Stellar fusion and supernovae are governed by quantum particle interactions (on massive scales).  In general there are many possible ways for particles to interact, decay, etc. at various probabilities.  To understand the physics of a system correctly you must take account of all possibilities (up to an error tolerance in actual practice).

One often sees phrases of the form "situation X favors reaction Y".  This means the conditions (X) make the the probability of Y greater than any other possible reaction, but it doesn't equate to Y having probability 1.  Some may prefer to use this phrasing for when Y's probability is, in some sense, much larger than any other event.

The proton-proton chain that governs the main sequence stage of a star is a prime example of why you cannot discount the reactions which aren't favored.  If you did, you would conclude that our own sun is incapable of fusion.  The 'favored' reaction for two interacting protons in a star like our sun is to simply bounce off of each other thanks to Coulomb repulsion.  The protons simply don't have enough energy to overcome this repulsion in order to undergo fusion.  Technically this statement is also not probability 1, but rather very close to it.  But an improbable quantum tunneling event is possible to bypass this potential barrier, allowing the interacting protons to fuse when they collide just right and the dice favor them.  But a diproton is unstable.  The heavily favored reaction in this state is for the diproton to immediately decay into two protons.  But there is a small chance that one of the protons will undergo a beta decay into a neutron, instead, at which point you have a stable helium nucleus.  That these things are so unlikely is a major factor in why our sun needs billions of years to exhaust enough of its hydrogen (aka protons) in order to initiate helium fusion and exit the main stage.  Even very large stars still need millions of years because of this.

So while the proton-proton reaction is rather unlikely, stars are simply so immense that it is not only a virtual certainty that some proton-proton reaction will yield helium, but in fact that it is a virtual certainty that the reactions will occur in large numbers and continuously until the available reactants dwindle.

At the other extreme, we have supernovae.  There are quite a lot of varieties of supernovae, but the commonly understood types--the outer layers collapse and then violently rebound off the core--feature a sudden loss of radiation pressure due to certain endothermic reactions becoming favorable.  Iron production from fusion is the culprit here: it takes more energy to fuse it than is released.  The final split second of this sort of star's life is when conditions favor the production of iron via fusion.  Ridiculously favors it, as I really do mean it is the final split second.  There is an inconceivably tiny chance that this iron production would fail to occur for some chosen period of time, and if it happens to stall long enough then the star would likely undergo some other event.  See also David's answer for another wildly improbable, but technically possible, example of random chance altering a star's death.


What is the average time between a nebula formation and a neutron star?

Short answer: It depends on the progenitor mass, but somewhere between 5 million and 30 million years.

I presume you are talking about the sum of the star formation timescale and the time it takes for the star to evolve through the various phases of its life and then explode as a supernova.

As LaserYeti has pointed out, the formation timescale is extremely short compared to the rest of a massive star's life. In fact, for the massive stars that will eventually yield supernovae, there really isn't a pre-main sequence phase at all.

The lifetime of the star then almost entirely depends on its mass, and to a lesser extent on its rotation and metallicity. The types of star that produce supernovae and neutron stars are probably in the range 9 to 40 solar-masses. Less massive progenitors produce white dwarfs, more massive progenitors probably produce black holes and there may not even be a supernova. Stars at 9 solar masses are about 30 times more commonly born than 40 solar-metallicity stars. To find their lifetimes we must appeal to theoretical models.

Schaller et al. (1992) provide a well-respected set of "vanilla" calculations for non-rotating, solar-metallicity stars. Their table 45, summing up the H-burning and He-burning stages (subsequent evolution happens comparatively instantaneously), suggests the evolutionary lifetimes range from about 30 million years for 9 solar-mass star, to as short as 5 million years for a 40 solar-mass star. These would be the numbers I would use, though rapid rotation can increase these lifetimes by perhaps 30%. Metallicity is a smaller effect.


You are barking up the wrong tree. The passage you quote is talking about how neutron stars cool after they have formed. Neutrinos are emitted in a burst lasting a few seconds after core collapse but then continue to be produced whilst the neutron star is hot.

Neutrino production is indeed fearsomely high in the first fraction of a second after core collapse. However, the hot, dense material is opaque to neutrinos (mean free paths of 10-100 m) and they diffuse outward in much the same way as photons from the Sun. But the timescale for this is only about 10 seconds. Once the neutron star has cooled to about $10^{10}$ K, the fermions (neutrons, protons, electrons) in the neutron star become degenerate and the neutron star becomes transparent to neutrinos because only the small fraction of fermions within $kT$ of their Fermi surfaces can interact with thermal neutrinos and the neutrino mean free paths increase as $\sim T^{-2}$ and the production rate of neutrinos falls as $T^6$.

Still, the neutron star would then radiate away all of its thermal energy within a further 10 seconds, but the main neutrino-generating reactions - cycles of beta and inverse beta decay (aka the Urca process) become blocked (except perhaps right in the core at the highest densities) by an inability to simultaneously conserve energy and momentum in the degenerate gas. Instead, neutrinos continue to be generated by the modified Urca process at a much slower rate, using "bystander" baryons to conserve momentum.
$$ n + n \rightarrow n + p + e + \bar{\nu}_e$$
$$ n + p + e \rightarrow n + n + \nu_e$$

Though the efficiency of this process is low, so is the thermal energy contained within degenerate gases. Thus rapid cooling still occurs. I am not clear where Wikipedia gets its information from, but cooling to even a surface temperature of $10^{6}$K (the interiors are 1-2 orders of magnitude hotter) is likely to take longer than a few years - somewhere between 100 and $10^4$ years, unless direct URCA processes are somehow permitted (e.g. in quark matter, see Yakovlev & Pethick 2004, for a thorough review of neutron star cooling).


What practical considerations are there for amateur observations of transiting exoplanets?

This is actually quite straightforward with digital CCD's (it used to be quite tricky with film cameras as you'd have to carefully develop film that moved past the lens and assess the width of the trail)

Get yourself a good telescope - a 12" Dobsonian or above if you want to give yourself a good chance of picking out the fluctuations against the noise background. Then select a decent CCD. Five hundered pounds gets a reasonable one, but expect to pay a couple of thousand pounds for a cooled CCD, which will also help to reduce noise.

(Buying in US dollars? A reasonable CCD runs about $1000. A cooled CCD will cost you at least $1500.)

You'll want a good quality equatorial mount, with computer controlled servos to track the target smoothly over long periods of time.

Ideally you will also slave a second telescope and CCD, pointed along the same path but slightly off - this will help you cancel out cloud and other fluctuations from our own atmosphere.

Oh, and get as far away from a city as possible - up into the mountains can be a good plan :-)

Then arrange your viewing for a series of full nights. The more data points you can get, the better the noise reduction. Imagine the exoplanet is orbiting every 100 days, in order to get any useful data, you will need to track it over some multiple of 100 days. So assume you set up to track your target star for 2 years, plan for 3 or 4 star shots per night to give you a range of data points.

These 600+ days of 4 data points per night gives you a reasonable stack of data - the challenge now is to work out whether there are any cyclic variations. Various data analysis tools can do this for you. As a first step, if you find a cycle around 365 days, it probably isn't the target, so try and normalise for this (of course this will make it very difficult to discover exoplanets with a period of exactly 1 year)


Are there real uses of abstract algebra in astrophysics?

There are certainly applications of number theory (as opposed to group theory/algebra - but you do also mention abstract mathematics in general) to astrophysics, though many would suggest these applications are on the boundary between real science and metaphysics.

One such application is the  measure problem - if we assume our universe is one of an infinite number of multiverses how can we draw conclusions about the fundamental physics of the many multiverses, for instance to say that intelligent life is possible in a given proportion.


Does the retrograde movement affect the true angular distance between 2 planets?

In our moving Earth-based point of view,
retrograde apparent motion
means a planet's geocentric ecliptic longitude is temporarily decreasing
instead of increasing as usual.
The word "apparent" avoids confusion with a
retrograde orbit.
All major planets' orbits are prograde,
with heliocentric longitude increasing at all times1.
Most retrograde-orbiting Solar system objects are comets.

As seen from Earth, the apparent angular separation
between two planets near the ecliptic
is approximately2 the difference
between their geocentric longitudes at that point in time,
regardless of their rates of change in longitude.
Even if you plot this value versus time,
your calculation need not break apparent motion down
into positive and negative cases.

Except where longitude wraps around from 360$^\circ$ to 0$^\circ$.
When longitude difference is small (near conjunction), latitude difference is also significant.


How long do solar eclipses last on Metis?

I am not sure of any direct measurement, but it can be figured out rather easily and accurately from theory with a little bit of maths. We have the following parameters for Metis and Jupiter (taken off Wikipedia):

Diameter of Jupiter: about 140 Mm (megameters, 1 Mm = 1000 km), so radius ~70 Mm. Compare Earth at ~6.5 Mm radius, 13 Mm diameter.
Orbital distance for Metis from Jupiter's center: 128 Mm
Eccentricity: effectively zero, so we can treat it very accurately as a circular orbit.
Orbital period: 25.4690 ks (kiloseconds, 1 ks = 1000 s). Compare against Earth's day, 86.4 ks, and orbital period of its Moon, about 2500 ks (more accurate: 2360 ks).
Also, one more relevant point is the distance to the Sun and its diameter, to find the apparent size: average distance to the Sun is 779 Gm ($7.79 \times 10^5\ \mathrm{Mm}$), and Sun's diameter roughly 1 Gm ($10^3\ \mathrm{Mm}$).

The way we will reason now is as follows. Jupiter "moves" in one circuit across the sky dome of an observer on Metis in one orbit. Hence, the time of eclipse is the time required for Jupiter to "move", from arrival of its leading edge to departure of its trailing edge, across the image of the Sun on the same sky dome. Equivalently, then, it is the time required for it to move through an angular distance equal to its own angular diameter plus two angular diameters of the Sun (one on first contact, the other before last contact):

$$\mbox{Eclipse time} := \frac{2 \cdot (\mbox{Angular diameter of Sun}) + \mbox{Angular diameter of Jupiter}}{\mbox{Angular speed of Jupiter on sky dome}}$$

Now the denominator is easy: a full revolution is $2\pi\ \mathrm{rad}$ or about $6283\ \mathrm{mrad}$, so we just divide that by 25.4690 ks, 4o get

$$\mbox{Angular speed of Jupiter on sky dome} \approx 246.7\ \frac{\mbox{mrad}}{\mbox{ks}}$$

Hence, now all we need is the angular diameters of the Sun and Jupiter in milliradians and we'll be done.

This is easily doable with a little bit of trigonometry: while I do not have a diagram on hand, you can easily find one. The angular size of an object of girth $D$ emplaced at a distance $d$ from an observer is

$$\mbox{Angular size} := 2 \cdot \tan^{-1}\left(\frac{D}{2d}\right)$$

Note that you'll often see the approximation

$$\mbox{Angular size} \approx \frac{D}{d}$$

but this is only good when the object is very far, i.e. $d \gg D$. In this case, though, that is not the case at least for Jupiter, so we have to use the full formula with the inverse tangent. Using the figures just given, we have

$$\mbox{Angular size of Jupiter} = 2 \cdot \tan^{-1}\left(\frac{140\ \mathrm{Mm}}{2 \cdot 128\ \mathrm{Mm}}\right) \approx 1.000\ \mathrm{rad}$$

or accurately(!) $1000\ \mathrm{mrad}$. The Sun is far enough that we can just use the approximation and get 1/779 rad or about $1.28\ \mathrm{mrad}$. Hence the total angular distance to cover is 1.28 + 500 + 1.28 ~ 503 mrad, and the eclipse time equals

$$\mbox{Eclipse time} = \frac{1000\ \mathrm{mrad}}{246.7\ \mathrm{mrad/ks}} \approx 4.08\ \mathrm{ks}$$

or for so many still not fully accustomed to the SI, a bit longer than an hour (3.6 ks), and in local measures, a bit less than 1/5 of a Metian day-month.

NOTE: I should point out a caveat of this is is that that this doesn't take into account the inclination of orbits - both Jupiter's and Metis's which may make the Sun in effect "encounter" Jupiter at a different point than simply cutting along its equator. But it should still give you an idea - 4 ks or so, plus-minus some amount that'd be too much work to get.


What is the range of the zenith angle between velocity and position vectors?

Partial answer:
What Braeunig calls gamma or zenith angle is actually beta!

...and γ is the angle between the position and the velocity vectors, called the zenith angle (see Figure 4.7))

According to @TomSpilker's answer

This is a problem that has plagued groups of people very knowledgeable about orbital dynamics but who learned using different textbooks: there are two different definitions of "flight path angle"!!
In addition to $\gamma$, the angle between the tangential direction and the velocity vector, there is $\beta$, the angle between the radial direction and the velocity vector. People often say "flight path angle" without saying which definition they're using. Confusing! (I just noticed that the diagram in Julio's answer also shows $\beta$)
If you work with $\beta$ instead of $\gamma$, $\beta$ is given by
$$\arccos\left(\frac{\mathbf{r \centerdot v}}{|\mathbf{r}| \ |\mathbf{v}|} \right) \tag{1} $$

So Braeunig's gamma is NASA's beta. It's positive from peri to apo, and negative from apo to peri. Because I'm not smart enough to do it with math, I'll do it with Python.
I'm doing this in dimensionless units, so the period is 2π and the semi-major axis is unity.  You can see beta's behavior over one orbit, positive moving towards apoapsis, then negative moving towards periapsis.

def deriv (X, t):
    x, v = X.reshape(2, -1)
    acc = -x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

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

time = np.linspace(0, twopi, 361)

r0 = 1.8
v0 = np.sqrt(2./r0 - 1)  # vis-viva

X0 = np.hstack([r0, 0, 0, v0])

answer, info = ODEint(deriv, X0, time, full_output=True)

x,    v      = answer.T.reshape(2, 2, -1)
rn = x/np.sqrt((x**2).sum(axis=0))
vn = v/np.sqrt((v**2).sum(axis=0))

x, y, vx, vy = answer.T

beta = np.arccos((rn*vn).sum(axis=0))

if True:
    plt.figure()
    plt.plot(x, y)
    plt.plot([0], [0], 'ok', markersize=8)
    x0, y0 = 0.5*(x.min() + x.max()), 0.5*(y.min() + y.max())
    hwx, hwy = (np.abs(x-x0)).max(), (np.abs(y-y0)).max()
    hw = 1.05*max(hwx, hwy)
    plt.xlim(x0-hw, x0+hw)
    plt.ylim(y0-hw, y0+hw)
    plt.show()

if True:
    plt.figure()
    plt.subplot(3, 1, 1)
    plt.plot(time, x)
    plt.plot(time, y)
    plt.plot(time, np.zeros_like(time), '-k')
    plt.title('x and y vs time')
    plt.subplot(3, 1, 2)
    plt.plot(time, vx)
    plt.plot(time, vy)
    plt.plot(time, np.zeros_like(time), '-k')
    plt.title('vx and vy vs time')
    plt.subplot(3, 1, 3)
    plt.plot(time, beta)
    plt.plot(time, halfpi*np.ones_like(time), '-k')
    plt.title('beta vs time')
    plt.show()


How well can the Sun's surface be resolved by an observatory?

Solar telescopes typically have filters to reduce the amount of light reaching their detector. Depending on their purpose they may use, for example, neutral density filters to reduce the amount of light to a level where they can resolve details without blocking any of the visible spectrum, hydrogen alpha filters to observe the Sun at a wavelength of 656 nanometres for observing features in the Sun's atmosphere such as solar prominences and the chromosphere, or blocking the solar disc to observe the Sun's corona, flares and other ejecta.

Hubble, on the other hand, is designed with a very different purpose in mind such as taking long-exposure images to obtain the deep- and ultra-deep field full spectrum images. It avoids the Sun as even a very short exposure to the Sun would damage its instruments.


Transform pixel coordinates (in FITS file) to equatorial

Assuming that each (square) pixel has the same angular scale (not a given if the field of view is large) of $\theta$ degrees/pixel.

Then the declination (in degrees): $\delta \simeq \delta_0 + \theta y$

The right ascension (in degrees): $\alpha \simeq \alpha_0 - \theta x/\cos \delta$

where $\alpha_0$ and $\delta_0$ are the RA and Dec at $x=0$, $y=0$. [The minus sign is there because right ascension increases towards the left of a sky image.]

This approximatimation becomes poor as the field of view gets larger.

A slightly more complex, but accurate, approach is described in http://gtn.sonoma.edu/data_reduction/astrometry.php


Does a planet's mass or gravity affect the height of it's mountains?

Yes, gravity definitely affects the maximum heigh of mountains.

Think on a solid bar of steel. It sticks straight because of the electronic forces. But when you make it larger and larger gravity makes it bend: gravity starts being considerable, but still smaller than electronic forces.

If you make the bar larger, there will be a moment in which the weight of the whole bar will be larger than the short-range electronic force: your bar will break purely due to gravity.

Exactly the same happens to mountains made of solid rock (as opposed to sedimentary ones quoted by Hobbes). There is a point,k depending on the strenght of planetary gravity, where it takes over short-range electronic forces, making the mountain collapse.

This is exactly the force that "rounds up" the planets, as opposed to the non-spherical asteroids.


Can Kepler's 3rd Law be applied in higher dimensions to describe orbits (quadric surfaces)?

On the risk of not fully answering your question:

Trajectories of a system through some solution space, are in physics always solutions of some set of differential equations. However using the right formalism, one can often derive properties of the solution, without actually knowing it.

For the Kepler-problem, there is such a trick.

Let us consider the Lagrangian $L=E_{\rm kin} - E_{\rm pot}$ of the Kepler-problem:
$$L = \frac{1}{2} m (v^2_x + v^2_y + v^2_z) - \frac{\alpha}{\sqrt{x^2+y^2+z^2}},$$ but to simplify the notation we shall abbreviate this to
$$L = \frac{1}{2}mv^2-\frac{\alpha}{r}$$
The fact that this simplifies to a 1-D problem is of importance later.
The equation of motion of the Kepler-problem can be derived from the corresponding Euler-Lagrange equation and results in Newton's law plus $E_{\rm pot}$.

Now we can multiply the Lagrangian by a constant $L'= c L$, and the resulting equation of motion, hence the solutions of the equation of motion, will remain identical. If we choose $c$ carefully, then we get the desired information about the solution of the equation of motion, by only touching $L$. This principle is called the principle of scale-invariance, and it doesn't apply to all Lagrangians or all systems in physics.

Now introduce a different idea, that of rescaling of space and time: Imagine a 'zoom' in space and a 'fast-forward' or 'slow-down' in time, such that $x$ and $t$ are rescaled according to
$x'\rightarrow c_x x$,
$t'\rightarrow c_t t$,
where $c_x$ and $c_t$ are some new constants, that are arbitrary initially. From those rescalings follow now, for the scaling of v that $v' \rightarrow c_x/c_t v $ and the potential, because it only depends on one length, scales as $\alpha /r' \rightarrow \alpha / (c_x r)$.

Let us plug-in all those results from the scalings into $L'$ and relate this to the unprimed, original quantities,
$$L' = \frac{1}{2}mv'^2 - \frac{\alpha}{r'} = \frac{1}{2}m v^2 \cdot \frac{c_x^2}{c_t^2} - \frac{\alpha}{c_x \cdot r}$$.

Now the latter looks a little like a mess. But we want to write this as $L'=cL$, which is only possible, if we can factor out those factors containing all the $c_x$ and $c_t$s. The condition for this is $\frac{c_x^2}{c_t^2} =  c \cdot \frac{1}{c_x}$ or  $$c_x^3/c_t^2 =c$$,
which is one statement of Kepler's third law, namely that the spatial-scale determining and the temporal-scale determining constants have the above relation w.r.t each other. Note that as long as the potential remains a central potential, it doesn't matter how many coordinates we have, i.e. if $r=\sqrt{x^2+y^2+z^2 + a^2+b^2+...}$ this statement retains validity.


There are (at least) a couple of problems with your hypothesis.

The first problem is that dark matter is just that - dark, in other words it emits no light that we can detect. The exploded remnants of stars are not dark at all. Indeed the winds and superwinds associated with exploding stars are often extremely hot and luminous in the UV and X-ray region. Even when these winds have slowed and cooled, they will then form warm dust (because they are enriched in heavy elements) that is then detectable in the infrared part of the spectrum.

The second problem is that we know that the majority of dark matter is not just non-luminous "normal" matter. Estimates of the cosmic abundances of helium, deuterium and lithium produced in the big bang are sensitive to the overall density of this "normal" matter. These estimates tell us that only about 20% of the gravitating matter that appears to be required to explain the dynamics of the universe, the dynamics of galaxy clusters and the dynamics of galaxies, can be in the form of "normal" matter. But the atoms and particles released from an exploding star are just part of this "normal" matter budget - they have already been accounted for and so contribute nothing to solving this second dark matter problem.


Table of dates for planet retrograde motion

EDIT: It turns out calsky.com will do these calculations for several years out, including other calculations commonly found in almanacs. Sample page:

https://www.calsky.com/cs.cgi?cha=7&sec=6&sub=3

The best place to start on this site isn't the home page, but rather the TOC:

https://www.calsky.com/cs.cgi/TOC?&f=

I now consider my answer to be better than @JohnHoltz's :)

Note: I saw @JohnHoltz's answer while I was typing out my answer, so some of the below may be redundant. Upvote/checkmark his answer, not mine.

This isn't really an answer, but too long for a comment:

I was disappointed how difficult it was to find this information, given how easy it is to calculate and how many other astronomical phenomena are listed in various places on the Internet.
When a planet starts or ends retrograde motion, it is temporarily stationary before it "turns around". Thus, some astronomical sources will say "Jupiter stationary" instead of "Jupiter starts retrograde motion".
Based on the above I google'd "jupiter stationary" "mars stationary" (as quoted; google tries to correct one of the stationary's to stationery, so I had to click one more time to do the search correctly), and finally found something: http://aa.usno.navy.mil/publications/docs/ap.php
The publication doesn't mention retrograde, but it does mention when a given planet is stationary and provides a brief description of each planet's position in the sky for the given year.

Because this seems like a fairly easy calculation to make, I may do it myself and update this answer.


Latitude and longitude of Eugene Shoemaker's final resting place?

A small portion of Eugene Shoemaker's ashes were placed on the Lunar Prospector mission. After completion of the mission's primary objectives:

Lunar Prospector was deliberately impacted onto the shadowed Shoemaker crater on the lunar surface at 09:52:02 UT July 31, 1999.
NASA Mission Description

No coordinates are given in that source, however, the Google Earth rendering of the Moon has a placemark for the crash site.  No coordinates are listed.
A source from before the impact occurred lists the planned crash site as 87.70 S 42.0 E, which is very close to the placemark, and gives an estimated crash time within a minute of the actual event.
Thus, the likely resting place of that portion of Shoemaker's ashes may be gives as: 87.70S 42.0E, Shoemaker Crater.
Note: The idea of crashing into the crater was put forth by the University of Texas at Austin, and they had a small site about the crash.  I was able to find a snapshot from the period that has working links on the Internet Archive here.
There was also an entire site devoted to the amateur astronomy aspect of the crash, also on the Internet Archive: lunarimpact.com.
Information from these sites indicates that the reason the crash time was different was due to a stronger-than-expected deorbit burn, and thus the actual crash site might be closer to 87.7S, 42.1E, which is exactly where the Google Earth marker is. The last mission report, however, lists it as 87.7S 42E.


With the axis of rotation of the earth matching the axis of rotation of the stars, and the equator of the earth matching the equator of the stars, any place along the equator would match what you have described. However due to closeness to the horizon, the polar-most stars wouldn't be visible. If you were able to get a hundred metres or so above the ground, with nothing in the surrounding landscape, you could see all of the stars in the sky over a year's time.


What exactly is "the rotating lighthouse model" in the context of a double pulsar?

It seems that the authors are just referring to the accepted model of a pulsar, i.e. a neutron star spinning and emitting beams of radiation at its poles. In that sense, the term is used here no differently than it would be in the context of an isolated pulsar. It's just a very simple way of visualizing why an observer far away appears to see periodic pulses rather than constant emission.
If the lighthouse model is correct, then the beams from one pulsar may interact with the other pulsar, which in turn would cause changes in the second pulsar's emission each time the beam sweeps across it. The modulation period would then be approximately equal to the rotational period of the second pulsar - not precisely, though, because of orbital motion; hence, the comparison between Earth's sidereal day and solar dat.
Modulation of pulse strength and features therefore isn't a feature of the lighthouse model but rather a consequence of it, present in some binary pulsar systems but of course not in isolated pulsar emission.


Does the Lagrange L1 point have a 'size'?

From this source I get:

  The size of these islands varies. Each planet in the solar system has
  its own Lagrangian points. The islands of stability get bigger farther
  from the Sun and also for more massive planets. The ones associated
  with Earth are roughly 500,000 miles (800,000 kilometers) wide. The
  biggest zones (at least in the solar system) are Neptune’s; they are
  about 2 billion miles (3.2 billion km) across.

L1, L2, and L3 are unstable. If a Sattelite ventures too far from the Lagrangian area it will fall towards the sun. As you can see from the image below. This image came from the source linked above.


Are black holes spherical during merger?

No need to guess. There's solid research done in this field. Even Wikipedia has some info:

  As two black holes approach each other, a ‘duckbill’ shape protrudes
  from each of the two event horizons towards the other one. This
  protrusion extends longer and narrower until it meets the protrusion
  from the other black hole. At this point in time the event horizon has
  a very narrow X-shape at the meeting point. The protrusions are drawn
  out into a thin thread. The meeting point expands to a roughly
  cylindrical connection called a bridge.

https://en.wikipedia.org/wiki/Binary_black_hole#Shape

There are research papers with images showing the results of calculations of the shape of the event horizons during merger. Here's an example:

The image above is taken from this paper:

On Toroidal Horizons in Binary Black Hole Inspirals

  We examine the structure of the event horizon for numerical
  simulations of two black holes that begin in a quasicircular orbit,
  inspiral, and finally merge. We find that the spatial cross section of
  the merged event horizon has spherical topology (to the limit of our
  resolution), despite the expectation that generic binary black hole
  mergers in the absence of symmetries should result in an event horizon
  that briefly has a toroidal cross section.


According to the standard interpretation of General Relativity (e.g. as presented in "Exploring black holes" by Taylor & Wheeler chapter 3, or "Black holes, white dwarfs & neutron stars" by Shapiro & Teukolsky, pp 343-345) then yes they do. But it depends on the frame of reference of the observer - there is no absolute answer.

According to an observer far from the black hole, the rate of change of radial coordinate with time (for an object that started falling radially inwards far from a non-rotating black hole) is given by
$$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}$$
where $r_s$ is the Schwarzschild radius and $r$ and $t$ are Schwarzschild coordinates.

If we call this the infalling speed as measured by a distant observer then we can see by differentiation that it goes through a maximum at $r=3r_s$ and that $dr/dt \rightarrow 0$ as $r \rightarrow r_s$.

However, an observer accompanying the falling particle would totally disagree. To them, their velocity is given by $dr/d\tau$, the rate of change of $r$ with respect to the time on their clock.
$$\frac{dr}{d\tau} = -\left( \frac{r_s}{r}\right)^{1/2}$$
which continues to increase up to and below the event horizon.

The latter appears to admit the possibility of faster than light travel, but no more so than me (correctly) saying that if you travel at close to the speed of light you can get to a star 10 light years away in much less than 10 years (as measured on your clock).

Finally we could have the point of view of stationary "shell" observers at fixed radial distances (outside the event horizon, because no stationary observer is possible below the event horizon). They would measure the speed of objects falling past them to be
$$\frac{dr_{\rm shell}}{d\tau_{\rm shell}} = -\left(\frac{r_s}{r}\right)^{1/2}.$$
This means that the reports of stationary observers (which is the gist of your question I think) at increasingly lower heights, is indeed that the velocity of the falling object is increasing as it falls.

There is no paradox to these apparently contradictory points of view. Measurements of non-local events and phenomena are not bound to agree in General Relativity, where there isn't even agreement on what is meant by "now" or whose coordinate system in what frame of reference should be used in any particular circumstance.


How do we convert other waves of the EMS to visible light?

It's just as simple as taking the flux at some wavelength (just a number) and using this number to represent a visible intensity.

If you only have one wavelength then you can only get a monochrome picture. However, if you have flux information at more than one wavelength, let's say three, you can use the flux at the longest wavelength to represent red (r), the middle wavelength to represent green (g) and the shortest wavelength to represent blue (b).

Put this together and you have 3 numbers representing a visible rgb signal that can be used to create a picture. Of course there may be lots of fiddling that goes on with the colour balance and the contrast to produce an effective picture, but this is the basic process.


Need advice on mirror-making for a home-made amateur reflector

I second the comment about Texereau -- originally written in French; it's the best, bar none. You won't need anything else.

Remember the old adage that is quicker to make a 4" mirror and then an 8" mirror than it is to make an 8" mirror first. Don't bother with borosilicate. You should be able to get mirror kits and all the abrasives that you need online. Join a club if you can find one.

How to Make a Telescope, 2nd edition by Jean Texereau:

http://www.willbell.com/tm/tm3.htm
https://www.amazon.com/Telescope-second-Texereau-published-Atlantic/dp/B00EKYQCQS


Terminology question: gas giant vs gas planet

"Gas giant" is the most standard term for planets in the Jupiter/Saturn class, and, as you said, "ice giant" (since the 90's) is the term for planets with compositions like Uranus and Neptune.

"Gas planet", on the other hand, is just not a commonly used term. Judging by Google n-grams, it is less than 1/12 as frequent. Also, if you do a Google book search, of the times it is used, it is very often as a part of the phrase "giant gas planet." I would consider "gas planet" simply to be a description rather than a technical term, at this point.

A lot of likely materials for Earth-sized non-rocky planets would be more likely to be classified as ices in the outer solar system. So planets that deserve to be called "gas planets but not gas giants" may be quite rare, so it's possible a distinction may not be needed.


Why does lithium fuse at lower temperatures than hydrogen?

The slowest reaction rate in the pp chain determines how quickly hydrogen can "burn" in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay.

The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of the PPII series of reactions that convert helium 3 into helium 4. It does not involve weak interactions and therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants. It is therefore initiated at lower temperatures - as you say, Li "burning" occurs in low-mass pre main sequence stars well before hydrogen "burning" commences (but after deuterium burning).


How to use a telescope to find a specific celestial body?

You have to set your equatorial axis to tilt to match your latitude eg if you were at 5 degrees North then the axis needs to tilt to 5 degrees. Then you need to find an object of known RA and declination (at my latitude I always used Polaris as it didn't move) and then set the setting circles (the dials) to those, then direct the telescope to the RA and declination you seek.


How are the compositional components of exoplanet atmospheres differentiated?

In theory one should just be able to determine the difference between the spectra during the star's eclipse of the exoplanet (starlight alone) and the spectra of the star and exoplanet together, but in reality equipment is not precise enough for this. To remedy this problem, the analysis is integrated over many eclipses. Some other calibrations are made as well which you can read more about in the paper A Spitzer Spectrum of the Exoplanet HD 189733b. This is called occultation spectroscopy.
There's another method used as well called transmission spectroscopy, which detects the change in light as the exoplanet passes in front of its star. You can read more about that in this paper: Infrared transmission spectroscopy of the exoplanets HD 209458b and XO-1b using the wide field Camera-3 on the Hubble Space Telescope.


What effects does the motion of the Sun have on the perihelion precession of Mercury?

A better way to phrase that question is "What effects do the planets have on the perihelion precession of Mercury?"
When calculating the perihelion precession of a planet, one is implicitly working in a heliocentric frame, one in which the Sun is viewed as fixed. Perihelion precession is defined as the precession of the motion of a planet relative to the Sun. The motion of the Sun is irrelevant. The motion of a planet with respect to the barycenter is significantly more complex than is the motion of a planet with respect to the Sun. To illustrate, I'll quote from an answer I gave on physics.SE five years ago:

The following plot shows the distances between Venus and the Sun (red) and Venus and the solar system barycenter (black) from January 1970 to December 2014. The horizontal (time) axis is in days from 12 Noon TT, 1 January 2000.

Note that the red curve, the distance between the Sun and Venus, exhibits a key characteristic of an elliptical orbit, which is a repetitive, nearly sinusoidal distance curve. The black curve, the distance between the solar system barycenter and Venus, does not. It exhibits beats and other nastiness.
One way to model the behavior of a planet orbiting the Sun while acknowledging the presence of other planets is to treat the center of the Sun as the center of an accelerating frame of reference. This results in what aerospace engineers and solar system modelers call "third body effects". The effective acceleration of Mercury toward Jupiter in a heliocentric frame is the gravitational acceleration of Mercury toward Jupiter less the gravitational acceleration of the Sun toward Jupiter.
This approach combined with numerical integration could be used to model the entire solar system. Doing so would have the advantage of not having to worry about where the barycenter is. It has the disadvantage of making an already highly coupled set of differential equations even more highly coupled. That disadvantage outweighs the advantage, making solar system modelers use a barycentric approach when modeling the entire solar system.
Neither approach (numerically integrating the solar system from a heliocentric vs barycentric approach) was used to discover the problem with Mercury's orbit by  Urbain Le Verrier in the 19th century. The numerical integration techniques currently used to model the solar system very much depend on digital computers, something that didn't exist in the 19th century. The number of calculations needed far exceeded the capabilities of the human computers available in the 19th century.
Instead, Le Verrier and others who followed used Lagrange's planetary equations, or variations of those equations to model Mercury's behavior. These equations yield the contributions of perturbing forces (or perturbing potentials) to the time derivatives of various orbital elements. In particular, what is $\dot\omega$, the time derivative of the argument of perihelion, for Mercury?
Le Verrier calculated that the planets would cause Mercury's orbit to precess by 526.7 arc seconds per century. By 1912, Doolittle (and others) had found some issues with Le Verrier's calculations and had refined the Newtonian effects of other planets on Mercury's orbit to a precession of 532.36 arc seconds per century.
Neither Le Verrier's value nor Doolittle's refinement agreed with observation. There was a 43 arc second per century discrepancy between Mercury's observed perihelion precession and the calculated values, which Einstein showed was very nicely explained by general relativity. Note that the relativistic effect is small, less than 10% of the combined planetary effects.

References:
Doolittle, Eric. "The secular variations of the elements of the orbits of the four inner planets computed for the epoch 1850.0 GMT." Transactions of the American Philosophical Society 22.2 (1912): 37-189.


At what rate is the observable universe expanding?

How fast does the observable Universe grow?

The observable Universe is $r = 46.3\,\mathrm{Glyr}$ (billion light-years) in radius, so by Hubble's law, galaxies at that distance recede from us at a speed
$$
v_\mathrm{rec} = H_0\,r = 9.6\times10^5\,\mathrm{km}\,\mathrm{s}^{-1},
$$
or $3.2$ times the speed of light, $c$.

At the same time, light from ever-increasing distances will have had the time to reach us. Since light travels at the speed of light, this adds another factor to the edge of the observable Universe. That is, its distance recedes at $v_\mathrm{rec}+c = 4.2c$.

Hence, today the radius of the observable Universe is
$$
\begin{array}{rcl}
dr & = & (v_\mathrm{rec} + c)\,dt\\
   & \simeq & 4.2\,\mathrm{light\text{-}days}\\
   & \simeq & 10^{11}\,\mathrm{km}\\
   & \simeq & 730\,\mathrm{AU}
\end{array}
$$
bigger than it was yesterday ($dt=1\,\mathrm{day}$).

But remember that most of this growth does not mean that our observable Universe encompasses more galaxies; only that they're farther apart.

How far apart were two photons emitted, if they arrive one day apart?

In reality, we don't see photons from the Big Bang. The oldest photons we can detect are from the cosmic microwave background, but humans don't see those. But for the sake of the argument, consider two photons, both emitted right after cosmic inflation, but at slightly different distances from the place you will once stand.

Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, and the Universe has expanded by a factor of $a^{-1} \simeq 2.3\times10^{26}$ since then. Hence, what we today call the observable Universe was, at the end of inflation, only
$$
r_\mathrm{then} = ar\simeq10\,\mathrm{m}.
$$

That is, the photon you saw yesterday, was emitted 13.8 Gyr ago, only 10 meters from you.

To find the distance at which the photon you see today was emitted, we can calculate the fractional distance between the two regions emitting the photons — but now we shouldn't use physical coordinates, but instead comoving coordinates, because the part of the extra distance that is due to expansion shouldn't be taken into account. In comoving coordinates, the observable Universe increases a distance of $r_\mathrm{com} =$ one light-day per day, or a fraction
$dr_\mathrm{com}/r_\mathrm{com} = dr_\mathrm{com} / r = 6\times10^{-14}$.

In other words, the region emitting photon #2 is, and always has been, a fraction $6\times10^{-14}$ farther away than the region emitting photon #1.
If region 1 was 10 meters away, then region 2 was
$$
\begin{array}{rcl}
\frac{dr_\mathrm{com}}{r_\mathrm{com}}\times 10\,\mathrm{m} & \simeq &  6\times10^{-11}\,\mathrm{cm}\\
& \simeq & 0.6\,\mathrm{picometers}.
\end{array}
$$

(I originally forgot to calculate in coming coordinates, resulting in a slightly too large value.)

A note on the calculations

In the above calculations, I've used cosmological parameter data from Planck Collaboration et al. (2016), because those are implemented in Python's astropy module. The newer data from  Planck Collaboration et al. (2018) would change the result only a little bit. If instead of $10^{-32}\,\mathrm{s}$ you assume $10^{-33}\,\mathrm{s}$ for the end of inflation, as some models predict, you'd get slightly smaller Universe at the end of inflation (a factor $\sqrt{10^{-32}/10^{-33}}\simeq3$).


Is looking through a telescope to do astronomy a thing of the past?

Scientifically there's little to gain when you look through a telescope with your own eyes. Attach a camera to the telescope and you immediately document what you observe and take out the subjective factor from it at the same time and allow careful analysis and more detailed. Additionally you gain improved sensitivity enhanced observation duration and broaden (or narrow) the range of possible wavelengths which allows to gain additional insights.

That said, many professional astronomers still enjoy looking at the sky with their own eyes, be that with or without telescope.


Between Mercury and Venus, which planet produces a longer transit? Assuming both planets describe same path on the solar disk

This is a cool question; I'll give it a go.

  Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?

No, nothing here is obvious!

tl;dr: Times for the transit of Mercury can range from 5.37 to 8.30 hours, and for Venus they range from 7.94 to 8.12 hours under assumptions of coplanarity but allowing for eccentricity.

Let's assume all orbits are in one plane. We can get the orbital speed of body $i$ with semimajor axis $a$ and current distanc $r$ from the vis-viva equation;

$$v_i = \sqrt{GM\left( \frac{2}{r_i} - \frac{1}{_i} \right)}$$

where $GM$ is the standard gravitational parameter of the Sun which is about 1.327124E+20 m3/s2.

               a          ecc        pei           apo       v_peri   v_apo
Earth    1.49598E+11   0.016709  1.470984E+11  1.520976E+11  30286.6  29291.1
Venus    1.08208E+11   0.006772  1.074752E+11  1.089408E+11  35258.8  34784.4
Mercury  5.79091E+10   0.205630  4.600125E+10  6.981695E+10  58976.4  38858.6

Imagine Venus and Mercury crossing a line fixed at one end to the Sun and the other end to the center of the Earth. The linear speed of a planet $i$ relative to that line will be $v_i$ minus the speed of the line at the distance of planet $i$ or $v_E r_i / r_E$, and the angular velocity seen from Earth will be that divided by ($r_E - r_i)$ or

$$\dot{\theta}_i = \frac{v_i - (r_i/r_E) v_E}{r_E - r_i} = \frac{r_e v_i - r_i v_E}{r_E^2 - r_i r_E}$$

(seconds)     Venus          Mercury
Earth       peri   apo      peri   apo
peri       28570  29242    19335  29884
apo        28060  28626    19384  29646

(hours)       Venus          Mercury
Earth       peri   apo      peri   apo
peri       7.936  8.123    5.371  8.301
apo        7.794  7.952    5.384  8.235


Practicing image processing and analysis

I know you've already accepted an answer, but I feel that it doesn't provide a true representation of the type of image processing astronomers actually do. Actual image processing in astronomy is rarely done with JPG, PNG, or TIFF files. Instead, astronomers mostly work with FITS files and cameras/astronomical instruments actually produce FITS files when taking images. The advantage here is that FITS files can record much more information than just image information such as where/when the image was taken, multiple layers of the image, encoded WSC information, etc.

As part the mission objective of various organizations such as the National Science Foundation and NASA, all raw astronomical data taken in the United States by observatories funded by the United States are free and open to the public (after a limited time period in which the observing scientist gets exclusive rights to the data for their science objectives - usually several months to a year). This includes both ground (e.g. Keck, SDSS, Arecibo) and space observatories (e.g., Hubble or Spitzer). This is also true of many observatories funded by other countries as well.

A great place to start would be to head over to the Hubble Legacy Archive. Enter the site and try searching for an object or coordinates. For example, you can search for M101 (otherwise known as the Pinwheel Galaxy). You'll see a list of images come up and next to them, you can download the FITS file generated by Hubble when it took the image.

Another great resource would be the Sloan Digital Sky Survey. This is a systematic astronomical observation of as much of the sky as possible using a variety of instruments. You can practice looking at images and spectra. I should warn you that it will take a bit more effort to figure out where the data is and how to interpret it, but feel free to ask questions here when you run into trouble.

In general though, if you're looking for a particular type of image or data, find an observatory that records it and look for their publicly available data sets.


Question about Telescopes , observations and explanations of the observations

I'll attempt to answer as many of these questions, to the best of my abilities, as I can at this late hour.

1) Lenses have an index of refraction that is different than vacuum, and air, and they are curved. ie: corners when light is concerned.

2) The entrance and exit pupils are the pupil size required to pass an "extreme" ray through an optic. Meaning that if the light ray from the optical axis was not refracted at all as it passed the optical plane, it would still pass within the pupil (entrance) and if reflected perfectly (exit). The diameters of the objective and the eyepiece are subjective, meaning that you can have, within reason, any size objective and use virtually any eyepiece. The restrictions to this would only be focal lengths.

3) You are looking through a thicker lens and thereby reducing the light incident on your eye. Think about the shape of the lenses, look at them and it might become apparent just how much more curvature the 4mm lens has than the 20mm lens. This added curvature comes at the cost of a higher angle of incident that a light ray has to take, and thus more optic to pass through.

4) Barlow lenses reduce the brightness a lot because you are adding additional optics. With each new addition you decrease the amount of light by a factor relative to the thickness and clarity of the optics. The Barlow lens is increasing the magnification and thereby reducing the field of view.

5) With your specific telescope (of which I have one with the exact same size/focal length) you should be able to see Jupiter as a disk, Mars too should be more than a "dot." Stars are point sources with angular diameters much too small for you to be concerned given the size of your telescope. Also, those planets appear reddish because they are-ish. Many stars will still appear colors other than blue if you can find them.

6) This is up to your discretion and desired results when observing. Please read your owners manual, or check out the telescope mfr website for recommended setups.


How to determine arrival time delay given dispersion measure?

As is the case with light traveling through any medium, radio waves traveling through space experience refraction, which reduces their speed. A wave of infinite frequency will experience no refraction, meaning that we can compute how much a given wave will be delayed compared to such an infinite-frequency wave as it moves through outer space. The group velocity of a radio wave is determined by $v_g=c\mu$, with $\mu$ the refractive index defined by
$$\mu=\sqrt{1-\left(\frac{\nu_p}{\nu}\right)^2}$$
with $\nu$ the frequency of the wave and $\nu_p$ the plasma frequency, a fundamental property of the medium, which depends on its density:
$$\nu_p=\sqrt{\frac{e^2n_e}{\pi m_e}}$$
with $e$, $m_e$ and $n_e$ the charge, mass and number density of electrons. (You're probably more familiar with refractive indices for light that humans can see, but they're important in a much wider portion of the electromagnetic spectrum!) In the ISM, we'd expect to see plasma frequencies in the kilohertz range - much lower than the frequencies of radio waves from whatever source we're observing, which are likely in the hundreds of megahertz to tens of gigahertz range. The delay a wave of frequency $\nu$ experiences over a distance $d$ is
$$t(\nu)=\int_0^d\frac{dl}{v_g}-\frac{d}{c}$$
where the second term requires no integral as the refractive index of a wave of infinite frequency is always $\mu=1$ regardless of the medium. We can use a binomial approximation for the refractive index of our physical finite-frequency wave, since $\nu\gg\nu_p$, and after some algebra, we find that
$$t(\nu)=\frac{1}{c}\int_0^d\left(1+\frac{\nu_p^2}{2\nu^2}\right)dl-\frac{d}{c}=\frac{e^2}{2\pi m_ec}\frac{1}{\nu^2}\int_0^dn_edl$$
Now consider two waves of frequencies $\nu_1$ and $\nu_2$. The difference in arrival times is
$$
\begin{aligned}t(\nu_1)-t(\nu_2)&=\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl-\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl\\
&=\frac{e^2}{2\pi m_ec}\left(\int_0^dn_edl\right)\left[\frac{1}{\nu_1^2}-\frac{1}{\nu_2^2}\right]
\end{aligned}$$
If we define the dispersion measure by
$$\text{DM}\equiv\int_0^dn_edl$$
and define it in units of $\text{cm}^{-3}\;\text{pc}$, write the chunk of constants in front in units of $\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$:
$$\frac{e^2}{2\pi m_ec}=4.15\;\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$$
and write out frequencies in units of $\text{GHz}$, then we indeed recover the expression you give - with the time delay in units of milliseconds. It would be a lot clearer if we rewrote it as
$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{cm}^{-3}\;\text{pc}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$
which has the proper units.
In short: If you write your frequencies in gigahertz and your dispersion measure in parsecs per cubic centimeter, the formula will give you a time delay in milliseconds.


Radiative Flux, Irradiance and Luminosity confusion

It looks to me like irradiant flux is the same as radiant flux is the same as irradiance, both are energy per second divided by area.  That interpretation would seem to be consistent with everything you quoted.  In more precise terms, these should all be "flux density", not "flux", because flux just means the amount of something per time, but that convention is never followed much any more.


R.A. and DEC. for Constellation area vertices

James Kilfiger has suggested me to expand my comment as answer, and here it is:

Searching I have found the data in the international astronomical union website. In http://iau.org/public/themes/constellations you can find a TXT for each constellation with vertex's R.A. and DEC.

At the end of that page there is an explanation about the fields on that text files. Their coordinates are J2000.


I need to study space, planets and black holes, how do I study them on my own

I am sorry to say, but the best resources are still books. If it is really a no-go, read further. It is very likely, that you can find and buy astronomy books on Arabic on the Internet. First start with popular science books. Buy multiple books. To learn something, books are still somehow far better than the Internet.
Check the Arabic wikipedia, like this article.
Just think freely. Imagine a question, anything, like "How far are we from the edge of Universe?"1. My experiences is that the resources on my first language (far lesser amount of speakers than Arabic) are significantly lesser, but they exist. Enter the question on Arabic into the google search.
Any time if something is not clear, come here and ask.

1It is actually a bad question, but you don't know yet, why. Here is the answer, why. This was only an example question.


When a rogue brown dwarf star hits a white dwarf star, will the collision at most cause a nova?

Collisions between a rogue brown dwarf and any other star would be very rare because the space between them is so vast.   I don't want to say it'd never happen, but it would be a rare event.   It's much more common for two stars that are already in the same system to collide by spiraling into each other, usually by tidal decay.

A collision with a brown dwarf and a white dwarf has been recognized.   The nova of 1670 didn't behave like an exploding star and recent research identified it as spiraling in collision between a brown dwarf and a white dwarf.  Original article and abstract here.

As for what kind of explosion is likely.  I would imagine something like a type-A supernova would be possible if there was sufficient mass to exceed the Chandrasekhar limit of about 1.4 solar masses.   Because brown dwarfs are much less mass than that, about .013 to about .08 solar masses, the white dwarf would need to be at least 1.32 solar masses prior to the collision.  Most white dwarfs are less than that.

CK Vulpeculae, the 1670 nova wasn't a type 1a supernova.  If it had been, it would have been over 100 times brighter than Venus (ballpark estimate) based on it's distance 2280 light years (plus/minus 490).   It would have been many times brighter than it's visible but not very bright peak of 2.6.   SN 1604, for comparison was about 20,000 light years distant and it was brighter than any star in the sky and visible during the day at it's peak.

The mass of the two colliding objects and energy of collision would have considerable effect on the brightness, but, unless the Chandrasekar limit is reached, and keeping in mind that direct collisions would be rare, and it wouldn't hit all at once, the result should be much less energetic than a supernova, but if it's close enough, it could still be a visible nova in the sky, like CK Vulpeculae in 1670.


What is the largest (angular) dark area in the sky?

Barnard 68 is the first thing that comes to mind for me, it is a little over 10' across and is opaque in the visible spectrum.

I'm sure there could be something much larger out there, but like I said, this was the first to come to mind.


Not strictly satellites/moons, but certainly companions are 2010 TK7 with a diameter of ~300 m, an Earth trojan at the L4 point, and the ~5 km 3753 Cruithne in a peculiar orbit locked to the Earth's.


What would happen if a planet is removed from the solar system?

This question isn't really very clear or practical. You need to add quite a bit of information, eg which planet?

If Sedna vanished, it would be detectable by instruments, but we would not notice.
Jupiter or Mars might make a bigger difference. We could spot it straight away by the naked eye.

Of course Earth life would be fine at first (unless it was Earth that was removed ...) but the orbits of asteroids and other bodies may be perturbed, which could lead to an influx of asteroid strikes - potentially - in the longer term.


What would stargazing be like at the edge of our galaxy?

Here is a render of what an observer might see near the edge of our galaxy. The "edge" itself being rather fuzzy, let's assume that we are in an area of the edge dense enough where a few occasional stars are still visible. (As far as we know, intergalactic space contains objects and stars, but so few that seeing even one naked-eye star in the sky there would be uncommon)

Looking at the sky towards the opposite side of the Milky Way, one could see the Magellanic Clouds, and a small quantity of naked-eye stars:

The Milky Way side of the sky would be interesting to look at:

Now, as interesting as these artistic renditions may be, there is very little that we know about actual objects in these regions. Seeing individual stars there is quasi-impossible from here; we have a general idea of how dense some regions of the galaxy are, but for the rendition part here, it is mostly procedural based on some general facts and statistics.

The software used for these renders is called Space Engine and contains all catalogued objects, and it procedurally generates objects to fill in the gaps based on what we know about space.


How many arms does the Milky Way galaxy have?

This is actually a really, really tough question.
Look at this diagram:

Purple: Norma Arm and Outer Arm.

Green: Scutum-Centaurus Arm

Pink: Carina-Sagittarius Arm

Cyan: 3 kpc Arm and Perseus Arm
So we can slightly modify this picture by saying that there are four arms, and calling them by the following names:
Norma-Outer Arm
This arm has one end at the center of the Milky Way; this end is called the Norma Arm. It's actually quite small. However, as you continue outward along the lanes of gas, dust and stars, the Norma Arm becomes the - wait for it - Outer Arm.
Scutum-Centaurus/Crux-Scutum Arm
This arm also emanates from the center (well, duh!) and is distinguished by large numbers of clusters of red supergiants near the center (where it is referred to as the Scutum Arm) of the Milky Way. The Scutum-Sagittarius Arm is one of the Milky Way's two major arms.
Carina-Sagittarius Arm
The defining feature of this arm is that is has many H II regions, where ionized gas is plentiful. It is though that stars can form there. However, H II regions are also present in many other parts of the galaxy. Observations from the Spitzer Space Telescope seem to support these theories and prior observations.
The Spitzer results are, I think, very important for understanding our galaxy and the things in it. It confirmed the existence of all four arms, but also confirmed that the Norma-Outer Arm and Sagittarius Arm are relatively minor compared to the other two.
Near/Far 3 kpc Arm and Perseus Arm
The Near and Far 3 kpc Arms (which really are just one arm) are very close to the galactic center - about 3,000 parsecs (hence the name). The Far Arm was only discovered recently, while the Near Arm was first observed about 50 years ago. These two mini-arms are expanding outward at an enormous rate. The combined 3 kpc Arm then becomes the Perseus Arm, which extends out to about 3.5 times the distance from the center of the 3 kpc Arm. It is the other major spiral arm of the Milky Way.

Okay, so that's simple enough. The Milky Way has four spiral arms, right? Well . . . sort of. The discovery of the "New Outer Arm"1 shook things up, because that would mean that the Norma-Outer Arm reaches nearly all the way around the Milky Way. That's strange and borderline inexplicable. Well, not really. But it's very interesting.
This image gives you a good idea of what it might look like:

By the way, someone at one point wondered, "Which are we in?" That, too, doesn't have a simple answer. Current observations state that we're in the Orion-Cygnus Arm, a/the midget of the galaxy (depending on whether or not there are other "mini-arms"). It's tiny - 1.1 kpc wide and 10 kpc long - and is between the Carina-Sagittarius Arm and the Perseus Arm.
Here's what it might look like:

Putting it all together, Robert Hurt made a very famous impression of what the Milky Way might look like:

NASA gives a good explanation of it here.
If you want a good overview on everything, check out this blog post, which is reasonably well-sourced. Another post explains the four-arms-vs.-two-arms issue:

Like the proverbial blind men describing an elephant, the two groups of scientists are examining very different parts of the Milky Way. The Spitzer study detected hot objects visible in infrared. The CfA study used radio telescopes which can also detect colder objects such as supernova remnants, very young star formation regions and huge clouds of hydrogen gas. So it seems as though the older established star formation regions are mostly concentrated in two spiral arms, but that the very new star formation regions and the hydrogen clouds from which they form are also developing in two additional arms.
If you were hovering above the Milky Way in a spacecraft using binoculars or a regular optical telescope, you would see two main arms. But if you also had a radio telescope with you, it would detect two more.

The other group of scientists referred to here used the Very Long Baseline Array to image the Milky Way with radio waves; their observations support the four-main arms model.
So at the moment, I think the answer is four, though two arms are more distinct that the other two.

1Dammit, paywall! Thank goodness for arXiv.


How is the diameter of a gas giant calculated?

I won't argue with the wikipedia definition (although the NASA Jupiter fact sheet lists it as the radius at 1 bar), but just to point out that the scale height of the atmosphere of Jupiter is given by, $h \sim kT/m g$, where $T$ is the temperature, $m$ is the mean molecular mass and $g$ is local gravity. Putting in some numbers: $ g \simeq 24.8 m/s^2$, $m \simeq 2.2 m_u$, $T \simeq 165K$, gives $h \simeq 25\ km$. Thus the pressure changes extremely rapidly compared with the actual radius of a gas giant,and is very small compared to the difference for instance between the polar and equatorial radius for a gas giant (like Jupiter) with significant rotation.

So, whether you give a planet's radius at 1 bar or 10 bar isn't going to make a lot of difference $<100\ km$.

However your comment about whether the definition is analogous to the photosphere of a star does raise an interesting question. The effective opacity radius of a gas giant can vary with the wavelength of light. It is this property that can be used in conjunction with transiting exoplanetary systems to learn something about their atmospheres. For instance, it requires less physical depth of a gas to block wavelengths corresponding to the strong 589nm sodium doublet and thus the exoplanet is opaque at larger radii at this wavelength and thus has a (slightly) deeper transit. This provides the opportunity for transit transmission spectroscopy. e.g. http://adsabs.harvard.edu/abs/2012MNRAS.426.1663S

Another example is that at UV and X-ray wavelengths there have been observations suggesting that hot Jupiters have very extended (and opaque to high energy photons) atmospheres and may be being photoevaporated. e.g. http://adsabs.harvard.edu/abs/2013ApJ...773...62P

EDIT: So this raises an important distinction. A definition based on the radius at a particular pressure is mostly used by theorists and solar system scientists. However, we still don't know what the atmospheric structures of exoplanets are and can't measure pressures. The radii quoted for exolanets are more akin to how the radius of a star is measured - based on its opacity at the wavelengths being observed.


What Right Ascension and Declinations are observable by a telescope from a particular location?

The declination limit, looking south, is about -52 degrees.  This is basically your latitude minus 90.

At 8pm, due south will be about RA 12hrs, and at 3 am it will be about 19hrs.  The eastern horizon will be about 6 hrs ahead, and that increases as you look further north, until you hit the circumpolar region, when all RAs are visible.


Largest body in the solar system that a baseball thrown by a professional pitcher could escape

Assuming an object with the same density as earth (not very good approximation) then a spherical object with a radius of 2.5 km, would have an escape velocity of about 96 mph. A velocity reached by several MLB pitchers. There are probably several asteroids with a radius near this.


What's the farthest point humans have explored?

This is a really interesting question to many and it deserves an answer post even if folks close it.
In Wikipedia's List of the most distant astronomical objects we can see that the distances that the light travelled to get to us are all around 13 giga light years, and it's no coincidence that the age of the universe is also about 13 billion years.
We can't really call that the actual distance to the object because the size of the universe is changing and the size of the observable universe is changing as well.
So while we like to say we are seeing really really far away, we should really think about it as seeing really really far back in time, because on the scale of the universe the speed of light is actually incredibly slow :-)
No matter how big of a telescope we build with unbelievable light gathering power, we'll never be able to see "farther" than a certain distance. There's a currently unknowably larger universe out there whose light and gravity has not and probably never will reach us
It's almost certainly based on current understanding, but that's subject to change of course ;-)


The stretching has no limit. Light can be redshifted to infrared, then to microwaves, to radio waves. There is no limit. This is because there is no medium that carries the waves to be stretched. There is no limit of the spectrum. It runs from arbitrarily long wavelengths, to arbitrarily short.

Instead of thinking of waves being stretched you could use a model suggested by Feymann. Each photon is like a small clock with one hand. THe colour of the light is the time it takes for the hand to go round. The photons in visible light turn a thousand trillion times a second. Radio waves only a million times a second. There is no limit to how slowly the hand can turn. Red-shifting is the slowing of the hand. There is no need to think of something stretching, and nothing to "break".


can i use heliocentric velocity as a rotation speed?

It's a little unclear what you're asking for, but...
"Heliocentric velocity" means measured radial velocity of an object relative to the Sun. (Which basically means relative to us, except you take out the variations due to the Earth's motion around the Sun.) For galaxies more distant than, say, the Local Group, the heliocentric velocity is the combination of a) the Sun's motion in our Galaxy; b) the local peculiar motion of our Galaxy; c) the local peculiar motion of the other galaxy; and d) the cosmic expansion of the universe ("cosmological redshift"). The first three never vary that much, but the fourth increases with distance, so for distant galaxies, it's (almost) all due to the cosmic expansion.
"Rotation velocity" for a galaxy generally means the speed of stars and gas clouds about the center of that galaxy; it is not one single number but a function of distance from the galaxy's center -- though in the outer parts of the galaxy it often settles to a nearly constant value, which may indeed be referred to as "the" rotation velocity.
The two have nothing to do with each other, though. For example, the rotation velocity in the outer part of M31 (the Andromeda Galaxy) is about 230 km/s, while its heliocentric velocity is about $-300$ km/s. If it were 100 megaparsecs away in the Coma Cluster, its heliocentric velocity would be something like 7000 km/s, but its rotation velocity would be the same.


Launch windows for Venus transit; how to convert heliocentric longitudes to date?

Heliocentric longitude is a direction, not a date.
A Heliocentric longitude of 0 is the direction from the sun towards the vernal equinox (currently in Pices)
You can ask for the date on which the Earth is at a particular Heliocentric longitude.  A careful calculation would require accounting for the eccentricity of the Earth's orbit, and also adjusting for the oddities of the calendar, but since when the Sun in on the vernal equinox (About March 21), the Earth will be at 180 degrees of heliocentric longitude, and longitude will increase by 360/365.24 degrees per day.


A star a black hole and planets around them

It's possible to have planets orbiting a binary pair of stars, your scenario of a close orbit, sometimes called "short orbit binaries".    See here, also posted above in comments.   In such a binary-system, nothing can orbit an individual star, but at some distance, plants can orbit and some systems like this have even been observed, listed in the link.  The orbital dynamics is the same for a star-black hole short-orbit binary.

Now, there are problems.   Black holes form out of very large stars and the formation is one of the biggest explosions in the universe, a Type II supernovas, and that's not very friendly to any planets in orbit.  A star might survive it, planets would be harder, though it might be possible for new planets to form from nebula material remaining after the nova (I'm just guessing there).

A black hole could also form from a Neutron star accreting matter, but you still have the problem that the formation of a Neutron Star also only happens out of a Type-II supernova, so such a system has a difficult beginning.

Edit:  While some planets have been observed around Neutron stars, these appear to be quite rare.   2 Neutron Stars have been observed with planets, out of over 1,600 Neutron stars observed.    A type II nova is very planet unfriendly.

Theoretically a close gravitational capture is possible, but those are very rare, as stars rarely get that close.   There's many stars that are known to orbit the super-massive black holes at the center of our galaxy (Andromeda galaxy too), but stars and stellar mass black holes are much more rare.   A few have been observed, but they don't appear to be common.   Such a system would be easy to observe, so the fact that there are only a few that have been noticed is evidence to them being rare.  Here's a few mentions of them.  One, Two, Three, Four.

From the 4th article, which is from 2011, so more may be known now, but it says:

  Only about 20 binary stellar systems are known to contain a black
  hole, out of an estimated population of around 5,000 in the Milky Way
  Galaxy.

A 2nd problem is that a star feeding a black hole would create an accretion disk which would be very radioactive and not ideal for life on an orbiting planet.   Maybe the planet could have a very thick atmosphere that might protect it, but that would also likely reduce sunlight reaching the surface.   The star feeding the black hole would also be losing mass, and over time, grow smaller and provide less light and heat to the planet.   Ideally, you'd want it to be a very slow feed.   It's pretty far from an optimal life on planet situation.

  So it is possible to have planets in this system not to be consumed by
  this black hole but just follow their unique orbits

This part is certainly possible.   Things can orbit a black hole at a safe distance without any problem.  As for life, we don't know how common life is in other solar-systems so nobody can say how likely it might be, but it's theoretically possible, but, in my opinion, pretty far from ideal.


What is the exact position of the Large Magellanic Cloud?

First of all: your coordinate system (as I understand it) is not "well defined". You only provide a line (Earth --> Milky way centre) and define that as the North-South axis. You don't give a reference for the east-west axis or the up-down axis. So, I'll assume that the East-West axis lies along the plane of the Milky way and up is where the Boötes Dwarf system is. This way, our coordinate system is equal to this image from the LMC Wikipedia site.

The problem with this picture is that the position of the Earth is only implicitly given. It is "behind" the center of the Milky Way. So from that position, the LMC is almost to our right (east) and a bit below the plane of the Milky way; about 30 to 40 degrees down.

This is also nicely validated by this image, acquired here (note that the grid is in 30 degree angles).

So, finally we can answer your question in the way you wanted: in the north-east to east segment, and below its plane.


Integral Field Spectroscopy vs Fabry-Perot Interferometry

I think you've got it pretty well summarized. In particular, your first three points are dead-on.

(I'm going to use "IFU" = integral field unit instead of "IFS" = integral field spectrograph, but that's just because I'm more used to the former term; they're basically synonyms.)

Let me combine the "data reduction" question with one other key advantage of IFUs that you may have missed, which is that they are (almost) plug-in substitutions for traditional long-slit spectrographs, which means they can take direct advantage of existing spectrograph-design expertise and data-reduction software.

As an example, one common form of IFU is a closely packed array of small input apertures attached to fiber-optic cables. Away from the IFU, the cables can be re-arranged so that their output ends form a 1D row, very much like the output of a long slit in a traditional spectrograph. (And even more like a multi-object spectrograph design, where multiple individual input apertures are scattered across the field of view rather than being organized into a contiguous 2D array.) This means that the rest of the spectrograph design -- gratings or prisms, collimation optics, imaging onto a 2D CCD detector, etc. -- is pretty much the same as traditional spectrograph. There are instruments on some telescopes which allow you to switch between long-slit, multi-object, and IFU modes, with the same set of gratings, filters, and detector being used for all modes (for example, the GMOS instruments on the Gemini North and South Telescopes).

This carries over into ease of data reduction: the spatial+spectral image on the CCD detector (or near-IR detector array) produced by an IFU is very similar to that produced by a traditional long-slit spectrograph, so the same data-reduction techniques can (with a little modification) be used. In essence, you just chop up the big 2D data image you get from the CCD into a bunch of little sub-images, one for each fiber that came from the IFU, and you can treat each one as though it were a standard single-aperture or single-slit dataset. This contrasts quite strongly with the very different (unique) form of data and data reduction that F-P requires.


How can I get/generate/extract PSF image from galaxy .fits image?

To answer the question: Yes, there is a way. Use a bright but unsaturated star. That might already be good enough, or you might want to do any number of corrections to this PSF model.
A better way is to figure out the parameters to use with TinyTim, and generate an approximately right PSF from those. Most parameters you know, or they are easy to look up, such as camera, chip, filter. For the pixel position, you need to look up the original exposures that went in to your (presumably) drizzled image; the object will be more or less in the same place in all of them, so you can use the average position.
With these, you can generate an overresolved PSF image. If your (presumably) drizzled image is north-aligned, you will need to rotate the PSF accordingly; ideally using drizzling, but a simple image rotation might work well enough if your PSF resolution is much higher than the HST resolution. The final result is a PSF that might not be useful for precision photometry, but should be good enough for modelling a galaxy. In any case, compare the PSF to the image of a star to make you it is roughly right (east-of-north angles in HST images have a way of being wrong).


What is the most oblate astronomical object known?

Saturn is the most oblate planet in the solar system. If the equatorial diameter is $a$ and the polar diameter is $b$ then its oblateness, $(a-b)/a = 0.1$.
We do not know the oblateness values for more than one or two exoplanets and even these are somewhat uncertain, but are thought to be lower than Saturn's value. For example http://adsabs.harvard.edu/abs/2010ApJ...709.1219C
The oblateness of a star depends on the ratio of it gravitational acceleration to centrifugal acceleration at the equator. So the most oblate stars ought to be the ones with big radii and fast rotation rates (i.e. types of stars that would have low surface gravities if non-rotating).
Giant stars come in two basic types - red giants, which are evolved intermediate and low-mass stars. These tend to be slowly rotating, though there is a class called FK Com variables which are fast-rotating red giants.
The other class is the blue giants, which are high mass main sequence, or slightly evolved, stars.
For maximum oblateness you want the star with the highest value of:
$$ f = \frac{R \omega^2}{GM/R^2} = \frac{R^3 \omega^2}{GM}$$
where $M$ is the stellar mass and $\omega$ its angular velocity. Often we don;t know $\omega$, but can estimate the rotation velocity at the equator $v=R\omega$. Hence
$$f = \frac{R^3 v^2}{GMR^2} = \frac{Rv^2}{GM}$$
A possible record holder is the O-type star VTFS 102 in the Large Magellanic Cloud.
This probably has a mass of $25M_{\odot}$ a radius of about $10R_{\odot}$ and has a measured rotation speed of $\geq 600$ km/s. Thus its value of $f$ is 0.75, compared with a value of $f=0.15$ for Saturn.
EDIT: Due to Michael B - what about "millisecond" pulsars? The shortest periods are about 1.4 ms, the radii are about 10 km and the masses are likely to be about 1.4$M_{\odot}$. Do the sums and you find $f=0.1$. So, (at least in Newtonian mechanics) they should be no more oblate than Saturn.
A further contender might by the dwarf planet Haumea that orbits at 40-50 au from the Sun. It is thought large enough to have achieved its shape via hydrostatic equilibrium between gravity and internal pressures (and rotation). Its mass is $4\times10^{21}$ kg, (mean) radius is 700 km and it has a rotation period of 3.91 hours. These numbers give $f=0.25$. It has been modelled using this, and matching its light curve, to be a triaxial ellipsoid (rather than oblate) with axis ratios 2 : 1.5 : 1 (http://adsabs.harvard.edu/abs/2006ApJ...639.1238R).


I won't go through your equations carefully nor your code, but since you haven't mentioned the exact starting positions that you used, my guess is that you haven't first solved for a stable horseshoe orbit state vector first.

The positions for the Sun and Earth are fixed in the synodic (rotating) frame. But where did you start your astroid? For a given starting position, you have to also give the asteroid the correct starting velocity (speed and direction) otherwise it may not be stable and just wander off or spiral around as you mentioned.

Have a look at this answer where I show the correct equations and then solve for stable horseshoe orbits before plotting them.

I start with a position on the x-axis, opposite the Earth, and a little farther or closer from the Sun than the Earth is. Then I launch it with a low velocity (in the rotating frame) exactly in the direction of the y-axis. I watch it drift towards Earth and then boomerang back to the region of the starting area. When it crosses the x-axis again, I check to see how close the velocity is to y. I use an optimization zero solver brentq to adjust the initial velocity until the return velocity is in exactly the opposite direction.

If that's true, then the orbit will be periodic and repeating, and we can call it a stale orbit under the constraints of the CR3BP.

From this answer (there's a lot more to read there, including several references!):

  $$\ddot{x} = x + 2\dot{y} - \frac{(1-\mu)(x+\mu)}{r_1^3} - \frac{\mu(x-1+\mu)}{r_2^3}$$
  $$\ddot{y} = y - 2\dot{x} - \frac{(1-\mu)y}{r_1^3} - \frac{\mu y}{r_2^3}$$
  $$\ddot{z} =  -\frac{(1-\mu) z}{r_1^3} - \frac{\mu z}{r_2^3} $$

  above: half-cycles of some wobbly horseshoe orbits

  above: times to first x-axis crossings of the same wobbly horseshoe orbits, used to calculate half-cycle times.

  above: cycle times from this calculation (black dots) versus from the synodic period estimation method (red dots). Good qualitative agreement. Also the starting y velocities at each starting point in x.

  below: Python script for these plots.

def x_acc(x, ydot):
    r1    = np.abs(x-x1)
    r2    = np.abs(x-x2)
    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))
    return xddot

def C_calc(x, y, z, xdot, ydot, zdot):
    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)
    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)
    C = (x**2 + y**2 + 2.*(1-mu)/r1 + 2.*mu/r2 - (xdot**2 + ydot**2 + zdot**2))
    return C

def deriv(X, t):
    x, y, z, xdot, ydot, zdot = X
    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)
    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)
    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))
    yddot = y - 2*xdot  -  ((1-mu)/r1**3)*y      - (mu/r2**3)*y
    zddot =             -  ((1-mu)/r1**3)*z      - (mu/r2**3)*z
    return np.hstack((xdot, ydot, zdot, xddot, yddot, zddot))

# http://cosweb1.fau.edu/~jmirelesjames/hw4Notes.pdf

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint
from scipy.optimize import brentq

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

mu = 0.001

x1 = -mu
x2 = 1. - mu

x = np.linspace(-1.4, 1.4, 1201)
y = np.linspace(-1.4, 1.4, 1201)

Y, X = np.meshgrid(y, x, indexing='ij')
Z    = np.zeros_like(X)

xdot, ydot, zdot = [np.zeros_like(X) for i in range(3)]

C = C_calc(X, Y, Z, xdot, ydot, zdot)
C[C>8] = np.nan

if True:
    plt.figure()
    plt.imshow(C)
    plt.colorbar()
    levels = np.arange(2.9, 3.2, 0.04)
    CS = plt.contour(C, levels,
                 origin='lower',
                 linewidths=2)
    plt.show()

ydot0s   = np.linspace(-0.08, 0.08, 20)
x0ydot0s = []
for ydot0 in ydot0s:
    x0, infob =  brentq(x_acc, -1.5, -0.5, args=(ydot0), xtol=1E-11, rtol=1E-11,
                           maxiter=100, full_output=True, disp=True)
    x0ydot0s.append((x0, ydot0))

states = [np.array([x0, 0, 0, 0, ydot0, 0]) for (x0, ydot0) in x0ydot0s]

times  = np.arange(0, 150, 0.01)

results = []
for X0 in states:
    answer, info = ODEint(deriv, X0, times, atol = 1E-11, full_output=True)
    results.append(answer.T.copy())

resultz = []
for x0ydot0, thing in zip(x0ydot0s, results):
    y     = thing[1]
    check = y[2:]*y[1:-1] < 0
    zc    = np.argmax(y[2:]*y[1:-1] < 0) + 1
    if zc > 10:
        resultz.append((thing, zc, x0ydot0))

if True:
    plt.figure()
    hw = 1.6
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x, y = thing[:2,:zc]
        plt.plot(x, y)
    plt.xlim(-hw, hw)
    plt.ylim(-hw, hw)
    plt.plot([x1], [0], 'ok')
    plt.plot([x2], [0], 'ok')
    plt.show()

if True:
    plt.figure()
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x, y = thing[:2]
        plt.plot(times[:zc], y[:zc])
    plt.show()

if True:
    plt.figure()
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x0, ydot0 = x0ydot0
        cycle_time = 2. * times[zc] / twopi
        ratio = abs(x0/x2)
        T_simple_model = twopi * abs(x0/x2)**1.5
        T_synodic_simple_model = 1. / (1. - twopi/T_simple_model) # https://astronomy.stackexchange.com/a/25002/7982
        plt.subplot(2, 1, 1)
        plt.plot(x0, cycle_time, 'ok')
        plt.plot(x0, abs(T_synodic_simple_model), 'or')
        plt.subplot(2, 1, 2)
        plt.plot(x0, ydot0, 'ok')
    plt.subplot(2, 1, 1)
    plt.xlabel('x0', fontsize=16)
    plt.ylabel('cycle times (periods)', fontsize=16)
    plt.subplot(2, 1, 2)
    plt.xlabel('x0', fontsize=16)
    plt.ylabel('ydot0', fontsize=16)
    plt.show()


What qualifies as a good place for stargazing, i.e. with least light pollution?

A good question, and in the early 2000s John Bortle published a categorization of a variety of conditions, with descriptions for each category. It is the commonly used scale to describe to others the sort of conditions at a location.

Probably one of the more significant factors provided by a dark sky site is: how faint do stars have to be for you not to see them anymore (overwhelmed by light pollution). There are specific stellar regions, each a triangle, and the idea is you could how many stars you can see in the triangle, and you look the number up in a table, and it will tell you the magnitude limit you are perceiving at that site.

Yes, high altitudes also help - less air between you and the stars means better seeing conditions/less atmospheric distortion.

Being able to see more stars IS better for star gazing. Living in a city, I am continually frustrated at the poor skies, and love being in dark sky locations like the International Dark Skies Reserve at Tekapo, New Zealand.


If the expansion of the universe is speeding up, how long until everything nears the speed of light, and what would happen?

The answer to your question is that if we assume the Universe is infinite then there will always be objects receding from us at the speed of light, simply because recession velocity depends linearly on distance. Indeed there are objects that we can see (i.e. they are within our observable Universe) that recede from us at the speed of light (this is a very misunderstood issue, so I'd advise anyone taking issue with this statement to make sure they've read this paper first).

If we assume dark energy takes the form of a cosmological constant, then the Universe asymptotically becomes like De Sitter space over time. This means that the sphere of objects that are receding from us at c (the Hubble sphere) settles down to a constant radius which is inversely proportional to the square root of the cosmological constant.

Edited to add: recession velocities though are purely a function of cosmological coordinates so don't confuse them with local velocities which are always constrained by c.


Is there a photo that shows a iron meteor or asteroid in space in raw form having no layers of fusion crust?

In short, no.

The reason is that while finding meteorites on Earth is hard, finding them in space is a lot harder. A small asteroid, weighing only a few kg can't be spotted while it is still in space. The first we know of them is when they make their fiery descent.

On the other hand, there are some large asteroids that are metallic, such as 16 Psyche, and we have direct images of other large asteroids. Metallic asteroids are relatively rare, and have a higher albedo than stony asteroids, and much more than comets. However they are not shiny balls of metal. After a few billion years in space they absorb 80% of the light that falls on them.


Current topics on Radio Astronomy and looking for advice

Here are some examples that you can read about within Astronomy SE and Space SE:
Time Domain Astronomy (link)
e.g. LOFAR and (as it turns out) CHIME. Think better FRB detection gravitational lensing effects on pulsar timing (needs several questions of its own)

Why did it take five years to "figure out" how to use astrometric calibration sources to deblur LOFAR images?
Open dataset from antenna array

and

How tightly clustered are the CHIME-detected FRB repeats?
What are the challenges for the building and data analysis of the CHIME telescope?
How does Canada's radiotelescope CHIME manage polarization-dependent apodization of the fast f/0.25 optics?
What will succeed the Arecibo Observatory?

also

What exactly is interplanetary scintillation; what was the Interplanetary Scintillation Array looking for? Did it successfully observe any?

Space based, and Satellite VLBI

Far Side of the Moon stuff, e.g. Chang'e 4: The NCLE on the orbiter and the LFS on the lander work in synergy performing low-frequency (0.1–80 MHz) radio astronomical observations.
What can be learned from low frequency radio astronomy available outside of Earth's ionosphere?
Has VLBI been done using any space-based receivers besides Spektr-R? (Space SE)
What's the status and timeline for Millimetron? (Russia's 10m Deployable Antenna cooled to 6 K Earth-Space VLBI) (Space SE)
What are Spectr-R's major contributions to radio astronomy that could not have been done from Earth? (Space SE)
How is Spektr-R doing these days? (Space SE)
Has VLBI been done using any space-based receivers besides Spektr-R? (Space SE)
What kind of ground-based radio astronomy is NASA's DSN used for? Who are the PIs? (Space SE)
Are images of exoplanets' surfaces technically possible? (Space SE)
Could a radio telescope orbit Earth-Moon L2 so that it is in the Moon's radio shadow?
Rotating liquid mirror on the Moon: this answer to Why can't we build a huge stationary optical telescope inside a depression similar to the FAST?

Fancy, complicated focal plane arrays and techniques

How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?
How do ASKAP's focal plane phased array feeds interact with the entire array phasing?
What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?
Would Adaptive Optics be Useful in Radio Astronomy?
Has radio astronomy ever been done on objects that appear very close to the Moon? Is this avoided?

Radar Astronomy (link)

What are recent discoveries with Earth-based radar telescopes?
What will succeed the Arecibo Observatory?
What causes "North-South ambiguity" when doppler radar imaging a planet surface equator on?
https://space.stackexchange.com/q/22493/12102
What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?
Farthest distance to a solar system object that's been measured by radar? (Space SE)
How did Arecibo detect methane lakes on Titan, and image Saturn's rings? (Space SE)
Why is Saturn invisible in this radar image of its rings? (Space SE)
Why was the 100m Green Bank dish needed together with DSN's 70m Goldstone dish to detect Chandrayaan-1 in lunar orbit? (Space SE, not exactly Astronomy but pretty cool from an instrumentation and signals point of view)
Why are Titan's lakes “black” in radar images rather than transparent? (Space SE)
Is delay-doppler radar imaging of NEO asteroids possible only if it spins fast enough? (Space SE)
Separating fact from fiction, how to get from this radar image to this GIF? (Space SE)


What is the length of time for each fusion process in a massive star?

My stellar astrophysics text, Francis LeBlanc's An Introduction to Stellar Astrophysics, gives the following quantities for the stages of burning in a $25M_{\odot}$ star (citing models by Arnould & Samyn 2001). This is somewhat less massive than a typical main sequence O5-type star, and with higher masses come higher core temperatures and shorter evolutionary timescales. Nonetheless, it should give you a reasonable order-of-magnitude idea of how long it takes for these processes to play out in a massive star.

Phase
Central temperature (K)
Timescale

Hydrogen
$6.0\times10^7$
$7\times10^6$ years

Helium
$2.0\times10^8$
$5\times10^5$ years

Carbon
$9.0\times10^8$
$600$ years

Neon
$1.7\times10^9$
$0.5$ years

Oxygen
$2.4\times10^9$
6 days

Silicon
$4.0\times10^9$
1 days

(One major caveat is that these models are two decades old, and stellar evolutionary codes are constantly being improved. However, I don't believe that these are old enough to be significantly out-of-date!)


How to find the Orbital Period of two Binary Planets around a Single Star?

I think that at the moment, all you have available for the pair's orbit around their star is a snapshot of their two distances; 964 and 989 seconds, or about 289 and 297 million kilometers. Your snapshot doesn't have any velocity information, so I think that there's no way to calculate the mass of the star from that data. Since the first planet is 2000 times heavier than the second, lets say that the pair's center of mass is 289 million km from the star.
However your second link states explicitly that the star is 1.0195 solar masses. That means that the standard gravitational parameter or $GM$ is about 1.352E+20 m^3/s^2, so using

$$T= 2\pi\sqrt{\frac{a^3}{GM}}$$

I get a year of about 972 days for the center of mass of the pair. Since the lighter planet oscillates by +/- 10 million km (its semi-major axis) every 80 days, that's a 7% amplitude change in distance to the star or roughly a 14% amplitude modulation of the brightness of the sunlight every 80 days, so that may have some impact on the weather for the smaller planet.

Interestingly, the smaller planet goes "around the world in 80 days" if we name the larger planet "world".


Does the sun now 'bypass" some of the original zodiac constellations?

The ecliptic still passes through the original 12 constellations, although Earth's precession has shifted the Sun's apparent position by roughly one constellation. That is, if your birthday is today, May 2, an astrologer would claim you were Taurus, whereas in reality the Sun is currently positioned in Aries.

Additionally, the ecliptic now passes through the constellation Ophiuchus (The Serpent Bearer). You can see the "historical" and the "true" constellations e.g. here.


What would happen if a rocket traveling at speed of light would collide with a planet (like earth)?

I don't know about a spaceship, but the XKCD guy wrote an interesting article about what would happen if Earth was hit by a solid asteroid, travelling at various different speeds:
what-if question: Diamond

The ship would be a lot smaller than an asteroid, so I think the damage would be a lesser version of these descriptions.  (The most extreme case described is the whole planet being vapourised and even neighbouring planets being affected by the radiation.)

As for the second part of the question - protecting the craft from minor collisions - I admit I have no idea how to work out the kinetic energy from collisions at such high speeds. But even a grain of sand would at least damage the front of the craft, and I guess some sort of thick shield would be needed to protect against erosion from interstellar dust...

Even at normal orbital speeds, Space Shuttle windows have had minor damage as this paper describes...


Moonrise after sunrise but moonset before sunset?

Right now, the moon is south of the ecliptic (for those of us in the northern hemisphere).  This means it encounters the horizon before the sun.


Solar Analemma - but with a star instead

An analemma shows the position of the Sun in the sky from a fixed location on Earth at the same mean solar time. It gets its north-south shape because of the tilt of Earth's axis makes the declination of the sun change across the year, and the sideways shape due to the orbit nonuniformity.
The star analemma would get a point at the same time every sidereal day (slightly different from the solar day). To a first approximation that fixes the star's right ascension: it will not move along the celestial equator. The celestrial equator of course rotates one turn per year as seen from the ground, so in one sense the analemma is just a circle around the pole of the sky. But if you look at it in the coordinate system of the celestial sphere the star is fixed.
...Except that there will be some stellar parallax due to Earth's different locations along its orbit during the year. The star will seem to move along a tiny ellipse when compared to distant stars in the background. I think the parallax ellipse is the closest counterpart to the sun analemma.


Multiple apparent lunar eclipses last night

This was clouds.

Eclipses occur at full moon, not last quarter.
Eclipses take several hours as the moon traverses the Earth's shadow, not a few minutes.
Eclipses don't look like the moon becoming a gleaming point of light. In a lunar eclipse the moon becomes a copper colour, as it is illuminated by red light that has passed round the Earth, through our atmosphere.

On the other hand, clouds can pass in front of the moon at any time, and when low in the sky would take a few minutes to pass in front of the moon (depending on wind), and could entirely cover the moon, giving a the impression of an eclipse, if you've not seen one before. But lunar eclipses look like this:

Photo Credit: Jimmy Westlake space.com