Which unit is most suitable in comparing needed light intensity for night reading on the Moon, vs on Earth?

Readability requires taking into account the sensitivity of the human eye to various wavelength. This is called Illuminance.
Physics knows Lumen (flux weighed by human perceptibility) and Lux (flux per unit area weighed by human perceptibility).
Quoting that, different light sources have roughly these values:
Source             Lux
Full daylight   10,000
Overcast day     1,000
Very dark day      100
Twilight            10
Deep twilight        1
Full moon            0.1
Quarter moon         0.01
Starlight            0.001


Cepheid variable stars and exoplanets transiting stars have very different light curves (the relation between brightness and time).

Exoplanet light curve from NASA:

Cepheid variable light curve from astronomynotes.com:

Also, as Rory Allsop points out, the scale of the change in brightness is very different. Cepheid variables can be seen at great distances, which allows us to use them to measure distances by their regular period-brighness relation, whereas exoplanets can only be found more locally by this method.


Dataset containing list of known globular clusters

So far we have about 150ish known globular clusters in the milky way. A list of all the known ones can be found HERE. This includes their locations (may be slightly shifted by now as they were correct as of 2011, the time of writing) and where applicable, the Messier/NGC/IC numbers are included for ease.


What is the cut-off point of $\frac{\mathrm{d}F}{\mathrm{d}r}$ beyond which no tidal locking will occur?

I can't comment here because I don't have enough reputation, but this is very related to your problem.

https://en.wikipedia.org/wiki/Tidal_locking#Timescale

Tidal locking will always occur albeit potentially not in a Hubble time. In order for there to be no tidal locking, you would need:

$$\frac{\mathrm{d}F}{\mathrm{d}r} = 0$$

Which any extended object will not have. Problems like this always involve timescales and not definite yes/no answers. I hope this helped.


How can I find a date when the moon at zenith over a specific spot on the equator?

Unlike the sun, which follows exactly the ecliptic and so is directly on the zenith on the date of the equinoxes, the moon is tilted by 5 degrees relative to the eclipic.  And the direction of this tilt changes over 18 years.
Moreover the motion of the moon is quite fast so even if it is rising due East, by the time it culminates it will have moved and so not be exactly on the zenith.
As you see the moon's orbit is actually one of the more complex orbits in the solar system. It will be quite close to the zenith once a day, and very close once a month, The only way to find if the moon is (within 0.5 degrees) of the zenith is to use planetatium software to model its motion. Possiblities include "stellarium" or if you have programming skills then "astropy".


Do astronomers generally agree that the distinction between comets and asteroids is not so clear?

If we're going to get technical, Asteroids are not really an official name anymore. In 2006, when the IAU redefined what a planet was (and thus demoted Pluto), they also decided to more formally define other terms to identify objects in our solar system. You can see a diagram of all the official terms and how they relate below. Notice the important factor that asteroids no longer made the cut. There is the collection of "Small Solar System Bodies" (SSSB) and within that falls comets and centaurs, but asteroids are not exclusively called out.

So if you want to get technical, asteroids are really SSSBs now and comets are SSSBs as well, but also happen to fit into the sub-category of comets. What defines an SSSB as a comet is its ability to form a coma based on the sublimation of icy surface volatiles. So you might distinguish an "asteroid" as an object which cannot form a coma (and orbits inside Neptune), but that still leads to an ambiguity because what happens when a comet runs out of surface volatiles? Does it now become an asteroid? Is it still considered a comet? There are no clear answers.


How are black holes doors to other universes?

It is correct that the Kerr black hole solution of GTR allows travel between universes. However, that does not mean that if you actually jump into any kind of black hole that you could go to another universe.

To motivate the resolution to this conundrum, let's start off very easy: suppose you stand on the ground with a ball in your hand, and you throw it with some initial velocity. For simplicity, let's ignore everything except a uniform gravity. Mathematics will then tell you that the ball follows a parabolic arc, and when and where the ball will hit the ground. And if you take the resulting equations too literally, then it will also tell you that the ball hits the ground twice: once in the future, once in the past. But you know the past solution isn't right: you held the ball; it didn't actually continue its parabolic arc into the past.

A morally similar kind of thing occurs for, say, a Schwarzschild black hole. If you look at it in the usual Schwarzschild coordinates, there's a problem at the horizon. Mathematics will then tell you that the problem is just with the coordinate chart, and that there's actually an interior region to the black hole that becomes apparent in different coordinates. And if you do this generally enough, it will tell you that there's more to it than even that: there's also white hole with a reverse horizon and its exterior region--another universe. This full "maximally extended" Schwarzschild spacetime has this other universe connect to ours via an "Einstein-Rosen bridge" and then "pinch off", producing separate black and white holes.

Of course, that too is an artifact of mathematical idealization: and actual black hole is not infinitely extended in the past and future; it was actually produced by something, a stellar collapse. (And the "bridge" isn't traversable anyway; one will be destroyed in the singularity if one tries.)

Finally, on to the Kerr solution, it's a bit better because formally the singularity is avoidable, unlike the Schwarzschild case. However, it's still physically unreasonable: in addition to the fact that actual black holes aren't eternal, the interior of the Kerr solution is unstable in regard to any infalling matter, which will perturb the solution into something else entirely. Therefore, it cannot be taken as a physically meaningful. Still, it is true that full Kerr spacetime contains a way into another universe--in fact, infinitely many of them, chained one after another.

If you're interested in the details of its structure, you could look at some Penrose diagrams of those black hole solutions.


It depends on what one means by "Ephemeris Time." Officially, the Newtonian-based Ephemeris Time has been superseded by the relativistically-correct Barycentric Dynamic Time (TDB) and Barycentric Coordinate Time (TCB). However, the concept of what was meant by Ephemeris Time carries on to this day. The intent of Ephemeris Time was to use the solar system as a timekeeping device. This is what TDB and TCB do, but in a relativistic rather than Newtonian setting.

TCB is a theoretical time scale whose clocks are infinitely removed from the solar system, are co-moving with the solar system barycenter, and that tick at the rate of one tick per 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom (at a temperature of 0 K). (Note well: This is the same definition currently used for an SI second.) That TCB clocks are well removed from the solar system means that a second measured by an atomic clock on the surface of the Earth ticks a tiny bit slower than does a TCB clock.

These slightly different tick rates was perceived as a problem by some. The solution was eventually to redefine TDB as a scaled version of TCB such that, on average, TCB ticks at the same rate as atomic clocks on the surface of the Earth. Atomic clocks on the surface of the Earth tick faster / slower than would TCB clocks based on whether the Earth is further from / closer to the Sun than average. But averaged over the course of years, TCB and TAI remain more or less in sync.

Regarding the more or less: TDB is updated every year. As technology improves, these subtle updates change the timing of past events. TAI does not do this. As technology improves, older atomic clocks are replaced with newer ones. Atomic clocks have a shelf life of a decade or so. This can result in discontinuities in atomic time. This is not a problem with TDB as it is re-issued annually. This also means that TDB is independent of atomic time.


Mass of black holes compared to parent star

There is no general consensus on this. Different evolutionary models give different results. The factors (in addition to the initial mass of the star) that effect the final black hole mass would be the rotation rate of the progenitor, its composition (or metallicity) and whether it was in a binary system or not and whether that binary system was able to transfer mass.

Rotation is thought to be important because it affects internal mixing and therefore the rate at which fuel is supplied to the core and the rat at which processed material gets to the surface, affecting the atmospheric composition. It can also enhance mass loss.

The composition is important because mass loss is driven by radiation and radiative opacities are higher for high metallicity compositions.

A set of calculations by Heger et al. (2003) are one of the canonical works on this subject. Below is a plot of initial mass versus remnant mass for stars with big bang primordial abundance (zero initial metallicity) and then the same again for stars of solar metallicity.

The ratio of the red line to dotted "no mass loss" line gives the fraction you are after.
In zero metallicity (primordial) stars it increases from 10-40% for initial masses of 25-100 solar masses and is perhaps even higher for supermassive population III stars. (I stress that these are theoretical results).

For solar metallicity stars the results are a bit different. The ratio of the red line to the dotted line varies from 10-25% for 25-40 solar masses, but then it is not clear whether black holes can even form at even higher masses because of the much higher mass loss rates (see the difference between the dotted line and the blue curve).


No, amateur astronomy equipment is not capable of doing such an analysis. Most color CCDs use something called a Bayer filter to detect color. Essentially, certain pixels detect only red photons, other pixels detect only green, and still others detect only blue. Each filter has a high transmission over some band of wavelengths (say, for red, 600 - 700 nm). Photons outside of this band are absorbed. The CCD pixels behind the red Bayer filter then count how many photons they see (they actually measure charge, but that's a detail). The point is, the CCD can't measure a photon's precise wavelength; it can only count how many it sees over some wavelength band.

Now, this is NOT necessarily the case when it comes to professional astronomy. To my knowledge, there do not currently exist optical CCDs that can do spectroscopy without a diffraction grating. For one thing, optical diffraction gratings work really well, so there is no reason. The other reasons are that optical photons don't generate a lot of charge in a CCD, so it would be hard to measure, and there are generally a LOT of optical photons, so you'd get a lot of pile up.

By contrast, X-ray CCDs often do spectroscopy, albeit not with Fourier analysis (for example, the ACIS CCDs on Chandra). There are a lot fewer X-rays than optical photons, and they deposit a lot more energy on a CCD. Unfortunately, you can't do X-ray astronomy from the ground, so even if an amateur astronomer could somehow afford one, it wouldn't do them any good. Even in the case of X-ray astronomy, you generally use an x-ray diffraction grating to get a spectrum if that is what you are interested in.


How to calculate a planet's current position within the solar system

Buy Astronomical Algorithms 2nd ed. by Jean Meeus. It is published by Willmann-Bell Inc. If you buy it direct from the publisher, you will get a corrected printing with all errata up to August 10, 2009. There are a number of systems for describing the location of a planet. This book provides a method to find the heliocentric longitude, latitude, and distance from the Sun. It also explains the different coordinate systems and how to transform from one to another.

The method you use depends on how accurate you need to be. If you state your needed accuracy some of us might be able to suggest other alternatives.


Is the moon moving further away from Earth and closer to the Sun? Why?

Yes, the moon is moving away from Earth at around 1.48" per year. According to the BBC:

The Moon is kept in orbit by the gravitational force that the Earth exerts on it, but the Moon also exerts a gravitational force on our planet and this causes the movement of the Earth's oceans to form a tidal bulge.
Due to the rotation of the Earth, this tidal bulge actually sits slightly ahead of the Moon. Some of the energy of the spinning Earth gets transferred to the tidal bulge via friction.
This drives the bulge forward, keeping it ahead of the Moon. The tidal bulge feeds a small amount of energy into the Moon, pushing it into a higher orbit like the faster, outside lanes of a test track.

So, tidal forces are ultimately what causes this to happen.
Also, there is a Wikipedia article on tidal forces:

Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. the Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary's rotation. The process eventually leads to tidal locking of first the smaller, and later the larger body. The Earth–Moon system is the best studied case.


Is it ever possible to see Earth's shadow on other planets?

First, the Moon is not a planet and therefore not applicable to this question.  Second, the inferior planets, specifically Mercury and Venus cannot have a shadow cast by the Earth because their orbits are inside those of Earth.  Notice I did not say they are between the Sun and the Earth, as they could be in superior conjunction with the Earth (opposite side of the Sun from Earth).

The superior planets (Mars out to Neptune) are candidates for the casting of shadow of the Earth by absence of light from the Sun, however we need to explain the difference between umbra (Latin:shadow) and penumbra. Let's first use the Moon as an example. The umbra (darkest part of the shadow) is the shadow cast by the Earth into space whereby an observer cannot see the Sun at all. This happens during a total Earthly eclipse of the Moon (not a lunar eclipse).  You would be in complete darkness on the Moon. If the Earth and the Moon were the same size, the umbra cast by the Earth would continue out into space past the Moon the same distance that the Moon is from the Earth. However, since the Earth's diameter is about 3.5 times that of the Moon, the umbra extends further, but not all the way to the nearest planet, Mars.  So the Earth's umbra will not even reach the nearest of planets.

However, the penumbra (Latin:partial shadow) in theory, extends out to infinity.  First, in order to have shadow, there must be a source of light.  So if you can "see" the light source, then there is also a penumbric shadow associated with that source anytime any matter comes between the observer and the source.  The term "see" in your question is the matter.  With the average unaided human eyeball, two of them in fact, there is not enough light sensitivity to detect a difference of penumbric shadow on a planet.  However, with a ultra sensitive telescope one might be able to detect changes in the intensity of light from one time moment to the next.  In fact, current research in finding planets orbiting nearby stars uses a similar method called the exoplanet transit or transit photometry method.  So, a very sensitive telescope could see the Earth's penumbric shadow when measured over a long time period, but the human eye is not sensitive enough.


Can Milky Way be seen in South hemisphere only?

That is not true. The band of the Milky Way can be seen from any latitude on the planet. It just requires the right time of year and a dark sky.


If all stars rotate, why was there a theory developed that requires non-rotating stars?

In a similar way, we could ask...

  No beams can be exactly 1 meter long. No beams can be exactly
  straight. The material making up a beam cannot be truly isotropic.
  So why should we bother calculating the stress in a 1 meter straight
  beam having isotropic material?

Because knowing how to perform this calculation is a building block for doing more complex calculations.

The non-rotating black hole calculation also provides a limiting solution.  The solution for a spinning star's collapse will approach this solution as the spin approaches zero.

Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line.  This is useful to know even though there is no place in our universe that doesn't have gravitational influence.


Why does moon's highest altitude change during the year

The moon's orbit around the Earth is inclined at about 5° to the ecliptic
So as the Earth progresses in its own orbit around the sun, the moon's altitude also changes

During summers the part of the orbit on the other side of the Earth(night side) is below the ecliptic. While during winters it is the upper part which is on the night side

Source


Could the Philae comet lander be recharged by laser?

Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?

No.

You could make a laser beam with a pretty low divergence - meaning a beam that remains tight and parallel over a very long distance. However, "a very long distance" is nothing on the cosmic scale.

The Lunar Laser Ranging experiment is using high-power lasers focused by telescopes, shooting a beam into the Moon. The light is reflected back to Earth by mirrors placed there by the Apollo program.

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

The problem is, even with all the tight focusing at the origin, of the initial 10^17 photons in a pulse, only 1 makes it back to the detector. It's very hard to tell that something makes it back at all, let alone charging any solar panels with it.

The comet and Philae are much, much further away than the Moon. You couldn't possibly project any useful energy density at that distance. Not with the current technology.


Meteor shower predicted to be best viewed BEFORE midnight (for reasons other than the moon)?

Not that I know of, and it would somewhat surprise me.

The number of meteors from a shower you will see depends on the time of night through a number of reasons (not considering the weather and the moon):

The distribution of meteoroids on Earth's orbit is not homogeneous. The highest potential occurs when Earth moves through the peak of the distribution. This is associated with a certain time on Earth.
Meteors from a shower appear tot come from the same spot in the sky. This spot is called the radiant and moves through the sky along with the stars. The higher the altitude (height) of the radiant, the more meteors you will be able see.
Meteors enter Earth's atmosphere because Earth moves into the meteoroid cloud. This means that you will see more meteors when the spot on Earth where you are standing is facing the meteoroid cloud. This is always the direction in which the Earth is moving. If you are on the other side, the meteoroid cloud is 'behind' you and possible meteors will have to move faster than the Earth in order to enter the atmosphere. As pointed out by user Florin Andrei in the comments, this effect is also called the "bugs on the windshield" effect, because

  Earth is like a car moving fast, and meteors are like insects that get splattered. Of course you get more bugs being splattered on the piece of glass that's facing forward.

Depending on your location, 1. and 2. could combine to maximise their effect some time before midnight. However, the effect of 3. is quite big. Because of the way the Earth rotates, this will always happen early in the morning, before sunrise.


Hypothetically, if the sun stays the same size forever, when will the Earth fall into it?

As has already been said, the major sources of change to Earth's orbit are interactions with other planets and passing stars.

We're ruling out mass loss of the Sun, so the next consideration is probably tidal interactions between the Earth and Sun. This paper suggests that the Earth is receding from the Sun by about 15cm/yr for this reason. This is 150 km per million years, so over a trillion years or so, would move Earth quite a bit further from the Sun. The effect would reduce as the Earth got further away from the Sun, but I think it would always be slowly receding. I haven't done the calculation, but I'm not sure it would ever reach tidal lock -- the angular momentum of the Sun is too high.

Independently of this, though, orbital energy and angular momentum would slowly be lost by gravitational radiation. This wikipedia page gives a formula

$$t={\frac {5}{256}}\,{\frac {c^{5}}{G^{3}}}\,{\frac {r^{4}}{(m_{1}m_{2})(m_{1}+m_{2})}}.$$

for the time to collision ignoring tidal effects.

We can plug in known values for the parameters $r = 1.5 \times 10^{11} m$, $c = 3\times 10^8 m/s$, $G = 6.7 \times 10^{-11} m^3 kg^{-1} s^{-2}$, $m_1 = 6\times 10^{24} kg$, $m_2 = 2 \times 10^{30} kg$ and get  about $3.3\times 10^{30} s$ as the time to collision. I suspect that's the time until the centres of the Earth and Sun coincide, so the actual collision would be a bit earlier, but probably not enough to notice at this precision. So that's about 10 million million times the lifetime of the universe so far.

I can't find enough information to work out how these two effects would interact. Clearly in the first $10^{12}$ years tidal evolution would be more important, but that conserves angular momentum and slows as the Suns rotation slows, whereas gravitational radiation actually carries angular momentum away, so might dominate in the long term. On the other hand the radiation drops as a high power of the separation, so we might have the Earth moving ever more slowly away from the Sun, while radiating gravitational waves of ever weaker strength, and never reaching tidal lock or starting to move back in.


Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years?

Saying why gets tricky beyond "because of Jupiter", but to clarify on the quote, the statement "Earth's eccentricity follows a 100,000 year cycle" is loosely true but it's also an oversimplification.    From Wikipedia.

  The Earth's orbit approximates an ellipse. Eccentricity measures the
  departure of this ellipse from circularity. The shape of the Earth's
  orbit varies between nearly circular (with the lowest eccentricity of
  0.000055) and mildly elliptical (highest eccentricity of 0.0679)2 Its geometric or logarithmic mean is 0.0019. The major component of
  these variations occurs with a period of 413,000 years (eccentricity
  variation of ±0.012). Other components have 95,000-year and
  125,000-year cycles (with a beat period of 400,000 years). They
  loosely combine into a 100,000-year cycle (variation of −0.03 to +0.02).

If the highest is 0.0679 and the variations are 0.012 and up to 0.03, that's over 50% variation peak to peak.  That's amplitude, not period, but if we look at the chart it's clearly not a neat and tidy cycle though it's somewhat close to a 100,000 year period (top line on the chart below).

Source of image.

The cause of Earth's eccentricity variation is the other planets, primarily Jupiter, with Venus and perhaps Mars (small but close) and Saturn also having effects.  Whether a particular planet is responsible for a particular period is a good question.   I'm not sure.

Ultimately this is a version of the 3 body problem or, more specifically, orbital perturbation, which is complicated mathematics.   Unfortunately, I can't give a good explanation of precisely why the gravitational perturbations work out to roughly 100,000 years or, more specifically, 413,000, 95,000 and 125,000.   Maybe someone else can.

  What would be the simplest set of conditions and/or fewest body system
  that would lead to this particular type of periodic variation in
  eccentricity?

The fewest bodies would be 3.   Sun-Earth-Moon is one example.  The Moon undergoes eccentricity variation on a much faster cycle than any of the planets.   Sun, Earth, Jupiter or any set of Sun and two planets would work too provided the planets were sufficiently large and/or sufficiently close to affect each other's orbits and (I think) you'd want the two planets to not be in orbital resonance.  That might create a different kind of pattern.


Can you explain the pattern of Hill sphere sizes of the objects of the Solar system?

Hill sphere is the region of space around a satellite where the satellite wins the gravitational tug-of-war with its primary.

If the mass of the primary object is $M$, mass of the satellite is $m$, semi-major axis of satellite is $a$, and eccentricity of the orbit of the satellite is $e$, then the radius $r$ of the Hill sphere for satellite is given by:

$$
r \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}
$$

Note that this formula does not take into account the other objects in the  vicinity.

The anomaly pointed out in the question is not really an anomaly. The contributing factor for the surprise values is the semi-major axis of the planets ($a$).

Take Jupiter and Saturn for example: Saturn has only around $30\%$ of the mass of Jupiter, and if the two gas giants had the same semi-major axis, this mass reduction will make Saturn's Hill sphere around $68\%$ than that of Jupiter. But Saturn is around $84\%$ farther from Sun than Jupiter. This is just enough to make Saturn's Hill sphere slightly larger than that of Jupiter.

Thinking along the same lines, we can also explain why Uranus, Neptune, Pluto, and Eris have surprisingly large Hill spheres.


The consequences and the mechanisms of a shift of the Earth away from the sun

It sounds more like world-building than astronomy to me, but since they punted it here, I'll give a crack.

There's a few ways you could approach this and stay within good science but making change relatively quick is difficult.  This might be possible if the planet is in a situation like mars, with no moon and between two more massive planets, so it's more vulnerable to orbital perturbations.   Mars axial tilt is believed to have changed quite a bit and fairly quickly.   While the total solar energy is constant, a variation in axial tilt can lead to a variation in ice buildup where, generally speaking a lower tilt is better for ice buildup, as the higher tilt leads to warmer summers and that melts ice more easily.   Mars axial tilt varies from 13 to 40 degrees, but that variation can take hundreds of thousands or millions of years, which is probably longer than you're looking for.

Nearly any orbital change, even a rapid one would probably take tens of thousands of years minimum.

Geological changes are probably a little faster.  While it's still in the direction of theoretical, and I think you're right to avoid a large volcano or something like that because the effect from the volcanic ash would be greater than any long term effect from climate change.

About 3 million years ago, the formation of the Isthmus of Panama is thought to have significantly changed ocean currents that cooled the Earth by a few degrees that lead to the Pleistocene epoch and ice age cycle.   You'd need to get creative to figure out a way for a land bridge like that to make a radical change, but it's possible.

Another method, if the oceans are shallow on your planet, you could have a valley like the Mediterranean valley about 5.3-5.6 million years ago, with a natural land bridge keeping the ocean water out.   That would be a hot, dry, sandy valley.   Then when the land bridge gives way the ocean quickly flows in.   If your valley is big enough the oceans could drop several feet and if you have land or small islands near the poles, the falling ocean level could increase polar land which could trigger ice formation.   Glaciers don't form over water, only land.   This is a bit of a stretch cause you'd need two geographical oddities, very shallow oceans over the poles and a very large dry valley ready to be flooded, but it could turn a planet towards a major ice age.   Ice tends to keep growing once it starts.   It's a natural feedback mechanism.

Probably the best method would be a passing rogue planet giving your planet a gravity assist.  You'd want a larger rogue planet and it would need to pass fairly close to your planet to tug your planet into a longer orbit.  You'd probably want a gas giant mass and as it passed, it would probably need to be visibly larger than our moon in the night sky.

You mentioned 20 million km, or a 13% increase in distance and about a 22% drop in solar energy received.  You wouldn't need that much because of feedback mechanisms.   Ice ages can be triggered by as small as a 1% difference and for radical, rapid climate change, pushing your planet just 4-6% further away should be enough.

You'd need to make your planet's year longer, using Kepler's 3rd law, you'd use the 3/2 power rule, so 4% further would make the planet's year 6% longer or an additional 22 days to a 365 day year.

As to how fast that would make the planet cold.  That's hard to say.   You'd probably notice longer colder winters and frosting pretty much immediately, but for ice buildup and enormous glaciers you're still probably looking at at least a few hundred years, but that's kind of a ballpark guess.

You could do the same thing with a theoretical primordial black hole but they've kind of fallen out of fashion since it's believed they don't exist.

If you wanted to get creative you could have the rogue planet pass in-between say your earth like planet and a mars like planet, pulling the Earth orbit more eccentric and longer and the mars planet shorter, possibly leading to the two planets having crossing or nearly crossing orbits - which could lead to all kinds of scenarios and predictions like a possible planetary collision in 6,000 years or something or a near pass and enormous tides.

It would also stir up the asteroids and comets in our system likely leading to an increase in impacts following the pass for the next 100 or more years.

The rogue planet would create tides too if it appeared larger than the moon in the sky and maybe an earthquake or two, but wouldn't need to be a global disaster.    Rogue planets passing that close would be very rare, but not impossible.

rftwqa


What aliphatic compounds were found on Ceres, and how where they identified?

OK, I did a bit of research on this.   The LA Times article you quoted says something a little bit useful.

  While the scientists aren’t sure exactly what the compounds are, the
  fingerprint is characteristic of material containing carbon-hydrogen
  bonds, and may include components like methyl and methylene.

Methyl means a carbon with at least 3 hydrogen bonded to it.  Because carbon has 4 bonds, that 4th bond can be part of a chain, so a methyl group offers a very large body of molecules to choose from.

More info

Methlyene is a carbon with 2 hydrogen.

This article, which is a pretty good summary, says:

  The organic molecules that have been detected on Ceres are complex
  aliphatic compounds that seem to be almost tar-like in nature.

Tar is a pretty nonspecific term, tar-like even more so, but tar is made by cooking coal or wood without oxygen, so I would assume this implies it's not Oxygen rich.  More "simple" complex hydrocarbon chains (and because "simple" complex is somewhat confusing, another way to say that it's probably mostly Cs and Hs in a fairly big molecule, probably 5, maybe 6 carbons minimum I would think, otherwise it would be too quickly sublimated into gas.  Tar like aliphatic compounds would still be vulnerable to photo-disintegration by UV light from the sun, which is common enough on Ceres Surface, but would be high enough melting points to not evaporate somewhat rapidly like shorter chains would.   That these compounds were found locally probably implies somewhat recent, localized hydro-thermal activity.

The Aromatic C-H bond and the Aliphatic C-H bond have different spectroscopy.   Source.

  There is a weak alkene CH stretch above 3000 cm-1. This comes from the
  C-H bonds on carbons 1 and 2, the two carbons that are held
  together by the double bond. The strong CH stretch bands below 3000
  cm-1 come from carbon-hydrogen bonds in the CH2 and CH3 groups.

and

  The IR spectrum for benzene, C6H6, has only four prominent bands
  because it is a very symmetric molecule. Every carbon has a single
  bond to a hydrogen. Each carbon is bonded to two other carbons and the
  carbon-carbon bonds are alike for all six carbons. The molecule is
  planar. The aromatic CH stretch appears at 3100-3000 cm-1 There are
  aromatic CC stretch bands (for the carbon-carbon bonds in the aromatic
  ring) at about 1500 cm-1. Two bands are caused by bending motions
  involving carbon-hydrogen bonds.

So, less than 3,000 cm wavelength implies what the LA-Times article said, Methyl and Methylene or CH3 and CH2 groups on fairly long chain carbon molecules, probably tar-like.   But that's hundreds, perhaps thousands of molecules that fit those criteria, some with single bonds between carbons, some with double bonds, some maybe more complex and from all the articles I've read, they don't know which ones.  They can't tell which hydrocarbons by orbiting spectroscopy, they can only see the signature of the CH3 and/or CH2 groups at the end of the chains.

To know which molecules, they'll need to test a sample.    So, a mars-like rover is probably going to go to Ceres at some point.

Why this is interesting is that these molecules aren't that long lived on the surface of Ceres, because UV light form the sun destroys them and they're location implies they came from inside Ceres, probably by hydro-thermal activity.   There's also ammonia rich clay and carbonate rich clay and those are 3 key ingredients to life.    None of this is proof that there's life underground in Ceres, but the ingredients are all there (or mostly there).  Warm water (below the surface - and Ceres has a lot of internal water based on it's density), and the basic building blocks.   It's still a far cry from proof of life as nothing has been found on the surface of Ceres that doesn't exist in comets, other than the implication of underground liquid water.   But having all the ingredients in the same place is interesting.

Small point to add, but the article above mention Ammoniated Clay and Carbonates.  Both of those are building blocks.  Neither is organic.

  This is strengthened by the fact that the molecules are found together
  with carbonates and clays containing ammonia. These have been observed
  in many regions of Ceres

Carbonates are simple.   A molecule or molecules bound to a CO3(2-) is a carbonate.

Ammoniated clay took some looking up, but I found this:

  The orbiting Dawn spacecraft has a more ideal perch to observe how
  molecules on the dwarf planet’s surface reflect various wavelengths of
  light. It was in those wavelengths that Dawn investigator Carle
  Pieters and her colleagues spied the signature of ammoniated
  phyllosilicates, which are minerals similar to clays on Earth, mixed
  in with other materials.

Phyllosilicates are just clay or dissolved rock, with one or more OH- molecules.   Ammoniated Phylosilicates, instead of having an OH- (Hydroxy group) have one or more NH2- (amides) or perhaps NH4+ (ammonium), which is a different charge so that wouldn't replace the OH but might still bond with silicates with hydro-thermal mixing.

From this article.

  The researchers also identified a mixture of minerals on the surface
  of Ceres, which they think are ammonia-bearing clay minerals and
  magnesium carbonate. The clay minerals could have been produced by
  silicates reacting with ammonia ice. However, if Ceres had formed
  where it is now, it would not have been able to pick up any ammonia
  ice to enable such a reaction

Ceres ammonia clay implies ammonia rich waters under it's surface, so Ceres may have formed further away from the sun than it's current position, perhaps in the Kuipebelt where ammonia ice can exist and be part of a planet or moon's formation.  Ceres lower density than most asteroid belt objects is also evidence of that.   While it's important that Ceres has abundant ammonia, where it formed isn't relevant to this question.

Somewhat complex Chemistry happening inside a planet with heat, pressure and liquid water isn't surprising.   None of these discoveries are even close to proof that Ceres has life, but they indicate that Ceres has all the basic building blocks of life.  Bound Ammonia, hydrocarbons, carbonates and liquid water.   It's still a huge step from building blocks to life, but these ingredients suggest it's worth taking a closer look.

Corrections are welcome.   That's all I could deduce from reading the articles.


What constellations touch the 9-degree wide Zodiac?

While I can't answer the 9 degrees part,
http://www.space.com/5417-ecliptic-zodiac-work.html notes, the band of
the zodiac actually passes through 22 constellations:

In fact, as pointed out by the well-known astronomical calculator,
Jean Meeus, along with Ophiuchus, there are nine other constellations
that occasionally can be visited by the Moon and planets: Auriga, the
Charioteer; Cetus, the Whale; Corvus, the Crow; Crater, the Cup;
Hydra, the Water Snake; Orion, the Hunter; Pegasus, the Flying Horse;
Scutum, the Shield; and Sextans, the Sextant.

So in truth, there really aren't twelve zodiacal constellations, but
twenty-two!


Is this object an asteroid or comet, and how can it produce so many tails?

A paper titled "Monte Carlo Models of the dust environment of a sample of comets from the Oort Cloud to the outer Main Asteroid Belt" (PDF also here) (2014) studied the behaviour of P/2013 P5 using models and concluded that

  The object has been subjected to an intermittent dust mass loss, most likely associated with a rotational disruption. This is confirmed
  from the analysis of both HST and GTC images. The total dust mass
  released was of the order of $10^7\ kg$, for particle density of
  $1000\ kg\ m^{−3}$ and geometric albedo $p_v = 0.04$.
  The model of rotational disruption, based on simulations of an object that loses mass from its equatorial region, and whose
  rotational axis is perpendicular to its orbit plane, reproduces to the
  last detail the observed complex brightness pattern at four different
  epochs of HST and GTC observations. For obliquities different from 0°
  or 180° , the fits get much worse. On the other hand, an isotropic
  ejection model does not fit the HST data because it produces much more
  diffuse tails than observed.
  The ejection velocities are very low, of the order of $0.02–0.05\ m s^{−1}$. This places the limit on the size of the object to be in the
  range $30–134\ m$ for assumed densities of $3000–1000\ kg m^{−3}$.

This seems to confirm what the original question cited as a possible explanation, i. e.

  The tails are suspected to be streams of material ejected by the
  asteroid as a result of a rubble pile asteroid spinning fast enough to
  remove material from it.

The asteroid is loosely bound (a "rubble pile") and spinning fast, so everytime some of the boulders which form it shift a bit, some material may escape into space and form a small tail.


Telescope buying guide for a beginner in India

If you are a beginning astronomer, there is not much point worrying about all the bells and whistles you can get with a good telescope.

Instead, the key points to look for are:

light-gathering diameter - this gives an indication of how dim an object you will be able to see
supports/stand/mount - a basic tripod will be fine if you want to see no further than Jupiter, but it will wobble too much for longer distance viewing, so a solid mount will be essential
motor/computer drive - at high magnification, objects will pass the eyepiece very quickly, so you will want to look at tracking drives that can follow objects smoothly

But to be honest, you could start with a $100 telescope and really enjoy yourself learning the sky.


Does CIBER Experiment from Caltech suggest that there can be lots of stars which are not in any galaxy?

Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?

Not really.  A key sentence in the article is "The best interpretation is that we are seeing light from stars outside of galaxies but in the same dark matter halo".  So the stars are still within the dark matter halo of a galaxy, but are outside the boundary of the galaxy if the dark matter halo is not considered.

Furthermore the "intrahalo light" explanation is just one of two possible explanations according to Updated analysis of near-infrared background fluctuations which explains:

  Two scenarios have been proposed to interpret the clustering excess. The first advocates the contribution from intrahalo light (IHL), i.e. relatively old stars stripped from
  their parent galaxies following merging events. These stars
  therefore reside in between dark matter halos and constitute a low-surface brightness haze around galaxies. The IHL
  is expected to come mostly from low redshifts (1 + z <
  ∼ 1.5)
  systems (Cooray et al. 2012b; Zemcov et al. 2014).

  The second scenario is instead based on the presence
  of a class of early, highly obscured accreting black holes
  of intermediate mass (∼ 10^4−6M⊙) at z >
  ∼ 13 (Yue et al.
  2013b, 2014). As a suitable mechanism to produce such
  objects does exist – the so called Direct Collapse Black
  Holes (DCBH, for a concise overview of the problem see
  Ferrara et al. 2014), and the interpretation of the super-
  massive black holes observed at z = 6 seemingly requires
  massive seeds (Volonteri & Bellovary 2011), such hypothesis seems particularly worth exploring.

  Both scenarios successfully explain the observed clustering excess, albeit with apparently demanding requirements. In fact, if the excess is to be explained by intra-
  halo light, then a large fraction of the stars at low-z must
  reside outside systems that we would normally classify as
  “galaxies” (Zemcov et al. 2014). On the other hand, in the
  DCBH scenario the abundance of seed black holes produced
  until z ∼ 13 must represent a sizeable fraction of the estimated present-day black hole abundance, as deduced from
  local scaling relations (Kormendy & Ho 2013) and recently
  revised by Comastri et al. (2015). However, it is important
  to outline that both scenarios are not in conflict with any
  known observational evidence


What would a red giant Sun look like from Proxima b?

The Sun will not become a red giant for approximately 7 billion years, so there is little chance that Proxima Cen will be anywhere near it.
[EDIT: To answer a comment, the current helocentric radial velocity of Proxima Cen is -22 km/s. The Sun's escape velocity at the distance of Proxima Cen is about 0.1 km/s. So the Sun and Proxima Cen are totally unbound. With a relative speed of 22 km/s (at least - I haven't calculated the additional tangential velocity difference) they will be at least 100 pc apart within 7 million years, let alone 7 billion years.]
The Sun goes through TWO red giant phases. In the first, hydrogen shell-burning phase, the Sun will achieve a peak absolute V magnitude of around -3 at the tip of the first ascent red giant branch. It wil have a temperature of around 2700 K, be 256 times its present size and 2800 times more luminous according to Schroeder & Connon Smith (2008).
From a distance of 0.77 pc (the current distance of Proxima Cen) it would have an apparent magnitude of about -8.5 (about 40 times brighter than Venus).
Its apparent size in the sky would be 1.54 arcseconds, so would have a disc that was resolvable using a small telescope at a good site and would be orange/red in colour.
According to Schroeder & Connon Smith's model, the asymptotic giant branch phase of the Sun, when it has an inert carbon/oxygen core and He + H-burning shells will be slightly smaller and less luminous.
The exact values for both these phases are somewhat dependent on the details of the mass loss model used for the strong stellar winds expected.


Which force makes planets to continuously revolve around their parent star?

Newton's First Law of Motion states that "an object in motion stays in motion with the same speed ... unless acted upon by an unbalanced force". On earth, an object in motion will slow down and stop due to air resistance (the unbalanced force), but in space, there is no atmosphere to cause air resistance, so the object will maintain its speed.


What is the shape and size of a gamma ray burst?

When "shape" is used in the context of gamma-ray bursts, it is typically used to refer to the jets emitted on opposite ends of an axis through the source. These jets contain huge amounts of energy, most in the form of photons so high in energy that they are classified as gamma-rays, some of the highest-frequency photons known.

The jets are typically emitted along an axis through the progenitor object (I'm not going to talk about the object itself, as the causes of GRBs are not yet fully understood). A quick search engine search (I prefer duckduckgo, but you might like Google better) can give you hundreds of pictures (mostly artists' impressions) of GRBs. Notice how thin the jets are. They look like long cylinders, but closer inspection shows that they aren't. They are actually cones (truncated ones, but cones nonetheless).

The reason that these jets spread out is relativistic beaming, which also affect the beam's luminosity. The angle at which the beams diverge is so slim - in fact, this pre-print says that one observed GRB had a jet emitted at an angle of one tenth of a radian! That's only 18 degrees. (Note: On page 21, this calls an angle of 3 degrees "typical" for some jets ).

So the answer to your question is that gamma-ray bursts are typically conical, although the jets are often nearly cylindrical. The first paper I mentioned also did an analysis of spherical models, but concluded that they did not fit observations.


This question is a bit tricky because it's uncertain as to what counts as "discovering" an exoplanet. Today, we count an exoplanet as having been discovered if it can be detected either by two separate techniques or three separate transits (if found via the transit method). Some of the very very early hunts for exoplanets did indeed find them, but many of their initial results were dubious or questionable (and sometimes wrong) and weren't really confirmed until much later. So do you count a planet as being discovered when it was first potentially detected or do you not count it until its been completely confirmed?

First Exoplanet Detection

The very first time someone could claim they discovered an exoplanet was in 1988. The astronomers Bruce Campbell, G. A. H. Walker, and Stephenson Yang announced that they had discovered a planet around the star Gamma Cephei using the radial velocity method. However, their results were not that great since, at the time, the detection was at the threshold of their technological capabilities and their results weren't entirely believable. It wasn't until years later that this planet was actually confirmed as existing. Whether you count this as the first real exoplanet discovery or not depends on what you define as a discovery, given that the actual confirmation didn't come until years after the first detection.

First Confirmed Exoplanet Detection

In 1992, Aleksander Wolszczan and Dale Frail managed to find and confirm a planet around a the pulsar PSR B1257+12, again with the radial velocity method. At this time, their measurements were good enough to be believed and so considered "confirmed".

First Confirmed Exoplanet Detection Around a Main-Sequence Star

This case is particularly called out because it was the first time an exoplanet was detected and confirmed to exist around a normal main-sequence star. This occurred in 1995 around the star 51 Pegasi. This discovery was made by Michel Mayor and Didier Queloz. Many people would consider this the first "true" exoplanet discovery and will cite this case if you ask when the first exoplanet discovery was made.


What's still needed before we can observe orbits of exomoons thereby weighing exoplanets?

My opinion (and I think there has to be a large element of that) is that the presence, and ultimately identification of the orbital period, of exomoons is going to come from very precise transit photometry.
If a sufficiently large moon orbits a planet, then this will leave its signature in the transit light curve. A "Fourier-type analysis" of the light curve might then reveal a periodic nature to the light curve shape during transit that could be attributed to the orbit of the moon. The requirements here would be a sufficiently large moon that its relative position alters the transit shape. The exoplanet also needs to be in a short period orbit so that you can observe lots of transits.
A related (and better) technique, that in practice would be used simultaneously, would be to look for the signature of a moon using changes in the duration and timing of transits that recur in a periodic way. The exoplanet-exomoon barycentre is what follows a Keplerian orbit; but the area-weighted "centre of opacity" of the system will, in general, not coincide with this barycentre because mass is proportional to $r^3$, while obscuring area is proportional to $r^2$. This will lead to a wobbling of both the time of  transit
and the transit duration that may again yield a periodic signal which could be identified with the exomoon period. These methods are discussed extensively by Kipping 2009, who point out that the transit timing and transit duration signatures have different dependencies on the exomoon mass and separation from the exoplanet and thus a careful measurement of both could yield the exomoon mass. i.e. Each method in its own has a degeneracy which means the exomoon mass/period could not be determined, but measured together, the degeneracy is broken.
There is a trade-off here. A wide exoplanet-exomoon separation will give bigger signatures, but of course by Kepler's third law, the orbital period will be longer and so you would likely need a longer dataset of transits to identify any periodicity and pin down the amplitude of these signatures.
At this stage, I don't think there is any likelihood of hearing about an exomoon period/mass in the next few years. Exomoon detection by these methods is though a possibility and may already have occurred (see Teachey & Kipping 2018) . Perhaps the PLATO mission, which will produce better light curves than TESS or Kepler, and have long datasets will stand a better chance (and indeed exomoon detection is one of the mission goals - see Rauer et al. 2014).


We can't. That is an over simplification only used in elementary treatments simply to provide the the flavour of the argument. If you see it done somewhere in the refereed literature, it is probably incorrect. Of course it may be true that the mass is almost spherically symmetric, especially if it is dominated by a spherically symmetric dark matter component, but that the visible light is not. Or, it may be true that at large distances from even an asymmetric mass distribution, a Keplerian potential is a good approximation for the orbits of distant objects (e.g. satellite galaxies to the Milky Way). A measured galaxy rotation curve makes no such assumption, it is merely a measurement of rotation speed as a function of radius. It is only the interpretation and modelling that needs to deal with the mass distribution.

The real situation is much more complex. See for example http://ned.ipac.caltech.edu/level5/March01/Battaner/revision.html

However, even if one assumed all the mass was concentrated into a disk-like shape, the only way you can get flat rotation curves is to assume that the mass in the disk does not "follow the light" - that the mass-to-luminosity ratio increases vastly with radius - which is essentially still saying that you have "dark matter", just in a disk.


Is there a gas giant orbiting TRAPPIST-1?

No such planet has been announced as having been discovered. The paper only shows evidence for the 7 (really 6 because the 7th can't be officially confirmed with only 1 observation) terrestrial planets and does not make the case for any other planets. The paper doesn't indicate that more planets could exist, but does remark that there are large error bars on some of their data, leaving room for uncertainty.

Ultimately, I think we can cross off a gas giant existing in this system for a few reasons.

We don't see transit timing variations due to a gas giant. A gas giant would have noticeable gravitational influences on the inner 7 planets and we would see this influence by the small variations in the orbital periods of the inner planets. The authors of the paper did see transit timing variations, but they were able to explain all of the variations as due to the gravitational influences of the 7 planets only. They never had to invoke an 8th unseen planet to explain what they say.
We don't observe such a (easily-detectable) transit. There are only two reasons we wouldn't see the transit of this planet. Either the gas giant is orbiting in a different plane than the rest of the planets (which are all remarkably close to the same plane, as pointed out by the authors), or else the orbital period is so long, it was never captured by any of the previous observations (which spans a number of years). Neither situation seems likely to occur.
The mass of the central star is only 8% of the Sun's mass. Smaller stars tend to form smaller planets. Its really hard for gas giants to form around small stars, primarily due to lack of material. From what we understand about planet formation, the chances of a gas giant even being able to form around this star is pretty small.

Of course, only continued observation will really be able to convince us that a gas giant doesn't exist.

  On these planets could exist life without a comet protector like Jupiter does for the Earth?

This is a great question. I think the answer is, we can't be sure. Jupiter does do a great job shepherding comets and protecting the Earth. Possibly this system is rife with comets that constantly bombard the planets. However, that's just one small part of the puzzle. Our Moon does a phenomenal job in protecting us as well.

I think, when it comes to these planets, your primary concern for whether life can exist or not is the central star. It is a low-mass, ultra-cool dwarf star. These stars tend to be very volatile, much more so than our generally quiescent Sun. This means these planets are likely to receive much more radiation and be hit by many more solar storms than we are. Furthermore, these planets are so close to TRAPPIST-1 that they're all tidally locked - one face is always towards the star and one is always away. This could make one side inhospitably hot and the other inhospitably cold. The climate/weather on such a planet would likely be unsuitable for life (but who knows for sure). The tidal locking could potentially be good though as it means only the outward-facing side would generally be hit by comets, thus protecting any life on the Sun facing side (assuming its not too hot on that side).


Whats the probability of a hazardous impact for human life on Mars?

We actually have a very good idea of this because the Mars Reconnaissance Orbiter has been orbiting Mars for over a decade. The MRO is, basically, a spy satellite around Mars and is continually taking high-resolution photos of the surface. It has revisted much of the surface, taking pictures multiple times over the years.

As a consequence, we have a very good idea of the rate of formation of craters on Mars. We have seen numerous craters that have formed between visits to a spot, but none are very large. The key thing is that we now have a very solid measurement of the flux of smaller objects onto Mars and can say with certainty, that that sort of impact is way down on the list of perils. (It's probably less than the odds of being killed by a tornado in the Midwest.)

Larger impacts are a bit harder since we haven't seen any, but we have two lines of evidence. First, space debris follows a distribution in size vs. frequency that we've measured, and it's not much different around Mars' orbit than it is around Earth's.

Second, the extensive photography we have of Mars' surface (which we now have mapped better than Earth's!) allows us to count craters and to make estimates of their age. We know that small craters a few yards to tens of yards across are much more common and we know the rate at which they hit, so astronomers count the number of small craters on the floors of larger craters and can make a very good estimate of the age of the larger crater.

From this we have developed measurements of the rate of cratering at all sizes, and, again, the hazard isn't that great.

I'd note that the risk of what on Earth would be civilization-killers is actually lower on Mars because of its lack of an atmosphere. The Chicxulub impact that killed the dinosaurs killed most of them through the superheated steam and air shockwave that went around the Earth followed by the years of Asteroid Winter that followed.  Mars' atmosphere isn't heavy enough to support either, so "all" you'd have to do is hunker down and wait out the fall of rock fragments... (Still, it's an event better watched from orbit than from the ground.)


Calculate telescope orientation based on RA, DEC and Lat/Long

PyEphem will get you Alt, Az from RA, Dec, Lat, Lng, Date, Time, and Elevation
http://rhodesmill.org/pyephem/


How to understand Kepler telescope under large aspherical wavefront

Not sure what you're asking here.  Pretty much any telescope will be unable to simultaneously produce infinity (magnified) output and an aberration-free intermediate focus.  Further, an aspherical wavefront can't be produced purely by a radiating point source (all objects are collections of point sources), so no optical system can refocus an ugly wavefront.

The way a Shack-Hartmann WFS works is basically not to care about aberrated images. Each subaperture's image is unresolved to begin with, and all the sensor needs is to determine the approximate centroid location.


The mass of a object does not increase when it collapses into a black hole. So a supermassive black hole must have started off quite small, and then grown.

The formation and growth of supermassive black hole is not settled science. Supermassive black holes probably started as large stellar mass black holes (The very earliest stars could have been very large, and were almost entirely free of any heavier elements, and could have collapsed to form black holes with a mass of about 100 suns) The black holes then grew as more matter fell into them.

There is a strong correlation between the size of a SMBH and its galaxy. Its not clear why this is, but it suggests that the environment that drove galaxy growth, also contributed to the growth of the black hole.

This article on the Growth of supermassive black holes across cosmic time may interest you.


Gas clouds with masses much higher than $10^3\,M_\odot$ are plentiful in galaxies; the typical star-forming cloud (the so-called molecular clouds) have masses of $10^3\,M_\odot$ to $10^7\,M_\odot$. When quasistars (hypothetical stars powered not by nuclear fusion, but by accretion onto a central black hole) cannot exist today, it is because all gas in the Universe has become polluted with metals.

Stars form from collapsing gas clouds. In order for a region of a cloud to collapse, it must be sufficiently dense, and sufficiently cool; if it's too dilute, there's not enough gravity, and if it's too hot, the energy of the individual atoms counteracts the collapse, making the atoms escape.

Jeans' mass

This criterion is captured in the Jeans instability equation. The relation can be expressed in several ways; one way is to say that the mass of the cloud — or a small region of it — must exceed the "Jeans mass":
$$
M_\mathrm{cloud} \gtrsim M_J \simeq 3\times10^4 \frac{T^{3/2}}{n^{1/2}}\,M_\odot,
$$
where $T$ (in $K$) and $n$ (in $\mathrm{cm}^{-3}$) are the temperature and the number density of the gas.

From this equation you see that the cooler the gas is, the smaller the threshold. In other words, the smaller stars you can form. If the gas is not able to cool, only the largest clumps will collapse, and hence such stars will be very massive.

Gas cooling

So, how does the gas cool? Hot gas means that the particles have large velocities. If the particles collide, they may excite each other, bringing an electron to a higher state at the expense of slowing down — i.e. cooling. When the electron de-excites, a photon is emitted, which may leave the system. Thus, the kinetic energy of the atoms is converted into electromagnetic energy which escapes.

However, an electron is only excited if the energy of the collision matches closely the energy needed for the excitation. If the collisional energy is too high, or too low, the atoms simply bounce off of each other, maintaining their total energy (although one may transfer some energy to the other).

The effect of metals

If the gas consists only of hydrogen and helium, there are only a few available energies for excitation. Hydrogen happens to be able to cool efficiently around $T\sim10^4\,\mathrm{K}$, while helium cools efficiently around $T\sim10^5\,\mathrm{K}$, but at other temperatures, the gas tends to stay at its given temperature.

However, as soon as there a some metals, the many electrons of these metals, with their many possible transitions, allow for atoms with many possible energies to be excited. Thus, before a gas cloud of $M\sim10^3\,M_\odot$ collapses to form a $10^3\,M_\odot$ star, it will fragment into smaller pieces, forming smaller stars.


Why do we keep orbiting through the Perseid meteors?

This is probably best understood by picture.

Comet Swift-Tuttle is both on a highly elliptical orbit, and a highly inclined orbit, though the inclination of the orbit doesn't matter to your specific question, but it helps with the overall picture.   What really matters for Earth getting a meteor shower is that the two orbital paths intersect.

What makes the periodic meteor shower is the dust trail, that follows (and precedes) the comet (not to be confused with dust tail - that's something else).   Any solid debris that gets blown off the comet at low velocity enters an orbit with a very similar orbital path to the comet and over hundreds or thousands of orbits, as the comet effectively gets broken up each time it passes the sun, the comet generates a dust or debris trail that largely follows it's entire orbit, at least, that's where the greatest concentration of particles is.   Not all meteor showers are periodic, but all periodic meteor showers happen in this way with a semi permanent orbiting ring of dust that the Earth flies through once a year.

Dust trail (generally not visible, too spread out, unless heated by a close pass to the sun)

Dust tail (can be visible), only exists as the comet passes close to the sun.

Every short period comet, unless it's very young, has a debris or dust trail that roughly spans it's entire orbit though the trail is most dense near the comet.  In fact, the density of the dust trail can tell us how long the comet has been on it's current orbit.   Most short period comets were deflected into their current orbits within the last 100,000 years or so.  Short period comets have relatively short life-spans.

Long period comets or near-parabolic comets are different because their orbits are much longer, though they can still generate trails during a pass close to the sun, they tend to be much more spread out given the enormous size of their orbits.

Because the Earth's orbit intersects Swift-Tuttle's debris trail once a year and the debris trail stays mostly in the same place, we get the Perseid meteor shower about the same time every year, and the meteors always come from the same direction in the sky.

Not every short period comet has an orbit that intersects with the Earth's orbit, which is why not all comets are associated with meteor showers, in fact, the majority aren't.   3 dimensions are hard to draw in 2-D pictures, but Pluto and Neptune can be used as an example.  It's often said that they cross orbits, but they actually don't.  Pluto gets closer to the sun than Neptune, but only when Pluto is near it's perihelion, at which point it's well below (or well above, depending on orientation), Neptune's orbital plane.

An interesting sidebar to this is that comets that give us meteor showers are also comets that are potential earth impacters.


Notable differences if all stars would disappear except the Sun

There are two main things we should look at here, because there are two main ways these distant stars can affect us: light and gravity.

Light

If all the stars in the universe disappeared (besides the Sun), the most of the luminous objects in the universe would simply vanish. There would still be objects emitting radiation - stellar remnants, accretion disks, etc. - but the notable sources of light would be gone.

Would this be an issue? I can't give you any numbers, but I would assume that there wouldn't be much of a difference. Night/day cycles would continue as normal, because the rotation of the Earth also would not be affected. The behavior of stoma in plants should remain the same, as should all the other natural rituals of life.

Gravity

This is where I'm slightly more concerned. Stars have mass, and so they influence other objects with gravity. There are varying estimates of how much of the Milky Way is made up of stars. For example, McMillan (2011) estimated the stellar mass of the Milky Way to be about 6.43$\times$1010 solar masses, while the virial (total) mass was estimated to be about 1.26$\times$1012 solar masses.

The extra matter consists of gas, dust, planets, stellar remnants, and, of course, dark matter (a lot of it). The mass of the Milky Way would be reduced by about 5% - not a terrible amount, but certainly non-negligible. I don't know how this would impact the motion of the Solar System; detailed simulations based on density models would be needed.

However, any effects from stars wouldn't reach us any time soon. Light and gravity both travel at the speed of light, so we'd have at least four years before we started seeing the effects (4 light-years is the distance to the nearest star, Proxima Centauri). We wouldn't feel the effects from the rest of the galaxy for many more years.


Fitting PSF to an image to get total flux (Python)

You can't really get a PSF for an extended object.  The Point-Spread-Function is a parameter of your imaging system which describes the shape of the image of a putative point source.

I don't have any knowledge of what your python package is intended to do, so I can't comment directly on the results.  I will point (sorry) out that the geometric centroid of the image is not generally the same as the energy(brightness) centroid of the image.

Further, the total flux is just the sum of all the pixels covered by the asteroid's image (less background noise and whatnot).  It's not clear why you need the PSF if you are just doing energy/power calculations.


I'd like to become an astronomer. What experience do I need?

Well, if you want to become a professional astronomer, there is no obvious answer, since education systems can be quite different depending on your country. But astronomy & astrophysics is physics in the end, so a good training and strong knownledge in physics are definitely required (that involves math for sure). And in countries/universities where you can find an astronomy-oriented degree (at master level most of the time, before that it does not really make sense), you won't be surprised that most of the lectures are actually just physics or math.

Then, to get a PhD in the field is the way to go.

As for the positions, you can find some statistics in this paper released by ESO (the European Southern Observatory) for jobs in astronomy in the 2009/2010 hiring season. Bottom line is there were 65 opened positions worldwilde, that cannot be consider as "a lot".


Lagrange points and the radius of the Hill sphere

Since the question is about inequalities and ratios, I took the script from this answer and made it better by normalizing it, now $a=1, M1=1$ and $m \equiv M2/M1$

For physical situations in the spirit of Hill spheres and Lagrange points, I think the answer is always going to be:

$$r_1 < r_H < r_2$$

but I can't prove it, that would need math and it's too late tonight for me for math.

Without normalizing $a=1$, the distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems are as follows:

a_Earth:     149598023  km
Sun-Earth L1:  1491524  km
Sun-Earth L2:  1501504  km
Earth r_Hill:  1496531  km

a_Mars:      227939200  km
Sun-Mars L1:   1082311  km
Sun-Mars L2:   1085748  km
Mars r_Hill:   1084032  km

We now apply the normalizations mentioned above:

Python 3 or 2

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

#unitless   a = 1, M1 = 1, m = M2/M1
# so for the Hill sphere and L1/L2 points of the Earth orbiting the Sun
# m = 5.9724E+24/1.9886E+30  = 4.88987E-07 and
# L1 = 0.0054525 AU, r_Hill = 0.0054625 AU, L2 = 0.0054724 AU

def solve_L1 (r, m):
    return m/r**2 + 1. - r*(1. + m) - (1.-r)**-2

def solve_L2 (r, m):
    return 1 + r*(1 + m) - (1.+r)**-2 - m/r**2

def r_Hill(m):
    return (m / 3.)**(1./3.)

ms = np.logspace(-8, -1, 61)

answers = []
for m in ms:
    r = r_Hill(m)
    L1 = brentq(solve_L1, 0.5*r, 1.5*r, args=(m,))
    L2 = brentq(solve_L2, 0.5*r, 1.5*r, args=(m,))
    answers.append([L2, r, L1])

L2s, rhills, L1s = np.array(list(zip(*answers)))

m_Jup = 1.8982E+27/1.9886E+30
rhill_Jup = r_Hill(m_Jup)
L1_Jup = brentq(solve_L1, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))
L2_Jup = brentq(solve_L2, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))

if True:
    plt.figure()
    fs = 14
    plt.subplot(2, 1, 1)
    plt.plot(ms, rhills)
    plt.plot([m_Jup], [rhill_Jup], 'ok')
    plt.text(0.001, 0.03, 'Jupiter', fontsize=fs)
    plt.ylabel('r_Hill / a', fontsize=fs)
    plt.xscale('log')
    plt.yscale('log')
    plt.subplot(2, 1, 2)
    plt.plot(ms, L1s/rhills, '-')
    plt.plot(ms, L2s/rhills, '--')
    plt.plot([m_Jup], [L1_Jup/rhill_Jup], 'ok')
    plt.plot([m_Jup], [L2_Jup/rhill_Jup], 'ok')
    plt.xlabel('m = M2/M1', fontsize=fs)
    plt.ylabel('L1 or L2 / r_Hill', fontsize=fs)
    plt.xscale('log')
    plt.show()


How do I modify redshifts to gain corrected line of sight velocities?

You need a model for the motion of the Sun with respect to the Milky Way centre. You then have to subtract the component of this that is resolved towards the galaxy in question.

The solar motion around the Galaxy is somewhat uncertain, but is roughly (11, 240, 7) km/s when expressed as a vector aligned with Galactic coordinates (i.e. towards the Galactic centre, tangential to the Galactic centre and perpendicular to the Galactic disk).


Invisible in visible light but visible in non-visible light?

While individual atoms or molecules can emit energy at discrete or narrow ranges of wavelengths, once you have larger chunks of matter (and they are above absolute zero) they will emit, at least, a continuous black body spectrum. While black body radiation can be observed at all wavelengths, far away from the peak wavelength it may be below our ability to detect its emission.


In geocentric models, did the Earth rotate?

In the historic geocentric models, and in particular in the Ptolemaic model, the earth is immobile in the centre of the cosmos, and the sphere of the fixed stars rotates around it once a day, carrying the sun, moon and planets with it. There were astronomers who realised that this apparent motion could also be explained, from a purely mathematical point of view, by assuming that the earth rotates on its axis, but they rejected this option. By the way, neither Ptolemy nor any other serious astronomer believed that the earth was flat.


Is lunar elevation at a given location for a given day unimodal?

Yes, it is unimodal.

Earth's movement is a simple rotation, thus if the Moon were fixed on sky it will describe a circumference like the fixed stars.

From that basic movement we need to introduce Moon's own movement on the fixed stars sphere, which makes Moon not to complete the circumference in 1 day. Instead, Moon performs an arc similar to that of the fixed stars but shorter.

This circumference-like movement does not wobble, so it has just one maximun and one minimum in elevation.


I have been looking to get into astronomy myself and I have been impressed with the reviews on the Celestron Firstscope, which is around $50. It has a decent optic size and is small enough to pack up and get out to a clear area if you live in a high light pollution area. The only drawback I find with the scope (without actually using it) is it doesnt have a viewfinder but you can find one of those easily.


"thick purple cluster of gas right in the center". I want to see that.

First off, the human eye is a wonderful device, but it's no match for modern cameras. We can't see colors at such low levels of light. We're all colorblind in those conditions. Only cameras continue to see color when illumination is low.

As a result, when looking at objects such as nebulae or galaxies, including our own, all you see is a black-and-white image. Even with telescopes, the human eye still doesn't see colors in these objects.

There is exactly one single exception - the Orion nebula, which is by far the brightest of all. Some observers, after a bit of practice, can begin to distinguish a few shades of color there.

Planets, and some stars, are different. These objects are much more bright. The human eye has no trouble seeing color where color exists here. E.g. Jupiter is a slight shade of yellow, Betelgeuse is quite clearly red, etc.

  purple

I'm pretty sure the Milky Way is not purple. That's probably just a processing artifact. Go on Astrobin and search for "milky way" for a lot more examples - they tend to not look very purple usually.

https://www.astrobin.com/

  I want to see this "thick cloud cluster, which I assume is the milky way". How do I see this? How on some days do I only see just stars and stars, when I really want to see this main beauty?

There are two main obstacles to doing astronomy at the bottom of an ocean of air: atmospheric turbulence (a.k.a. "seeing") and light pollution.

Seeing affects observing high resolution objects such as planets and double stars when seen in telescopes. It's caused by air eddies and vortices that muddy the image. It's of no interest here so we can ignore it for this discussion.

Light pollution affects observing faint objects such as galaxies (including our own), nebulae, and star clusters. It's caused by light from artificial sources (city lights, industrial lights, cars, etc) going up, being reflected on particles in the air, then going back down. It makes the sky seem to "glow". It drowns out those objects which are not very bright - and galaxies tend to not be bright.

There's only one way to avoid light pollution: go far away from human settlements and facilities. The desert and the open sea are not light polluted.

Google "light pollution map" for some websites that will help you choose a dark site. Here's one example out of many:

http://www.jshine.net/astronomy/dark_sky/

Recently I've visited Death Valley. The surrounding area has essentially zero light pollution. Even driving 30 minutes out of Ridgecrest you're already under a very dark sky. I took the opportunity to show my kids how a dark sky really looks like.

The Milky Way is very clearly visible. Wherever you look, there are stars upon stars. Take a pair of binoculars, point them at random places in the sky, and even more stars are visible. There are star clusters that you could chance upon just by sweeping the sky with the binoculars, and they look like someone threw a handful of diamonds on black velvet. Little shooting stars are visible often. It's extraordinary.

It should be noted that the Milky Way is not visible during the winter months (December – February). It's best to try to watch it in June – August. Even later in the year (September – November) it's still visible in the West, just before it sets, a short time after nightfall. If you're an early riser, you can see it before daybreak in the spring (March – May).

You need to wait until the sky goes completely dark after sunset. That means 1.5 hours after sunset, or more. To determine the precise moment of sunset, go on Google or Wolfram Alpha and search for "sunset today". Then add 90 minutes to the result.

You will also need to time your visit to a night when the Moon is not visible. The Moon must be below horizon, otherwise it would cause some "light pollution" of its own. A few days before and after the new moon are the best time for this.

You could go on Wolfram Alpha and search for "next new moon" or even "new moon in september" and it will give you the date. Several days plus or minus that date should be fine.

http://www.wolframalpha.com/

TLDR: A moonless night in the summer, in a desert or some other unpopulated area, 1.5 hours after sunset or later.


There is one known pair of moons in the solar system that seemingly swap orbits every couple of years. That's Saturn's Epimetheus and Janus. Their orbits are so close together that they interact gravitationally every couple of years (when the inner moon catches up to the outer moon), so that the outer moon is slowed down, and the inner moon is accelerated.

Each moon changed direction from prograde to retrograde and vice-versa in two co-orbital horseshoe orbits, never crossing paths as they exchange a bit of momentum when closer but Saturn's gravitational pull prevails and pulls each of these two moons back in the opposite direction where they came from, swapping from lower to higher orbit. It's a bit tricky to explain in words:

[From: The Rotation of Janus and Epimetheus - Tiscareno, Matthew S. et al. Icarus 204 (2009).]


Calculating the distance of objects using the Gaia DR1

The early 1990s Hipparcos mission
yielded parallaxes for 118000 stars (Hipparcos catalog) and positions without parallax for another 2.4 million (Tycho-2 catalog).
The Tycho-Gaia astrometric solution (TGAS) combines those data with preliminary Gaia observations to get 2 million parallaxes, "only" 17 times as many as Hipparcos, with better precision.
If this is a small fraction of the total number of star positions, it's not a deficiency; it's just easier to get a position than a parallax.
Lindegren et al. 2016 explain what they could or could not do with the first 14 months of Gaia observations.
Future Gaia data releases should have still more parallaxes.


Is something wrong with my luminosity calculation?

Yes it would.
It is that way because the effective temperature is defined to be $(L/4\pi \sigma R^2)^{0.25}$.
The radius of a neutron star is about 10 km $(1.4\times 10^{-5}R_\odot)$. They are born with surface temperatures of around $10^8$ K. The coldest white dwarfs have effective temperatures of about 3000 K.
The luminosity ratio is
$$ \frac{L_{\rm NS}}{L_{\rm WD}}  = \left(\frac{10^{8}}{3000}\right)^4 \left(\frac{1.4\times 10^{-5}}{0.01}\right)^2 = 2.4\times 10^{12}\ .$$
Thus the most luminous neutron stars are (briefly) much more luminous than the coldest white dwarfs. They would remain more luminous until they cooled to $\sim 10^5$K, which could take as long as $10^7$ years (Lattimer & Prakash 2004).


You can derive the comoving distance formula without the Robertson-Walker metric, but not using "the photon being like a boat on a water stream". You need to use that the speed of light is constant as peculiar velocity of the photon.
So let $R(t) = a(t) \, r(t)$ be the proper distance. The rate of change is $$\begin{aligned}R'(t) &= a'(t) \, r(t) + a(t) \, r'(t) \\ &= v_{\rm rec} + v_{\rm pec}\end{aligned}$$ where $v_{\rm rec} = a'(r) \, r(t)$ is the recession velocity, and $v_{\rm pec} = a(t) \, r'(t) = \pm c$ is the peculiar velocity. The formula follows.


Tests for code that computes two-point correlation function of galaxies

A very simple example is to fill a 3d density with a regular plane wave, which should result in a single peak in the power spectrum. Here is a simple python scripts that does exactly that.

Computing the two point correlation function should be only one inverse fourier transform of the Power spectrum away. (Depending on which FFT library you use, you might need to renormalize the results. FFTW and numpy.fft for example have unnormalized fourier transforms: $F^{-1}[F[f(x)] = Nf(x) $, where $N$ is the number of samples.)

#!/usr/bin/python3

import numpy as np
import matplotlib.pyplot as plt

#==================================
def main():
#==================================

    nc = 128                # define how many cells your box has
    boxlen = 50.0           # define length of box
    Lambda = boxlen/4.0     # define an arbitrary wave length of a plane wave
    dx = boxlen/nc          # get size of a cell

    # create plane wave density field
    density_field = np.zeros((nc, nc, nc), dtype='float')
    for x in range(density_field.shape[0]):
        density_field[x,:,:] = np.cos(2*np.pi*x*dx/Lambda)

    # get overdensity field
    delta = density_field/np.mean(density_field) - 1

    # get P(k) field: explot fft of data that is only real, not complex
    delta_k = np.abs(np.fft.rfftn(delta).round())
    Pk_field =  delta_k**2

    # get 3d array of index integer distances to k = (0, 0, 0)
    dist = np.minimum(np.arange(nc), np.arange(nc,0,-1))
    dist_z = np.arange(nc//2+1)
    dist *= dist
    dist_z *= dist_z
    dist_3d = np.sqrt(dist[:, None, None] + dist[:, None] + dist_z)

    # get unique distances and index which any distance stored in dist_3d
    # will have in "distances" array
    distances, _ = np.unique(dist_3d, return_inverse=True)

    # average P(kx, ky, kz) to P(|k|)
    Pk = np.bincount(_, weights=Pk_field.ravel())/np.bincount(_)

    # compute "phyical" values of k
    dk = 2*np.pi/boxlen
    k = distances*dk

    # plot results
    fig = plt.figure(figsize=(9,6))
    ax1 = fig.add_subplot(111)
    ax1.plot(k, Pk, label=r'$P(\mathbf{k})$')

    # plot expected peak:
    # k_peak = 2*pi/lambda, where we chose lambda for our planar wave earlier
    ax1.plot([2*np.pi/Lambda]*2, [Pk.min()-1, Pk.max()+1], label='expected peak')
    ax1.legend()
    plt.show()

#==================================
if __name__ == "__main__":
#==================================

    main()


Can a frost line exist around a protoplanet?

Yes they can exist. The effect of ice lines in circumplanetary discs has been studied, see for example Heller & Pudritz (2015), who propose that the ice lines may act as a barrier preventing moons from being lost as a result of inward type I migration.


Searching dataset for Constellation detection

If actual photographs are not required, Stellarium could render a suitable set of reference images.
Use the Perspective (gnomonic) projection for the closest match to a photograph, and display only the features you want.
Its scripting capability may help you avoid repeating some manual operations.
Here are a sample pair of Stellarium images.
The first has only synthetic stars, which the online plate-solving service Astrometry.net successfully recognizes:

The second is the same part of the sky with only constellation labels and boundaries:

The unlabeled region between Ophiuchus and Scutum is Serpens Cauda,
part of the only discontinuous one of the 88 standard constellations.


Ephemeris data mismatch between Nasa Horizonz and Astro.com

You might be comparing an ecliptic longitude to a right ascension.
These are measured around different circles, the ecliptic and the equator respectively.
Their values coincide at the solstices ($\lambda$ 90$^\circ$ or 270$^\circ$, RA 6h or 18h) but generally not elsewhere.

With target body Mars, a geocentric observer location, and output quantities 2 (apparent RA & Dec) and 31 (observer ecliptic lon & lat) selected, JPL HORIZONS gives:

Date__(UT)__HR:MN     R.A.__(a-apparent)__DEC    ObsEcLon    ObsEcLat
2018-Jul-22 00:00     20 38 23.96 -24 58 44.6 305.5424192  -6.2964464

RA 20h 38m 24s is 309.60$^\circ$, but the quantity to compare is the ecliptic longitude, 305.54$^\circ$.

The Astrodienst ephemeris table, based on JPL DE431, gives ♒ 5$^\circ$ 33', i.e. ecliptic longitude 300$^\circ$ + 5.55$^\circ$ = 305.55$^\circ$, for Mars on 2018-07-22.


What galaxy looks the most like the Milky Way?

According to this article on space.com the most similar looking galaxy to our own is NGC 6744. The source of this article comes from a statement by the ESO (European Southern Observatory) which says:

If we had the technology to escape the Milky Way and could look down on it from intergalactic space, this view is close to the one we would see — striking spiral arms wrapping around a dense, elongated nucleus and a dusty disc.
There is even a distorted companion galaxy — NGC 6744A, seen here as a smudge to the lower right of NGC 6744, which is reminiscent of one of the Milky Way’s neighboring Magellanic Clouds

Included below is an image of the galaxy (NGC 6744) in question, taken with the Wide Field Imager on the MPG/ESO 2.2-meter telescope.

However it is also noted that while the Milky Way and NGC 6744 are similar in appearance, NGC 6744 is almost twice the diameter of the Milky Way.


How do I get started in astronomy data analysis?

A possibility might be a course at a university, youtube, or a self-study-course via a book.

What we've used in our astrostatistics class was "Wall & Jenkins: Practical Statistics for Astronomers" which presents theoretical sections with exercises as follow-ups. No answer sections exist, as you need to do some programing for some of them.

But you can maybe lend this or similar books at a local university library, without the need to spend money.


Do black hole singularities actually merge?

Infinities are hard to bend the mind around, but in this case, the merge is not impossible. Yes, the distance between them must reach zero in order to make the black holes merge, but the rate of which energy is lost to gravitational waves also increases when they get close to each other.

We are therefore dealing with a $\frac{\mathrm{infinity}}{\mathrm{infinity}}$ problem, where a finite limit may, and in this case does exist.

Do not forget your $\lim$'s!


Photometry extinction correction magnitude

If an object has a measured brightness, then it's extinction-corrected brightness will always be higher. Hence the extinction-corrected magnitude will always be smaller than the uncorrected (observed) magnitude.


Can mechanical angular momentum be transferred by gravity?

does the change of $\omega$ alter orbit of moon only when that change is attributable to the moon or in any case?

Angular momentum is a conserved quantity (more on this later). What this means is that if some object is undergoing a change in angular momentum, that angular momentum must necessarily be transferred to some other object or objects. The Earth is losing rotational angular momentum. It's rotation rate is slowing down. That means some mechanism must exist for transferring that angular momentum.
The Earth's angular momentum would change if the Earth was struck by an asteroid, but that would be a very, very small change. The loss of gases from the top of the atmosphere is also represents a negligibly small transfer of angular momentum, as does the interaction of the Earth's magnetic field with the solar wind. The Earth is far too large to be subject to the YORP effect, and The only mechanism left is gravitation. Almost all of the Earth's loss of rotational angular momentum due to a slowing rotation rate is via gravitation. A small amount is transferred to the Sun and to the Earth's orbit about the Sun. All that is left is the Moon. That vast majority of the Earth's lost angular momentum goes into making the Moon recede from the Sun.
Angular momentum is a conserved quantity. The conservation laws are very important in higher level physics (Lagrangian and Hamiltonian physics); they are assumed rather than derived. Even higher level physics derives the conservation laws from Noether's theorem. However, angular momentum can also be derived from the strong form of Newton's third law. Proving this is standard fare in every undergraduate and graduate classical mechanics text. For example, it's the subject of chapter 2 in Classical Dynamics Of Particles And Systems by Marion et al. and it's the subject of chapter 1 (starting on page 5) in Classical Mechanics by Goldstein et al. There are some nice online derivations as well, such as the week 1 lecture notes for a class on Central Forces at Oregon State University.
Another name for a force that obeys the strong form of Newton's third law is a central force. A central force is one in which the force

Acts between pairs of particles that comprise a system of particles,

Obeys the weak form of Newton's third law (equal but opposite forces),

Is directed along or against the line connecting the two particles, and

Is a function of the radial distance between the particles only.

Examples include Hookean spring forces, Coulomb forces, and Newtonian gravitation. The gravitational force between a pair of particles is given by $\vec F = G M_1 M_2 / r^2 \hat r$, where $\hat r$ is a unit vector pointing along the line that connects the two particles.

would there be tides if there were no moon?

Yes. The Sun also raises tides, but smaller in magnitude than those raised by the Moon. The heights of the tides raised by the Sun are a bit less than half those raised by the Moon.

what is the mechanism through which angular momenta can exchange magnitude? This is apparently impossible since there is no connection or, we we consider the obsolete Netwonian gravity, the connection is radial and never tangential?

There are two issues with this thesis. One is that Newtonian gravity is not an obsolete theory; it remains widely use in space exploration and in astronomy. A better way to look at it is that Newtonian gravity has a limited domain of applicability.
The other issue is the claim that Newtonian is purely radial. This is only true in the case of point masses or objects with a spherically symmetric mass distribution. It is not the case for objects with an aspherical mass distribution. For example, the Earth has a significant equatorial bulge due to its rotation. This equatorial bulge exerts a torque on satellites orbiting the Earth that causes the satellites' orbits to precess. Scientists take advantage of this. There's a special class of satellites, the sun synchronous satellites, whose orbits by design precess by 360 degrees over the course of a year.
With regard to the Moon, the rotation of the Earth makes the tides raised by effectively lead the Moon by a small amount. While this isn't Newton's tidal bulge, it has a similar effect on the Moon. Averaged out over the course of time, the waters on the near side of the Earth pile up slightly in front of the Moon while the waters on the far side of the Earth pile up slightly behind the Moon. Because those far side waters are further from the Moon, the near side waters dominate gravitationally, causing the Moon to accelerate.


It is very hard to tell whether we are in a unique Universe. The difficulty is the closer we attempt to probe in the fractions of a nano second after the Big Bang our current forumlations of Gravity and the three our fundamental forces do not work. So it is very hard to tell whether our Universe is unique.

However, with all that said, my own personal belief is that our Universe is not unique. The estimated time for random quantum fluctuations to generate a new Big Bang as quoted in the page that you have referred to is an UNIMAGINELY large time scale.

My question I will pose back to you is, how could we ever tell we are a unique Universe, or part of a collection of Universes?


Question about redshift and effective gravity at the rotating neutron star surface

The redshift factor is given in (Asaoka & Hoshi 1987) and depends on latitude and radial coordinates. The equilibrium surface may however be an ellipsoidal surface of constant redshift and frame dragging (Marsh 2014).


I think what you are envisioning is having one pole always pointed towards the primary star. So the planet would rotate about that pole, but then the direction of the pole would change over the course of a year to keep the pole pointing towards a star.

This phenomenon of the direction of the pole changing is called "precession". The Earth does it over a period of 26,000 years, but the pole only traces out a fairly small circle in the sky, rather than twisting about a perpendicular line like you envision.

I think your scenario is improbable at best. Planets are like gyroscopes and resist having their axes of rotation be twisted about a perpendicular axis like that. Even a planet with mass concentrations at both poles (maybe a long cigar shape) to maximize tidal force would probably end up losing the perpendicular rotation to tidal friction, and just end up rotating slowly about an axis nearly parallel to the axis of the orbit.


22nd is shortest day in some places, but the 21st is shortest in other places. Can this be true?

Sure it can be (and is) true. Everyone is going through the Earth's "zone of lightness" at different times. Presumably whichever of those times is "nearest to" the actual moment of solstice will be the "shortest period between sunrise and sunset" for any given person (longitude) (ignoring natural and time-zone variations as noted in other comments). As you note, for your friends and everyone west of them up to the Date Line, that will likely be 12/21. And for you and everyone east of you around to the Date Line, that will likely be 12/22.

Maybe it helps (or maybe not) when you realize that people very close to the Date Line itself but on opposite sides experience 12/21 and 12/22, resp., as "the same day".


When Phobos collides with mars (or breaks up), will it have any noticeable effects on Earth?

No. Phobos is small - just 11 Km across - the size of a small city. Mars (and Phobos) is so far away that a Phobos impact will not affect Earth much. (Mars ranges from 100 billion meters away to nearly 400 billion meters depending on its and Earth's position in their orbits.)

When Phobos hits the Roche limit as it will break up and become a thin light-grey planetary ring around Mars for a few million years.

When it breaks up, (almost) all the debris should stay near Mars because the escape velocity from its orbit around Mars is greater than the velocity imparted to boulders of Phobos as it self-destructs.


Help with Converting Right Ascension and Declination into 3D Vectors

The stars should fit $r = \sqrt{x^2 + y^2 + z^2}$ but seem to be plotted in the $x = r$ plane instead.

Conventionally the x axis is at (α=0h, δ=0°),
the y axis is at (α=6h, δ=0°),
and the z axis is at δ=+90°.

Using right triangles instead of isosceles triangles,

Then $h = r \cos \delta$, and
$$\begin{align}
x &= r \cos \delta \cos \alpha \\
y &= r \cos \delta \sin \alpha \\
z &= r \sin \delta
\end{align}$$

Most math libraries use radians for angles.
Passing angles in other units to trig functions may result in a celestial sphere covered somewhat evenly but incorrectly.


attractors of Mercury's spin-orbit relation

Over enough time there is a tendency of all orbiting objects to fall into resonance with the object they're orbiting.   1:1 being is most common.

1:1 usually called Tidal Locking, is quite common. The Moon to the Earth.   Pluto and Charon to each other.  All four of the Galilean moons are tidally locked to Jupiter, likely others.

Mercury's 3:2 ratio is more unusual but not necessarily uncommon, especially for a planet with an eccentric orbit.

See Here

An eccentric orbit likely helps create a spin-orbit resonance other than 1:1 and resonance other than 3:2 should certainly be possible (the Wiki article implies that) though there are no known examples that I know of.

For a really crazy orbital spin - check this out:  http://www.businessinsider.com/pluto-moons-weird-behavior-2015-6