said planet and the tidal forces exerted on the planet relative to Earth:
$$\begin{array}{d|c|c|c|}
\text{Spectral Type} & \text{Orbital Distance (AU)} & \text{Orbital Period (days)} & \text{Orbital Velocity (km/s)} & \text{Radial Velocity (m/s) } & \text{Tidal Forces} {(\oplus)} \\ \hline
\text{M8V} & 0.0207 & 3.82 & 59.166 & 2.167 & 9166\\ \hline
\text{M5V} & 0.0547 & 11.68 & 50.929 & 0.956 & 977.34\\ \hline
\text{M2V} & 0.163 & 36.51 & 48.812 & 0.333 & 100.09\\ \hline
\text{K8V} & 0.281 & 70.95 & 43.134 & 0.219 & 26.5 \\ \hline
\text{K5V} & 0.406 & 114.84 & 38.518 & 0.17 & 10.11\\ \hline
\text{K2V} & 0.58 & 182.93 & 34.525 & 0.133 & 3.98\\ \hline
\text{G8V} & 0.82 & 280.06 & 31.877 & 0.101 & 1.7\\ \hline
\text{G5V} & 0.942 & 337.48 & 30.375 & 0.093 & 1.17\\ \hline
\text{G2V} & 1 & 365.56 & 29.973 & 0.088 & 1\\ \hline
\text{F8V} & 1.425 & 572.18 & 27.102 & 0.068 & 0.407\\ \hline
\text{F5V} & 1.853 & 799.11 & 25.231 & 0.057 & 0.209\\ \hline
\text{F2V} & 2.236 & 1018.01 & 23.901 & 0.049 & 0.128\\ \hline
\text {A8V} & 3.049 & 1505.21 & 22.041 & 0.039 & 0.058\\ \hline
\text{A5V} & 3.793 & 1984.29 & 20.799 & 0.033 & 0.033\\ \hline
\text{A2V} & 4.611 & 2526.01 & 19.859 & 0.029 & 0.02\\ \hline
\end{array}$$

These calculations were done with the planet being at such a distance that it recieves exactly the same flux from its star as Earth does from the Sun. In reality a planet can be significantly farther away or closer and still remain in the habitable zone.
You can see a very clear correlation: As spectral type decreases (star becomes cooler, smaller, and less massive),

Orbital distance decreases,
Orbital period decreases,
Orbital velocity increases,
Radial velocity of the star increases,
Tidal forces on the planet increase exponentially.

I'm sure there are other factors that I didn't take into account but these are the most obvious ones I could think of.

1. I got the data for the mass, radius, and temperature of each star from here. This is an average relationship and not meant to be exact.
2. The luminosity was calculated using the Stefan-Boltzmann law, assuming a perfect sphere and a blackbody radiator, which most stars approximately are.


Any tip on maintaining a log of celestial objects?

I suggest you to take a look at the Amateur Astronomy Observers Log Web Site, where everybody can share their astronomical logs.

The logs contain:

Instrumentation used
Sky condition (seeing, light pollution, ...)
Accurate date and time of the observation

Specific informations that you could add depend by the kind of object that you are watching. For example you could try to estimate the magnitude of a variable star, or you could describe colour variation of Jupiter bands.

Some software can give you more support. There are specific astronomical log software for smartphone (for example Stargazing Log for Android).
Morover in the Linux planetarium software KStars when you open the detail of an object you have a specific tab where you can register logs in form of simple text.


Following a reference to Darley et al., ApJ 746, 61 (2012) from your Wikipedia link gives a (very technical) discussion of nova progenitors, including distinctions between nova systems where the secondary stars are main sequence or supergiant stars, and distinctions among white dwarfs with different chemistries.
The first sentence of that paper is

  A classical nova (CN) outburst occurs in an interacting
  binary system comprising a white dwarf (WD, the primary) and
  typically a late-type main-sequence (MS) star (the secondary)
  that fills its Roche lobe (Crawford & Kraft, 1956).

That suggests the 1956 paper is the original proposal for the Roche overflow model of the classical nova.
Like many original-idea papers, it's a pretty clear read.
But for your question, Crawford and Kraft seem to hedge about whether the "blue star" in their particular pair needs to be a white dwarf:

  [T]he
  observed
  luminosity
  of
  the
  blue
  star
  is
  essentially
  due
  to
  the
  energy
  released
  by
  the
  accreted
  material.
  This
  view
  is
  strengthened
  also
  by
  the
  fact
  that
  the
  blue
  star
  occupies
  a
  peculiar
  position
  in
  the
  H-R
  diagram.
  It
  lies
  10.5
  vis.
  mag.
  below
  the
  main
  sequence
  but
  about
  4
  mag.
  above
  the
  most
  luminous
  white
  dwarfs,
  whose
  effective
  temperature
  it
  exceeds
  by
  about
  8000°
  K.
  Unless
  the
  blue
  star
  is
  essentially
  degenerate,
  it
  can
  readily
  be
  shown
  that
  the
  small
  radius
  implies
  such
  a
  high
  internal
  temperature
  that
  electron
  scattering
  is
  the
  principal
  source
  of
  opacity.
  A
  simple
  calculation
  based
  on
  the
  standard
  model
  then
  yields
  a
  luminosity
  8
  mag.
  brighter
  than
  is
  observed.

In other words, Crawford and Kraft don't come out and say "definitely a WD," but if it's a non-degenerate star, it's a very strange one.
More modern observations of novae are compared to detailed models of the dynamics of the surface dynamics, models which have been debated vigorously for decades; the current generation of comparisons to data are sensitive to details like the amount of helium accumulating on the white dwarf's surface during the nova event. It seems unlikely that such details could even come close if the underlying assumptions about the basic physics of the erupting star were wrong.

Note that a classical nova system can be thought of as a type of contact binary star.
For any reasonable estimate of the size of the giant star, a distance of 10 AU between the two members of the pair seems like an overlarge estimate.
Ten astronomical units of separation viewed from a distance of 50 parsecs is already a gap of 0.1 seconds of arc. I wouldn't expect to see visible-light photographs showing both the giant star and the white dwarf, but rather that all of the information about the binary systems comes from spectroscopy.


Do galaxies have a preferred orientation?

There is no evidence for that. That's either a systematic observation error, or a random result within usual statistical fluctuations, compare this Q&A site.


Computing orbit positions of Jovian Satellites / Moons using JPL data

An excellent book by Jean Meeus, "Astronomical Algorithms", provides the calculations for planetary positions and Jupiter's Galilean Moons (Io, Europa, Ganymede, and Callisto). Meeus does not really explain the math to a great degree, so it is relatively simple to follow it as a formulaic approach. He uses VSOP for planetary positions and Lieske's E5 theory for the moons (you can get an explanation of both on Wikipedia). As a bonus, Meeus shows how to make calculations for both parameters with a lesser degree of accuracy which might improve your calculation speed. Though this book can be a bit dense, you might want to read up on the difference between "ephemeris time" and "terrestrial time", as this will affect your accuracy as well.


What should be the declination of a star to be marginally circumpolar given the latitude & height from ground?

Just to be clear, the "height from sea level"  (mentioned in the title) does not change what star is circumpolar. Whether the ground is at 0 units above sea level or 1500 units above sea level does not make a difference. I think what you intended to write is the height above the ground makes a difference ("angle of dip from the height" mentioned in the question). At 0 units above the ground, a circumpolar star has a declination of D1. At 1500 units above the ground, a circumpolar star has a declination that is farther from the pole, or D1-angle. (I assume that you are in the northern hemisphere.)

Refraction raises the apparent altitude of an object, so you can see a circumpolar star that is farther from than the pole than is possible without refraction. Thus, a circumpolar star has declination = D1-angle-Refraction. The adopted value for refraction of an object on the horizon is 34 arcminutes (Ref 1: The Astronomical Almanac, 2001, page A12). The actual amount of refraction depends on the atmospheric pressure and temperature. An approximate formula for refraction (applicable for observed altitudes below 15 degrees) is R=P(0.1594 + 0.0196a + 0.00002a^2)/[(273+T)(1 + 0.505a + 0.0845a^2)] where T is the temperature (degrees C), P is the barometric pressure (millibars), and a is the altitude (degrees) (Ref 1: page B62)


Why can't we point the centre of the universe from inflation graph we see?

That diagram does not depict the entire universe. At most, it depicts the history of what is now our observable universe (specifically, a 2D slice through it), with us at the center only because we're observing it. Someone at the furthest reaches of that portion of the universe would see us at the furthest reaches of their observable universe, and themselves at the center.
And like us, they would observe the universe to be expanding away from them, and doing so equally in all directions. This, as far as we can determine, is true for anywhere within the observable universe. The simplest assumption is that it continues to do so outside it as well, and there is no evidence that it does anything else.


How do/did we figure out that planets move in orbits?

I've also heard that people in the past knew about orbits even when they thought that Earth was at the center of the solar system. How did they figure this out in their times with their technology?

The same celestial objects (stars, planets, the Moon) could be seen every year. So, people figured out there was a pattern to it.
At first, geocentrism was prevalent due to prejudice and the  stars "orbiting" the Earth in a clean manner (no strange effects). If you stare up at night, stars move in a circle with the center approximately at the Pole Star. This can be mistaken for the Star moving around the Earth.
Objects like visible planets (with retrograde motions) were relatively small in number — we have the inner solar system plus Jupiter and Saturn, so people gave plausible explanations for these. Ptolemy theorized that the planet orbits around a "ghost point", for example:

(This is called the epicycle theory)

how we know for certain today that planets really move in orbits around the Sun (i.e., rather than moving but not around the Sun or moving around the Sun in a shape that is not a regular path).

The heliocentric model came up because of various smaller discoveries:

Jupiter has moons, so not all celestial bodies orbit the Earth
Venus has full phases, but in the geocentric model only a small subset of the phases should be visible
Theories like epicycles were not as accurate as they were expected to be, and people started realizing this over time.

With all this in mind, the simple solution of heliocentrism explains everything in a much cleaner fashion, without leaving much doubt in mind. Kepler's laws make much cleaner predictions.
Remember that people had catalogs of astronomical data to verify things with.
Later on, once Newton came up with his theory of gravitation, Kepler's laws made even more sense (though I believe that Newton formulated his theory with some inspiration from Kepler's laws). I'm pretty sure that there were on-Earth experiments to verify the inverse-square as well.
Nowadays, sending probes to space and getting a different perspective has confirmed this with 100% certainity for us. More or less.


Particles are generated and destroyed all the time. This is obvious for photons, but also holds for massive particles. Today most particles are stable and long-living, but shortly after the big bang the universe was so hot that particle generation and destruction were in some equlibrium and all particles participated in this process (for photons this continued until the epoch of re-ionisation at redshift $z\sim1000$. There is definitely no particle around today that existed all the time since $t=0$.


How deep is the "A" Fraunhofer line in the solar spectrum? Is it from the Sun's or Earth's atmosphere?

Where can I see - or even download - a fairly high resolution (~0.1 nm) solar spectrum at zero air mass - in other words from space without absorption and other features from the Earth's atmosphere?

You can download a normalized solar spectrum with 0.1 nm resolution from here. However this is not at zero airmass.

A solar spectrum with zero airmass can be found on this website with the direct link to their data in an excel sheet here. This spectrum appears to be at a 1 nm resolution.

  Of particular interest is the near infrared around 750-780nm. Is the "A" Fraunhofer line due to absorption by oxygen in the Sun's atmosphere, or in Earth's?

If you look up the Fraunhofer A line (really it's a "band"), you'll see it is caused by molecular oxygen, namely $\mathrm{O_2}$. This should tell you right away that it is caused by our atmosphere as molecules do not easily exist in the atmosphere of the Sun. In our Sun's atmosphere, most species are ionized forms of single atoms such as $\mathrm{O\:IV}$.

In Astronomy, spectral lines which are caused by the Earth's atmosphere and not by the source object being observed, are referred to as Telluric lines (from the latin word tellus meaning Earth). I found this source which describes, in detail, the various telluric lines that you may encounter.


Why does Titan have lower surface gravity than the Moon when Titan is more massive?

Surface gravitational acceleration on an object with mass $M$ and radius $R$ is given by
$$
g = \frac{GM}{R^2} \propto G\rho R
$$
where $\rho \propto M/R^3$ is the density of the object.
If one body has smaller surface $g$ than another, it must have smaller density $\rho$, smaller radius $R$, or both.
Titan is larger than Earth's Moon, so your observation about its surface gravity means Titan must be less dense than the Moon.  Wikipedia confirms:

$R_\text{Titan} = 1.5 R_\text{Moon}$, but
$\rho_\text{Moon} = 3.34\rm\,g/cm^3$ while Titan has only $\rho_\text{Titan} = 1.88\rm\,g/cm^3$.


Can there be an object with planetary discriminant between Ceres and Neptune?

I know this is an old question, but recent developments in astronomy have caused this to become relevant again. Early in 2016 there was a paper published by Konstantin Batygin and Michael E. Brown (here) which indicates the possible existence of a new planet, dubbed "Planet 9". Subsequent work by other authors have lent more evidence to its possible existence, e.g., a paper by Fienga et al. (here) which concludes that including Planet 9 in orbital dynamic models decreases residuals.

This being said, the original paper specifically states

  ...our calculations suggest that a perturber on a $a'\sim700\:AU$, $e'\sim0.6$ orbit would have to be somewhat more massive (e.g. a factor of a few) than $m'=10\:M_\oplus$ to produce the desired effect.

Such a planet is certainly massive enough to have cleared out its orbit under normal circumstances. For reference, Neptune has a mass of $17\:M_\oplus$ while Uranus has a mass of $14.5\:M_\oplus$. The special feature of this planet is that it is located $700\:AU$ distant from the Sun which is in a very ambiguous and unknown region of our solar system. It is not clear at all how much, if any, debris in the form of comets resides out there. There are models suggesting a disk (referred to as the Hills Cloud) exists at those distances, implying that this new planet could reside inside an extended cometary disk. Of course this is only speculation, but it is a distinct possibility.

If the above-mentioned disk exists, we have to consider a few scenarios.

Planet 9 was "inserted" into this disk by being expelled from the inner solar system early on in its formation. This is a possible cause for its current position and suggests one of two outcomes.

It has existed here long enough that it has cleared out its orbit, thus making its planetary discriminant well within the "planetary" side of the definition.
It has either not existed in the disk long enough to clear the orbit, or has achieved some sort of equilibrium such that it can't now clear the orbit (considering most of the clearing occurs during formation). Here is where you may run into an ambiguous case. For all intents and purposes we may consider this a planet, but there remains the possibility that, despite it being more massive than the Earth, it could formally be considered a dwarf planet or else exist in this fuzzy region as you suggest.

Planet 9 formed at its current orbit (plus or minus some migration).
In this scenario I think it is clear that it would have cleared its
orbit and resoundingly be considered a true planet.

Now these are all wild suppositions. I don't have any of the math to back it up, and I'm not sure the numbers to plug into the various equations are well known, or even exist. I'm merely remarking on the possibility that the situation you asked about now seems like it could exist.


Gravity of other objects in black holes?

Gravity, according to the majority of published and non-published scientists, is able to propagate information from one point to another at the speed of light. This means that as soon as any massive body "physically" interacts with a black hole (enters the Schwartzchild radius) there is no longer a way for anything outside of the black hole to know anything else regarding the status of the body. Therefore the new addition to the black hole simply becomes part of the system and can no longer be considered a separate entity. The body's gravity becomes that of the black hole's.

So... No


Can the Keck Observatory be improved with a third telescope?

Original Overview
The Keck telescopes mainly work independently of one another. Therefore, adding a third telescope wouldn't really have much of a point (aside from the benefits of having a telescope on top of a mountain with low light pollution and excellent atmospheric conditions). In the past, this might have been a slightly more appealing idea, because the Keck Interferometer (KI) ran from 2003 to 2012, and was a combination of the two telescopes (an effective increase from two 10-meter telescopes to one 85-meter telescope).
KI had some problems, though, that prevented it from living up to its full potential. For instance, four secondary "outrigger" telescopes to improve the interferometer were never built, even though they were an important part of early plans. It was eventually "mothballed" after several of the main scientific objectives were completed. Obviously, the interferometer could be reactivated, but you'd have to make a strong case for doing so. If you could get that to happen, and get permission to build another telescopes, many might choose to use the funds for these outriggers, rather than a whole new instrument.
Finally, building telescopes on Mauna Kea is controversial because of possible environmental impacts, as well as arguments that it endangers Hawaiian cultural and religious sites. There are objections enough over the planned Thirty Meter Telescope, never mind another Keck (which would, admittedly, probably be cheaper, although the two would use different wavelengths).
In short, a third telescope would help only if the interferometer was brought back on line - and even then, you'd have to make a case for both building a new telescope and for choosing it over the original outriggers.
Resolution
The angular resolution for a single telescope, to an order of magnitude, is
$$\theta\sim\frac{\lambda}{D}\tag{1}$$
where $\lambda$ is the wavelength of observations and $D$ is the diameter of the dish (adaptive optics, by the way, can make things a bit better). Each Keck telescope has a diameter of $10\text{ m}$ and observes at the optical and near-infrared ranges - so about $4\times10^{-7}$ to $4\times10^{-6}\text{ m}$. This approximation yields angular resolutions of about $0.0083$ to $0.083$ arcseconds. This approximation is accurate to within an order of magnitude.
Likewise, for an interferometer with two telescopes separated by a baseline of distance $b$, the angular resolution is approximately
$$\theta\sim\frac{\lambda}{b}$$
The Keck interferometer had a baseline of $85\text{ m}$ and observed at wavelengths of $2.2\times10^{-6}$ to $10^{-5}\text{ m}$. The approximation gives us resolutions of $0.0053$ to $0.024$ arcseconds - pretty close to the actual values.
What if we add a third telescope, identical to the other two and placed in a line next to them, $85\text{ m}$ from the closer one and $170\text{ m}$ from the farther one? The longest baseline is now $2\times85=170\text{ m}$, twice the baseline of the two-dish setup. We've now made the angular resolution twice as small as before. That's excellent. At $2.2\times10^{-6}\text{ m}$, the resolution is $1.3\times10^{-8}$ radians.
Let's think about observations at $2.2\times10^{-6}\text{ m}$, at the shortest wavelengths the interferometer used. Assume we want to build another telescope with the same angular resolution. Rearranging $(1)$ gives us
$$D\sim\frac{\lambda}{\theta}$$
Plugging in our numbers ($\lambda=2.2\times10^{-6}\text{ m}$ and $\theta=1.3\times10^{-8}$ radians), we find that $D$ should be about $170\text{ m}$ - which is of course the longest baseline of the three-dish interferometer. This is simply implausible at these sort of wavelengths. The largest single-dish optical telescope - observing in shorter wavelengths than this by a factor of about three - is the Gran Telescopio Canarias, at $10.4\text{ m}$ (although the Large Binocular Telescope is slightly better - $11.8\text{ m}$).
If you want to use the interferometer at optical wavelengths, you've still looking at a $\sim55\text{ m}$ dish, and that's currently maybe feasible. There are several extremely large telescopes being planned, notably, the Thirty Meter telescope I mentioned earlier and the European Extremely Large Telescope (E-ELT), coming in at around $40\text{ m}$. The E-ELT should be online as soon as 2-24 (!), which would be fantastic, and would have a comparable resolution to this three-dish interferometer.
Costs
Let's look at how much all of this might cost. The E-ELT is expected to cost a little over one billion euros (1.083 billion, to be precise), which comes out to around \$1-\$1.25 billion, in US dollars (that sounds like a lot, but space-based telescopes can be up to ten times that - though often with higher-quality images). Let's assume that our putative $\sim55\text{ m}$ optical telescope costs \$1.5-\$2 billion to build.
The cost of the instruments for the Keck Observatory ran about \$80 million. This includes the telescopes proper, as well as the cameras and other equipment needed to actually capture the images. Making improvements and other developments has run the total project to about \$200 million (not counting observing costs) We can be generous and say that building a third telescope - identical to the other two - and getting the interferometer system up and running might cost \$50-\$60 million. Maybe that's off, but it certainly can't be wrong by over an order of magnitude. At that point, the entirety of this new Keck Observatory has cost only about one tenth of what a single dish would cost.
Yes, it's likely that I'm wrong somewhere above - that I've made an error in the size of the single-dish telescope, or in estimating the total cost. But the discrepancy is still extraordinary.
Here's the problem, though: Nightly observing costs are high. One night using on Keck telescope costs \$54,000. The interferometer uses both telescopes, and, as an article I cited earlier notes,

it's complicated and expensive to link the giant eyes through a system of optical pathways for just a few dozen nights each year.

Interferometry isn't as simply as looking at an object with two different telescopes. It's complicated and costly. If we assume that operating the interferometer would cost around \$200,000 per night (is that gratuitous?), we find yearly operating costs of $\sim$\$73 million - assuming that it's used every night, which isn't likely. But even accounting for all the nights when the interferometer wouldn't be in use, that's probably still more than the estimated yearly costs for the E-ELT (50 million euros, or \$54.33 million). The single-dish telescope - again, operating a shorter wavelengths than Keck - would probably cost less per night than the new Keck Interferometer.
Perhaps the relatively low construction costs of the third telescope would offset all of this in the short-term (it likely would).
Here's the really obvious thing, however, that I've been ignoring: More dish means more light. An interferometer with an $170\text{ m}$ baseline is not the same as a single dish telescope of the same diameter. The three dishes are much, much smaller than the single one. While the resolutions are the same, the size of the dishes really does matter.
A dish with a diameter of $10\text{ m}$ has an area of $25\pi\text{ m}^2$. A dish with a diameter of $170\text{ m}$ has an area of $7225\pi\text{ m}^2$ - a difference of a factor of 289! If you want to make true comparisons between the new interferometer and a large telescope, you'll need more than three dishes. And that will make costs skyrocket.
Just for fun, imagine that we end up having 50 total dishes, with a baseline. If each costs \$30 to \$40 million to build (and here I'm being generously low), we quickly find that building 48 more dishes runs us to \$1.9 billion, or at least the cost of building our $55\text{ m}$ optical telescope.
Okay, so let's again assume that I made an error somewhere, and this figure is too high. Also, if you ended up building this interferometer (Where would you put it, by the way? Mauna Kea's crowded enough!), you could probably make each dish simpler than Keck I or Keck II. You still have to deal with high nightly operating costs - really high operating costs.
Science!
So, why isn't the interferometer up and running? According to NASA, it's because the interferometer's main target, observing certain circumstellar disks, is done:

NASA’s primary goal in developing and operating the two 10-m telescope KI was the characterization of faint dust disks around nearby main sequence stars.

For an overview, see Millan-Gabet (2011). As the authors note in section 3, the interferometer's primary observing wavelengths (about $10^{-6}$ to $10^{-7}\text{ m}$, the micrometer range) encompass the peak emission wavelengths of circumstellar dust. Additionally, the resolution at these wavelengths can resolve features of circumstellar disks.
As we determined before, it's hard to build a telescope of comparable resolution operating in the near-infrared. Remember, our single-dish telescope operates largely in the optical range, although it would of course have near-infrared capabilities, to an extent. Therefore, an interferometer like the Keck Interferometer is our best choice - and perhaps our only choice, in the foreseeable future. However, it's certainly possible to build extremely large telescopes operating at slightly shorter wavelengths, with roughly the same angular resolution (to an order of magnitude, really). Plus, they'll gather more light, and - as I found shortly before posting this, in an excellent answer by Rob Jeffries - limitations of interferometer instruments and atmospheric issues makes them less versatile than large, single-dish telescopes.
TL;DR
There are a few takeaways here:

Adding a third telescope would give you a resolution unmatched at those wavelengths by any other land-based telescopes.
This would likely be a lot cheaper.
However, you'd gather only a small fraction of the light that a much larger telescope would.
To gather more light, costs would skyrocket - as would technical problems.

So, economically and technically, it's really not feasible to add a third disk to Keck - at least, compared to building a single extremely large telescope.

This is a pretty long answer, and I'm guessing I made an error somewhere. I'll go back and do one more check in the next day or so. If anyone finds one, do let me know!


Why dimmer high-redshift supernovae means the expansion is accelerating, if the dilated region pertains the distant past?

They are further away than they "should" be according to a decelerating universe model (one without dark energy). That is because the expansion has accelerated.

The Hubble parameter isn't a constant, it was larger in the past and is getting smaller. The Hubble parameter does not equate to the expansion rate of the universe. It is defined as the rate of change of the scale factor divided by the scale factor. A Hubble constant would imply exponentially accelerating expansion. e.g. Evolution of the Hubble parameter and Is hubble constant dependent on redshift?

More detail

The light travel time distance to a distant supernovae is measured by the time it takes light to travel from the supernova (when it went off) to us, now. A relatively nearby supernova is barely affected by any acceleration/deceleration of the expansion. It's distance is given approximately by a straightforwad application of a "coasting universe", one that is expanding at a constant rate  - the flat line on your top plot.

If the light has to travel further across the universe then the effects of the cosmological parameters start to become apparent. In a decelerating universe the rate of expansion slows, so a simple application of a "coasting universe" would result in an underestimated distance (the dashed line curving down in your top plot).

I attach a better plot, from the Riess et al. (1999) paper that you reference, with lots of lines representing different combinations of $\Omega_M$ (matter density, a decelerating influence) and $\Omega_\Lambda$ (dark energy, an accelerating influence). Thus a coasting universe has neither and that is the flat line in the middle, representing a universe that has always been expanding at a rate that can be measured locally.

In a decelerating universe (e.g. a flat universe with $\Omega_M=1$, $\Omega_{\Lambda}=0$), the expansion rate was much faster in the past. If however we lived in an "empty" coasting universe, the rate of expansion would have been constant and we would now see the supernovae as further away. If the universe is accelerating (e.g. $\Omega_{\Lambda}=1$, $\Omega_M=0$), then this effect becomes more extreme; the supernovae would be much further away and even fainter.

We appear to live in a universe somewhere between coasting and extreme acceleration (well we know there is some decelerating matter in the universe right!). This is the $\Omega_M=0.28$, $\Omega_\Lambda=0.72$ curve. This universe has been accelerating for the last few billion years, which covers the distances to the supernovae on this plot. The distant ones suffered a lower rate of expansion than those closer to us and as a result are further away than predicted by a coasting model.

However, in such a universe, if you go even further back in time to even more distant supernovae, then the expansion was being decelerated, because whilst the dark energy density stays the same, the matter density was much higher. That is why you can just see the $\Omega_M=0.28$, $\Omega_\Lambda=0.72$ turning down again at the highest redshifts on the plot.

And here indeed is a more recent plot (from Riess et al. 2007), which appears to show this turn-down at higher redshifts quite clearly and delineates the turnover from a decelerating to accelerating universe at $z \sim 0.8$ (e.g. Daly et al. 2008).


A quick skim of a few papers on Earth's tropical year, e.g. M.J.White, suggests to me that existing formulas are best-fit to data, not analytic.  How Saturn's ellipticity of orbit & axial precession(s) play into this is beyond my learning, but I'm getting the feeling that "it just is," and that any number of cofactors such as planetary layers, densities, and so on might affect what's being observed.


Energy can indeed transform into matter - for example pair production; but this creates equal amounts of matter and anti-matter.


Why the Milky Way appears to be arching across the sky?

The Earth is a sphere (or is nearly a sphere). So to make a map of the whole Earth you would need to project it onto a flat surface. When you do this you create distortions. For example, on many maps, the straight line between New York and Japan looks like a long curve. It is not actually a curve, but when you stretch the surface of Earth to make it flat, some straight lines become curved.

The sky is also a sphere. If you ignore distance and just think of the direction of each point in the sky it would be a huge sphere that surrounds you. When we take a picture of the sky we must project that sphere onto a flat surface. That causes distortions, and it means that what is a straight line on the sphere of the sky is distorted by the projection into a curve.

If you choose a projection in which the milky way goes straight across the middle, you can see that the milky way goes straight across the sky.

The appearance of an "arch" is a distortion caused by stretching the image to fit on your flat screen. It is not real.


Calculating obital velocity from radial velocity

You can't without assuming something about the overall velocity.

The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could assume that as open clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on roughly circular orbits. NB. This cannot possibly work for globular clusters, which are an entirely different population, many with non-circular orbits and orbits out of the galactic plane.

Having made that assumption, your radial velocity is the orbital velocity resolved through an angle that is formed between your line of sight to the cluster and a tangent to the Galactic centre at the cluster's Galactic position.

This method will not work well for clusters with a Galactic latitude much different to zero, or clusters with small Galactic longitude (which should have zero radial velocity in this model).

For all this to work you need the radial velocity, the Galactic latitude and longitude of the cluster, the distance to the cluster and an assumed distance to the Galactic centre.

I've had a go at sketching the situation. The cluster is on a circular orbit around the Galactic centre with a true orbital velocity $v$ (shown as a vector). We measure the component of this in our line of sight as the radial velocity, where
$$v_r = v \cos \alpha.$$
The trick is to find $\alpha$ using the triangle formed by: us, the Galactic centre and the cluster. If the distance to the galactic centre is $d_{gc}$ and the distance to the cluster is $D$ and the galactic longitude of the cluster (angle measured from the direction of the Galactic centre) is $l$, then I think that the distance of the cluster from the Galactic centre $d_{Cl}$ is given by the cosine rule as
$$ d_{Cl} = \left( d_{gc}^{2} + D^2 - 2d_{gc}D \cos l \right)^{1/2}$$

Then the angle $\beta$ at the apex defined by the cluster id given by
$$\cos \beta = \frac{d_{Cl}^2 + D^2 - d_{gc}^2}{2d_{Cl}D}$$
which leads to
$$ \cos \beta = \frac{D - d_{gc} \cos l}{d_{Cl}}$$

You then look at the diagram and note that $\beta = 90 - \alpha$, so that
$\cos \beta = \sin \alpha$. Hence you can calculate $\alpha$ and hence calculate $v$.


What happened before, Big Bang or inflation?

The big bang (a singularity 13.8 billion years ago that was the beginning of time) didn't happen at all. Inflationary cosmology consists of:

ΛCDM (aka "big bang") cosmology starting at a cosmological time of some small fraction of a second, preceded by

an inflationary epoch lasting for an unknown amount of time (but at least 50-60 e-folds), preceded by

there may be no way to know, because inflation erases any hints of what came before.

Times quoted as "after the big bang" are ΛCDM model times and don't count from the true beginning of time. There's a common misconception that if inflation ended $10^{-32}$ seconds "after the big bang", that it must have lasted for no longer than $10^{-32}$ seconds. In reality, it has to last longer than that in order to accomplish its goal of producing smooth starting conditions for the ΛCDM epoch, and in most (all?) specific models it lasts for far longer than the minimum time. There are eternal inflation models in which it lasts forever.
We don't know how old the universe is. We only know that the current era of (non-inflationary) expansion started about 13.8 billion years ago.


Is T Tauri a compulsory stage for stellar evolution?

Yes, only certain stars that are less than about 3 solar masses were T Tauri stars. Here are some reference and quotes that I saw to answer this question. There are all review papers.

  Our planetary system was formed in early stages of the suns evolution. What the sun was like at that time can be learned by observing young solar-mass stars.

https://link.springer.com/article/10.1023/B:ASYS.0000003267.35552.f7

  T Tauri stars are newly formed low-mass stars that have recently become visible in the optical range,

https://www.annualreviews.org/doi/abs/10.1146/annurev.aa.27.090189.002031

  This review presents the main aspects of the body of knowledge on T Tauri variable stars, which are presumed to be young, gravitationally contracting low-mass stars.

http://iopscience.iop.org/article/10.1088/0034-4885/47/2/001


Are all solutions of the 3-body problem restricted to 2 dimensions?

The five stationary solutions to the three body problem are restricted to a plane. That does not mean that all solutions are restricted to a plane.

Has any work been done on 3-dimensional orbits?

NASA is proposing to build its Lunar Gateway as a vehicle in a "Near-Rectilinear Halo Orbit" (NRHO) about the Moon. These halo orbits only exist in the presence of a third body, in this case, the Earth. The orbit is nearly orthogonal to the Moon's orbit about the Earth.


When will the next summer solstice and a full moon coincide? What day?

The Summer Solstice in 2016 will occur at 22:34 UT on the 20th June, and the full moon will occur at 11:02 UT on that day (but since they are 12 hours apart that won't be on the same day everywhere).

Solstice data from here (GMT- Grenwich Mean Time =UT)

New and full moon data from here (UTC- Coordinated Universal Time =GMT=UT)


Why don't all perseid meteors follow the same trajectory?

They are all following parallel paths. They are not on exactly the same trajectory but they are all travelling parallel to each to other in their orbit around the sun, in a stream of particles.

What you see is then multiple meteors on parallel paths. When you look in the direction that the meteors are coming from you see them appearing to radiate from a point

These meteors are all travelling parallel to each other. It is only perspective that makes them appear to be radiating from a point.

This is similar to the "flying through space" screensaver that used to be popular on windows.


Why do constellation names change (eg from Cancer to Cancri) when used as part of a star's name?

These are the Latin word endings.

Latin nouns change their endings according to how the word is being used. English doesn't have this grammar. But note how the word "I" changes to "my". In Latin you would call "my" the genitive form of "I" The good people at https://latin.stackexchange.com/ can give more explanation.

For example, "Leo" is a third declension noun and its genitive "Of the lion" is Leonis. Third declension nouns have a genitive that ends in -is.  On the other hand "Cancer" is a second declension noun with Genitive "Cancri". A Latin student spends much of their first year learning all these word endings.

When you write "Alpha Cancri" you mean "the alpha star of the Crab", because "of the crab" is cancri.


Is it possible to completely see the edge of a half moon at night?

As uhoh has written in the comments, this phenomenon is known as "planetshine", and in this case more specifically "earthshine". The sunlight hits the earth and is reflected to the moon, illuminating what cannot be directly reached by the sunlight.

https://en.wikipedia.org/wiki/Planetshine
https://en.wikipedia.org/wiki/Earthlight_(astronomy)


What are the challenges in finding Earth-like planets via the radial velocity method?

Assuming a circular orbit of a planet of mass $m_p$ around a star of mass $m_*$, with an orbital period of $P$ and we can assume that $m_* \gg m_p$ for the case of an Earth-like planet.
Newton's second law tells us that
$$ m_p a \frac{4\pi^2}{P^2} = G\frac{m_* m_p}{a^2}\ , \tag*{(1)}$$
where $a$ is the orbital radius.
The observed radial velocity amplitude of the star will be
$$ K \simeq \frac{2\pi a}{P} \left(\frac{m_p}{m_*}\right)\sin i\ , $$
where $i$ is the orbital inclination. Replacing $a$ using eqn. (1)
$$ K \simeq \left( \frac{2\pi G}{Pm_*^2}\right)^{1/3} m_p \sin i\ .$$
Putting in some typical numbers
$$ K = 0.09 \left(\frac{P}{1\ {\rm year}}\right)^{-1/3} \left( \frac{m_*}{M_\odot}\right)^{-2/3}\left(\frac{m_p}{M_{\rm Earth}}\right)\sin i\ {\rm m/s}\ . $$
This equation shows that to detect an Earth-like planet in an Earth-like orbit around a sun-like star we would need to detect a sinusoidal wiggle of amplitude 9 cm/s, with a period of 1 year. Things become more favourable if (i) $m_*$ is lower or (ii) $P$ is shorter.
Indeed Earth-sized (Earth-mass) planets have already been found around very low mass stars (e.g. a planet with $m_p \sin i = 1.35$ Earth masses in a 10-day orbit around the M-dwarf Ross 128; Bonfils et al. 2018).
The state of the art fibre-fed, temperature and pressure-controlled high resolution spectrograph at the moment is known as ESPRESSO at the VLT. It has a demonstrated precision capability of 25 cm/s and an instrument stability that is likely to be at the 10 cm/s level over several years (Pepe et al. 2020). ESPRESSO may be able to start picking up many Earth-mass planets, with orbital periods of up to 1 year around solar-type and less massive stars.
However, just achieving the necessary instrument stability is only part of the problem. The spectra of sun-like and lower-mass stars are not well behaved because of (i) convective motions in their photospheres and (ii) the presence of dark spots and bright plages that are inhomogeneously distributed over their surfaces. Both of these effects can caused velocity "jitter" that are an order of magnitude (or more) larger than 10 cm/s. The two approaches to deal with this are avoidance - where you try to choose targets that are old, quiet and magnetically inactive; or mitigation - where you try to work out what perturbations are being inflicted on your velocity measurements by looking for other signatures of the jitter like asymmetric spectral line profiles or by decorrelating with magnetic activity indicators (e.g. Wright 2017).


How many earths fit in the observable universe?

Without checking the numbers in detail, according to Wikipedia, the volume of the observable universe is about $3.5\cdot 10^{80} \mbox{ m}^3$, and the volume of Earth is about
$1.08321\cdot 10^{21} \mbox{ m}^3$.

By dividing the two volumes we get a factor of $3.2\cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about
320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.


Why does the latitude of an observer affect the hour angle of a body in Stellarium?

I do not have Stellarium, so I cannot confirm what is causing it with 100% certainty. But by using JPL's Horizons website, I can confirm that the position on the Earth's surface changes the apparent Right Ascension (R.A.) and Declination (DEC) of Venus. The following output is for an observer at 80 degrees W longitude and the latitude listed. The time was chosen so that Venus was approximately 7 hours east of the meridian; that is, the hour angle was approximately 7 hours.

Date__(UT)__HR:MN     R.A.__(a-apparent)__DEC
2020-May-17 01:00     05 22 07.26 +27 00 24.9 Equator
2020-May-17 01:00     05 22 09.23 +27 00 01.6 North Pole
2020-May-17 01:00     05 22 09.23 +27 00 48.4 South Pole

These results show exactly what you were noticing in Stellarium:

The R.A. of Venus changes by approximately 2 seconds from the equator to the poles.
The change is the same going to the North or South pole. I thought the values would not be the same because Venus is north of the equator, but the shift is identical in R.A. for those two locations. (Note that the shift in DEC is not the same.)

Observing from the Earth's surface (topocentric position) is important for the Moon if you are looking for occultations or close conjunctions or precise hour angle, but not as important for other objects which are much farther away. Stellarium probably has a setting to change the calculation from topocentric to geocentric, in which case the hour angle discrepancy will disappear.


How many satellites does the Milky Way have?

Starter answer:

Wikipedia's Local Group lists the following items as galaxies, dwarf galaxies, or possibly galaxies. You should refer back to the Wikipedia page, but here is a summary:

Galaxies: Sagittarius Dwarf Galaxy, Large Magellanic Cloud, Small Magellanic Cloud

Dwarf Galaxies: Ursa Minor, Draco, Carina, Sextans, Sculptor, Fornax, Leo I, Leo II, Ursa Major I, Ursa Major II

Disputed Dwarf Galaxy: Canis Major

other: several additional ultra-faint dwarf spheroidal galaxies

However: There is a much longer list of "galaxies" (over 60 items) Wikipedia's Satellite galaxies of the Milky Way

Hopefully another answer will be posted discussing just how galactic these need to be to be called galaxies, and if they are all known to be gravitationally bound to the Milky Way or just in its proximity.


Have we ever seen the formation of a blackhole?

“We think that ‘The Cow’ is the formation of an accreting black hole or neutron star,” said Northwestern’s Raffaella Margutti, who led the research. “We know from theory that black holes and neutron stars form when a star dies, but we’ve never seen them right after they are born. Never.” ... Another team of astronomers, led by Paul Kuin from, an astrophysicist at University College London, came to a different conclusion. They think that the Cow is a star that’s been ripped apart in what’s called a “tidal disruption event.” Information on that here.

There are only 25 Stellar black holes and candidates.

If 'The Cow' was a dwarf star being consumed by a blackhole then the most likely occurrence that has been observed is N6946-BH1.
"Candidates outside our galaxy come from gravitational wave detections. The disappearance of N6946-BH1 following a failed supernova in NGC 6946 may have resulted in the formation of a black hole." The list of other ones is here and for information on the disappearance of N6946-BH1 here.


You seem to be confusing the simple mathematical relationship between radius and surface area, and the more complex relationship between mass and size.

If you double the radius of a sphere the surface area quadruples.  This is pure maths, and is not particular to stars. The volume is multiplied by 8.

But in a star, increasing the amount of matter by a factor of 8 will not result in the volume increasing by a factor of 8. The relationship between the mass of a star and it's volume is not simple and depends on factors like age, and the particular combination of elements in the star.

Doubling the radius does quadruple the surface area.

Doubling, or even increasing 8 fold, the mass does not quadruple the surface area.


It turns out that if you hold the mode key for about five seconds and have the 1244 focuser plugged into the mount then a screen will pop up that allows you to use the arrow keys on the top of the remote to adjust the focus. Once you are done focusing, you simply press mode to exit.


You've confused its rotation and revolution. Venus revolves around the Sun in the same way as all the other planets, but its rotation is in the opposite direction of theirs. It has retrograde rotation, but not revolution. I think this solves your paradox.


Rotational speeds of the material forming planets should increase, not decrease

You need to consider which material the planets are formed of.

Terrestrial planets are formed from less material (way less), and with less momentum (since they were orbiting near from the Sun). Gaseous planets (which is another name for external planets, due to their material), also known as gaseous giants, are formed with a lot more of material with a lot more of angular momentum.

Angular momentum formula is $\vec L=\vec r\times m\vec v$.

Orbital speed formula is $\vec v=\vec\omega\times\vec r$.

Thus: $\vec L=\vec r\times m\vec\omega\times\vec r$

Since all vectors are perpendicular:

$L=r^2 m \omega$

Thus a higher mass $m$ makes a higher momentum $L$, and a higher distance to the Sun $r$ makes a quadratically higher momentum $L$.


Distance parallax and cosmological distance in universe in expansion

Yes, exactly. The angular diameter distance is defined such that it is compatible with "regular" trigonometric rules, just as the luminosity distance is compatible with regular relations between flux, distance, and luminosity.

Only for at flat and non-expanding universe would these distances coincide with each other, and with the "true" distance, which is what you would measure if you lay out meter sticks (after freezing the Universe in time, so it doesn't expand more before you're done).


Why pulsar pulses are so extremely regular?

Pulsars are precise because... in essence, there isn't much to cause them to be imprecise. It's as if you spun a top in intergalactic space and left it there for a hundred years - it would likely still be spinning (at pretty much the same rate).

In essence, pulsars are super-dense neutron stars that are rotating really fast, and their emission axis happens to line up correctly with Earth so that we see it as a pulsating source of radiation. Because there isn't really much to slow their rotation down, it stays relatively constant (though there can be what we call pulsar glitches). However, pulsars do gradually "spin-down", and their period decreases over time (very slowly, mind you), and we can calculate this - we call it the p-dot, or period-derivative - the rate of change of the period, if you will. (As James K pointed out in his comment, this emission of radiation is also part of what causes this spin-down.)


Is it possible that only one hemisphere of a planet has an atmosphere?

That situation would lead to heat transfer between the cold and the hot zones just like it happens here at the Earth.

There is a paper on this that may help you:

  [...] heat transport from the day side to the night side of the
  planets efficiently reduces temperature contrasts [...]

Source: http://arxiv.org/abs/1001.5117


According to Caltech there is possibly a planet 10 times the mass of earth in a 15-20 thousand year orbit at an odd angle. It hasn't been observed yet. There's a search for it ongoing.


Why don't we detect planets around OB stars and no terrestrial planets around A or early F stars?

There are a number of reasons and you will see that my answer is subtly different to HDE226868's.
OBA stars are less common than FGK stars, but they are much brighter and any magnitude limited sample would contain plenty. However they have not been observed as intensively and planets are much more difficult to find around them. And the latter has actually led to the former!
Transits: OBA stars are bigger than the Sun by factors of 2-10. This increases the probability of seeing a transit by a similar factor for a planet of a given size and orbital radius. However, the amplitude of the transit is reduced by the square of that factor. This makes transits harder to detect - more or less impossible in OB stars - and they are harder to validate by observing radial velocity variations (see below).
Radial velocities: OBA stars are 2-30 times the mass of the Sun. For a given planetary mass and orbital radius this decreases the radial velocity amplitude observed in the star as $M_*^{-2/3}$. However, given the precision of current spectrographs this would not prevent the detection of hot Jupiters/Neptune's, or even cool Jupiters. No, the big problem with hot stars is the lack of spectral lines and the broad width of these lines due to the intrinsic rapid rotation of these stars. This prevents achieving the precision necessary to detect exoplanets.
More success has been obtained in identifying planets around the giant stars that evolve from A-stars. That is because as they expand, they cool, spin down and they have many sharp spectral lines. Many exoplanets have been confirmed around giant stars that would have been A stars on the main sequence.


It depends on where we observe it from. There is no absolute velocity reference.

For example, the most distant quasars are moving away from us at nearly the speed of the light. They "see" our Moon going away from them at nearly light speed. So, for them, the Moon travels around a light year each year (its other movements are negligible, compared to that).

If we observe it from the Earth, then we see a Moon on a circular orbit around us. Consider the following well-known facts:

Its orbital radius is $\approx$ 385000 km,
Its orbital time is $\approx$ 29 days,
A year is $\approx$ 365.25 days,

, then we can easily calculate the distance: $385000km \cdot \pi \cdot \frac{365.25}{29} \approx \underline{\underline{3\cdot 10^7 km}}$.

If we observe it from the point of view of the Sun: as we can see detailed in this answer, the orbit of the Moon around the Sun is practically circular: the 400000km orbital radius around the Earth is just a minor perturbation on its $\approx$ 150million km orbit around the Sun. Thus, we can calculate the length of the orbit of the Earth, which is $2 \cdot \pi \cdot 150 000 000 km \approx \underline{\underline{9.4245\cdot 10^8 km}}$.

Well... it is not entirely true that there is no absolute velocity reference: the WMAP probe, measuring the minor perturbations of the cosmic microwave background radiation, found also a relativistic Doppler effect, from which you can read more here. In essence, this is what it measured:

However, the results actually measured were these:

The deviation is due to our galaxy's movement of around 160km/s in the direction of the constellation Hercules, in an reference frame of the Cosmic Microwave Background, that is the closest thing there is to an absolute reference frame.

The answer of your question in this frame: all other movements (of the Sun, the orbit of the Earth around the Sun, and the orbit of the Moon around the Earth) are negligible or cancel themselves out. The result can be gotten by multiplying:

the speed in the frame of the CMB (160km/s)
the seconds of a year ($\approx$ 31.5million)

It is $\approx 160\frac{km}{s} * 31.5\cdot 10^7 s = \underline{\underline{5\cdot 10^{10} km}}$.


How are the newly discovered Janus/Epimetheus rings different from the other rings of Saturn?

http://saturn.jpl.nasa.gov/science/moons/epimetheus/

Scientists believe the ring formation was due to the break up of one moon. Cassini has taken photographs of the moons and Voyager was the satellite to allow scientists to discover the moons and the rings which follow their orbit. What is seemingly special about these rings is the structure.

http://saturn.jpl.nasa.gov/photos/imagedetails/index.cfm?imageId=2277


What makes the coming comet increase/decrease in brightness?

The nucleus of the comet is small and dark. We never see comet nuclei with the naked eye. The visible part of a comet is the coma and sometimes the tail which are made of dust and gas that have vaporised off the comet's nucleus by the heat of the sun.

Now comets are unpredictable.  They get more active, producing more dust as they get nearer the sun, because it gets hotter.  But sometimes there might be a sudden jet (as part of the surface of the comet gives way) and releases lots gas and dust.  Or the comet may develop a "crust" that prevents much dust from escaping. These events can cause a comet to become much brighter or remain dimmer than expected.

Comets are active and dynamic bodies. And so they are intrinsically hard to forecast.  Extreme examples of this are 17P Holmes, which went from magnitude 17 to magnitude 2.8 in two days, probably due to a crust layer cracking.


Do telescopes harm while observing sun through them?

Observing the Sun through a telescope is very dangerous, whatever the telescope you use, if you don't use the appropriate tools.

A telescope a basically a light collector: its purpose is to collect all the light that is arriving on his primary mirror and focus it on a point.

You may have already tried to make the Sun light converge through a little magnifying glass (which works quite similarly to a telescope). If you make it converge on something like paper, it will simply burn it. This can eventually be the cause for big fires (example, a glass bottle in a dry forest, concentrating the Sun's light on leaves). Here is an example to convince you: http://www.youtube.com/watch?v=N7nJ3wIxt3o

A telescope works the same, except that the collecting surface is usually bigger, so it will be more effective at burning ... your eye for example (astronomers make this joke that you can observe the Sun with a telescope twice in your life: once with your left eye, once with your right eye). Or, for bigger telescope, it could even start a fire, or burn instruments...

To observe the Sun, you will need a proper filter that you will put usually "before" the telescope. You may also observe it indirectly on small telescope by projecting the image of the Sun.You can see examples here: http://www.skyandtelescope.com/observing/objects/sun/Observing_the_Sun.html.


Are we looking for brown dwarfs where the dark matter has to be?

The dark matter has to have a roughly spherical distribution and a smooth radial distribution if it is to account for the dynamics of gas and stars in our and other Galaxies.

The exact shape of any required dark matter halo is still the subject of debate and research. Whilst computer simulations of galaxy formation predict mildly non-spherical or triaxial profiles (e.g. Kunting et al. 2018) and some stellar kinematic data appear to confirm sphericity (e.g. Smith et al. 2009 ); others disagree and suggest there is some significant prolateness ( Bowden et al. 2016) or oblateness suggested by large scale surveys (e.g. Loebman et al. 2014).

Any concentration of dark matter into a disk is totally ruled out by observations of the dynamics of stars above and below the plane.
e.g. The latest word, using Gaia DR2 data, is that there can only be about 10% of the local mass density in the form of dark matter of any kind (Buch et al. 2019, agreeing with lots of previous work. For instance this was established fact at least 30 years ago (Kuijken & Gilmore 1989).

The amount of dark matter required at different radii (assuming Newtonian gravity is the correct formulation)  is deduced from the rotation curve. The image below, courtesy of Nick Strobel at www.astronomynotes.com, shows a sketch of the current thinking on how much dark matter needs to be at what radius (labelled as "Corona") and takes the dark matter to be distributed spherically symmetrically.

Dark matter already "outweighs" luminous matter by 2:1 at the solar radius, yet there is room for only 10% of the local density to be in a "dark" form of any kind.

Brown dwarfs are not actually "dark objects" that cannot be found. They emit infrared light and surveys near the Sun reveal that stars outnumber them by 4:1 and that they contribute less than a few percent the local mass density.
They can also be found via gravitational microlensing, in surveys that have been performed towards the Magellanic clouds (sampling the halo) and towards the Galactic bulge (sampling the inner disk). In neither case has evidence been found for huge numbers of brown dwarfs (or cold white dwarfs or black holes) and these have been ruled out as significant contributors to the Galactic dark matter problem (e.g. Pietrzynski 2018). In particular, the numbers of microlensing events towards the Galactic bulge is perfectly consistent with the brown dwarf/star fraction in the local neighborhood (Niikura et al. (2019).

As a final remark, dark matter (whether in the form of brown dwarfs or non-baryonic particles) doesn't clump very easily because it cannot readily dissipate it's kinetic energy. There is little evidence for any clumping of dark matter on scales smaller than dwarf galaxies. In particular no evidence for dark matter is found in globular or open clusters of stars.

The stellar disk forms because the stars formed more-or-less in a disk that had already formed from dissipative baryonic gas that had already collapsed into a disk.  As brown dwarfs must also form from baryonic gas then there are really only two alternatives for their distribution. Either they follow the disk and should be associated with disk stars or they should be very old and associated with halo stars that formed in a spherical gas cloud before it collapsed to form a disk and therefore still have a spherical distribution.

The dark matter needs to be much more extended than the luminous matter of the Galaxy. In neither scenario described above do we end up with tens of trillions of brown dwarfs (for that is what would be required) distributed a long way outside the solar circle and unassociated with any luminous matter.


Can anyone one show how speeds greater than c cause a paradox on a space time diagram?

Suppose there was a magic gun that fired a bullet at ten times the speed of light relative to the firer.

If I have the only such gun, and I don't move then there is no paradox.

But now suppose these guns are generally available.

At time zero, an observer flies past me at velocity 0.5 c and I give them one of these guns. 7 years later, I fire the gun after them, so in my frame of reference the bullet will catch them at about time  7.36 years when they are about 3.68 ly away.

To get these numbers I just need to solve $10*(t-7) = 0.5*t$ to calculate when the bullet overtakes the spaceship.

In their frame of reference though, that occurs about
time 6.38 years after they pass me, at which point, I am about 3.19 ly away from them (we both agree on our relative velocity).

This is calculated using the Lorentz transformations. $$t' = \gamma(t - vx)$$ taking $c = 1$. Here $t = 7.38$, $\gamma = \frac{1}{\sqrt{1-v^2}} \sim 1.15$ $v = 0.5$ and $x = 3.68$.

They now fire back and we can do a similar calculation, but in their frame of reference. So in their frame the bullet will pass me at time $t_1$ when
$(t_1 - 6.38)*10 = 0.5 t_1$ which tells us that $t_1 = 6.71$.

Now we use the Lorentz transformation again to determine what time that is in our frame. We find that it is 5.81 years. A bit more than a year before we fired our bullet.

The key thing is that $n$ times the speed of light in one frame will be backwards in time in another frame.

The diagrams below illustrates this scenario.

Both diagrams show exactly the same events. The one on the left uses the time and space coordinates of the original observer ("you") the one on the right uses the coordinates of the other observer. In bith cases time is on the y axis, space on the x axis. Units are years and light years.

The black diagonal lines are some possible world lines of light rays.
The red lines are lines of constant x (solid) or t (dashed) as determined in "your" frame of reference. The thick red arrow (at x = 0) is your world-line.

The blue lines are lines of constant x (solid) or t (dashed) as determined by the ship. The thick blue arrow is the ship's world-line. As can be seen you meet at the event that you both agree is x = 0, t= 0.

The green line is the bullet that "you" fire. It crosses three solid red lines and goes a bit less than half way from one dashed red line to the next (travels 3+ ly in one 0.36 y). The cyan line is the return bullet. It crosses 3 solid blue lines and goes rather less than half way from one dashed blue line to the next (travels 3+ ly in about 0.3y).

Assuming I've got this right, all the intersections and crossing patterns of red, blue and black lines, and the four world-lines should agree between the two diagrams, since they really do depict the same events.  In particular both diagrams show a causal loop, in which the cyan bullet arrives back at the red observer before the green bullet is fired. They disagree about which bullet travels back in time, but that's relativity for you.

What this tells us practically is that any physics which allows FTL travel but not time travel must not be Lorentz invariant, so your gun will work differently in different frames of reference. We know of no such physics at present, and see no clues which suggest it might exist.


Observing the night sky with a 6-year old

Your issues seem to be rather in line with what one would expect from a 6-year old getting in touch with observational astronomy. Actually, adults can sometimes experience these difficulties too when they first get in touch with it. I'm speaking out of experience of having shown objects of the night sky through a telescope to visitors. Some people just don't understand that when they touch the telescope, it moves. I would recommend the following guidelines to "fix" your proposed problems (expanding on hartacus' answer, which appeared while I was writing this):

Someone else could keep her busy while you set up the telescope. Since it is computerized, you won't waste time looking for objects and the only downtime will be due to the moving of the telescope. You can minimise this by planning your observation ahead of time to minimise the distance between two following objects.
As for the eyepiece, just tell her how to use it. She'll learn it in time. Even adults who have never used a telescope before generally need to be told how to look through it.
Most telescopes have some mechanism to lock the primary axes of the telescope mount. You can use this to make sure the telescope always points to the same spot. If you find that it is the tripod that is moving, you could try to find a way to lock that as well. Maybe surround its legs with loose bricks?
To minimise the last two difficulties at the same time, the best position to look through a telescope is to have both hands behind your back, move one eye as close to the eyepiece as possible without touching it, and close your other eye.

As a final remark on looking through the telescope, she might not be seeing the same thing as you when she looks through the telescope, because her eyes are different from you. I don't know whether this is applicable, but most telescopes have a way to adjust their focus to make the image as sharp as possible. People with near - or farsightedness generally require the focus of a telescope to be different.

These, however, are difficulties you will definitely overcome in time. When these are fixed, the most important part of the experience will be in the objects you show her. As hartacus mentioned, show her things she can relate to (planets are great), and tell her about the things she is looking at.
Apart from planets, star clusters can be impressive too. She'll surely be astonised if you tell her how many stars there are in one cluster. Some open clusters can even be thought of as 'micro-constellations' (more specifically, this one).
There are some nebulae / galaxies that are nice to watch, but I would restrain from them in the beginning because most will appear as blurry grey spots through an average telescope, which is not that impressive.

One last thing: you mentioned a binoculars. You can already see a lot with binoculars, although most impressive objects would be too faint. The main issue with the binoculars is that you can't aim them for her (unless you have a mount for them, which exists for some). If you can teach her to aim them herself however, there are some spots in the night sky which have 'mini-constellations': sets of stars in which you can recognize images, but they are too small to see with the naked eye, and too big to see with a telescope.


Can someone please explain it to me whether the M-theory can be used
  to explain what is happening in the centre of the black hole where the
  singularity exists?

In Layman astronomy - yes, M-Theory (and probably a few string theories) do explain in detail what happens at the singularity, but it's a mathematical construct, not a testable hypothesis, so what it tells you might be wrong and little more than creative but irrelevant mathematics.

Looking inside a black hole, well, that's not a good idea either.  Even if you were able to see what happened, you wouldn't be able to send a message back telling anyone, so that's also no help.

Quantum physics is all a model anyway, but it's a good model because it provides accurate answers and it predicted the existence of particles that were later discovered.   Relativity is a good model too that stood up to testing.   Good models are useful and highly accurate and provide answers.

M-theory is a mathematical construct.  That's a different thing entirely.   It could still end up being right, but until it's testable, it's really just a mathematical exercise.   They really shouldn't call it a theory, it should be M-Hypothesis and String-Hypothesis.  They aren't real theories.

If I may quote Shakespeare from Henry IV, just for fun:

  GLENDOWER

I can call spirits from the vasty deep.

  HOTSPUR

Why, so can I, or so can any man;
But will they come when you do call for them?

I'm personally much more comfortable with the "both accepted theories, relativity and quantum theory, break down at the singularity, so we don't know what happens".

If you prefer that M-Theory says what happens, preferring that answer is entirely optional, but I think the majority of people in this field are more comfortable with the we don't know answer.


Does a meteor's direction change between country or latitude?

If you see the same meteor, it of course only has one path - even while the apparent movement differs for different observers. What you basically do in that case is triangulation - which you could use with your known geographic location (and time for the observations, and the direction horizontally and vertically) reconstruct the flight path. That's a method which actually is used to identify meteors and and tracking them, e.g. via the fireball network.

Is it possible that you saw the meteor given your geographic separation?
(note 11 March: This paragraph is heavily revised and comes to the opposite conclusion with the right trigonometry)
Given the mentioned separation between Norway and Spain, we talk of a few thousand kilometres. The typical height at which you see a meteor is between 50 and 100km. The curvature of Earth then makes sure you don't see the same meteor when you are too far apart - but it might be still just right here:
Madrid and Oslo are about 2000km apart, the Earth's circumference is about 40000km. The earth's radius is $r_E = 6371$km which allows us to calculate the angle $\alpha$ between those as seen from Earth's centre: $\alpha = 0.31$ (radians). With that angle we can calculate the height an object must be at when we look horizontally and just see it at the horizon (laying on the floor makes it easy to calculate), given it's separated by the angle $\alpha$:

$$
d_s = r_E\cdot tan\alpha
$$
and
$$
\frac{h+r_e}{\sin 90°} = \frac{d_s}{\sin\alpha} = \frac{r_e\tan\alpha}{\sin\alpha} = \frac{r_E}{\cos\alpha} = r_E \sec\alpha
$$
so that
$$
h = \frac{r_E}{\cos\alpha}-r_E
$$
Now if you enter the angle between Oslo and Madrid you get a height of about 79km for an object passing half-way (thus at half the angle to each). So it might have been that very same meteor if it was very close to the local horizon. Given that this estimation is pessimistic as it ignores observer height, things get a bit better if you assume that you stand on your feet, on a building or even a mountain. Either way, the meteor will have been close to the horizon for both of you, and very likely for one travelling left to right, for the other observer right to left.


Pair together light curve and radial velocity data of specified star

An excellent page to get most or all names for a star is Simbad which also happens to know the Kepler IDs (KID).

Ont the other hand, every of the RV data files contains info on the star's identifier it belongs to in the header, e.g:

  \STAR_ID='HD 4628'

So using Simbad you should be able to do a cross-matching. SimBad allows scripted query and includes all alias names for stars; thus query for one known star name and look for identifiers found in the list of RV-observed stars or vice versa. Simbad allows to be queried via scripts, so you can make a short programme to do this task for you.

However I'm not 100% convinced that you necessarily find many matches between those two data sets; it's easier to do photometry than spectroscopy.


Do all the stars visible to the naked eye belong to Milky Way?

All stars visible individually to the naked eye belong to the Milky Way. A handful of galaxies are visible to the naked eye, but here you only see the combined light of millions to billions of stars.

The only exception is when a massive star ends its life, exploding as a supernova. This can be visible to the naked eye for a brief perdiod of time. Most notably, the gamma-ray burst GRB 080319B was visible to the naked eye for roughly 30 seconds in 2008, despite its distance from us being $\sim10^5$ times the size of the Milky Way.