Likely more Neurinos are produced billions of years ago than today because there were more very large stars in young galaxies.   I'm not going to touch the first few seconds after the big bang.   Someone else can if they want, but post star formation, it's a surprisingly simple problem to work out.

Neutrinos are produced primarily in nuclear reactions in stars.   Large stars produce far more than smaller stars.  A star with 10 solar masses, would, very roughly, produce about 10,000 more neutrinos.   Stars that size are uncommon because they only live about 10-15 million years.   Only about 1 in 800 stars are B-type or greater in terms of neutrino production and most of those are only a few solar masses, producing between maybe 50 and 500 times the number of neutrinos our sun produces.   Very high neutrino producing stars are sufficiently rare that they don't throw off the numbers.  (I'm ignoring supernovas, but I'll get to that).

If we estimate 100 billion stars in the milky way, the larger ones give off far more neutrinos than our sun and the smaller ones, far fewer.   And our sun gives off 1.79 x 10^38 neutrinos per second.   Doing some very bad estimating, all the stars in the Milky way together give off probably in the range of 1 x 10^49 neutrinos per second.  Figuring our sun, being on the large size, gives off more neutrino to mass ratio than average, but not enormously more).   Again, we're ignoring supernovas.  You'll see why soon.

So, bad estimate, 1 x 10^49 per second from all the stars in the galaxy, or 3 x 10^56 per year.   Or 1 x 10^58 in about 30 years.   (Why do I care about 10^58 you ask?)

Because that's the number of neutrinos that the type II supernova observed in 1987 released.  Source.    So one type II nova produces the same number of neutrinos as all the stars in the milky way currently produce in about 30 years, give or take.

The milkyway has an estimated one billion neutron stars and about 100 million stellar mass black holes.   Each of those can only form by type 2 supernovae.   Given the age of the Milkyway, about 13 billion years, if we average it out, that's one type II supernovae every 12 years or so.   The current rate is much slower, so the young milky way had significantly more type 2 supernovae than it currently averages, and, it would be consistant that there were significantly more large stars.   Even with significantly fewer smaller and main sequence stars, the large stars and type II nova are the clear winners in neutrino production.

The current supernova rate (Type 1 and II) in the milky way is one every 50 years, so the rate had to be several times higher when the galaxy was young.

Peak Neutrino production would coincide with peak type II supernova explosions, which would only be 5-15 million years or so after peak very large star formation.   I don't mention Type 1 supernova, because, while more common now, they (I think) produce far fewer neutrinos.  Still a great deal, but well fewer than 10^58, though I had some difficulty finding a specific number.

It's very likely that young spiral galaxies like our milky way produce (very bad guess), maybe 10 times more neutrinos than they produce some 13 billion years after formation, so that would be the most likely answer.   Once large star formation in galaxies was well underway, Neutrino production probably peaked, whether that was when the universe was a few hundred million years old or a couple billion, I don't know, but neutrino production likely peaked early with the formation of many large stars and began a steady gradual decline after the peak.

Corrections welcome.


Stopping down a telescope to reduce brightness by 5 magnitudes

Treating as a school question

Each change of 1 magnitude changes brightness by how much? So by what factor will a change of 5 magnitudes give? (5 magnitudes is a very convenient number for this question)

You will need to reduce the area of the telescope by the same factor.


There are a variety of models of stars collapsing into black holes without proper supernova explosions. These are often called failed supernovae or direct collapse black holes - although the former seems more common. Failed supernovae can happen if the initial shock wave rebounding from a collapse loses enough energy, causing it to fizzle out. Neutrinos produced in the core often supply this energy, but a small enough burst may not be enough to keep the shock going. There might then be a simple collapse.

The minimum mass for the progenitor of a failed supernova is unknown (and as Rob Jeffries pointed out, this is metallicity-dependent). Some early models (e.g. Fryer (1999)) found that stars above $40M_{\odot}$ could produced failed supernovae, while newer ones suggest that this could be as low as $25M_{\odot}$. This lower limit would include a sizable portion of red supergiants - which could be a possible solution to the red supergiant problem.

A number of searches, largely targeting supergiants, have been carried out. Just a few are

Adams et al. (2016), using the Hubble Space Telescope to examine N6946-BH1, a disappeared red supergiant of $25M_{\odot}$.
Reynolds et al. (2015), looking at past Hubble data. They found one possible candidate, a $25\text{-}30M_{\odot}$ red supergiant in NGC 3021.

So far, we have no definite confirmation that these stars were failed supernovae. However, they do seem plausible candidates, and there are many more possibilities out there. One group proposed monitoring one million supergiants in an attempt to catch failed supernovae. Perhaps they'll be successful.


Does matter accelerate to the speed of light as it approaches the singularity?

There is indeed a sense that infalling objects accelerate to the speed of light as it reaches the event horizon, but you need to be careful what you mean by the speed of the object because the speed is observer dependent.

This is explained in detail on the Physics site in my answer to Will an object always fall at an infinite speed in a black hole? An observer watching from far from the black hole would see the falling object initially accelerate towards the black hole but then decelerate to a halt at the event horizon. However an observer hovering a distance $d$ above the event horizon would see the falling object pass them at a speed:

$$ v = c \sqrt{\frac{r_s}{r_s + d}} \tag{1} $$

where $r_s$ is the radius of the event horizon. As the distance above the horizon $d$ goes to zero the speed calculated from equation (1) goes to the speed of light $c$.


How to calculate the argument of perihelion for exoplanets with incomplete orbital elements?

Ford et al. (2008) give the following relationship for eccentric transiting planets:

$$\frac{t_\mathrm{D}}{P} \simeq \frac{R_\ast}{\pi a \sqrt{1-e^2}}\sqrt{(1+r)^2 - b^2} \left(\frac{d_\mathrm{t}}{a}\right)$$

Where $t_\mathrm{D}$ is the total transit duration, $P$ is the orbital period, $R_\ast$ is the stellar radius, $a$ is the semimajor axis, $e$ is the eccentricity, $r \equiv R_\mathrm{p}/R_\ast$ is the ratio of planetary radius $R_\mathrm{p}$ to the stellar radius, $b \equiv d_\mathrm{t} \cos i / R_\ast$ is the impact parameter, $i$ is the inclination and $d_\mathrm{t} = a(1-e^2)/(1+e \cos \omega)$ is the distance between the star and the planet during the transit. By inverting this you would be able to estimate $\cos \omega$ from the parameters listed on the site, which unfortunately won't give you a unique value (and if you're unlucky, the various uncertainties may conspire to give you a value $\left|\cos \omega \right|>1$).

In the case of Kepler-444, the website gives a link to the paper where the orbits were published, Campante et al. (2015). Their table 4 tabulates both $e \cos \omega$ and $e \sin \omega$ for the planets. Unfortunately it is not possible to simply determine the value from these estimates without having access to the full distribution of samples, so this isn't entirely helpful.

The good news is that the statistical significance of the non-zero eccentricities doesn't appear to be too high in this case, so you can probably treat the orbits as circular and not worry too much about it: tightly-packed systems like this usually have very low eccentricities otherwise they would become unstable.


Is Earth's rotation Slowing down or Increasing

Earth's rotation is slowing down, but it's a little bit irregular, which is why leap seconds can't be predicted in advance. There's provision for negative leap seconds, but they've never been used.
You can get the current list of leap seconds, maintained by The Internet Engineering Task Force (IETF), at https://www.ietf.org/timezones/data/leap-seconds.list
The decision to insert a leap second is made by the International Earth Rotation and Reference Systems Service (IERS), who coordinate the operations that gather the actual Earth rotation data. There is a wealth of information on their very extensive website.

Measuring the irregularities of the Earth's rotation
The variability of the earth-rotation vector relative to the body of the planet or in inertial space is caused by the gravitational torque exerted by the Moon, Sun and planets, displacements of matter in different parts of the planet and other excitation mechanisms. The observed oscillations can be interpreted in terms of mantle elasticity, earth flattening, structure and properties of the core-mantle boundary, rheology of the core, underground water, oceanic variability, and atmospheric variability on time scales of weather or climate. The understanding of the coupling between the various layers of our planet is also a key aspect of this research.
Several space geodesy techniques contribute to the permanent monitoring of the earth's rotation by IERS. For all these techniques, the IERS applications are only one part of their contribution to the study of planet earth and of the rest of the universe.
The measurements of the earth's rotation are under the form of time series of the so-called Earth Orientation Parameters (EOP). Universal time (UT1),
polar motion and the celestial motion of the pole (precession/nutation) are determined by VLBI.
The satellite-geodesy techniques, GPS, SLR and DORIS, determine polar motion and the rapid variations of universal time.
The satellite-geodesy programs used in the IERS give access to the time variations of the earth's gravity field, reflecting the evolution of the earth's shape, as well as the redistribution of masses in the planet. They have also detected changes in the location of the centre of mass of the earth relative to the crust. This makes it possible to investigate global phenomena such as mass redistributions in the atmosphere, oceans and solid earth.
Universal time and polar motion are available daily with an accuracy of 0.5 mas [milliarcseconds] and celestial pole motion are available every five to seven days at the same level of accuracy - this estimation of accuracy includes both short term and long term noise. Sub-daily variations in Universal time and polar motion are also measured on a campaign basis. Past data, going back to the 17th century in some cases, are also available.
EOP series are provided by the IERS in bulletins and as permanently updated series (long term earth orientation data).


Is there any residual oscillation left from the Moon rotation?

This is called "physical libration". The Moon’s physical librations and determination of its free modes (2011) estimates it using the Apollo mission retroreflectors. Their result is very small. They note that the "damping times for these two modes are estimated as $2$x$10^4$ and $2$x$10^6$ years", so it is a mystery why it is even as large as their small result. They say:

  Some possible excitation mechanisms have been explored in the past without satisfactory explanation. It has been shown that a recent meteoroid impact is an unlikely source of such excitation Peale (1975). Eckhardt (1993) proposed an excitation process related to a resonance crossing of the longitude normal mode (of 2.9 years) and a close forced frequency. During the evolution of the lunar orbit, the free and forced frequencies change slowly and can cross. However, the mechanism excites only the libration in longitude mode. Yoder (1981) proposed an alternative mechanism, based on turbulent fluid core interaction, to excite the wobble mode. The new determination of the amplitudes of the free librations invites new investigation of their excitation mechanisms.

I did not search to see whether anyone has taken up their invitation.


Why do we use solar filters instead of simply reducing aperture of the telescope?

Suppose we want to make the Sun appear only as bright as the full Moon does in a 5-inch telescope.
A difference of 14 magnitudes is a factor of 400 000; we can either add a filter transmitting 2.5 millionths of the light (including UV and IR), or reduce the aperture diameter by a factor of 630.
We can achieve that by covering the 5-inch aperture with either a large #14 welding shade, a similarly dark solar filter, or a mask with a 0.2 mm pinhole.

However, angular resolution is inversely related to aperture.
For a wavelength of 500 nm and an aperture of 0.2 mm, the Rayleigh criterion gives 1.2 * (500 nm / 0.2 mm) = 0.003 radian = 10 arcminutes, about 1/3 the angular diameter of the Sun.
Even large sunspots would be blurred beyond recognition.
A 1.0 mm hole would give 2 arcmin resolution but make the Sun too bright for direct viewing.


What is the upper and lower limit of temperatures found on stars?

The answer depends on what you'd want to consider as a "star." If you're just thinking about stars on the main sequence, then you can just refer to the classical stellar type letters, "OBAFGKM" (which has relatively recently been extended to accommodate the coolest brown dwarfs with the letters "LTY"), where O-stars are the hottest stars (~30,000 K) and Y-stars are the coldest, so-called "room-temperature" stars (~300 K).

Self-gravitating, gaseous objects are incapable of fusing deuterium below about 13 Jupiter masses, and thus simply collapse and cool perpetually (as is the case for all the giant planets in our solar system). These objects can be colder than 300 K but are not technically stars as they do not undergo nuclear fusion.

For stars that leave the main sequence, two possible outcomes are a white dwarf star or a neutron star, both of which are born extremely hot: White dwarfs are born with surface temperatures of ~10^9 K, whereas neutron stars are born with surface temperatures of ~10^12 K. However, both white dwarfs and neutron stars cool as they age, with the coldest known white dwarfs being ~3,000 K, and neutron stars cooling to ~10^6 K.

So to answer the first part of your question: The coldest known stars are Y-stars (i.e. brown dwarfs) and the hottest known stars are either O-stars or young neutron stars, depending on whether you consider objects that have left the main sequence or not.

And as for strict lower and upper limits, the coldest stars possible are likely black dwarfs, which are what white dwarfs become after cooling for a very long time (>10^15 years). The hottest stars are likely the newly-born neutron stars I previously mentioned, it is very difficult to get much hotter than 10^12 K because any excess energy is carried away via neutrinos.


What planets or exoplanets orbit the Sun’s elder twin HIP 102152?

No planets have been detected in orbit around HIP 102152.

That does not mean that no planets exist, but that our current techniques are no able to detect them.  Most planet are detected by the transit method. This observes the very small dip in light when a planet goes in front of the star.  However if the planet's orbit doesn't line up exactly with Earth, then the planet will not be detected.

Other methods can detect large planets that orbit close to the star, or very large planets in orbit very var from the star. Solar systems like ours are harder to detect.

Given what we know about the abundance of planets, it is likely that HIP 102152 has a planetary system. But actual detection might no be possible with current technology.

Moreover there is no obvious reason to think that life or intelligent life is particularly likely just because the star is similar to the sun. The nature of the star may rule out intelligent life. But as far as we know, life just needs a reasonally stable energy source and a lot of luck. It doesn't need a solar twin.


The one from 2014 is still the record holder I believe - in the sense that it is reasonably convincing that the unseen companion of the pulsar PSR 2227-0137 is consistent with being a white dwarf with a surface temperature below 3000 K.
It is worth considering why such objects might be difficult to find.
(1) It is only the highest mass white dwarfs that have had time to cool below 3000 K in the lifetime of our Galaxy. The reason for this is that more massive white dwarfs are smaller and have much higher densities. This enables them to crystallise in their interiors at high temperatures and the heat capacity of a crystalline solid becomes much lower than a gas at the same temperature, meaning that they have less heat left to radiate away.  High mass white dwarfs are rare; they are the endpoints of the evolution of 6-8$M\odot$ stars, which are much less common than the 1-6$M_\odot$ stars that produce most of the lower mass white dwarfs in the galaxy.
(2) They are almost invisible. Temperatures below 3000 K are getting down into brown dwarf territory. Now we've found thousands of brown dwarfs, but these are (a) likely to be more common than massive white dwarfs (about 1 for every 5 stars) and (b) they will be about 20 times the radius of a massive white dwarf and therefore emit 400 times the flux at the same temperature. Thus it will be very difficult to find free-floating cold white dwarfs, especially if they are intrinsically rare.
(3) But can't we find them in binary systems? If the companion star is $>1 M_\odot$ then it will have lived and died in the time it took the massive white dwarf to cool and the system would have to be a double-degenerate. i.e. A neutron star + white dwarf or two white dwarfs. Many such systems are known, but I think we come back to the fact that massive white dwarfs are rare, and if they are more massive than say $1.15M_\odot$ it becomes difficult to distinguish them from neutron stars if all you have is the dynamics of the system. A more promising line of attack would be to look for them as binary companions to old K and M dwarfs. But this would imply that the original binary system would have had a very high mass ratio - a fair fraction of these sytems would have become unbound as the white dwarf progenitor lost its (and most of the system's) mass, and very high mass-ratio binary systems are not that common. My guess is that the Gaia astrometric survey may be a good source of candidate cold white dwarfs because it will identify the binary motions and secondary (really, the unseen primary) masses for many, many K and M stars. Any K/M star with an unseen companion that is more massive than $1M_\odot$ will be of interest.


Is diffuse energy distribution considered in the galaxy rotation curve explanations?

You appear to be asking whether gravitational potential energy acts as a significant contributor to the total mass-energy of a galaxy - a question that is only relevant within the context of General Relativity, where indeed other forms of energy besides the rest-mass energy contribute to the gravitational effects.

It turns out that GR is not required to interpret galaxy rotation curves because the influence of the other forms of mass-energy (or more formally, other contributors to the stress-energy tensor) are very small.

The contribution of gravitational potential energy can be assessed using the ratio $GM/rc^2$, which is approximately the ratio of gravitational potential energy to rest mass energy. For example, if $10^{12}$ solar masses exists within a 100 kpc radius (reasonable numbers for the Milky Way); then the ratio is $5\times 10^{-7}$. Of course, if there is no dark matter then this would be even smaller.

I should also add that this tiny correction, like the gravitational potential energy, is negative. That is, it reduces the gravitational influence of the mass/energy in a galaxy. The (positive) total kinetic energy of the Galactic components (also tiny) would also need to be added, but since bound systems have a total energy (kinetic plus gravitational potential) $<0$, there is still a net reduction in the gravitational mass.


The stars are so immensely far away that to the human eye there would be no noticeable difference. The nearest stars are moving roughly 1.5 arcsecond wrt. the background when viewed from Earth's position with half a year's interval (i.e. half a revolution around the Sun). Mars' orbit is only 50% larger, but the human eye cannot resolve better than roughly one arcminute.

However, the sky would look different for another reason: Although stars are point sources, on Earth stars are smeared out due to the atmosphere. On Mars, there's almost no atmosphere, but on the other hand there's lots of dust whirling around. Dust tends to absorb blue light more than red and thus redden objects. However, Martian dust happens to scatter red more than blue (Ockert-Bell et al. 1997), so the stars should appear more blue. Also, the dust will scatter the Sun's light so much that no stars are visible during the day (as opposed to e.g. the Moon).
Note also that the north pole of Mars points not toward Polaris, but toward Deneb in Cygnus.

Since a shift in position of 1 AU (the distance from Earth to the Sun) corresponds to a shift in position of 1 arcsec for a star which is 1 pc (3.26 lightyears) away, if you want to go to a place where the nearest star (Proxima Centari; 1.3 pc away) has moved a full moon diameter, i.e. ½ a degree — you would need to travel roughly 2340 AU away, i.e. much farther away than Pluto, but still inside the Solar system.

Most stars are much farther away, however. The brightest stars that give the constellations their appearance are tens to hundreds of pc away, so to really change the constellation, you should go 10 or 100 times farther away, i.e. roughly 1 pc. In other words, the constellation will look (slightly, but conspicuously) different from out nearest neighbor stars.


Where exactly does the Moon flip, given that it appears the other way up in the other hemisphere?

The moon doesn't flip. Is us who flip.

Those in the antipodes have their heads pointing in the opposite direction. So the moon appears inverted. But there is no sudden point where it "flips". We don't live on a discworld.

You can see this yourself. Because as the Earth spins you will change the direction in which your head is pointing. Look at the full moon over the course of one night. As it appears to travel from East to West in the sky, it's orientation will slowly change. There is no mystery here. You are being rotated by the Earth.


How do we define temperature in outer space?

First of all, the medium they are mentioning is far from empty even though it is much less dense than what we do know on Earth (see this question for more details).

Then, the problem with temperature in astrophysics is that the medium you study is in general far from thermodynamic equilibrium (thermodynamic equilibrium means that there are no flows of enery or matter within the system and with the outside of the system).

Most commonly, however, we can use another flavor of equilibrium to estimate temperature. This is the kinetic equilibrium: since most collisions are elastic (meaning that the energy is conserved), particles velocities will follow a Maxwell-Boltzmann distribution. Depending on the ionization fraction (how much is your medium ionized?) this temperature will either the kinetic temperature of electrons or the kinetic temperature of hydrogen.

Then, if you want to understand these high temperatures, you have to take into account the photoionization rate and the photoelectric heating (a process in which you eject electron from the matter of the studied medium due to UV absorption). The thermal state of the medium then depends on the equilibrium between the energy absorption by the matter and the re-emission of this energy in thermal radiation. In this framework, the energy that can be absorbed comes from the interstellar radiation field (ISRF). In the solar neighborhood, it is dominated by six different types of radiation: galactic synchrotron emission of relativistic electrons (electrons accelerated to tremendous speed in the galaxy), the cosmic microwave background, the infrared emission of the matter heated by the light emitted by stars, emission from ionized plasma, the light emitted by stars itself, and lastly X-rays from hot plasma.

References:

Klessen & Glover (2016)
Ferrière (2001)


Most stars are born in groups or clusters of anywhere between 100 and millions of stars. Most of these groups and clusters do not survive more than about 10 million years or so and the fraction that survive decreases further with time.

There are two basic reasons. First, it seems that most young clusters and associations may be born in an unbound state. Second, those that are gravitationally bound can become unbound through the expulsion of gas or through tidal interactions with the Galaxy, spiral arms and giant molecular fields. A supernova is one event that might be a culpable agent in expelling mass from a young cluster and consequently unbinding it.

It is therefore highly unlikely that a 4.5 billion year old star will be found in a cluster. There are very few examples of clusters at this age or older.

As an aside, the theory that the Sun was born from the debris of just one or even a few stars is totally false and easily disproved by looking at the isotopic composition of dust grains formed in other stars before the solar system was born. These can be found inside meteorites and demonstrate that the solar system contains debris from many, many stars; of both high and relatively low masses.

See How can there be 1,000 stellar ancestors before our Sun? and Parent stars of our Sun - Where are its remains?

Edit: It seems that you are just asking why we can't see siblings of the Sun, including this massive star that may have been involved in the formation of the solar system.

First - it isn't clear that the Sun was born in a cluster from material seeded by a supernova from another star in that cluster. There is a timing problem. The massive star that exploded would have to form more than 10 million years before the Sun. Clusters we see today do not seem to have such an extended star formation history.

Second - even if that weren't the case, a supernova very frequently gives rise to a "kick". The velocity dispersion of pulsars (recent supernova remnants) is exceedingly large (100s of km/s). Many remnants would escape the Galaxy and there is no chance of predicting where they would be in billions of years.

Thirdly - even if the first and second problems were not applicable, as I mentioned above, even bound clusters do disperse. Once free of any binding potential stars can undergo a process called "disc heating" - the action of encounters with spiral density waves and giant molecular clouds gradually increases their velocity dispersions (changing their kinetic energies and exerting torques that can change their angular momentum wrt the Galactic centre). This allows the solar siblings to go their separate ways. A Galactic orbit at the solar radius is more than 200 million years. A velocity difference of just 1 km/s (typical in a cluster even before it unbinds or is heated) could result in a separation by 200 pc in that time. The Sun has executed more than 20 such orbits.


I agree that the common explanation of the longitude of periapsis, as the question gave it, appears at first sight to make no sense -- because it constructs a sum of two angles in two different planes, and that seems at first sight to lack geometrical meaning.

In fact the fault, if any, is in the explanation rather than in the geometry. It is not so difficult to show a geometrically sensible construction for the longitude of periapsis, without adding any angles in different planes. But before showing this, it may be worth answering the question 'Why do it at all, what's the point?'.

At this point the present answer links up with the answer given by James K.  One of the uses of such a construction came to attention in the early days of astrodynamics. When programmed orbit calculations are to be done for artificial space objects, the orbits to be calculated can in principle have any values whatever of the inclination and eccentricity, in principle down to zero.  When either or both of these orbital quantities approaches zero, then the position of the node and/or the periapsis, or else computations involving them, can become in practice numerically ill-defined, or anyhow show some undesired numerical behavior. As a result of the difficulty, searches were made a long time ago now, for alternative systems of orbital elements that do not generate undesired numerical behavior at zero or near-zero values of inclination or eccentricity.

The resulting formulae for orbital equations, usable for low-inclination, low-eccentricity orbits, typically made use of devices that included the equivalent of replacing the anomaly of an orbiting object by its longitude from a defined initial point, and replacing the argument of periapsis by a corresponding longitude from the initial point.  The development of orbital equations in terms of such quantities can usefully be generated, or at least illustrated, by the following construction that also provides an in-plane account of the longitude of periapsis, the subject of the question.

Sources for the geometry and formulae connected with this kind of development include R M L Baker (1967) 'Astrodynamics', especially Appendix C, and applications are shown for example in A E Roy, 'Orbital Motion' (especially 4th ed., 2005) (or google for other sources of the same title), especially chapter 8, Special Perturbations, section 8.5, for perturbational equations and their derivation.

Baker's 'Astrodynamics' illustrates the definition for these purposes of an alternative ('F G W') coordinate system.  See the figure below.

One starts with ordinary rectangular coordinate axes defined by points X, Y, Z on a unit sphere with center O, where the xy plane is for example the ecliptic plane, and OX is in a chosen initial direction, e.g. towards the equinox of date.  Define an orbital plane through F N G, intersecting the xy-plane at node N with inclination angle $i$ .  Angle XON is then the longitude $\Omega$ of node N. Define axis OW perpendicular to orbital plane F N G. Let angle FON = XON, and let angle FOG = 90°. Then make F the initial point of coordinate system FGW. In this coordinate system, the longitude of the orbital periapsis starting from the initial point is FOP, which is the sum, as usual, of $\Omega$ the longitude of the node FON and $\omega$ the argument NOP of periapsis, but now they are measured in the same plane -- which was the thing to be illustrated.

This FGW system can be conveniently oriented by the coordinates of point W, the pole of the orbital plane,
which has coordinates (in terms of the x,y,z axes and the quantities $i$ and $\Omega$) of

$ W_x =  + \sin (\Omega) . \sin (i) $ ,
$ W_y =  - \cos (\Omega) . \sin (i) $ ,
$ W_z =  \cos (i) $ .

Then initial point F can be shown (with no difficulty but at a little length) to have coordinates relative to initial axes x, y, z, of

$ F_x = 1 - W_x^2 / (1 + W_z) $ ,
$ F_y =  - W_x W_y / (1 + W_z) $ ,
$ F_z =  - W_x $ .

Quantities Wx, Wy, Wz are sometimes identified as $h_x/h_m, h_y/h_m, h_z/h_m,$ and can be called direction cosines made by the orbital angular momentum vector (in direction OW) with axes x, y, z.

You might also find some of the formulae useful in Preliminary orbit-determination method having no co-planar singularity by R M L Baker and N H Jacoby, Celestial Mechanics 15 (1977) 137-160.


If "9 th planet" is ejected by Jupiter or Saturn, why does it have a very far perihelion?

This is certainly a problem for any hypothesis that form planet 9 by ejecting it from the inner solar system. The issue (for those wondering) is that if you simply scatter an object out of the inner solar system, but it is retained in a bound elliptical orbit, then it should come back! The proposed planet 9 needs to have a perhelion distance beyond 200 au and at least a moderate eccentricity.

The solution might be found in a paper by Bromley & Kenyon (2016), who address this very point. In their model they scatter a "super-Earth" from 5-15 au in the inner solar system, at an early stage of the solar system formation when the Sun is still surrounded by a gaseous disc. They run through a suite of simulation with different mass planets, different gas surface densities and different rates of evolution and dissipation of the gas disc.

The outcome is that there are sets of parameters that lead to eccentric orbits with large perihelion distances that are caused by damping by dynamical friction from the gas disc. What is required are long-lived, low-density gas discs, or more short-lived, massive gas discs that clear from the inside-out.

The same paper also summarises a number of other possibilities that could explain the large perihelion distance and eccentricity of putative planet 9. These are: (i) in situ formation from a ring of solids - though the orbit would tend to be more circular; (ii) the nearby passage of another star could perturb the orbit, provide a boost to the perihelion distance and increase the semi-major axis, perhaps not by enough, but this mechanism could be more effective if the planet was in an eccentric orbit to begin with; (iii) tides, from the Galaxy, or more likely, from the birth cluster of the Sun could have influenced the orbit of a scattered planet and produced its high eccentricity and perihelion distance.

The gist of the discussion is that whilst the authors think these other things are possible, they favour their own gas-drag model (naturally!).


Is it possible to find what crater the Mars meteorites come from?

Scientists think they have - the Mojave crater. Also see  https://science.sciencemag.org/content/343/6177/1343


Where did the Sun get hydrogen to work with if it is in the 3rd generation of stars?

Most of the galaxy's gas is not incorporated into stars and remains as gas and dust. This is not really my area of expertise, but papers such as Evans et al. 2008 and Matthews et al. 2018 seem to suggest that in the Giant Molecular Clouds where most stars in the Milky Way Galaxy form, the star formation efficiency is about 3-6%. So the vast majority of the gas (94-97%) is not made into stars. In very dense environments such as globular clusters, which were formed much earlier in the Milky Way's history, the star formation efficiency get as high as approx. 30%. The canonical quoted rate for "regular" spiral galaxies like the Milky Way is about 1 solar mass of new stars are made per year, which is very low summed across the whole galaxy.

Stars also give off a fair amount of their outer, hydrogen rich outer layers during the later red giant phases when the stellar wind is stronger and the atmosphere expands a huge amount (radius of the Sun during the red giant phase will be about what the Earth's orbit is now). Also in the end state when the white dwarf is formed, it's only the core and inner layers that form the white dwarf. The typical white dwarf mass is about 0.6 times the mass of the Sun (S. Kepler et al. 2006) and so there will be a fair amount of unfused hydrogen-rich outer atmosphere left over after the star dies. For higher mass stars, even more of the mass goes into the (ejected at high speed) envelope than goes into the remaining neutron star. These high mass stars are much rarer though; most of the Milky Way's stars are faint, cool M dwarfs.


No; it's real - observations of the movement of galaxies indicate that there is an unusual concentration of mass (probably not a massive "object" of course, but totalling as much as tens of thousands of galaxies) in a place around 200 million lightyears from our galaxy.

It's hard to study since the view in that direction happens to be blocked by our own galaxy, but given that it has been studied for decades, it seems unlikely that the phenomenon turns out to be based on some sort of methodical error.


It's a big question, but kind of a favorite subject of mine, thinking about exoplanets, so I can give a ballpark answer, and I invite anyone to give correction or give a more technical answer if they like.

Ice(s) formation

An ammonia-water ocean wouldn't be friendly towards ice formation because water ice would sink in the ammonia-water solution and Ammonia ice would sink in liquid ammonia.  There's no range where the ice floats unless you remove nearly all of the ammonia and had it a high percentage of water.

That said, oceans are likely to have dissolved salts in them, especially on rocky worlds that have a very high rock to water ratio.    A water world might have much less salty oceans, but, lets not get too sidetracked.

If there's enough dissolved salts or iron in the water-ammonia mix, then the density might be sufficient that water-ice could remain on top.  Ice tends to form out of nearly pure water, with very little ocean salt, which increases it's buoyancy.   It's possible, with enough saltiness, that ice could form in an ammonia-water ocean.  It's also possible that ice would sink.

Density

I want to point this out now, because it's important.    Using our oceans as a model, higher salt concentration sinks.

Salt concentration by density

That's kind of common sense and is likely to happen with a water-ammonia ocean as well.   Higher ammonia concentration near the surface, higher water (and higher salt) concentration as you go lower.  That's speculation on my part, but it seems reasonable.

Varying ratios of water-ammonia based on temperature or other factors

Higher water concentrations would sink and higher ammonia concentrations would rise and flow towards the lower.  That's simple physics.   Like on Earth, the variation of salinity of surface water isn't all that much.  The same thing would likely happen with an ammonia-water ocean, the surface currents would try to equalize the different ratios.  There would be some variation, but likely not as pronounced as the variation with depth.   You're not likely to get a 90% water ocean in one region and a 40-60 ratio in another region because gravity abhors an imbalance like that.   The 90% water region would be heavier and as a result, lower.  You can have that kind of density imbalance with a solid crust, but not a flowing ocean.

The varying density of the ocean is important for another reason.   On Earth, the salinity being relatively consistent across the ocean surface, the cold polar ocean is more dense and it sinks, and this is what's thought to drive the oceanic conveyer.

There's a range of uncertainty on whether an ammonia-water ocean would conveyer or not.   You'd need a sufficient temperature and/or density variation, keeping in mind that if you begin with a ratio imbalance the ocean would work to correct it, so the planet needs an engine to maintain the conveyer.   On Earth, that engine is the warm equator and cold poles.  Different planets would have different temperature variations.  Venus, for example, has practically no temperature variation at it's surface.

I looked, but couldn't find a table on ammonia-water solution by temperature.  That would be a factor if anyone can find it.

Evaporation

Ammonia in water has a saturation point, based on temperature and pressure.   If the ammonia-water solution exceeds that saturation point, say, on a hot day, then ammonia would bubble out of the ocean fairly quickly.  Perhaps quickly enough to generate a weather system and trigger a density change, and in this case, warmer water sinking and a kind of reverse conveyer, driven by sinking water at the planet's warmest regions.

The phase transition, liquid to gas would remove heat from the water as well.

A planet with an ammonia-water ocean and a surface temperature that exceeds the saturation point could be quite dynamic.

If the planet is slightly colder and/or the ammonia percentage stays below the saturation point, then it's more of an equilibrium system.   Water is a good solvent and it takes in many gases, such as CO2, in small amounts.    Ammonia and Water, unlike CO2 and Water, mix very happily, so evaporation of an ammonia-water ocean below saturation wouldn't be that different to water on Earth, where evaporation is driven primarily by wind and sunlight.  (contrary to logic temperature isn't as big factor in evaporation as direct sunlight).  Wind is also a big player and on Earth, running water from rain or snow melt and transpiration from plants all return water to the atmosphere, which quickly rains or snows back to the surface.   At any given time, Earth's percentage of water vapor in the atmosphere is about 1%, but it varies locally based on air temperature and relative humidity.

Ammonia would enter the atmosphere similar to water, but how long it would stay in the atmosphere is a factor.    Air temperature and 100% relative humidity are the upper limits for how much water vapor can exist in the atmosphere.  There are no such upper limits for Ammonia at Earthlike temperature and pressure because Ammonia, unlike water, is a gas at standard temperature and pressure.

As a result, a theoretical Earth-like planet with an ammonia-water ocean is also, very likely, going to have a lot of ammonia in the atmosphere, not the trace 1% of water that's in our atmosphere, but a lot more.   You can't have an ammonia ocean without ammonia in the atmosphere at well, in high concentrations.

For example, to have ammonia raindrops at 1 ATM, the air temperature needs to be colder than -33 C because that's ammonia's boiling/liquification temperature.   Unless the planet was really cold, a lot of the ammonia would just stay in the atmosphere, though, presumably the planet would have water-rain and that water rain would pull some of the ammonia back out and return it to the oceans.   That's the equilibrium, and it's a complicated thing to try to calculate, but the question isn't just how fast would it evaporate, but also, how fast would it return and the return rate would depend on the method of return.   Ammonia, being a gas at what we think of as normal planetary temperatures, means it would likely remain in the atmosphere as well, it wouldn't all go into the ocean.  It would be some of both.

Perhaps over time, the ocean ratio and the rain-water/ammonia ratio would equalize, but again, that's just speculation.

If the planet has very cold poles, where the ammonia could rain out of the atmosphere, or perhaps tidal locking, and a colder night-side of the planet, then the ammonia could return from the atmosphere more quickly.

I'd like to know if this has been studied and if these behaviors have been described in papers, books, or in-depth articles on this topic.

I don't know of any papers.  Theoretical planetary formation is fun to think about, but to do it right, requires some pretty serious computer modeling.   What I've read on the subject is much less ambitious than your question.   Models are done on water worlds and on tidally locked worlds and on plate tectonics for different mass planets.   I've never seen a study on ammonia-water oceans, but I think it's a fun subject.

Earth had to have fairly abundant ammonia at some point because it's one of the primary "ices" in comets.   Comets are mostly ammonia, water, CO2 and CH4, with smaller amounts of other elements.  It's worth asking where Earth's ammonia went.  It may be more chemically reactive than the other elements.   If Earth lost it's ammonia, other planets may lose their ammonia as well in the process of planet formation and the chemistry that goes with it.  It's also possible that ammonia-water oceans are common, perhaps more common than the pure water oceans like the ones we have on Earth.

Perhaps time and better telescopes will tell us more about this subject.

Take my answer with a grain of salt as I'm a hobbyist.


How dark would the sun become if we traversed a Bok globule?

Not noticeably darker.

Assuming such a globule has a mass of 50 solar masses and a diameter of 1 light year, that would make it's average density about $2.2\times 10^{-16}\,kg\,m^{-3}$ which is fairly close to not being there at all in human terms.

An imaginary tube of this stuff 1 AU tall and of area 1 meter squared would contain about $3\times 10^{-5}\,kg$ of mass.  That is an imaginary tube from Earth to the Sun.

That much matter won't block much light.

For comparison the density of air at ground level is about $1.2\,kg\,m^{-3}$.  So less than a millimeter of air at ground level blocks more sunlight than the dust cloud would.

Also


There are a few ways that a gigantic ring system outside the Roche limit might be possible.   Ring systems can extend well past the Roche limit if they are very faint or very young.   A faint/light ring system doesn't have enough mass to coalesce and a young system might not have enough time.   Young small moons with lots internal heat of formation can be volcanically active too, feeding a ring system.

Saturn's E ring is well outside it's Roche limit.

https://upload.wikimedia.org/wikipedia/commons/f/f7/Saturn%27s_Rings_PIA03550.jpg

The most likely explanation for it's enormous and very dense rings, however is that J1407b is very young and a lot of it's ring system will likely to coalesce into moons in time.  The breaks in it's ring system suggests that some moons are already clearing paths in it's rings.

It's mass is also very significant.   While difficult to estimate, it's thought to be about 10-40 Jupiters in mass.  Source.  That would give it a mass in the range of 33 to 130 Saturns and a Roche limit some 3 to 5 times (cube root of the mass), distance to Saturn's center, perhaps 6-10 times the distance from the surface surface.   Because it's ring system is about 200 times the size of Saturn's rings, the mass alone wouldn't explains part of it.    Mostly likely it's a combination of very massive planet/perhaps a brown dwarf star and young plant/star with a satellite system still in formation.

There might also be a very powerful magnetic field at play that helps prevent the ring debris from forming into satellites, but that's just speculation on my part.


You can get moon centered position and velocity vectors at the Horizons Ephemeris Generator.

Here's a screen capture:

Code for moon is 301, Code for earth is 399


How often are there lunar eclipses on Jupiter

Every Galilean moon and inner moon go into lunar eclipse once per orbit.

How often is there a lunar eclipse of the Jovian moons?

I set up a little animation of the Galilean moons (I didn't include the inner moons), and made a few assumptions (negligible moon size, sun at infinity, perfectly circular orbits) and found that there is a lunar eclipse in the Jovian system just under 10% of the time.

Here's a bigger version of the second graph.


I think the outer edge of an accretion disk is not well-defined, and observationally the radius will depend on which wavelength you consider, since the farther you get from the BH, the softer the radiation will be. But if you look in the UV, then Morgan et al. (2010) find the following relation between $R_{2500}$ (the radius when observed at $\lambda = 2500$ Å) and the mass $M_\mathrm{BH}$ of the black hole:
$$
\log\left( \frac{R_{2500}}{\mathrm{cm}}\right) = 15.78 + 0.80
\log\left( \frac{M_\mathrm{BH}}{10^9M_\odot}\right),
$$
(modulo some uncertainties that you can look up in the paper).

That is, if your BH has a mass of $10^8 M_\odot$, its radius will be $R_{2500}\sim64\,\mathrm{AU}$, or roughly 1/3 lightdays.

For comparison, its Schwarzschild radius is $\sim2\,\mathrm{AU}$, so your estimate was actually pretty good.

This result is consistent with Edelson et al. (2015), who find 0.35 lightdays, also in the UV. However, in you look in longer wavelengths, the disk is much, much larger. If you're interested beyond your homework assignment, take a look at the accretion disk theory review by Armijo (2013), who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.


Heliocentric distance from galactic x y and z coordinates

The galactic coordinate system is centred on the sun with the xy plane in the plane of the galaxy and the x coordinate pointed towards the centre of the galaxy.

Since it is a heliocentric system the Euclidian distance is $\sqrt{x^2+y^2+z^2}$.


Does the dust around SMBH's protect habitable planets from jets?

Most galaxies have enough dust to hide their cores from investigation in visible light. But the jet from an AGN is composed of large amounts of relativistic particles, and is easily powerful enough to clear any dust out of its path. We see in M87 a jet punch 5000 light years from an active nucleus.

The dust that hides Sagittarius A* is not localised around the black hole, it is spread through the bulge, bar and disk. You would not want to be anywhere near an active galactic nucleus. Large amounts of radiation are produced by the accretion disc surrounding the black hole. Distance and dust may provide some protection to a star orbiting an active galaxy at the distance of the sun from Sag A*, but not if the star is directly in the jet.


Where you're wrong:

  even little bit of orbital speed tangentially prevents something from falling into the sun

You need to have huge tangential velocity to escape the sun attraction! The closer you are, the higher the velocity must be for celestial body to avoid falling on the sun.

  While Venus is orbiting the Sun at 35.02 Km/s, the Earth is also orbiting the Sun in the same direction at 29.78 Km/s. [Reference]

It's not easy to understand, because there is no clear analogy of how stars form at our level on earth.

Interstellar gas clouds eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The star is at the beginning a little portion of the cloud that is more dense than the rest - it results of local and large scale instabilities (some call it an element of chaos).

Little by little, the density at the center of the cloud increase - at the same time, it has to overcome opposite forces arising from more probable friction. If the gas is cool enough compared to the density, at some point you start having more and more matter, which results in more and more gravitational attraction for surrounding gas.


What are the conditions for an ionosphere to form?

According to the paper "Simultaneous ionospheric variability on Earth and Mars" (Mendillo et al. 2003), the overall structure of any ionised region of a planetary atmosphere depends on

  a blend of in situ production and loss
  processes, plus effects of transport of ionization into or out
  of the local region of interest

Specifically, according to Mendillo et al., the only significant influence is from the photon flux from the parent star - so, the further the planet is from the star, the less the incoming photon flux.  The other mechanisms are suggested to be planetary-based, including (from the article):

planetary rotation rate
orbital obliquity

which both also affect the photon flux reaching that particular part of the planetary atmosphere. Other mechanisms are to do with the planetary atmospheres themselves, with factors including:

the thermal structure, chemistry and dynamics of the planetary 'neutral' atmosphere itself.
by how much energetic particles (from the stellar wind and/or magnetosphere) affects the planetary atmosphere.
diffusion and electrodynamics from coupling from above the layer; and tides, waves and electrodynamics from coupling with below the layer.

A further study in the paper "Ionospheric photoelectrons: Comparing Venus, Earth, Mars and Titan" (Coates et al. 2011) compared the sunlit portion of planetary ionospheres, which is sustained by photoionization of primarily neutral atmospheric constituents by solar EUV (Extreme UV), a process originally identified for Earth, but recently:

  the Mars Express, Venus Express and Cassini-Huygens missions have revealed the importance of this process at Mars, Venus and Titan, respectively.

It should be noted, that for planets without magnetic fields (Venus and Mars for example), the Max Planck Institute for Solar System Research page "Research about the planetary plasma environment",

  The ionosphere is assumed to display two distinct states, "magnetized" and "unmagnetized", depending on the penetration depth of the solar wind magnetic field.

One more factor is suggested - dust, theorised from observations reported in the paper "On the role of dust in the lunar ionosphere" (Stubbs et al. 2011). According to the NASA page summarising the paper "Mystery of the Lunar Ionosphere", ubiquitous grains of dust from the Lunar surface may be the source of the Lunar ionosphere.  The research from Stubbs et al. is summarised as:

  UV rays from the sun hit the grains and ionize them.   According to their calculations, this process produces enough charge (positive grains surrounded by negative electrons) to create the observed ionosphere.

The process with the lunar ionosphere is illustrated below (from the NASA page linked):


What is the characteristic time of the evaporation of the galaxies?

The standard treatment can be found in (Binney & Tremaine 2008), but see also (Adams & Laughlin 1997) for a good treatment.

The overall timescale for galactic evaporation is $$\tau_{evap}= 100\tau_{relax}\sim 10^{19}$$ years.

The relaxation timescale $$\tau_{relax}=\frac{R}{v}\frac{N}{12 \ln(N/2)},$$ where $R$ is the size of the system, $v$ is the typical random velocity, and $N$ is the total number of stars. This corresponds to the time it takes to completely randomize the velocity of a star by interactions with other stars.


Relation between molecular cloud mass and galactic mass

The mass of an average galaxy appears to be totally dominated by dark matter, so your calculation would not give the galaxy mass.

Even if all you wanted was the baryonic (non dark matter) mass then what you suggest will be very much a lower limit. For example you can look at this paper by Chabrier (2001), who estimates that gas forms less than half the baryonic mass in the Milky Way. This fraction will be even lower in elliptical galaxies that are largely gas-free.

I am no expert, but I also think that even the gas component is dominated by ionised and atomic gas where you wouldn't expect to see CO emission and which are better traced by recombination lines or 21cm radiation respectively. In other words, the molecular gas is a fraction of the gas, which is a fraction of the baryonic mass, which is a small fraction of the total mass.

There is an extensive review of the "CO method" by Bolatto et al. (2013). The conclusion appears to be that the molecular gas mass can be estimated to around 30%, but that the conversion factor can depend on the star forming environment and on the metallicty of the gas in poorly understood ways.


Where did the nickel in the Earth's core come from?

Nickel is formed in supernovae.
There are several different types of supernova.  Wikipedia suggests that about 70% of nickel was formed in exploding white dwarfs, and 30% in explosions of massive stars.  Unlike say gold, merging neutron stars do not produce a substantial proportion of the nickel on Earth.
We know about the composition of the core by Earth Science techniques (studies of Earthquake waves that pass through the core tell us about its density, we also know about the mass of the Earth, and these are all consistent with an iron-nickel core)  And from meteorites.  Metallic meteorites have come from the cores of disrupted asteroids, and these asteroids formed at the same time as the Earth. They are made of the same mix of iron and nickel.


Around what apparent magnitude can the naked eye observe an object during full moon

It may shave off 1 or 2 magnitudes, but it depends on many factors: nature of the object (star, nebula, galaxy), altitude (higher altitude has less light scatter), transparency, etc. There's no One Single Answer To Rule Them All.

Light pollution from artificial sources has a greater impact in most cases.


What was the nature of the signal received from the direction of HD164595?

The page you link to contains what is, to my knowledge, the most descriptive graph of the signal from the RATAN-600 data that is open to the public. Bursov et al. have not yet presented their results, which will be explained at the International Academy of Astronautics SETI Permanent Committee meeting late this September. Chances seem to be good that we won't see much more information until then.
Here is the graph at a frequency of 11.1 GHz, as will be presented by Bursov et al. and as used in Croft et al. (2016), who very quickly put together a paper with an analysis of some basic follow-up observations using the Green Bank Telescope:

The main blip lasted only for two to three seconds - two seconds seems to be the popularly cited figure. Since then, no other group looking at HD 164595 has detected a repeat of the signal, and Croft et al. were unable to find anything similar in records in the NVSS catalog. Continuum imaging revealed nothing, and the authors concluded that the signal may have been due to observational error - although a one-time event is not out of the question.
Again, though, there is no more information forthcoming from the RATAN-600 group, and I expect this will be the case until the results are formally presented in full this September. Until then, this is all the data we have, but it appears to be the most important bit of data.

Update, 14 September 2016
The Russian Academy of Sciences recently released a statement saying

Subsequent processing and analysis of the signal revealed its most probable terrestrial origin.

I'm not aware if the "processing and analysis" was done by the RATAN team or an outside group, but it seems more and more like the signal is not in fact from an extraterrestrial civilization. We'll probably find out more details two weeks from now.

Update, 25 October 2016
I have yet to see any published results from the RATAN group after the IAA meeting about a month ago, and I haven't seen any articles on the findings. The IAA has some information on the meeting, but it has yet to add details to its Minutes page. Perhaps that will be released at a later date - and I assume it will be - but that could be a while. In the meantime, I'm taking the lack of information to mean that the RATAN group agrees with the analysis by Croft et al., and that the signal was a false positive.


What astronomical observations would give conclusive proof of alien life?

Lots of frontier science lectures to watch on nearby topics here:
https://astrobiology.nasa.gov/seminars/

Your very question seems to be the hottest topic in astronomy today.

My impression is that if oxygen gas and methane were on the same exoplanet, then it certainly must be living biological processing creating that there. Nothing else could explain the coexistence of the two. And even if one of those two gasses could occur abiologically, and a range of exoplanets with one of them would be found, statistically one could deduce a frequency of planets which must have life, because the abiotic processes could only explain a certain maximum part of them.

It is a pitty that most exoplanets are in size between Venus and Neptune, which are at the same time the least explored planets in our Solar system. When a sub-Neptune or a super-Venus has its atmosphere examined for the first time, we won't have as good references as we could've had.


What I think you're trying to do is workable, but approximate. You first find the Sun's geographical position (GP), the point on the Earth where the Sun is in zenith at the time of the observation. You start by finding the Sun's GP latitude (its declination). It's approximately 23 degrees times the sine of ((days since the vernal equinox)times 360/365). Then you find its longitude as the ((hours since local apparent noon)times(15 degrees) + observer's west longitude). You will have three points, vertices of a spherical triangle: observer's location, Sun's GP, and the north pole. The angle at the pole is the difference between the longitudes of the Sun's GP and the observer. The sides between the pole and the other two points are 90 degrees minus their latitudes. All that's needed to solve a spherical triangle is the arc length of two sides and their included angle. Using the law of cosines for spherical triangles, find the arc distance between the observer and the Sun's GP. The altitude of the sun will be 90 degrees minus that number. Find the angle at the observer using the law of sines for spherical triangles. That will be the Sun's azimuth.


Large radio telescope have pretty good pointing accuracy:

the individual dishes of the VLA are accurate to about 10 arcsec.
the giant Lovell telescope at Jodrell Bank has a similar accuracy.

The second critical parameter is the beam width. Beam width depends highly on frequency:
beamwidth = wave length/dish diameter
When you use several dishes in an interferometer, you can increase their effective accuracy and decrease their collective beam width.
A bit more info on the mechanics of the Lovell telescope:

A control computer calculates the required drive rates to follow each radio source. The drive motors are servo-controlled, so there is a continuous check that the correct rate has been achieved. The position of the telescope is constantly monitored and fed back to control computer to ensure that the telescope is pointing correctly.
For good tracking the pointing accuracy should be about a twentieth of the resolution. Since the resolution is proportional to the wavelength being received (see below), it follows that the pointing accuracy is more critical at shorter wavelengths. The control computer is able to correct for pointing errors caused by the telescope bowl sagging under its own weight as it moves up and down. In this way the pointing errors can be kept to about 10 arcsec.

So, servo motors and presumably, calibration are what enables this accuracy.
The Lovell telescope has 2 elevation drive motors with gearboxes on each side of the dish. These can be driven so that one motor pulls the dish, and the dish drags along the second motor. This eliminates gear backlash in the system: the two gear trains are "wound up" in opposite directions. This means they can make adjustments without gear backlash interfering. (source: a video presentation that runs in the visitor center at Jodrell Bank)
Further reading: the story of Jodrell Bank, and a Radio Electronics article.


Can I make a black hole with one or two atoms?

There are two answers: yes and no.

Yes because every mass M has a Schwarzschild radius given by $\frac{2GM}{c^2}$ (where G is the gravitational constant (about $6.7\times10^{-11}$ and c is the speed of light (about 300 000 000 $\mathrm{m/s}$). If something is compressed to its Schwarzschild radius it becomes a black hole. You can do this for an atom. An atom of carbon (for example) has a mass of $2\times10^{-26}\mathrm{kg}$ so its Schwartzschild radius is
$$ \frac{2\times (6.7\times10^{-11})\times (2\times10^{-26})}{300 000 000^2}\approx 3\times10^{-53}\ \mathrm{metres}$$

So the actual answer is no as there is no feasible way of compressing an atom to this size. Of significance here is the fact that this size is so small that objects this small don't behave like small balls but as quantum mechanical objects. But a black hole is a gravitational object modeled by General Relativity, and Relativity and quantum mechanics don't work well together. In other words, we don't have a scientific model for describing how an atomic mass black hole would behave.

Stephen Hawking has shown that small black holes are unstable, so an atomic mass black hole would be very unstable, evaporating in a very short time.


Why is today the shortest day but the solstice is actually tomorrow?

The other possibility is https://sunrise-sunset.org is inaccurate.

An app that I use, LunaSolCal, show the length of day on Dec 20, 21, and 22 in Ashland OR are 9:05:45, 9:05:44, 9:05:49.
The website https://www.timeanddate.com does not have Ashland OR in its database, so I used nearby Medford OR. The length of day on Dec 20, 21, and 22 are 9:04:50, 9:04:49, and 9:04:53.

Although those two sources have a difference of 1 minute in the length of the day, the pattern is correct and expected: the shortest length of daylight is on the day closest to the winter solstice. Therefore, I think sunrise-sunset.org is off by one day with its calculations.


Magnetic fields of peculiar HgMn A type stars

One thing that seems to be clear is that HgMn stars have only an extremely weak net longitudinal magnetic field component, if any. Shorlin et al. (2002) did an early survey of HgMn, Am, and Ap stars, and detected no longitudinal magnetic fields in the former, with a median 1$\sigma$ uncertainty of 39 Gauss. Makaganiuk et al. (2010) also found $B_z$ values of 0 in the stars they surveyed, with a higher precision - a 1$\sigma$ uncertainty of 0.81-10 Gauss, varying between stars. Other studies also yielded precisions of less than a few Gauss for some stars (see mentions by Makaguniak (2011)).

Some reports have found longitudinal values in the 10s to 100s of Gauss, but as Kochukhov notes, subsequent inveistigations have failed to confirm these findings, which have had extremely high uncertainties. One example is Hubrig et al. (2012), the paper you cite, which claimed to have found weak longitudinal and quadratic fields in several stars, including HD 65949. Kochukhov et al. (2013) then found no longitudinal fields on the star, to within a few Gauss, and Bagnulo et al. (2013) attributed to 2012 findings to instrument error, leading to flawed data.

Non-longitudinal magnetic fields have not been observed in much detail (small-scale longitudinal fields have not yet been ruled out, either, by large-scale global ones appear to be nonexistent), and complicated ones could still exist. Kochukhov et al. (2013) do say that they have ruled out large so-called tangled magnetic fields, but small-scaled ones are still possible, according to Hubrig.

One thing worth noting is that the vast majority of these studies, including the one you referenced, which has been disputed, are focused on B-type HgMn stars, in part because fewer A-type HgMn stars have been discovered.


This is actually a very subtle question, much more so than the answers to the similar questions provided in the comments give it credit for.  When I was in graduate school at Ohio State I routinely asked this question to visiting dynamicists and invariably got different answers.

The very basic answer is that if you have two sufficiently strong resonances sufficiently close together, then the resonance will be unstable.  Otherwise, the resonance will be stable.  But what determines "sufficiently strong" and "sufficiently close" is where things get very complicated quickly.  A basic criterion is the Chirikov criterion.  (The Scholarpedia article is somewhat more detailed.)  However, the Chirikov criterion is not universally valid.

If you have overlapping resonances, then an object gets bounced back and forth between these two resonances chaotically.  These different resonances perturb the orbit in different ways, and eventually they will perturb the orbit into an unstable orbit, thus leading to depletion of the resonance.  If a resonance is "distant" from other resonances, then the resonance tends to keep objects locked in place, leading to an excess of objects in the resonance.

Most of the resonances in the asteroid belt are fairly close together, which leads to them being unstable.  The Kirkwood Gaps are the most prominent manifestation of these instabilities.  For example, the Alinda family of asteroids are in a 1:3 resonance with Jupiter, and are very close to a 4:1 resonance with the Earth.  This leads to instability, and hence very few asteroids in this family.  However, in the outer Solar System, the resonances are generally far apart, and so are mostly stable.  The plutinos are one example of such a stable resonance, being in a 3:2 resonance with Neptune.


A three-torus is the boundary of the solid three torus, just like the two-torus is the surface of a solid donut. You can imagine it as a cubical room where each wall/ceiling/floor is a portal to the opposite-facing wall (i.e. the wall to your right is a portal that sends you to your left), but preserves orientation (when you walk out of the portal, your heart is still on your left).

You can also think of it as a world where your position is described by three coordinates (like x, y, and z in Euclidean space), but each coordinate corresponds to an angle on the unit circle. If you go far enough in one coordinate, you loop back to where you started (360 degrees is the same as 0 on a circle).


Very near GAIA DR2 sources - has Proxima Centauri been dethroned?

These are spurious data points. They are likely genuine objects, but the parallax value is incorrect. There are a small number of these in the Gaia data set.

There are 59 object with a parallax greater than Proxima Centuri. These don't represent genuine objects, instead they are when two sources are closely aligned (within about 0.2 arcsec). Most of these are found in the milky way, where there are lots of stars close together.

See
https://www.cosmos.esa.int/web/gaia/forum#!/thread/1688565/1688576

And section 9 of https://arxiv.org/pdf/1804.09365.pdf


What was earth's mass before the collision with theia?

I'm not sure if there was a significant amount of debris, however the calculation you mentioned would at least give an upper bound on the value:

$
M_{piE}=\text{Greatest possible mass of pre-impact Earth}\\
M_E=\text{Mass of Earth today}=6.0\times10^{24}kg\\
M_M=\text{Mass of the Moon}=7.3\times10^{22}kg\\
M_T=\text{Mass of Theia}\approxeq6.4\times10^{23}kg\\
$
$$
\begin{eqnarray*}
M_{piE}&=&M_E+M_M-M_T\\
&=&(6.0\times10^{24})+(7.3\times10^{22})-(6.4\times10^{23})\\
&=&5.4\times10^{24}kg
\end{eqnarray*}
$$
So it wouldn't have been more than about $5.4\times10^{24}kg$.


No the sun is not part of a cluster.

There are several types of clusters that we see in the sky. The most familiar is the "open cluster", like the Pleiades. These are a group of stars that formed together and have remained close.

As the stars drift apart they can become part of a "moving group", a collection of stars that don't appear to be a cluster, but since they share the same age and direction of motion we can tell they used to be a cluster.  Many nearby stars are part of the Ursa Major moving group, but the sun is not one. It just happens to be in the same part of the Milky way. The sun was probably part of a cluster shortly after it formed (4.6 billion years ago) but that cluster has long ago broken up. We don't (yet) know of any other stars that seem to have come from the same cluster.

Globular clusters, like M13 and Omega Centauri, are larger and have many more stars tightly packed together. They are all rather distant, and the brightest look like slightly fuzzy stars (in fact Omega Centauri was originally thought to be a star).

Of course a galaxy is a group of 100 billion stars. We don't normally think of galaxies as a star cluster, because they are so much bigger and the stars in them don't form at the same time.

Then there are clusters of galaxies, the Virgo cluster is a cluster of galaxies, and the local group of galaxies is on the edge of this cluster. But the Virgo cluster is not a star cluster.


Why are there no stars visible in cislunar space?

It is a matter of exposure and dynamic range. A sensor like a camera can only handle inputs in a certain range of intensities, and much of photographic skill (or smart presets) is about mapping the outside light onto this range so the details you care about show up rather than turn into white or black.

If you take a picture of a brightly lit scene, in order to make out the details of the bright parts (such as a lunar landscape, the Earth, the ISS etc) you will have to adjust the exposure making faint objects like the stars too dim to see against a dark sky background. You could try to set the exposure to show the stars instead, but now the landscape and Earth would be too bright (and likely also mess up the picture by causing flaring).

One can try to work around it by taking several pictures at different exposure levels and later digitally compositing them together. But this requires a lot of extra work.


Is a black hole made from a anti-matter star the same as a ordinary black hole?

To put a formal answer to this question, a black hole made of anti-matter would be indistinguishable from a matter black hole. Black holes are dubious objects, but it seems likely that it makes no sense to talk about an "anti-matter black hole" as once the matter goes into the black hole, be it matter or anti-matter, it ceases to have properties that define it as matter or anti-matter. The No-Hair Theorem for black holes states that, among a few other properties, a black hole is defined only by it's mass and net charge. The matter state of individual particles which fell into the black hole play no part in that description.

In fact, there is a vigorous debate as to whether or not "information" is conserved for black holes. E.g., if an anti-matter particle fell into a black hole, could you, at a later time, pull out a particle of the same properties or would you get out something of completely randomly new properties, not tied to the original matter. This in part plays into your question because it indicates that there is still a lot of uncertainty regarding the loss of information (including if the particle is anti-matter or matter) as particles fall into a black hole and whether not feeding anti-matter into a black hole actually makes it an anti-matter black hole or just simply a black hole.


Are ocean water planets at Mars' atmospheric pressure possible?

There's a couple reasons why it would be difficult, but not necessarily impossible (and I'll toss in a hypothetical at the  end).   I don't know how to answer this questions with a lot of "in theoreys" and "this is unlikelys", but I'll give it a shot.  Maybe when the James Webb Space Telescope gets up and running we'll get a better look at planet's atmospheres and maybe even identify a planet somewhat like your scenario.  For now, exoplanet atmospheres are incredibly hard to measure, so there's mostly just models and educated guesses to work with.
One reason it would be tricky is because the vapor pressure alone approaches the triple point air pressure, so this means that if the planet got too warm too often, say, mid-day noon or during summer, assuming it has seasons, then the vapor pressure, just from the ocean would probably rise above the triple point and that's before you add any additional gases.   CO2, CH4, N2, O2 if you want to get creative and get rid of a lot of the CO2 by photosynthesis and lets not forget Argon which forms from decaying potassium 40 in the planet's crust.  Earth's atmospheric Argon alone is about twice the triple point pressure.
So you'd need a planet with oceans but virtually no other gases and that might be hard to do.   Planets can lose atmosphere to space by solar wind or jeans escape, but water is a light molecule and arguably gets removed from a planet more easily than gasses like CO2, N2, O2 or Argon.  That's not to say it's impossible that a planet might be sufficiently water rich that it loses atmosphere but retains water.  I just think it would be a rare occurrence.
A second potential problem is temperature variation, which means, if the planet is too warm, too much ocean becomes atmosphere and adds too much vapor pressure and if the planet is too cold, the oceans would be largely covered by ice.   Planets are dynamic and a thin atmosphere planet should have a wide temperature range, just as Mars does.   Latitude, day/night cycle or day side/night side if tidally locked and eccentricity could all play roles in significant temperature variation, and lets not forget, the planet's internal heat and potential volcanism.   There would be a narrow range where the planet could maintain both oceans and a very thin atmosphere.
A final consideration is what you mean by triple point.   Oceans are likely briny and astronomers as a rule, are more interested in the "triple point" of briny water not pure water, though a chemist would say that triple point applies only to pure substances not salt water or possibly ammonia water.
If you consider the triple point of salt water oceans, then it goes below freezing where ice sheets become possible, and with an expected natural variation in the planet's temperature, ice sheets become very likely.  That doesn't make your scenario impossible, but I think you'd need a cold planet mostly covered by ice with narrow equatorial regions of exposed ocean - just a thought.
The triple point of briny water is discussed here - a fun read.
And a chart with some thing like the triple point of salt water in this research gate question here.
If, in theory, Mars was given oceans, by some terraforming efforts, as a result, initially the water would evaporate but over time, it's combination of water vapor and CO2 would push it's atmospheric pressure further above the triple point, and as it's south pole releases more CO2 as it presumably gets a little warmer, Mars would move away from this scenario if it had enough liquid water.  It has too much CO2, though for now, the majority of it's CO2 is frozen.
And Ammonia/Water ocean is especially interesting as the freezing point of the correct ratio can dip below the freezing point of CO2, this drops the triple point significantly, below -80 C° but it may solve the CO2 problem.  Frozen CO2 at the poles with Ammonia/Water oceans and lots of water ice.  Most of the oceans would have a permanent water-ice cover.
Mars, for example, undergoes a variation in atmospheric pressure as it goes through it's seasons by as much as 25%.  (Scroll down to the Mars section).   That's mostly eccentricity driven and happens because Mars is further from the sun during it's Southern Hemisphere winter and this allows some atmospheric CO2 to freeze, which thaws again as the planet moves back towards it's perihelion and Southern Hemisphere summer.
Certainly a planet with oceans and triple-point like conditions could be designed if you could create your own planet, but in actuality, the likely mix of compounds in the atmosphere and temperature variation and general dynamics of a planet would, I think, make such planets rare, if they exist at all.   That said, observations have shown astronomers things that they didn't expect to see many times before, so there's a chance.
Hypothetical: a Mars like planet has ice at it's poles, but warms up enough to lose it's atmosphere.  Ice is stable even in a vacuum, provided it stays below freezing temperature.
After losing it's atmosphere, the planet warms up and it develops a magnetic field.  The ice melts and now it has oceans and a very thin atmosphere.   Perhaps not impossible.  It requires a few things to go just right, so, still unlikely, I think but who knows.


Two luminous points disappearing in the sky

Edit 1: @JohnHoltz correctly pointed out two errors - geosats are on the equator, not ecliptic. My brain was thinking "equator" but my fingers had other plans. I read about geosat flares quite some time ago and thought it was solstices when they could be seen but it's around equinoxes, not solstices.  I've changed my mind and think planes are the most likely answer as to what was seen.  However, I do stand by the statement that geosat flares end due to the Earth's shadow. They are unlike other sat flares in this regard according to CalSky which I've always trusted above all other sources for info like this. The link is in the first sentence of my original post below. End of edit.
Edit 2: (by @uhoh) answers to How bright are geostationary satellites due to reflected sunlight? explain that flares from GEO can be bright enough to be visible, and when you look at how big those "mirrors" are it's not hard to imagine!
The heat radiators are dark in thermal IR but mirror-like in the visible to prevent absorbing heat from sunlight. The solar panels are larger but have fairly low reflectivity.

I'm 90% sure that you saw geostationary satellites catching the Sun just right causing a flare.  It's the right time of the year for them to be seen, too. Jupiter is sitting on the ecliptic at the moment which is where geosats are parked in clusters (the reason there were two dots visible at 1-2 degrees in separation). Their brightness increased because they were moving towards opposition (reflecting more sunlight), then faded because of the Earth's shadow exactly as you thought.  Geosats move slowly against the background stars since their job is to remain over the same point on the Earth; this is how people can get satellite TV with satellite dishes on their rooftops pointing at the same spot all the time. It'd be hard to notice their motion without keeping a steady eye on them for 15 mins or so.
Your description matches perfectly with a geosat flare and I'm a bit jealous because I haven't caught one yet!  Congratulations, you witnessed an event that isn't often visible to the naked eye!  They show up in my astropics all the time, but I have yet to see one with only my eyes.
A comment on the video in the link:  Geosats stay in the same position over the Earth but this video makes them appear to be moving left as the stars set behind them.  I was puzzled by this until I realized the ground objects visible in the bottom of the video are slowly moving, so the time lapse video was made with a camera slowly panning the to the right.  The video shows the clusters well but with the wide angle the individual geosats in each cluster aren't visible.  I think the ones you saw were in the same grouping.


Yes neutrons can exist outside the atom (or nucleus). In free space a neutron will beta decay into a proton, and electron and an anti-neutrino on a timescale of 10 minutes. However, in the dense interiors of a neutron star, the electrons form a degenerate gas, with all possible energy levels filled up to something called the Fermi energy.

Once the Fermi energy of the electrons exceeds the maximum energy of any possible beta-decay electron, then beta decay is blocked and free neutrons become stable. This is what happens inside a neutron star and you end up with mostly neutrons with a small fraction perhaps a few per cent electrons and protons.

In the outer parts of the neutron star, the protons and neutrons can still arrange themselves into nuclei (but not atoms), but these nuclei are extremely neutron-rich (they would not normally exist in nature) and are only stabilised against beta decay by the process I described above. The very outer envelope may consist of completely ionised iron-peak element nuclei and there may be an ultrathin (few cm) layer of recognisable ionised hydrogen, helium and carbon (e.g. Wynn & Heinke 2009).

Once the density reaches about $3 \times 10^{16}$ kg/m$^3$ it becomes more favourable for the neutrons and protons to organise themselves into "macro-nuclei" - long strings and sheets of nuclear material, known colloquially as nuclear pasta.

At higher densities still, the pasta dissolves into a soup of mostly neutrons with about 1 per cent protons and electrons.

The diagram below (from Watanabe et al. 2012) shows roughly how these layers are arranged. It should be stressed that this is based on theoretical modelling, with the theory becoming less certain the further into the neutron star you go. Testing these ideas involves nuclear and particle experiments, observations of pulsars, of neutron star cooling, of X-ray bursts, mass and radius estimates in binary systems, pulsar glitches, etc., etc. None of the details have been observationally confirmed beyond dispute, but the basic picture below fits what we know. In particular, the the crust and the n,p,e fluid regions are well understood in theory. The details of the nuclear pasta phases are still the subject of a lot of theoretical work, as are the details of superfluidity in the interior, and what happens in the very central regions (solid neutron core, extra hadronic phases, boson condensation, quark matter) is still theoretically difficult and observationally untested except perhaps to say that the softest equations of state have been ruled out by the existence of $2M_{\odot}$ neutron stars.


Do stars of a galaxy change their positions relatively to each other?

Stars do in fact move relative to one another within galaxies of all types. The orbital period of stars in a typical spiral galaxy (at around the same distance as the Sun is from the center of the Milky Way) is on the order of hundreds of millions of years. For the Sun it's something like 230 million years (source). A particularly old person (~100 years old), will have been around long enough to see something like one-tenthousandth of 1% of the orbit of stars in this region of the galaxy! Most of these stars simply do not move much relative to us in that amount of time.

The exception to this view of the motions of the stars is the Sagittarius A* (which is not a star at all, but is in fact a supermassive black hole at the center of the Milky Way). The period of stars traveling around it are on the order of 10's of years.

Also, to give you a rough idea of how galaxies look dynamically, here's a link to a particularly nice galaxy simulator implemented in the browser (created by Adrian Price-Whelan).


Are the compact objects in supernova remnants isolated neutron stars?

There are plenty of hard-x-ray binaries that involve a compact object with a companion.  Some should be in supernova remnants, but I don't know how many.  Are you sure there is a problem?


How many stars are there max and average per galaxy?

The term galaxy can not be given an approximate number of stars since our galaxy has a 27,000 light-years radius while a dwarf galaxy is typically 130 light-years. On the other hand, for and old elliptical galaxy, it can happen that some of the stars are already dead and the creation rate is almost zero, but since those are probably made out of 2 spiral galaxies colliding, the number would be around the sum of the two initial spirals. For some young spiral galaxies, 2-3  stars per year are being created and very few die.

So there are many factors that play a role. Ironically, for our galaxy the number of stars is pretty uncertain since we are inside of it, and we can't infer it from the apparent mass knowing how it rotates, because the total mass and the dark matter both play a role on it. Plus the star mass range can be as wide as 0.1 to 150 (!) solar masses. So, an order of magnitude for our galaxy? $10^{11}$  assuming the population of stars is consistent with the H-R diagram and that half of the population are binary stars.


Does a kilonova leave a high mass remnant?

I think it is generally expected that the merger of two neutron stars will lead to the formation of a black hole. What is more uncertain is whether that black hole forms straight away, or an intermediate state of a hyper(/supra)massive neutron star forms (see, e.g. Sec IIC of Hotokezaka et al, PRD, 044026, 2013 for a definition of hypermassive and supermassive neutron stars), which then collapses to a black hole after a short time. A few papers looking at the stability of such a hypermassive neutron stars formed from a binary merger are - 1, 2, 3 and 4 (check out the references and citations to them for more info). In the first three reference I think the collapse to a black hole is expected within a second (see Table 2 in Hotokezaka et al, PRD, 044026, 2013 for expected lifetimes before collapse times for various neutron star equations of state and masses), whilst in the final one a range of collapse times from fractions of a second up to several tens of thousands of seconds are found.
A search for a post-merger remnant from the binary neutron star merger GW170817 is now available here. In the introduction of this paper it describes in more detail some of the post-merger remnant possibilities that I mentioned above. The searches don't find any evidence for a signal, but given their sensitivities this isn't surprising (they'd require more than the total mass of the system to have been converted to gravitational waves to see anything!)


Does having an event horizon imply the existence of a singularity?

An event horizon is not an inherent component of any given object. It's not like once a star turns into a black hole, it suddenly gets an event horizon. The event horizon is merely a mathematical boundary which defines the distance from a mass $M$ where the escape velocity equals the speed of light. I can calculate such a boundary for a black hole, for the Sun, the Earth, or even you. So I guess the answer here is no, having an event horizon does not imply the existence of a singularity.

  Could it be that you have a neutron star massive enough to reach a escape velocity equal to the speed of light but not strong enough to make the matter collapse?

The answer here, technically, is no. The reason being that once it requires a speed greater than or equal to the speed of light to escape your object, it is necessarily a black hole. That is the definition of a black hole. So that means this neutron star you propose is actually a black hole. Another equivalent definition of a black hole is any object whose mass is concentrated inside that object's event horizon.

But you might still ask, could you have a black hole where the mass inside the event horizon is not a singularity. This would require some sort of support to prevent the matter from collapsing down to the singularity. The answer to this is that it is currently unknown. The problem is that inside event horizons, suddenly you need to work with both GR and quantum field theory but those two theories don't play nice. Instead you should be using a Quantum Gravity theory but this theory hasn't been developed. So ultimately any answer to this would be a guess until this theory gets fully fleshed out.


Does natural satellite(s) of a planet affects its orbital velocity around a star?

Yes, it does affect the planet's orbital velocity. Both planet and satellite move around the planet-satellite center of mass. That center of mass is orbiting the star with some constant velocity (let's assume circular orbits to keep things simple) but if you're only looking at the planet, then the orbital speed around the star will wobble up and down around the because of its additional motion around the planet-satellite center of mass.

For most planet-satellite systems this effect is very small since usually the satellite is very, very light compared to the planet. Only when you have a genuine double system like Earth-Moon or Pluto-Charon does this effect start to matter. For transiting exoplanet systems they're trying to infer the existence of exomoons by exploiting the effect this will have on the transit timing of the planet. So far no exomoons have turned up though.


State vectors of "interesting" multiple stars

The USNO Sixth Catalog of Orbits of Visual Binary Stars (ORB6)
has the orbital elements you need.

$\gamma$ And is WDS 02039+4220
$\alpha$ Gem is WDS 07346+3153
$\zeta$ UMa is WDS 13239+5456

Double star expert Bruce MacEvoy explains orbital elements in maybe enough detail to help you work out some state vectors.
My search also turned up a book chapter by Andreas Alzner.


70-year star in Talmud confusing sailors?

Comets can look like stars. Sometimes the outside of the coma and tail of the comet are too faint to see with the naked eye, and only the central part of the coma is bright enough to be seen. In this case the comet will appear like a (slightly fuzzy) star.  Look at the Great Nebula in Orion. It appears star-like, perhaps a little fuzzy.  Other clusters and nebulae were initially classified as "stars".

Moreover in the language of the first century AD (and until relatively recently) the word "star" meant "any heavenly light" and included comets, planets, and meteors. A navigator at that time would consider a comet to be just a "hairy star", while the learnéd debated whether comets were astronomical or meteorological stars.

At the time, many people also believed in "astrology" (the belief that the stars could have magical influences on life on Earth). It would seem quite reasonable, then, to think a star that appears every 70 years could affect navigation in mysterious ways.

As with many such ancient texts, the correct interpretation is probably lost. I'd consider it to be a parable: A man may be knowledgable and wise (as evidenced by his knowledge of the stars) but still be poor. I see no reason to try to identify the "star", no more that I would try to identify the farm in Isaiah 28. Trying to prove the story is "scientifically correct" seems to miss the point.


Which worlds have unique words for a "day"?

Word are created in response to need. There are only two planets on which we have any significant presence on the surface: Earth and Mars.

When running operations on Mars, the day/night cycle is significant, so a word was needed for the length of this cycle. Especially since the sol is so close to the day, there is greater potential for confusion. Other planets either have very long rotation periods or no solid surface.

On, for example, Mercury you can use "day" until there is a need for a new term.


What star was the asteroid 2014 MU69 predicted to occult on 2018-08-04?

A VizieR query within 15 arcseconds of Buie's coordinates returns a 13th magnitude star, variously identified as:

2MASS 19042146-2035362
UCAC4 348-170642
Gaia DR2 4082342058110504064


Use Wien's displacement law - as you suggest.
Let's assume that the spectrum you have been given incorporates almost all the flux from the star. This might be ok, so long as the flux is heading towards small numbers at each end of the spectrum? If so, then you can use the temperature-independent shape of a blackbody function to argue that some fixed fraction of the flux is contained below a wavelength which depends only on the temperature. This is a more robust version of Wien's law.

For example 50% of the flux should lie below a wavelength of $4107/T$ $\mu$m.

Thus you sum up your 20 points, using Simpson's rule or Trapezium rule if they are unevenly spaced or have gaps. Then you work out how much flux there is up to the wavelength of each of the wavelength points in turn. When you reach the one where 50% of the total flux is below it, then the temperature is $4107/\lambda$, where $\lambda$ is in $\mu$m.

In principle method 2. is more robust, because you are always trying to find the effective temperature - which is the temperature  of a blackbody that gives the same observed integrated flux. Since method 2 is an integral method, it is less sensitive to the detailed shape of the spectrum. Which is good, because stars are not blackbodies.

In contrast, method 1 is really at the mercy of how close to a blackbody the spectrum is.

A possible problem with method 2. though, is that your spectrum does not contain all of the flux. In which case your temperature will be an overestimate (assuming the missing flux is longward of 1000 nm).


How bright are the outer planets, when at close range?

There is a rule about average surface brightness: it is conserved as you change distance or magnification.  For the gory details see.  Telescopes make unresolved objects brighter, but for resolved objects, a telescope shows you the actual surface brightness (assuming that the exit pupil diameter is well matched to your eye's pupil (~7mm) and is not throwing away light, dimming the view).   The surface brightness of any resolved object through a telescope is the same as it would be if you were there.
Also, the 0.01 of earth sunlight received at Saturn is way more than the lighting provided by a full moon which is 400,000 times fainter than the Sun.
What you also need to take into account, is that when you look at Saturn through a telescope, it is generally (not always) done at night when your eyes are dark adapted.


For gravity the force is :

$$F=\frac {G M_1 M_2} {r^2}$$

But the acceleration :

$$a=\frac F {M_2} =\frac {G M_1 } {r^2}$$

and this is the same for all bodies as it depends only on the primary's mass ($M_1$).