How do OGLE-III and GAIA measure the mass of free microlensing black holes?

The introduction of the Wyrzykowski & Mandel paper gives the following information about estimating the lens mass.

In order to obtain the mass of the lens (Gould 2000a), it is
necessary to measure both the angular Einstein radius of the lens
($\theta_\mathrm{E}$) and the microlensing parallax ($\pi_\mathrm{E}$)
$$M = \frac{\theta_\mathrm{E}}{\kappa \pi_\mathrm{E}}$$
where $\kappa = 4G / (c^2\ \mathrm{AU}) = 8.144\ \mathrm{mas/M_\odot}$; and $\pi_\mathrm{E}$ is the length
of the parallax vector $\mathbf{\pi_\mathrm{E}}$, defined as $\pi_\mathrm{rel}/\theta_\mathrm{E}$, where $\pi_\mathrm{rel}$
is relative parallax of the lens and the source. The microlensing parallax vector $\mathbf{\pi_\mathrm{E}}$ is measurable from the non-linear motion
of the observer along the Earth’s orbital plane around the Sun.
The effect of microlensing parallax often causes subtle deviations and asymmetries relative to the standard Paczynski light
curve in microlensing events lasting a few months or more, so
that the Earth’s orbital motion cannot be neglected. The parameter $\mathbf{\pi_\mathrm{E}}$ can also be obtained from simultaneous observations of
the event from the ground and from a space observatory located
∼1 AU away (e.g., Spitzer or Kepler, e.g., Udalski et al. 2015b,
Calchi Novati et al. 2015, Zhu et al. 2017).

In particular, the Gould 2000a paper gives a good summary of the various relationships between the quantities. The Udalski et al. 2015b notes that the distance between the Earth and Spitzer (which would also apply to Gaia) means that Spitzer would see differences in the light curve, allowing the parallax to be determined.
Note that things get more complicated if the source is a binary, in which case a "reverse parallax" effect from the source's orbital motion, usually called "xallarap" needs to be taken into account — but that's a matter for another question...
The other relevant quantity is the angular Einstein radius of the lens. In their discussion of measuring $\theta_\mathrm{E}$, Wyrzykowski & Mandel reference Rybicki et al. 2018. That paper notes that precision astrometry can help measure $\theta_\mathrm{E}$ because microlensing also changes the apparent position of the source:

The positional change of the centroid depends on the
$\theta_\mathrm{E}$ and separation $u$. Contrary to the photometric case, the
maximum shift occurs at $u_0 = \sqrt{2}$ and reads (Dominik & Sahu 2000)
$$\delta_\mathrm{max} = \frac{\sqrt{2}}{4} \theta_\mathrm{E} \approx 0.354 \theta_\mathrm{E}$$
Thus, for the relatively nearby lens at $D_l = 4\ \mathrm{kpc}$, source
in the bulge $D_s = 8\ \mathrm{kpc}$ and lensing by a stellar BH with
the mass $M = 4M_\odot$, the astrometric shift due to microlensing will be about 0.7 milliarcsecond.

The bulk of the paper goes on to determine that these shifts should be observable by Gaia.
Another way to measure the size of the lens is to measure the lens-source proper motion by searching for the lens several years after the event, this has been done for a couple of exoplanet-hosting lenses but would not be possible for a dark lens like a black hole.


If two white dwarfs collided, would they become a star?

The answer to your question is both yes and no, depending on the circumstances.

Two white dwarfs colliding would likely yield a Type Ia supernova, assuming the combined mass exceeded the Chandrasekhar limit ($\sim1.4$ solar masses). The unstable object resulting from the collision could not be supported by electron degeneracy pressure; when the temperature rises rapidly due to the collision, there is nothing to stop it (compare this to thermal pressure in a "normal" star, which can make the star expand or contract to compensate for changes in temperature).

The rise in temperature triggers fusion, which then increases the temperature, which triggers more fusion . . . and so on and so on, in a very short time span. This is the same process, more or less, as in an accretion-powered type Ia supernova. The resulting explosion destroys the object, ejecting matter into space.

As with normal Type Ia supernovae, this sort of collision would likely happen in a binary system, with two white dwarfs in close orbits losing energy to gravitational waves and spiraling in (the odds of two unrelated white dwarfs colliding is very, very low). I'm not sure how luminous this event would be in the gravitational wave spectrum; likely fainter than a neutron star-neutron star inspiral but still strong. The speeds of the components could be quite fast, meaning that a lot of energy would be released in the collision.

Things are slightly more complicated if the combined mass is less than the Chandrasekhar limit. An example system that is predicted to undergo a collision is SDSS J010657.39-100003.3. The total mass of the two white dwarfs is $\sim0.6$ solar masses, certainly below the limit. In roughly 37 million years (Kilic et al. (2011)), they will collide and form a subdwarf - fusing helium, interestingly enough (compare this to the carbon/oxygen fusion in Type Ia supernova, which in those circumstances is unstable). Similar scenarios could play out in other systems.

This sort of a star would not be a "normal" star in the sense that it wouldn't truly be on the main sequence - just a bit below it. It would be an O- or B- type subdwarf, less luminous than main sequence stars of the same spectral type. I saw a couple of articles on these objects claiming that they would be stars just like the Sun - it's not true! Given that white dwarfs are highly depleted in hydrogen - remember, they are the degenerate cores of old stars - there can be virtually no hydrogen burning, even in a shell. Core helium burning is the only option.

On a final note, one explanation for the formation of R Coronae Borealis variables is the collision of two white dwarfs. I don't think we have solid evidence yet for preferring this over other models (e.g. something involving hydrogen loss from the outermost shell), but if it is the correct pathway, then the answer to your question could be a definitive yes. Note that this would require intermediate-mass white dwarf progenitors.


Do other supercluster like Laniakea also get pulled by the Great Attractor?

In short, not much gravitational pull at all, not relative to the distance and relative velocity.

Everything in the observable universe is gravitationally attracted to everything else, at least within the appropriate horizon.  There should be some objects far enough apart, taking into account cosmic inflation, that gravity can't travel that far over the age of the Universe.  For example the farthest galaxy we can see in one direction and the farthest one we can see in the opposite direction, those two galaxies can't see each other and they likely experience no gravitational pull from each other, but that's not really relevant to your question.

The Great Attractor is a comparatively small but very dense part of Laniakea, with an estimated mass thousands of times the mass of the Milky Way.

The Laniakea supercluster has a mass of about 100,000 Milky Ways, so any gravitational attraction one supercluster experiences would primarily be from other entire superclusters, and the great attractor would be a small percentage of that.

Superclusters experience gravity from other superclusters but the Universe is very uniform, so that gravitational pull is largely balanced out because it's coming from all sides.

This touches on one of the modern puzzles in physics, called the Horizon problem. This question goes back to the big bang and why the young universe was very close to, but not precisely uniform.

It might be a fun exercise to calculate the gravitational pull neighboring superclusters have on each other.   I imagine it's pretty negligible compared to the expansion of space between them.


What causes fast moving pulsars to move so fast?

The most likely scenario is that an asymmetry in the supernova explosion  imparts momentum to the proto neutron star at its core. The issue is not settled.

A recent study by Verbunt et al. (2017) models the
speed distribution of young pulsars as the sum of two Maxwellian distributions, one with an average speed of $\sim 130$ km/s and the other with an average speed $\sim 520$ km/s. About 30% of the population are in the low speed  population and only a few percent have speeds below 60 km/s.. A plot of the (model) distribution is shown in Fig.9 of that paper.

Why do some pulsars move so fast? Conservation of momentum. Pulsars are neutron stars with mass about $1.5 M_{\odot}$, born at the heart of core-collapse supernovae. A 500 km/s speed corresponds to a kinetic energy of about $4\times 10^{41}$ J and a linear momentum of $1.5\times 10^{36}$ kg m/s, but during the supernova, approximately $10^{46}$ J is made available when the core collapses from about the size of the Earth to a ball of neutrons 20 km across. Thus only a tiny fraction of this energy needs to be given to the neutron star to accelerate it to high speeds.

Most of the energy in a supernova is released as almost massless neutrinos from the core. In order of magnitude terms, the neutrinos carry away a linear momentum of $\sim E/c = 3.3\times 10^{37}$ kg m/s. If there were a small asymmetry in how these neutrinos escaped - i.e. more in one direction than another, then conservation of momentum would demand that the core moves in the opposite direction. To match the observed linear momentum of fast moving neutron stars would demand
a directional asymmetry of neutrino emission of $\sim 5$%, (which seems unlikely and may require magnetic fields of $10^{12}$ T -  Nardi & Zuluaga 2001 ). A better idea might be that in the initial stages of the collapse, the core material is opaque to neutrinos and it could be that there is an asymmetry in the way the neutrinos break out of the absorbing material (e.g. Nordhaus et al. 2012).

Alternatives are that there are instabilities in the final phases of nuclear burning, just prior to the core collapse. These instabilities cause the core to wobble back and forth and when the supernova happens, the core just happens to be travelling one way or another. There have also been suggestions that asymmetric radiation from the pulsar itself could gradually accelerate it to the speeds observed.


How do I apply a velocity shift to a wavelength array with uniform logarithmic spacing?

If you have bins of equal log wavelength increment then a velocity shift corresponds directly to a pixel shift. That is the reason for binning in equal log wavelength increments.

See  this question for a short proof. All you need to do is work out what each pixel corresponds to in terms of a velocity.
$$\Delta V  = c\, \Delta \log \lambda$$


Computing the Sérsic profile of a galaxy from jpg images

Is there any standard software or algorithm for computing the Sérsic index from an image?

I don't think it is standard but Vika et al (2013) have used a modified version of GALFIT to extract Sérsic profiles. EDIT> But I see you already found it ;-)

  Are there reference images I can work from that would let me check if my algorithm is reasonable?

There are references in literature that provide images with Sérsic profiles, but I haven't been able to find a database which I think is what you are looking for.

  Any recommendations on how to proceed would be welcome.

For your purposes it is not really necessary to calculate the Sérsic index so that you can compare it with literature. If your Sérsic index, which is just one feature for your classification algorithm, is distinctive enough to be able to distinguish between different morphological classes than that is fine.

I imagine that it might be difficult to calculate reliable Sérsic indices from the images provided by GalaxyZoo. The images I remember are often low resolution images of galaxies.

So I would proceed by calculating your Sérsic index for your training and test sets and see how well this feature performs in a classification task.


Planets are really small!  (when seen from Earth)

Deepskywatch has an example of what can been seen with a 150mm reflector and a magnification of x180 (ie slightly more than your Astromaster which has a 130mm main mirror, and comes with a x65 eyepiece)

You should be able to see the 4 moons, and one or two bands on Jupiter. This assumes good "seeing" (the amount of turbulence in the air, not the same as the clarity of the air), and a properly collimated and aligned telescope lens.

There are lots of other good deep-sky targets that benefit more from a dark sky than the planets.


How was an infrared picture of the entire universe taken?

This picture was taken by the WISE (Wide-field Infrared Survey Explorer) space telescope. It is in deed an infrared picture of the entire universe but what you see in this line is our home galaxy, the Milky Way.

This telescope has taken infrared pictures of the entire Universe (as we see from the Earth) and joined them in this full-sky view.

More info: http://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer


Is it likely that intergalactic stars would still retain their planets?

While near encounters with supermassive black holes are bad for the stability of solar systems [citation needed?], many intergalactic stars become intergalactic without a close encounter. They are part of streams and tidal tails that emerge when galaxies or globular clusters get tidally disrupted by another galaxy.

Still, even while encounters with black holes are usually disruptive simulations suggest that some planets may remain around the star. This is most likely for planets real close to the star, like hot jupiters.


That's called New Moon and happens roughly once a month.
It's when (viewed from earth) the moon, circling the earth, is in the same direction as the sun. So, by rotation of the earth, both become visible (sunrise/moonrise) and invisible (sunset/moonset) at roughly the same time.
Moon is then hardly visible for two reasons:

The back side of the moon is lit by the sun, the side facing to us does not get sunlight.
The moon is visibly rather close to the sun, so sunlight will blind the human eye when looking in that direction.

Two weeks later, we have Full Moon, the moon is opposite to the sun, it rises when the sun goes down and goes down when the sun rises, meaning that it's visible at night. And the side facing us gets full sunlight, so the moon becomes a very bright object on night sky.


Which planet or moon has all resources that can be used to sustain life in a controlled biosphere?

No known planet except Earth can be colonized by a human civilization. There are at least three serious issues: temperatures at around 300K, an atmosphere of appropirate pressure, and damaging cosmic radiation (low gravity is also a worry for long-term human presence). Minerals are less of a problem (and water can be synthesized). Mars and the Moon are close enough to the Sun for pressurized greenhouses to provide an environment where humans could stay without space suit. However, the problem of cosmic radiation remains and such "biospheres" must forever dependent on supportive technology. Venus's inhospitable atmosphere makes it completely unfeasible.

Using robots to mine planets and moons in the Solar system is certainly a realistic and interesting option. Sending humans there is very expensive and not necessarily sensible. Moreover, it is not clear whether humanity should send any live (including microorganisms) to such places, as this would inevitably forever change their nature (like sending rabbits/cats to Australia); all ongoing space missions are sterilized before being sent on their way.


It depends on your definition of "layman" but Patrick Wallace has a slide presentation on what changed and why with the new reference systems. There is also the USNO Circular 179 which covers all of this and is pretty readable.

In a lot of ways it boils down to the realization that the wobbly (in space and rotation) Earth was a lousy thing to base co-ordinate systems on. The new systems are what's known as "kinematically non-rotating" which mean they are fixed in space and don't move. This makes it much easier to make more precise theories (which were needed) to describe how things like the orientation of Earth's pole (the precession and nutation you may have heard about) move in the Solar System without having to reference it to something like the equator and ecliptic which are always moving about.


Is there a physical limit to how far we can go?

Your 3 points are spot on.   There is a finite number of galaxies we could theoretically reach for the reasons you say.  The further away the distant galaxy the greater the expansion of space between us and that galaxy and beyond a certain distance, galaxies can't be reached, even by the speed of light.   We can see galaxies that we couldn't possibly travel to.

  Does this mean there is a limit to how far we can go, and furthermore,
  we will experience the “big rip”?

No.  It only means that some galaxies that we can see, we could never travel to.   There's no limit to how FAR we can go but there's still a finite number of galaxies that we could reach with close to the speed of light travel.

This article, using the current size estimate for the observable universe of 46 billion light years in radius, says that 14.5 billion light years distance is the limit that we could theoretically reach, about 3% of the observable universe or about 3 billion galaxies.

As for the big rip.   Nobody knows if that's an accurate prediction.  It's a possible outcome, but we don't know enough about dark energy to say if the big rip will happen or not.   Nobody knows.

And it's a pretty psalm, but I don't see the prediction of dark energy in it.   Only that it says the Universe will age, but life is eternal - which is a nice thought.


As the Sun dies, does the Earth's orbit change?

The answer is yes. As the Sun ages, it will become a red giant and the mass loss rate from its surface will increase. This effect will increase (dramatically) further when the Sun enters the asymptotic giant branch phase, where thermal pulsations drives a cool wind that may carry away a millionth of a solar mass per year, eventually leaving a burned-out core in the form of a white dwarf with about half a solar mass.

At any point in this evolution we can model the evolution of the Earth's orbit using some simple approximations - that the wind from the Sun escapes to infinity, that a negligible proportion is actually accreted by the Earth and nor does it exert a torque,  that the mass loss takes place on a timescale much longer than the Earth's orbit and that the mass of the Earth $m$ is always much less than the time-dependent mass of the Sun $M(t)$.

In which case we consider the orbital angular momentum of the Earth:
$$ m a \omega^2 \simeq G\frac{M m}{a^2},$$
where $a$ is the semi major axis. So the angular momentum $J = m a^2 \omega$ is given by
$$ J^2 = m^2 a^4 \frac{G M m}{m a^3} \propto M a$$

As the angular momentum of the Earth's orbit is conserved, the $M(t) a(t)$ is constant and as the Sun loses mass, the semi major axis increases by the same factor.

Coming to the specifics - when the Sun is a half solar mass white dwarf, the semi-major axis will be 2 au (assuming the giant Sun did not quite engulf it - it will be a close-run thing) and Kepler's third law $(P^2 \propto a^3/M)$ can be used to estimate an orbital period of 4 years.

The tidal effects of the Sun on the Earth's orbit are quite negligible compared with these mass loss effects.


Instrument aperture sizes on Hubble Telescope

EDIT: You can't have a very wide field, and simultaneously a very high resolving power, in an instrument with a large focal ratio (or any instrument, really), unless you use sensors with an unrealistic pixel count. Details below:

The linear size of the focal plane is large - one document says it's the size of a "dinner plate". But the instrument is an f/24, which is a large focal ratio. At that ratio, a big linear size in the focal plane doesn't mean much in terms of angular size.

From what I've seen, the Wide Field Camera has a total field of 4.7 arcmin, or 280 arcsec according to this document:

http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf

Anyway, yes, that's a small field, if you compare it with ground-based amateur astrographs, which can have a field as wide as several degrees of arc.

But the Hubble telescope has a very high resolving power - 0.05 arcsec. Let's take your information and assume the WFC has a total field of 3 arcmin.

3 arcmin / 0.05 arcsec = 3600 pixels in theory, as a linear size of the WFC

If you wanted the WFC to have a bigger total field, you'd have to either sacrifice resolving power, or use a sensor with an enormous pixel count, which is not feasible.

The WFC uses several sensors, each one of them 800 x 800 pixels.

TLDR: Wide field, or great resolving power - pick one. They chose resolving power, and they compensate the narrow field by doing multiple passes.


What is the significance of minus sign in the expression of apparent magnitude?

So if I read your question correctly, you're asking why as stars get dimmer, they are given higher magnitudes?

The reason is purely historical. The ancient greeks assigned stars with 6 brightness levels or magnitudes. The brightest stars were of first magnitude and the least bright stars (to the naked eye) were of 6th magnitude.

In the 1800's this system was formalised by an astronomer named Norman Robert Pogson. By then they were (presumably) able to more accurately measure the brightness of a star, and he decided to use a system that fitted well with the ancient measurements, but could be used to interpolate and extrapolate to brighter and dimmer objects. In order to do this, the formula has to have a negative in front of the log.

More information here


How can a mirror see things that the naked eye can’t?

Your eye has two functional parts: a lens (that is about ¼cm²) and a light-sensitive surface (the retina) that is covered in rod-cells that detect light. It takes about 10 photons arriving at about the same time (0.1sec) for a rod to react to the light and send a signal to the brain.

This means that, unless about 400 photons arrive from a star onto every square centimetre then there isn't enough light to see the star, even under ideal conditions.  In practice, many more photons are needed for your brain to be able to interpret the signal from your eye as a point of light.

In contrast the mirror of a telescope is much larger. A small amateur telescope could have a mirror that has an area 1000 times larger than the lens of your eye. This means that a star can be 1000 times dimmer, but still be visible when you look at it in the telescope. Large, professional, telescopes have mirrors that are millions of times greater than the area of an eye.

Moreover a telescope can be fitted with a camera, and film (or digital sensors) that are both more sensitive to light than your retina and can integrate the light arriving over a long period of time.  If you make a long exposure, of several minutes, or even hours, then even dimmer objects will become visible.  This can increase the sensitivity by a factor of many thousand again.

The combination of the large light gathering area of the mirror, with the sensitivity and long exposures make possible to "see" things using the mirror of a telescope that can't be seen with the naked eye.


How far are we from the edge of the Universe?

In the standard models of the Big Bang, there is no such thing as the "edge of the expansion front". The universe, as far as we can see and as far as standard cosmology assumes, is homogeneous and isotropic on the large scale, so there is no edge or anything analogous to an the shock front of an explosion.

According to the seven-year WMAP results [pdf], the proper distance to the surface of last scattering that emitted the cosmic microwave backgrounds is approximately $46.0\,\mathrm{Gly}$, which is basically as far as we can actually see. The true cosmological horizon slightly more distant that this, closer to $46.6\,\mathrm{Gly}$. What's beyond the horizon is not known.

Another cosmologically significant distance scale is the Hubble radius, at which the galaxies comoving with the Hubble flow recede from us at the speed of light. It is $c/H_0 = 13.9\pm0.3\,\mathrm{Gly}$.


When can I see the Milky Way from the Earth?

Wherever you have a dark place, you can see the Milky Way. Anywhere in the countryside, far from large cities might work. And it doesn't need to be a very far away - a few km may do the work, specially if there are some hills in between.


How could visible light be in pitch-black? Can it be?

Black is not a color; it is the absence of light.

I think you interpret the sentence as "the light is black", but it's just a way of saying "there is no light (at optical wavelengths, but there is at radio wavelengths)".

In my opinion this is not a good way of explaining the CMB. The sky is not black at any wavelengths, but has a background radiation at all wavelengths. In the optical, this background originates (primarily) from the integrated light from stars; in the microwave it originates (primarily) from the afterglow of Big Bang; in the infrared from dust, etc.

See the figure in this answer to a question about the number density of CMB photons, to see how large the background is in various wavelength regions.


Can it be assumed that every star in an amateur astrophotograph has been cataloged?

Variable stars might challenge this.  A distant Mira type variable or recurrent nova could have been at its minimum when the (for example) Gaia catalogue was being assembled, but appear in an amateur photographs.
These stars undergo very large variation in brightness.  Mira can increase in brightness by 8 magnitudes, so it is not inconceivable that an uncatalogued Mira type variable could be found in an amateur photograph. Yet have been omitted from other catalogues by bad luck.
Recurrent novae are also "variable stars". They also undergo large changes in brightness. As they spend many years at minimum brightness and then suddenly flare, there is a greater chance that they could have been missed in the Gaia and other catalogues.
However, these are rather special cases.  In general, any star that an amateur could image with reasonable equipment will have been photographed, identified and catalogued by one of the sky surveys. Automated image processing makes this possible. Many of the things that amateurs did in the 1980s and before are now done by robots and computers.
Supernovae have often been found in other galaxies by amateurs, with no catalogued progenitor (though supernovae might not be considered a "star")


Classical mechanics approach to this would be gravitational force as we know F=GMm/r2 and Centripetal force F=mv2/r.

In free space we when two things are kept some distance apart they accelarate towards each other (along straight line from center of on body to center of another body) due to force of gravity but if they are under some influence with other bodies they does not get accelarate exactly towards each other(deviation along the path), it causes the dirrection of bodies to change. In this case the planets(cause of relatively small mass).
If this deviation is large body(planet) fly by without any effect. If deviation is small, bigger body continuously change smaller one's direction, thus orbit.

http://curious.astro.cornell.edu/about-us/57-our-solar-system/planets-and-dwarf-planets/orbits/243-why-do-the-planets-orbit-the-sun-beginner

http://www.windows2universe.org/our_solar_system/formation.html

fun Q:why does astronauts on ISS feel 0 gravity when ISS is revolving under earth's gravity?
because they experience gravity but the revolution causes them to free fall, which they feel like 0 gravity.


What would be the scientific impact of locating the Thirty Meter Telescope at its alternate site?

The Canary Islands have slightly poorer seeing on average, but the main difference would be the higher column density of water vapour. Mauna Kea is a  higher and "drier" site, which means you can observe into the mid-IR.

Looking at the first light instrument package I would guess that WFOS (wide field imaging at 0.3-1 micron) would not be badly affected. But there would be some impact on IRIS (near-IR spectroscopy) towards the long-wavelength end of its range.

The near-IR adaptive optics would also be somewhat affected by the slightly poorer natural seeing, but I'm not sure I can quantify this.


Is there any way to seperate astronomical and instrumental objects?

A priori it is very difficult to distinguish the origin of any particular feature from just one image.

For that reason it is established workflow, especially in astronomical context, to create 4 kind of images:

the light frame R. That's the actual image of what you are interested
in
the dark frame D. That's an exposure of identical length and at
identical conditions, but with closed apperture. That takes care to
assess the sensor reading in the absense of anything visible
the bias frame B. That's a short-duration image which allows to assess the
read-out noise
the flat field F. That's an image of a perfectly white /
grey area which allows to assess the different sensitivity of the
individual pixels.

Then you can obtain the corrected image C via $ C = (R-D-B) \cdot \frac{<F - D - B>}{F - D - B}$

https://en.wikipedia.org/wiki/Flat-field_correction


Can a person at a photon sphere of a black hole decide where the black hole is?

For an observer hovering [I say hover because by definition there are no orbits for an observer at the photon sphere] at the photon sphere for a non-maximal de Sitter-Schwarzschild black hole, a simple accelerometer will tell an observer which horizon is which. The reason for this is that in order to hover at the photon sphere the observer must take up a non-inertial local frame with a directional bias decided by the direction of the BH event horizon.

In the maximal de Sitter-Schwarzschild spacetime the BH singularity disappears and the observer will not be able to tell the difference between the two horizons as there is an automorphism of that spacetime which will map one horizon to the other.


I wrote some JavaScript to help answer my question. It's in terrible shape right now, but it plots the hourly position of Aldebaran in azimuth & elevation when viewed from Dayton, Ohio, USA, every fortnight for a year:

http://fasiha.github.io/starpath/

The source code is at https://github.com/fasiha/starpath but everything in it is hardcoded. However, when I have a bit of time, I can make it into a nice little webapp where you can input your position and select a celestial object (or inputs its RA/dec) and see the same plot—possibly along with information about when the sun obscures it.


Which celestial body's length of day is closest to that of Earth?

Methone, a tiny moon of Saturn has an orbital period of 1.01 days. This is approximately 24 hrs and 14 minutes. Ref. 1

https://www.windows2universe.org/our_solar_system/moons_table.html


The helium core becomes significantly denser and hotter after a star has left the main sequence.
Providing the overall mass of the star is greater than about 50% that of the Sun, the helium core will become hot enough to begin fusion whilst the star is a red giant.
From there, the star will pass through phases of burning helium in its core, then hydrogen and helium in shells around an inert core of carbon and oxygen.


There definitely is no definition of what is low and what is high redshift — it is always seen in relation to something. An astronomer working in, say, the interstellar medium of the Milky Way ($z\sim0$) might talk about the ISM of a $z\sim2$ galaxy as being high redshift, whereas another astronomer working the reionization of the Universe (which took place around $z\sim10$) would consider $z\sim2$ to be low redshift. A cosmologist working in the CMB or inflation, however, would probably call $z\sim10$ low redshift. It all depends on the context.

Nevertheless I think that most people would consider $z\lesssim0.1$ to be low redshift.


How to calculate luminosity in g-band from absolute AB magnitude and luminosity distance?

Here is my attempt to reconcile your calculations.

If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by
$$f_{\nu} = 10^{(-48.6-22.5)/2.5} =  3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$

If the distance is 1991 Mpc, then the absolute g magnitude is
$$ M_g = m_g - 5\log d + 5 = -19.0$$

The absolute AB magnitude of the Sun in the g-band (from your source) is 5.12.

The latter tells us that the ratio of the "g-band luminosites" of the galaxy and the Sun are given by
$$\frac{L_g}{L_{\odot,g}} = 10^{0.4(M_{{\odot},g} - M_g)} = 4.45\times 10^{9}$$

One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.

Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does not mean that all the flux from the Sun emerges in the V band!

Your second method requires a figure for $\Delta \nu$, but you haven't said what you have used. The g-filter has a width of around $\Delta \lambda =100$ nm. Using $\lambda \nu = c$ we can say
$$ \Delta \nu  = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$
$$ L_g = 4\pi d^2 f_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$

The two figures would agree precisely if the "g-band luminosity" of the Sun were $4.9 \times 10^{32}$ erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.

Additional: Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m$^{-2}$ nm$^{-1}$ at 467 nm (just search on 'solar spectrum' for many examples) at a distance of $1.5 \times 10^{11}$ m. Putting these numbers together, I calculate a "g-band luminosity" for the Sun of
$$ L_{\odot,g}=4\pi d^2 f_{\lambda}\Delta \lambda = 4.8\times 10^{25} \rm W$$
which is indeed 13% of the solar bolometric luminosity.


What Would it Look Like Inside Jupiter's Great Red Spot?

You would not actually be able to see that much, the sunlight is less than 4% of the strength at Earth's distance to the Sun. That is still enough to see though (Much brighter than Moon light at night in fact). The problem is the transparency of the atmosphere; we can not look in, you can not look out. You would only see a weak brown or red glow below the cloud cover.

However, there is plenty of lightning!
Strong winds and thunder exist too.

Lightning on Jupiter.


Trouble with plotting a graph for the Planck function

Here's what I got for $\nu B_{\nu}(T)$ using Python, which is easy enough to read that maybe you can check against your Excel calculation. I also plotted just $B_{\nu}(T)$ for comparison.

Neither looks like a straight line and they shouldn't. If you print out $\log_{10}(k_B T / c)$ you get about 16.2 which is right in the middle of your range. They've specified a range of frequencies centered roughly on the peak of the Planck distribution.

def Bnu(nu, T):
    top = 2 * h * nu**3 / c**2
    bot = np.exp(h * nu / (kB * T)) - 1.
    return top/bot

import numpy as np
import matplotlib.pyplot as plt

h  = 6.6261E-34 # m^2 kg / s
c  = 2.9979E+08 # m / s
kB = 1.3806E-23 # m^2 kg / s^2 K

T   = 7.3E+05  # K

print "log10(kB T / h): ", np.log10(kB * T / h)

nu  = np.logspace(15.5, 17.5, 201) # Hz

lam = c / nu

print nu.min(), nu.max(), 'Hz'
print 1E+09*lam.min(), 1E+09*lam.max(), 'nm'

B = Bnu(nu, T)

plt.figure()

plt.subplot(1, 2, 1)
plt.title('B(nu, T)', fontsize=16)
plt.plot(nu, B)
plt.yscale('log')
plt.xscale('log')

plt.subplot(1, 2, 2)
plt.title('nu x B(nu, T)', fontsize=16)
plt.plot(nu, nu * B)
plt.yscale('log')
plt.xscale('log')

plt.show()


An incomplete shortlist would be:

(i) If you can measure the maximum possible mass of a neutron star (the most massive neutron star gives a lower limit to that), then it tells you how "hard" or "soft" the equation of state (the relationship between pressure, density and composition) is in the interior (especially at the core), and therefore tells you something about the matter composition and the nature of the strong nuclear force at small separations. Nuclear pasta is unlikely to have a dramatic effect on this.

(ii) Measuring the masses and radii of neutrons stars would directly tell you about the equation of state at different depths inside a neutron star. Again, I think nuclear pasta does not have a major bearing on these diagnostics.

(iii) Measuring the rate at which neutron stars cool, tells you about the interior composition. Quark matter and boson condensations in the core would allow for faster cooling. Abrupt changes to the cooling rate can tell you about superfluid phase transitions. Cooling can also be accelerated by nuclear pasta phases by allowing the URCA process to operate.

(iv) Measuring the response of pulsars to glitches tells you about superfluidity in the interior and crust and the coupling of crust and core will depend on the properties of what is in between - i.e. nuclear pasta.


The Pole star, and all other stars, are a long long long long way away.  Space is big. Also, the pole star just happens to be very close to the axis of the Earth's rotation.
As the Earth goes around the sun, it would cause the position of the pole star (and other stars) to move with a regular annual motion.  The more distant a star is, the less it appears to move. The pole star in particular moves by about 2.1 millionth of a degree.  That is not enough to be noticed by eye.  It is mind boggling (to me) that we can measure that with telescopes!
There is another motion, and that is the apparent motion of stars across the night sky each night.  That is not due to the orbit of the Earth but the rotation of the Earth. The Pole star is close to a line drawn through the Earth from pole to pole, this means that, (unlike most other stars) it also moves relatively little from hour to hour each night; it always appears in the Northern sky.
So the motion due to the Earth's orbit is very small (for all stars)  The apparent motion due to the Earth's rotation is small (due to the Position of the pole star) and so the pole star does not appear to move, it is always in the same position.


How is it possible that the CMB approaches the earth from all directions?

You are comparing apples and oranges. The CMB is like the light from a distant galaxy, which while the galaxy is receding the light approaches us. CMB is the light emitted by universe at recombination. The matter that emitted the CMB is/was receding from us with a very high red-shift.


What are the present, and potentially the future roles of data scientists in Astronomy?

Astronomer is moving towards the big(ger) data era because of many sky survey technologies. The coming ones include, e.g., LSST, JWST, and WFIRST.

By the meaning of survey, it normally means observing the whole sky over a few days, and keep repeating over and over. Also, since most of the surveys are imaging technologies, every pixel in an image is important.

So, you can imagine how much data is incoming per day, and how much data science is important in many fields of astronomy (and cosmology). So, data science is, kind of, taking over when it comes to analyzing data from surveys.


How do they know the newly-spotted gas near the center of our galaxy is molecular without knowing what gas it is?

They know there's molecular gas because they observed emission from the CO molecule in two of the previously identified (atomic-hydrogen-emitting) gas clouds. From the paper:

... we targeted two objects (hereafter, MW-C1 and MW-C2), highlighted
by red boxes in Fig. 1, in the 12CO(2 → 1) emission line at 230.538
GHz with the 12-m Atacama Pathfinder Experiment (APEX) telescope....
We mapped both clouds in 12CO(2 → 1) emission over a 15′ × 15′ field
centred on the peak of the H I emission,... These data revealed
molecular gas outflowing from the centre of our Galaxy.

CO emission is always assumed to come from molecular gas clouds which are, by mass, mostly H$_2$, though the conversion factor needed to turn a CO measurement into an H$_2$ measurement is a matter of debate (this is discussed in the paper).


What are these data in light curve generator

The file format for submitting observations to AAVSO sheds some light on this.

"Visual" data are estimates by visual comparison to nearby reference stars of similar brightness.

U, B, V, R, I data come from CCD cameras using standard filters in the Johnson-Cousins photometric system, with passbands ranging from near ultraviolet to near infrared.
The J and H filters pass longer wavelengths in the infrared.

"Unfiltered w/(V|R) zero pt" are also from CCD cameras, not using one of the above filters, and calibrated using published V- or R-band magnitudes of reference stars.

Tri-R, Tri-G, Tri-B refer to the RGB channels of a color (e.g. consumer DSLR) camera.


whose gravitational pull is more stronger , Neutron Star or Black hole?

The black hole has greater pull.

If you stand beside a massive object (planet, star, black hole or anything else with mass) you will feel a gravitational pull, and unless you do something about it you will start to accelerate towards the object.

So the gravitational pull (or more precisely the gravitational acceleration) depends on two things: How massive the object is, and how far you are from its centre.

Neutron stars are all between about 1.2 and 3 times the mass of the sun, any larger and they become black holes. Objects that are believed to be black holes are know with masses of about 4 times the mass of the sun. A black hole can be of any mass, there is no known way for smaller black holes to form.

Now if you fix the distance (say 150 million km), then the gravitational pull depends only on the mass of the object. A neutron star can be at most about three times the mass of the sun, black holes are nearly all larger than that, so the gravitational pull of the black-hole is greater. But if you find a way to make a small black hole, then it would have lower mass and so less gravitational pull, at the same distance.

You can also look at surface gravity. Black holes don't have a proper surface, but you could use the event horizon instead. Here the answer is easy: The gravitational pull becomes infinite at the event horizon (from a certain point of view), Neutron stars have very strong, but not an infinitely strong pull so the gravitational pull of a black hole is greater than a neutron star.


Is our Solar System moving independently through our galaxy?

The arms of a spiral galaxy are "gravity waves"  Stars orbiting the galaxy are pulled towards them and speed up as they approach, and then are slowed down as they leave. So an arm of the galaxy is not composed of the same stars.  The arm is, in some ways, like a fountain, because it can maintain its shape even though the stars in it are constantly changing.
So the sun will move in its orbit out of the arm and into another, just the same as other stars.


A wormhole, or Einstein-Rosen bridge, is inherently unstable. It will collapse before anything can traverse it. Keeping it stable enough for even a photon to get through requires negative energy densities. And that requires exotic and suspicious new physics in most cases. Technically there are well-known effects which can produce negative energy densities, such as the Casimir effect. But being able to harness such an effect to stabilize a wormhole is beyond current technology (to say nothing of creating the wormhole in the first place). We have no observational evidence to suggest wormholes exist, especially not on the scales you see in Sci-Fi.

So in the rapidly collapsing case it looks like nothing except possibly a spray of subatomic particles (which you might not even know aren't the usual background ones), and in the stable case it's impossible to say because you'd have to know what the exotic matter or other negative densities look like.

As for what it's like to traverse a stable wormhole, as long as the exotic matter doesn't mess with you then it would be pretty much the same as any other spaceship travel. You could just have interesting time travel effects and beat light rays that went around the wormhole. A number of these are mentioned on the wiki page.


Is it unlikely for a meteor to hit the day side of the earth?

Remember that the Earth, and the meteoroid, are both in orbit around the sun. As the meteoroid approaches Earth, it will also curve due to Earth's gravity. If you imagine the Earth to be stationary and the meteoroid to be moving in a straight line your intuition is likely to be wrong.

The result of the complex dynamics is that there is no time or direction from which it is impossible for a meteoroid to have come from. On average the direction of impact is 45 degrees, and the timing is somewhat more likely in the morning when the rotation of Earth combines with its movement through space. A direct impact at noon, from the apparent direction of the sun, is unlikely but possible.

When you look at lists of meteors, many are at night. Except the brightest fireballs, a meteor is much more noticeable at night.

Nasa hs a database of boloids: http://neo.jpl.nasa.gov/fireballs/ By combining the UT time with the latitude information it is possible to get an estimate of the local solar time of the fireball:

This histogram of time of impact of about 250 fireballs since 2007 shows no particular decrease around midday. The documentary graphics may show an apparently fully lit earth, and a perpendicular impact as it is easier to recognise the Day side of the Earth, and it looks more dramatic.


Perseid Meteor Shower Viewing events near Chicago

The Astronomical League's list of astronomy clubs leads to these opportunities:

The Skokie Valley Astronomers have a meeting at the Ryerson Forest Preserve in Riverwoods on Friday the 10th.
Some Naperville Astronomical Association members will be at a stargazing picnic at Cantigny Park in Wheaton on Saturday the 11th.

Like most meteor showers, the Perseids produce more meteors in the early morning, but there should be some in the late evening.


Moving lights in the sky looks like they were moving coordinated

These are almost certainly satellites. Keep in mind that satellites look like slowly moving stars; well, they move more slowly than meteors do at least! If they were moving somewhat slowly and constantly, and if they had no additional flashing lights and no bright colors, they were definitely satellites. NOSS satellites used to be known for flying in groups of three, but it looks like none of the generations of those triplets are still flying in formation. It looks like the Chinese Yaogan satellites are the only ones flying in triplet formation these days; you probably saw one of them.

See here.


As far as I'm aware, there's no direct evidence of an atmosphere at the current time. The abstract of Sicardy et al. (2011) "A Pluto-like radius and a high albedo for the dwarf planet Eris from an occultation" gives a limit of ~1 nanobar on the presence of methane, argon or nitrogen atmospheres during the stellar occultation observed in November 2010.

Hofgartner et al. (2019) "Ongoing resurfacing of KBO Eris by volatile transport in local, collisional, sublimation atmosphere regime" argue that at aphelion there would be a local atmosphere over the warmest region, which they state is a similar situation to the atmosphere of Io. They predict that this local atmosphere could still have significant effects on the volatile distribution on Eris:

  The model results indicate that volatile transport (VT) on Eris, even at
  its aphelion distance of nearly 100 AU, can be significant. The nitrogen-ice
  temperatures are < 30 K and vapor pressures are < 10 nbar but the significant pressure gradients in the local atmosphere regime result in transport of
  nitrogen mass, that integrated over the long timescales associated with such
  a distant orbit, can be significant as compared to the column mass of the
  atmospheres of Triton and Pluto.

So depending on where you are on Eris, the atmosphere (if there is one) may or may not last for the entire orbit. It will take a long time to confirm this, though.


What are the exact physical parameters used to calculate Mercury precession with Einstein theory?

In very a general post-Newtonian metric for a two-body system with the first body oblate, where $M\equiv m_1+m_2$ is the total mass, $\mu\equiv m_1m_2/M$ is the reduced mass, and $p\equiv a(1-e^2)$ is the semi-latus rectum of the orbit, the perihelion advance per orbit is
$$\small\delta\varpi = 6\pi\frac{GM}{pc^2}\left[\underbrace{\frac{2-\beta+2\gamma}{3}}_{\text{GTR}=1} + \underbrace{\frac{2\alpha_1-\alpha_2+\alpha_3+2\zeta_2}{6}}_{\text{GTR}=0}\frac{\mu}{M} + \frac{J_2R^2c^2}{2GMp}\right]\text{.}$$
According to GTR, $\beta=\gamma=1$ exactly. The second term contains various parameters relating to preferred-frame effects and energy-momentum non-conservation. In GTR, they are all identically zero, while experimentally, $\alpha_1\lesssim10^{-4}$ and the rest are many orders of magnitude less than that. Finally, in the third term, $J_2$ is the quadrupole moment of the first body; $J_2R^2 = (C-A)/m_1$ where $C$ and $A$ are moments of inertia about rotational and equatorial axes, respectively. A derivation can be found Will §7.3.

  I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.

This is just the front coefficient in the above: if $T$ is the orbital period of Mercury, then
$$6\pi\frac{GM/c^2}{a(1-e^2)}\frac{1}{T} = \frac{42.98''}{\mathrm{century}}\text{,}$$
and we can restate the GTR prediction of Mercury as
$$\dot\varpi_{\text{☿}}^{\scriptsize\text{GTR}} = \left.\frac{42.98''}{\mathrm{century}}\right.\left[1+2958J_2\right]\text{,}$$
where $J_2$ is the gravitational quadrupole moment of the Sun.

  NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.

If you're referring to one of Anderson et al.'s analyses, the value is $43.13\pm0.14$. Under GTR, this value is compatible with any $J_2$ up to about
$$\begin{eqnarray*}\frac{J_2R^2c^2}{2GMp}\lesssim 0.0067 &\Longleftrightarrow & J_2\lesssim 2.3\times10^{-6}\text{.}\end{eqnarray*}$$
But what is the Sun's quadrupole moment? This is a somewhat sticky question, but the results of lunar laser ranging experiments are incompatible with $J_2$ of more than about $3\times10^{-6}$, while using helioseismology, Pijpers (1998) estimates $J_2 = (2.18\pm0.06)\times 10^{-7}$. Using Pijper's value yields a GTR prediction of $43.01''$ per century, which is compatible with Anderson et al.'s analysis.

  With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.

I have no idea what Plamondon's model is, but I would be rather suspicious of any model that gives "exactly" $43.13$ when that's just the middle of an interval of statistical uncertainty according to one of the many prior analyses of Mercury's orbit.

  When I put my parameters into Einstein metric, I got something unexpected.

If we ignore the post-Newtonian parameters other than $\beta$ and $\gamma$, the static post-Newtonian metric is
$$\mathrm{d}s^2 = -\left(1+2\Phi+2\beta\Phi^2\right)\mathrm{d}t^2 + \left(1-2\gamma\Phi\right)\mathrm{d}S_\text{Euclid}^2\text{,}$$
where the Newtonian potential of the Sun is modeled up to the quadrupole term
$$\Phi = -\frac{M_\odot}{r}\left[1-J_2\frac{R_\odot^2}{r^2}\frac{3\cos^2-1}{2}\right]\text{.}$$
Deriving the perihelion advance of Mercury with this metric is Exercise 40.5 in MTW, and it matches the general result referenced above with $M\equiv M_\odot+M_{☿} \approx M_\odot$ and sans the middle term. For the Sun-Mercury system, $\mu/M\sim 10^{-7}$, so the middle term is even more irrelevant than bounds on the PPN parameters would suggest.

References:

Will, C. M., Theory and experiment in gravitational physics, (Cambridge University Press, Cambridge, 1993), 2nd edition.
Anderson, J. D., Campbell, J. K., Jurgens, R. F., Lau, E. L., Newhall, X. X., Slade, M. A., Standish Jr, E. M., in Proceedings of the Sixth Marcel Grossmann Meeting on General Relativity, ed. by Sato, H. and Nakamura, T., (World Scientific, Singapore, 1992).
Pijpers, F. P., Mon. Not. R. Astron. Soc. 297, L76 (1998). [arXiv:astro-ph/9804258]
Misner, C. W., Thorne, K. S., and Wheeler, J. A., Gravitation, (Freeman,
San Francisco, 1973).


What uncertainty does an error bar signify in astronomy?

In a published scientific paper, the significance of any error bars should be explained. If this has not been clearly stated then peer review is not doing it's job. However, in some fields there may be unstated conventions which slip through without explanation.

In my own field (star formation) plots which shows averages over a large data set would show means with error bars representing one standard deviation, while plots which show single measured values might have error bars representing calculated uncertainties on the measurement, for example due to detection limits or instrumental uncertainties.


How was the Earth-Sun distance originally calculated?

The book The Transits of Venus, by Sheehan and Westfall, describes how Aristarchus used Hipparchus' calculation of the Earth-Moon distance, who in turn used Eratosthenes' calculation of the Earth's circumference, to calculate the Earth-Sun distance.
Aristarchus of Samos was the first to seriously calculate the distance to the Sun, using geometry.  When the Moon is exactly half illuminated when seen from the Earth (first or last quarter phase), then there is a right triangle between the Earth, Moon, and Sun, with the Moon at the right angle.  Then he could measure the angular distance in the sky between the Sun and the Moon, plus the Earth-Moon distance and geometry, to get the Earth-Sun distance.

The most famous ancient estimate of the earth's circumference as made by Eratosthenes of Cyrene (c. 276-196 BCE), the librarian at the great library at Alexandria.  By using a simple gnomon, he found that at Syene, ... the sun at the summer solstice cast no shadow at all: it was exactly overhead.  ... At the same moment, at Alexandria, the shadow cast by the sun shows that it stood 7.2 degrees from vertical.  This difference is equal to 1/50 of a circle.

Using the distance between the cities, Earth's circumference could be calculated.

Once the earth's radius is known, the earth itself can be used as a baseline for determining still greater distances -- the distance to the moon.
[I]t becomes possible to work out the earth-moon distance indirectly from the geometry of [lunar] eclipses.  Using this method, Hipparchus of Rhodes (fl. 140 BCE) worked out that the distance of the moon was 59 earth radii.  It's a good approximation - with 1 1/2 or 2 earth radii of the modern value.

Using the Earth-Moon distance and the separation of the Moon from the Sun in the sky when the Moon was at exactly half-phase, Aristarchus calculated the Earth-Sun distance.

Aristarchus put forward a geometrical argument, based on determining the sun-earth-moon angle at the time the moon's phase is exactly half.  For this angle, which is actually 89.86 degrees, Aristarchus used 87 degrees; the disagreement is more significant that it might appear because the critical quantity is the difference between the angle and 90 degrees.

Because of this Aristarchus only got a value of the equivalent of "5 million miles", much too small.
Phil Plait has, on his old Bad Astronomy site, an article answering a question about how astronomers originally calculated the distance from the Earth to the Sun (the AU, or astronomical unit).
Huygens was the first to calculate this distance with any kind of accuracy at all.

So how did Huygens do it? He knew that Venus showed phases when viewed through a telescope, just like our own Moon does. He also knew that the actual phase of Venus depended on the angle it made with the Sun as seen from the Earth. When Venus is between the Earth and Sun, the far side is lit, and so we see Venus as being dark. When Venus is on the far side of the Sun from the Earth, we can see the entire half facing us as lit, and Venus looks like a full Moon. When Venus, the Sun and Earth form a right angle, Venus looks half lit, like a half Moon.
Now, if you can measure any two internal angles in a triangle, and know the length of one of its sides, you can determine the length of another side. Since Huygens knew the Sun-Venus-Earth angle (from the phases), and he could directly measure the Sun-Earth-Venus angle (simply by measuring Venus' apparent distance from the Sun on the sky) all he needed was to know the distance from Earth to Venus. Then he could use some simple trigonometry to get the Earth-Sun distance.
This is where Huygens tripped up. He knew that if you measured the apparent size of an object, and knew its true size, you could find the distance to that object. Huygens thought he knew the actual size of Venus using such unscientific techniques as numerology and mysticism. Using these methods he thought that Venus was the same size as the Earth. As it turn out, that is correct! Venus is indeed very close to being the same size as the Earth, but in this case he got it right by pure chance. But since he had the right number, he wound up getting the about the correct number for the AU.

Basically, Huygens used good methods, except for using "numerology and mysticism" to determine the size of Venus.  He was lucky that Venus was almost the size of Earth; that made his estimate for the AU pretty close.
Not long after, Cassini used the parallax of Mars to determine the AU.  (Same article as linked above.)

In 1672, Cassini used a method involving parallax on Mars to get the AU, and his method was correct.

Parallax is the apparent difference in angle observed due to the differing observing positions.  The smaller the parallax, the larger the distance.
However, the precision of the resultant calculation depends on the precision of the observations, and measurements of the parallax aren't that precise.
In 1716, Edmond Halley published a way to use a transit of Venus to accurately measure the solar parallax, i.e. the difference in the Sun's position in the sky due to observers at different latitudes.

Because of the latitude difference of the observers, Venus would appear to move along chords of different length over the disk of the sun.  The motion of Venus being nearly uniform, the length of each chord would be proportional with the duration of the transit.  Thus, observers would not actually have to measure anything; they would only have to time the transit.  Fortunately, existing pendulum clocks were more than sufficiently accurate for this purpose.

They could time the transit, which would last hours, with great precision.  But they had to wait until the next transit of Venus in 1761.  Then, observers observed the black drop effect, which made it very hard to time the event from start to finish precisely.

The black drop effect cannot be eliminated altogether, but it is much more ever in observations made with telescopes of imperfect optical quality (as many of those used at the 1761 transit were) and in boiling or unsteady air.  Confusion about the times of the internal contacts ... yielded contact times that differed among observers, because of the black drop, by as much as 52 seconds.
In the end, there as a wide range of published values, from 8.28 arc-seconds to 10.60 arc-seconds.

But then there was the transit of 1769.  Observations in Norway and in the Hudson Bay were made for the northern observations, and Captain James Cook was sent to what is now Tahiti to make a southerly observation.  Jérôme Lalande compiled the figures and calculated a solar parallax of 8.6 arc-seconds, close to the modern figure of about 8.794 arc-seconds.  That calculation yielded the first fairly accurate calculation of the Earth-Sun distance, of 24,000 Earth radii, which given the Earth's radius of 6,371 km, about 153,000,000 km, the accepted value being about 149,600,000 km.


angle to the equator of the Sun on sky at a given time

Depends on the precision you need.. The sun's plane of rotation coincides with the plane of rotation of the solar system.
And for the latter, there is already a name in astronomy, the Ecliptic. Depending on your latitude on the planet $\delta$, the ecliptic plane cuts through your local horizon in east and west at an angle of $\alpha = 90°-\delta - 23.5°$.
This is the maximum height of the sun at midday and the sun's height varies thus between 0 and $\alpha$.

I think this would then be the angle you're looking for.

Another recent answer pointed out the angle between solar axis of rotation and the Ecliptic is $7.25°$. Translating that into the angle unter which one sees the solar equator, it is still $\alpha$, up to a small correction factor of $- \frac{R_{\odot}}{1 AU} \sin(\delta + 23.5°)$ which can be maximally $-0.05°$.


What is the naming convention for temporary moons?

These are just the standard minor planet designations.
Provisional designations
As described on the MPEC page on New- and Old-Style Minor Planet Designations, the new-style provisional designations for minor planets (like 2006 RH120) are in the form year, letter indicating the half-month, letter indicating the sequence within the half month, and number indicating the cycle of the second letter (no number = 1st cycle, 1 = 2nd cycle, etc.)
The letters for half-months omits the letter I and doesn't use Z as there are only 24 half-months in the year. The second half-month in a month starts on the 16th for all months.
The letter indicating the order is the alphabetic sequence, again skipping I but this time including Z.
Old-style provisional designations were used in the 19th century and are not relevant here.
Temporary designations
These are assigned by observatories according to their own rules, rather than by the MPEC. I think three of the ones you list (YX205B9, WT1190F and XC83E0D) were assigned by the Catalina Sky Survey, but I haven't seen any documentation of what their internal rules for assigning identifiers is. These objects were confirmed to be artificial before they had sufficient orbit determinations to be assigned minor planet designations, so never received provisional designations.


Why is the moon that special? Is it really a well built coincidence?

If the "perfect rotation" you describe is the fact that same face of the moon always faces the earth, this is explained by the 'Giant Impact Hypothesis' which also explains the formation of the moon. The molten earth was impacted by a large mass causing a part of the liquid earth to be flung out into space creating the moon. For a time, the two celestial bodies to be (the earth and moon) were connected to one another by some bridge of mass until eventually the centripetal force of their rotation finally pulled them apart. In fact, the two bodies are still moving apart today.

https://en.wikipedia.org/wiki/Giant-impact_hypothesis


What can the total stellar mass of galaxies tell us?

Actually, the light of most galaxies is dominated by red giants, which are predominantly not very massive stars (more massive than the Sun sure, but less than a few solar masses). These are much more common than short-lived high mass stars, which in any case don't exist in many galaxies (e.g. old ellipticals).

In galaxies that not gas-rich - again most ellipticals, the stars form most of the baryonic mass and most of the stellar mass is in low-mass stars. Even in galaxies where there is lots of gas (the dust is a negligible mass component), there are ways of measuring this too. Estimating the total mass of baryonic versus dark matter is essential in understanding the formation and evolution of galaxies, since they behave differently.

Stars less massive than about 0.9 solar masses have had insufficient time to evolve. The baryonic mass in these stars is locked away and cannot be recycled into other stars. Knowing the fraction of mass in this situation is important in understanding the chemical evolution of galaxies.


How does one measure velocities of far-off, bright objects

In general the total velocity of a faraway object is not obtainable, only the component along the line of sight. However, if a star is not too far away, and its position on the sky is measured very accurately with an interval of some years, the velocity component perpendicular to the line of sight can be determined, yielding together with the redshift the total velocity. The recently launched Gaia spacecraft is expected to be measure the proper motion of millions of stars in this way.

In distant galaxies, this is not possible. If one is interested in, say, the 3D velocity dispersion of the gas, the 1D dispersion is measured, and assuming (fairly) an isotropic dispersion, this number is then multiplied by $\sqrt{3}$ to get the total velocity.

The distance to which parallax is measurable depends on the angular resolution of your instrument. For milliarcsec resolution, parallax measurements are good out to roughly 1 kpc. If you wait long enough, the motion around the galactic center increases your baseline such that larger distance can be probed.


Is it possible to witness a star's death?

Naked eye nova are fairly common, several per year. Here's one.
Naked eye supernova are far rarer. SN1987a in the large Magellanic cloud was naked eye visible (vid). From this list, it appears the supernova in 1987 was the most recent naked eye supernova.
There was a naked eye gamma ray burst in 2008, but I don't think anyone actually got outside in time to see it.

If you have 50 years to look at the stars, you might see a supernova.
If you have a small telescope, you can pick them up pretty regularly in nearby galaxies.


Running the math for a 5 meter long pendulum and 1 kg mass, I get an amplitude of 0,017 mm.

You are off by quite a bit. There is essentially no horizontal deflection when the Moon is at the horizon. The maximum horizontal deflection occurs when the Moon is about 45 degrees above or below the horizon.

The tidal acceleration at some point on the surface of the Earth due to some external body (e.g., the Moon) is the difference between the gravitational acceleration toward that external body at the point in question and the gravitational acceleration of the Earth as a whole toward that body:

$$\vec a_{rel} = GM_\text{body} \left(\frac {\vec R - \vec r} {||\vec R - \vec r||^3} - \frac {\vec R} {||\vec R||^3} \right)$$
where

$M_{body}$ is the mass of the external body,
$\vec R$ is the displacement vector from the center of the Earth to the center of the external body, and
$\vec r$ is the displacement vector from the center of the Earth to the point in question on the surface of the Earth.

The resultant force looks like this:

Note that the tidal force is away from the center of the Earth when the Moon is directly overhead or directly underfoot, toward the center of the Earth when the Moon is on the horizon, and horizontal when the Moon is halfway between directly overhead/underfoot and on the horizon.

The tidal force is maximum when the Moon is directly overhead, and even then it's only about 10-7 g. You need a sensitive instrument to read that. A simple pendulum or a simple spring will not do the trick.


http://spaceplace.nasa.gov/review/dr-marc-solar-system/gas-giants.html

We think of a gas as something very . . . well, airy. After all, air is the gas we all know and love. We breathe it and fly planes right through it with no trouble. So it makes sense to think that a gas planet must be like a big, airy cloud floating out in space.
Saturn in true color.

The bigger a planet becomes, the heavier is the material weighing down on its center. Think of how it feels to dive under water. If you are wearing a face mask, you notice that as you dive deeper, the mask presses harder and harder on your face. Also, your ears start feeling the pressure even at 2 or 3 meters (5 or 10 feet) below the surface. The pressure you feel on your body is due to the weight of the water above you. The deeper you go, the heavier the water above you and so the greater the pressure on your body. Even on Earth's surface, each square inch of your body experiences 14.7 pounds of pressure due to the weight of the atmosphere above you. If you could dive down to the center of Earth, the pressure on your body would be about 3.5 million times as great! The center of Jupiter is more than 11 times deeper than Earth's center and the pressure may be 50 million to 100 million times that on Earth's surface!

The tremendous pressure at the center of planets causes the temperatures there to be surprisingly high. At their cores, Jupiter and Saturn are much hotter than the surface of the Sun!

Strange things happen to matter under these extraordinary temperatures and pressures. Hydrogen, along with helium, is the main ingredient of Jupiter's and Saturn's atmospheres. Deep in their atmospheres, the hydrogen turns into a liquid. Deeper still, the liquid hydrogen turns into a metal!

But what's at the very center of these planets? The material becomes stranger and stranger the deeper you go. Scientists do not understand the properties of matter under the extreme environments inside Jupiter and Saturn. Many different forces and laws of nature are at work, and the conditions inside these planets are very difficult to create in a laboratory here on Earth. But you can be sure that you wouldn't be able to fly through these bizarre materials! As we now know, the gas giants are much more than just gas.


How can you determine the distance of a star using the magnitude equation?

This is all absolutely true (pun intended). Finding absolute magnitudes is hard.  For many types of star, we don't really know their absolute magnitude and so we don't really know their distance.
For close-by stars we can get the distance by measuring the parallax (how far the star appears to move over a year due to the orbit of the Earth. Nearer stars appear to move more) From this we can work out absolute magnitude.
For other stars we can get their absolute magnitude in other ways. For example, a type of large bright star called a "Cepheid variable" will have a pulse rate P that is connected to it's absolute magnitude $M_\mathrm{v}$ by the following (empirically derived) formula:
$$M_\mathrm{v} = -2.43(\log_{10}P - 1) - 4.05$$
With this you can easily measure the period in days, and use that to find the absolute magnitude.  The apparent magnitude is easy to measure, it is how bright the star appears to be on Earth.
And with apparent and absolute magnitudes you can find the distance of the star.
Stars like Cepheid variables are called "standard candles", since their brightness can be determined exactly


What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?

Source: JPL Article

"The gravity anomalies along the border structures are interpreted as ancient, solidified, lava-flooded rifts that are now buried beneath the surface of the dark volcanic plains, or maria, on the near side of the moon."

Because of the lava under Oceanus Procellarum, it's heavier/massive so it has a stronger gravitational pull than the area surrounding it.

Hope that helps!


L2 point ephemeris (celestial mechanics)

Bad news, this type of SPK file has a different sort of interpolation that is not supported by  the jplephem package (Hermite interpolation vs Chebyshev polynomials). You can find this out by doing:

In [1]: print(len(kernel.segments))
1
In [2]: print(kernel.segments[0].describe())
2415020.50..2506696.50  Earth Barycenter (3) -> Unknown Target (392)
  frame=1 data_type=12 source=Sun-EarthMoon L2

which shows that the segment is of data type 12. Looking this up in the NAIF SPK documentation shows that this is Type 12: Hermite Interpolation --- Equal Time Steps; looking in the jplephem code in the spk.Segment._load() routine, shows the code that produces the error you are seeing and the supported types (2 and 3). According to the NAIF SPK docs these are Type 2: Chebyshev (position only) and Type 3: Chebyshev (position and velocity).

I think your options are either:

add support for this type of interpolation to jplephem (the math is at least documented in NAIF docs),
see if SpiceyPy has support for this more unusual type of SPK/BSP,
find an alternative source for an L2 ephemeris.


Is there a limit to the size of a black hole?

General Relativity does not limit the size of a black hole, and we have good (indirect) observations of black holes with a mass of more than a billion suns.

Black holes are not object that are held together by gravity, and which could be ripped apart if forces opposing that gravity were increased. They are (in GR) singularities: a complete collapse of the matter to a point. There is no "object" inside a black hole that could be emitting radiation.

Now if all the matter in the universe were to be a black hole, then that singularity would be in all our futures. We wouldn't "see" the singularity because it is in the future, never in the past. In fact a universe in which is one black hole could look rather like the one in which we are living.


Do planets migrate suddenly or gradually?

Thommes et al. (2001) ran simulations and found that, at optimal conditions (namely, a planet of ~ 10 Earth masses), migration can be complete with ~ 100,000 years. Note that this was done before in-depth research was done on the Nice model, which is very similar. However, the mechanisms are different, as are the planet masses. The difference in timescales is dramatic.

Levison et al. (2007) did explore their own model - the Nice Model. They found that it took 60 million years to 1.1 billion years for Jupiter and Saturn to break their resonance. The period of encounters and scattering lasted for 878-885 million years, followed by a period of eccentricity damping lasting for 0.3 million years to 4 million years.

What was relatively quick was the ejection of the hypothetical 5th giant planet. The change in the other giants' orbits, however, was not.

So yes, planetary migration on this scale takes a long time. A very long time.

For some really interesting results, see the graphs of semi-major axis vs. time from the various four-, five-, and six- planet models of Nesvorný  & Morbidelli (2012). There are some incredible oscillation among the orbits of Uranus and Neptune in some of the simulations, which is eventually slowly damped.

  Fig. 14.— Orbit histories of the giant planets in a simulation with five initial planets. See the caption of Fig. 1 for the description of orbital parameters shown here. The five planets were started in the (3:2,3:2,2:1,3:2) resonant chain, Mdisk = 20 MEarth and B(1). The fifth planet was ejected at $t =$ 6.1 Myr after the start of the simulation.

There's another relevant passage, if you're looking at setups with six giant planets:

  The instability typically occurred in two steps, corresponding to the ejection of the two planets. Sometimes, as in Fig. 18, the ejection of the two planets was nearly simultaneous, but most of the times there was a significant delay between ejections. This was useful because the first planet’s ejection partially disrupted the planetesimal disk and reduced its capability to damp e55, which was then excited by the second planet’s ejection. While this mode of
  instability can be important, we would need to increase the statistics (>100 simulations for each initial condition) to be able to properly resolve the small success fractions in the six-planet case.


Sunrise - from earth - sun position or when sun rays reaches earth?

It's when we see the Sun's rays and includes refraction. It also is the moment we see the first glimpse of the sun. Think of it this way: we humans need to know sunrise and sunset for our human activities; the fact that the sun is about eight light-minutes away doesn't concern us, doesn't affect the things we do. BTW, this is also true of solstices and equinoxes.


Is a black hole a three-dimensional or four-dimensional object?

It would help to clarify your question on what exactly you mean by a 4th dimension.  Space-time for example is 4 dimensional everywhere, but it's 3 spatial dimensions and 1 time dimension.

Quantum theory allows for extra dimensions, but they're very small, smaller than an atom and impossible to measure by normal means.  CERN has looked for those folded dimensions but has found no evidence of them, and string theory defines 10 or 11 dimensions, most of them we can't see or measure, if they exist at all.

If you mean 4 spatial dimensions, then the answer is no, with a capital N.   There's no evidence that an extra spatial dimension would form in a black hole.  There's nothing in relativity that suggests that might happen.  An extra time dimension . . . probably not either, but it's fun to think about what 2 time dimensions would look like.

What happens inside a black hole is that one of the dimensions becomes like time.  It only moves in 1 direction, but that doesn't make it a 4th dimension, it just changes the nature of 1 of the dimensions.

If you actually had a 4th dimension (and the book flatland discusses this in pretty good detail), but if you actually had that, very strange things can happen, for example, you can be flipped over, like flipping a card over, which can represent 2 dimensions, from face down to face up.  Your left side would become your right side.  Your heart would be on the other side of your chest and all the left handed molecules in your body would become right handed molecules, which would make your ability to absorb and use nutrients that you digest problematic.   It might even make things taste and smell different.

You can't tie your shoes in the 4th dimension, but even if you could, your shoes could slide off your feet, no matter how well you tied them.   Food could fall out of your stomach, not through one of the two normal exits, but it could just slip out, like a ball rolling off a table.   Anything not attached could slip apart.   Sex would be much more difficult.  Bolts and nuts could just move sideways and detach, unless there was some electrostatic charge keeping them in place, in fact, the pressure that forms natural between a screw and a nut might cause all nuts and bolts to fly apart at considerable velocity once you entered 4th dimensional space.

Two people could also stand in exactly the same place, just in a different 4th dimensional space.   That brings a whole new aspect to the how many people can you fit in this Volkswagen game, as you could fit theoretically any number of people in the same Volkswagen.

A 4th spatial dimension is probably impossible, because it breaks too many rules and laws as we know them, and it's never been observed.

Black holes, by comparison, can do strange things like slow down time, and they aren't fully understood, but a 4th spatial dimension probably isn't possible, unless you mean at quantum distances, in which case . . . maybe.

Feel free to read this for a different approach to a somewhat similar question.


How fast can and do gravitational waves travel?

Gravitational waves travel at the same speed as the speed of light in a vacuum, in general relativity. In special relativity, this is the maximum speed for any interaction in nature.

If gravity was faster than the speed of light, then predictions we make according to the bending of light would be incorrect.
If gravity was faster than the speed of light, then the orbits and positions of the planets would be completely different.

Those are my two main points, to help me prove to you that gravity travels at the speed of light.

Also, light is not the only thing in the universe which travels at the speed of light. It is about all massless particles! Photons and gluons are two good examples I can think of of massless particles, I'm not sure about gravitons though - but I'll still include them.


Was comet ISON in a Kepler orbit prior to disintegrating?

To your first question, yes, it was in a Kepler orbit, (this interactive model shows just how it appeared). ISON never completed a full circuit around the Sun since this was it's first time coming out of the Oort Cloud and it disintegrated.

Further answering, the orbit of a body around another in space is not affected by it's mass. Remember the experiment that proves two object fall at the same speed regardless of it's mass (here the experiment is made on the moon with a feather and a hammer) and remember that orbiting is just falling at a certain speed and angle in respect to another body.


How do higher numbers affect orbital resonance?

General Explanation
In the history of our Solar System, the orbits of Solar System bodies have gradually evolved, passing through many "higher number" orbital resonance ratios. Only the stabilizing effects of the most stable "lower number" resonance ratios have been enough to overcome the forces causing orbital migration.
When we look at the orbital resonances in the current Solar System, we get a snapshot in time that just shows us the resonances that have been powerful enough to overcome other forces that cause orbital migration.
A rigorous mathematical explanation for why lower-number resonances have a greater effect than higher number resonances may be resonance-type specific and most have enough math to choke a cat.  Intuitively, if you push a child on a swing 28 times for every 29 times they swing, you aren't going to have much of an effect.  Orbital resonances are similar.
Specific Example (Mercury Spin-Orbit Resonance):
Common wisdom holds that orbiting bodies will eventually achieve a 1:1 orbital/spin resonance with a main body, like the Earth's Moon.  Though the Moon was spinning faster in the past, tidal torque caused it's spin to slow until the spin rate was equal to the orbital period.  As the Moon spun down, it passed through many other ratios of orbital periods.
The case for Mercury's orbital/spin resonance around the Sun is a lot more complicated.  Scientists and astronomers are still working out how and why Mercury was caught in a 3:2 spin/orbit resonance as it spun down, rather than passing through the resonance. From this excellent news article:

Tidal dissipation in the planet and the ensuing deceleration of its
spin inevitably carry the planet through a sequence of the spin-orbit
resonances. The question then becomes in which of these resonances the
planet should eventually get trapped...
the international team revisited the problem of tidal evolution of
Mercury's spin and found that the 3:2 resonance is indeed the most
probable end-state. The frequency-dependent tidal torque acts as an
efficient trap for the planet trying to traverse a resonance. The
efficiency of the trap strongly depends on the value of orbital
eccentricity, as well as on the temperature and viscosity of Mercury's
mantle.

Note:
Forces causing migration of planetary orbits include (but are not limited to):

Loss of planetary angular momentum due to gaseous drag in the early proto-planetary circumstellar accretion disk.

Tidal forces resulting in exchanges of angular momentum.

Collisions.

Various n-body gravitational interactions.

The Yarkovsky effect.

Resonances with other bodies.


Why is our solar system "tipped" about 63° with respect to the plane of our galaxy?

What you could think at first, regarding the orientation of any planetary system, is that it should be roughly in the plane of the galaxy, simply by angular momentum conservation.

But, when you take a look at observations, you see that protoplanetary disks orientation is not what you would expect, with no preferential orientation (protoplanetary disks are embryo of planetary systems, that makes them interesting). In the following figure, the orientation corresponds to the inclination between the line of sight and the rotation axis of the disk.

Why there is this distribution of orientation?

The angular momentum scenario is nice but far to simple: star formation occurs in gas clouds in the interstellar medium, and these clouds are known to be turbulent (Larson, 1981). Turbulence simply disrupts the gas, and is dominent over the global angular momentum of the cloud. Actually, you can even test that with numerical simulations of star formation: put an initial angular momentum consistent with observations, and some turbulence (subsonic or slightly supersonic, also in accordance to observations), and you will get a misalignement of the rotation axis, due to turbulence.


What are realistic and unrealistic values for the high energy gamma ray spectral indices of pulsar wind nebulae?

You might find these websites for your answers:

https://www.cv.nrao.edu/course/astr534/SynchrotronSpectrum.html
http://astronomy.swin.edu.au/cosmos/S/Synchrotron+Emission
https://ui.adsabs.harvard.edu/abs/2006MNRAS.371.1441S/abstract

We have a large amount of high energy emission these days, and the third reference shows the distribution from 394 GRBs.


At what gravity does a body's gravitational lensing become observable?

The article 'A practical relativistic model for microarcsecond astrometry in space' (Klioner 2003; caution, very heavy duty maths...) describes the framework which was developed to support the data processing for the ESA Gaia astrometric space mission. This mission has the goal to measure the position (astrometry) of stars and other objects to approx. 20 microarcseconds (one millionth of an arcsecond, which is 1/3600th of a degree). In order to do this, you need to account for all the sources of error and deflection, such as light bending near massive objects, to a level that is well below you expected accuracy (this level of systematic error was chosen to be 1 microarcsecond, hence that part of title).

Section 6 of the paper discusses gravitational light bending and Table 1 gives the size of the effect for various bodies in the Solar System. The column headed by $\delta_{pN}$ gives the size of the effect in microarcseconds ($\mu$as). As expected the Sun has the largest effect at $1.75 \times10^6\,\mu$as or 1.75", with Jupiter second etc. The next two columns gives the angular separation a light ray has to pass to the body to cause a $1 \mu as$ or $10 \mu as$ effect. So for example, any light ray from a distant object passing within $11.3^\circ$ of Jupiter will cause at least a  $10 \mu as$ effect. The smallest bodies that have an effect larger than $1 \mu as$ are the dwarf planet Ceres (mass $8.958 \times 10^{20}$ kg), and the moons Dione ($1.05 \times 10^{21}$ kg) and Umbriel ($1.27 \times 10^{21}$ kg) - these masses are several thousand times smaller than the Earth for example.

Now the level of precision needed by Gaia is far more precise than and the deflections are tens of thousands times smaller than you would see in an optical image with a regular telescope e.g. Eddington's confirmation of general relativity during the 1919 total solar eclipse (space.com story). The Klioner (2003) paper includes a formula to estimate the size of the effect of the light deflection as a function of the density but you need to know what level of deflection you care about to get an answer. The formula (number 35 on page 11 of the paper) for the radius $L$ of a body producing a light deflection than $\delta$ is:
$$
L\geq \left(\frac{\rho}{1\,\textrm{g cm}^{-3}}\right)^{-1/2}\left(\frac{\delta}{1\,\mu\textrm{as}}\right)^{1/2}\times 624\,\textrm{km}
$$
where $\rho$ is the mean density of the body. Once you have chosen a level of deflection/lensing you want ($\delta$), you can chose suitable densities (e.g. asteroids are $\sim2\,\textrm{g cm}^{-3}$, Earth and other rocky planets are  $\sim5.5\,\textrm{g cm}^{-3}$)


What is the naming convention for craters on Mercury?

A quick search returns that wikipedia has an extensive list how different things are named on various celestial bodies.
For Mercury's craters it quotes "Famous deceased artists, musicians, painters, authors". As source of its wisdom it gives the IAU working group on planetary system nomenclature as authoritive source for them - the source which decides on official naming.


Can a solar eclipse be directly overhead?

They could be directly overhead if you live in the tropics.

The tropics are the regions near the equator, between 23 degrees North and 23 degrees south. In the tropics, the sun and the moon can pass directly overhead. If both pass overhead at the same time there will be a solar eclipse directly overhead.

The eclipse of July 11, 1991 was visible directly overhead, (to within a degree) and in totality, for observers in the Nayarit province in Mexico


Yes

Tidal locking is a mutual process, and both the satellite and the parent body will eventually
permanently face each other. However, while tidal locking of a satellite can happen relatively quickly (that is, within the life span of the solar system), tidal locking of the parent body is extremely slow and will in a practical sense never happen (the current age of the universe is not enough in most cases).

The relative mass of the satellite and the parent body is one of the most important factors:

Charon and Pluto with a mass ratio of just 8.1 are already mutually locked.
The Moon and Earth has a relatively high mass ratio of 81. This causes a measurable effect on the Earth.

Compare that to the mass ratio of Juptier and Europa of 39,600 (or Io: 21300, Ganymede 12,800, Callisto: 17,600). While the time required for a locking of Jupiter itself is difficult to measure due to the fact that we don't have the necessary data about the moons, this is not a noticeable effect before the Sun dies.

Tidal locking of the parent body can most likely be justified by conservation of energy, but conservation of momentum gives a more intuitive view: As the satellite slows down and moves away from the parent body, its angular momentum increases (angular momentum is proportional to distance times velocity, and the orbital velocity is proportional to the square root of distance). This angular momentum must be supplied from somewhere. The only source for that is the rotation of the parent body. Even if all the moons of Jupiter were to completely escape, the angular momentum drain is not enough to stop the rotation of Jupiter.


Why is every year the same number of days despite the gravity in the solar system?

Derived from this paper the interval between March equinoxes for 1989/1990 has been 4 minutes 28 seconds longer than the corresponding interval for 1987/1988.

So the length of a year can vary several minutes, e.g. by gravitational effects of other planets. For calendars a mean year is used. Atomic clocks run far more precise than the more or less periodic motion of Earth.

All planets aligned don't cause anything grave. A similar kind of sysygy has been 1982.


What physical features determine if a planet is a major, minor or dwarf planet?

Planets

For a body to be classified as a planet it must have a few physical characteristics:

Mass

It must have enough mass to have a strong enough gravity to overcome electrostatic forces to bring it to a state of hydrostatic equilibrium.

Hydrostatic equilibrium is important because early in a planets life it is nearly entirely fluid, crust and all

Internal Differentiation

The life cycle of a planet essentially leads to denser heavier metals being at the centre of a planet, surrounded by a mantle which must at some point have been fluid (it can still be solid and called a planet as long as it used to be fluid.)

Atmosphere

This is usually driven by its mass but a planet should have an atmosphere. This means it should be massive enough to have a strong enough gravity to hold some gasses to its surface.

More massive planets are capable of keeping lighter gasses, such as hydrogen, bound to them too. I.e Jupiter.

Magnetosphere

A magnetosphere suggests that the body is still geologically active. This means they have flows of elements that conduct electricity in their interiors.

As specified by the IAU:

  A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

Minor Planets

Minor planets  are things such as asteroids, they are usually small mass, rocky bodies. This is different from small mass bodies that might be predominantly ice or water.

Dwarf Planets

Size and Mass

The IAU have not specified an upper or lower limit for mass to be considered a Dwarf Planet, therefore it is predominantly determined by one other feature called:

Orbital Dominance

Orbital dominance is achieved when a body has cleared it's orbit of all other bodies.

For example planets are able to remove small bodies out of their area through impact, capture, or gravitational disturbance.

Any body that is incapable of doing so is therefore classified as a dwarf planet. So if Jupiter was incapable of clearing its neighbourhood of bodies then it too would be a Dwarf planet.


Electrical potential difference between planets

I dunno how or why a planet would have a significant net charge, but supposing it did, here's some initial calculations.

If the planets have a different net charge, you can treat them as capacitor plates.
Capacitance in its simplest form is

$C = \frac{\epsilon *A}{d}$
The permittivity $\epsilon$ of vacuum space is 8.85 x 10-12 farad per meter .
Let's choose a nearby (nonexistent) planet at 5 LY = 5 * 9.461e+15 m .

Looks like the capacitance is pretty close to zero.  So the energy available,

$E = \frac{CV^2}{2}$  is going to be nonexistent unless there's a seriously large voltage.


How could we search for life in TRAPPIST-1

As far as I know, and aside from direct communication, there's only one way to "find life" on another planet. That method is to look for molecular oxygen in the atmosphere. Molecular oxygen hates existing by itself and loves to combine with other atoms and molecules through oxidation. The Earth's atmosphere is currently ~30% $\mathrm{O_2}$, but that's only because we have a huge amount of plant life (primarily algae) constantly producing it. If you removed all life from the Earth, the oxygen in the atmosphere would quickly oxidize out and you'd be left with an atmosphere primarily of nitrogen and a few smaller constituents.

What this means is that if we do see a substantial amount of oxygen on a planet, the best bet is that that oxygen was artificially produced by life. It's not ironclad evidence, but its a very strong indicator. We can search for oxygen through a few techniques that already exist. They're still being developed, but the technology is very quickly coming to fruition. Ultimately, you'd need to look at the spectrum of the planet's atmosphere. This can be achieved by either looking at the star's light as it passes through the planet's atmosphere, or else observing at numerous wavelengths and seeing how the transiting planet's radius changes (due to changing atmospheric opacity at different wavelengths).


The Great red spot used to be bigger. Over the course of twentieth century, it lost half of its size and between 1996 to 2006, its area diminished by 15%. If the shrink rate is constant, then the spot will vanish within 20 years. However, computer simulation showed that Jovian weather is stable enough to sustain smaller red spots as Hubble space telescope showed two smaller red spots besides the great red spot indicating that small red spots will persist even if the Great red spot vanishes. The reason for this shrinkage is unknown but scientist suggested it has something to do with Jovian equivalent of global warming and climate change.
Unlike Earth, Jupiter does not have solid landmass so energy lost due to friction is out of question. This explains the long lasting nature of Jovian storms. The Great Red Spot is indeed long-lived, even by Jupiter standards. Researchers are able to pinpoint that Jovian storms that are located in cloud bands with the same direction of rotation tend to be longer lasting. So, the shrinkage of this spot might indicate that the spot is drifting from the clouds bands.
The Great Red Spot is confined by an eastward jet to its north and a westward jet to its south, confining the storm to a constant latitude. However, it has undergone considerable changes in longitude over time, and recent evidence suggests that its rate of westward longitudinal motion is increasing. The bands have also undergone little change in latitude over the time during which they have been observed. So, researchers suggested that this dynamic change of the cloud bands and drifting of the storm might be the reason for the change in the shape of The Great spot.
Although the reason has not been proved yet, researchers are trying to trace the evolution of the Great Red Spot, analyzing its size, shape, color, drift rate and storm’s internal wind speeds from the information available from spacecraft and hope they will provide an explanation in future.
Reference:

The Things that Nobody Knows: 501 Mysteries of Life, the Universe
and Everything William Hartston, Atlantic Books Ltd, 2011.
Celestial Calculations: A Gentle Introduction to Computational Astronomy, MIT Press, 2019.
https://cosmosmagazine.com/space/jupiter-s-great-red-spot-is-shrinking/
https://earthsky.org/space/jupiter-great-red-spot-getting-taller-as-it-shrinks
Historical and Contemporary Trends in the Size, Drift, and Color of Jupiter's Great Red Spot by Simon et.al., The Astronomical Journal, Volume 155, Number 4, 2018 (link)