Is the vernal equinox always in zenith somewhere on the equator?

The Vernal Equinox is defined by the point where the sun's path across the sky, the Ecliptic, crosses the Celestial Equator. It is always going to be on the Celestial Equator.

Given the various ways that objects in orbit over Earth are perturbed, any set of keplerian orbital parameters that describe the orbit of any object over the Earth (including the Moon!) will need updating long before the drift of the Vernal Equinox along the celestial equator becomes significant.


In the end what is the ultimate matter/element in the universe, due to fusion process in stars?

If we don't wait for too long, where we don't know for sure, whether protons or atomic nuclei stay stable forever, there won't be just one single element.

The most abundant element in the universe will probably stay hydrogen, since it will stay in the intergalactic medium, and thin out by accelerating cosmic expansion.

The second-most abundant element will probably stay helium, mostly for the same reason.

The most frequent stars are probably red dwarfs.
They'll burn hydrogen and helium to oxygen and carbon, and end up as white dwarfs.

Only a small fraction of stars is large enough to burn oxygen and carbon to silicon and iron.

Very heavy elements, like uranium and beyond, as formed in supernovae, will decay to lead, helium (alpha particles), some hydrogen (via neutrons decaying to hydrogen), and a couple of less abundant elements formed by spontaneous fission. Other stable (isotopes of) elements heavier than iron are also formed in, or a consequence of, supernova explosions.

Part of the dead stars will collide, and merge to heavier stars, which will end as intergalactic dust (mixture of elements, e.g. iron and silicon), including planets, after supernova explosions, and as Black Holes. Others will be scattered out of reach of the supermassive black hole in the center of the galaxy (gravitational relaxation).


You are missing the PC1_2 and PC2_1 elements of your pixel transformation matrix. If your coordinate transformation is a rotation, you would use

PC1_1 = cos(t)
PC1_2 = sin(t)
PC2_1 = -sin(t)
PC2_2 = cos(t)

for the PC matrix. The pixel scale (in degree/pixel, according to CUNITi) goes into CDELTi.

As pointed out, you can combine the PC and CDELT steps into a single CD matrix.

See here and here for more details.


The size of the radius of the event horizon of a black hole created by the merger of a black hole binary system

The "radius" (there is no physical surface) of the event horizon of a rotating black hole, depends on both it's mass $M$ and angular momentum $J$, and is given by the equation
$$ r = \frac{GM}{c^2} +\sqrt{\left(\frac{GM}{c^2}\right)^2- \left(\frac{J}{Mc}\right)^2}.$$

It is thus difficult (for me anyway) to give a straightforward answer to your question. When two black holes merge, they will each have their own masses and angular momenta, plus there will be angular momentum in the orbit. The gravitational waves emitted during the merger can take away mass from the system as a whole (e.g. the final mass of the first observed black hole merger was three solar masses less than the summed masses of the merging components).

So, in general yes, for non-spinning black holes the event horizon grows as the total mass. But if the merger results in a black hole with maximal spin, where $J = GM^2/c$, then the final event horizon could be half the size given by the Schwarzschild radius ($2GM/c^2$), even though the total mass is larger than the black holes that contributed to it.

However, I think what is certainly true, is that it is not possible to add mass (even in the form of another black hole) to a given black hole and for the event horizon to get smaller, whatever the angular momentum contributed by that mass (section 4.2 of "Black Holes", by Raine & Thomas, 2015, Imperial College Press).


In which direction does the ecliptic plane make an angle of 60 degrees with galactic plane?

The line defining where the ecliptic plane and the Galactic plane coincide lies (approximately) between the points defined by right ascensions  and declinations  of 6h +23 and 18h -23.

This can be seen below - the plot is in celestial coordinates with the ecliptic plane marked (thicker solid line) and the Galactic plane clearly visible in the background map of dust extinction.

The second of these coordinates is close to, but not coincident, with the Galactic centre (marked Sgr A*). So the relative tilt is sort of at right angles to the Galactic centre at present.

Of course this will change over tens of millions of years as the solar system orbits the Galaxy.

Edit: Another way of specifying the direction is just to ask where the ecliptic north pole is. It is at RA=18h, Dec=66d 34m, in the constellation of Draco.

A further way of thinking about it is to consider at which points in the sky do the ecliptic and Galactic planes have their maximum angular separation. These are 90 degrees away from the two points I defined above and are almost on the celestial equator at about 0h and 12h.


Derivation of the formula for longitude of ascending node for a satellite

I think you're not too far from understanding the formula. However, there are a couple details you're missing:

From what I understand from the document you linked to, $t_0 = t_{oe}$ is the time at the beginning of the week, or reference epoch.
When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t_0)$ or $\dot{\Omega}_E(t - t_0)$.
The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.
$w_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}_E t_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.

To summarize: the formula would have been more explicit and comprehensible written as:
$$
\Omega(t) = \Omega_0 - \dot{\Omega}_E t_{oe} + (\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})
$$
with:

$\Omega_0$ = LAN wrt the vernal equinox at $t_{oe}$
$\dot{\Omega_E}$ = angular velocity of the Earth wrt the vernal
equinox
$-\dot{\Omega}_Et_{oe}$ = Longitude of Greenwich at $t_{oe}$
$\dot{\Omega}$ = angular velocity of the satellite wrt the vernal
equinox
$(\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})$ = relative angular
displacement between the satellite and Greenwich, between time $t$ and
$t_{oe}$

That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t_0) + \dot{f}\times(t - t_0)$


How can I obtain raw observation data of stars surrounding Sgr A*?

Observations of stars near Sgr A* are done in the near-infrared, usually with adaptive optics. Most of these are done by two groups: Andrea Ghez's group at UCLA, using the Keck Telescopes in Hawaii, and Reinhard Genzel's group at MPE (Max-Planck-Institute for Extraterrestrial Physics), using the European Southern Observatory's Very Large Telescope (VLT) in Chile.

Both telescopes have public data archives, so in principle you can retrieve the raw data for these observations, as long as the observations are at least a year old (which most of them are). You need to know which are the relevant instruments; NACO has been used for most of the images at the VLT, and I think NIRC2 has been used at Keck. (There are also spectroscopic observations using both telescopes.)

So you could, for example, search using the NACO-specific search form in the ESO archive. I would suggest entering "Sgr A*" in the "Target name" field, restricting the "Search Box" field to something like 30 arc seconds (change the default "00 10 00" to "00 00 30"). Note that these observations involve lots of short exposures in different filters, and with no other restrictions on the search, you will get more than 10,000 hits.

(In order to actually retrieve the data, you will need to register as a user of the archive, which is free.)

Something similar might work for the Keck Observatory
Archive, though I've never really used that.

You'll need to be familiar with the reduction and analysis of near-IR imaging to get anything from the data, of course. ESO does provide "pipeline" software that can do some of the reduction (search on their site).

You can, in principle, apply to use the VLT yourself, but unless you really know what you're doing, it's very unlikely you'd get time in competition with, say, Genzel's group. Also, NACO has been taken off the VLT; its replacement instrument, ERIS, won't be available until sometime in 2020. (You can't generally apply for Keck time unless you're at one of its member institutions.)


Estimates of exoplanets distribution consistent with current data

As you've correctly mentioned observation biases play a huge role in current understanding of planetary populations. Let's have a look at a slightly outdated plot (age 2 months) that shows us the semimajor-axis vs mass distribution of currently known planets including KOI's (unconfirmed planets).

Here we see two things:

Hot Jupiters seem to make up a significant part of the confirmed planets (confirmed = RV + Transit)
The farther out we go, the less likely it is to have a measured planetary radius (black dots)

Now knowing how our methods and instrumentation work, we can try and debias this data.
Transit measurments pose the easiest ones to do this: as for transits the detection probability is a simple geometric function of planetary alignment, radius and stellar radius one can overplot the found planets with this probability. So this is usually of not very massive interest to the planetary community.
edit
The probability usually is given as $\sim r_{stellar}/2 a$, with $r_{stellar}$ being the stellar radius and $a$ the planet's semi-major axis. A simple derivation is here. Discussion how everything gets more complicated when combining with RV measurments here.
/edit

Any instrument one uses however, will have their own quirks and biases and starspots or eclipsing binaries can lead to false positives. Petigura+ 2013, fig. 1 try to asses those effects for Kepler transits and come up with the following detection probability map.

The corrected population then would be found by dividing $\rm\frac{found population}{completeness}$. Mathematical issues like dividing by zero are easily interpreted as issues stemming from the incomplete data. As where the detectability is zero, we can't know if there are 0 or 1 or 20 planets hidden in a bin.
When projected on one axis or the other those corrected data-plots then result in corrected Radius- or Period occurence rates.
I won't show those for the transit method, as I want to encourage to check the original reference.

I show one of those plots for the radial velocity method however. From Mayor+ 2011, fig. 12 that analyse their HARPS-spectrograph data from the preceeding decade:

In black: Found planets, red: debiased radial-velocity population
Note: A Jupiter-mass is roughly 320 Earth-masses, so in this plot the Hot Jupiters are on the far right. This now closes the thematic circle to my initial note about Hot Jupiters: While in black Jupiter-like planets seem very common, in the debiased version this is not the case anymore. Mayor et al. estimate the Hot Jupiters to make up only 0.5% of the total planetary population with this method.

Note also that although the debiased number of terrestrial planets increases significantly, their error bars explode in size. This stems simply from doing statistics with low numbers. Also for reasons of simplicity and clarity the error bars on the first two plots are not shown.

Concluding:
It is important to debias the data we have about exoplanets by method.
When doing this, we get rough ideas about how the true planetary populations look like.

Recent results and non-stop public announcments about "hundreds of newly found planets" can lead to a certain, totally understandable euphoria that we now know everything about planets.
This is however very far from true. The Number distribution in full mass-radius-semimajor axis space is still very, very thinly spread. So in the end we do low-number statistics when debiasing, which is scientifically simply unsatisfying.
Or said differently: There is still a lot of exciting work to be done!


What is the density profile within the Sun's photosphere? Which one of these is wrong?

I usually don't answer my own questions, but sometimes when the question itself is called into question I make an exception.
The density of the photosphere at $\tau_{5000}=1$ is predicted to be $3 \times 10^{-7} \text{g/cm}^3$ in the Holweger-Müller Model Atmosphere7.
As pointed out in comments there is a spread in values here.
The 1E-06 g/cm^3 density value (plot) is more consistent with  = 1 or "bottom" of the photosphere, while the density in the quote is more consistent with the cooler "top" of the photosphere (circa 4300 K).
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 ).


How does one calculate inclination from a graph of radial velocity?

From its RV graph alone, with no other information, you cannot calculate inclination.  This is why RV velocity measurements are typically reported as "$v \sin{i}$", because what you're actually measuring is the orbital velocity projected along the line of sight.  Without other information, you cannot disentangle the orbital velocity from the viewing angle.

What additional information could be had that would allow you to disentangle?  The following list is probably not exhaustive.

If you can observe the two stars as separate objects, then you can track their relative positions through an orbit and can measure inclination that way.
If you have "light curves" (brightness measurements over time) for the star(s) and you see periodic eclipses corresponding to the period and correct epoch from the radial velocity data, then you know the stars are aligned such that they periodically have their light blocked by one, the other, or both.  In such a case, you know that the inclination must be close to 90 degrees else the eclipses could not occur.
If you have reliable distance measurements to the stars and determine somehow that they are main sequence stars, you can determine the star's masses via theoretical stellar evolution models and calculations given a measurement of absolute luminosity (from a measurement of the apparent luminosity coupled with the distance measure).  Once you know the star's masses, you can calculate the expected orbital velocity, compare with your $v\sin{i}$ measure, and determine the inclination.
If the stars are not main sequence stars in the above scenario, then you'll need some additional information (their ages should be sufficient) to correctly extract a mass from their apparent luminosity and color.


Gas Planets composition and gravitational effects

The reason the hydrogen and helium (that make up most of the mass of planets like Jupiter) remain as gases and don't condense is essentially the same as any other gas. So I could ask "why don't the nitrogen and oxygen in the atmosphere of the Earth condense?"

The reason is that the temperature and pressure in the planet's atmospheres do not favour the formation of a solid. Hydrogen and Helium both have very low boiling points, and even under extreme pressure they tend to remain as a gas.

Further inside the planet the pressure does become sufficent to cause the hydrogen to condense into a liquid, or more properly a supercritical fluid (as at these high pressures the distiction between a gas and a liquid is blurred.  The temperatures are too high for solid hydrogen to form.

There are a couple of things I don't fully understand about your question, which may point to misconceptions. For example most of the mass of Jupiter is in the form of gas, and so most of the gravity comes from the gas, and not from the core (solid or otherwise).

There are layers in the atmosphere of Jupiter. The outer layers are actually rather similar to those in Earth's atmosphere. There is a turbulent troposphere and a stable stratosphere, with a warmer exosphere extending into space. Inside there are layers of supercritical hydrogen and helium and deep inside the pressures cause the hydrogen to behave like liquid metal.

Studying the gravity field (by careful observations of orbiters) is one of the principle ways that we can investigate the internal structure of Jupiter. This suggests that the heavier elements are not concentrated in the core, but mixed in the internal layers, suggesting that the planet has undergone some kind of extreme event in during its formation.


Does the observed color of the sun change based on the observer's location?

The Sun of course looks less yellow/red if viewed through less atmosphere; i.e. when it is overhead, which it never is in many countries (like the US and UK for instance).


If Kepler-444 planets existed for 11.2 billion years, why fear for life on Earth after six billion years?

The rate of evolution of main sequence stars is highly dependent on their mass. Roughly speaking, the time on the main sequence is proportional to $M^{-5/2}$, where $M$ is the mass of the star. Thus if Kepler 444 has $M=0.75\ M_{\odot}$, it can live for $0.75^{-5/2} = 2.05$ times as long as the Sun on the main sequence.

Another way of saying the same thing is that the Sun evolves 2.05 times as fast as Kepler 444. This is important because even whilst on the "stable" main sequence, burning hydrogen in their cores, stars are still changing because the average mass of a particle in their cores is increasing due to the build up of a Helium ash. This slow change means that the temperature of the core must gradually increase, the nuclear burning rate increases and the star becomes more luminous.

In terms of the Sun this means, depending on the atmospheric composition at the time, that the Earth could become much hotter in a billion years or so, and probably too hot for life (as it currently exists on Earth). Thus it is not the existence of Earth that is in jeopardy, it is the surface temperature that may destroy life.

None of the discovered planets in the Kepler 444 system are (or ever were) far enough from their parent star that they are in the "habitable zone" where liquid water could exist. The significance of Kepler 444 is that it shows that rocky planetary systems can have formed as long ago as 11 billion years, giving such systems more time, and therefore perhaps more chance, to have formed life.


How can the 13.8 billion years old universe have a radius of 46 billion light years?

You assumption is wrong. Universe can( and is ) expand faster than the speed of light.

The photon emitted towards our planet in the early universe, had to traverse a universe that is expanding. That photon experienced redshift, that means his wavelenght increased and frequency decreased. We measure that photon's wavelength, compare it to a photon that was not redshifted and calculate the scale, which is then applied to the radius of the observable universe.

Reading this Wikipedia page should give you more complete understanding of the problem.

http://en.wikipedia.org/wiki/Observable_universe


Can you identify this planet seen in the night sky?

This Community Wiki answer was created to preserve knowledge while other answer(s) are under review.

From Dr Chuck:

  It was Venus. It should have been 10 - 15 degrees above the moon.

From Wayfaring Stranger:

  Venus is whiter than Jupiter, Saturn, or Mars, which have yellowish, yellow, and reddish tints instead. It's also usually much brighter than the other planets, including Mercury. Mercury always hangs closer to the sun than did Venus in early Feb 2017.


It is 'From Nothing', made by cosmicspark, a.k.a Tim Barton.

  This picture was painted from nothing, in roughly 12 hours while I listened to a lot of interesting lectures on physics, including Dr. Lawrence Krauss' famous lecture "A Universe From Nothing".


Long term temperature effects of an eclipse

It was suggested that we could cool down Venus by putting a powder cloud in the Lagrange point between it and the Sun. This will cool Venus in a few months to the point where its atmosphere will start liquify, creating a global CO2 ocean. The surface of an ocean is much easier to colonize than creating flying cities.


Is the diameter of the observable universe a relative quantity?

If the observer has always been moving at a high speed, then yes, s/he will measure a different value than you. But if s/he accelerated to a high speed just now, you would obtain the same result.

The reason is that the radius of the observable Universe is defined as the distance light has had the time to travel since the Big Bang, and that does not depend on your instantaneous velocity.

However, if an observer started moving at, say, 0.866 times the speed of light when the Universe was born, then in the reference frame of that observer, the Universe would only be $\sqrt{1 - 0.866^2} = $ 1/2 the age measured by you, i.e. 6.9 Gyr. At this age, the radius of the observable Universe was only ~39 Gly, i.e. ~0.8 times the present value.

Note that in order for this result to be exact, the observer would have to slowly accelerate. The reason is that in an expanding Universe, a given "peculiar" velocity wrt. the surrounding matter will slowly decrease, because the comoving coordinate system expands with the Universe. For instance, an observer starting out with $v = 0.866c$ at the time the cosmic microwave background was released 380,000 yr after the Big Bang, would only be moving at 80 km/s today if s/he didn't keep accelerating.

If the observer were clever, s/he would probably realize that s/he did't live in an asymmetric universe where all matter seemed to be whirling by at close to the speed of light, but rather that the Universe were quite isotropic, but s/he had somehow acquired a high speed with repect to everything else. It would then be possible to deduce the "true" age of the Universe.


Initially non-flat space-time makes dark matter obsolete$\dots$

I think there a few problems with this idea.

First off, as Stan Liou indicates, the early universe expanded extremely rapidly, potentially through hyper-inflation during the inflationary epoch. While we're still looking for observational evidence that this period occurred (e.g., the failed BICEP2 results a few years ago), the existence of such an epoch early in the universe neatly explains away several problems with the standard Big Bang theory (see the horizon problem, monopole problem, and the flatness problem). Assuming such a period existed, then any region of space-time that was "crumpled" would have been quickly and cleanly smoothed out during the hyper-inflation. Such crumpled-ness would not have been around long enough to affect cluster growth and structure. What's more, we very clearly measure that the Universe is flat over large, cosmic scales so it seems unlikely that there are constant and pervasive wrinkles as you suggest there could be.

A second issue I potentially see with your proposal is that it would imply a top-down approach to galaxy formation. That is, first you get huge clusters of mass (conglomerated by your "crumpled" space-time), then you get fragmentation and individual galaxies can form, then galaxies fragment into individual stars, etc. In other words, big structures form first, then smaller structures derive from the bigger structures. The opposite formation model would be "bottom-up" formation where first small things clump, then those clump into larger bodies and soon you have galaxies clumping into clusters. I don't think there is definitive evidence that galaxy formation is either top-down or bottom-up, but simulations seem to me to suggest bottom-up is more likely (of course those simulations use dark matter). I'll leave you to think about that one.

The last point of contention I have is that dark matter plays more of a role than just providing extra gravity. It contributes to the total mass-energy of the universe. Current theoretical models perform extremely well when we include dark matter. I believe a strong piece of evidence for this fact is our ability to theoretically match the CMB power spectrum using a Lambda-CDM model. I don't see that crumpled space-time could reproduce this result nearly as well.

I think you'd also have a very hard time describing a mechanism that could allow "crumpling" to exist in the space-time of our universe which exactly matches the observed gravitational effects in galaxies. We can easily see that they appear to have a halo of matter which follows something akin to an NFW halo or else an Einasto profile. If you can find away that crumpled space-time would naturally form to match these profiles and somehow not be smoothed out by hyper-inflation, I'm all ears.


I found a text that suggests a Population III star can have a mass between 200 and 10^5$M_{Sun}$. There you can read:

  "...This would apply for VMOs larger than $M_c$ $\approx$ 200 $M_{Sun}$. Stars larger than 150$M_{Sun}$ are termed supermassive objects (SMOs)...".

It seems supermassive stars like R136a1 (with masses in excess 150$M_{Sun}$, the accepted upper limit for a non-pop III star) can be formed by the collision of massive stars.

So, It is not difficult to imagine, say, two 150$M_{Sun}$ monsters colliding and forming a 300$M_{Sun}$ object.


Is the American Astronomical Society 230 years old?

The first meeting of the AAS was held in September 1899 in Williams Bay, Wisconsin. The meetings were held once a year initially. The frequency varied somewhat for the next 60 or so years. The meetings appear to have settled down to twice a year about 40 years ago, one in January and another in late spring.

Source: https://aas.org/meetings/past-aas-meetings


Can there be an infinity of humans in the Universe?

IMO, this sounds more like a philosophical question rather than an astronomical question, but since I like both subjects, I'll give it a shot.

  Note: This is hypothetical question, so this answer is purely subjective (and hypothetical).

Infinity

If the number of stars is infinite, so would the numbers of humans.

I believe that in every system where an entity of infinity is introduced, all other entities will (eventually) be infinite. But, (and this is the fun "part") you cannot measure infinity. It would give you infinite results over a time span of infinity.

Take your example for instance. If there were infinite numbers of stars. Picture youself flying around in a spaceship with the goal to find out if this question is true or not. What do you see? "Worse case" scenario; you see only stars. Wherever you go. Forever. The opposite;  only humans.

One might think that it's the "in between" that is the most likely scenario. But given the probability theory, we should flatten out at approx. 50%.

Finite

If there is a finite numbers of stars then I think it's "safe" to conclude that there ain't an infinite number of humans as we're all made of starstuff.

Conclusion

  Yes: If there's an infinite numbers of stars.

  No: If there's a finite numbers of stars.


Why is this HST WFPC2 image so much fuzzier than the WFC3 image of the same field?

As Rob Jeffries suggested, in the WFPC2 observations the progenitor star fell located on one of the wide-field ("WF") chips of WFPC2. (I verified this by downloading one of the preview images from the Mukulski Archive.) This was purely by chance, since the WFPC2 observations were taken years before the discovery of the failed supernova. It's good luck that the region of the galaxy where the failed SN took place was imaged by WFPC2, and only slightly bad luck that it wasn't in a region imaged by the planetary-camera ("PC") chip.

WFPC2 is/was an array of four equal-sized CCD chips, each with 800 x 800 15-micron-wide pixels. However, the optics were set up so that 3 of the chips (the WF chips) were imaged using a focal ratio of F/12.9, while the fourth (the PC chip) was imaged separately with a focal ratio of F/28.3. In essence, a magnified optical beam was focused on the PC chip, giving a scale of about 3.0 milli-arcseconds per micron, or 0.0455 arc seconds per pixel. Meanwhile, the WF chips received a scale of 6.67 milli-arcseconds per microns, or 0.097 arc seconds per pixel. This undersamples the actual resolution of the telescope (which has a point-spread function with a full-width half-maximum of about 0.09 arc seconds in the I-band).

The UVIS (UV-visual) part of WFC3 has two 2k x 4k CCDs (with, I think, 15 x 15 micron pixels) and a focal ratio of F/31, which works out to 0.0395 arc seconds per pixel.

So in practice, WFC3 will always resolve things better than the WF chips of WFPC2 can, because its pixels can properly sample the telescope resolution and the WF pixels cannot.

You can see the difference in the upper "progenitor" (WFPC2) images, compared with the lower "2015" (WFC3/UVIS) images from the actual paper:

Note the stronger visible pixelation in the WFPC2 images (top two panels), due to the fact that the WF pixels are larger in angular size.

I suspect any other apparent differences in the press-release images are probably due to the processing used to create the color image. In particular, the side-by-side (WFPC2 vs WFC3/UVIS = "before and after") color image from the press release shows both images at the same spatial scale, as part of a single image file, which means that the same pixel density is present throughout the combined image. But since each original WF pixel is about 2.5 times the (angular) size of a WFC3 pixel, the WFPC2 image had to be rebinned to match WFC3 pixel scale (1 WF pixel = 6.25 WFC3 pixels) before the combined (JPEG/PNG/TIFF) color image could be made. The interpolation used for this rebinning process may be what introduced the extra smoothing in the WFPC2 image.

[Edited to add comment about rebinning used to generate color press-release image.]


How to calculate planetary position over vast time spans

Jacques Laskar, at IMCCE, has been working on describing planetary motions on long term time scales.

Quoting and translating a paper by him:

Motion in the solar system is chaotic, meaning that trajectories diverge exponentially: an error on the location of a planet gets multiplied by 10 every 10 million years
This means that a 15 meter error on the initial location of the Earth yields an error of 150 meters after 10 million years, and 150 million km afer 100 million years. Long term predictions involving a single trajectory are impossible.

For predictions over longer periods, it becomes necessary to study multiple possible trajectories. Laskar and Gastineau have published results of a simulation of the solar system involving 2501 variations of the location of Mercury. In this simulation, "initial conditions  differ by only 0.38 mm in the semi-major axis of Mercury". Some of those trajectories involve "possible collisions of Mercury, Mars or Venus with the Earth".


What is the size of the universe.is it infinite?

"No one knows if the universe is infinitely large, or even if ours is the only universe that exists." - NASA

Still, the "observable universe" is a sphere roughly 92 billion light-years in diameter.


Information speed and our view of the cosmos

The effect is small, but not negligible. It is not accounted for in astronomical catalogues.

Let's work it out. We can start with the visible stars. Most of these are closer than 1000 light years; let's say 100 light years. The typical velocity of these stars with respect to the Sun is tens of km per second; let's take an extreme example of 100 km/s in a direction at right angles to the star-Sun line.

100 years at 100 km/s is 300 billion km. This sounds huge, but it is a tiny fraction of the size of the Galaxy and small compared with the distances between stars; but by how much does it change the apparent position in the sky? The change in angle (in radians) is given by the tangential distance moved in 100 years, divided by the distance to the star. This is the ratio of the speed of the star to the speed of light and is actually independent of distance.

For the example given, the change in position is $3.3\times 10^{-4}$ radians or 69 arcseconds. This is an enormous change in astronomical terms (not an angle you could easily resolve with the naked eye), but it is not accounted for in astronomical catalogues, since we are interested in where to look at stars now and neither do we know the accurate tangential velocity of everything in the sky.

What is accounted for in telescope pointing software is relativistic aberration, which is a change in the apparent position of a star due to changes in the relative motion as the Earth orbits the Sun. These corrections can be as large as 22 arcseconds.

As for galaxies - you have your dimension estimates wrong. A large galaxy is 100 thousand light years across, with stellar speeds limited to a few hundred km/s. I don't understand what you mean by "warping". Individual stars will have moved whilst the light travels towards us, others will have taken their place, stars will have died and others born, whole galaxies will have disintegrated or merged. All this is known. We can only observe what we can see now.

A dynamical timescale is typically tens of millions of years in a Galaxy and far longer than it takes light to travel across the galaxy. To all intents and purposes, when we look at a distant galaxy we are seeing everything at roughly the same epoch in that galaxy's history.


If we found evidence of life on Mars, how would we know that it originated on Mars rather than Earth?

Why would we assume that the early martian life originated on Mars,
  rather than Earth?

There's still a whole lot we don't know.  As Wayfaring stranger points out in the comments, Origin is a whole different question.  It's possible that life originated outside our solar system and came to either Mars and/or Earth from outside the solar-system.   I don't think anyone who studies this idea is "Assuming" life originated on Mars, only that the idea has a chance of being true.

  Is it not likely that somewhere between 2.5-3 billion years ago, a
  major volcanic eruption or meteor impact could have hit an earth
  teeming with simple-life and sent it hurdling through space to Mars?

As I understand it, volcanic eruptions are unlikely to send anything into space unless it's a smaller sized moon.   The escape velocity (young earth, maybe 10 km/s, young mars, maybe 3-4), volcanic eruptions, as far as I know, don't shoot things out at 10,000 - 20,000 MPH.   But meteor impacts of sufficient size can do that.

Mars is a better meteor debris making target than the Earth cause it's smaller, so the gravity is lower and presumably it's mostly had a thinner atmosphere too.  We've found martial meteors on Earth.   We might not find any Earth meteors on mars cause it takes a much bigger impact to knock bits of rock off Earth and because the Atmosphere slows objects down, both coming in and going out.

  I understand that my theory has the gravity of the sun working against
  it, but I also see it not being entirely impossible.

The sun isn't as big a factor as you might think.  Once something is knocked off a planet and it gets into orbit around the sun, gravitational assists can move it around further out or further inside the solar system.  What presumably happens is that, with a big enough impact, many thousands if not millions of bits of debris get into solar-system orbit and from there, some of them land on other planets - probably much less than 1% of those hit Earth, but if it carries life that can survive the trip, all you need is one rock.

  I mean, 100 years ago, one could assume that life originating on Mars
  and then being sent this was was ridiculous and impossible but now
  it's a rational, viable theory.

While that's true, the "People used to think this was impossible" isn't a scientific approach for what might be true.   We should determine what is possible and/or thought to be likely, based on physical evidence, not what wasn't understood 100 years ago.   Your example is a good point on why it's important to keep an open mind about the unknown.   You can still make theories based on evidence, and keep an open mind on the unknown.  There's really no conflict between the two.

  Considering such a scenario, how would scientist verify that these
  life-forms didn't originate from Earth, rather than Mars? It seems to
  me that this question is the first question that would need to be
  asked and answered after the fossil-records were found and verified.

It's a good question.

The simple answer is that Mars cooled first and Mars (likely) had oceans first, so it's a better candidate to have developed life first though extremophiles can live in hot oceans, so . . . time will tell.


I would bet against lightning due to the very thin SO2 atmosphere; under normal conditions the pressure is  0.3 to 3 nbar, with plumes reaching 5 to 40 nbar. That does not sound like it could convey lightening. But given that the environment also has a lot of ionization there might also be all sorts of plasma and vacuum breakdown discharges. In short, I suspect this could go either way, but if there were lightning it would be rather different than on Earth since it would be a near-vacuum discharge.
As far as I have been able to find, there are no empirical detections of lightning at Io. I suspect the radio measurements that could detect it would be swamped by the overall noise of the Jupiter-Io environment.


How exactly can the Hubble Space Telescope see up to 15 billion light years?

Hubble is a telescope.  This means it has a big mirror to gather light and focus it onto a small area, and a system of lenses and mirrors to magnify the image formed by the big mirror.
Now,  how far can you see? 100 miles? 1000?  You'd be way out!  The stars are already trillions of miles away.  Without a telescope you can see the Andromeda galaxy, which is 14 billion billion miles away.
You can see it because it is intrinsically very bright. Because it is so bright you can see it from far far away. If you had bigger eyes, you would be able to see things that are further away and dimmer.  And the Hubble telescope has a big "eye"
Hubble has one more trick: It does have a "digital camera" and it can use this to make a long exposure. This means that the light from a long period of time is allowed to accumulate on the light-sensitive chip. This means that even dimmer objects can be observed.
By a combination of a big mirror, and long exposures, Hubble is able to get a very deep view of very dim and distant objects.


Do the magnetic fields of stars and/or other planets reverse?

Sure. It even happens with the Sun, as part of the solar cycle.

Earth's magnetic field is described by something called the dynamo theory, which states that the movement of fluids in the core generates the magnetic field. A modification of this is used to model the magnetic fields of stars. Given that the cause of geomagnetic reversals is related to the movement of currents deep on Earth's core, it makes sense that the same would be the case in other bodies.


How do you carry errors for brightness magnitudes

You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function $f(x)$ that depends on variable $x$, then the uncertainty $\sigma_x$ on the quantity $x$ is related to the uncertainty on $f$ by
$$
\sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2}
$$
Here $f$ is your relative magnitude $m_1 - m_2$, and $x$ is your flux ratio.  So taking that partial derivative will give you the factor that relates the two uncertainties.


Why doesn't the earth's surface collapse onto itself?

For the same (an more prosaic) reason milk skin floats over boiling milk - because it is lighter than the layer below. During the formation of the planet it went through a process called Iron Catastrophe, were a large parcel of the denser materials like iron and nickel sunk to the core, leaving the lighter materials close to the surface.

Nice graph with relative densities follow:

Fission contributes to keep our internal engine super-hot and well-oiled, allowing for our metallic core to spin and generate our magnetosphere. Eventually we'll run out of heat, the outer core will solidify and stop spinning.


What error bar / confidence level is generally reported in galaxy distances?

The error bars are typically very large.
There are several ways of determining the distance of galaxy:
Cepheid Variables
Cepheid variable stars have a known relationship between luminosity and period.  If we can observe these variables as individual stars in a galaxy, we can determine distance.
But individual stars can only be observed in the nearest galaxies. This method can't determine the distance of galaxies in which we can't resolve individual stars.
There is about +-7% uncertainty, plus possible systematic errors (wikipedia gives a distance modulus uncertainty of 0.16 for an individual galaxy, and states that the uncertainty in distance can be calculated as 0.461 x distance modulus uncertainty = 0.07)
Type 1a supernovae
These have a known and fixed luminosity (of about magnitude -19.3) If we can observe a supernova of the right type in a galaxy, we can tell how far it is.
But to use this method we have to wait for a supernova, and even in the larger galaxies they are rare.  Also the "known luminosity" has to be calibrated against Cepheid variables, so there might be systematic errors.
A well-observed supernova can fix the distance with uncertainty of about 5%, however there may be systematic errors.
Association with other galaxies
If two galaxies are interacting then we know that they must be about the same distance,  if we can measure one, (by supernova) we get an estimate of the distace of the other.
Redshift
Distant galaxies are receeding from us and E. Hubble observed that there is a correlation between the redshift caused by the galaxy moveing away from us, and the distance.  So for more distant objects we can use the red-shift as a proxy for distance.
However the peculiar motion of the galaxy also causes red (or blue) shift and this is not know. The calcualation of distance depends on the observed constant of proportionality, and while this is about 70 the exact value is not known (and probably not constant)  This method can't be used for local galaxies.
Other methods
There are other methods but they are less reliable, See Extragalactic_distance_scale These methods may have uncertainties five times greater than measurements of supernovae.
Systematic errors are due to uncertainties in the models being used.  For example the nearby cepheid variables tend to be from stars with lots of carbon, nitrogen and other elements (besides H and He). These have a different luminosity from stars formed with little except H and He.  Cepheid variables were used to measure the size of the milky way, assuming that the stars being measured were like the nearby stars.  They were not.  Correcting this systematic error resulted in distances that were double what they had been previously.  The calculated size of the milky way doubled!
So if someone says This galaxy is 50 million light years away, you can be reasonably confident that it is somewhere between 30 and 80 million light years away.  And perhaps more confident if there has been a recent supernova there.


Are ascending node of sun, the point of intersection of prime meridian and equator and center of earth all collinear?

Not exactly.

During either of the equinoxes there is a moment when the line between center of earth and the sun aligns with the equator. This doesn't coincide with prime meridian in any way though; it may happen at any meridian whatsoever that happens to coincide with the line.

Of course if it happens the sun is in zenith above prime meridian at that moment, your conditions will be satisfied. But depending on precision you allow, that may be a very short time window. It's all about how precisely you want that to be.

Near Equinox Earth tilts by about 0.25 degree of latitude per day, meaning that is the maximum angle by which the line will be off from equator while crossing the prime meridian on the same day as when Equinox happens. 0.25 degree of latitude is about 27 kilometers, so this is the maximum error. Of course on some years, that will be much less, it's just that Earth tilt is correlated with length of year, zenith line crossing of prime meridian is correlated with time of day, and time of day is not really correlated with length of year.


The farthest Orbital Point from Earth is regulated by the fact that at what point the gravitational pull of the Sun overcomes the gravitational pull of the Earth.

I found a post for this on Quora

This was the most up voted answer by user Paul Olaru on Quora:-

  Approximately 1.5 million kilometers. It is limited by the point at which the influence of the sun's gravity becomes sufficiently stronger than Earth's that any object at that distance would become unbound from Earth & end up in independent orbit around then sun.

  That limit applies in general to any system of two bodies, and is called the Hill sphere. It's not actually perfectly spherical, but close enough for most practical purposes, and lies between the L1 & L2 Lagrange points. For circular orbits, the radius of the Hill sphere for any planet or moon can be calculated from

  r = a*(m/(3M))^(1/3)

  Where r is the radius of the Hill sphere, m is the mass of the planet or moon, M is the mass of the sun a planet orbits, or the planet that a moon orbits, and a is the distance from the sun (or planet) to its planet (or moon).


Do white dwarfs lose mass as they fade to black dwarfs? Is there a correlation between temperature, mass, and radius?

The mass of a white dwarf continues to be that which it was "born" with. It will not change significantly unless it accreted material from a companion.

The radius of a white dwarf is, to first order, given by the "mass-radius relationship" and this relationship does not involve the temperature.

The mass-radius relationship is appropriate for a "cold" star. "Cold" in this context mean that the pressure that supports the white dwarf is only dependent on density, which is the case for the degenerate electrons in the interior, which have kinetic energies much greater than their thermal energy.

However, degenerate electrons also have excellent thermal conductivity, so white dwarf interiors are isothermal. Yet, they have a density gradient - denser in the middle and less dense answer move outwards. At some point close to the surface the electrons are hot enough to stop being degenerate and the gas pressure becomes temperature sensitive.

What this means is that the radius of a hot white dwarf is bigger than that of a cold white dwarf of the same mass. The effect depends on both the mass of the white dwarf (bigger for lower mass white dwarfs with lower surface gravities) and its age (since white dwarfs cool as they get older).

This effect is not negligible and needs to be properly modelled in order to understand the luminosities of white dwarfs.

The plot below is from Parsons et al. (2017) and shows (as points) measurements of masses and radii of white dwarfs in eclipsing binary systems. The lines are model curves for white dwarfs with surface temperatures ranging from zero (dashed line) to 60,000 K (appropriate for a very young white dwarf) in steps of 10,000 K. Clearly, there is not a unique mass radius relationship and the radii do apparently depend on temperature (and core composition), as the model curves suggest.


Do unattached objects contribute to a planet's gravity?

Yes, it would change the attractive force by 10^-14 if there was a world without people.

Depending on the distance, you can measure gravity of people and buildings, in fact they used gravity to measure gaps in the pyramids of Giza once.

Today gravity surveys can detect caves and reservoirs of 2m to 50m at 2m to 150m depth, so in the future, we will be able to discover many caves and fossils with nanotechnology gravimeters.
 http://physicsworld.com/cws/article/news/2016/mar/30/new-gravimeter-on-a-chip-is-tiny-yet-extremely-sensitive


How to calculate parallactic angle from a fixed alt-az position?

I found an equation on this website,

$\require{cancel} \cancel{\sin{\theta_p}=\sin{\theta_{az}} \cos{\theta_{lat}}/ \cos{\delta}}$,

with $\theta_p$ being the parllactic angle, $\theta_{az}$ being the azimuth, $\theta_{lat}$ being the latitude of the observing stations, and $\delta$ being the declination.  The website cites this equation as being from "Spherical astronomy, Small pg. 49."  I am unable to find such a book (based on my searches, I'm wondering if the correct reference might actually be "Textbook on Spherical Astronomy, Smart", which I do not have ready access to), but in my experience so far this equation has been correct.

The above equation, my original answer, seems to provide something similar to the parallactic angle (something like the complement in each quadrant, with an extra negative sign, but I haven't investigated in detail).  However, when testing against values gotten from astropy, it did not match the astropy values.

Instead, I found the following, taken from Astronomical Algorithms by Jean Meeus, which does match the astropy values:

$\theta_p = \text{atan2}( \sin{\theta_H}, \tan{\theta_{lat}}\cos{\delta} - \sin{\delta}\cos{\theta_H}),$

with $\theta_H$ being the hour angle, calculated via

$\theta_H = \text{atan2}( \sin{\theta_{az,mod}}, \cos{\theta_{az,mod}} \sin{\theta_{lat}} + \tan{\theta_{el}}  \cos{ \theta_{lat} } ),$

Where $\theta_{el}$ is the elevation and $\theta_{az,mod}$ is the azimuth, but defined differently from $\theta_{az}$ as used originally and elsewhere in my answer.  The zero-point $\theta_{az}$ is due north, increasing clockwise, and the zero-point $\theta_{az,mod}$ is due south, increasing clockwise.

Declination can be gotten from latitude, elevation, and azimuth:

$ \sin{\delta} = \sin{\theta_{el}} \sin{ \theta_{lat} } + \cos{ \theta_{el} } \cos{ \theta_{lat} }  \cos{\theta_{az}} $,

Thus indeed, given an alt-az (or az-el depending on your preferred terminology) position and the latitude of the observing station, it is unnecessary to use RA or an independently determined hour angle in a calculation of parallactic angle. I say "independently determined hour angle" meaning that although hour angle is used in the calculation, it can be determined entirely from az/el and latitude, so one does not need to provide an hour angle for this calculation to work.


Is there a ceiling for stable L4 or L5 masses?

The Giant Impact Hypothesis, a theory of how the moon was created, says that once it exceeds 10% of the mass of your 'J', the orbit of an L4 or L5 (your 'T') destabilizes.

Possible origin of Theia

  In 2004, Princeton University mathematician Edward Belbruno and astrophysicist J. Richard Gott III proposed that Theia coalesced at the L4 or L5 Lagrangian point relative to Earth (in about the same orbit and about 60° ahead or behind), similar to a trojan asteroid. Two-dimensional computer models suggest that the stability of Theia's proposed trojan orbit would have been affected when its growing mass exceeded a threshold of approximately 10% of the Earth's mass (the mass of Mars). In this scenario, gravitational perturbations by planetesimals caused Theia to depart from its stable Lagrangian location, and subsequent interactions with proto-Earth led to a collision between the two bodies.


Converting orbital elements from equatorial reference frame to ecliptic

The orbital elements $\omega$, $i$ and $\Omega$ are Euler angles in the sequence $(3, 1, 3)$.
The easiest way to transform them is to convert them to a representation that's easier to manipulate, e.g. unit quaternions or a matrix, apply the required transformation then convert back to Euler angles.
A useful guide to converting between the systems is given by James Diebel (2006) "Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors".
For example, using the formulae in §8.10, the rotation matrix for the orbital elements is given by:
$$R_{313}(\Omega,i,\omega)=R_3(\Omega) R_1(i) R_3(\omega) =
\begin{bmatrix}
c_\Omega c_\omega - s_\Omega c_i s_\omega & c_\Omega s_\omega + s_\Omega c_i c_\omega & s_\Omega s_i \\
-s_\Omega c_\omega - c_\Omega c_i s_\omega & -s_\Omega s_\omega + c_\Omega c_i c_\omega & c_\Omega s_i \\
s_i s_\omega & -s_i c_\omega & c_i
\end{bmatrix}$$
Where $s_\omega = \sin \omega$ and $c_\omega = \cos \omega$, etc.
You can then apply the transformation from equatorial to ecliptic coordinates (or whatever other transformation you want to do) to this matrix. The equatorial and ecliptic coordinates are related by a rotation about the x-axis by $\epsilon$, the obliquity of the ecliptic, so combine the rotation matrix using matrix multiplication. Depending on the precise details of the coordinate systems of the input elements and the ones your software expects, you may also need to invert the axes, which is another matrix multiplication.
This yields a new matrix $R'$ with elements $r_{ij}'$. You then convert this back to Euler angles:
$$
\begin{bmatrix}
\Omega' \\ i' \\ \omega'
\end{bmatrix}
=
\begin{bmatrix}
\operatorname{arctan2}(r_{13}', r_{23}') \\
\arccos(r_{33}') \\
\operatorname{arctan2}(r_{31}', -r_{32}')
\end{bmatrix}
$$
If $i' \in \{0, \pi\}$ then $r_{13}' = r_{23}' = r_{31}' = r_{32}' = 0$ and the above formula for $\Omega'$ and $\omega'$ no longer works as it requires evaluating $\operatorname{arctan2}(0, 0)$. In this case, the angles $\Omega'$ and $\omega'$ are in the same plane and therefore cannot be uniquely separated. Choosing $\omega' = 0$, the transformation from the rotation matrix to Euler angles becomes:
$$
\begin{bmatrix}
\Omega' \\ i' \\ \omega'
\end{bmatrix}
=
\begin{bmatrix}
\operatorname{arctan2}(-r_{21}', r_{11}') \\
\arccos(r_{33}') \\
0
\end{bmatrix}
$$

Note that the two-argument arctangent function in the above uses the convention $\operatorname{arctan2}(y, x)$ giving the angle between the positive x-axis and the point $(x, y)$. Some systems such as Microsoft Excel, which provides the function as ATAN2(x, y) have the arguments reversed, consult the manual for the software you are using.
Depending on whether you pre- or post-multiply the rotation matrix with a vector, the rotation matrix elements may be transposed from the ones given above, in which case you need to switch the element indices in the reverse transformation.


The likelihood of a reaction occuring at a given energy is a product of the number of particles with that energy (the Maxwell Boltzmann distribution), which decreases with energy, and the tunneling probability, which increases with energy. This product forms the Gamow window.
The total reaction rate (for a non-resonant reaction) is proportional to the area under the Gamow window - i.e. the product of its width and height.
For resonant reactions, that occur over a narrow energy range, all that really matters is how close to the peak of the Gamow window that energy is.


The book by Dvorak et al. 2006 "Chaos and Stability in Planetary Systems" is a good place to start (as with many titles by Dvorak), but reviewing and studying recent(ish) published work about the topic can greatly enhance understanding of the topic.  For example:

Malhotra et al. 2011, Chaos and stability of the solar system, PNAS (full text online) - this article explores the notion that there may be a significant element of chaos in the planetary orbits. A small snippet from the abstract:

  Over the last two decades, there has come about a recognition that chaotic dynamics is pervasive in the solar system. We now understand that the orbits of small members of the solar system—asteroids, comets, and interplanetary dust—are chaotic and undergo large changes on geological time scales. Are the major planets' orbits also chaotic? The answer is not straightforward, and the subtleties have prompted new questions.

In a similar vein, the article: Barnes and Quinn, 2004, The (In)stability of Planetary Systems, The Astrophysical Journal (full PDF online),

  present results of numerical simulations that examine the dynamical stability of known planetary systems, a
  star with two or more planets.


Would a spaceship have to evade interstellar clouds?

Yes.

As has been commented, the amount of damage taken by an interstellar spaceship depends on its velocity $v$, as well as the number of gas and dust particles that it encounters on its way. This number is usually measured per area, in which case it's called column density $N$, and is equal to the total distance $d$ traveled times the particle density $n$, i.e. $N = nd$. For instance, if a spaceship travels 1 lightyear ($10^{18}\,\mathrm{cm}$) through a region with a density of $10\,\mathrm{cm}^{-3}\!$, each square centimeter of the spaceship will encounter $10^{19}$ particles.

That is, the faster you go, the farther you go, and the more dense regions you go through, the more your spaceship is damaged.

The Breakthrough Starshot project aims to reach our nearest-neighbor stellar system $\alpha$ Centauri in ~20 years, with a gram-sized satellite reaching $0.20c$ by means of a light sail. Today, there was a paper by Hoang et al. calculating the amount of damage taken by such a satellite. The total column density of gas from Earth to $\alpha$ Cen is $\sim10^{17.5\mathrm{-}18}\mathrm{cm}^{-2}$, and assuming (fairly) a dust-to-gas ratio of 1% and a carbonaceous/silicate dust grain population with a Weingartner & Draine (2001) size distribution, they calculate that this journey to $\alpha$ Cen will erode the surface of the spacecraft to a thickness of the order of 1 mm.

Most of the damage is caused by dust, not gas, but in principle gas may slowly heat up the spaceship. However, at $v=0.2c$, as long as the density is $\lesssim10\,\mathrm{cm}^{-3}$, the temperature is insufficient to cause any melting.

Molecular clouds — the dense clouds where stars are born — have densities from $10^2\,\mathrm{cm}^{-3}$ and even up to $10^6\,\mathrm{cm}^{-3}$, i.e. many orders of magnitude higher than the roughly $1\,\mathrm{cm}^{-3}$ found in the more dilute interstellar medium. To reach even more distant stars in a tolerable time, you would have to go faster than $0.2c$, and thus it seems that it is in fact a good idea to evade these clouds.


Can Fermi Bubbles be produced in non-spiral galaxies?

The discovery paper (Su et al. 2010) mentions that bipolar shells on the order of $0.5\text{ kpc}$ were found on both sides of the nucleus of Centaurus A (Quillen et al. 2006). Centaurus A is a galaxy of contested morphology, likely either elliptical or lenticular - certainly not a spiral. This would seem to be a strong indicator that similar bubbles can indeed be found in non-spiral galaxies.

Now, the nuclear shells Quillen et al. detected were seen at mid-infrared wavelengths (5.8, 8.0 and 24 $\mu\text{m}$), rather than via the gamma rays with which Su et al. discovered the Fermi bubbles. However, the authors believe they may have been caused by a burst of star formation a few million years ago, driven by winds. This is one of two main scenarios considered by Su et al. for the formation of the Milky Way's bubbles (the other being AGN activity), meaning that the same processes could indeed have been at work in both galaxies. I don't know if corresponding gamma-ray emission from the nuclear bubbles in Centaurus A has since been detected, but it certainly seems plausible.

I find it a little strange that Pshirkov et al. were so careful to focus on spiral galaxies. It's clear that starburst galaxies and the host galaxies of AGN can obviously have non-spiral morphologies. Perhaps the group was just interested in searching for other nuclear bubbles on galaxies similar to the Milky Way which might have similar conditions and evolutionary histories.


No.

Lunar eclipses are caused when the Moon is in opposition to the Sun. Normally this produces a full moon, but if the Moon is in exact opposition (considering incline of the Moon's orbital plane), all direct sunlight will be blocked from the Moon:

So if all the sunlight is blocked, how can we see the Moon? Well, the main reason is that sunlight is often dispersed in the atmosphere, and so it can reach the Moon. In addition, there's  airglow, which is when sunlight hits Earth's upper atmosphere and causes multiple chemical reactions, scattering light throughout the night sky.

Thus, all this light from Earth, called earthshine, reflects off the Moon and illuminates it. In addition, Earth removes and blocks parts of the sunlight's spectrum, leaving only the longer wavelengths. This causes the Moon to appear red. Lastly, because the Earth blocks off all the direct sunlight from the Sun (only diffracted sunlight and airglow reach the Moon), we can actually see Earth's shadow on the Moon.


Which focus point does the sun occupy for each planet?

In addition to the other answers, that are right, a different doubt seem to arise from the wording of the question:

Let us say that the left focal point is f1 and the right f2.
Is the sun at the same focal point for each planet?

If the question is if the Sun is in f1 or f2, the answer is that it doesn't matter. Ellipses are symmetrical, and therefore both foci are equivalent and any naming of them is just conventional. Just by saying that planets move in elliptical orbits and the Sun occupies one focus, Kepler's first law is unambiguously established.
If you need to mention a particular focus, one is just the Sun position and the other one is often referred as the empty focus.


How fast would Earth need to rotate to fling someone off due to centrifugal force?

If it would rotate fast enough to throw a person off, it would throw everything off - dirt, houses, cars, etc., as the mass of an object is irrelevant - the centrifugal force (pulling out) and the gravitational force (pulling in) are both proportional to an object's mass.

Basically the whole thing would explode out at the same time. To calculate we equate the force of gravity $mg$ with the centrifugal force $mr\omega^2$, where $\omega$ is the angular velocity in radians per second

$$F = mg = mr\omega^2 $$ $$\Rightarrow g = r\omega^2 $$ $$\Rightarrow \omega = \sqrt{g/r} \approx 0.0012/s.$$

Insert the radius of the earth $r$ and the gravitational constant $g$, take the square root, and you have the necessary angular speed $\omega$.

with $\omega = 2 \pi/T$, you get $T = 2 \pi/\omega \approx 1.5 \mathrm{hours}$ [I did that in my head, so it might be wrong].


Can the difference between a star and a galaxy which are point sources be detected?

To distinguish galaxies from stars, you can use the spectrum. Roughly, stars have a black-body like spectrum with features depending on the absorption and emission on the line of sight and in the chromosphere of the star.

Galaxies on the other hand of a spectrum that is the composite of tons of stars. The spectrum will for example be much wider (ranging from smaller to larger wavelengths) because of the diversity in the spectra of stars.

Take a look at http://www.atnf.csiro.au/outreach/education/senior/astrophysics/spectra_astro_types.html if you want a quick overview of the differences.

I don't have a precise number about the number of galaxies we see as point source, but the answer varies greatly from one instrument to another one. If you try to observe a galaxy using radiotelescopes in interferometry, you can resolve much better scales than an Earth-based small visible telescope, etc…


How is stellar mass of a galaxy obtained?

In short, you count stars. Of course you can't count or even see all the stars, so you have to count certain types of (bright) stars and then assume that the ratio of these to all the stars in their neighbourhood is similar to the ratio in a local part of the Galaxy where we can do a more complete census.

Once we have an idea of a number for the stars in any region, we can use an assumed mass distribution for the stars to calculate the total stellar mass.

There are details and nuances (e.g. the form of the stellar mass distribution depends on the age of the population), but that's the basic process in our own Galaxy.

In other galaxies, where you don't resolve individual stars, you work on the basis of mass to luminosity ratios for typical populations to convert a luminosity to a stellar mass.


Apparent size of the super blue blood moon

Yes, they probably use a telescope if it's much bigger.

The 'super' refers to the fact that the moon is closer to the earth than normal during a full moon (it's near to the perigee in its orbit), but the human eye cannot see the difference between the moon size an hour before and an hour after the perigee.

Also, there's a phenomenon called moon illusion which makes the moon appear larger near the horizon, which might or might not be the case in your situation.

Incidentally, the fact that it's a blue moon has nothing to with it; that's just the moment it takes place on the calendar.


Is the universe older than 13.7 Billion years?

No, the oldest light we can see is as old as the Universe$^\dagger$. This is simply because light started traveling at that time, so after one year you'd se light that had been traveling for one light-year, after 1 million years you'd see light that had been traveling for a million light-years, and so on. There  is (most likely) more Universe beyond the horizon, but light from those regions just haven't had the time to reach us yet.

You might think that the distance to the cosmic horizon today is 13.7 billion light-years, but because the Universe is expanding, the distance is much larger (in fact 43.6 billion light-years).

$^\dagger$Because the Universe was opaque the first 380,000 years, we can't see all the way, but that's a technical detail.


The mass of the sun is 1.989 × 1030 kg.

Abundance in the Sun of the elements gives a percentage 1 × 10-7 % for gold *, so that leaves you with a mass of 1.989 × 1021 kg of gold.

HowStuffWorks states that there is 1.26 × 1021 kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 1021 kg.

This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans.

* They cite WolframAlpha as their source. Executing SolarAbundance "Gold" there confirms this (mass) percentage.


Random references from the internet:

Telescope Field of View - How much can you see?
Determine Your True Field of View
What is a Field of View in Astronomy and Why is it Useful?
Celestron YouTube video: Absolutely Everything You Need to Know about Astronomical Eyepieces & Visual Observing

As long as we need moderate accuracy only, there are several ways to do this.
The stated focal length of your eyepiece is approximate, 10 mm might really be 9.6 mm or 10.4 and they simply didn't bother with the details, each little lens' focal length inside the eyepiece will have some manufacturing tolerance. If your telescope's focal length is stated to be 1000 mm it might be 960 mm or 1040 mm and there'd be no reason for them to measure every mirror and give you it's own focal length to several digits of accuracy. So let's call these methods estimates +/-10%.
Examination of the eyepiece itself.
At the "input" end of the eyepiece there will be a field stop, an aperture that defines the your field of view (FOV) in the sky. It's likely a metal ring with a circular hole in it.
If the ring has a diameger $d$ of 10 mm and your telescope's focal length $f_T$ is 1000 mm, then you can ignore all of the other math and say that your FOV is $d/f_T$ = 10/1000 = 0.01 radians or 0.57°. Technically it is
$$FOV = 2 \arctan \left(\frac{r}{f_T} \right)$$
where $r=d/2$ is the radius of the hole. That doesn't matter much here but it does next.
Since your question asks for the apparent FOV of the eyepiece, you can get it by remembering that the plane of the aperture is what the eyepiece of focal length $f_E$ focuses on with a conjugate at infinity (for your eye to refocus to your retina), so the eyepiece's apparent FOV (aFOV) is given by
$$aFOV = 2 \arctan \left(\frac{r}{f_E} \right).$$
IF for some reason your eyepiece didn't have an aperture stop, you could bring a ruler into focus when looking through it and trying to focus your eye at infinity (perhaps by keeping both eyes open and focusing across the room with the other eye) and count the number of millimeters visible across the diameter. This doesn't sound easy though.
Looking through the telescope
At night you can measure the time it takes for a star to move across the FOV of your eyepiece and get the  real FOV of the system, then divide by the telescope's focal length $f_T$ to get the apparent FOV (aFOV). The Earth rotates 360° degrees on its axis every 23 hours 56 minutes 4 seconds and change (86164 seconds), so at a declination of 0 stars move 0.00418°/sec or 0.251°/minute. If you use something at a different declination, just multiply those speeds by $\cos(dec)$ to slow it down.
This gives you the real FOV of your system, again multiply by the magnification ($f_T/f_E$) to get aFOV.
You can also just look at something of a known or measurable or estimatable size far away, like a house or yardstick or meterstick to estimate the system FOV and multiply by magnification to get eyepiece FOV.


What does oxygen on comet 67P/Churyumov-Gerasimenko mean?

It's not that oxygen should not be in this state, but rather we had not discovered it (e.g. on other comets). One of the reasons for this is that interstellar O2 is very reactive and quickly reacts to form e.g. H20 and O3.

Current models also favoured O2-less comets.

Interestingly, the paper you're referring to also mentions the O2 abundance was rather constant over the half year of observation, which hints to the O2 actually being primordial, i.e. from when the solar system condensed.

To reconcile these  observations with models, a couple of possible solutions are offered: a slightly warmer primordial cloud (from which the solar system formed), or reactions with ices.


Question from Introduction to Modern Cosmology by Andrew Liddle

From the definition of the rms (e.g. here),
$$
\mathrm{rms}^2(x) = \langle x \rangle^2 + \sigma_x^2,
$$
where $\langle x \rangle$ is the mean value, and $\sigma_x$ is the dispersion.

For galaxies with random velocities, the mean velocity should be $\langle x \rangle = 0$, unless they drift in some direction. Hence, the velocity dispersion should be $600\,\mathrm{km}\,\mathrm{s}^{-1}$.

For randomly distributed velocities, the speed $V_{||}$ you measure along your line of sight (LOS) is a factor $\sqrt{3}$ smaller, i.e. $\sim350\,\mathrm{km}\,\mathrm{s}^{-1}$.

Thus, to get a 10% accuracy, the mean value should be 10 times larger, i.e. $\sim 3500\,\mathrm{km}\,\mathrm{s}^{-1}$.

For a Hubble constant of $H_0 = 100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, that corresponds to a distance of
$$
d = \frac{v_{||}}{H_0} \simeq 35\,\mathrm{Mpc},
$$
while for $50 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ you get $70\,\mathrm{Mpc}$.


is pan-starrs' data available to public users?

Looks like you've to be member of the PS1 Science Consortium to be able to log in.
And you'll need some database query knowledge, at least by using the PSPS Query Builder, or more advanced programming capabilities (XML, SOAP, Python, Perl). A short documentation of the database tables can be found here.
If you like to dig deeper into the database, this FAQ page may be a starting point.

The image processing pipeline (IPP) is described in this paper.


Is there a radius inside of which objects are (doppler) blue shifted?

No, there is no spherical volume of space within which objects all tend to approach us and thereby show blue-shift. The fact that the Andromeda galaxy is blue-shifted is because at these relatively small distances the cosmological red-shift is much smaller than further away. Andromeda's own (non-cosmological speed) is larger then the cosmological speed away from us and it happens to move towards us and is therefore blue-shifted. It could also have been moving away from us, and be red-shifted as a result, but that red-shift would not be cosmological.

I'd expect that the chance that a nearby (Local Group) galaxy is either moving away from us or moving towards us is 50%.

Furthermore, the Andromeda galaxy, and all other galaxies belonging to the Local Group, are gravitationally bound and would therefore not show any cosmological red-shift anyhow.

NB. The Andromeda Galaxy is actually 2.5 million light years away from us.


What does the velocity dispersion of a galaxy mean?

Definition of the velocity dispersion

From the title of your question, I'm unsure whether you actually know what "dispersion" means: The dispersion of some numbers is the spread around their mean, usually taken to mean their standard deviation. If you measure the velocity of several light sources (from the Doppler shift of their spectral lines) that are gravitationally bound, they will have some average velocity$^\dagger$. But since the light sources — which could be stars or gas clouds — move around each other, some move a little faster, and some move a little slower. This will cause their combined emission line to be broadened, instead of being a very narrow line. The central area of this line that emcompasses roughly 68% of the area of the line has a certain width. If the distribution is Gaussian (which is typically the case), then this width, divided by two, is the standard deviation. This is the usual definition, I think, but note that other definitions could be used, e.g. the full width at half maximum which differs from the standard deviation by a constant factor of $2\sqrt{2\ln2}$ for a Gaussian.

A relation between the mass, size, and velocity dispersion

The velocity dispersion $\sigma_V$ of gravitationally interacting "particles" such as stars in a galaxy, or galaxies in a cluster, provides a measure of the total mass and its distribution. The virial theorem, which describes the total energy of a bound system, can be expressed to show that the components of a galaxy of mass $M$ and radius $R$ will have characteristic velocities of the order of the velocity dispersion
$$
\sigma_V = \sqrt{\frac{GM}{CR}},
$$
where $G$ is the gravitational constant, and the observationally determined factor $C$ depends on the geometry and the actual mass distribution, as well as on whether the system is rotation-dominated (such as disk galaxies) or dispersion-dominated (such as elliptical galaxies) (Binney & Tremaine 2008). For dispersion-dominated galaxies, $M$ refers to the total, dynamical mass of the system, and $C \simeq 6.7$ for various galactic mass distributions (Förster Schreiber et al. 2009). For rotation-dominated galaxies, the appropriate mass is the mass enclosed within $R$, and $C \simeq 2.25$, again averaged over various galactic mass distribution models (Epinat et al. 2009).

The velocity dispersion is rather easy to determine with spectroscopy, since it broadens the spectral lines (other processes also broaden the lines, so one must be careful to disentangle different effects). With photometry, the size of the galaxy can be determined (note though that the size can differ when observed in different bands). Thus, even if you don't know whether a galaxy is rotation- or dispersion-dominated, a moderately precise measure of its mass — one of the main characteristics of a galaxy — can be obtained (e.g. setting $C=5$ won't be too far off, and in astronomy we're usually happy with such small errors).

Other relations

Because of various scaling relations, $\sigma_V$ can also be used to probe other quantities, or other quantities can be used to probe $\sigma_V$. For instance, the M-sigma relation relates the mass $M_\mathrm{BH}$ of a galaxy's central, supermassive black hole to $\sigma_V$ (through $M_\mathrm{BH} \propto \sigma_V^4$, roughly), and the Faber–Jackson relation relates the luminosity $L$ of a galaxy to $\sigma_V$ (through $L\propto\sigma_V^4$, more roughly).

$^\dagger$In astronomy, this average velocity is called "the systemic velocity"), and is given by the expansion rate of the Universe (the Hubble law), plus in general a small velocity in a random direction (the "peculiar velocity" due to local dynamics).


Seems a little contrived, essentially you just have to dig deep enough to find a specific enough category that matches the letter. But here you go :

K - KBO (Kuiper Belt Object) , if you still don't like that you could use

K - Kreutz Sungrazer (A type of comet, https://en.wikipedia.org/wiki/Kreutz_sungrazer)

T - T-Tauri Star (A type of variable star, https://en.wikipedia.org/wiki/T_Tauri_star)

Z - Z Camelopardalis star (https://en.wikipedia.org/wiki/Z_Camelopardalis_star redirects to Dwarf Nova, scroll down)


Which space telescope is the most distant?

Currently New Horizions is temporarily hibernating; it's last activity was two months ago. So I'm going to post a supplementary answer here because it is "operational" in the sense that it still works and will be used again, even though it is not "active" at the moment.

The most recent and farthest-from-earth telescopic observations that I know of are from the New Horizons spacecraft and it's Long-Range Reconnaissance Imager (LORRI), a 208 mm (8.2 inch) diameter Ritchey–Chretien telescope equipped with silicon-carbide optics and a cooled silicon CCD imager with a resolution of about 1 arcsecond:

The Long-Range Reconnaissance Imager (LORRI) is a long-focal-length imager designed for high resolution and responsivity at visible wavelengths. The instrument is equipped with a 1024×1024 pixel by 12-bits-per-pixel monochromatic CCD imager giving a resolution of 5 μrad (~1 arcsec)] The CCD is chilled far below freezing by a passive radiator on the antisolar face of the spacecraft. This temperature differential requires insulation, and isolation from the rest of the structure. The 208.3 mm (8.20 in) aperture Ritchey–Chretien mirrors and metering structure are made of silicon carbide, to boost stiffness, reduce weight, and prevent warping at low temperatures. The optical elements sit in a composite light shield, and mount with titanium and fiberglass for thermal isolation. Overall mass is 8.6 kg (19 lb), with the optical tube assembly (OTA) weighing about 5.6 kg (12 lb), for one of the largest silicon-carbide telescopes flown at the time (now surpassed by Herschel).

In December 2017 LORRI captured the images below of two Kuiper belt objects. From the February 9, 2018 NASA News item New Horizons Captures Record-Breaking Images in the Kuiper Belt:

NASA’s New Horizons spacecraft recently turned its telescopic camera toward a field of stars, snapped an image – and made history.
The routine calibration frame of the “Wishing Well” galactic open star cluster, made by the Long Range Reconnaissance Imager (LORRI) on Dec. 5, was taken when New Horizons was 3.79 billion miles (6.12 billion kilometers, or 40.9 astronomical units) from Earth – making it, for a time, the farthest image ever made from Earth.
New Horizons was even farther from home than NASA’s Voyager 1 when it captured the famous “Pale Blue Dot” image of Earth. That picture was part of a composite of 60 images looking back at the solar system, on Feb. 14, 1990, when Voyager was 3.75 billion miles (6.06 billion kilometers, or about 40.5 astronomical units [AU]) from Earth. Voyager 1’s cameras were turned off shortly after that portrait, leaving its distance record unchallenged for more than 27 years.
LORRI broke its own record just two hours later with images of Kuiper Belt objects 2012 HZ84 and 2012 HE85 – further demonstrating how nothing stands still when you’re covering more than 700,000 miles (1.1 million kilometers) of space each day.

Here is a 25-Feb-2018 screen shot of a map of the Solar System with New Horizon's current position plotted. You can read more about New Horizons at:

Where is New Horizons

NASA's New Horizons main page

Instrumentation on New Horizons is extensively described/explained in Spaceflight 101's Instrument Overview.
The LORRI telescope is described for example in the ArXiv preprint Long-Range Reconnaissance Imager on New Horizons (click "PDF"). It has a spectral range of about 350 to 850 nm. The limiting magnitude for high precision positional measurements is about +11.5 for a short exposure, but with some binning and a roughly 10 second exposure that increases to about +17:

For optical navigation, LORRI is required to be able to image a star of visual magnitude V=11.5 at SNR>7 in a single 100 ms exposure, with full width at half maximum (FWHM) >1 pixel. It is not desirable for too great a fraction of the energy from a point source to be imaged onto a single pixel, because stellar images become too undersampled. LORRI has a 4×4 pixel binning mode, for which its limiting magnitude requirement is V>17 in a single exposure of 9.9 s. This 4×4 pixel-binning mode will be used to search for the target KBO and to perform optical navigation on approach.

below: Image and caption from here:

With its Long Range Reconnaissance Imager (LORRI), New Horizons has observed several Kuiper Belt objects (KBOs) and dwarf planets at unique phase angles, as well as Centaurs at extremely high phase angles to search for forward-scattering rings or dust. These December 2017 false-color images of KBOs 2012 HZ84 (left) and 2012 HE85 are, for now, the farthest from Earth ever captured by a spacecraft. They're also the closest-ever images of Kuiper Belt objects.


Do planetary rings always follow rotational axis?

Nearly always.

The asymmetric gravity of a spinning (and hence oblate) planet will induce tides that can pull small moons and ring particles into orbit around its equator in a few million years.

If you had a moon that was for some reason not orbiting around the equator, and it broke up (for example by passing the Roche limit) then the resulting ring would not be around the equator at first. However it is unlikely that a moon would be in such an orbit in the first place.

Very large moons (such as the the Moon) and those that are captured minor planets (Triton) can be in an inclined orbit. Triton is large, and was captured from the Kuiper belt. There has not been enough time for Triton to correct its orbit. Triton orbits backwards too. Such moons are comparatively far from their host planets, and so not at risk of being broken up to form a ring.


The detection of Alpha Centauri Bb unfortunately did not survive follow-up analysis and it is no longer thought to exist, see Rajpaul et al. (2016). This planet candidate was in any case too close to the star to support liquid water.

There has been a claim of a single transit of a longer-period planet orbiting Alpha Centauri B, although this has not been confirmed. The relevant paper is Demory et al. (2015). Once again, this planet candidate was inferred to be too close to the star for liquid water.

Recently it has been confirmed that Proxima Centauri is a bound member of the Alpha Centauri system (Kervella et al., 2017). The planet Proxima b has been found in the liquid water zone (Anglada-Escudé et al., 2016) and has a minimum mass of 1.27 Earth masses, so if the orbit is close to being edge-on it is likely to be a terrestrial planet.

Of course, just being in the liquid water zone is not sufficient for habitability (obvious counterexample: our Moon). A major issue is that Proxima Centauri is a flare star and the space weather environment experienced by Proxima b is extremely hostile. This brings into question whether or not Proxima b can maintain an atmosphere under such conditions, for example see Garcia-Sage et al. (2017).

There is also the issue that low-mass stars like Proxima Centauri spend billions of years in a luminous pre-main sequence stage of their evolution. This would likely trigger a runaway greenhouse on planets in the main sequence liquid water zone and result in the loss of significant amounts of water from the planet, as discussed in Ramirez & Kaltenegger (2014).

So while Proxima b appears to meet some requirements for being able to support life, in several other ways it appears to be far less than ideal and we should not be too surprised if it turns out to be non-habitable.


Can a single impact event trigger a runaway greenhouse?

What your calculations don't account for is the dynamics of the meterological system.  An atmosphere with $10^{18}$kg of water vapour would be a powerful greenhouse, but at the temperatures on Earth, there is no way that the atmosphere could hold so much water vapour, it would rapidly condense.
Instead, in the case of a major impact into the oceans, a lot of water vapour is produced, which rapidly condenses to clouds because the Earth isn't hot yet.  The clouds will tend to cool the Earth (by raising albedo).
The climatic effects will be significant, but you would expect all the water vapour generated to condense fairly rapidly, and far faster than the greenhouse effect of the water vapour.
The Earth has certainly experienced such impacts into the oceans, (this is a Chicixlub level event, expect once every 100 million years or so, and ⅔ will fall into the ocean) and yet no runaway greenhouse has occurred.


The Sun outputs several different kinds of things.

Electromagnetic radiation

The Sun is (partially) a black-body radiator at a temperature of near 6000 K, and therefore emits all sorts of electromagnetic energy, including UV and X rays.

UV is stopped in the upper atmosphere. X rays are absorbed by the whole atmosphere, and are pretty weak anyway.

The Sun's EM activity in terms of harsh radiation increases greatly if there's a solar flare blowing up. Even then, we're pretty safe on Earth.

High energy particles

The solar wind is basically a flux of particles (nuclei of hydrogen, etc) shooting out of the Sun at pretty high energies. These are charged particles. When approaching the Earth, the Earth's magnetic field for the most part deflects them. If you're on Earth, you're safe. Even in space, the flux is not very great; you'd be safe inside a spacecraft.

However, during a solar flare, the flux of particles increases greatly. The Earth is safe, but an astronaut en route to Mars would be in a pretty unsafe situation.


Accessing SDSS server SQL server via python

You can do it via the astroquery SDSS module; there is a function called query_sql.


I too saw a star like object moving across the sky it appeared to be a star moving!

This is very probably another artificial satellite, there are hundreds of them

references :
http://www.pixalytics.com/sats-orbiting-earth-2016/
https://www.universetoday.com/42198/how-many-satellites-in-space/


Sun's apparent motion above the arctic circle during summer solstice

Instead of walking in a circle, spin. Take a tiny step forward with your right foot, backward with your left. Then repeat. You are spinning counterclockwise when viewed from above.

Next, imagine holding a landscape photograph at arms length, and you initially facing a floor lamp. Now take that tiny step forward with your right foot, backward with your left. From your body fixed perspective, the landscape photograph doesn't move. It's the lamp that appears to be moving, and it moves to the right.

Next, imagine a helmet cam on your head that rotates to keep the floor lamp in the center of the camera image. From this perspective, it's the landscape photograph that appears to be moving, and it moves to the left.

That's what you are seeing in that youtube video.


So, what, exactly, is a tropical period?

Is the US Naval Observatory wrong somehow here? Have mistakes been made? Or are there misquotes?

That's a matter of opinion. In Duncan Steel’s opinion, yes, they made a mistake. In the US Naval Observatory's opinion (and also that of Her Majesty's Nautical Almanac Office), there is no mistake. The Nautical Almanac defines a tropical year as the time from one mean vernal equinox to the next. Steel is using a more pedantic definition. Increased pedantry does not always equate with increased correctness.

  Is it possible to explain in everyday English the geometrical difference between a "Vernal Equinox year" and a tropical year or a mean tropical year?

Right now, Northern Hemisphere spring and summer (Southern Hemisphere autumn and winter) last longer than do Northern Hemisphere autumn and winter (Southern Hemisphere spring and summer). This situation will be reversed in about 13000 years. These changes are due to the Earth's axial precession. That the Earth undergoes axial precession has multiple ramifications, three of which are that:

The length of a tropical year (no matter how one measures it) is less than is the length of a sidereal year.
The times from one March equinox to the next, one June solstice to the next, one September equinox to the next, and one December solstice to the next are not constant.
The times from one March equinox to the next, one June solstice to the next, one September equinox to the next, and one December solstice to the next are not equal to one another.

Elaborating on the third point, the calculated length of one mean December solstice to the next was 365.242740 days (ephemeris days) on 1 Jan 2000 while one mean June solstice to the next was 365.241626 days, or about 1.6 minutes shorter than the duration from one mean December solstice to the next.

So what is a "tropical year"? The US Naval Observatory, which has a much longer history than does Duncan Steel, defines the tropical year as the passage from one March equinox to the next. The US Naval Observatory inherited this definition from Her Majesty's Nautical Almanac Office, which in turn inherited it from the Church of England, which in turn inherited it from the Catholic Church. One of the key drivers of the Gregorian Calendar was getting the date of Easter correct. (Getting the date of Easter correct was very important to many Renaissance minds.)

Duncan Steel defines the tropical year otherwise, as an average value of the times it takes for the Sun's longitude as perceived from the Earth to advance by 360°. That's fine, but to argue that people who use a different definition is not just wrong. It's "wronger than wrong". The fact is that there are multiple definitions of what constitutes a "tropical year", all of which are valid.


Lunar illumination (percentage) data on an eclipse day needed

I found some chart regarding this, which is here:
http://www.mkrgeo-blog.com/remote-observation-of-the-total-lunar-eclipse/
The light level drops from about 50-60% prenumbral phase. It becomes noticeable.

If you want to get the percentage illumination of the Moon during the eclipse, install the Cartes du Ciel and turn on the Earth's shadow. Then you will be able to gauge the percentage value.
https://xjwfriends.com/2018/02/18/astronomical-diary-vat-4956/


Temporal path through Hertzsprung-Russell diagram?

This is typically a bit tricky to show because (1) some phases (e.g. the main sequence) are much, much longer lived than other phases and (2) a number of phases are spaced very close together in HR-space.

Be that as it may: any elementary text book on stellar structure will guide you through the path of a star through the HR diagram, usually accompanied by pictures. The book by Prialnik on this subject matter, or Ostlie & Carroll, are both excellent.

Here is a very crude version of the temporal path through the HR diagram I made some time ago. Numbers show the different phases (2 is MS), colorcoding shows the ages.


How was Cassini able to measure and use the Earth-Mars distance to get the Earth-Sun distance?

Giovanni Cassini used parallax with a friend, from two different sides of the Earth.

It was generally accepted that the stars didn't move relative to each other or relative to the motion of the Earth.   The planets moved, the stars didn't.

Cassini was before Newton but after Kepler, so he had access to Kepler's 3rd law which is the law of periods, which tells you the ratio of Mars' distance to the Sun and Earth's distance to the Sun.     (Estimates like that could also be done with angles before Kepler), But, if Cassini could measure a single distance, for example, Earth to Mars when Mars passes at it's closest point, then the other distances can easily be worked out.

Cassini sent a friend and fellow astronomer, Jean Richer, to French Guiana and they both took the position of Mars at the same time from different sides of the world, nearly 12,000 km distance.

Mars' distance at closest pass is about 54 million km in distance, so the variation in angle from one side of the Earth to the other is a ratio of 1 to about 4,500, which is a bit under 1 minute in angular variation.

I don't know how Cassini and Jean Richter were able to measure to such a high degree of accuracy, and whether there was some lucky guessing in there, or if they had equipment that good (if anyone knows, please feel free to make a more detailed answer).  But that's the gist of it.   Mars is in a different place relative to the background stars when viewed from different sides of the Earth and by measuring how much Mars moves from those two points on Earth, an estimate of distance can be made.

Cassini's estimate was, it turns out, surprisingly accurate, within about 7% of the actual distance.  His astronomical unit was 140 million km or 87 million miles.   His method works, though the elliptical orbit of Mars makes it a bit harder.   The real problem, as was often the case in early astronomy, was how accurate the measurement was.

Short article here

Longer article here

Also discussed here, under 1671-1673

The 2nd article says the angle is 20 arc seconds.   When I do the math, I get slightly about 46 arc seconds.   So, I invite correction if I made a math error.     20 arc seconds is even harder to measure, so any details on Cassini's margin of error would be welcome too.     I suspect there was concern that his method lacked accuracy because the Venus transit 100 years late seems to get a lot more attention as the first accurate measurement of the orbital distance of the planets.


Which stars have been named after astronomers?

You probably need someone better versed in the history of astronomy than me, but I'll give it a stab.

As you've already noticed, the Wikipedia article on stars named after people has some entries, and that's probably about it.  In terms of modern naming procedures, we tend to use catalogue names. Unlike, say, asteroids, there's no systematic way of naming stars after people or anything else, so stars usually continue to be referred to only by catalogue number. Those rare stars that are named after people (or something else) are usually because they are now somehow historically associated with that system.  Unless, of course, the catalogue is named after someone!  Another exception are those very bright stars with pre-telescope names, but those aren't named after people (e.g. many scientific papers refer to Betelgeuse as such).

I guess in short, there's just no standard modern naming procedure by which a star can end up with a person's name, so few exist, and probably most are listed in the Wikipedia article.


Is there any other theory, apart from the Big Bang paradigm, which describes the birth of universe?

This is not an answer but a comment which is too long to post as such.

There is no such thing as a scientifically approved theory, there are theories consistent with the evidence some of which are simpler in some sense than others, and philosophy of science tells us that we should prefer those that are simpler.

Also I doubt that the big-bang theory is what you think it is, which is simply that at a certain time in the past the Universe was in a hot dense state which then expanded, forming the lightest chemical elements... The first part of this is a simple extrapolation of the observation of the way that galactic Doppler shift increases with distance and the observed cosmic microwave background. The second part is the consequence of the application of well know nuclear properties of matter.

Now there are plenty of more-or-less widely accepted additions to this, mainly to explain the uniformity and fluctuations in the microwave background, and the emergence of first stars and galaxies etc. But all of these are to some extent speculative since we have no generally accepted unification of Quantum Mechanics and General Relativity or require physics that has no predicted consequence (so far) beyond the phenomena it has been proposed in order to explain.

To make this approximate an answer I will point you to the Wikipedia page on Non-Standard Cosmologies


Can cosmic rays alone produce noticeable aurorae in bodies far away from the Sun?

You may be interested to learn that Jupiter experiences aurora on its poles nearly (if not totally) non-stop. These aurora never cease, because they do not rely entirely on the sun as their source of charged particles, instead receiving particles from the volcanic ejections of its moon Io, as well as from cosmic rays. Granted, these aurora are occurring in the ultra-violet, but they are still 'noticeable'.

In short, yes aurora can be caused by non-solar sources, and conceivably cosmic rays at that. As to your other question concerning 'how much' of the aurora is due to which sources... I couldn't tell you how much of the auroras visibility comes from the sun, and I doubt there is much (or any) information on that subject, since determining the source of such convoluted light is likely outside the realms of both interest and empiricism. Perhaps you could pioneer that research? In any case, an intriguing subject to be sure.