When was the parallax of a star first measured?

Telescopes were apparently invented in 1609, but didn't become advanced enough to measure stellar parallax until the 1830s.
Observation of stellar parallax would be a big step in proving the heliocentric theory, and I think that the lack of detectable stellar parallax was used as an argument against the heliocentric theory in ancient times.  It was certainly used as an argument against the heliocentric theory in early modern times.

Stellar parallax is so small that it was unobservable until the 19th century, and its apparent absence was used as a scientific argument against heliocentrism during the early modern age. It is clear from Euclid's geometry that the effect would be undetectable if the stars were far enough away, but for various reasons, such gigantic distances involved seemed entirely implausible: it was one of Tycho Brahe's principal objections to Copernican heliocentrism that for it to be compatible with the lack of observable stellar parallax, there would have to be an enormous and unlikely void between the orbit of Saturn and the eighth sphere (the fixed stars).1

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After the Copernican theory gained popularity astronomers made many attempts to measure stellar parallax.

In astronomy, aberration (also referred to as astronomical aberration, stellar aberration, or velocity aberration) is a phenomenon which produces an apparent motion of celestial objects about their true positions, dependent on the velocity of the observer. It causes objects to appear to be displaced towards the direction of motion of the observer compared to when the observer is stationary. The change in angle is of the order of v/c where c is the speed of light and v the velocity of the observer. In the case of "stellar" or "annual" aberration, the apparent position of a star to an observer on Earth varies periodically over the course of a year as the Earth's velocity changes as it revolves around the Sun, by a maximum angle of approximately 20 arcseconds in right ascension or declination.

The Copernican heliocentric theory of the Solar System had received confirmation by the observations of Galileo and Tycho Brahe and the mathematical investigations of Kepler and Newton.[10] As early as 1573, Thomas Digges had suggested that parallactic shifting of the stars should occur according to the heliocentric model, and consequently if stellar parallax could be observed it would help confirm this theory. Many observers claimed to have determined such parallaxes, but Tycho Brahe and Giovanni Battista Riccioli concluded that they existed only in the minds of the observers, and were due to instrumental and personal errors. However, in 1680 Jean Picard, in his Voyage d’Uranibourg, stated, as a result of ten years' observations, that Polaris, the Pole Star, exhibited variations in its position amounting to 40″ annually. Some astronomers endeavoured to explain this by parallax, but these attempts failed because the motion differed from that which parallax would produce. John Flamsteed, from measurements made in 1689 and succeeding years with his mural quadrant, similarly concluded that the declination of Polaris was 40″ less in July than in September. Robert Hooke, in 1674, published his observations of γ Draconis, a star of magnitude 2m which passes practically overhead at the latitude of London (hence its observations are largely free from the complex corrections due to atmospheric refraction), and concluded that this star was 23″ more northerly in July than in October.[10]

Consequently, when Bradley and Samuel Molyneux entered this sphere of research in 1725, there was still considerable uncertainty as to whether stellar parallaxes had been observed or not, and it was with the intention of definitely answering this question that they erected a large telescope at Molyneux's house at Kew.3 They decided to reinvestigate the motion of γ Draconis with a telescope constructed by George Graham (1675–1751), a celebrated instrument-maker. This was fixed to a vertical chimney stack in such manner as to permit a small oscillation of the eyepiece, the amount of which (i.e. the deviation from the vertical) was regulated and measured by the introduction of a screw and a plumb line.[10]

The instrument was set up in November 1725, and observations on γ Draconis were made starting in December. The star was observed to move 40″ southwards between September and March, and then reversed its course from March to September. [10] At the same time, 35 Camelopardalis, a star with a right ascension nearly exactly opposite to that of γ Draconis, was 19" more northerly at the beginning of March than in September.[11] These results were completely unexpected and inexplicable by existing theories.

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So as a result of searches for stellar parallax, the aberration of light was discovered by James Bradley.
Bradley continued to research the aberration of light, and made another unexpected discovery, the nutation of the Earth's axis.

Nutation was discovered by James Bradley from a series of observations of stars conducted between 1727 and 1747. These observations were originally intended to demonstrate conclusively the existence of the annual aberration of light, a phenomenon that Bradley had unexpectedly discovered in 1725-6. However, there were some residual discrepancies in the stars' positions that were not explained by aberration, and Bradley suspected that they were caused by nutation taking place over the 18.6 year period of the revolution of the nodes of the Moon's orbit. This was confirmed by his 20-year series of observations, in which he discovered that the celestial pole moved in a slightly flattened ellipse of 18 by 16 arcseconds about its mean position.3
Although Bradley's observations proved the existence of nutation and he intuitively understood that it was caused by the action of the Moon on the rotating Earth, it was left to later mathematicians, d'Alembert and Euler, to develop a more detailed theoretical explanation of the phenomenon.5

https://en.wikipedia.org/wiki/Astronomical_nutation[1]
As it turned out, the changes in the apparent positions of stars due to the aberration of light and nutation of the Earth's axis are many times larger and easier to detect than even the largest stellar parallax of even the nearest star.
It was not until the 1830s that astronomical instruments became advanced enough that the first stellar parallaxes were detected and measured, after centuries of failed attempts.  And the parallaxes of only three stars were measured during the the 1830s.

The large proper motion of Alpha Centauri AB was discovered by Manuel John Johnson, observing from Saint Helena, who informed Thomas Henderson at the Royal Observatory, Cape of Good Hope of it. The parallax of Alpha Centauri was subsequently determined by Henderson from many exacting positional observations of the AB system between April 1832 and May 1833. He withheld his results, however, because he suspected they were too large to be true, but eventually published them in 1839 after Friedrich Wilhelm Bessel released his own accurately determined parallax for 61 Cygni in 1838.[62] For this reason, Alpha Centauri is sometimes considered as the second star to have its distance measured because Henderson's work was not fully acknowledged at first.[62] (The distance of Alpha Centauri from the Earth is now reckoned at 4.396 ly or 41.59 trillion km.)

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In 1804 Piazzi reported that 61 Cygni had a very large proper motion and was probably one of the closest stars to Earth, and thus a good candidate for parallax observations.  There were many unsuccessful attempts to measure the parallax of 61 Cygni.

When Joseph von Fraunhofer invented a new type of heliometer, Bessel carried out another set of measurements using this device in 1837 and 1838 at Königsberg. He published his findings in 1838[31][32] with a value of 369.0±19.1 mas to A and 260.5±18.8 to B, and estimated the center point to be at 313.6±13.6. This corresponds to a distance of about 600,000 astronomical units, or about 10.4 light-years. This was the first direct and reliable measurement of the distance to a star other than the Sun.[27][33] His measurement was published only shortly before similar parallax measurements of Vega by Friedrich Georg Wilhelm von Struve and Alpha Centauri by Thomas Henderson that same year.[34] Bessel continued to make additional measurements at Königsberg, publishing a total of four complete observational runs, the last in 1868. The best of these placed the center point at 360.2 ±12.1 mas, made during observations in 1849.[27] This is close to the currently accepted value of 287.18 mas (yielding 11.36 light-years).[35]

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The distance to Vega can be determined by measuring its parallax shift against the background stars as the Earth orbits the Sun. The first person to publish a star's parallax was Friedrich G. W. von Struve, when he announced a value of 0.125 arcseconds (0.125″) for Vega.[37] Friedrich Bessel was skeptical about Struve's data, and, when Bessel published a parallax of 0.314″ for the star system 61 Cygni, Struve revised his value for Vega's parallax to nearly double the original estimate. This change cast further doubt on Struve's data. Thus most astronomers at the time, including Struve, credited Bessel with the first published parallax result. However, Struve's initial result was actually close to the currently accepted value of 0.129″,[38][39] as determined by the Hipparcos astrometry satellite.[4][40][41]

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So the first three measurements of stellar parallax are almost tied for first place.


Did we discover 10 or 12 new moons of Jupiter?

Per the Carnegie Science article that Magic Octopus Urn linked from NASA in the comments, a Carnegie Science team led by Scott S. Sheppard noticed something new in spring of 2017 (though some observations occurred as early as 2016). It took a year to confirm the discovery of the new moons.

Ten of the moons orbit in the outer swarm of moons, which is one of the ways to divide the groups of moons. Nine of these follow the pattern of the other moons which orbit Jupiter in retrograde (the opposite direction of the planet's rotation). One they call "oddball" because it orbits prograde (the same direction as the planet's rotation) and has a more inclined orbit than the inner prograde moons.

Two of the moons orbit in the inner group, which is what brings the total number of discoveries up to twelve, as referenced in the first set of articles in your post. Like the other inner moons, these orbit prograde.

One of the retrograde moons and the "oddball" were first noticed in 2016. One of the inner group and the rest of the retrograde moons were discovered in 2017. The second inner group moon was discovered in 2018, then most were announced together on July 17, 2018. (Dates collected from Wikipedia.)

The retrograde moon discovered in 2016 and one of the retrograde moons discovered in 2017 were announced in 2017, this means only ten moons were announced on July 17, 2018, which is why the Nature article refers to ten new moons. The EarthSky article talks some about the ten and the twelve, and it does clarify both groups (divided by inner/outer and year of announcement).


Are the planets in our solar system in free fall?

Yes. Free fall means the only force acting on a body is gravity.

The only force acting on the Earth and every other body in orbit is gravity, so all the planets are in free fall.

(There are very small forces from, for example, solar wind but these are minute in comparison to the gravitational forces)


Spicules are much smaller, more dynamic, and possibly hotter. They are also much more abundant. While the term spicules is strictly defined for objects off the solar limb, their disk counterparts (let's call them fibrils for simplicity) cover the whole Sun, all the time. Spicules have lifetimes of a few minutes and extend up to ~15 Mm, with the most energetic of them being ejected with speeds of around 100 km/s.

While prominences/filaments have a relatively stable temperature around 10,000 K, spicules/filaments are also seen in filters sensitive to temperatures around 100,000 K, meaning that some of them will undergo a thermal evolution to at least the transition region.

Spicules/fibrils are a candidate for energy and mass deposition in the corona.


If a star were to suddenly lose nearly all of its stored heat, would it be able to return to its normal state?

If you suddenly sucked all the heat out of a normal star like the Sun, nuclear reactions would cease, the gas pressure would fall to zero. The star would collapse, initially on a freefall timescale (about 1 hour for the Sun).

The released gravitational potential energy would half be radiated away and half would go into re-heating the interior. The temperature would rapidly rise, the collapse would be halted, nuclear reactions would restart, the star would slowly expand again and a thermal and hydrostatic equilibrium should be restored on a Kelvin-Helmholtz timescale (about 10 million years for the Sun).

One way to see this, is to compare the current thermal content of a star with its capacity for generating new heat by nuclear reactions. e.g. approximate the Sun as a uniform ball of ideal gas at an average temperature of a few million degrees. The thermal content is roughly $10^{41}$ J. The Sun generates this much energy in nuclear reactions in only 8 million years - coinciding with the timescale to reestablish an equilibrium. Or you could compare it with the gravitational potential energy of around $2\times 10^{41}$ J. i.e. There is enough gravitational potential energy to reheat the Sun.

However, not everything would be a continuation. The collapse and reheating would be similar to the star repeating its pre main sequence phase. During this phase the star becomes fully convective and so the core material would get completely mixed with the rest of the star again. This would have the effect of rejuvenating the star as a new zero age main sequence star, but with a slightly higher helium abundance.


The relation between absolute magnitude $M$ and luminosity $L$ for stars
$$\frac{L_{Star}}{L_{Sun}}=10^{(M_{Sun}-M_{Star})/2.5}$$
should also be appliable to i-band luminosities of galaxies.

Taking the absolute magnitude 4.08 of the sun on the I-band the luminosity of a galaxy with absolute magnitude −18.06 would be
$$\frac{L_{Galaxy}}{L_{Sun}}=10^{(4.08−(-18.06))/2.5}=0.7178\cdot 10^9.$$
An order of magnitude estimate for the mass of the galaxy would be
$0.563\cdot 0.7178\cdot 10^9=0.404\cdot 10^9$ solar masses.

But $M/L$ isn't necessarily biased the same way for the sun to $M_i/L_i$ as for the galaxy.
Therefore you'll probably need to compare the spectrum of the sun with the spectrum of the galaxy to find out the ratios of the i-band fraction of the total emission. Absorption and extinction at different wavelength may be different, therefore determining kind of mean temperature of stars in the galaxy could help finding a more realistic estimate for the stellar mass.

A similiar approach has been used in this paper.


You are right that helium around a gas giant is no kind of surprise, and oxygen on an Earthlike planet would be spectacularly more important.  However, the excitement is simply that if you can detect one element, it means you might be able to detect another.  It just gives you more ways to learn about the planet, and offers promise for more new ways to learn.  One might find analogy with the first detection of exoplanets, which were "hot Jupiters."  No one was surprised stars can have planets, and no one thinks hot Jupiters are conducive to life, but if you can discover one kind of planet, it bodes well for the later discovery of others.  Anything we can learn about composition is an important step in the right direction.


Why are we building larger land-based telescopes instead of launching larger ones into space?

It's cheaper.

(1) With adaptive optics you can get 0.1 arc second resolution on the ground (admittedly only on a mountain top with particularly good air flow, but still!). This eliminates one of the major advantages of space until you get above several meters mirror diameter.

(2) Rocket fairings are the shrouds which protect payloads during the supersonic atmospherics speeds reached during launch. A 5 meter fairing is about the largest that can be flown, which limits the size of the one-piece mirrors which can be launched.  (The Dreaded Webb Telescope's mirror is in pieces which will assemble themselves in space -- a very scary and very expensive piece of design.)

(3) Servicing a telescope on the top of Mauna Kea or in the high Chilean Andes is a difficult and expensive process. Servicing a telescope in orbit makes that look like small change.  (Cost comparable to the cost of building a new giant scope on Earth.) And in-orbit servicing can't even be done with current technology except in low earth orbit.

(4) While high resolution is one frontier in astronomy, going deep is another, and going deep requires big mirrors.  A 30 meter mirror on Earth gathers much more light than a 5 meter mirror in space.  The giant terrestrial telescopes simply do a better job of being light buckets for spectroscopy than anything we can yet put into space.

The bottom line is that with the development of adaptive optics, space-based telescopes of currently buildable and launchable size lost their main advantage over ground-based telescopes.  And since they're 10x to 100x the cost, they are simply not worth building for many purposes.

Space based telescopes still hold a significant edge in parts of the spectrum blocked by the atmosphere such as UV and IR (Webb), and for certain tasks involving long-term high accuracy photometry (Kepler) and astrometry (Gaia).  But for general purpose use, the balance seems firmly on the ground's side for large telescopes.

This will change if space flight becomes cheaper -- the SpaceX BFR, for example, with its 9 meter fairing and dramatically lower launch costs, offers great hope for space telescopes.


Very likely not.

If it's a gravitational process like galaxy lensing, then a "mirror" would require a very large deviation. That's extremely unlikely with this process.

With other processes, e.g. classic reflection, it seems even more unlikely. We don't know any natural processes that could craft mirrors at cosmic scale.


On orbital mechanics of evaporating planets

The mass of a planet does not affect its orbit significantly unless it is a very large planet.  On the opposite end of the spectrum would be a very small mass planet that was evaporating, and that is essentially what a comet is.  It's all Keplerian, though comets can have their orbits affected slightly by the tiny push they get from the evaporating gases.  So it's not the change in mass, but how the evaporation actually happens, but this is a small effect.  Even a large planet like Jupiter does not affect the Sun much (the Sun orbits around a point that is not even external to the Sun), so losing mass would not affect the orbit of Jupiter in any significant way either.


Earth orbit around the Sun and its position in the Galaxy

The radio source Sgr A* is at RA 17h46m Dec -29°, or about 6° south of the ecliptic at longitude 267°.
The Sun appears most nearly opposite that position, ecliptic longitude 87°, about three days before the June solstice.
At that time of year, the brightest part of the Milky Way is highest in the sky around local solar midnight.

However, the distance from here to Sgr A* is about 7.9 kpc or 1.6 billion au.
The Earth is only infinitesimally closer to the galactic center in June than in December.


Extent of knowledge of potential future gamma ray bursts

Although not being an expert on the field, I found a multitude of studies dealing with events which may be used as indicator that a GBR is about to happen somewhere. Research on GBR precusor events range from smaller gamma ray precursors, over gravitiational waves, to thermal signatures, and neutrinos:

Sarp Akcay: Forecasting Gamma-Ray Bursts using Gravitational Waves (2018)
Thomas Koshut et. al: Gamma ray precursor activity as observed with BATSE (1995)
L. Piro et al.: Probing the Environment in Gamma-Ray Bursts: The Case of an X-Ray Precursor, ... (2005)
E. Troja,  S. Rosswog, and N. Gehrels: Precursors of short gamma-ray bursts (2010)
Frédéric Daigne, Robert Mochkovitch: The expected thermal precursors of gamma-ray bursts in the internal shock model (2002)
Coppin, Nick van Eijndhoven for the IceCube Collaboration: IceCube search for high-energy neutrinos produced in the precursor stages of gamma-ray bursts (2019)


Why do these Hubble images of Neptune look like Cheela?

Neptune has an axial tilt of 28.3 degrees and an orbital period of 164.8 Earth years; currently, the north pole is tilted away from Earth.
The maps in your question were each constructed from twelve exposures of Neptune: four each in 845 (red), 547 (green), and 467 nm (blue) wavelengths.  Each of the exposures was then converted from an effectively-orthographic view of a hemisphere of Neptune to the equirectangular projection used in the maps.  The precise technique is cited to https://iopscience.iop.org/article/10.3847/1538-3881/aaa6d6 and https://iopscience.iop.org/article/10.1088/0004-637X/812/1/55.
The bumpy appearance comes from using four discrete observations rather than a single continuous one; the black area is portions of Neptune that were never visible.  Each of those spikes is the result of projecting a single color of a single pixel on the edge of Neptune's visible disk onto an equirectangular map.


What's the smallest scale at which dark matter has been measured?

It looks like Segue 2, found by the Keck Observatory, is a candidate for the lightest galaxy, but I don't know if its the smallest one.


Is there a maximum size for a black hole?

As David Hammen commented, the power emitted through Hawking radiation is proportional to $M^{-2}$. Thus the evaporation timescale for a black hole is proportional to $M^3$. This means that a more massive black hole is much more stable against evaporation than a lower mass black hole.

The other issue you mention is the limited rate that you can "feed" a black hole. There is inevitably a feedback; as gas is compressed towards the event horizon it gets hot and emits radiation. The pressure of this radiation can eventually balance the inward gravitational infall. For spherically symmetric accretion this leads to the Eddington limit, which sets the maximum spherical accretion rate, where $\dot{M}_{\rm max}\propto M$. That is, the maximum accretion rate is proportional to the black hole mass.

If accretion proceeds at the Eddington limit then the black hole mass grows exponentially with time and with a characteristic doubling timescale of around 50 million years (independent of the original mass - see this Physics SE page for some mathematical details).

If black holes were limited to this accretion rate (though there is some evidence from the presence of very luminous quasars at high redshift that they may exceed it), then the maximum black hole mass will depend on the age of the universe and the size of the initial "seed" black holes. If we assume an initial mass of 100 solar masses, a doubling timescale of 50 million years and that the seed black holes formed 400 million years after the big bang (all plausible, but contestable), then there have been 266 doubling timescales since and the black hole could have grown by a factor of $10^{80}$ !

Clearly there are no black holes with anywhere near this mass in the observable universe - the largest appear to be of order $10^{10}$ solar masses. Their growth is limited by their food supply. Supermassive black holes are found at the centres of galaxies. There is a poorly understood relationship between the black hole mass and the mass of the bulge of the galaxy it is in. The ratio peaks at about 1 percent for the most massive bulges (e.g. see Hu 2009; McConnell & Ma 2013). Since the most massive massive elliptical galaxies are of order $10^{12}$ solar masses, this appears to set the maximum mass of a black hole in the present day universe.

The future is speculation. If the cosmic expansion rate continues to accelerate, then galaxy mergers will become increasingly uncommon and the opportunities for further black hole growth will be limited.


Determine spectral type of star from its properties

There are no exact boundaries in temperature, luminosity, surface gravity etc. for spectral classes because the classification system works in a different way - it is fundamentally an empirical system, with classification based only on the appearance of features in the spectra.
The Yerkes or Morgan-Keenan (MK) system is based only on a set of standard stars and their spectra, and the type of any other given star (that is not one of those standards) is based only on where that star’s spectral lines fit into the two-dimensional matrix of standards. The introduction of this Annual Reviews article by Morgan and Keenan spells out the principle.
Of course those lines reflect physical effects
Iike temperature or luminosity (really primarily surface gravity), but those physical properties are not the basis of the classification.  And there are other factors that can change the underlying properties while preserving the appearance of the spectral lines.  For example in the intro of this paper, Gray discusses how stars of the same luminosity class can have different luminosities due to rotation.
Divorcing the classification from the physical properties is an interesting philosophical approach, but one that has stood the test of time even as our understanding of stars and their physical properties has improved in the past 70 years.  If you have time, I encourage you to read the first couple of pages of the review article linked above - the quote from Dimitri Mihalas (who was a leading theorist on stellar atmospheres) is especially interesting, I think.


Did Tycho Brahe improve measurements twentyfold?

Accuracy relative to other authors is hard to estimate.

However, here's a scholarly article that shows a modern determination of the actual accuracy of Brahe's measurements in absolute terms:

http://adsbit.harvard.edu/full/1978JHA.....9...42W

The conclusion is that it depends on the instrument and it depends on the actual measurement, but as a general rule he was able to obtain an accuracy better than 1 arc minute, often down to 0.5 arcmin.

For comparison, the resolving power of the healthy, young, average human eye is around 1 arcmin. So Brahe was able to somehow make measurements with an accuracy better than the smallest details resolved by the human eye, even though there were no magnifying instruments available in his time. I believe this was achieved by averaging multiple measurements.

In other words, he was very near the limit of what was theoretically achievable before the invention of the telescope. He was certainly the best of his era by a large margin, and far better than distant predecessors such as Ptolemy. What mattered here was the methodology, and the unprecedented scale of the instruments (the armillary sphere, etc).


Small changes to Earth's orbit caused by a large impact on the Moon

... and still remain within the habitable zone?

Everything perturbs all the time.  Every change in the distribution of mass in the solar system (or in the universe) perturbs the orbit of the Earth... so the literal answer is "yes".
But suppose someone asked instead:

Can an 'invisible' impact perturb the orbit enough to have a measurable effect on the the climate

The answer to that question would be "no"  The Earth's orbit already takes it closer and further from the sun each year, reaching its closest in January. To move the Earth enough that it has a climatic effect would require a huge force on the Earth-moon system, and an impact on the far side of the moon that is large enough to cause such a perturbation would cause other effects (like melting the moon, disrupting it completely, blasting enough of the moon into space and having it fall to Earth....) These would affect the climate much more.
An undetected asteroid impact on the far side of the moon might change the orbit of the Earth-moon system around the sun by millimeters.
It is just really really hard to change the orbit of a planet.


which or what kind of star has a very stable luminosity?

Hipparchos numbers of the 26 most stable stars known:
2021, 2854, 5542, 16611, 19747, 24927, 32537, 38414, 42913, 45556, 50191, 57363, 71053, 73555, 74666, 76440, 74946, 90139, 94648, 96052, 10239, 102488, 104732, 111169, 116631, 118322.

Compare Hpmax and Hpmin: They differ 0.01. Standard error e_Hpmag is 0.0003 mag.
There are 681 stars in the Hipparchos catalog with an amplitude of 0.01 mag.

More detail in this paper, including a link to the appropriate VizieR query.


How will the recent appearance of a supernova, the closest in 27 years, help reduce the uncertainties in our measurements of dark energy?

The current supernova is a supernova of type Ia. Supernovae of type Ia are used as standard candles for distance estimates, especially used to determine the Hubble constant. Hence by a better calibration of this kind of supernovae, more about the reliability and accuracy of distance estimates can be learned.

The expansion rate (in relation to the distance) of the universe, expressed by the Hubble constant, is an important incredient for the estimation of the fate of the universe.


Can hawking radiation be eaten by another blackhole?

Black holes can "eat" Hawking radiation, but the radiation has to be directed towards the black hole before it can be "eaten". Black holes are pretty small, and space is very very very big.

So as a black hole evaporates, and gives off" Hawking radiation, some may eventually fall into another black hole, but most will not. It will just travel out into space, travelling forever (or until it is red-shifted to such low energy that it loses an individual identity - the so-called "heat death").  We can't picture this because our brains are not equipped to deal with the massive amounts of time. But on these time scales there is nothing very mysterious about a black hole evaporating and so vanishing. It is no more mysterious than a candle flame going out.

Here is a metaphor:

There are some children in a field, each child is carrying some balls (the children represent black holes, the balls are the mass of the black holes and the field is the universe).

The children are walking away from each other (the represents the expansion of the universe*)

Roughly every minute each child throws a ball in a random direction. (This is Hawking radiation)

If the ball passes another child, he catches it and adds it to his collection (Radiation falling into the black hole).

If a child runs out of balls, he sits down, he is out of the game and can't catch any more balls (this is the black hole running out of mass, and so evaporating).

Will the game end?

Most of the balls will never be caught by other children because they are not thrown in the right direction. Eventually, all the children will be sitting down and all the balls will be lost (This is the "heat death")

*This isn't quite right because it is an expansion, not walking from a centre, there is no centre to the universe.


What is the schedule for science runs of aLIGO (and VIRGO)?

There is no set date due to the complexity of the machines and their upgrades but this is the closest you will get to an answer:

  O1 is planned to last for four months, closing mid-January 2016. Then work will start on upgrading the instruments for our second observing run, which is called O2; those upgrades will be informed by what we have learned about the instruments during O1. O2 will start in 2016 and last around six months. Hopefully, around this time Advanced LIGO will be joined by Advanced Virgo.

They have no fixed date for the second observing run (O2) but it should be up an running later this year (my guess would be June-July time).


Due to the oblateness of the Earth (the Earth's equatorial bulge), a geocentric satellite's orbital plane will precess (rotate) relative to inertial space.
The rate at which the line of nodes moves owing to this bulge is given by
$$
\dot\Omega = -\frac{3}{2}J_2\left(\frac{r_E}{ℓ}\right)^2 n\cos\iota
$$
where $J_2$ is the is the zonal harmonic coefficient ($1.08262668\times10^{-3}$ for Earth), $r_E$ is the body's equatorial radius ($6\,378\,137$ m for Earth), $ℓ$ is the orbit parameter (the semi-latus rectum), $n$ is the mean motion, and $\iota$ is the inclination of the orbit.

For a given orbit parameter ($ℓ$) and mean motion ($n$), the inclination of a geocentric satellite orbit can be selected to obtain, for example, a Sun-synchronous orbit ($\dot\Omega=360^\circ$ per $365.26$ days, or $0.9856$ degrees per day).

Coordinate System

This Wikipedia article (and diagram from that article) describes the coordinate system used.

The orbital plane (yellow) intersects a reference plane (gray). For Earth-orbiting satellites, the reference plane is usually the Earth's equatorial plane.  The intersection is called the line of nodes, as it connects the center of mass with the ascending and descending nodes.  This plane, together with the vernal point ($\gamma$), establishes a reference frame.

Two elements that define the size and shape of the elliptical orbit
are the semi-major axis, $a$, and the eccentricity, $e$. The
semi-latus rectum is related to $a$ and $e$ by $ℓ=a(1-e^2)$.
Two elements define the orientation of the orbital plane in which the
ellipse is embedded: the inclination, $\iota$, and the longitude of the ascending node, $\Omega$.


applying nutation to ecliptical coordinates (apparent position of star)

The standard methods of applying nutation retain equatorial components of position, and for near-polar objects they avoid the drawbacks of the abbreviated approximation by converting the equatorial components of the position into rectangular components and applying the nutation matrix for the date.  In principle this method need not involve approximation.

The topic is explained in principle in chapter 3 of 'Explanatory
Supplement to the Astronomical Almanac' (ed. P K Seidelmann, 2nd ed, 1992/2006), where the full nutation matrix is specified on p.115 in terms of classical components of nutation in longitude and obliquity, dpsi and deps.

The calculations to evaluate these components has undergone much development over the years, the latest version has a very long list of >1000 trigonometrical components, but a selection of specifications and an approximation is given in the suite of algorithms and codes available at the website for the IAU SOFA software ('Standards of Fundamental Astronomy'), at  [http://www.iausofa.org] , and of course any user can apply further shortening approximations to the series.

The IAU suite has three versions of the calculation of the nutation components in longitude and obliquity, routines iau_NUT80 (older standard),  iau_NUT00A (full and very long post-2000 standard calculation), and iau_NUT00B (abridged version of the modern calculation).  There is also a routine iau_NUMAT for calculating the nutation matrix.

(An intermediate approximation for the components of nutation, useful for applications where abridgment 2000B is too drastically approximated but 2000A is too long, is given in the US Naval Observatory's 'NOVAS' software suite (Naval Observatory Vector Astrometry software) [http://aa.usno.navy.mil/software/novas/novas_info.php]. The NOVAS suite also offers a more integrated approach to calculating astrometric problems , where IAU-SOFA provides individual short modules to be selected and assembled by the user.)

Lastly, the question doesn't mention applying a precession correction as well, but in case that is wanted (which it very often is, in conjunction with nutation), the references already given above provide the needed resources.


Why can I see the Andromeda Galaxy despite Milky Way's rotation?

As barrycarter already pointed out, it is due to the huge time it takes for the milky way to rotate.

For once, as he pointed out, the stars in an orbit similar like the sun take about 250 million years for one full revolution around the center of the galaxy. If you put that in a relation to an earth year and take our civilization of roughy 6000 years of keeping records in comparison, we exist roughy 12 minutes of our galactic year. If you only take the last 150 years of the modern era, it's down to 18 seconds.

So if you compare that, how much does Earth move around the sun in 18 seconds? That is about how much Andromeda has apparently moved relative to the constellations in the night sky since we can capture visual images on film.


Are there any known planets whose axis is oriented such that one pole always faces its star?

Would this even be theoretically possible?

No. The law of conservation of angular momentum prevents this for a planet with reasonable inner structure in a gravitational field like that around the sun.

The angular momentum of the planet points in a certain direction. Changing this direction requires a torque perpendicular to the rotation axis.

If a torque existed that could force the rotation axis of the planet to always point towards the sun, the corresponding force had to have a component perpendicular to the line from the sun to the planet. The force also had to change periodically with the orbit of the planet. Obviously such a force does not exist in the planet-sun-system.


If a star were travelling at 99% of the speed of light, directly at Earth would we see it?

Processes energetic enough to accelerate a star to 99% of the speed of light are not known. A close interaction and ejection from a multiple black hole system could perhaps provide enough energy, but such close interaction with black holes would probably rip the star apart before it could reach such excessive speeds. Hypervelocity stars are known. They have speeds less than about 1% of the speed of light.

A star travelling through space at such speeds would also have problems.  It would be slamming into interstellar gas. I have no idea if it could even survive, or would it be eroded by the interaction with the thin gases in interstellar space.

A star at that speed would be significantly blue-shifted. This would make its spectrum very odd.  A small dim star could be visible, catalogued and observed to have a very odd spectrum at a distance of about several hundred light-years. It is likely that we would first observe such a star several hundred years before it got close to us. However such high speed objects probably don't exist.


Is there Kuiper belt/Oort cloud like structure in gas giants?

The gravitational region around the planets isn't hard to calculate, sometimes called the Sphere of Influence, sometimes called the Hill Sphere.  They're calculated differently but they define pretty much the same idea.

The actual long term stable region is somewhere around 50% of the Hill Sphere.   A gas giant like Jupiter simply doesn't have a large enough stable region that it has gravitational control of to have it's own Kuiper Belt or Oort Cloud equivalent.  Also, proximity to the sun and the gravitational stability of the trojan points is a factor.  What actually happens with a Sun-Jupiter system is that the stable regions are the trojan points and (because of Saturn) the Cis-Resonance orbits but not trans-resonance (Saturn is too big and too close to Jupiter, so it disrupts trans-resonance).    Jupiter's crowded Trojan points and Hildas (mostly Cis-resonance) are a little bit like it's kuiper-belt equivalent and that's a result of the sun-jupiter system, the sun being dominant and jupiter, a thousandth the mass of the sun, but large enough to be dominant among the planets.

It's worth pointing out that at Jupiter distance from the sun, the solar system is quite a bit more crowded and has largely flattened out into a plane, unlike the Kuiper belt, so it's not a real comparison to the kuiper belt, but that's where orbiting objects tend to collect, cause those are the most stable regions.   The Kuiper belt and Oort cloud aren't really regions as they encompass the entire sphere around the sun at certain distances.

It might be possible, I would think, for a gas giant in the right situation to have something like it's own Kuiper belt and Oort Cloud, but it would need to be hugely distant from it's star and far from other large planets and preferably quite massive.   The limitations would still be no more than 1/2 of that planet's hill sphere, but it's theoretically possible, I would think, under the right circumstances.   Even so, such a situation, the Trojan points and trans and cis orbitals might still be more common than any possible kuiper belt or oort cloud equivalent, so I'm just speculating that it might be possible.   I'm not 100% sure.


Why does mass naturally move closer toward's the center of other masses?

Think at the surface of a ball. A straight line within this curved surface is a great circle.
Take some point on the ball. Draw two great circles (apparent straight lines in the surface) starting in this point, with some non-zero initial angle. The two apparent straight lines intersect at the opposite side of the ball.
Wrongly assuming the surface of the ball being flat would result in the conclusion, that the lines must be curved.

The same applies to our three-dimensional space. We intutively assume it as flat, such that two lines starting at an observer with some angle should never intersect elsewhere. But in some cases they do; so we intuitively think the lines must be curved. But according to general relativity, it's actually our space-time which is curved by masses.

As an example, a ballistic curve looking like a parabola is actually a straight line in the curved space-time. So, two objects moving in different directions in e.g. Earth's orbit can eventually hit each other despite lack of drag or propulsion. They both move on a straight line (geodesic) in a curved space-time. The curved space-time allows these to lines to intersect later. Our three-dimensional Euclidean intuition interpretes this as curved trajectories.


Replacing Meade ETX with computerized mount

Meade make the 497 auto star hand controller, which plugs in where your existing hand controller goes. This turns your mount into a fully computerised one. Alternatively you should be able to use a laptop to control the mount via the aux ports. It’s all in the manual.

If your telescope did not originally come with the auto star hand controller, here is a quote from weasner.com:

  Q. How do I upgrade a non-AutoStar ETX to an AutoStar capable ETX?

  Short answer: you can't. Meade does not offer an upgrade for the
  original model ETX (also known as the ETX-90RA or ETX Astro M). But
  you can mount the OTA on a different mount (see the Telescope Tech
  Tips page) or (if still available) get a JMI computer system (see the
  "NGC-microMAX/NGC-MAX" on the Accessory Reviews - Miscellaneous page).
  Or you could sell it and by a new AT model.


There's a whole Wikipedia page about it.

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

If you don't want Alpha Cen A, then 18 Sco might be the one.


It is possible in a Trojan configuration:

In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.

However, as this link shows,

  In unnormalized units, this criterion becomes

  $$\frac{m_2}{m_1+ m_2} < 0.0385$$

  We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
  equilibrium points, in the co-rotating frame, provided that mass $m_2$
  is less than about $4\%$ of mass $m_1$.

Thus, the mass of the second star should be at most 3.85% of the central star.

As far I know, no such known star system exists, but if it would, it would be stable.

Stable planetary orbit is possible to

either very close to one of the stars (compared to the size of the triangle)
in the other Lagrange point
or very far from all of them.

If the triangle is big, then a planet even in the habitable zone is possible.


Voyager 1 and 2 detected a 20-fold increase in plasma density, significantly different refractive index?

The index of refraction of a plasma is the square root of the permittivity:
$$n=\sqrt{\epsilon}=\sqrt{1-\frac{\omega_p^2}{\omega^2}},\quad \omega_p\equiv\left(\frac{n_ee^2}{\epsilon_0 m_e}\right)^{\frac{1}{2}}$$
where $n_e$ is the electron number density and $e$ is the elementary charge. Note that $n$ is frequency-dependent, and so the difference in refractive index is wavelength-dependent. A plot at small $\omega$ indicates that at $\omega\gtrsim 10^5\text{ rad s}^{-1}$, this difference is fairly small, and presumably negligible for optical light, where we have $\omega\sim10^{16}\text{ rad s}^{-1}$:

The area of the spectrum where there is a significant change appears to be beyond even the interest of most radio observations.


Why does higher surface gravity broaden spectral lines?

In order to be in hydrostatic equilibrium, high surface gravity atmospheres have higher pressures at the same sort of temperatures. That means the density of atoms and ions is higher.

If the particle density is higher then there is more chance of interactions between the particles. In particular, collisions between atoms perturb their energy levels and reduce the lifetimes of atoms in particular states. The natural frequency width is proportional to the reciprocal of the radiative lifetime. If this lifetime is truncated due to frequent collisional (de)excitation, then the lifetime is shortened and the frequency width increased. This is known as collisional broadening.


Referring only to the image you present, the radio part of the image (I think this is a stacked optical and radio image) shows something called the Magellanic stream. Despite what the wikipedia page says, I believe it was discovered by Dieter (1971) using 21 cm observations of neutral hydrogen atoms.

The radio emission in the case of the picture you show arises from the so-called "spin flip" transition associated with changes in the relative spin of the proton and electron in a hydrogen atom. A configuration where the spins are anti-parallel has a lower energy than where they are parallel. (Quantum mechanics only allows these two possibilities). The energy difference is small - only $5.87 \times 10^{-6}$ eV. In hydrogen gas in space it is possible for the atom to make a transition from the higher level state to the lower level, in the course of which it emits radiation with a wavelength of 21 cm (i.e. radio waves). This is not seen in terrestrial laboratory samples because the gas is collisionally de-excited before it has the chance to emit radiation. Hence 21cm radiation is an example of a forbidden line, with a long radiative lifetime and which is only observed from the rarefied neutral hydrogen gas found in the interstellar medium.

The intrinsic width of the emitted spectral line is very narrow. This means that the observed wavelength of this 21 cm line is an excellent probe of the motion of the emitting gas, since the observed radio waves will be doppler shifted from the 21 cm wavelength at which they were emitted according to the line-of-sight velocity of the gas.

The gas in question here has been stripped from the two satellite galaxies of the Milky Way, known as the Magellanic clouds. Although the picture also shows copious 21 cm radiation arising from neutral hydrogen gas in the plane of our Milky Way.

There are many other ways that gas can be a source of radio emission. In addition to atomic and molecular transitions that give radio emission at discrete wavelengths (including the radio equivalent of lasers, called masers), there can be continuum emission caused by the acceleration of electrons in the electric fields of partially and completely ionised ions - so called bremsstrahlung or free-free emission. There is also recombination radiation (free-bound) radiation and continuum radiation emitted by electrons that are accelerated in electromagnetic fields - cyclotron radiation; synchrotron radiation (if the electrons are relativistic); curvature radiation (a variety of synchrotron radiation) or even just the Rayleigh-Jeans tail from cold blackbody emission (e.g. in the cosmic microwave background).

A decent introductory summary can be found here.


Cooling of astrophotography camera sensor chip

The practical length of the time exposure will depend on your equipment. Cameras used in astrophotography are generally able to be used with exposure times ranging from milliseconds to hours. As your question stands, cooling the chip will have no effect on the length of the exposure time. What it can affect is the amount of noise in the image. Cooling the sensor generally reduces the amount of noise. Cooling beyond -20°C may have limited value, particularly if you are not using research-grade instruments. While you may further reduce the noise, you may still be at the mercy of the quality of the optics, drive mechanisms, etc. and you may have less noise in the image but the limiting factor is not your image noise but the system limit for drift and distortion.

Some professional instruments use liquid nitrogen cooling while others use thermoelectric cooling, depending on the manufacturer and the intended use. The instruments that use sub-zero cooling also usually have a high quality sensor with a pixel well depth that supports longer exposures, without bleeding effects from adjacent pixels. It all depends on the sensor, camera, and other parts of the system, so your mileage may vary.


Measuring misalignment between two positions on sky

Assuming you mean the angle between the meridian line through A and the great circle that goes through points A and B, then it goes something like this.

Define vectors from the origin to A and B assuming they lie on a unit sphere, such that $x_A= \cos \delta_A \cos \phi_A$, $ y_A = \cos \delta_A \sin \phi_A$ and $z_A= \sin \delta_A$, and similar for B. Here $\phi$ refers to right Ascension and $\delta$ is declination.

The great circles in question define planes that pass through the origin. A normal to the plane defined by OAB is given by the vector product $\vec{n_1}=\vec{A}\times \vec{B}$. Similarly a great circle passing through the O, A and the NCP $(0, 0, 1)$ has a normal of $\vec{n_2}= \vec{A}\times (0,0,1) = (y_A, -x_A, 0)$.

The angle you are looking for is the angle between these two normal vectors, which can be found from scalar product in the usual way.
$$\cos \theta = \frac{\vec{n_1}\cdot \vec{n_2}}{|n_1||n_2|}=\frac{\cos \phi_{A}(y_{A}z_{B} - y_{B}z_A) + \sin \phi_{A}(x_Az_B - x_Bz_A)}{|\vec{A}\times \vec{B}|}$$


How would light from a blue or red star affect the way we see?

I guess it would be influenced. Right now, our vision is already influenced by that massive blue sky above us. The human visual system corresponds to the large influx of blue light that gets scattered in the atmosphere. Having a red or a blue star would probably also have some influence, as would have a different kind of atmosphere (which is not mainly nitrogen and oxygen). The brain would probably compensate, but the rods in the eye would have a different composition.


Is there a strong galactic magnetic field?

No, the galactic magnetic field is very weak, about 0.1nT.
It is able to bend the trajectory of highly-energetic charged particles and also to align dust grains across the magnetic field.

However, is too weak to affect the rotation of a galaxy.

Although the origin of galactic magnetic field is not clear yet, the supermassive black holes do not significantly influence the galactic magnetic field far from them.


What was the first astronomical measurement which demonstrated that "the Earth is surrounded by vacuum"?

Torricelli, the inventor of the Mercury Barometer (~1644) argued that the height of the column of mercury was governed by atmospheric pressure (the "weight of the atmosphere" as he would have put it).   He asserted that the space above the mercury in his tube was a vacuum, a totally anti-Aristotlean concept at that time.
To test this he enlisted the help of Blaise Pascal and Florin Perier in carrying one of two barometers up a mountain, the Puy de Dome in central France. His prediction was that the pressure at the top of the mountain would be less than at the base because of a lesser weight of air.
The experiment showed a drop in height (ie of pressure) and Torriceli himself concluded that the height of the atmosphere would be about 20km, above which there would be vacuum - the same vacuum as in the top of the barometer tube.
The experiment was IMHO one of the greatest experiments in the  history of physics.  See for example  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3768090/


Does the focus of terrestial telecopes need adjustment for the distance of astronomical objects?

This is incorrect in any operational sense, and no astronomer will tell you that:

  Any astronomer will tell you focus changes are needed between stars, stars farther away, planets, and the Moon. ... If focused on the Moon, you need to refocus slightly when looking at stars. When looking at Jupiter you need to refocus if the target is the Moon. Close stars vs far away stars, the same thing, a need to refocus. ... To the tune of: Jupiter and it's moons, which are close to Jupiter , need focus adjustment between them. Either Jupiter or it's moons will be in focus. Ask any astronomer. That's a difference and it's well past the supposed "infinity focus" point.

It's true that Jupiter and the Moon are at different distances and thus geometrical optics says that the focus needs to be adjusted, but this is only true in mathematically in an idealize instrument. In any real instrument, it is not possible to discern any difference between the various focus points.  They are all at infinity.

So one size really does fit all.

With exceptional quality optics, it's possible that a distant streetlight might require a very, very slightly different focus than the Moon or Jupiter. Nearby streetlights, sure.


How much more massive would Pluto have to be to clear its neighborhood?

This Wikipedia page does a decent job of describing the orbit-clearing criterion, based on the original paper by Stern & Levison (2002), which can be found here (PDF).

In order to have cleared its orbit over a period of billions of years, an object needs a "Stern-Levison parameter" $\Lambda$ which is $> 1$; Pluto has $\Lambda \approx 3$-$4 \times 10^{-3}$. Since $\Lambda$ depends on the square of the object's mass (see the Wikipedia page, or Equation 4 of Stern & Levison), Pluto would need to be about fifteen to twenty times its current mass in order to cross the threshold.


Can a planet with a hydrogen atmosphere have a water ocean with dissolved oxygen?

I'm going to say no, or at least, very unlikely.   Short answer, no sunlight, no oxygen formation.    Longer answer below.

A water ocean on a large Rogue planet with a thick enough atmosphere is possible, at least for while.  If there's enough heat coming from the surface or under the oceans through volcanism, perhaps with it being a a younger planet with higher internal heat and more radioactive elements, and/or if the atmosphere was thick enough then liquid oceans are certainly possible.  You run into problems with too think an atmosphere, because the pressure and perhaps temperature would be enormous where liquid water could form.

But if we assume somewhat Earth like, with extensive volcanic activity to maintain enough surface heat, Oxygen is more unlikely for a few reasons.   The hydrogen atmosphere would prevent significant amounts of free Oxygen from ever being present and without oxygen in the atmosphere, disolved oxygen in the oceans would be temporary if it was formed there, not equilibrium.

Oxygen is also very reactive.  It readily binds or reacts with all kinds of common elements like hydrogen, methane, Iron, Silicon, Magnesium, Aluminum, Sulfur and inside the Earth, trapped Oxygen wouldn't escape as a gas but it would likely bind with silicates pretty much permanently.

Oxygen in it's pure form is rare on planets because a whole lot of it needs to be produced before it's abundant enough to begin collecting in the atmosphere (or in liquid form if the planet is cold enough . . . but that's not relevant with a liquid water question).  Photosynthesis is a great way to produce Oxygen from water and CO2 but it only works with sunlight as the process is very energy intensive.   On a rogue planet without sunlight, I can't think of any way to produce any statistically relevant amounts of oxygen inside the oceans.

Plants on Earth produce oxygen in abundance because of free energy from sunlight and easy access to CO2 and H2O, but without all that free energy, I don't see it happening.

Extremophiles come in may types, so I don't want to say it's impossible, but life that we know of relies on chemical energy and the laws of chemical energy are pretty straight forward and the production of free oxygen as a by product is difficult without a significant supply of energy.

Extremophiles, loosely speaking, eek out a living on not very abundant chemicals from volcanic gas escaping into the ocean floor or in other difficult situations.    There's many different kinds and more possibilities if you factor in Astrobiology theories, but I have a tough time seeing a method where oxygen is released.   There's plenty of heat in ocean vents, but heat isn't easily converted into usable energy by bacteria.  Chemical energy and exothermic chemical reactions to create stored energy, or photosynthesis is generally how the life we know creates and stores it's energy.   The chemical energy required to release oxygen in significant amounts seems very impractical without significant sunlight.

Now, granted, the biology and chemistry behind what extremophiles can produce, gets pretty complicated and is closer to biology and perhaps chemistry than astronomy (though some might say astrobiology is part of astronomy), but in addition to that, exploring what chemical processes might be possible would get very long and above my paygrade so, there's a chance it's possible, but I feel pretty confident in saying your scenario is unlikely.


How do you figure out the magnitude of stars?

You must remember that there are Absolute magnitude (that measured from a fixed distance of 10pc) and Apparent magnitude (which is measured from Earth).

Historically magnitudes ranged from First to Sixth and were assigned by guess by Ptolomaeus. His First Magnitude was the equivalent of saying first size, so it was the brightest, and Sixth Magnitude was the assignment he made for the dimmest seen with the naked eye.

Nowadays, we use roughly the same scale but with real numbers, ranging from -26.74 (Sun) up. We defined the star Vega to be the exact 0.0 apparent magnitude (and 0.0 colour in all filters, for the case), and then we use the formula

$m_1-m_{ref}=-2.5log{I\over I_{ref}}$

to define all other apparent magnitudes from that one by measuring Intensities.

So to answer you, yes, we measure apparent magnitudes rather accurately. For stars other than Sun, they range from -1.46 for Sirius up to +8 in a perfect night for a trained naked eye (typically +6 for the normal eye on a normal, far from city lights, night).

Reference http://en.wikipedia.org/wiki/Magnitude_(astronomy)


How can I find the width of an orbit knowing some values?

To find the width (semi-major axis, see Wikipedia) of an orbit, you only need two of the following: perihelion, aphelion, and eccentricity, represented by the variables $P, A, $ and $e$, respectively, and $a$ is the semi-major axis.
If you are given perihelion and aphelion, you can just take the average of the two: $a=\dfrac{A+P}{2}$
If you are given perihelion and eccentricity, you can calculate $a=\dfrac{P}{1-e}$. Similarly, given aphelion, the width of your orbit is $a=\dfrac{A}{1+e}$.
If you are talking about the other kind of width (Semi-minor axis), represented by $b$, then the formulas are:

Given $P \text{ and } A: b = \sqrt{A \cdot P}$
Given $P \text{ and } e: b = \sqrt{-\dfrac{P(1+e)}{e-1} \cdot P}$
Given $A \text{ and } e: b = \sqrt{-\dfrac{A(1-e)}{e+1} \cdot A}$

I hope this helps.


Why does the Gaia color-magnitude diagram have this shape?

Try plotting absolute G magnitude (i.e. corrected for the fact tthat all the stars are at different distances) on the y-axis.


Why do astronomers like green laser pointers?

Mostly general purpose laser pointers are used for pointing things at smaller distances eg. Diagram or Equations in Powerpoint presentations, so the power of such laser pointers is quite limited/restricted to 5mw ( Class 3A or IIIa) or 10mW in some regions.
Because of the low power and small aperture of laser pointers if you point them through empty space, you can only see them impact spot where it hits a surface.

So to point stars/planets as in astronomy club during observation sessions these low power laser pointers are not useful as there is no surface where the beam can hit & you can see the spot.

Some higher-powered laser pointers project a visible beam via scattering from dust particles or water droplets along the beam path.
Higher-power ( Class 3B or IIIb lasers: greater than 5mW) and higher-frequency (Green or Blue) lasers may produce a beam visible even in clean air because of "Rayleigh Scattering" from air molecules, especially when viewed in moderately-to-dimly lit conditions. The strong wavelength dependence of the scattering (~λ^ −4 ) means that shorter (green & blue) wavelengths are scattered more strongly than longer (red) wavelengths.

[ Rayleigh scattering is the mainly the elastic scattering (in lay term collision) of light or other electromagnetic radiation by particles much smaller than the wavelength of the light.
Rayleigh scattering results from the electric polarizability of the particles (locally separating positive & negative charges by small distance apart inside the atom. i.e. temporarily separating centre of positive charge and centre of negative charge). The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency (hence the green colour). The particle therefore becomes a small radiating dipole (or small sources of light) whose radiation we see as scattered light (the beam passing from pointer to star pointed). ]

The intensity of such scattering increases when these beams are viewed from angles near the beam axis. (That's the reason, if you are standing close to the person pointing stars in astronomy club, you can see the beam bright and clearly whereas the people standing away can barely see it. So always stand near the presenter But avoid contact with your eye. :-) )

The apparent brightness of a spot from a laser beam depends on the optical power of the laser, the reflectivity of the surface, and the chromatic response of the human eye. For the same optical power, Green Laser light will seem brighter than other colours because the human eye is most sensitive at low light levels in the green region of the spectrum (wavelength 520–570 nm). The most sensitive pigment, rhodopsin, has a peak response at 500 nm. Sensitivity decreases for redder or bluer wavelengths.

Further, Green laser pointers are of moderate power (sufficient for Rayleigh  Scattering in clean air), they are compact and relatively cheaper than blue ones (even though human eye is less sensitive to blue colour, blue laser are a little bit more expensive than green one.)

Avoid their contact with your eyes and don't point them at aircraft.
People have been given up to five years in jail for aiming a green laser at an aircraft.


Distribution of Stars in Milky Way and globular cluster analogy

Particles in a gas approximate to point-like objects that interact roughly elastically through short-range forces when they collide, but are otherwise non-interacting.

Stars interact gravitationally over long ranges, occasionally with each other, but always with the overall gravitational potential of the system.

Sometimes people do talk thermodynamically about star clusters. You can discuss the "temperature" of stars when you are referring to the velocity dispersion in a cluster. The concept of heating or cooling a cluster also has some merit.


Why does the sun appear 13x bigger through diffraction in Voyager 1 image from 1990, but not from Earth?

Don't base the size of the Sun on what you see in that image because the Sun is over-exposed.  The big spot in the center of the diffraction spikes are saturated pixels because they all appear to have the same value (the spot looks uniform) so you don't know what their real values are; you would need to take a shorter-exposure image (or put filters in front of the lens) if you wanted to take a proper picture of the Sun.  Since Venus is a lot dimmer than the Sun, they set the exposure long enough to see it and accepted the fact that the Sun would be over-exposed.  You see the same thing in some of the Hubble images where to bring out the background stars, the brighter ones are over-exposed.

Another way of looking at it, if you see an image where you can very clearly see the diffraction spikes, that part of the image is over-exposed.


Both formulae are correct. The discrepancy is because the formula from the eccentric anomaly article uses the centre of the ellipse as the origin, but the formula from the Kepler's equation article uses a focus of the ellipse (i.e, the central gravitating body, eg the Sun) as the origin. Note that $c = ae$ is the distance from the ellipse centre to a focus


What possible effects of life replicating on gas giant atmosphere?

Yes, but probably not by a huge amount. There are two obvious ways this could happen:

Direct change in atmospheric composition. If our microbe does something like eat methane and excrete longer-chain hydrocarbons, that's going to eventually change the atmosphere enough to have some effect. (If the hydrocarbons weren't already forming, they probably aren't stable under local conditions, so there will be some equilibrium reached, the way the Earth's atmosphere isn't 100% oxygen despite hundreds of millions of years of photosynthesis.)
Change in solar heat balance. The change in atmosphere will probably also change the amount of greenhouse effect (one way or the other), and second-order effects or even just enough microbial biomass can change the albedo (again, either way). On Earth, these effects can combine to change the temperature by tens of Kelvins, which would be a few percent different in atmospheric volume.

In neither case is the added life going to change the total mass of the atmosphere.


Is it possible for a solar flare in one star to influence the chances of a solar flare in another star?

I'd say it's an open question awaiting some observational evidence that it can happen.  It presumably would depend a lot on what kind of situation you are looking for, and what would you call a flare on one star setting off a flare on the other.  Certainly you'd need two stars that both have active atmospheres and are in a close binary, such that one mass ejection could interfere enough with the magnetic configuration in the other star to set off a flare.  Some stars may be so close together that the magnetic configuration of their atmospheres actually connect both stars, so then when get a flare you might see brightenings on both stars but it would be essentially the same flare.  There are even stars called "contact binaries" where it's not even all that clear if you should think of that as two stars that touch (and share a magnetosphere), or just one star that has two separate cores.  But the bottom line is, it sounds like you'd need something unusual going on, involving two stars that were unusually close together and are interacting in some way.  You might want to read up on "RS CVn stars," for example.


What is a UBV source for stars or stellar objects?

According to the associated paper (to get to this, click on the Bibcode link near the top right in VizieR), this refers to the source of the UBV photometry.
I believe the relevant references are the ones in the V/19/clusters table, e.g. for the second cluster (NGC 188) the reference is to Upgren, Mesrobian & Kerridge (1972). This does provide cross-references, but only to another catalogue of the cluster by Sandage (1962). It isn't particularly surprising that there aren't cross-references to more general catalogues, as identifying which star is which in such crowded fields can be challenging depending on how accurately positions and stellar properties are given.
Tracking some of the other references down is challenging, this is not helped by the fact that the ADS seems to have Romanized several of the names in various different ways, and searching for one Romanization does not return results for the others (at least, nothing came back for "Lavdovskij", although there are papers attributed to a "V. V. Lavdovsky" and "V. V. Lavdovskii"). I'm not sure how complete ADS coverage of Soviet publications (e.g. Izv. Glav. Astron. Obs. Pulkovo) is.


Where will the planned Deep Space Atomic Clock orbit "after" Earth?

"The instrument will be tested in Earth orbit for one year, ..."
What would it then orbit?

It will continue to orbit the Earth, at least for a while. It will eventually end up not orbiting the Earth because it will be placed in low Earth orbit, where orbits decay due to atmospheric drag.
The project is a technology demonstrator, which means a limited amount of funding.


You can use the Purdue 'Impact Earth!' website to estimate the effects of impacts of objects of various size, speed and composition. As detailed in the accompanying documentation, which was published in Meteoritics and Planetary Science, the energy calculation and atmospheric drag during entry is relatively straightforward physics; things like crater sizes, overpressure and thermal effects are scaled from nuclear tests.

Plugging the numbers in for 2019 OK involves some assumptions since there is considerable uncertainty (~50%) on the diameter since we don't have an albedo and this leads to a large uncertainty in the volume. In addition, since nobody got a spectrum during the close approach, we don't have a good idea on the NEO's taxonomy and therefore what it's made of. This makes the mass, and therefore the energy of the impact, somewhat uncertain (since $Energy\sim\frac{1}{2}mass \times velocity^2$).

If we assume 80m diameter and density of 2000 kg/m$^{-3}$, which is mid-way between that of C- and S-type asteroids (most common types among NEOs, making up ~65% of NEOs; Binzel et al. 2015, in Asteroids IV), and an encounter velocity of ~25 km/s (rounded up from JPL Small Body DB close approach data) gives ~40 Megatons of TNT equivalent. ImpactEarth! also reports:

  The projectile begins to breakup at an altitude of 71800 meters = 235000 ft
  The projectile bursts into a cloud of fragments at an altitude of 3860 meters = 12600 ft.
  The residual velocity of the projectile fragments after the burst is 4.94 km/s = 3.06 miles/s.
  The energy of the airburst is 1.61 x 10^17 Joules = 3.85 x 10^1 MegaTons.
  No crater is formed, although large fragments may strike the surface.

If you put in a distance from the impact point, it will calculate the effects of the airblast for you, which will be the main effect (as it was for the ~0.5 Megaton 2013 Chelyabinsk impact)


Could spy satellites use laser guide stars (for adaptive optics)?

The guide star laser is used to adapt the optics at a specific time and location, for specific atmospheric conditions. It is not, as your question implies, used to calibrate the optics for long-term use.

Therefor, if order to calibrate the satellite with a laser fired from Earth, the reconnaissance target would have to provide the laser. This is useless for non-cooperative targets, which I would imagine make up the bulk of the reconnaissance targets.

I suppose that one could fire a laser from the satellite itself, but this would have the obvious disadvantage of making it even more obvious when a spy satellite was overhead.

In any case, the guide star image does not penetrate the entire atmosphere, I believe that it produces an image just a few tens of kilometers up. This is useful because it is the thicker, denser and more turbulent lower air of which is the most concern to optics. Firing the laser from above would thus see the laser energy absorbed in the thinner, stiller, more predictable upper atmosphere, thus it would have must less advantage.

One final reason not to use a guide star laser on a reconnaissance satellite would be power budget. These satellites are as small as possible, with as small a heat and reflectivity index as possible. The added power source, be it RTG, conventional batteries, or solar, would likely increase the reflectivity of the satellite.


SOFA is the Standards Of Fundamental Astronomy, approved by the International Astronomical Union (IAU) as being the canonical standard for calculating positions and time in astronomy. It is available in C and F77 versions as standard, and wrappers have been written in other languages, most notably Python as part of AstroPy. For your purpose, the thirdparty part of the website mentions the World Wide Astronomy (WWA) library which is C# so that would be worth a look for your application. I would recommend reading the SOFA Cookbooks, particularly the ones on Earth Attitude and Time Scales to get a better idea of how the (many) various time and co-ordinate systems fit together.


What percentage of visual stars are actually binary stars?

This is a difficult question, It's been changing through the years and it's difficult to calculate accurately. Now it's know that this percentage changes depending on the star type and ranges from 50% for Sun like stars up to a 80% for type O stars.

Fonts:

http://www.space.com/1995-astronomers-wrong-stars-single.html (2006)

http://www.space.com/22509-binary-stars.html (2013)


Is it possible that a star system ejects an object out of the galaxy?

Intergalactic planets have never been observed.  They're too small and it's too great a distance, but intergalactic stars have, and if stars can be ejected from a galaxy, planets can to.  The orbital mechanics is the same for both.

The escape velocity of the galaxy is impressive.  For solar-systems, where most of the mass is in the star in the center, the escape velocity is just about the orbital velocity times the square root of two.   Source.

For the Milky way excape velocity calculation is larger and more complicated because most of the mass is in the dark matter halo and very much spread out, not localized in the center.   The center is still more dense, the mass is far more distributed so the square root of two rule doesn't apply.  Using the Sun as an example, the orbital velocity of our sun is about 220 km/s and it's escape velocity is 537 km/s.  Source.   As objects moves further towards the edge of the Milky way, the escape velocity decreases and the orbital velocity generally increases, so at close to the edge of the Milky way, you'd get something close to the square root of two ratio again.  That means an object close to the edge of the Milky way only needs to add 41.4% of it's velocity to escape.

Gravity assists depend on the relative velocity of the larger object to the "stationary Milkyway", so, if our solar-system was to fly close enough to very massive star and get a big gravity assist, assuming the massive star has a similar orbital velocity around the Milky way, 220 km/s, that could add as much as 440 km/s to our solar-system's velocity which would be enough to escape the Milky way.   Only large stars have enough gravity for that kind of kick, but there's enough large stars in our galaxy that they shouldn't have a hard time ejecting objects that fly reasonably close to them.   You probably need at least a half dozen or dozen or so solar masses and a close fly-by, but, statistically, it should happen from time to time.

The Wikipedia article points out that galactic collisions could be responsible for these stellar ejections, and galactic collisions probably do eject a great deal more material compared to a galaxy in a peaceful spiral arm rotation, but it's still possible for gravity assists to cause objects to be ejected even from a galaxy not currently colliding with anohter, because twice the orbital velocity of the large star plus the orbital velocity of the approaching star or planet, when combined is greater than escape velocity, at least for the outer parts of the spiral galaxy.   Needless to say, most gravity assists don't line up perfectly, and half of them push an object further in, not further out, but statistically, and with multiple assists, it's very likely that stars and rogue planets get ejected from the Milky-way, it's probably happening as I write this.

By some estimates, there might be a HUGE number of objects that doesn't belong to any galaxy.


Could this estimate of the size and mass of the Chicxulub Impactor be accurate?

Well, that article was never accepted for publication in any peer-review journal apparently.

That said, estimates may vary widely depending on assumptions about the composition and velocity of the asteorid. One could estimate the mass of the object assuming compositional ratios similar to a certain class of objects and integrating the whole Iridium deposits but that's still just a very indirect method with high uncertainties and a lot of assumptions to be made.

One may also estimate the impact energy by using the size of the crater. Since that is related to kinetic energy it is a way to estimate the mass and velocity of the impactor, but this are coupled. Not only that, the kinetic energy depends on $m$ and on $v^2$ thus is very sensitive to the velocity of the object. The impactor might have had $m = 1$ and $v = 1$, or $m = 0.25$ and $v = 2$ (in arbitrary units) and the kinetic energy would have been the same. As you can see this dependence is a source of high uncertainty in the mass. The assumption of the impact velocity is needed and the precission of this assumption is very important if you don't want a wide variety of masses for the impactor. Once you got the mass of the impactor you need to know the density of it to estimate its size (another assumption). If it was a comet then the density is as low that for that mass the object would have been gigantic (but then one have to explain the low impact speeds since a comet would have been traveling faster in general). A purely rocky body is denser and thus for the same mass you need a smaller diameter. We in fact have done geochemical measurements on the impact site and there is strong evidence that the object was an asteroid and in particular a carbonaceous chondrite, which means that we have a good estimate of the density of the object. Detailed analysis of the structure of the impact crater can also decouple the mass-velocity constraint whithin the $E_{kinetic} = \frac{1}{2}mv^2$ relationship. Overall there best estimate is still around 12 km in diameter (+/- 3 km).


A system of coordinates requires a fixed plane to define "zero latitude" and a direction chosen in that plane to define "zero longitude".

For directions to the stars, the distances are so great that it matters not if you base your coordinates at the Sun or at the Earth, because the distance between the Sun and Earth is so small in relation to the distances to the stars.

There are a few coordinate systems in use.  The first uses the Earth's Equator as the fixed plane and the direction of the sun on March 21 (vernal equinox) as the direction of zero longitude. For historical reasons the longitude in this system is called Right Ascension and latitude is called Declination. It would be quite suitable for your adventurers to use.

You can also use the plane of the solar system, or the plane of the galaxy as the fixed plane. In these coordinates systems Proxima is located at

-44:45:47.62 Latitude(deg)   +239:06:53.08 Longitude(deg) Ecliptic
-01:55:37.90 Latitude(deg)   +313:56:23.96 Longitude(deg) Galactic

But remember that Proxima is moving, so these figures are not fixed.


What could have made a satellite appear to move side to side while near the horizon?

Your link shows estimated magnitude, and it gets quite dim when it gets close to the horizon.

above: from heavens-above, (from url: lat=52.2681&lng=-113.8112)

As a spacecraft in low earth orbit appears to approach the horizon, the line-of-sight distance from it to you can change from a few hundred to a few thousands kilometers, which is why the brightness in your table changes so much.

The path in the sky shown in the link is going to be almost vertical (reaches 70 degrees). Wikipedia says GEOS-3 is a satellite put into orbit in 1975. These days there is a big effort to de-orbit the "leftovers" so they burn up in the atmosphere (space junk mitigation) but back then it was common for each launch to leave other stuff up there for decades or longer.

Wikipedia says the COSPAR-ID for the actual GOES-3 satellite is 1975-027A. That means it was the 27th internationally recognized launch in 1975. "A" is assigned to the GEOS-3 satellite itself, but if you look up 1975-027 in the satellite catalog (SatCat) at Celestrak.org you can see there are were four other objects associated with the launch being tracked. One of them has burned up, but the others are still there.

You can double check if 1975-027B is the one that your link is talking about. Anyway, since they are so old, none of them is capable of a spectacularly sudden engine burn and change in orbit. So I'd have to say that whatever you did see, it isn't associated with the GEOS-3 launch.

Another point to consider; if you just coming out of a restaurant, the chances are low that you could see a 5th magnitude object. Near the horizon the apparent speed would be much slower than that of a satellite overhead.


Why is this shooting star wavy or squiggly toward the end of it's visible streak

It's shutter vibration. It begins at one end of the trail (which I'm guessing is the start of the exposure) and quickly damps out.


Array of telescopes with single orbiting collector?

If I understand your question, what you're talking about is one of the things that has been on the astronomical wish list for ever.  There are two ways to combine separated telescopes.  One of them (the easy way) is to just overlay the images from the two telescopes.  In this way you get to double the light you collect, but you have to take care to overlay the two images very well, such that they have the same magnification, rotation, etc.  The resolution you get doing it this way is only the resolution you get from either individual telescope, so the main reason for doing this is to increase you signal-to-noise (it doesn't make sense to do this if you're looking at a bright scene).

The other way to combine multiple telescopes, which is along the lines of what you are thinking, is to phase the two telescopes, meaning that not only are you laying the images on top of each other, you are also doing it so that the two images have the same phase, meaning they have traveled exactly the same distance from source to collector.  What you gain by doing this is that the resolution you get this way is what you would get if you had a telescope aperture that circumscribed the two telescopes.  For instance, if you had two 10 cm telescopes separated by 10 meters, you'd end up with an image that had a resolution of a 10 meter diameter telescope; however it would only have the light collection capability of the two individual telescopes, so though you have the resolution of a 10 meter telescope, you have the light collection capability that is only $\frac{2*{0.05}^2}{5^2} = 0.02\%$, meaning that you'd need to take an exposure of 5000 seconds to get the same signal you'd get with a 1-second exposure with a 10 meter telescope.

The reason you can easily make these virtual telescopes with radio telescopes is that the level you need to match path length goes as a fraction of the wavelength you're looking in.  Radio waves are many many meters long, so if you need to match the path length to, say, 1/50th of that, you're talking distances that are on the order of many centimeters or even meters.  What this means is that the radio telescopes can measure the incoming radio waves and time-tag them with atomic clocks; the ability to time-tag and the level that we can synchronize separate clocks results in phasing errors that easily fit within the phasing requirements.  Radio astronomers can record the data at each telescope then after the fact combine their signals in a computer.  However, for visible light, a fraction of a wavelength is now on the order of nanometers.  We currently don't have the ability to time-tag optical data to be able to go back later and do the phasing, which means the phasing has to be done real-time.

So the challenge to do what you are suggesting is that we need to phase the separate telescopes in real-time.  This means that we need to hold the optics in the separate telescopes so that they are not only matched to the nanometer level, but you need to hold them to that level as you are imaging.  We can easily do that on an optics table in the lab, which is basically what you are doing when you make a Michelson interferometer, but doing it on separated satellites that are moving around, vibrating, etc., it is a bear of a challenge.


How to interprete function value of get_sun in astropy

Looking at the astropy documentation, the Sun's position is calculated in GCRS co-ordinates these are intern expressed in J2000.0 and not JNow.

The difference is therefore due in part to precession between J2000.0 and JNow.

You will also find that the time of the equinox is not exactly 4:30 but has been rounded. The algorithm used to get to 4:30 itself may also have precision limitations due to all sorts of reasons. The same will be true of the calculated position of the Sun at a given time.

These errors will all add up to give a slightly incorrect figure.

Cartes du Ciel (a well known planatarium application) calculates the Sun's location at 4:30 as being:
JNow
RA: 00h00m01.71s
DE:+00°00'01.4"
J2000
RA: 23h59m11.85s
DE:-00°05'23.6"

So it looks like most off the error is due to limits of the Astropy library.


Computationally intensive technique to search for TNO's within the noise of historical photographic data?

Dark Energy Survey data have yielded several TNO discoveries.
Bernardinelli et al. 2020 describe their method in detail.
First they compute a trial orbit from images where a transient object was detected. Then

Final validation of the reality of linked orbits uses a new "sub-threshold confirmation" test, wherein we demand the object be detectable in a stack of the exposures in which the orbit indicates an object should be present, but was not individually detected.

Their figure 10 shows a candidate object passing that test; figure 11 shows one failing.
Rice and Laughlin 2021 use a baseline subtraction and shift-stacking technique to search TESS data for TNOs without prior detections.
After verifying that they recover a few known TNOs this way, they list a few unknown candidates for further observation.
If neither of these is what you're thinking of, ADS's References and Citations links may help you find it.


Where should I look at if I had a 25m parabolic radio reflector?

On the fast radio burst side of things:
The FRB distribution is isotropic so far, which isn't surprising - you'd expect cosmological sources of this sort to be more or less everywhere in the sky. This means that you could point your telescope at random and still have a roughly equal chance of detecting one - albeit an extremely low chance. Fortunately, in the last couple of years we've discovered that at least a small subset of FRBs repeat; pointing your telescope at one of those known repeaters could substantially increase your odds of a detection.
I don't think there's been a substantial effort to look for FRBs solely with small dishes, but it's been attempted with a 25-meter dish. Spitler et al. 2018 used the 25-meter Stockert radio telescope, operated by the amateur astronomy group Astropeiler Stockert e.V., in conjunction with main observations by the 100-meter dish at Effelsberg. They were looking at FRB 121102, the first known repeating FRB. While Effelsberg detected three bursts from that source during their observations, Stockert turned up nothing - not overly surprising. As it turned out, the bursts Effelsberg picked up had flux densities much lower than you'd need for a reasonable signal-to-noise ratio with Stockert. There have been a couple other non-detections with small dishes, including the Nanshan 26-meter dish (Men et al. 2019).
On the plus side, you might have better options than FRB 121102, depending on your latitude. The ASKAP folks recently detected a bright repeat burst from the source FRB 20201124A, and say that the first burst could have been detected at SNR > 10 with a 20-meter telescope (with adequate specifications, of course). So depending on your precise setup, you might be in business - assuming, of course, you've got the proper setup.
I know less about exoplanet radio emissions, but I'd say your odds are probably much worse, partly because this is an area which has only had success even more recently than repeating FRBs. There was a detection from Tau Boötis this past year (Turner et al. 2021), and some previous tentative signs from elsewhere, but we have yet to get follow-up observations confirming things. You could point your telescope at Tau Boötis and hope for the best, but I wouldn't get your hopes up. Even that signal wasn't exactly strong.


Leonid meteor showers and the Tempel-Tuttle comet

When material is shed from a comet, it mostly continues to follow the same orbit as the comet, but drifts out ahead or behind the comet in a complex way due to interactions with solar wind and the gravity of other objects.

If you imagine the elliptical orbit of 55P/Tempel-Tuttle, it will have patches of denser dust in clumps spaced out around the ellipse corresponding to different perihelion approaches.  The 1466 clump will be in a different place than the 1500 clump, and they will both be travelling more or less around the comet's orbit.

Since the comet's orbit only crosses Earth's at one place, and we only cross that spot once a year, we will only encounter whatever clumps of dust happen to be at that point at the same time the Earth is.

Calculating exactly where these clumps of comet dust are is non-trivial.  And that is a big reason why predicting how good a meteor shower will be is a rather tricky business.


What's the equation for moon-moon tidal heating?

This is for instance the case for the Saturn moons Enceladus and Dione.

This is not the case for Enceladus and Dione. Enceladus's orbit about Saturn has an eccentricity of 0.0047, which while low, is more than enough to result in tidal heating. Dione does play a role in the tidal heating of Enceladus, but that role is secondary. Dione is what keeps Enceladus's eccentricity non-zero via Enceladus's 2:1 mean motion orbital resonance with Dione.


Looking at the Expanding Universe through the lens of relativity

If you work out the equations of motion for the gravitational free-fall collapse of a clouds of particles (or galaxies), you will find that, from the point of view of any of these particles, all other particles appear to be approaching. This is true in both Newtonian theory and general relativity.

The calculation is found in numerous text books and lecture notes, e.g. Susanne Höfner's, but a qualitative argument goes like this:

From your point of view, sitting on one of the particles, a particle farther from the center of the cloud does "feel" a smaller gravitational pull from the central particles. This is, I think, the reason that you think it should accelerate more slowly. However, any particle feels the gravity from all particles that are closer to the center, so the farther you are from the center, the more particles will attract you. On the other hand, the gravity from all particles that are farther away from the center cancels out, so you don't feel those.

This is, in essence, Newton's shell theorem, and it turns out to imply that all particles reach the center at the same time. The distance between each particle is getting smaller and smaller, until the cloud collapses into an point after a a time inversely proportional to the square root of the density.

This is exactly opposite to what is observed.


Well, there might be.

I strongly disagree with your statement that circumstellar discs are what differentiate B(e) stars from other related stars. In fact, there are four primary criteria for stars that satisfy "the B(e) phenomenon":

Strong emission lines in the Balmer series
Emission of lines of certain (low ionization) metals such as Fe II
"Forbidden emission lines" in the visible part of the electromagnetic spectrum such as O I
Hot circumstellar dust

The dust is not the key to the B(e) phenomenon, though it can change the spectrum slightly.

HAeB(e) stars are a class of stars (and a subclass of B(e) stars) related to Herbig Ae/Be stars, which have not yet left the main sequence. There are additional criteria that a star must satisfy to be considered an HAeB(e) star, too. These stars can be of spectral type A, meaning that you can have B(e) stars that are not of spectral type B, if you use a certain definition.

This answer is based largely on Lamers et al. (1998), a fascinating paper on the classification of B(e) stars.


Estimating the Rate at which Moon is Moving Away from the Earth

This is such a complex problem that I think your chances of getting an accurate answer are almost nil.

First, the equations are likely to be highly complex if you want to be accurate down to the tenth of an inch. For example, you'd need to model the exact shape of the Earth due to the Moon's tides, most likely accurate down to the meter or so. Scientists have done this on average, but finding how these types of models minutely change on an hourly basis is near impossible. The data just doesn't exist.

Second, filling in the variables you'll have in your equations is likely going to be impossible because many won't be known. For example, do you know the flexure/elasticity/tensile strength/etc. of the entire Earth's crust at every point on Earth over time? My guess is you don't. You'd need to know that though to do really accurate calculations.

The only way we know the recession rate of the Moon is through measurements. Crude models might get you crude order of magnitude numbers, but to match the precision of observations would require models that are far more informed than we currently are.


Difference between first (focal length (fl.) 500 mm, ocular (o.) 10 mm) and second telescope (fl. 1000 mm, o. 20 mm)? (both 50 x)

The larger the entrance aperture, the greater the light-collecting capability.  Now, you didn't specify that, but that does matter in your final choice.

Next,  as the comments suggest, the choice of optics depends on the cost vs. quality trades.  You want not only a given magnification but also as little coma, distortion, spherical aberration, and so on.   At a very simple (and probably unrealistic) level, the longer lens pair (1000,20) might be easier to fabricate with low aberrations.  In real life, we quickly descend into multi-element systems, or cemented doublet designs, and so on.

I'm guessing you are talking about a Keplerian design (two positive lenses), in which case consider how much longer the physical telescope is in the 1000-20 case.


Yes, there are quite a lot of them, but ISS is by far the brightest. The other satellites can be visible with magnitude fainter than 3. "Can be" means in dusk/dawn, when the satellite is in the sunlight.


Is the difference between LIGO & Virgo and their "Advanced" versions really generational, or were these just planned incremental upgrades?

I think this is largely a matter of what you decide is incremental as opposed to a generational change.
The basic location and overall size and detection principle of the interferometers did not change. An important change to the interferometer configuration was the addition of the signal recycling mirror that allows a certain amount of "tuning" of the frequency response. A homodyne readout scheme was employed instead of heterodyne - and the interferometer is run slightly off the dark fringe as a result. The test mass suspension system changed from being a single to a quadruple pendulum and the suspension wires are now silica rather than metal. The test masses themselves were increased from 11 kg to 40 kg which reduces the low frequency noise significantly.
I believe most of the other changes were essentially upgrades to the materials and technology being used. New mirrors, better lasers, improved vacuum, better seismic isolation etc.
I don't think it is anywhere near the generational change between aLIGO and the proposed Einstein Telescope, which will be in a different location, have a different size, different number of arms and geometry and will be operated cryogenically etc.
However, as pointed out by @mmeent, the improvement in sensitivity is about an order of magnitude in both cases, which is probably what justifies the "next generation" tag for going from LIGO to aLIGO and from aLIGO to the 3G detectors.


Definition of Rotation versus Revolution

A planet that is not rotating with respect to the stars is not rotating. In your case this would give it a solar day length of 100 Earth-days. Such planets are not likely to exist in reality.

This is in contrast to the actual case when a body is tidally locked, such as our moon.  The moon rotates once per (lunar) month at exactly the same rate as it orbits the Earth.

There are different ways of defining "length of a day".  There is the (average) time from midday to midday (this is 24 hours, or one solar-day) or there is the time it takes for the stars to return to the same position that they were the previous night. This is the "sidereal day" and is about 23hours 56 minutes. The Earth takes 23hours 56 minutes to rotate (and not 24 hours)

For your planet, the sidereal day length is undefined and the solar day is  100 Earth-days. We would say that the planet is not rotating.