Does the observed period of a pulsar change with the time of year?

Yes. In terms of pulsar timing measurements, this is a massive effect! A +/- 30 km/s doppler shift changes the pulsar frequency by +/- 1 part in 10000. This sounds small, but the accumulated phase shift over many periods is readily apparent. In addition, the light travel time across the solar system has to be taken into account, as well as the Earth's rotation and some other smaller effects - such as Shapiro delay.

If the question refers to the specific annually varying difference in clock rates caused by the different gravitational potential experienced by an Earthbound telescope on an elliptical (as opposed to circular) orbit -the answer is still yes.

This is item 4 in the list of applied corrections given on p.52 of "Pulsar Astronomy" by Lyne et al. The maximum effect is a rate change of $3\times 10^{-9}$ leading to a maximum lead or lag of 1.7 ms.


Radius inflation seems to mainly occur in planets with irradiation in the range 105–106 W/m2, see Sestovic et al. (2018). In contrast, at 1 AU from the Sun, the irradiation is 1361 W/m2, which is well below the threshold at which the inflated gas giant radii are observed. It seems unlikely to me that there would be a significant increase in the radius of Jupiter if it were moved to a 1 AU orbit.

As regards the hot/warm Jupiter classification, the term "hot Jupiter" seems to mainly be used to describe the population of giant planets with orbital periods below about 10 days or so. Your Jupiter at 1 AU would not fall into this category.


How much infrared radiation is emitted by a bow shock?

Has the amount of energy being emitted from R Hydrae's bow shock been studied?

Yes, here are a few papers:

"Detection of a Far-Infrared Bow-Shock Nebula Around R Hya:
the First MIRIAD Results" (July 13 2006), by T. Ueta, A. K. Speck, R. E. Stencel, F. Herwig, R. D. Gehrz, R. Szczerba, H. Izumiura, A. A. Zijlstra, W. B. Latter, M. Matsuura, M. Meixner, M. Steffen, & M. Elitzur.
"Detached shells as tracers of AGB-ISM bow shocks" (July 21 2006), by C. J. Wareing, Albert A. Zijlstra, Angela K. Speck, T. J. O’Brien, Toshiya Ueta, M. Elitzur, R. D. Gehrz, F. Herwig, H. Izumiura, M. Matsuura, M. Meixner, R. E. Stencel, and R. Szczerba
"No Shocks for This Bow: IBEX Says We’re Wrong" (May 14 2012), by Susanna Kohler

  Does this mean the CSE (bow shock?) is about 57% (19.7 compared to 33.9) the flux of the star at the 70 microns (mosaic) band?

The CSE is not a "bowshock". An envelope surrounds an object, while bowshock preceeds the direction of movement.

Reference: "The chemistry in circumstellar envelopes of evolved stars: Following the origin of the elements to the origin of life", by Lucy M. Ziurys, PNAS August 15, 2006 103 (33) 12274-12279:

  "Chemically rich circumstellar shells are found around asymptotic giant branch (AGB) stars and red supergiants (RSGs). ...
  Given sufficient contraction, the helium core will eventually ignite as the H shell continues to burn. The fusion of helium produces $^{12}$C, and alpha capture on $^{12}$C yields $^{16}$O in the core. Eventually, the central helium is exhausted, leaving an oxygen/carbon core surrounded by He-burning and H-burning shells. The appearance of this two-shell structure marks the beginning of the AGB phase. The primary source of luminosity in an AGB star is the H-burning shell, but periodically the He shell ignites for short periods of time. The energy released in this so-called “thermal pulse” creates a convective zone extending from shell to shell, which mixes nuclear products from the He-burning zone to the stellar surface. This process, called the “third dredge-up,” brings carbon and other elements into the detaching envelope. The thermal pulses also aid in driving the outer material from the star, creating a circumstellar shell. In fact, AGB stars lose up to 80% of their original mass in the form of an envelope, with typical rates of 10$^{−6}$ to 10$^{−4}$ M$_\odot\cdot$ yr$^{−1}$.".

A useful but rough analogy is: a plane's paint fumes are the CSE and the bowshock is caused by a difference in the speed of travel that is supersonic.

The image you included is used on Wikipedia's webpage "Bow shocks in astrophysics - Around other stars" and appears to have originally come from: Caltech's website, dated 2006-11-30. Another copy is on JPL's website: "Red Giant Plunging Through Space" (December 8, 2006), Spacecraft: Spitzer Space Telescope, Instrument: Multiband Imaging Photometer (MIPS). Simbad query for R Hya.

Within the webpage's text it says: "This bow shock is 16,295 astronomical units from the star to the apex and 6,188 astronomical units thick (an astronomical unit is the distance between the sun and Earth). ... The mass of the bow shock is about 400 times the mass of the Earth.".

The webpage: "About the color mapping table", which explains three of the wavelengths used in the image,
says: "$\bbox[#E8D8D0, 0.15pt]{\color{yellow}{\text{Infrared}}\, 3.6 \mu m}$, $\bbox[#F8E8D0, 0.15pt]{
\color{orange}{\text{Infrared}}\, 4.5 \mu m}$, $\bbox[#E5DBD2, 0.15pt]{
\color{red}{\text{Infrared}}\, 8.0 \mu m}$". They don't explain the purple band on that webpage. The raw data is far more useful than a color mapped image prepared for public consumption.

The paper that you linked to says on page 6:

  "At 165 pc (Zijlstra et al. 2002) with the -10 km s$^{-1}$ radial
  velocity (Knapp et al. 1998), the space velocity of R Hya is 50 $\pm$  1 km s$^{-1}$ into the direction 12° away (receding side) from the plane of the sky. The asymmetric CO profiles (Knapp et al. 1998; Teyssier et al. 2006) can then be explained by this inclined bow shock, in which the receding and approaching shock fronts intersect with the line of sight at distinct angles.
  The physical dimensions of the observed bow shock (0.03 pc thick located at 7.9 x 10$^{-2}$ ahead of the star) are consistent with the stellar-wind bow shock models for AGB stars (Wareing et al. 2006a). Adopting the mass-loss rate of 3 x 10$^{-7}$ M$_\odot$ yr$^{-1}$ (Zijlstra et al. 2002), wind velocity of 10 km s$^{-1}$ (e.g., Knapp et al. 1998), and formula for $l$ derived
  from the momentum conservation across the shock (Van Buren et al. 1990, eq. 2), the ambient $H$ density is 0.4 cm$^3$ and the
  shock surface density is 2.5 x 10$^{-7}$ g cm$^2$. Thus, the estimated amount of matter contained in the bow shock is about 1.3 x 10$^{-4}$, assuming a paraboloidal bow shock of $y = x^2 /3l$ with a 300" opening radius.".

In the paper: "Exact Analytic Solutions for Stellar Wind Bow Shocks" (March 1996), by Francis P. Wilkin, Astrophysical Journal Letters v.459, p.L31:

  "Stellar wind bow shocks are cometary structures due to the supersonic passage of wind-blowing stars. They sweep up interstellar matter into thin, dense shells, which may be revealed by their postshock emission or by scattered light, and provide a means of studying winds that might otherwise go undetected. ...
  The theory of momentum-supported bow shocks was first developed by Baranov, Krasnobaev, & Kulikovskii (1971, hereafter BKK), who were motivated by the problem of the interaction of the solar wind with the local interstellar medium. ...
  This method allows one to obtain simple, exact solutions for all quantities in the numerical model of BKK: the shell’s shape and the mass and velocity distributions within it. The shocked fluid in the shell has momentum that is precisely the vector sum of the momenta imparted to it by the wind and ambient medium, integrated over the surface of the shell, so there is no need to divide the problem into normal and tangential parts in curvilinear coordinates. Moreover, the tail of the bow obeys a simple, momentum-conserving snowplow model in two dimensions.".

In particular, on page 3, it says:

  "The tangential velocity and mass surface density are plotted in
  Figure 4 for several values of $a$. For most systems, $a$ is expected to be small, and the tangential velocity should prove a more sensitive way of obtaining a than the mass column.".

  FIG. 4.—Internal shell velocity in units of V$_*$ (top) and mass surface density in units of $R_{0\rho_a}$ (bottom) for several values of $a$. Curves for $a$ = 0 and 2 are labeled, and intermediate curves correspond to $a$ = $\frac{1}{4}$, $\frac{1}{2}$, and 1. The mass per
  unit area $\sigma$ is defined for a line of sight that is normal to the shell. The normal direction to the shell may be obtained from eq. (6), using the relation $\hat{e}_n = \hat{e}_t$ x $\hat{e}_\phi$. If viewed from angle $\vartheta$ to the normal, the mass column $\sigma$ must be modified by a projection factor sec $\vartheta$ so long as the path length of the line of sight through the thickness of the shell is short compared to the shell’s curvature.

The exact amount of infrared radiation emitted is a function of speed and resulting temperature, it's not a means to calculate the energy from the bowshock nor is it consistent. It's a great means to visualize that there is bowshock, much as smoke or clouds help to visualize a plane's bowshock; at best it would be an indirect means to measure speed and energy.

In the monthly notices of the Royal Astronomical Society, Volume 441, Issue 1, 11 June 2014, Pages 809–812 article titled: "Detecting floating black holes as they traverse the gas disc of the Milky Way", by Xiawei Wang and Abraham Loeb they explain that bowshock can be unaccompanied by CSE and that the radiation isn't confined to the infrared bands.

  "In this paper, we propose an additional observational signature of floating BHs, using the MW gas disc as a detector. As the BHs pass through the MW disc supersonically they generate a bow shock, which results in synchrotron radiation detectable at radio and infrared frequencies. ...

  Observations of the Galactic disc can be used to infer $n_0$ and $T_4$. The BH speed $V_•$ can then be estimated from the Mach cone angle. The maximum Lorentz factor $\gamma_{max}$ can be inferred from the peak of the synchrotron spectrum. This, in turn, yields $B_{−5}$ based on equation (3). From the slope of the synchrotron spectrum, the power-law index $\rho$ can be estimated. Finally, with the above parameters constrained, the synchrotron flux can be used to calibrate $M_•$. The above interpretation can be verified by observing the properties of the star cluster carried by the floating BHs (O'Leary & Loeb 2009, 2012). The diffuse X-ray emission from the BH and synchrotron emission from the bow shock are supplemented by stellar emission from the star cluster around it. Since the total mass of the star cluster is much smaller than $M_•$, gravity is dominated by the BH, and thus the stars do not affect the bow shock. One can measure $M_•$ spectroscopically from the velocity dispersion of the stars as a function of distance from the BH, and verify consistency with the synchrotron flux estimate.".


Frequency analysis on Semiperiodic object(s) using Lomb-Scargle

This is a partial answer based on the discussion in comments below the question.
If one already has an "amplitude spectrum" and one wants to convert to a power spectrum, all you have to do is take the absolute value of the amplitude and square it. In Python (introduced in the question) that's just np.abs(amplitude)**2.
When you take the Discrete Fourier transform of a time series it is returned as a complex array, you need both real and imaginary components to retrieve both the absolute value of the amplitude spectrum and its phase spectrum.
Sometimes you can see "amplitude" used for the absolute value of amplitude, but it doesn't usually hurt if you take the absolute value again to be sure.
I've combined the signals of random noise with $\sigma$ = 1, a sinusoidal 50 Hz electronic interference noise signal, and a slightly phase-jittery square wave with an amplitude of 0.5 and a period of 12 milliseconds or a frequency of about 83 Hz.
I've kept the average near zero so we don't get distracted by a giant peak at 0 Hz, and I've shifted zero frequency to the center of the spectrum using np.fft.fftshift so we don't forget that negative frequencies exist in Fourier space.
We can see that the FT has both real and imaginary parts and squaring the absolute value gives us power.
For the time series plots I only show the first 200 milliseconds for clarity, the series has 1,000 time intervals of 1 millisecond each.
I've also used np.fft.fftfreq to automatically generate the frequency assigned to each bin in the FT.

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(42) # this way you can compare your output to mine directly

N = 2000
nt = np.arange(N)
dt = 0.001 # one millisecond per bin
time = nt * dt

twopi = 2 * np.pi

noise = np.random.randn(N) # normal distribution with sigma = 1
frequency_1 = 50
signal_1 = 0.2 * np.cos(twopi * time * frequency_1)

# jitter = (7 * np.random.random(N)).astype(int) - 3
interval_2 = 12
q_int = int(interval_2 / 4)
ints = np.arange(0, N, interval_2)
jittered = np.random.randint(-2, 2, len(ints)) + ints
signal_2 = np.zeros_like(nt, dtype='float')
for j in jittered:
    signal_2[j-q_int:j+q_int+1] += 1 # I suppose one could convolute or sum over np.shift
signal_2 -= signal_2.mean()

combined = noise + signal_1 + signal_2

ft = np.fft.fft(combined)
freqs = np.fft.fftfreq(N, dt)

ft_shifted = np.fft.fftshift(ft)
power_shifted = np.abs(ft_shifted)**2
freq_shifted = np.fft.fftshift(freqs)

things, names = ((noise, signal_1, signal_2, combined),
                 ('noise', 'signal_1', 'signal_2', 'combined'))
m = len(things)
fig, axes = plt.subplots(m+2)
for i, (thing, name, ax) in enumerate(zip(things, names, axes[:-2])):
    ax.plot(time, thing)
    ax.plot(time, np.zeros_like(time), '-k', linewidth=0.5)
    ax.set_xlim(0, 0.2)
    ax.set_ylabel(name)
axes[-2].plot(freq_shifted, ft_shifted.real)
axes[-2].plot(freq_shifted, ft_shifted.imag)
axes[-1].plot(freq_shifted, power_shifted)
axes[-1].set_xlabel('frequency (Hz)')
for ax in axes[-2:]:
    ax.set_xlim(-150, 150)
plt.show()


Here's a Wikipedia page discussing various theoretical methods of propulsion. It has some very interesting ideas.

The one that piqued my interest was the Alcubierre drive. For all intents and purposes it is a warp drive - well, as close as science can get to the fictional engine of The Enterprise anyway. It has been nicknamed the warp drive, I believe, by NASA scientists.


There are a couple of atmospheric phenomena that can create rings around the moon.

Corona are caused by water droplets diffracting the moon's light, they are fairly small, and close to the moon. They are coloured but not as brightly as a rainbow. Our eyes are also not so good at seeing colours in dim light.

The 22-degree halo is a larger ring, as big as two hands stretched out at arms length. Solar halos are clearly coloured, lunar halos are dimmer and I haven't seen lunar halos that are bright enough to appear coloured to me.

An eye condition, astagmatism, can also create rings, such as the "thin blue ring" (though I am not a doctor) You may be seeing a combination of all three at various times.


How can I make the fringing pattern visible in my flat (twilight) images for filter i?

I was finally able to make the fringing visible by block averaging after taking the steps above. If anyone else has any helpful tips I'd love to hear them!


Questions about time and space (from beginner)

A)

  Is it right that what we see/feel in every moment is projection of
  some 4D reality into our 3D world noticed by our senses.

It's more similar to a slice, hyperplane or hypermanifold, less to a projection.

  If my eyes were able to "somehow" see in 4D and I pointed my eyes to a
  person, would I see that person in all stages of his/her life from
  birth to death as a single "entity"?

In a classical world, yes.

B)

  Can I put infinite number of 3D spaces into a single 4D space?

Yes, a possible mathematical description of such a space-time is a 3+1-Minkowski space-time.

  If we considered that 5th dimension exists and that we could put
  infinite number of 4D spaces into 5D space, then we would end up with
  infinite number of parallel realities, in which Caesar did not crossed
  Rubicon, or did not even exist. Is it right or complete nonsense? Can
  be somehow proved that this is right/nonsense?

That's close to the many-world interpretation of quantum theory.
But it's thought of as an infinite-dimensional Hilbert space, as one of the simpler ways of thinking.

Thinking of a 5D space-time as a collection of all possible 4D space-times is less suitable, since it would be a mere set of 4D space-times without meaningful structure. Take Ceasar crossing Rubikon a day earlier or day later or a meter to the left or a meter to the right with or without a leaf falling from a tree, or a huge number of other versions. Which would be the correct order to pile up these versions to a 5D world? The infinite-dimensional Hilbert space model solves this ambiguity, and provides a meaningful metric for the set of all possible worlds.

The 10-dimensional models of some string theories are something else: The 6 extra-dimensions are thought to be curled up, and tiny.

We have no access to parallel worlds, so it cannot be proven by observational evidence. It's a mathematical model. It's possible to prove theorems within the mathematical theory, but that's usually based on axioms, or assumptions. Mathematical theories may, but don't need to match with observation, depending on how the axioms match to physical "reality".


Question about orbital direction on binary systems

Stars don't "orbit the centre of mass", they orbit each other.
What is true is that in a two-body system, the orbit of each star is an ellipse with the centre of mass at the focus.  What is not true is that the acceleration is towards the centre of mass.  There may be nothing at the centre of mass to attract!  The acceleration of each star is directed towards the other star. This is the "law of gravity". So it is quite accurate to say "Each star orbits the other star".
An orbit in which one star moves clockwise and the other moves anti-clockwise would break the law of gravity, because the acceleration would not be directed towards the other star.


Are the planets Trappist-1 in the solar system?

During high school, I have learned that we are in the solar system and there are some planets orbiting the Sun.

Yes that is correct. According to current definitions of what counts as a planet, there are 8 planets orbiting our Sun $-$ Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. There are also a collection of smaller "Dwarf Planets" which include (in no particular order): Pluto, Ceres, Eris, and Makemake. There's also a lot of other minor solar system bodies out there like asteroids, comets, etc.

  but there are stars, planets, constellations

Yes, there are many other stars out there besides our own (as you can see by simply looking up at the night sky). Each of those stars has the potential for having its own planets, much like ours does. Any planet outside our own solar system is referred to as an exoplanet (exo- meaning outside or external). Within the past 20 or so years we've developed the technology to find exoplanets around other stars and this TRAPPIST-1 discovery was simply the latest and most sensational discovery. To date we've discovered ~3500 exoplanets.

  What confused me a lot are these two pictures. [...] And this definition of an exoplanet.

Those pictures, while seeming to show the the TRAPPIST-1 planets in our solar system, are not indicating that those planets actually reside in our solar system. Instead, what they're trying to do is compare the new TRAPPIST-1 system to our own so you can see the relative sizes. In your first image, you see Jupiter along with the orbits and distances of a few of its Moons. Next to it, you see the TRAPPIST-1 system with the seven planets we know of around it. Next to that is the Sun where you can just begin to see the orbit of Mercury (with the other planets beyond the bounds of the image. The point here is just to compare sizes of orbits. Clearly this TRAPPIST-1 system is much smaller than our own since all 7 planets orbit closer to the main star that even Mercury does for us. However, it is still bigger than Jupiter and its moons. They're not trying to make the point that these systems reside in the same place, just putting them together for comparison.

As for the second image you have, what they're trying to show here (I assume) is the habitability zone. Around every star, depending on the star's size and temperature, there will be a good distance to be from the star that is optimal for life to exist. You don't want your planet to be too hot (by being too close) or too cold (by being too far). This image appears to be showing you the regions where a planet can be at a good distance from TRAPPIST-1 and still be at a good temperature. There is the Inner B line which I take to be the Inner Boundary of this zone. In other words, if a planet is any closer than this, it will be too hot. There's also the Outer B (or Outer Boundary) where beyond that, any planet would be too cold for life. Finally they have the Earth B. I can't be certain but I believe this point is actually the distance for the "optimal" temperature that we have here on Earth. In other words, the best chance a planet has of being at a good temperature for life is to orbit between Inner B and Outer B and ideally be at the distance of Earth B. This image also shows that the orbit of the planet TRAPPIST-1e is close to Earth B and still inside Outer B, meaning it is likely to be at a hospitable temperature for life (which is good news for us!). I won't go into what an AU (aside from mentioning it is a unit of distance) or the frost line, since they're not pertinent to the point of the image. Feel free to ask other questions for that.

Lastly, the definition of an exoplanet you found describes it an "extrasolar planet". That just means it is beyond our solar system $-$ "extra-" meaning "outside" or "beyond" and "solar" referring to our own Sun.


Difference between quiet, enhanced and active networks

I'm not sure what you mean by "network" in this context, but on the SOHO website, you can find pictures of the Sun for every day since 2006.

Quiet: there aren't any visible structures at the surface of the Sun

Active: the black spots are sunspots


Does the Perseid peak cause enough light pollution to be a problem?

Tought to answer, because the amount of airborne dust due to meteor showers and its stability in the upper atmosphere will vary significantly enough. We had many meteor showers, Perseids included, in the past that were a bit of a letdown and didn't produce as many shooting stars as initially predicted, and also the other way around of course.

But here's the thing tho, that discounting for any additional atmospheric glow during a few crucial hours of dusk and dawn, this airborne meteor dust might even improve the average atmospheric refraction with mostly extremely fine powder silicate dust settling onto the lower atmosphere and dispersing water molecules and binding to other solids (read: decrease humidity and pollutants), by being highly electromagnetically charged from the atmospheric reentry and electrostatically binding to larger atmospheric or pollution molecules, or simply for being hygroscopic and binding really good to water molecules, making them heavier enough to fall in a light spray back on the ground and remove the pollutants from the lower atmosphere in the process, too.

There might be other physical phenomena going on that I'm not aware of, but what I wanted to illustrate is, that there might even be positive side-effects in terms of light pollution vs increase in meteors, which is how we call meteoroids entering Earth's atmosphere.


I know that the radius of the observable universe is about 46.6 billion light years and is expanding at every instant of time due to dark energy.

There's some ambiguity about dark energy. It's said to be responsible for the accelerating expansion of the universe as opposed to the expansion. See the Wikipedia dark energy article where you can read that the term "dark energy" was coined by Michael Turner in 1998. This was in a paper that referred to the type 1a supernovae observations made by the Supernova Cosmology Project and the High-Z Supernova Search Team. Their observations indicated that the expansion of the universe is accelerating, which was contrary to expectations and led to Nobel prizes in 2011.

  So is it possible that we our entire universe could be inside a black hole

No. We aren't in a black hole. A black hole is a place where the vertical light beam can't get out because the "coordinate" speed of light at the event horizon is zero. As Stephen Hawking said, the universe is like a black hole in reverse, but it isn't the same as a black hole.

  and that the edge of our observable universe represents the event horizon which in this case is expanding faster than the speed of light

The edge of our observable universe is a horizon of sorts, see Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Davis and Lineweaver. But it isn't the same sort of horizon as the black hole event horizon.

  due to matter being constantly accreted inside it.

That isn't supposed to be the reason why the universe is expanding.

  Where that black hole is a part of another universe and that universe of another black hole and the pattern goes on. Is this possible ?

Reputable people have talked about black holes as baby universes but there's no supporting evidence whatsoever. My understanding of general relativity and cosmology is such that I think such hypotheses are flat out wrong. So I would say no, it is not possible.


How much of the Milky Way is visible to the naked eye from earth?

At any one time, an average observer can see about 2,500 stars in a clear dark sky. Note that eyesight varies and sharp-eyed individuals may be able to see a half-magnitude dimmer stars than the average eye (apparent magnitude is a scale in which each integer is $2.51$ ($100^{0.2}$) times brighter or dimmer than the next consecutive integer.) A very dark sky may enable magnitude $+7.5$ or even $+8$ stars to be seen, but in a typical "dark" non-urban sky the limit is often $+5.5$ to $+6.5$.

Supernovae can potentially be seen as far as 13 billion light-years (ly) away, essentially from the 'edge' of our observable universe. So, it's not saying much to say that a supernova might be seen across a distance of 100,000 ly.

Some types of supernovae can be over $-22$ in absolute magnitude — where absolute magnitude is defined as their apparent magnitude if they were observed from a 32.6 ly distance (to be honest, I'm not sure if supernovae absolute brightness is defined in exactly this same way). By way of comparison, the Sun's abs. mag. is $+4.8$ (lower numbers indicate brighter stars).

Because of the dust and interstellar medium, the possible distance we can see is just a small fraction of the size of the Milky Way. In reality, few stars are bright enough to be seen over 400 ly away. Deneb which has an estimated distance of 2,600 ly (but may be as close as 1,550 ly, the large uncertainty is due to its variation in brightness). Only 6 visible stars are thought to be farther from us than 1,000 ly.

So, to sum up, looking inwards into the Milky Way, our visibility is very restricted to the nearest 1,000–2,000 ly (while the MW disk's radius is 50,000 to 90,000 ly and we're about 27,000 ly on this side of center, but only the brightest stars are visible from more than about 400 ly away.

While looking away, we can see the Andromeda Galaxy (but not its individual stars) which is 2,500,000 ly from us. In other words, most people blame dust for the poor visibility, but that's mostly only relevant for telescopes. For us, our limited eye-sight is the real barrier (not to mention the dearth of really dark skies).


How can pre-main sequence stars radiate more energy than main-sequence stars?

Although pre-main sequence stars have lower temperature, they are essentially huge clouds of gas, often as large as 1 pc wide. The luminousity being proportional to square of the radius is essentially large for pre-main sequence stars.

Plus, the problem with gravitational contraction is not the amount of energy that can be generated per second. That can actually be larger than that produced from fusion if you insist on really massive objects. The logic, when we ruled out gravitational contraction as a source of energy for Sun, was that gravitational contraction cannot sustain it for long. Geologists knew from their dating science that Sun had to be older than grav. contraction accounts for.

Thus, admit it. Gravitational contraction makes pre-main sequence stars radiate more energy than main sequence starts, but that's just luminousity and not really energy.
Luminousity is energy per second.


Why do sun spots appear near the equator of the Sun?

Indeed the regions where sunspots  form tend to be equatorial, with sunspots concentrated near the regions at +/- 40° latitude.

Sunspots are regions at temperature lower than that of the surrounding as for they are closing point of the sun magnetic field. At sunspots, the magnetic field is much higher than average and that hinders heat transport to the photosphere due to a reduced convection.

While the magnetic field of the sun is loosely speaking and as seen by far described as a dipole field (similar to that of Earth or of a magnetic bar), its local distribution is complicated.

Picture from https://www.windows2universe.org/sun/sun_magnetic_field.html showing the magnetic field and spots forming at the loops closing areas.

Besides the fact that the magnetic field flips every roughly 11 years, and keeps changing in meantime, sun, which is not rock solid, has a latitude depending rotational speed. This, superimposed to the dynamo mechanism for stars magnetic field generation, is responsible for "carrying the local magnetic field  along with rotation" and forces the field closing before the geometrical poles.

In the words of Wikipedia under the entry stellar magnetic field https://en.m.wikipedia.org/wiki/Stellar_magnetic_field

"spots are regions of intense magnetic activity on the surface of a star. (On the Sun they are termed sunspots.) These form a visible component of magnetic flux tubes that are formed within a star's convection zone. Due to the differential rotation of the star, the tube becomes curled up and stretched, inhibiting convection and producing zones of lower than normal temperature".

This doesn't provide a clear picture of why the magnetic field behaves as such, but at least does tell that sunspots are a consequence of the magnetic field behaviour.

Understanding the magnetic field of stars and even that of Earth is indeed ongoing research. A little more articulated answer is given by the above linked Wikipedia page, particularly by sections Field generation and Surface activity.


Apart from Io, will any Galilean moons have oceans and land if their ice melts

We're not even sure the Galilean moons will still be orbiting Jupiter in 5 billion years.   They probably will, but it's hard to say with certainty.   It's difficult to model their orbital stability over that long a time-frame.

But in the spirit of your question, it takes much more than oceans and land to make a planet or moon livable.   You also need atmosphere, proper gravity (Jupiter's moons have very low gravity) and the right mix of elements.   If, somehow, one of the Galilean moons had continents, it's "ocean water" would be full of ammonia and other elements and we'd find it both stinky and likely toxic.

But the answer to your question is no, there are no continents.   The ice/water to silicate and metal-oxide ratio for the Galilean moons is much too high for it to ever form continents, and even if it could, a moon that light would have some difficulty retaining it's oceans.   Some would boil of into an atmosphere and with that low gravity, the atmosphere would be stripped from the moon over time.

From a practical standpoint, there aren't any good options for mankind to move to in 1-4 billion years.   Terra-forming mars is possible but difficult and only a temporary solution.  In time the Sun would get too hot for Mars as well.   Titan is an interesting option because it has about the right atmospheric pressure.   That's actually very useful, though living on Titan would be kind of dreary.  It's under a constant thick blanket of clouds.

The good news is, there's a lot of time to find a solution and living in space doesn't require living on the surface.   Living under the surface of an icy body (like Pluto) would be workable too.  In fact, there's several advantages to living inside an icy world rather than trying to construct one where we live on the surface.   Much less work to set it up.

If someone wants to move this to world-building, feel free, but I felt the question was asked in good faith.


Calculating length of summer solstice day using tilt and latitude

If you are reducing everything to plain geometry, you can compute the fraction of the "sunny" arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is

$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$

The angle of this segment is

$$a = 2\times \arccos(1 - h)$$

I get

New York: 14.9297092807291
London: 16.4195095586004

which is fairly close to the "expected" numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.


How do we detect if a planet has a liquid core?

A good way to show how this is achieved is to show you a case study, below I will detail how we came to make the theory that Mercury has a liquid core.

It is first important to note that we are never definite of the properties but we are able to use the data to infer and create a theory that we believe to be an accurate representation of the truth.

The MESSENGER spacecraft was launched for the purpose of analyzing MErcuries Surface, Space ENvironment, GEochemistry and Ranging (hence MESSENGER).

Messenger had a number of pieces of equipment on board to allow it to conduct it's experiments, as detailed in the below image:

Mercury Laser Altimeter

So how does the MLA aide us in determining if Mercury has a liquid core? Well the purpose of the MLA is to:

  track the planet’s slight, forced libration – a wobble about its spin axis – which will tell researchers about the state of Mercury’s core.

How does a Laser Altimeter work?

  A laser altimeter is operated from a plane, a helicopter or a satellite. It determines the distance to the [Planet's] surface by measuring the time-of-flight of a short flash of infrared laser radiation. The instrument emits laser pulses which travel to the surface, where they are reflected. Part of the reflected radiation returns to the laser altimeter, is detected, and stops a time counter which was started when the pulse was sent out. The distance is then easily calculated by taking the speed of light into consideration.

Using this technology one can build up a 3D visualization of a planets surface

What scientists discovered from this mapping is that:

  The planet's crust is thicker in low latitudes and thinner at the poles, a distribution that suggests the planet could have a liquid outer core. Its core is also large relative to the planet, comprising 85 percent of the planet's radius, much more than Earth.

How did they come to this conclusion? This paper explains in depth. (The full version is free, but requires signing up to the site, signing up is also free and has no costs and has the ability for you to turn off receiving emails and newsletters from them)

The most notable part is this excerpt and image:

  Resulting moment of inertia parameters for more than 1 million Monte Carlo models that include a silicate crust and mantle, as well as an Fe-rich core that may contain solid and liquid layers are shown in

(A) Outer radius of Mercury’s liquid core. (B) Average density of the solid shell that overlies the liquid core

Conclusion

Essentially, specific readings taken by a Laser Altimeter allow us to map out the three dimensional shape of a planets surface. Analysis of the shape shows us that the crust is thinner at the poles which is likely due to a liquid core.

On top of this, Mercury is believed to have a small magnetic field (1% of our own) and this might be due to a liquid outer core circulating a solid inner core, acting like a dynamo. Though there is still much speculation given over what causes the magnetic field so this is unreliable to use for a solid footing.

Aside from that there is the logical thought process that larger planets are capable of creating huge amounts of heat and pressure at their cores which create a molten core, or at least a molten layer. This is why there is much skepticism over Mercury because it is such a small planet that its core should have solidified long ago.


Earth and moon gravitational ratios and proportionalities

The first part is just replacing $M$ with $M_E$ and so forth. You get two statements of proportionality, one with the $E$ values and one with the $M$ values.
The second part is algebraic division.  Mathematically if $a\propto  b$,  and if $a_1,a_2,b_1,b_2$ are specific values that are in this proportion. $a_1=kb_1$ and $a_2=kb_2$ for some constant $k$. Division gives $a_1/a_2 = b_1/b_2$.  The constants  cancel
By division, you turn a statement of proportionally into an equation.


When was it first determined that the Sun is a star?

I'm going to be vague on the timing because it depends on what you mean by "determine". Which in turn depends on what is meant by "know" in science. There is a very good model for stars which predicts that balls of plasma that are prevented from collapse by fusion in their cores will look like the sun up close and will look like stars from a distance. So people believe that both the sun and the stars have the same nature. However, while there is direct evidence of nuclear fusion in the sun (neutrinos) there isn't the same evidence of fusion in stars.  So it can be argued that this hasn't been "determined", and given the nature of scientific discovery, might never be.
It also depends on what you mean by "the same nature". What if we suppose that two things have the same nature, but are completely wrong about what the nature of the object actually is.  We didn't have any real understanding of the core of stars until the 1930s, and we are still filling in the details.
The early claims that the sun is a star were based on general evidence and some incorrect, or at least odd beliefs.  Later discoveries showed how starlike the sun actually is but there was no Eureka momement; there is no single event that you can point to in which someone discovered this.
Anaxagoras in the 400s BC suggested this or something like this.  He suggested that the sun was a ball of molten metal, and the stars were fiery stones, so something of "the same nature"  He proposed that the sun and the stars were torn from the Earth, and then ignited.  He probably supposed that meteors were "falling stars" and might have linked meteorites to meteors.  He almost certainly believed that the sun was bigger than the stars. So "of the same nature" but the sun was the greatest of the stars. He is significant because he is probably the first to give a physical theory for the nature of an astronomical body.  He made the first steps to bringing the study of the heavens within natural philosophy, or "physics".
Anaxagoras was the first to hypothesise that the Sun and the stars might have the same nature.
The notion that the sun is not special is suggested by Copernicus, and there were mutterings about it in the period after Copernicus' publication; it is implicit in some of Galileo's work. However, it was Giordano Bruno who gave full weight to this idea. His evidence seems merely to have been that, if the sun was very distant, it would appear as a star. The observation was naked eye viewing of the sky. He does not seem to have estimated how distant the stars would be, as he had no way of measuring the actual brightness of stars of the sun, but he most clearly declared that:

“The composition of our own star and world is the same as that of as many other stars and worlds as we can see.”

(And by "our own star" he means "the Sun")
It wasn't until the middle of the 18th century that any actual evidence for stars being of the same nature as the sun was found. Angelo Secchi used spectroscopy to analyse starlight, and found that it was very similar to sunlight. By this time though Secchi was confirming the orthodox position rather than proposing a new discovery.
With the measurement of stellar parallax, another piece of the puzzle settled into place, now that the distance to stars could be measured, the were found to comparable to the sun in brightness as well as in spectrum. And developments in nuclear physics in the '30s-'50s showed how stars and the sun could be powered by similar nuclear engines in their cores.
So this question doesn't afford a simple single answer.  This wasn't a sudden paradigm shift. Rather it was a gradual process.  However, if you need just one name to hang on this idea then Bruno would seem to be the first to clearly assert that stars have exactly the same nature as the sun.
(Much taken from http://www.astronomytrek.com/who-discovered-the-sun-is-a-star/)


How do astronomers find interesting events?

The answer to your question is yes. Many times discoveries in astronomy are serendipitous in nature - some of the most well-known discoveries fall into this category: the discovery of the cosmic microwave background, the discovery of some of the planets in our solar system, are but two examples of this.

However, many times they are not serendipitous. Astronomers plan to discover things - look at the Kepler satellite which finds nearby exosolar planets by detecting planetary transits. The Sloan Digital Sky Survey was a planned galactic survey which gave us new insights about the properties of galaxies as well as structure formation.

One also has to consider time scales. Many things astronomers care about have timescales which are millions or billions of years (merging clusters, the evolution of main sequence stars, etc..), however, some phenomena are on very short time scales, namely supernovae. Microlensing events can also be somewhat serendipitous in their discovery, though researchers still devise experiments to look for them (see OGLE).

It is certainly a combination of luck and planning. This answer was intended to be a bit vague, since the question doesn't specifically ask about one phenomenon. Software does play a big role in identifying and classifying objects and events in the universe. I can talk a little bit more about this in the context of gravitational lensing if people care to know about it.


What is difference between horizon and comoving distances?

The comoving distance is a distance. You can calculate the comoving distance from your head to your toes (though it's more useful in cosmological contexts). It is defined such that today it coincides with the actual, physical distance you would measure if you froze the Universe and laid out meter sticks. In 11 billion years from now, when the Universe has doubled in size, physical (cosmological) distances have increased by a factor of two, but the comoving distances are, by definition, the same.

A "horizon" is a term used for a (non-physical) boundary, e.g. the "particle horizon", which is the boundary between the observable and the un-observable Universe. Or the "(cosmological) event horizon", which marks the boundary between regions from which we may at some point receive a signal emitted today, and regions from which we cannot.

Just like on Earth it makes sense to talk about "the distance to the horizon", you can say, e.g., "The comoving distance to the particle horizon is 46 billion lightyears", or "The comoving distance to the event horizon is 17 billion lightyears".

As you say, to calculate the comoving distance you use the Robertson-Walker metric (assuming here a flat Universe for simplicity)
$$
ds^2 = -c^2 dt^2 + a(t)^2 \big[ dr^2 + r^2 d\Omega^2 \big],
$$
and set $ds=d\Omega=0$ to measure along a radial null geodesic. Then $dr = c\,dt/a(t)$, or
$$
r = c \int_{t_\mathrm{em}}^{t_\mathrm{obs}} \frac{dt}{a(t)},
$$
where $t_\mathrm{em}$ and $t_\mathrm{obs}$ is the time when the light is emitted and observed, respectively.

To calculate the comoving distance to the particle horizon, you set $t_\mathrm{em} = 0$ and $t_\mathrm{obs} =$ today, because the particle horizon corresponds to light emitted at Big Bang and observed today.

To calculate the comoving distance to the event horizon, you set $t_\mathrm{em} =$ today and $t_\mathrm{obs} = \infty$, because the event horizon corresponds to light emitted today and observed in the (almost) infinitely distant future.


There is no average value. The large scale magnetic fields of main sequence stars are both a function of spectral type and rotation. In turn, rotation is age-dependent. In addition, the gas pressure at a distance from the star depends on conditions in the stellar wind and this too will depend on spectral type, rotation and age.
There are no universally agreed models for the behaviour of these quantities.


How do large solar flares compare to flares on other stars?

This is a part answer, a comparison of sun-like star superflare behaviour with our sun.

According to the article Superflares on Solar Type Stars Observed with Kepler
I. Statistical Properties of Superflares (Shibayama et al. 2013), observations were made on solar-like stars (type G), over 500 days.

One key observation was that they

  found 1547 superflares on 279 G-type dwarfs

Despite this seemingly massive amount, they deduced that

  the occurrence frequency
  of superflares with energy of 10^34 − 10^35 erg is once in 800-5000 years.

and

  in some G-type dwarfs the occurrence frequency of superflares was
  extremely high, ∼ 57 superflares in 500 days (i.e., once in 10 days). In the case of
  Sun-like stars, the most active stars show the frequency of one superflares (with
  10^34 erg) in 100 days.

These are associated with very large star-spots, far larger than those on our sun.

There was an earlier theory that the presence of Hot Jupiters were a major contributor to super flares, hence the reason that our Sun did not exhibit this phenomenon often. However, there is some evidence of a possible past superflare occurred from our Sun:

  an occurrence of a energetic cosmic ray event in 8th century
  recorded in a tree ring of Japanese cedar trees. There is a possibility that this event was
  produced by a superflare (with energy of ∼ 10^35 erg) on our Sun.

and no hot Jupiters were detected around many of the stars observed either, so this theory is largely ruled out by the authors. Rather, they postulate that G-type stars 'store' magnetic energy.


I'm having trouble achieving sharp telescope focus

Your telescope has a 2000 mm focal length, and 200 mm aperture. With the 20 mm eyepiece, you get 100x magnification. With the 10 mm lens, you get 200x. Several things could cause what you describe:

Seeing (turbulence) might be bad. It is common in many places that less than great seeing means everything at 200x and over might start looking blurry. But seeing changes with the seasons, or day to day, hour to hour, or indeed from one second to another. It's random. If you get the exact same results all the time, and the image never improves even after many weeks of trying, then perhaps it's not seeing that's the culprit.

Your telescope might not be collimated. Look in the user's manual. Is there a collimation procedure you need to follow? (EDIT: yes, there is - page 20) A miscollimated scope indeed behaves as you described - seems "sharp" at low-ish magnification, but quickly becomes blurry if you push magnification up.

Point the scope at a bright star and defocus. Are the rings perfectly circular? If not, it might be miscollimated.

http://www.astrophoto.fr/collim.html

Another factor is optics quality. Optics with lots of aberrations cause "mushy" images and a difficulty to find the true focus. High quality optics produce a scope that is very "snappy" when achieving perfect focus, and make very sharp images when seeing is good. There's nothing you can do about this, but the good news is - this problem is not very common nowadays; optics quality, while not always great, tends to be at least passable in many cases. Quality could be evaluated either on the test bench (which requires equipment), or via direct observation of stars (which requires a very experienced observer and takes time).

Finally, have you compared views with another instrument of similar aperture? The image at higher magnification will always seem a bit more soft, compared to lower magnification - even in perfect seeing with a collimated instrument. Also, the contrast always decreases as you increase magnification. This is why there is always an optimal magnification for each view - sometimes higher, sometimes lower, depending on many factors. The best scope in the world will look mushy and blurry if you push magnification up too much.

Can you easily see the Cassini division in Saturn's rings? If yes, then you're probably in okay shape (collimation and seeing), you're just not used to the softness at higher magnification. I have a very high quality reflector (self-made), similar aperture class to yours (a bit smaller), and Cassini is immediately visible at 180x and looks sharp; it's also visible at 255x but the whole thing starts to look mushy and faded. It looks super-sharp and high-contrast at 136x but it's kind of small. This is in perfect collimation, perfect thermal equilibrium, high quality optics throughout the stack (primary and secondary mirrors, eyepieces), and good seeing typical of N. California.

When seeing is bad, Cassini starts to fade no matter what I do.

Does Saturn look equally soft in the center of the image, as well as near the edge? Good optics (mirrors, eyepieces) make a good image everywhere, cheap optics make an okay image in the center and a blurry image at the edge.

My bet is that it's either bad collimation, or you haven't compared the view at 200x with another scope, or both. But it's hard to diagnose things over the Internet.

You should probably get a 15mm eyepiece as well. You'll use it a lot, probably more than the 10mm.

Light pollution is never a factor for: planets, the Moon, the Sun, most double stars. These things are just too bright to care about it, so observing them from the city is fine. What does matter is seeing (turbulence). Also, it's important that the scope is in perfect collimation.

Light pollution only matters for the "faint fuzzies": nebulae and galaxies. These are low-brightness, low-contrast objects, that are difficult to see from the city. Being low contrast, your eye will not perceive a lot of detail anyway, so for these objects seeing doesn't matter.

Congratulations for being aware of, and following, the rule that says you need to acclimate the scope to ambient temperature. A lot of people are not aware of it. At least you're removing one unknown from the equation. It's always a good idea to let the scope "breathe" outside for an hour before you even begin to observe.

One way to be 100% sure that your scope is focused perfectly is to use a Bahtinov mask. Just make (or buy) one to match the scope's diameter, put it over the scope's mouth, point scope at a bright star, and tweak the focuser until all spikes intersect exactly in the center of the star. Then remove the mask - the scope is in perfect focus. This will probably not help in your case, but it's just one way to remove another uncertainty. It also helps with regular observations - I use it almost every time I observe, it makes focusing so much easier.

The mask could be made from a piece of cardboard and a sharp knife in like 20 minutes. It doesn't need to be a marvel of engineering precision to work well.

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

One thing that is certainly NOT a factor is the central obstruction. You'll often hear on the Internet how Cassegrain instruments, or other scopes with a large central obstruction (secondary mirror) are supposedly less "sharp" than scopes without a central obstruction, or ones with a very small obstruction.

Yes, the obstruction does matter a little, but not to the extent that Internet mythology would have you believe. A Cassegrain instrument, with good optics, in perfect collimation, can be a superb telescope even at a 40% obstruction. It simply performs like a somewhat smaller instrument, that's all (the physics of the central obstruction are very complex, and would require an entirely different discussion on this forum to fully elucidate).

(Added in reply to a comment below)

Miscollimation is quite likely the #1 performance killer of all amateur telescopes. Some well-made refractors are more or less immune to it, but most reflectors need periodic collimation.

It's a good idea to google around for various collimation techniques, there's so many of them. Some require the observation of stars, some require special devices known as collimators. The technique is also different for different types of telescopes (newtonian, cass, etc).

If you own a reflector, it's a good idea to collimate it once in a while, just in case, like changing the oil on a car. Dobsonian telescopes probably require most frequent collimation.

Check Howie Glatter's site - he makes some of the most precise collimators currently available:

http://collimator.com/

Regardless, just learn any method you can and apply it. If the telescope is massively miscollimated, you'll get an improvement immediately.

Not being able to see Cassini is not a good sign. Maybe seeing is persistently bad (it can happen), or maybe the scope is off-collimation. Can anyone else in your area see Cassini these days in a similar aperture, in a scope known to be collimated well?

Surface detail on Saturn is not easy to see, but should be doable with apertures over 100mm. It's like zones of latitude similar to Jupiter's equatorial belts, but very, very pale and faded. You have to keep looking for a while to see it. Don't worry if you don't see that stuff for now.

Jupiter's Great Red Spot is harder to see nowadays. It's been shrinking and fading for decades now. Go on Wolfram Alpha and type in Great Red Spot. It will tell you if it's visible at the moment. Usually it looks more like a dent in the equatorial belts, rather than an actual visible spot.

http://www.wolframalpha.com/input/?i=great+red+spot

Jupiter works best at not so high magnification, otherwise the contrast gets too low. Around 140x ... 180x should give you decent contrast in your aperture.


Can one calculate the size (i.e. the radius) of our universe?

If I understand the question the answer is, no, you can't estimate the final size of the Universe by calculating the schwarzschild radius of the universe.

The mass of the Universe is not known, and may be infinite.

It is not know if the observable universe will ever stop expanding, however the evidence suggests that it won't, and so won't have an "ultimate" size.

The mass of the observable universe is of the order $10^{53}kg$, mostly comprised of dark energy and dark matter.

A black hole with that mass would have a radius of $2MG/c^2 = 10^{26}m$, which as has been observed is on the same order of magnitude as the (comoving) radius of the observable universe. Whether this has any cosmological implications is not clear. It doesn't imply that the Universe is a black hole. It doesn't say anything about the Universe at all, as this calculation was for the observable universe. It doesn't say anything about the "ultimate size" of the universe or the observable universe (which is expected to continue expanding without limit)


What's the map on Sheldon & Leonard's refrigerator of?

It's the Beethoven basin region on Mercury.
The MESSENGER image gallery says:

A digital terrain model (DTM) derived from Mercury Dual Imaging System (MDIS) stereo images.
The lateral spacing is 330 m and the map is in stereographic (conformal) projection.
The image is taken from abstract number 1913 submitted to the 2012 Lunar and Planetary Science Conference, by Frank Preusker and colleagues.

NASA / JHU APL / Carnegie Institution of Washington
A collage of MESSENGER photos from 2011 debuted on the Apt. 4A freezer door around the same time.
A statement released 2012-04-05 quotes PI Sean Solomon:

The MESSENGER team is thrilled by the decision of the producers and writers of 'The Big Bang Theory' to weave some of the spacecraft's latest images of Mercury into this week's episode...
We look forward to seeing how the images figure in the story line and camera shots, and we hope that interest in the MESSENGER mission is broadened by this exposure on one of the most popular series now on television.


Is it possible to image a meteor so the rock is visible?

Meteors are very small, typical ones are the size of a grain of sand, and bright fireballs are only a few grams. So they are nowhere near as big as "rocks" and there is no way to image them.

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

http://curious.astro.cornell.edu/physics/73-our-solar-system/comets-meteors-and-asteroids/meteors/303-what-is-the-typical-size-of-a-visible-shooting-star-intermediate

https://www.amsmeteors.org/fireballs/faqf/

But let's say you could fly next to a shooting star in a UFO and image it that way. The next problem you would face is that the meteor is incandescent and and it would be difficult to get anything but an overexposed blob in a regular camera.


Can two neighboring galaxies move apart at steady speed?

The key word here is "eventually."
The strength of the gravitation felt between two objects is inversely proportional to the square of the distance between the two bodies. As a result of this, raising an object from a nonzero distance from object A to an infinite distance from object A requires a finite amount of energy.
In a Keplerian/Newtonian trajectory, the velocity of an object solely under the influence of gravity is governed by an exchange between gravitational potential energy, and kinetic energy.  As a result of that, no matter how long their mutual gravity acts, there's a limit to how much kinetic energy gravitation can take from the object.
If the equations are balanced such that the object would come to rest at an infinite distance, the speed that the object has is called "Escape Velocity."
If the object is moving faster than escape velocity, it would wind up with a finite relative velocity at an infinite distance, and this value is called Hyperbolic Excess Velocity.
For instance, if the Sun and Voyager I were the only objects in the universe, and there were no other forces between them other than mutual gravity, the relative velocity between the two of them would never fall below 16.6 km/s.
A similar case holds for any two objects, regardless of mass; If they're moving relative to each other higher than their mutual escape velocity, considering only the  two bodies, and only the gravitation between them they wind up asymptotically approaching a finite relative velocity.


Pluto will never be a planet.

There are a number of technical papers that give more precise meaning to the concept of "clearing the neighborhood". It's not just now, it's can the object in question clear the neighborhood of its path while the Sun is still a star.

In the case of Pluto, Ceres, Eris, and a host of other not-quite-planet objects, that will not happen. Ever. Or at least so long as the Sun shines. After that, does it matter?


Why does the moon looks like having different filters sometimes?

TL;DR: because of the Earth's atmosphere.
When the Moon is low on the horizon, the light we receive from the Moon has to go through a lot of layers of the atmosphere. Rayleigh scattering makes this light more yellowish/orangish, for the same reason sunsets look orange.
A very thin layer of clouds can also change the appearance of the Moon.
Sometimes, the Moon ends up in the Earth's shadow. In this case, the light that hits the Moon first went through the Earth's atmosphere. Because of Rayleigh scattering (again), the light that went through the atmosphere is red. This gives the Moon a reddish tint.


What does the milky way look like from earth to the human eye without light pollution?

It is pretty obvious from a dark sky.

Sky and Telescope have a simulation (using stellarium) of the sky with a limiting magnitude of 6.5 (about the limit with excellent eyesight and a very dark sky). The Milky Way is very clear. Light and dark tracks are visible. It is clearly something that every person would know about and be able to see nightly. It is no surprise that it is mentioned in the myths of many ancient peoples.


What is the meaning of "bright unresolved sources"

"Unresolved" here means a source of light that appears as a point, not as a disc or a cloud.

For example, the planets can all be resolved to discs by even quite moderate telescopes. Galaxies also can be resolved, they don't appear as point sources.

However, stars, quasars, and most asteroids cannot be resolved by amateur equipment. In the context of the LSST paper, resolved sources are "galaxies" and unresolved sources are "stars and quasars".

  The photometric measurements for resolved sources (galaxies) have to include several standard magnitudes, such as Petrosian magnitudes, as well as appropriate model magnitudes.


Accuracy of Mercury transit calculations

This got me interested to know if there is a gold standard for such calculations

For the orbital mechanics and geometrical aspects, the NASA JPL Development Ephemerides are commonly used. These have errors for the Sun and inner planets of tens of kilometers or less.

How to pronounce “Ephemerides”?

  ...and what is the range of error for it. For forecasts n years out is the error range n times more?

That depends on the specific ephemeris you use and a lot of other details. The ephemerides are build from a huge body of observations and calculations and it's not a simple thing to say. I think the errors will be dominated by the definition of the edge of the Sun for several decades or even centuries. Farther out in time, the ephemeris itself will have some error that may increase with time in either direction.

Below I calculate that the apparent radial speed at which mercury crosses the "edge" of the Sun's disk is about -70.6 and +70.6 km/sec, assuming a "radius" of 695,700 km, using the Python package Skyfield which in turn uses the NASA JPL Development Ephemerides. For this calculation I used DE421.

I used Caracas as the location since the whole event was visible from this location. The main calculation happens in these three lines. The .observe() method automatically does the light-time correction for the positions.

sun_astrometric     = caracas.at(times).observe(sun)
mercury_astrometric = caracas.at(times).observe(mercury)
angular_separation  = sun_astrometric.separation_from(mercury_astrometric).radians

I put the words "edge" and "radius" in quotes earlier because these are not well defined for things like stars and gas giant planets. There are definitions that can be used to define nominals for these, in the case of the Sun it's when the opacity reaches 2/3 or something like that.

70 km is a pretty small distance at the edge of the Sun. From this answer you can see that there is a region of a few hundred kilometers where things are changing. It's hard to say exactly what defines the "edge" of the Sun in order to say when the hard edge of rocky Mercury will be seen to touch it. I can't be sure, but I have a hunch that you can get several seconds of ambiguity based on how the edge of the Sun is defined.

Wikipedia even lists **two different values for the radius that differ by 642 km*!* At 70 km/sec that's almost ten seconds right there!

Equatorial radius
  695,700 km  [7]
  696,342 km  [8]

From this answer:

  From Chapter 2: The Photosphere of Timo Nieminen's thesis Solar Line Asymmetries: Modelling the Effect of Granulation on the Solar Spectrum

    Figure 2-3: The Holweger-Müller Model Atmosphere

    7 Holweger, H. and Müller, E. A. “The Photospheric Barium Spectrum: Solar Abundance and Collision Broadening of Ba II Lines by Hydrogen”, Solar Physics 39, pg 19-30 (1974). Extra points have been cubic spline interpolated by J. E. Ross. The optical properties (such as the optical depth and the opacity) of a model atmosphere are, obviously, very important, and will be considered later. See table C-4 for complete details of the Holweger-Müller model atmosphere including all depth points used.

    8The height scale is not arbitrary. The base of the photosphere (height = 0 km) is chosen to be at standard optical depth of one (i.e.  5000Å = 1 ).

Python script using the Skyfield package to calculate some things about the 11-Nov-2019 transit of Mercury across the Sun as seen from the city of Caracas, Venezuela.

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader

# https://astronomy.stackexchange.com/questions/33918/accuracy-of-mercury-transit-calculations

load        = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data        = load('de421.bsp')
ts          = load.timescale()
bodies      = ('sun', 'mercury', 'earth')

seconds               = 12*3600. + np.arange(0, 390*60+1, 10)
hours                 = seconds/3600.
times                 = ts.utc(2019, 11, 11, hours)

sun,  mercury, earth  = [data[x] for x in bodies]
caracas               = earth + Topos(latitude_degrees  = 10.48,
                                      longitude_degrees = -66.90,
                                      elevation_m = 900.)

sun_astrometric     = caracas.at(times).observe(sun)
mercury_astrometric = caracas.at(times).observe(mercury)
angular_separation  = sun_astrometric.separation_from(mercury_astrometric).radians
distance_to_sun     = sun_astrometric.distance().km
distance            = distance_to_sun * np.tan(angular_separation)
r_sun               = 695700*np.ones_like(hours)
thing               = distance < r_sun
i_in                = np.argmax(thing)
thing[:i_in+1]      = True
i_out               = np.argmax(~thing)
hours_in, hours_out = hours[[i_in, i_out]]

dt                  = seconds[1] - seconds[0]
ddistance_dt        = (distance[1:] - distance[:-1])/dt

if True:
    plt.figure()
    fs = 12
    plt.subplot(2, 1, 1)
    plt.plot(hours, distance)
    plt.plot(hours[[i_in, i_out]], distance[[i_in, i_out]], 'ok')
    plt.plot(hours, r_sun, '-k')
    plt.ylim(0, None)
    plt.title('projected distance from Sun center (km,)', fontsize=fs)
    plt.subplot(2, 1, 2)
    plt.plot(hours[:-1], ddistance_dt)
    plt.plot(hours[[i_in, i_out]], ddistance_dt[[i_in, i_out]], 'ok')
    plt.xlabel('hours UTC 2019-11-11', fontsize=fs)
    plt.title('d projected distance/dt (km/s)', fontsize=fs)
    print(ddistance_dt[[i_in, i_out]])
    plt.show()


Infalling observer could never cross Black Hole event horizon?

It is often pointed out that the infinite time dilation at the EH is what is known as a "coordinate singularity", rather than an essential singularity that says something is physically singular there (like never reaching the EH for the infaller).  A coordinate singularity is essentially a hiccup in the language we are using, like the way a north-south-east-west coordinate system is of no use at the North pole.  The problem is that if you want to talk about "whether something has crossed the EH yet", you must match up your own concept of "now" with some time at the EH, and that matching up is the job of a coordinate system.  So saying "it hasn't reached the EH yet," and repeating that every day for infinity of time here at Earth, is not the same thing as a physical statement about what is happening at the EH, it's just a reflection of the way we are matching up our "nows" with theirs-- it is purely a matter of the chosen coordinates.

To know what is physically happening to the observer at the EH, the ideal solution  would be to simply ask them, and they would say "yes I am crossing the EH now".  Here it gets tricky, because the normal coordinates we use to talk about the EH (the one you were using also), says that we could never receive that communication.  But those coordinates make physical assumptions that are violated by having something falling into the black hole.  At this point it would be best to have a GR expert, but I believe what happens is that the very act of having mass fall into the black hole in effect moves the EH out to swallow that mass, and we should therefore be able to receive the communication that the observer has crossed the EH in a finite time.  This must be true because, as pointed out, the very existence of the black hole is like information coming to us that stuff has crossed its EH, an EH that formed and moved out as the black hole was created by mass infall.


Extracting kinematics from stellar spectra. Why are template stellar spectra necessary?

(Optically thin) Emission lines in nebulae are generally unresolved and have simple shapes that might be approximated by Gaussians. Stellar spectra are complicated - there are blends, there is rotational broadening, the lines are not Gaussian, there are molecular bands, there may be two stars!

By cross-correlating (or fitting) a template you are effectively modelling your spectrum with what you already think is the best representation of the spectrum.

If emission lines are not unresolved and show velocity structure, as you say, then fitting a simple Gaussian will give an incorrect result. A more complex model would be required, but as far as I know there are no template spectra that can help you out here. The situation might be thought analogous to a stellar spectrum with rotational broadening, but there the template can simply be convolved with the appropriate rotational broadening kernel.


Why does disorder (entropy) increase in the same direction of time as that in which the universe expands?

Entropy is the only quantity in the physical sciences that requires a particular direction for time, sometimes called an arrow of time. As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease, Universe can seen as one big isolated system in which total energy is always constant. Hence, from one perspective, entropy measurement is a way of distinguishing the past from the future.
https://en.wikipedia.org/wiki/Entropy_(arrow_of_time) - wiki for relation between time and entropy.

https://www.youtube.com/watch?v=870y6GUKbwc - Basics of entropy.

Highest state of entropy is achieved when the system is in equilibrium, to have this equilibrium, energy entities (mass and other energy radiating sources) should be enough apart from each other that no more energy interaction(give and take) could happen and energy is distributed evenly in the universe, thus expansion.

https://www.youtube.com/watch?v=xZTb6sfHEX8- Will the Universe expand forever
https://www.youtube.com/watch?v=26_GN-uL4Jc - The Laws of Thermodynamics and                      Fate of The Universe


The Luminosity of two stars with same brightness but at different PC away

The absolute luminosity of A is bigger than B. Like two lamps.
$$\frac{120}{24}=5,$$
so A is five times farther B, but as the luminosity decays with the square of the distance, then  A is $5^{2}=25$ times more luminous.


Why do we use FITS format for scientific images especially in astronomy? How is it different from formats such as JPEG, PNG etc?

File formats tend to be industry/field-specific, with the format, tools, and expectations of the field coevolving to become more dependent on each other over time. JPEG co-evolved with amateur digital photography, PNG co-evolved with the web. Likewise, FITS co-evolved with astronomical data processing, and so is naturally more suited for that purpose than formats that had entirely different goals and communities involved in their development.

(Pedantic nit: JPEG is the name of the compression method and the group that designed it. The file format is technically "JFIF", but everybody colloquially calls it JPEG.)

As an image format, FITS has a number of desirable qualities that are lacking in JFIF and PNG, that are crucial for scientific data:

Storage of more bits per pixel (CCDs can records 12, 14, or more bits), and also floating point values.
Storage of arbitrary number of data channels (scientific data may have many, or other, frequency bands than the RGB that JPEG and PNG are limited to).
No lossy compression as is typical (though not strictly required) for JPEG.
Higher resolution (JPEG/JFIF, for example, is limited to 65,535 pixels in each direction), and FITS is also capable of storing 3D data volumes.
Support for unlimited metadata in the header, for example the sky coordinates, information about the telescope, etc. JPEG and PNG don't have the fields and aren't set up to record arbitrary metadata.

Also, the use of FITS for astronomy predates the existence of both JPEG/JFIF and PNG -- FITS was standardized in 1981, JPEG in 1992, PNG in 1996. So even if those formats were suitable (which they are not), by the time they were invented there was already widespread use and existence of astronomical image processing tools that were geared to FITS files (and growing archives of astronomical data in FITS format), so it would have been a major effort to switch formats that would never be undertaken unless a new format not only met but exceeded all the prior requirements of the field.


Trying to categorize moon phase using fraction illuminated

I used pyephem to calculate the fraction illuminated at each time

Program

import ephem
m = ephem.Moon
d1 = ephem.Moon('2017') # d1 is the date of the first new moon in 2017
days = [3.69132375*i for i in range(1,8)]
for day in days:
    m.compute(d1+day)
    print(day, m.moon_phase)

Output

3.69132375   0.1482372306829685
7.3826475    0.5259582805101828
11.07397125  0.8929835990614957
14.765295    0.9932503040700423
18.45661875  0.8004766309895268
22.1479425   0.4698580677294108
25.83926625  0.15043495343265506

Note that this is dependent on the particular month. There is an asymmetry as the motion of the moon is not perfectly even, it moves in an elliptical orbit.


Why do not we use optical telescopes to study cosmic rays directly?

The telescopes in your picture are optical telescopes. They are used to detect cosmic rays using the   Cherenkov radiation caused by charged particles moving faster than the speed of light (in air) in the atmosphere. These are normally secondary particles that have been produced in collisions between cosmic rays and nuclei in the upper atmosphere.

Radiation produced by particles moving at close to light speed is beamed in the direction of motion. An array of optical telescopes picks up the beam, and the orientation of the image is a projection of the original track of the particle. By combining images taken with telescopes covering a wide area, and taken at almost the same time, one can reconstruct the paths of the secondary particles to work out which direction the original cosmic ray came from. The intensity of the images tells you something about the original energy of the cosmic ray.

Possibly this is the answer to your question - a single optical telescope might detect the Cherenkov radiation from a single secondary particle, but it would be difficult to reconstruct the path or energy of the original cosmic ray on its own - at least a pair of telescopes, or better still, a network is preferable. The properties of these "cosmic ray" telescopes are also quite different from "ordinary" astronomical telescopes. Cosmic ray telescopes need as much collecting area as possible, because the Cherenkov radiation is faint - but they can do this at the expense of image quality. They use very wide angle cameras (5 degrees or more) and use huge faceted/sgemented mirrors that give massive collecting area, but image quality (maybe 3 arminutes full width half maximum for a point source) that would be completely unacceptable for conventional optical light astronomy.

Or (thanks Conrad Turner) you are asking why we cannot use the mirrors of an optical telescope to focus the cosmic rays onto a detector and image their source? The reason for that is simply that cosmic rays are not reflected from glass/silvered surfaces in the same way that light is. They are extremely energetic particles that either pass straight through the mirrors or are absorbed within them. i.e. They do not reflect or refract in the same way as light.

In other circumstances, cosmic rays are a nuisance when conducting observations with optical telescopes. They are a source of background noise in CCD detetor images. Cosmic rays (or the secondary particles) are capable of liberating electrons in the silicon and therefore simulating small intense light sources in the sky. Often these are seen as small bright pixels, groups of pixels or trails on the CCD image, confusing what you were originally trying to look at. However, these cosmic rays are not focused by the telescope, they are essentially being picked up by the detector itself and would have been so if the detector was not even attached to a telescope at all. They are nearly impossible to shield against because they have very high energies and your detector needs to have a hole to let the optical light in!


Which are the best American universities/colleges for observational astronomy?

Out of curiosity: where does this limit of 70 lightyears come from? Exoplanets can be detected to way beyond that. This strikes me as a strangely restrictive limit.

Be that as it may. There are several great astronomy departments in the US, and it really depends on what you want to do exactly. For exoplanets, Harvard would be a great choice. Professor David Charbonneau is a heavy weight in the field.

Similarly, many of the other renowned Universities have good astronomy departments. Yale, for instance, or Berkeley (although since the Geoff Marcy incident I imagine the exoplanet department has taken a hit).

So at this point it is not really a matter of choosing 'the best' University. I suggest you have a look at some papers by the professors at different Universities, to get a sense of what you'd like to work on.

And don't forget: for your research to be fruitful the environment is also very relevant. Just have a look and see what city you'd like to live in.


Why does Neptune only have one major moon?

It seems possible that Neptune once had a system of moons similar to other outer planets. However Triton, a Kuiper belt object was captured by Neptune, and this disrupted the existing moons. It would have initially had an eccentric orbit, intersecting with the orbits of other moons, setting them into unstable orbits and ultimately either colliding with each other, hitting the planet or being ejected.

The minor moons formed from the debris of these collisions after Triton had entered a more stable orbit. So the disruption was the capture of Triton.


How does science knows what is the early universe?

I assume you're referring to the recent press release about the quasar J043947.08+163415.7, observed recently using Hubble. The paper about the observations details how the authors measured the distance to the quasar, by calculating its redshift, a quantity that describes how the wavelength of light appears to change based on whether the object is moving relative to an observer. On cosmological scales, redshift can then be converted to distances. We usually see quasars at $z>0.1$.

As is usually the case, the quasar's redshift was determined by looking at spectral lines - in this case, Mg II emission around 21000 angstroms (see in particular the inset):

This emission line was observed to have a different wavelength than it would if the quasar was at rest, enabling the astronomers to calculate its redshift: $z=6.511\pm0.003$.

This is the typical process used to determine redshift, and therefore distance - well, not necessarily using that particular Mg II line, of course, but nonetheless the use of spectroscopy to measure how the location of spectral lines changes. The difference between "early" and "late" isn't quite clear, and probably varies depending on who you're talking to. For instance, someone studying the Big Bang might refer to the period before recombination as "early" - although we're seeing the quasar as it existed much later than that; it wasn't around for recombination.

I'd like to note that it's somewhat misleading to call this quasar the brightest (or even brightest known) quasar in the universe, because - as that press release notes - the gravitational lensing by a foreground galaxy, which enabled the object to be discovered, also magnified its brightness. It appears to have a brightness of 600 trillion solar luminosities, but without the lensing, that number is reduced to 11 trillion solar luminosities - a much more modest figure.


Is it be correct to say that we live in a young (only 14 billion years old) universe?

An alternative way of looking at it is that we live in a dying universe. It may have a long, drawn out end, but the peak of star forming activity happened about 10 billion years ago, both in our own Galaxy (see Is there a stellar database that indicates how long ago stars in our Galaxy formed?) and in the universe as a whole (see Will new stars stop forming at some point of time? ) and most quasar activity petered out about 5 billion years ago (e.g. see  Why are all quasars so far away? ).

On the other hand, it is true that the number of (low-mass, very low luminosity) stars in the universe is still increasing (e.g. When will the number of stars be a maximum?), thanks to their very long lifetimes. So although the fireworks are over, the bonfire is still smouldering (to use a UK metaphor that will have others baffled!).

I guess it all depends on whether you are a glass half-full or half-empty kind of person.

Another thing to consider from a philosophical point of view is the Carter catastrophe. The basic idea here is that it is unlikely for us to be alive at a special time just after the beginning of the human race, and that this indicates we are shortly to be doomed. This is normally framed in terms of humans and our survival on this planet, but I suppose could also be applied to the universe as a whole. The arguments that the universe could last trillions of years are based on concordance cosmology with the parameters for dark matter and dark energy fixed. The problem is we don't know what dark energy is or whether it varies in time or exactly what the relationship is between dark energy density and the (negative) pressure it exerts. There are big rip scenarios that may see the end of the universe within a mere 100 billion years!


Could someone please explain to me how I can extract velocity component values from Stellarium, for a given planet?

To follow up on the helpful comments below your question, here's a program that will do what you need, but without using Stellarium.

If you really need to use Stellarium then hopefully someone will find a way to help you.

This uses the Python package Skyfield and you can see a lot of documentation on that site and elsewhere. Right now I've specified the DE421 ephemeris, but there are many to choose from. This one only works +/- 100 or 200 years I think, others go for thousands of years in the past or future.

If you would like to use NASA JPL's Horizons web interface online, here are some instructions. It uses some of the same Development Ephemerides that Skyfield downloads for you.

Here's some output for roughly 2019-10-02 07:18:59 UTC which is the moment when you posted your question.

                    relative to solar system barycenter
               x            y           z            vx        vy          vz
Sun        -453731.5    1039601.2    451057.1     -0.015      -0.002       0.0
Earth     1.4764e+08   2.1455e+07   9.3005e+06    -4.9362     26.9295     11.6751
Jupiter  -2.2857e+07  -7.2214e+08  -3.0898e+08    12.9012      0.3382     -0.1690
Mumbai    1.4763e+08   2.1454e+07   9.3025e+06    -4.8369     26.5009     11.6749

                      relative to Earth's geocenter

Sun      -1.4809e+08  -2.0416e+07  -8.8494e+06     4.9215    -26.9312    -11.6754
Earth     0.           0.           0.             0.          0.          0.
Jupiter  -1.7050e+08  -7.4360e+08  -3.1828e+08    17.8374    -26.5912    -11.844
Mumbai   -5873.8303   -1362.2932     2072.3226  9.9338e-02    -4.2861e-01 -1.9206e-04

                           relative to Mumbai

Sun      -1.4809e+08  -2.0414e+07  -8.8515e+06     4.8222    -26.5026    -11.6752
Earth       5873.83     1362.29     -2072.32      -9.9338e-02  4.2861e-01  1.9206e-04
Jupiter  -1.7049e+08  -7.4359e+08  -3.1828e+08    17.7381    -26.1626    -11.8439
Mumbai    0.           0.           0.             0.          0.          0.

Here's a Python 3 script for that output as an example

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite

load    = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data    = load('de421.bsp')
ts      = load.timescale()

sun     = data['sun']
earth   = data['earth']
jupiter = data['jupiter barycenter']
mumbai  = earth + Topos(latitude_degrees  = 18.98,
                        longitude_degrees = 72.83,
                        elevation_m       =  20.0)
time    = ts.utc(2019, 10, 2, 7, 18, 59)

things     = (sun, earth, jupiter, mumbai)
names      = ('Sun', 'Earth', 'Jupiter', 'Mumbai')
positions  = [thing.at(time).position.km for thing in things]
velocities = [thing.at(time).velocity.km_per_s for thing in things]

for name, position, velocity in zip(names, positions, velocities):
    print(name, position, velocity)

print("relative to Earth's Geocenter")
for name, position, velocity in zip(names, positions, velocities):
    print(name, position-positions[1], velocity-velocities[1])

print("relative to Mumbai")
for name, position, velocity in zip(names, positions, velocities):
    print(name, position-positions[3], velocity-velocities[3])


Seeing a line rather than a dot when looking through refracting telescope

First off I want to say I am a complete amateur when it comes to stars and even bigger when it comes to optics.

I was looking at sky from indoors and my windows have double glass. It turns out that this was the problem or maybe the glass on the window is not some good quality.

In any case, all works well now and I am thrilled at what I'm seeing in the night sky.


You have to remember that the alignment for an eclipse is a very rare situation. It's not as simple as the occurrence of a new moon. The planet and the moon have to align just so in relation to the sun for an eclipse to occur. Sorry, no numbers here but just trying to emphasize that several factors have to occur between 2 fast moving bodies.


No energy is in transit as GRBs, just like Gamma rays that started in a GRB.

If you remember that all that energy is in the form of radiation, it obeys the same laws as ordinary light: the farther, the dimmer (inverse of square of the distance law).

So if you see stars dimmer as they are farther from you, same happens to the RGBs. Radiation from the stars expands infinitely (becoming dimmer and dimmer) until absorbed by clouds or just becoming dimmer than the background.

Same happens to gamma rays from GRBs. It is just that since rays are more energetic, they are less easily absorbed, and since GRBs themselves are more energetic, the becoming dimmer than background distance is bigger.


How to calculate the heliocentric velocity of an object?

The heliocentric velocity $V_\mathrm{H}$ of an object is its velocity wrt. the Sun. When you measure an object's velocity, you measure it in the reference frame of Earth, which revolves around the Sun with ~30 km/s (varying a bit from aphelion to perihelion), so convert to $V_\mathrm{H}$ you need to know the time of the year of the observation (unless the line of sight toward your object is exactly perpendicular to the ecliptic plane), as well as the angle between the line of sight, and the line of sight toward the Sun. This involves a number of sines and cosines that you can find in e.g. Barbieri (2006).

If you further want to convert from $V_\mathrm{H}$ to the reference frame in which the Milky Way's center is at rest, the Local Group is at rest, or the Cosmic Microwave Background is isotropic (the "cosmic" frame), then you use the formula you link to, adding a similar term as described above, but instead using the velocity (i.e. speed and direction) of the Sun wrt. to the given frame.


How thick was Earth's primordial atmosphere?

We have no way of knowing how much hydrogen and helium was in the Earth's primordial atmosphere. What is more important is what the atmosphere consisted of when the primeval soup from which life arose was formed, just over 4 billion years ago. We have a fairly good idea what this atmosphere consisted of: mainly carbon dioxide and nitrogen, rather like Mars at the present time, but with substantial amounts of ammonia and methane. Also very small amounts of hydrogen and helium, most of which had by then been blown away by solar radiation. There was hardly any oxygen. As for pressure, all we can say for sure is that on Earth it was a good deal higher than it is today. Conditions at that time were similar, though not identical, on Venus, Earth and Mars, perhaps on Titan as well, so it is possible that all three terrestrial planets developed a version of the Earth's primeval soup, which was essential for the creation of life. Whether life emerged on any planet other than Earth, we don't yet know. Because all three planets had oceans, water vapour would also have been an important atmospheric constituent.


Why can't we determine the center of the universe

There is no point of origin! It is a misconception that the universe started as a single point within space. Indeed, measurements since the 1990s seem to show that the universe is infinite in size, which means that the very early universe must have also been infinite.

It has not really been established that the universe originated from a "singularity" at all. We know that in the early universe everything was much closer together, and hence the universe was much denser. The "Big Bang Theory" goes back to the universe being extremely dense, but it is not required that the universe started from a single point.

Really, the Big Bang happened everywhere. Even at these early times it looks like it was infinite in size with basically constant density, and so there's no sensible way for it to have a centre.


If the earth was not tilted, what would be the north star?

It turns out that 0, 0 on ecliptic coordinates is not the "winter" direction, nor the "summer" direction.  And it turns out to be a good thing.  Zero longitude is defined using the "spring" direction, and this means that converting these points to another coordinate system is really easy.  The earth's axis is tipped 23 degrees, but imagine the axis on which it is tipped.  Thats the spring/fall axis.  This means that if we tip the whole coordinate system 23 degrees, two points stay unaffected (the spring/fall points).  The tipped coordinate system is called equatorial coordinates and its more useful so everything uses it.  The point 0, 0 is the same in both coordinate systems, as is the antipode of 0, 0 which is 0, 180.  So you can get the spring and fall stars right away by browsing something like http://www.sky-map.org/

Note that the longitude coordinates are in hours (0 to 23) instead of degrees (0 to 360), and they are shown in the order longitude, then latitude.  Also note that stars you can see without a telescope have an apparent magnitude of 6 or less.

Zero longitude is the direction the sun is in when it is spring, so the 0, 0 star is actually the fall star.

Now the other four stars: because the difference between the two coordinate systems is a single rotation around the spring-fall axis, the winter/summer points actually keep their longitude the same during the rotation.  And because the amount of rotation is 23 degrees:

0, 90 becomes 23, 90
and 0, -90 becomes -23, -90

You can check this here: https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm Note that equatorial coordinates are called "celestial" on this page.  Note that you have to enter a year because the tip of the earth's axis slowly changes.  Just enter the current year in both epoch boxes.

It should make sense that the untipped north star would have an equatorial latitude of 90 minus 23 degrees, and the untipped south star would have an equatorial latitude of -90 plus 23 degrees.  Using the same tool to do the conversion:

90, 0 becomes 66, -90
-90, 0 becomes -66, 90

If you go back to sky map and poke around in those locations, you find:

The untipped north star is in Draco (18h 66d)
The untipped south star is in Dorado (6h -66d)
The fall star is in Pisces (0h 0d)
The spring star is in Virgo (12h 0d)
The winter/summer stars are in Gemini (6h -23d) and Sagittarius (18h 23d)

Note that the latter four stars are in zodiac constellations - this is because those constellations lie on the ecliptic.

For winter/summer which is which?
Well the sun comes up in the east, so earth rotates counter clockwise, so it travels around the sun counter clockwise.  We know that after the line from the sun to the earth points toward fall, it points toward winter.  So whichever is left of Pisces in the sky would be the winter one:

The winter star would be in Gemini
The summer star would be in Sagittarius

Note that the four zodiac constellations we got are evenly spaced throughout the year, and that the months associated with them are 6 months off from the seasons we mapped them to.  This is because the zodiac months are assigned based on where the sun is in the sky.  You never see a zodiac's constellation at night during the month associated with it.


How do stars change in size over their main sequence lifetime?

Very roughly.
The star is in hydrostatic equilibrium so
$$ \frac{dP}{dr} = -\rho g$$.
Replacing $dP/dr$ by $P_c/R$, where $P_c$ is the central pressure and $R$ the radius, and letting $\rho \propto M/R^3$ and $g \propto M/R^2$, then we get
$$ P_c\propto \frac{M^2}{R^4}$$
But the central pressure is proportional to $\rho T/\mu$, where $T$ is the interior temperature and $\mu$ is the mean molecular mass. So
$$ \frac{\rho T}{\mu} \propto \frac{M^2}{R^4}$$
$$ \frac{MT}{R^3 \mu} \propto \frac{M^2}{R^4}$$
$$ R \propto \frac{\mu M}{T}$$
Now during the main sequence lifetime, the mass stays (roughly) constant and $T$ stays (roughly) constant because the temperature sensitivity of hydrogen burning is high.
However, $\mu$ increases because 4 hydrogens (4 protons + 4 electrons) are getting turned into 1 helium (1 nucleus + 2 electrons). So $\mu$ increases (in the central regions) from 0.5 to 4/3. Hence the radius increases.
The radius increase is indeed about a factor of 2 over the whole main sequence lifetime.
A different way to think about it, more appropriate for solar mass stars and above, is in terms of the radiative energy transport. The increase in $\mu$ is also accompanied by a decrease in the number of free electrons per mass unit. This reduces the opacity in the central regions and decreases the temperature gradient. This means the star is bigger for roughly the same central temperature.


What is the physical, geometric shape of the universe?

Ok, maybe you have some misconceptions.

The Universe has no center at all. It looks the same from avaery point, to wherever you look to. It is approximatedly as follows:

This image represents a very large scale box on our universe on current time (not on perceived time, based on received light). Of course, it is just a computer simulation. Each dot represents a cluster of galaxies.

So you need to imagine an infinite tridimensional space filled with filament-like structures like these. And infinite means it has no bounds, so it has no "external" shape. No ball, rugby-ball nor cone there. Also not irregular external shape, just infinite. Any of these shapes have a 2D boundary on a 3D space, but the universe has no boundary.