How do I find out when a constellation reached its zenith?

The key number you need to find is the "right ascension" This tell you the "longitude" of the constellation, relative to the other stars in the sky.  It is measured in hours and minutes
Bootes has a right ascention of about 14:30  This means that on the day of the spring equinox (March 21) it will culminate at 02:30  (in local solar time, you'll need to make adjustments for daylight saving time or if your local noon isn't at 12:00) Note that 14:30 is 2:30 pm
Every month it will culminate two hours earlier So on April 21 it will culminate at about 00:30  (conveniently there are 12 months and 24 hours)
Working the other way, On January 21st it would culminate four hours later, or at about 06:30, and now, in Early February it culminates at about 06:00
This kind of calculation is good enough for a rough estimate.  More accurate calculations can be done, but are more convenient to get a computer to do them.


This is a tough question to answer because the dimensions of your proverbial cube would be affected by the mass inside (as dimensions can only exist in space, so anything that affects space will also bend your cube), and mass does not bend space, it bends spacetime. You have to keep in mind that when talking about special and general relativity, you are talking about four-dimensional spacetime. This is what is curved by mass. For example, due to the curvature of spacetime due to Earth's gravity, you feel the same effects on its surface that you would if you were accelerating upwards at 9.8 m/s^2 through deep space (where we assume there is no gravity). This doesn't mean the physical space is curved - if you draw a straight line on a piece of paper in space and then travel down to Earth's surface, it will remain straight. Instead, spacetime is curved, which can affect objects' paths of motion, but not the objects themselves. Now to your questions:

The volume inside of the cubes would be exactly the same, but two identical objects flying through each cube would follow two different paths, and if either looked at the other, it would see that the other's clock is moving at a different speed than its own due to time dilation (due to the curvature of spacetime towards mass, time moves more slowly closer to massive objects). We can see evidence of the effect of gravity on paths with gravitational lensing. Light travels in a straight line through spacetime, but when looking at the space around a star, you will actually see objects that should be directly behind the star and blocked from sight. This is because light rays that, in the absence of gravity, would pass next to the star and continue off at an angle are pulled towards the star by it's gravitational field (in reality they're just following the curvature of spacetime around the massive star), so they curve around the star, becoming visible to us, even though, in space alone, they should be blocked by the star.
The distribution of the mass has no effect on the gravitational field around it (same with electromagnetic fields). If you look at any equation that has to do with gravity (gravitational potential energy, gravitational force, etc.), you will find a term for the total mass of the object, but unless you are within the bounds of the control volume containing the mass, the distribution does not matter. This is one of the reasons why black holes are so hard to study - we can tell how much mass is inside of them, but without the ability to see inside, we have no other way of detecting their properties.
Like I said above, the volume is not affected because space is not warped by gravity, only spacetime. If you made a physical box (or we'll say a cube frame) around empty space and then moved it to a space that contained a star, it would remain the same size and shape (assuming that it was strong enough to withstand the force of gravity pulling it towards the massive star).
Neither, it simply curves spacetime.


Reason for disqualifying Pluto as a Planet?

Pluto:

It was disqualified as a planet because orbital dominance was not achieved in the case of Pluto. Orbital dominance means that a the planet candidate should have remove all the small bodies from its orbit, by impact, capture of gravitational disturbance.

Planet:

According to the IAU in its resolution B5 (IAU is the International Astronomical Union; in particular it is in charge of naming celestrial bodies), a planet is a celestial body:

in orbit around the Sun
with a sufficient mass so that self-gravity overcome rigid body forces (that assumes a hydrostatic equilibrium)
has cleared its neighbourhood.

Exoplanets:

As you see in the resolution B5, this definition is specified for "Planet in the Solar System". It is mainly because it does not really make sense now that it is already challenging to observe planets, so the distinction between a planet and a dwarf planet is not yet crucial for exoplanets. So the official definition of an exoplanet by IAU makes mainly a difference between a planet and a brown dwarf.

However, I guess the IAU definition of a planet should be kept for the other planetary systems if we are able to detect objects such as dwarf planets.


Can Astronauts Aboard ISS See the Shadow of The Moon on The Earth During an Eclipse?

Yes. Here's an ISS photo from the 2002 eclipse over the Indian ocean.
This year, 2017, the ISS will see the moon's shadow on earth once, over southern Illinois, at 43.9% totality.


To add to Rob's answer, I wanted to expand on where this naming convention comes from.

The International Astronomical Union (IAU) is the organization which generally sets conventions and definitions. They're the ones who demoted Pluto to being a dwarf planet in 2006. Anyway, before any exoplanets were found, there existed a convention for naming multiple-star systems. The rule was that you gave the system a name, for example Alpha Centauri, and then the brightest star in that system is designated by "A". Any other objects are then given the letters "B", "C", etc. The Alpha Centauri system has three stars in it and so they're named Alpha Centauri A (The brightest of the three), Alpha Centauri B, and Alpha Centauri C (a.k.a. Proxima Centauri).

This convention then continued on to naming exoplanets. The star, according to this convention is "A", whereas all the exoplanets are given the letters "b" and onwards. Note that generally the exoplanet letters are lowercase, whereas they'd be uppercase for stars. Also note that if you have a star (say TRAPPIST-1) around which planets are found, the star's name doesn't then become TRAPPIST-1 A. The "A" is implicit.

Just to point out another point of interest with this naming convention, the letters of the planets don't always signify their distance from the star. That is, the "b" planet isn't necessarily the closest, followed by "c", etc. The planets are labeled at their time of discovery. Someone could find a planet, label it as "b" and then later find a second, closer planet and label it "c". It just so happens that the lettering of TRAPPIST 1 is in order because the first three were found at the same time and labeled according to distance, then the last 4 were found at the same time and labelled according to distance.

You can read more about the full set of rules concerning this on Wikipedia's exoplanet naming convention page. There are a lot of caveats such as what to do when your planet is orbiting a binary system.


How do astronomer measures the size of any celestial objects?

The main tool to measure the diameter of a star is interferometry combined with a parallax-based distance measurement - a brief review by Kervella (2008) might be useful. The principles behind interferometry are described here.

Interferometry involves measuring the light from a star using two (or more) telescopes that are separated by some distance. Together, the signals from these telescopes can be combined to give an angular resolution that can be (in the best circumstances) equivalent to a telescope with a diameter equal to the telescope separation. These measurements give the angular size of the star, which must then be multiplied by their distances to get a physical diameter.

One of the most successful experiments is the Chara array, which has yielded diameters for many nearby stars. Precisions can be as good as a few percent, but more usually 10% and of order 100 (predominantly nearby) stars have had their radii measured in this way.

A second main direct technique is to use eclipsing binary systems. The measured light curve can be used in an almost model-independent way to estimate the radii of the two stars involved. Of course most eclipsing binaries are close pairs with short orbital periods and with orbital inclinations that allow us to see the eclipse. They are therefore highly prized objects. Radii can be measured with precisions of 1%. A reasonably complete catalogue of the $\sim 100$ known eclipsing binaries with precise radii can be found here.

Another technique is lunar occultation. The passage of a star behind the limb of the moon results in a changing diffraction pattern that can be used to estimate the angular size of the star. Again a distance is required to convert this into an actual diameter.

More distant stars are inaccessible - their angular diameters are simply too small. At the moment only indirect estimates of their radii are possible. For example, if we were to assume that a star radiates as a blackbody, then its luminosity ($L$), radius ($R$) and temperature ($T$) are related by Stefan's law.
$$ L = 4\pi R^2 \sigma T^4,$$
where $\sigma$ is the Stefan constant. If the star has a measured flux at the Earth and we know how far away it is, then $L$ can be estimated. If we take a spectrum and estimate its temperature, then the equation above can be rearranged to give the radius in terms of the measured luminosity and temperature. Real stars are more complicated than blackbodies, but the principle is the same.

Neither of the above techniques can work for black holes, and the sizes (event horizon or Schwarzschild radius) of black holes have not yet been directly measured. The physics of a black hole is relatively simple(!) and so there is a direct relationship between their Schwarzschild radii and their masses (modified somewhat by rotation). Basically it is 3 km multiplied by the mass in solar units. Therefore a measurement of the black hole mass gives its "radius". The masses of black holes are measured by looking at the motions of stars and gas around them and applying our knowledge of how gravity works.


Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers?

Dimensionally speaking, the luminosity of a gravitationally radiating binary system, consisting of two objects of mass $M$, separated by $R$, goes as $(M/R)^5$.  The timescale of the chirp for such a system goes as $M^{-3} R^4$. (Schutz 1999).
Thus the total energy released goes as $M^2/R$, i.e. it is proportional to the gravitational potential energy of the system.
Because of the $R^{-1}$ dependence, it is basically the mass and radius of the "final" state that determines the energy lost. For the black hole case, the final mass is just less than $2M$ and the final configuration event horizon is $4M$ (with $G=c=1$). So $M^2/R = M$. Thus I would expect the mass-energy released in gravitational waves to be a fixed fraction of the combined mass of the black holes (note that unequal mass black holes will lead to complications).
Looking at the data for the list of mergers this model seems reasonable, with the fixed fraction being about 5%.
Extending this to neutron stars, well the "final" radius is going to depend on the physics of the neutron star material and so will be model dependent. However, that radius will be $>4M$ (i.e. probably several times the Schwarzschild radius). Another way of saying this, is that neutron stars can't get as close together before the merger takes place.
So from that point of view I would expect $<$5% of the combined mass-energy is radiated as gravitational waves.
Observationally, there is no accurate estimate of the final remnant mass for GW170817.


Could the Moon naturally capture a large asteroid and keep it as a subsatellite? For how long?

I'm going to discuss only our moon, to keep this on topic.

This sort of orbit could be stable on the sort of timescales that you want.

Objects in Low Lunar Orbit will likely suffer from the irregular gravity of the moon. There are "mass concentrations" as a result of ancient meteorite impacts. These "mascons" distort the orbits of objects in low orbit, and cause them to collide with the moon in a relatively short time. However at 24000km the gravity field is smoothed by your distance from the surface. You should not have problems with mascons at this distance, at least not over 100,000 years or so.

The Hill sphere of the moon extends to about 60000km, so your orbit is well within the Hill radius, and so can be stable.

The moon spins slowly, so tidal effects would tend to cause your asteroid to lose altitude, and eventually hit the moon. The timescale of this would be longer than a few 100,000 years

Getting your asteroid in orbit is very tricky. You would need to hypothesise double asteroids transferring momentum between each other and the moon, and one being lost while the other ends up in lunar orbit.


How to interpret this old degree notation?

This doesn't really answer your question, but I read https://github.com/profound-labs/calculating-the-uposatha-moondays/blob/master/references/Rules%20For%20Interpolation%20in%20the%20Thai%20Calendar.pdf and noted that footnote 13 (which is referenced by equation 8, although it sort of looks like 3^13) reads:

The routine subtraction of 3 arcmins is a causing Ashadha to be month
10; but by this Wat geographical longitude correction for the sun, as
is the Tapotharam reckoning Vaisakha is month 7 (Keng subtraction of
40 arcmins for the moon (sec. C13). Tung style), causing Ashadha to be
month 9.

which suggests the calculation is being done in arcminutes, not degrees.

Unfortunately, this doesn't help, since the result is 76.5283 arcminutes, which is nowhere near 2 degrees 19 minutes 28 seconds (which is 139.467 arcminutes).

134*sin(0:32) may refer to 32 seconds of arc, and the result is 0.02 which is 1 minute or arc (0:01). Terrible use of notation here if that actually is the case.


Relationship between Color Index and Temperature

They are mostly empirical. Found by measuring the $B-V$ for stars of known $T_{\rm eff}$ (which are in turn measured by knowing the luminosity and radius of a star, and this is only known for a small number of stars). The relationships also depend on stellar surface gravity and composition.

An alternative approach is to derive "synthetic" relationships by folding spectra calculated from model atmospheres through the appropriate filter responses.


Why does gravity increase in star formation?

The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.

This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.

Radiation pressure is a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.


Where will I be able to see the Lyrid meteor shower?

You should definitely be able to see this meteor shower. It will be in the north east from your viewpoint near the star vega.

http://www.astronomy.co.uk/skymap

http://earthsky.org/astronomy-essentials/earthskys-meteor-shower-guide

You may find these websites useful. If you're interested in amateur astronomy, there is free software for interactive star maps. I use one called stellarium. I'm also an amateur astronomer and I've learned so much with it. You may enjoy it as well.


The idea is to create a specific diffraction pattern so that you know if you image is in focus.  The pattern is so that you end up with 3 lines crossing the source.

The mask consists of three separate grids, positioned in such a way that the grids produce three angled diffraction spikes at the focal plane of the instrument for each bright image element (star). As the instrument's focus is changed the central spike appears to move from one side of the star to the other. In reality, all three spikes move but the central spike moves in the opposite direction to the two spikes forming the 'X'. Optimum focus is achieved when the middle spike is centered on the star and symmetrically positioned between the other two spikes. Small deviations from optimal focus are easily visible.

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

You can construct a mask with some materials:

http://www.deepskywatch.com/Articles/make-bahtinov-mask.html

There is also a Hartmann Mask that splits the object into multiple when not in focus.  This seems a little easier to construct.

http://www.cloudynights.com/item.php?item_id=518


What does G/IG signify in the galaxy names ESO 012- G 014 and ESO 293-IG 034?

These identifiers correspond to the ESO/Uppsala Survey of the ESO(B) Atlas (1982). The identifier column (ESO) is associated with the following note:

Note (1)  : ESO designation is written FFF-oooNNN, where
FFF = Field (plate) number, see file "plates.dat"
ooo = Object class with
* = star ** = double or multiple star
SC = star cluster N = nebula
A = asteroid C = comet
G = galaxy IG = interacting galaxy
EN = emission nebula PN = planetary nebula
RN = reflection nebula SNR = supernova remnant
? = questionable object
NNN = number within field

So the "G" means "galaxy" and "IG" means "interacting galaxy".


I'd like to add onto what James K said, but I'll try and simplify things too.

A singularity may sound like everything was at a "single point", but that's not what the word means in physics/math. A singularity, loosely speaking, is just a point where a function reaches infinity. It's not a physical object, but an input to a function. As James mentioned, 1/x has a singularity at x=0. At the moment of the Big Bang, several functions have singularities, as I discuss below.

So with that set aside, let's explain the Big Bang. Our modern theory of gravity, general relativity, indicates that the universe is currently expanding (i.e. space is being created between all matter). By that logic, we expect the matter in the universe to be closer together as we look further back in time. Our physical theories indicate that at earlier periods, temperatures were so high that even atoms couldn't form.

Our theories can handle these conditions pretty well, but eventually they start to fall apart. At a point around 13.8 billion years in the past, our calculations predict that the amount of space in the Universe should be zero. As a result, density, gravity and temperature would skyrocket to infinity — in other words, they have singularities at that moment. This moment is known as the Big Bang.

These infinities are a problem, since our theories break down at such extreme conditions. The Big Bang is an unsolved problem that modern physics has yet to explain. It's not the beginning of the universe per se, but rather we do not know what happened around, or before, that moment.


What's the distribution curve for meteor sightings?

The meteor sightings in a meteor shower can be approximated to a normal distribution. In fact, research papers use a normal distribution to model meteor showers.

The exact characteristics of the shower will depend on the location, local weather conditions etc. The number of sightings and the number of actual meteors impacting the earth's atmospheric conditions are different as it is the local conditions that determine the number of sightings.

The International Meteor Organisation maintains data on the number of meteor sightings which shows that the sightings can be approximated to a normal distribution.

However, this data has to be used carefully because these refer to the reported sightings and the errors are significant.


The other thing(s) involved are:

Contraction and the release of gravitational potential energy. Gas giants are born bigger than they are now. They gradually contract towards a completely degenerate configuration, and as they do so half the released potential energy is radiated away, whilst the other half heats the planet.
Gravitational diffusion and separation also releases gravitational potential energy. Heavier elements tend to settle towards the centre, falling through the majority hydrogen.
Radioactive decay of heavy elements that may form part of the giant planet's core.
If the "planet" is more than about 13 times the mass of Jupiter, then it will likely attain a central temperature sufficient to fuse its deuterium. Some would refer to such objects as brown dwarfs at that point.


Are there celestial bodies that are about as far away as the Sun?

Asteroids with an orbit of between 1.8 and 2 Au (So their closest approach is is about the same distance to the Earth as the Earth is to the sun) are called Hungaria group asteroids, named after 434 Hungaria.

Hungaria itself has an orbit of 1.94 AU, so would be a good example of body. Such asteroids are rather rare. There is a gap in the asteroid belt caused by Jupiter at 2.06 AU, and tighter orbits tend to be disrupted by Mars.

Of course, at any given time, the actual distance to Hungaria will vary between about 1 and 3 AU


In Orion, how close in local distance are the closest of the member stars?

The three stars in Orion's belt, along with Sigma Orionis and the Orion Nebula cluster, along with several other young clusters and star forming regions in the central Orion region are all part of the Orion OB1 association at a distance of 350-420 pc.

A review of Orion OB1 can be found in Bally (2008). The three belt stars are probably a similar age and distance and form the brightest stars of the OB1b sub-association. Measured distances to these stars are individually rather uncertain, but better estimates come from the low-mass, co-moving stars around them. It is therefore true to say that the individual distances to each star are not known well enough to be sure that they are close together in distance.

The other bright stars in Orion - Betelgeuse, Rigel, Bellatrix, Saiph are significantly closer than this by 150-300 pc and probably not at a common distance. It has been speculated in the past though that these stars and the Orion OB1 association are part of one, large star forming complex, extending 200-300 pc along the line of sight (Bouy & Alves 2015).

The recent paper, claiming the discovery of a "Radcliffe wave" gas structure in the solar vicinity, does include the young stars in Orion as part of this structure.

So to spell it out: The belt stars are within 70pc of each other in terms of distance. If they are at a similar distance of 400pc, their angular separation on the sky of around 3 degrees, translates to a tangential distance separation of about 20 pc. Other members are more separated than this.


Why does the closest approach of star S2 to Sgr A* not appear to be near the focus of its elliptical orbit?

The orbital elements are on wikipedia:

$$e=0.884\ a=0.125'',\ i=134^\circ,\, \Omega=228^\circ$$

(At an assumed distance of 8kpc, $0.125'' = 1000au$)

It is the inclination that means that the black hole is not at the focus of the projected ellipse.

Imagine a circular orbit, with a central body.  If viewed from a distant but high inclination, the orbit is projected as an ellipse, with the central body at the centre of the ellipse (not the focus)

A distant projection of an ellipse is another ellipse, but the projection of the focus is not the focus of the projection.

Below you can see another example.  The ellipse is in the grey plane, with foci at A and B.  But the apparent ellipse has a completely different axis, and the focus is nowhere near the axis.  These image have been generate with geogebra and no perspective has been used.


I have found the answer. I just found that venus has an axial tilt of nearly 180 degrees, so I realised that the >90 degree tilts are due to the fact that they are the planets with retrograde rotation, which is equivalent to an 'upside down' (>90 degrees) counter clockwise, normal, rotation.


What are the exact causes that differences in sunrises are shorter around june and december but long in march and september?

If I understand you correctly, you ask why does the length of the day not change much around the winter solstice.

There is no mystery here: The rate of change of day length has (almost) nothing to do with the elliptical orbit of the Earth.

In autumn the days get shorter; the rate of change of day length is negative. In spring the days get longer; the rate of change is positive.
At midwinter solstice the rate of change of day length is zero: The length of the day doesn't change.  Exactly the same is true in midsummer. Mathematicians would say that the "day length function" is stationary on those dates.

This has nothing to do with the elliptical orbit. Every smooth function is stationary at its maxima and minima. The reason is not astronomical but mathematical.

Note that the Earth is furthest from the sun (and therefore moving slowest) on July 4th.


Theoretically could the expansion of the universe affect acceleration due to gravity?

Right now the expansion of space is only observed on scales like that of galaxies and the entire universe. We say that with the fancy term globally. Expansion of space does not take place within galaxies at least for now.
As pela has rightfully noted in the comments, for now the gravitation of galaxies, stars etc. is strong enough to overcome the expansion of the universe. But as General Relativity tells us gravity is local, that is, it only affects objects that are considerably close. So, on large scales expansion continues. (I am no expert, but I think that the reason for expansion not happening within galaxies out smaller systems is that the matter and energy is more dense, resolution by in stronger gravitational fields).
It is hypothesized that the universe may end with a big rip: this means that the expansion eventually would start to affect galaxies. Galaxies, star systems, planets and eventually atoms will be ripped apart due to the expansion of space. When and if that happens, not only gravity but not even electromagnetism or the strong force would work to overcome the mighty expansion.
But this situation is highly hypothetical and maybe it won't even happen. So for now, the answer is no: the expansion of the universe could not affect gravitational acceleration.


Understanding Earth Tilt, Sun's Position and Lattitude Calculation

I've been trying to figure out how to calculate the latitude of my
  location with just the shadow of a stick cast by the sun. I am doing
  this under the assumption that I am stranded in some place with no
  almanac or any other data. (not even day or month).

You really cannot find your latitude with only the sun and sticks at your disposal and no declination data or time data. Maybe if you could record your shadow over a period of a day, you could tell if you are in the northern or southern hemisphere by observing which way the shadow moves. If it moves clockwise, you are in northern hemisphere.

And if you could repeat and observe for 365 days (from the day you are stranded, assuming you are very accurately measuring the angles) you could find if you are within the tropic of cancer or above it, as you might not have the sun right on top of your stick over the year if you are above the 23.5 N parallel.

And that's as far as you can get. To determine the precise location, you will need the sun's declination at any given time. May be you will get lucky, if the day you are measuring the angle and the equinox matches, but you wouldn't know it.

  Could you help me understanding the concept and give me a definitive
  picture of angles that can be calculated.

A picture is worth thousand words. And an interactive demonstration... It's worth at least a hundred 2D diagrams.

Additional Interactive media - http://astro.unl.edu/naap/motion1/animations/seasons_ecliptic.html


How to derive the depth in my image (limit Magnitude )?

It's the magnitude where the average signal-to-noise ratio (I assume that's what j_snr represents) is 3.


Was the progress of astronomy in the 1800s surprisingly slow, and if so why?

I feel this question is so broad, it might well be the topic of an interesting book on the history of astronomy, should someone be inclined to write it. :)

Anyway, I think a few points could be made briefly.

1. Collecting data

In astronomy, that means observing the cosmos. That means using an instrument of some kind, typically a telescope, and gathering information through it. Telescope performance is dictated by many factors, but the most important one is size (or aperture).

Telescope size grew rapidly through 1600s and 1700s, from Galileo's 1.5 cm refractor in early 1600s, surpassing the 1 meter aperture in early 1800s - Herschel's 40 foot reflector. There was a steady stream of improvements regularly throughout that 200 year period. One could say that the first golden age of the telescope aperture race both culminated and ended with Herschel and his giant telescopes.

Then there was a lull, briefly interrupted by Lord Rosse's 1.83 meter telescope, the Leviathan of Parsonstown, in mid-1800s. Then nothing again.

The aperture race was resumed only in early 1900s, with the 2.5 meter reflector on Mt. Wilson, the Hooker telescope. Afterwards, throughout the 20th century, and now in early 21st, the race is going strong, with the 10.4 meter Gran Canarias segmented reflector currently in the lead, and the 39 meter E-ELT reflector being under construction at Cerro Armazones.

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

2. Interpreting data

The year 1900 marks the boundary between classic physics and the new physics. After that year, relativity and quantum mechanics took off. This is what enabled the new cosmology to emerge in the 20th century.

In other words, with the 1800s science, even with tons of data, there would have been no way to figure out, basically, everything. Supernovas? The expansion of the universe? Dark matter and the rotation of galaxies? This is all based on 20th century physics. 19th century physics would have been clueless.

Astronomy used classic physics to derive interpretations from data pretty quickly, and that process had achieved great success already well into the 1700s. That's when the structure of the solar system was figured out, way back to Kepler in the 1600s. Herschel found Uranus in late 1700s.

There are some exceptions here. Stellar parallax was detected in early 1800s, which enabled an estimate to nearest stars. Spectroscopy showed that distant stars are made of the same elements like the Earth in the 1850s. Around that same time, Neptune was discovered.

So the 1800s was not quite a completely dry period, in terms of theoretical progress.

In any case, a limit was reached anyway in late 1800s, because what was needed was new paradigms in physics to give new life to the interpretation process. That boost occurred after 1900, with relativity and quantum mechanics.

Cosmology is highly dependent on physics (and vice-versa).


How would one figure out the rate of axial precession of a planet?

Wikipedia's page on Axial Precession has a good deal of mathematics on it and, unless you have absolutely got to have precise numbers for some obscure story reason (which is probably what I'd call the tail wagging the dog - change the story to avoid that problem) then making up the numbers or just forgetting all about Axial Precession is the way to go.    The periods involved will be very long by the standards of any story you're likely to write, so why burden yourself with something you probably don;t need.

The maths that follows only gives a rough approximation anyway, and you don't want to even contemplate the kind of things you need to do to get a better one : it's not worth it.

That said let's have a look at the very basic theory result that Wikipedia gives :

There are two component to axial precession that matter (for Earth) : the one due to the Moon and the one due to the Sun.  The Moon's effect is actually larger, but these numbers are very sensitive to the values you use.

The Solar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_s}{a_s^3\left(1-e_s^2\right)^\frac 3 2} \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Lots of symbols so what do they mean ?

$G$ - the Universal Gravitational Constant also famous from $F = \frac {GM_1M_2} {r^2}$ Newtons law for gravitation.
$M_s$ - The Sun's mass - in your case you need the mass of your planet's star, of course.
$a_s$ - The semi-major axis of the orbit of the planet around it's star
$e_s$ - The eccentricity of the planet's orbit around it's star.

Now that second term in square bracket, which is also in the expression for the Lunar contribution.  This one is trickier.

$C$ - moment of inertia (of Earth) around the axis of rotation
$A$ - moment of inertia around the equator
$\epsilon$ - the angle between the equatorial plane and the ecliptic plane (see below)
$\omega$ - Earth's angular velocity (due to it's rotation, not it's orbit)

Now this expression is really poorly dealt with in Wikipedia because of two problems.

$epsilon$ in Wikipedia is assumed to be the same for both the Solar and Lunar contributions.  This is not (AFAIK) correct.  The angle should be the angle between axis of rotation of the body and the plane of the orbit of the other body (which means it's different for the Sun and Moon).

The $C-A$ and $A$ terms are really hard to deal with for mere mortals (and frankly just messy for anyone else).  For your purposes I would propose the following compromise term instead.  It's based on modeling the planetary bulge as an ellipsoid of constant density compared with the $A$ value for an ideal sphere - both objects have the same mass and density, which I'm taking as constant .  I'll spare you the derivation :

$$\frac {C-A} C \approx 1 - \frac{R^2} {a^2}$$

where in this case $R$ is the average radius of the planet and $a$ is the equatorial radius of the planet.

The Lunar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_l}{a_l^3\left(1-e_l^2\right)^\frac 3 2} \left(1-\frac 3 2 sin^2i\right) \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Not much change here expect that the masses and so on refer to the Moon and not the Sun (hence the different subscripts).  There is one additional term which is the factor :

$$\left(1-\frac 3 2 sin^2i\right)$$

This corrects for the effect that the angle of inclination of Moon's orbit to the ecliptic is not zero.  The ecliptic being the plane with the Sun and the Earth's orbit in it.

You have to decide these numbers for yourself.

The total effect :

The total effect is simply the sum of the two other effects so :

$$\frac {d\psi}{dt} = \frac {d\psi_l}{dt} + \frac {d\psi_s}{dt}$$

If you had multiple moons you would need multiple lunar correction terms.

Just for clarification that $\frac {d\psi}{dt}$ means the rate of change of the angle $\psi$ with respect to time $t$.  To get how big an angle you'd move through in a century you do this :

$$\Delta \psi \approx \frac {d\psi}{dt} \Delta t$$


Best chance would be the Hipparcos catalogue. The first set of Gaia data will be released Mid-September 2016, but I don't know if it will be more accurate than Hipparcos already.

  All stars within 11,462 light years of Earth.

That won't be easy. These catalogues have a magnitude (brightness) cutoff. Brighter, more massive stars can be seen further away than smaller, darker ones.

  Their stellar classification.

Spectral type should be in there.

  If they have any planets around them (hypothetical or otherwise)

Searching for Exoplanets is an ongoing project, and there is no complete catalogue in any sense of the word yet for any radius. Most Exoplanets were found with the Kepler mission, but that one looked in a fairly small part of the sky. What exactly do you mean by hypothetical planets?

  Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)

You'll have to convert that yourself. If you know how to program, there are libraries for coordinate conversion. I would argue, though, that an Earth-centred spherical coordinate system is preferential for most applications.

Distances are hard to measure. Hipparcos and Gaia use the parallax method. This gets less accurate with increasing distance. Also, distances are naturally measured in parsec, but the conversion to light years is trivial.

The Gaia catalogue will have $10^9$ stars with positions and velocities. It will also be more accurate than Hipparcos. But the measurements are ongoing, and the final data release is planned for 2022.


If blackholes are spherical, what is the prespective of the blackhole image?

One can never actually see the event horizon, one sees the gas that is about to enter the event horizon and gets hot along the way.  But this gas has a history, it doesn't just fall straight in.  The history imparts some angular momentum to it, like the water in a draining bathtub.  So it spirals in, not falling straight in, and this creates a disk shape called an accretion disk.  The disk does not have to be as thin as a pancake, but it's far from spherical, and that gets further distorted by the strong gravity (creating the brightness variations and the central dark sphere).

The disk does have a hole in the center because at some point the orbits go unstable and fall quickly below the event horizon.  It could be argued we are not actually seeing the event horizon where it goes black, we are seeing the drop in gas density where the orbits are no longer stable and the  slow spiral turns into a very rapid fall, but the way the strong gravity bends the light rays is always very important as well.

The distortion by the strong gravity is what provides the test of general relativity that everyone is so excited about.  The strong gravity produces a "shadow" effect, so you don't have to see the accretion disk from above to get a spherical blackness at the center.


Determining effect of small variable force on planetary perihelion precession

You may want to use perturbation theory. This only gives you an approximate answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to integrating the perturbation along the unperturbed original orbit. Writing your force as a vector, the perturbing acceleration is
$$
\boldsymbol{a} = K \frac{GM}{r^2c^2}v_r\boldsymbol{v}_t
$$
with $v_r=\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}$ the radial velocity
($\boldsymbol{v}\equiv\dot{\boldsymbol{r}}$) and
$\boldsymbol{v}_t=(\boldsymbol{v}-\hat{\boldsymbol{r}}(\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}))$ the rotational component of velocity (the full velocity minus the radial velocity). Here, the dot above denotes a time derivative and a hat the unit vector.

Now, it depends what you mean with 'effect'. Let's work out the changes of the orbital semimajor axis $a$, eccentricity $e$, and direction of periapse.

To summarise the results below: semi-major axis and eccentricity are unchanged, but the direction of periapse rotates in the plane of the orbit at rate
$$
\omega=\Omega \frac{v_c^2}{c^2}
\frac{K}{1-e^2},
$$
where $\Omega$ is the orbital frequency and $v_c=\Omega a$ with $a$ the semi-major axis. Note that (for $K=3$) this agrees with the general relativity (GR) precession rate at order $v_c^2/c^2$ (given by Einstein 1915 but not mentioned in the original question).

change of semimajor axis

From the relation $a=-GM/2E$ (with $E=\frac{1}{2}\boldsymbol{v}^2-GMr^{-1}$ the orbital energy) we have for the change of $a$ due to an external (non-Keplerian) acceleration
$$
\dot{a}=\frac{2a^2}{GM}\boldsymbol{v}{\cdot}\boldsymbol{a}.
$$
Inserting $\boldsymbol{a}$ (note that $\boldsymbol{v}{\cdot}\boldsymbol{v}_t=h^2/r^2$ with angular momentum vector $\boldsymbol{h}\equiv\boldsymbol{r}\wedge\boldsymbol{v}$), we get
$$
\dot{a}=\frac{2a^2Kh^2}{c^2}\frac{v_r}{r^4}.
$$
Since the orbit average $\langle v_r f(r)\rangle=0$ for any function $f$ (see below), $\langle\dot{a}\rangle=0$.

change of eccentricity

From $\boldsymbol{h}^2=(1-e^2)GMa$, we find
$$
e\dot{e}=-\frac{\boldsymbol{h}{\cdot}\dot{\boldsymbol{h}}}{GMa}+\frac{h^2\dot{a}}{2GMa^2}.
$$
We already know that $\langle\dot{a}\rangle=0$, so only need to consider the first term. Thus,
$$
e\dot{e}=-\frac{(\boldsymbol{r}\wedge\boldsymbol{v}){\cdot}(\boldsymbol{r}\wedge\boldsymbol{a})}{GMa}
         =-\frac{r^2\;\boldsymbol{v}{\cdot}\boldsymbol{a}}{GMa}
         =-\frac{Kh^2}{ac^2}\frac{v_r}{r^2},
$$
where I have used the identity
$(\boldsymbol{a}\wedge\boldsymbol{b}){\cdot}(\boldsymbol{c}\wedge\boldsymbol{d})
=\boldsymbol{a}{\cdot}\boldsymbol{c}\;\boldsymbol{b}{\cdot}\boldsymbol{d}-
 \boldsymbol{a}{\cdot}\boldsymbol{d}\;\boldsymbol{b}{\cdot}\boldsymbol{c}$ and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}_p=0$.
Again $\langle v_r/r^2\rangle=0$ and hence $\langle\dot{e}\rangle=0$.

change of the direction of periapse

The eccentricity vector
$
\boldsymbol{e}\equiv\boldsymbol{v}\wedge\boldsymbol{h}/GM - \hat{\boldsymbol{r}}
$
points (from the centre of gravity) in the direction of periapse, has magnitude $e$, and is conserved under the Keplerian motion (validate all that as an exercise!). From this definition we find its instantaneous change due to external acceleration
$$
\dot{\boldsymbol{e}}=
\frac{\boldsymbol{a}\wedge(\boldsymbol{r}\wedge\boldsymbol{v})
     +\boldsymbol{v}\wedge(\boldsymbol{r}\wedge\boldsymbol{a})}{GM}
     =\frac{2(\boldsymbol{v}{\cdot}\boldsymbol{a})\boldsymbol{r}
       -(\boldsymbol{r}{\cdot}\boldsymbol{v})\boldsymbol{a}}{GM}
     =\frac{2K}{c^2}\frac{h^2v_r\boldsymbol{r}}{r^4}
     -\frac{K}{c^2}\frac{v_r^2\boldsymbol{v}_t}{r}
$$
where I have used the identity
$\boldsymbol{a}\wedge(\boldsymbol{b}\wedge\boldsymbol{c})=(\boldsymbol{a}{\cdot}\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}{\cdot}\boldsymbol{b})\boldsymbol{c}$
and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}=0$. The orbit averages of these expression are considered in the appendix below. If we finally put everything together, we get
$
\dot{\boldsymbol{e}}=\boldsymbol{\omega}\wedge\boldsymbol{e}
$
with [corrected again]
$$
\boldsymbol{\omega}=\Omega K \frac{v_c^2}{c^2}
(1-e^2)^{-1}\, \hat{\boldsymbol{h}}.
$$
This is a rotation of periapse in the plane of the orbit with angular frequency $\omega=|\boldsymbol{\omega}|$. In particular $\langle e\dot{e}\rangle=\langle\boldsymbol{e}{\cdot}\dot{\boldsymbol{e}}\rangle=0$ in agreement with our previous finding.

Don't forget that due to our usage of first-order perturbation theory these results are only strictly true in the limit $K(v_c/c)^2\to0$. At second-order perturbation theory, however, both $a$ and/or $e$ may change. In your numerical experiments, you should find that the orbit-averaged changes of $a$ and $e$ are either zero or scale stronger than linear with perturbation amplitude $K$.

disclaimer No guarantee that the algebra is correct. Check it!

Appendix: orbit averages

Orbit averages of $v_rf(r)$ with an abitrary (but integrable) function $f(r)$ can be
directly calculated for any type of periodic orbit. Let $F(r)$ be the antiderivative of $f(r)$, i.e. $F'\!=f$, then the orbit average is:
$$
\langle v_r f(r)\rangle = \frac{1}{T}\int_0^T v_r(t)\,f\!\left(r(t)\right) \mathrm{d}t
= \frac{1}{T} \left[F\left(r(t)\right)\right]_0^T = 0
$$
with $T$ the orbital period.

For the orbit averages required in $\langle\dot{\boldsymbol{e}}\rangle$, we must dig a bit deeper. For a Keplerian elliptic orbit
$$
\boldsymbol{r}=a\left((\cos\eta-e)\hat{\boldsymbol{e}}+\sqrt{1-e^2}\sin\eta\,\hat{\boldsymbol{k}}\right)\qquad\text{and}\qquad
r=a(1-e\cos\eta)
$$
with eccentricity vector $\boldsymbol{e}$ and $\hat{\boldsymbol{k}}\equiv\hat{\boldsymbol{h}}\wedge\hat{\boldsymbol{e}}$ a vector perpendicular to $\boldsymbol{e}$ and $\boldsymbol{h}$. Here, $\eta$ is the eccentric anomaly, which is related to the mean anomaly $\ell$ via
$
\ell=\eta-e\sin\eta,
$
such that $\mathrm{d}\ell=(1-e\cos\eta)\mathrm{d}\eta$ and an orbit average becomes
$$
\langle\cdot\rangle = (2\pi)^{-1}\int_0^{2\pi}\cdot\;\mathrm{d}\ell = (2\pi)^{-1}\int_0^{2\pi}\cdot\;(1-e\cos\eta)\mathrm{d}\eta.
$$
Taking the time derivative (note that $\dot{\ell}=\Omega=\sqrt{GM/a^3}$ the orbital frequency) of $\boldsymbol{r}$, we find for the instantaneous (unperturbed) orbital velocity
$$
\boldsymbol{v}=v_c\frac{\sqrt{1-e^2}\cos\eta\,\hat{\boldsymbol{k}}-\sin\eta\,\hat{\boldsymbol{e}}}{1-e\cos\eta}
$$
where I have introduced $v_c\equiv\Omega a=\sqrt{GM/a}$, the speed of the circular orbit with semimajor axis $a$. From this, we find the radial velocity $v_r=\hat{\boldsymbol{r}}{\cdot}\boldsymbol{v}=v_c e\sin\eta(1-e\cos\eta)^{-1}$
and the rotational velocity
$$
\boldsymbol{v}_t = v_c\frac{\sqrt{1-e^2}(\cos\eta-e)\,\hat{\boldsymbol{k}}-(1-e^2)\sin\eta\,\hat{\boldsymbol{e}}}{(1-e\cos\eta)^2}.
$$

With these, we have [corrected again]
$$
\left\langle \frac{h^2v_r\boldsymbol{r}}{r^4}\right\rangle =
\Omega v_c^2\,\hat{\boldsymbol{k}}\,
\frac{e(1-e^2)^{3/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta}{(1-e\cos\eta)^4}\mathrm{d}\eta
=\frac{\Omega v_c^2e}{2(1-e^2)}\hat{\boldsymbol{k}}
\\
\left\langle \frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle = \Omega v_c^2\,
\hat{\boldsymbol{k}}\,
\frac{e^2(1-e^2)^{1/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta(\cos\eta-e)}{(1-e\cos\eta)^4}\mathrm{d}\eta=0,
$$
in particular, the components in direction $\hat{\boldsymbol{e}}$ average to zero. Thus [corrected again]
$$\left\langle 2\frac{h^2v_r\boldsymbol{r}}{r^4}-\frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle
=\frac{\Omega v_c^2e\,\hat{\boldsymbol{k}}}{(1-e^2)}
$$


How did Arecibo detect methane lakes on Titan, and image Saturn's rings?

Titan "lakes":

Published Open Access in Science: Radar Evidence for Liquid Surfaces on Titan Campbell, D. B., Black, G. J., Carter, L. M., and Ostro, S. J., Science 302, 5644, pp. 431-434, 17 Oct 2003 DOI: 10.1126/science.1088969

This was a really elegant experiment! A continuous, unmodulated, circularly polarized 13 cm wave was broadcast from Arecibo towards the Saturn/Titan system, and the Doppler shift was used to isolate the returned signal from Titan.

Most of the surface is rough, so there is signal returned from areas all over Titan's disk, and since the moon rotates, albeit slowly, returned power from the "left" and "right" sides are shifted to higher and lower frequencies.

However during some observing times there was a very strong and pronounced reflection with zero Doppler shift with respect to Titan's known radial velocity, and this peak is attributed to specular reflection. Checks on received polarization confirm that while the power from the rough surface is returned in both circular polarization state, the presumed specular component is only in the expected circular polarization state.

As pointed out in @Martin Kochanski's thoughtful answer there is no determination from the radar observation that the returned specular reflection comes from methane. This is simply a presumed component of the presumed lakes, based on known information about Titan's chemistry at the time (2003).

  We observed Titan on 16 nights in November and December 2001 and on 9 nights in November and December 2002, transmitting at 13-cm wavelength with the 305-m Arecibo telescope and receiving the echo with Arecibo. Titan's rotational and orbital periods are 15.9 days, and our 2001 observations were obtained at a uniform 22.6° (∼800 km) interval in longitude. The 9 observations in 2002 did not provide uniform coverage. The latitude of the subearth track was 25.9°S in 2001 and 26.2°S in 2002, its farthest southern excursion. The round-trip light time to the Saturn system during the observations was 2 hours 15 min, and the limited tracking time of the Arecibo telescope meant that signal reception was restricted to ∼30 min per day, corresponding to 0.5° of Titan rotation (20 km of motion of the subearth point). On one night in 2001 and for most of the 2002 observations (as well as others when we were attempting ranging measurements to Titan), the 100-m Green Bank Telescope (GBT) was also used to receive the echo for the full round-trip time. These data have lower signal-to-noise ratios than those obtained with Arecibo receiving the echo, but the longer receive time corresponding to 2.1° of Titan rotation allowed more subearth locations to be studied.

Here is some of the Titan data:

  Fig. 3. The OC radar echo spectrum at 1.0-Hz resolution for the 2002 observation at the subearth longitude of 80°. The normalized cross section for the specular component of the echo and the RMS slope are 0.023 and 0.2°, respectively.

 Click for full size

  Fig. 1. Arecibo radar echo spectra from the 2001 data for five subearth longitudes on Titan. Spectra are shown for both the expected (OC) sense of received circular polarization and the cross-polarized (SC) sense. The ordinate is in standard deviations of the noise. The limb-to-limb Doppler-broadened bandwidth for Titan is 325 Hz. Four of the OC spectra show evidence of a specular component at 0 Hz.

Saturn's rings "imaged" (delay-Doppler):

From Radar imaging of Saturn’s rings Nicholson, P. D. et al., Icarus 177 (2005) 32–62, doi:10.1016/j.icarus.2005.03.023

The "image" below is not a conventional image, since the Arecibo dish has no way to spatially resolve the transverse expanse of Saturn and its rings. It is a "delay-Doppler" image, using 12.6 cm, ~500 kW radar broadcasts transmitted by Arecibo. The round-trip light time was about 135 minutes. Because Arecibo has limited steering away from the zenith (<19.7 degrees) max Saturn was only available to the dish for 166 minutes even under ideal conditions.

The vertical axis shows a delay of about +/- 800 milliseconds which demonstrates spatial resolution, but in the radial or depth direction. The horizontal axis is Doppler shift. The +/- 300 kHz shift represents the orbital velocity of particles in the rings.

While the Titan specular reflection above was done with a continuous or CW beam, the delay-doppler imaging technique requires a frequency modulation of the beam with a frequency-hopping pattern. By applying a correlation function using the known  pattern to the recorded received signals, components with different return times and different Doppler shifts can be extracted, and the results is then hzstogramed, producing the delay-Doppler image below.

This is a standard technique and has been used to image other planets and asteroids:  See the following items and refereces within:

this answer to What causes “North-South ambiguity” when doppler radar imaging a planet surface equator on?
Why do radar maps of the surface of Venus have missing slices?
What is the physical geometry of this apparent “eclipse” of a tiny moon of Asteroid Florence?
Is delay-doppler radar imaging of NEO asteroids possible only if it spins fast enough?

  Fig. 2. Delay–Doppler images constructed from data obtained in (a) October 1999, (b) November 2000, (c) December 2001, and (d) January 2003.
  Both OC and SC polarizations were combined to maximize the signal to
  noise ratio. Note the four bright regions in each image where the delay and
  Doppler cells are parallel and where the A and B rings appear to cross over
  one another.


The solution as pointed out by samcarter in the above comment was to remove the tab in conto.con so that it had linebreak + linebreak instead of linebreak + tab + linebreak


In which two constellations, were the two gravitational wave events of 2016?

The data is indeed very uncertain. As for GW150914, LIGO posted this image of the probable area of the sky:

The most probable banana-shaped region stretches from Crux to the Large Magellanic cloud. Comparing the image with a constellation map, there are several small constellations that may have hosted the event. Those are Mensa, Chamaeleon, Musca and Crux.


Moment of inertia factor of Venus and Pluto

Measuring the moment of inertia of a planetary body is not straightforward and simple, but instead it requires a lot of data about how the body rotates, models for how mass is distributed across the body, and the gravity field of the body (these are related things, i.e. the gravity field can tell you how mass is distributed).  We have so many satellites in orbit of Earth and so much ground research that we know this well enough to measure the moment of inertia of our own planet.  The same is reasonably true for the Moon, and more recently for Mars.
The MESSENGER spacecraft had a highly elliptical orbit around Mercury for several years and so was also able to provide data to reasonably estimate Mercury's moment of inertia, too, but to far less precision:  Based on the Wikipedia article from the question, the Moon's moment of inertia has a relative uncertainty of ±0.2%, while Mercury's is ±4% (for completeness, Mars is ±0.1%).  Mars' is better known, despite having had fewer orbiting spacecraft, because it has two moons that help.
Reading through the paper on Mars' moment of inertia gives some idea of everything you have to do to really get at this value:  Love number, second-degree gravity field, correct things for atmospheric drag (including tides within the atmosphere), account for orbital changes in the moons, map the precession rate, measure the solar torque since Mars is oblate, look at all of Mars' orbital parameters (eccentricity, obliquity, mean motion, spin rate, precession) ... it's a lot.  Once they had all that, with their associated uncertainties, they could calculate the moment of inertia.  But, they suggested this could then be modified based on mass asymmetries across Mars, such as the giant Tharsis volcanic region providing a load on that side of the planet, which could tweak one of the terms that goes into the calculation and change the moment of inertia from 0.3644 to 0.3652.
I didn't provide links to what all the terms mean in the above paragraph (someone can edit if they wish), the real point is that we need to know a lot to actually measure the moment of inertia of a planetary body, and we simply don't have those for most bodies in the solar system.  The Wikipedia article that's linked in the question does list values for 17 (not 18) bodies, noting that Venus is unknown (hence 17 bodies), but then the other 13 are listed as "not measured" but rather calculated through other means.
For example, several list the Darwin-Radau equation method, which relates the body's shape and spin to moment of inertia.  Since we can relatively easily see those, we can guess at the moment of inertia via those means, but it makes a lot of assumptions about the interior structure.  Since the interior structure of Venus is likely more complicated than something like Io, whomever wrote the Wikipedia article chose not to list them in the table, though two references are given with values of 0.327–0.342.
The internal mass distribution of Pluto is also quite complicated given what we know about Sputnik basin and the glacier within it.  We also don't even have a full shape model for Pluto or Charon because of the lighting geometry during the New Horizons flyby in 2015, where south of about 45° was in permanent night for this part of Pluto's orbit (same with Charon).  Without even the shape, you can't use the Darwin-Radau equation to estimate the moment of inertia, though based on limb profiles no flattening was detected so I suppose some estimates could be made.  But, like the Venus estimates, how accurate those would be is debatable might not be at all useful.
Keep in mind that the moment of inertia for a perfectly uniform, solid sphere is 0.4, so a planetary body is going to be less than that (unless it's perfectly uniform and solid).  Values less than that indicate the mass is more concentrated toward the center, but even with Earth's large metallic core, the value is still 0.3307, which is not hugely different from 0.4 when you're trying to precisely measure all of these different values that go into the calculation.  So, if you calculate something like 0.35±0.05 for Pluto (to make up numbers) based on your limited data, is that really useful?  Maybe.  Or, maybe not.


Understanding energy loss in a telescope aperture due to atmospheric dispersion

I'm going to make the assumption that the "aperture" of the instrument refers to the width of a fibre that feeds a spectrograph (I can't think of any other plausible scenario).
What you want is to enclose the most power you can within a 2 arcsec diameter in the telescope focal plane at the wavelengths of interest. One confounding effect is the seeing. If 80% of the energy is enclosed within 1 arcsec diameter, then it is a reasonable assumption that you almost get all of it ($>95$%) within a 2 arcsec fibre.
The above consideration applies to light of any single, monochromatic wavelength. If you wish to observe a spectrum over a range of wavelengths, then you must take account of atmospheric dispersion. Light of different wavelengths is refracted by the atmosphere and thus light of different wavelengths is brought to a different focal point depending on its wavelength. The amplitude of this differential displacement in the focal plane depends on how much atmosphere is in the way and so increases with zenith distance.
The graph in your question shows you what that displacement is (in arcseconds) with respect to light at 5000A. What this means is that the range of wavelengths you can get successfully into your 2 arcsec fibre will depend on the central wavelength and the zenith distance.
For example, trying to get a spectrum between the atmospheric cut-off at 3200A and 5000A in a single observation will not be possible above a zenith distance if 50 degrees because the centroid of the images at these two wavelengths differ by more than 2 arcsec. And in fact you will start to lose signal before that because the displacement in the graph is just of the image centroid and obviously if that centroid gets near the edge of the aperture then the blurring effects of seeing will lead to light loss.
If instead the "aperture" refers to a slit width on a more traditional spectrograph, then you can get around this dispersion problem by rotating the slit to be at the parallactic angle, so that the slit lies along the dispersed image of the star and so collects light at all wavelengths (though they will appear in different pixels on the camera).
If you are doing imaging the effects of atmospheric dispersion will depend on the bandwidth of your filter and the zenith distance. With a narrow band filter the size of your images will be limited by the seeing. If you are using broadband filters, especially U and B, then your images can become vertically elongated at large zenith distances due to the differential dispersion.
A solution is to introduce an atmospheric dispersion corrector prior to the focal plane, which reverses the calculated effects of atmospheric dispersion (at the expense of some loss of signal and image quality).


How was Sugar Grove Station supposed to work?

I can't comment in detail but Earth-Moon-Earth communication at radio wavelengths is well established.

At wavelengths longer than a few cm (frequencies below 10 GHz), the Moon appears smooth and acts like a reflective sphere as far as radio waves are concerned.


How do astronomers know when the Sun will die?

Yes, astronomers could be wrong.  Part of being a scientist is always having to be ready to admit that you were wrong.

Astronomers have developed models of how the sun and other stars work. We understand them to be nuclear furnaces. These models may not be perfect, but they are well supported by the evidence and can predict much of what we observe from the sun and other stars.

Using these models we can expect the sun to exist for about 10 billion years, and it is currently about 4.7 billion years old. While the model could be wrong, it is probably not very wrong, and the sun will go on shining for a long time yet.


What is the current accepted theory as to why Titan has retained its atmosphere?

It's not really known why only Titan has a thick atmosphere while moons like Ganymede don't.

One thought is that temperatures may have been too high (well above ~40K) in the Jovian subnebula of our solar system due to the greater gravitational potential energy release, mass, and proximity to the Sun, greatly reducing the $NH_{3}$-hydrate inventory accreted by both Callisto and Ganymede. The resulting $N_{2}$ atmospheres might have been too thin to survive the atmospheric erosion effects that Titan has withstood.

EDIT: some links:

http://perso.utinam.cnrs.fr/~mousis/papier3.pdf (though it's from 2002).

http://adsabs.harvard.edu/abs/2012ttt..work...25N (this paper critiques the previous idea of a $CH_{4}$ supported atmosphere)

I would also think the NASA/JPL Cassini pages would have any up-to-date info in regards to Titan.


Coordinates of known black holes in our Galaxy (Milky Way)

If I go from the list of black holes to the page of a particular object like this there is a link to SIMBAD database like this. This SIMBAD database has galactic coordinates in this format

Gal coord. (ep=J2000) : 209.9563 -06.5399 [ 100 12 0 ]

where 209.9563 is a Galactic longitude, which changes from 0 to 360 degrees. -06.5399 is Galactic latitude, which is always around zero (+/- 10 degrees), because our galaxy is almost flat. Good explanation for galactic coordinates is on this wikipedia page.

To put the object on 2D map like this only Galactic longitude and distance to the object is needed. For example, this is the map of all black holes which I could find in wikipedia in our Galaxy.


If light has no mass, why is it affected by gravity?

Another way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good).  The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead.  So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now).

Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally.  In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that raising effect is accelerating.  So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser.  Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward.  Ergo, light "falls."

Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because it's nothing happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating.

Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit.  Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass.  All objects fall the same under Newtonian gravity.   So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit.  Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.


Your best bet for finding relevant information on this is to look up actual published papers. I'll walk you through my research process to help in the future, as well as provide the results I found.

Step 1: Google Scholar is your friend

I start out using Google Scholar. This is much like Google, but rather than returning any old website, it specifically returns published papers and other scholarly articles. Part of this magic is of course knowing what to search. I went with "multiplicity of stellar systems". Immediately I found two sources which seemed promising.

Stellar Multiplicity and the Initial Mass Function: Most Stars Are Single (Lada 2006)

A Catalogue of Multiplicity Among Bright Stellar Systems (Eggleton & Tokovinin 2008)

I've lucked out because both articles happen to be free so I can see the full contents!

Step 2: NASA ADS

NASA Astrophysics Data System is a very large catalogue of just about every astronomy (and some physics) related paper published in any journal. What's even more amazing, very often the papers here are free, even if they aren't free on the journal's actual web page. If you ever run into a paper from a journal that only gives you an abstract, try looking for the paper here.

It was already free on the journal's webpage, but I was able to look up and find the Lada paper I cited above.

I specifically point you towards ADS for a few reasons.

It has a large collection of free papers.
You may have heard of or even use arxiv, which is a great resource, but very often the papers on arxiv are versions which are pre-publication and not the officially peer-reviewed and published papers. ADS usually has the official published papers (and links to arxiv versions!).
ADS has another hugely important feature. It lists article citations and articles which have cited that article.

Step 3: Backtracking and Forwardtracking

Once you've found a good source (such as the two papers above), you don't want to stop there. There's likely more or better papers and you can use the papers you have to find new ones. What you want to do now is take a look at all the papers your current source has cited, and look at all the papers which cite your current source. ADS very conveniently gives you all this information.

The Lada 2006 source is 10 years old at this point so I wanted to see if there was anything more recent. I was able to click on the "Citations to the Article" link on ADS for that article, and found the list of 217, more recent papers which cited the Lada paper. Scanning through this list I found two promising papers:

Stellar Multiplicity of the Open Cluster ASCC 113 (Guerrero et al. 2014)

An Adaptive Optics Multiplicity Census of Young Stars in Upper Scorpius (Lafrenière et al. 2014)

From here, just rinse and repeat until you've got the information you want.

Results

I found four papers which appear to give some or all of the information you want.

Lada 2006 - This paper is more or less a conglomeration of past research on the fraction of stars which are single star systems. It focused on lower mass stars (G to M) and found that the fraction of single star systems ranged from ~43% for G type stars to ~75% for M type stars (which are the most populous type).
Eggleton & Tokovinin 2008 - I think that this is the best source for your particular question. Within their abstract they state We identify 4559 such bright systems (including the Sun), and the frequencies of multiplicities 1, 2,..., 7 are found to be 2718, 1437, 285, 86, 20, 11 and 2. This implies the fraction of multiplicities is: 59.62%, 31.52%, 6.25%, 1.88%, 0.44%, 0.24%, and 0.04%. However, note they say their measurements have "substantial" uncertainties which they describe.
Guerrero et al. 2014 - This group looked at and around a specific cluster and found a ratio of the number of single to binary stars to be 27:7 within the cluster itself. Including stars around the cluster, the following multiplicities were found (ranging from 1 to 8 companions) 125:27:4:1:0:0:0:1 These two sets of ratios indicate that within the cluster they saw 79% single star systems and 21% binary star systems. Overall they saw multiplicity fractions of 79.1%, 17.1%, 2.5%, 0.6%, 0%, 0%, 0%, and 0.6%, respectively.
Lafrenière et al. 2014 - A study that looked at 91 stars and found 57 single stars (63%), 29 binaries (32%), and 5 triple systems (5.5%).

What this suggests is that the multiplicity of stars is highly variable, still somewhat unknown, and that it depends on the environment you look at (and I found a few papers discussing this very point). The Eggleton paper had the largest sample and so may be the most trustworthy in regards to a true average, but make sure to understand what their uncertainties are.


About the reference for equation of state

You cannot just add the ideal gas pressure and degeneracy pressure together for any particular species of particle. (Well, you can, but it would be a poor approximation to the actual pressure).

The trouble is that they are not separate things. In general, the electrons in the gas are partially degenerate and always obey Fermi-Dirac statistics. Unfortunately there is no analytic solution that gives you pressure as a function of temperature and density; you either have to use Tables of pre-calculated results, or there are analytic approximations that work well in certain regimes.

Clayton's Principles of Stellar Evolution and Nucleosynthesis contains some of these tables and also some better analytical approximations that work in certain regimes of (partial) degeneracy.

What is true is that you are fine to add to this pressure, the radiation pressure due to photons and the ideal gas pressure of the ions.

EDIT: I realise that what may be troubling you is not the above, but why you can add the separate pressures together at all (eqn 122 in the notes you reference). The reason can be traced to eqn 31 in the same notes. This says that the pressure is an integral over the number density and momenta of individual particles. Note that this equation does not contain anything that identifies the particles (i.e. their mass). The integral can therefore be treated as a summation over all particles, which in turn can be separated into three integrals that cover the ions, electrons and photons separately.


how does Kepler stop exposure after a 5 minuate or 30 minute cadence time?

They don't stop. The Kepler CCDs read out the signals collected during the  accumulated time of 6.02 seconds. The fixed read out time is 0.52 seconds. So each CCD gets one frame every (6.02 + 0.52) = 6.54 seconds. Then Kepler sums up every 9 frames (short cadence) and 270 frames (long cadence). The time between two short cadences is (6.54 x 9) = 58.9 seconds, and (6.54 x 270) = 1766 seconds between two long cadences, but the exposure time is (6.02 x 9) = 54.2 seconds for a short cadence, and (6.02 x 270) = 1625 seconds for a long cadence.

There are totally 95 million pixels available on Kepler CCDs, but the spacecraft can store only 5.4 million. The maximum number of targets is about 170,000. see http://archive.stsci.edu/mast_faq.php?mission=KEPLER


Since Proxima Centauri orbits the other two Alpha Centauri stars, will it eventually be farther away from us than they are?

In about 26,000 years, Proxima Centauri will no longer be the closest star. Instead, Alpha Centauri AB will be the closest, but only for about ~7000 years, soon replaced by Ross 248 for 10000 years, and Gliese 445 for 6000 years. Then, Alpha Centauri will continue to be the nearest star for another ~28000 years, then briefly becoming Ross 128. Refer to the diagram below (sourced from Wikipedia).

Correct me if I'm wrong, and feel free to edit my answer if it needs improvement.


What happened to NS spectral types after OBAFGKM?

They still exist, but are perhaps less used because they don't indicate a particular temperature, or mass.
The stellar classification is based on the presence of particular lines in the spectrum:  A type stars have a strong hydrogen line. B type stars have Helium.  F and G  and K stars have various ionised or neutral metals. M type stars have compounds like TiO2 in their atmosphere.
But these are associated with temperature, and B type stars are hotter than A type (and O type stars are hotter still) hence OBAFGKM.
Asymptotic branch supergiants can have processes that dredge up carbon from their interior, producing carbon lines.  These are the R and N type stars, but R type stars are similar in temperature to K type stars. and N type are like M type.
S type stars have heavy elements produced by s-process neutron capture stars.  So if you are ordering the stellar classification by temperature, these don't have a natural place to go, and since the purpose of "Oh Be A Fine Girl Kiss Me (Right Now Sweetheart) is to remember the order of the letters then RN and S don't fit, since they are not ordered by temperature.
As for political correctness, many alternatives have been suggested, but none has caught on, perhaps because a good mnemonic should be a bit "naughty".


Why does gas form a star instead of a black hole?

Essentially gas does, it just happens to form a star first.

Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from reaching a critical density needed for black hole formation. As these energy producing processes run out of useable fuel the star will eventually collapse onto itself creating a black hole.

So you can't just take a large volume of gas and create a black hole. Other physical processes occur.


Yes the earth rotates around its axis,  which gives rise to the motion of the sun as we see it.
To understand sun rise you'd have to start with a reference frame. For us,  the earth is considered to be the inertial frame for any apparent motion in the celestial sphere, it simplifies the computations. Thus any motion of celestial bodies is defined with respect to that frame.
So in that context Sunrise would be the time when for a given latitude, when the sun first appears above the local horizon, and sun set would become the time when the sun disappears below that local horizon.


Using Tully Fisher to measure Distance Problem

With the Tully Fisher relation, the inclination angle is determined from the major-minor axis ratio where $sin(i) = \sqrt{1-q^2} $ and q is the inverse of your ratio.

You need to apply the sine of the angle to the rotational velocity because the measure is based on our line of sight observation which must be converted to the true 3D velocity.


Why not take a picture of a closer black hole?

I was surprised too when I first heard they were trying to image M87's black hole.

The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*, which they are also attempting to image.

A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.

A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.

There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.

Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.

Comparison of the apparent size of the largest nearby black holes

And of course obligatory XKCD to remind us how small these objects really appear.


What's the closest Venus gets to the moon?

Venus is, on occasion, occulted by the Moon (that is to say, the Moon passes in front of Venus as seen from our point of view).  Exactly when that happens is dependent on your location on the Earth, since the location of the Moon in the skies varies by about a degree or so depending on your location on the Earth, and of course, it might be on the other side of the Earth when it happens (it only lasts about an hour or less).  From North America, it happened in December 2015; it will happen again in September 2017, but only from Australia and neighboring countries.

If you google "Venus occultation" you should get something like what you want.  ETA: Here's one link: http://www.lunar-occultations.com/iota/planets/planets.htm


Does the Zwicky Transient Facility only run when triggered by a cell phone?

The ZTF is operating on most clear nights on an automated queue of pointings. When they discover transients, they inform TNS which may trigger other observatories to point at the same object, just as ZTF may respond to transient events from other observatories.


Why does Earth have a dip in the CO2 absortion spectrum from 14 to 16 micron?

Eric Jensen has already provided a nice link to a description of the basic structure of the ${\rm CO}_{2}$ spectrum, so I'll focus on the question of why there's a "spike" at 15.0 microns in the Earth spectrum, but not in the Venus or Mars spectra.
If you look at the link in Eric's answer, the very first image shows a high-resolution version of the ${\rm CO}_{2}$ absorption spectrum, including the main features to the right and left. There is also a very strong absorption peak in the middle (the "Q-branch") at about 15.0 microns (667 ${\rm cm}^{-1}$ in wavenumbers). The question then becomes: why does this feature show up as an extra dip in the Venus spectrum (and, barely visible, in the Mars spectrum) -- as one might expect -- but apparently as a peak in the Earth spectra?
The answer hinges on the fact that the spectra shown in the question are not, strictly speaking, absorption spectra. They are actually radiance/intensity measurements of infrared light emitted by different layers in the atmosphere of the planet in question, as seen from orbiting spacecraft. (In the first figure, it's narrowed to the atmosphere above certain locations on the Earth.) They thus show a combination of the molecular absorption spectra and the temperature structure of the atmosphere.
In simple terms, the spectra show thermal (blackbody) emission from layers in the atmosphere where upward-going photons can escape to space without being absorbed by higher layers in the atmosphere. The altitude of these layers is determined by the amount of the absorbing gas (e.g., ${\rm CO}_{2}$) and the strength of the absorption at a given wavelength. The stronger the absorption, the higher up is the layer that can emit freely to space.
(Note that for wavelengths with effectively zero absorption from any molecule, the "emitting layer" is actually the surface of the planet.)
How much infrared light is emitted at a given wavelength then depends on the temperature of the emitting layer, and it is here that the vertical temperature structure of the atmosphere becomes important. For Venus and Mars -- and for most of the emitting altitudes for Earth -- the temperature decreases with altitude, so that higher emitting layers have lower temperatures and thus emit less radiation. The net effect is to reproduce the absorption features, even though the whole thing is not, technically, an absorption spectrum.
But in the case of the Earth, the strongest absorption -- e.g., in the Q-branch part of the ${\rm CO}_{2}$ absorption, at 15.0 microns -- forces an emitting altitude that is so high it is actually in the stratosphere. And in the stratosphere, the temperature increases with altitude. So the Q-branch absorption peak forces the emitting layer to be at a higher temperature than is the case for the rest of the ${\rm CO}_{2}$ absorption regime.
This University of Chicago MODTRAN web page, which plots predicted emission from the Earth's atmosphere as seen from an altitude of 70 km (default setting), shows this rather nicely. The default parameters include a modern ${\rm CO}_{2}$ concentration of 400 ppm, which gives the feature seen in the Earth observations (i.e., the local emission peak in the middle of the 15 micron trough, which I've labeled "Q-branch reversal"):

But if you dial the ${\rm CO}_{2}$ concentration way down (e.g., to 10 ppm), the emitting altitudes are lower for all ${\rm CO}_{2}$-absorbing wavelenths, so much so that the emitting altitude for the Q-branch is down below the stratosphere, and so is at a lower temperature than the rest, and thus emits less flux, giving you the expected profile (i.e., with the weakest emission at the Q-branch wavelength, instead of a reversal):

(Finally, I believe the peculiar spectrum for the "Antarctic" case in the first figure is due to the unusual temperature profile of atmosphere in arctic regions, which includes the temperature increasing with altitude in the first few km above the ground; the fact that the interior of Antarctica is high altitude might also be relevant.)