Why is a giant planet around a tiny star unusual when binary stars are normal?

It's important to realize that binary stars form much differently than planets do. Assuming that both stars form in situ (i.e. excluding scenarios where one is captured from outside the system), there are several main ways for a binary star system to form from a molecular cloud. The most widely-accepted model at the moment is the fragmentation hypothesis, where the protostellar cloud splits during its collapse, forming two clouds which will each collapse into separate protostars.

Planets, on the other hand, form not from collapsing clouds but in protoplanetary disks around stars. The dynamics of these systems are substantially different, as there are interactions between the gas, dust, and protoplanets - perhaps a more complicated system than a collapsing gas cloud. We therefore see planets arise through different processes, such as gravitational instabilities within the disk or pebble accretion.

The paper being referenced found that the GJ 3512 system appears incompatible with the pebble accretion models the team tested - that is, rocky planetary cores being built up through collisions. In a low-mass star system, the cores should move inwards too quickly to accrete enough mass, and tweaking other initial conditions didn't help. That's what the BBC article focuses on.

What the BBC ignores is that the authors were able to reproduce the system by assuming that the planet(s) arise through gravitational instabilities within the protoplanetary disk, and this model is borne out by the simulations. It may not have been widely considered, but it still works. That's one of the key results to pay attention to.


How to calculate exact angle to the moon?

For a one-off calculation, the easiest solution would be to use planetarium software, such as stellarium.

For a programming solution, a python package such as pyephem is an efficient way to calculate the position of the moon at multiple times, in a way that can be imported into a spreadsheet.

The apparent motion of the moon is more complex than you might think.  The moon moves about 15 degrees per hour due to the rotation of the Earth. But on top of this fairly simple motion is the Moon's actual orbit around the Earth, which is elliptical (and so this motion is not even, the moon moves faster when it is close to the Earth) and the position and inclination of the orbit is perturbed by the Sun. The perturbations are fairly regular, and can be calculated, but these effects (rotation of the earth, the orbit of the moon, the eccentricity of the orbit, perturbation of orbit) combine to make the actual calculation more complex  than "a formula" that can fit easily in a spreadsheet, which is why a package like pyephem is recommended.


Why do the magnetic field lines of the earth go from south pole to north pole?

That was a curiosity that I also had once.
The reason for that is that the magnetic pole near earth's geographic north pole is actually the south magnetic pole.
You can find a more detailed answer on https://wtamu.edu/~cbaird/sq/2013/11/15/why-does-a-magnetic-compass-point-to-the-geographic-north-pole/

Happy Thanksgiving!


The CNO cycle in first generation of stars

First generation stars didn't do the CNO cycle initally.

It is estimated, that after the Big Bang three quarters of the matter is hydogren, one quarter helium, and trace amounts of heavier elements.  This is then also the composition of (as of now unobserved) first generation stars, having zero metallicity.  Even though those stars probably were very massive, they could not run the CNO cycle initially.

Luckily, the CNO cycle is not the only nuclear fusion process in stars.  In the proton-proton chain reaction, hydrogen is fused into helium.  The star is in equilibrium until the hydrogen in the core is "used up".  Then it may collapse and heat until the triple-alpha process begins to produce carbon.  From there on there are more and more processes producing many different elements.  Due to convection and shell-burning, the CNO cycle might have been running in those first generation stars at later stages.

Interestingly, the most massive of the first generation (population III) stars may have ended up as black holes, gobbling up all of the produced metals (elements heavier than helium).  Many others may have exploded in pair-instability supernovae ejecting all material into interstellar space, leaving no compact remnant behind.

A Heger and S. E. Woosley: The Nucleosynthesis Signature of Population III


shouldn't contracting objects show a redshift?

Yes, they should. But do they, actually? That's the better question.

To have any appreciable redshift, the speed of the object needs to be huge. That will be a very short implosion. Also, there aren't many mechanisms that can accelerate implosions that much.

What I'm saying is - it's theoretically possible, but in practice you won't see it that often, if ever.

  And if so it should be less towards the edges than towards the centre,
  right?

Sure. If the object is transparent to its own radiation, then you'd see blueshift from the other side, too. Again, this is very theoretical.


What is a "jumping trojan"? And what do their orbits look like?

In short, most of the trojans stay orbiting arround L4 or L5. These can be called tad-pole orbit asteroids.

There are some trojan asteroids, however, that their orbit never get's too close to L4 nor L5 to get trapped and have a tad-pole orbit, or they do but they have too much energy to get trapped.

These trojan asteroids have larger orbits, following a path like:

L4 --> L3 --> L5 --> L3 --> L4 and so on. This kind of orbit is called horseshoe.

There is, however, a third kind of trojan asteroid. This one appears to follow a tad-pole orbit (I.E. arround L4), but numerical analysis proves they will eventually adopt a horseshoe-like orbit. When they reach the other stable point (L5) they  will get trapped again in a tad-pole orbit. This kind of orbit is very rare (rarity decreases as planet mass increases) and could be considered (personal opinion) as an hybrid tadpole - horseshoe orbit.

For more information, see:

"Jumping mechanisms of Trojan asteroids in the planar
restricted three- and four-body problems" from: Kenta Oshima and Tomohiro Yanao.


Would bagging an asteroid destroy valuable science about it?

The simple answer, unfortunately, is yes.

In fact, it is a resounding "yes". Option "A" uses a large "bag" to enclose the asteroid, and then to tow it to lunar orbit (or another feasible location). There's a problem, though, which is that simply surrounding an asteroid with a cylinder won't capture it. You have to "tighten" the "bag". This is implied in a graphic here and is explained in an article in Popular Mechanics ("How to Mine an Asteroid", Aug. 2012), and shown in a video on NASA's website. Basically, a series of "arms" constrict the "bag" to form-fit the bag to the asteroid's surface. The procedure could create many indentations on the asteroid's surface, and, as you said, destroy some of the most valuable information. The footprints of the astronauts who walked on the moon will stay there for many years; any indentations on an asteroid will be impossible to fix.

However, this entire idea is far from finished. NASA will not even decide between options "A" and "B" until "late 2014". So we won't know the precise details for a while.

So why go ahead with the whole thing if we're going to lose some of the best information we could get? Well, the Initiative isn't motivated purely by surface features. Part of it is spurred on by plans to pioneer new technologies needed to get humans to Mars. Other tangential benefits include diverting an asteroid that could be headed for Earth (although the asteroid candidates for this mission are much smaller than any of their more dangerous cousins; also, the Asteroid Initiative is under "Asteroid and Comet Watch" on NASA's website - right near plans to blow up a dangerous asteroid!), or even mining an asteroid for metals or resources for astronauts (such as water).

Also, this mission isn't the only way to study an asteroid (although it might be the best). We could always send a lander to an asteroid - after all, we can land on a comet (erm, well, we're trying - I'm rooting for Rosetta and Philae!), which would make it simple to collect samples before heading home . . . although the distance to travel would be a lot longer than just to lunar orbit.

All of this comes together to mean that NASA won't worry itself over losing data on the surface of the asteroid. Would that information be nice? Oh, yes. But given the other good things coming out of the mission, the pros here outweigh the cons. It's simply one of the trade-offs that might have to be made for this mission.

Sources:

Planetary.org

Space.com

NASA

"How to Mine an Asteroid". Popular Mechanics. August, 2012.


What you may be missing is that no two telescope, filter and detector combinations are the same. A colour index defined by your method would be measured differently by different observers.

The solution is to define standards that  have defined magnitudes and colours that all observers can use to calibrate their observations.

In actual fact, the basic measurement is
$$ B-V = \alpha (b-v) + z + \beta X,$$
where $b$ and $v$ are native measured magnitudes (-2.5 log the detected adu with a correction for exposure time) and $X$ is the airmass. The constants $\alpha$, zeropoint $z$ and extinction coefficient $\beta$ are determined by observing a network of secondary standards with a range of colours at a range of airmasses.


Not that reddit is the most reliable of sources, but here's a Q&A with a response that is at least consistent with orbital mechanics.

  [Q} Oh. But why doesn't [the original particle set] just scatter
  around the planet, creating a big "bubble"?

  [A]Let's say there's a particle that moves around in an inclined
  orbit, so it has both a horizontal and vertical component to its
  velocity. If it collides with something, it'll either get more
  velocity, or it'll cancel out. If the vertical component is cancelled
  out, its orbit doesn't change, it just becomes less inclined. If the
  horizontal component is cancelled out, it'll fall to a different
  orbit. If you've got lots of particles going in different directions,
  their vertical components will cancel out, while those not going in
  the predominant horizontal direction will collide with others, lose
  their velocity, and fall inward. The net result will be a ring.

  [Q]That makes sense in two dimensions. But it doesn't explain why the
  rings are thin.

  [A]If the rings are thick, it means the particles still have a
  vertical component to the orbit, each being slightly inclined in a
  different way so as to make up thickness. They would have to move very
  precisely for them not to collide and cancel out.


Three lunar rovers and the Chinese Yutu rover. None of which are operational. https://en.m.wikipedia.org/wiki/List_of_artificial_objects_on_the_Moon


Are the stars outside of the galactic plane in the galactic halo?

The galactic disk, as Riley Jacob wrote, has a definite thickness. It's actually composed of a thin disk $\sim0.3\text{ kpc}$ thick and a thick disk $\sim1\text{ kpc}$ thick, at least (McMillan (2011) has models with data from the Sloan Digital Sky Survey). There's also a central bulge that is even thicker, as the following diagram (from here) shows:

Pollux is $\sim10\text{ pc}$ away from the Solar System, which is about 1% of the thickness of the thick disk (Capella is barely farther). Essentially, it's in the same plane as the Solar System; that's an insignificant distance.

There are stars in the thin disk, thick disk, and halo, which compose different populaitons based on metallicity. In the thin disk are Population I stars, which are high in metals. In the halo (and thin disk, to some extent) are Population II stars, which are lower in metals and on average older than Population I stars. The Sun is a Population I star in the disk. There may also be a sub-population of Population II stars in the thick disk.

Halo stars certainly do not compose the majority of stars in the Milky Way galaxy. Most stars are contained within the thin disk, thick disk, and bulge.


Why smaller black holes are considered to be more dangerous than bigger black holes?

All black holes are dangerous if you get close enough, but smaller ones are more dangerous because they cause stronger tidal effects, so they can cause spaghettification well before you reach the event horizon, whereas you can cross the event horizon of a supermassive black hole without the tidal effects becoming noticeable.

Very small black holes would also be dangerous due to the large amount of Hawking radiation they'd emit. E.g., a 1 million metric ton hole has a luminosity of over 356 terrawatts. However, there's no known mechanism to produce black holes that small, apart from waiting for a large hole to get small by evaporating via Hawking radiation, and that won't even start to happen for a very long time, since the cosmic microwave background is currently around a billion times hotter than even the smallest stellar black holes.

Actually, the main danger in the neighborhood of a black hole is if the hole is active, that is, if it is currently surrounded by matter that it's absorbing. That matter gets accelerated to very high speeds due to the gravitational acceleration and when it collides with other matter large amounts of dangerous radiation is released.


Space-time is not "made" of anything, it is merely a medium or coordinate system. Think about the grid lines on a map, they aren't "made" of anything, they're just a representation of the geometry of the Earth. Space-time is a concept envisaged by Einstein when he wrote his theory of Special Relativity that the properties of space and time become intrinsically linked at relativistic speeds. One of the consequences of this is that you cannot go forward in space without going forward in time, this can be visualized in a light cone:

This is by no means a representation of what space-time looks like, only how it behaves in 2 dimensions of space and one of time. Gravitational waves are caused when two massive objects (binary stars) orbit each other, and they cause "ripples" in space-time. Again another 2D visualization is needed:

This shows how the gravitational potentials of two stars interact in the fabric of space-time as they orbit, causing the radiation of gravitational waves.

Both of these are only visual guides, we will never be able to "see" gravitational waves or space-time itself with our own eyes, that is why we need detectors such as LIGO or VIRGO to infer their existence.


Do novas create a nebula? Could a planet survive if its star exploded in a nova?

A nova may leave behind a nova remnant: a shell of gas, expanding at about 1000km/s. In recurrent novae, they may be lit by light echoes from subsequent eruptions. They are much less massive than planetary nebulae, and at much lower energy than supernova remnants.

Novae displaying nebula shells or remnants include: GK Per, RR Pic, DQ Her, FH Ser, V476 Cyg, V1974 Cyg, HR Del and V1500 Cyg.

Planets in such a system would be bathed in high energy gas, this would not be enough to vaporise the planet, but it would make life on such a world extremely difficult.


Could we carve a large radio dish in the Antarctic ice?

Great question!
There are many open and active antarctic permanent bases and several antarctic astronomical observatories. There are even major large projects that involve substantial drilling and removal of ice such as the IceCube Neutrino Observatory. The biggest construction project is probably the Amundsen–Scott South Pole Station.
This is certainly a valid idea to consider and should not be dismissed out of hand by "Antarctica is hard".
There's no bedrock on which to affix your dish; it will be floating in creeping ice and subject to drifting snow.
From Amundsen–Scott South Pole Station:

The new station included a modular design, to accommodate rises in population, and an adjustable elevation to prevent it from being buried in snow. Since roughly 20 centimetres (8 in) of snow accumulates every year without ever thawing the building's designers included rounded corners and edges around the structure to help reduce snow drifts.

Basically I think @PeterErwin's comment summarizes the problem nicely. For more details see their link and also Challenges That Face South Pole Architecture which mentions the following:
WIND

Constant winds result in snow pile-up on buildings.
The new station faces into the wind, and is airfoil-shaped.
The airfoil forces air into a compressed space where it accelerates.
The fast wind scours out built-up snow.
Years later, if snow still builds up, the building can be lifted two more
stories on its columns.

ICE FLOW

The station sits atop a 2-mile deep layer of ice.
Each year, the geographic South Pole is marked.
Ice (cold water), slowly drips down to the ocean with gravity.
The trail of yearly South Pole markers shows that the ice moves 33 feet per year.

ICE CREEP

The weight of the building also causes the ice to move locally.
Ice compresses and shifts away from sources of pressure.
Resulting variable rates of sinking make keeping the building level a challenge.
Architectural elements built into the design will help meet that challenge

Basically everything is built up on stilts above the surface. This lets wind blow under things and prevents snow drifting which would quickly burry a surface structure.
The South Pole Telescope consisting of a 10 meter point-able dish is also built on an elevated platform sitting on stilts to prevent drifting (see below).
You could propose heating the entire dish to de-ice and de-snow it regularly, and pumping out the melt water, but power is at a premium at the South Pole. Fuel is flown in on airplanes during a short summer season and burned in electrical generators for power. Building something that requires electrical power to constantly melt drifting snow has serious practical hurdles.
Radio telescope dishes built into fixed surface depressions
At least two very large, single dish telescopes have been built into naturally occurring depressions in local rock: Arecibo at 18.3° N and FAST at 25.7° N latitudes. Both telescope designs have large primary focus structures (small buildings where people can work) suspended high above the ground at the primary focus of the dish, and in order to view the celestial equator and to track objects as the Earth rotates, moveable cables suspended from very tall towers were designed allow these structures to move around by about in two directions by roughly +/-19° and +/-26°.
For more on that see this answer to Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work?
Arecibo succumbed to deterioration of its cables caused by environmental factors and fell down.

Is there any role today that would justify building a large single dish radio telescope to replace Arecibo?
What will succeed the Arecibo Observatory?
How could the loss of the Arecibo Observatory have been prevented?
Building a floating, ocean-going giant radio telescope?

Maintaining these kinds of necessary tall structures in the high winds and snow storms at the South Pole would be quite a challenge as well.

Sources: nsf.gov news and South Pole Station Destination Alpha

Source: nsf.gov news

Sources left and right

Source

Sources for both: Antarctic Bases and Buildings - 3 Building at The South Pole via @PeterErwin's comment.
left: "The South Pole radio telescope, the very clear and dry air at the pole make it the best place on earth to have a space telescope.
Photo Christopher Michel from San Francisco Creative Commons Attribution 2 Generic license" right: "In the 6 months of night time of the South Polar winter with the Aurora Australis overhead, July 2008 Photo: Patrick Cullis - National Science Foundation"


What is 'surface acceleration' of a planet or other body? As opposed to its 'g' (gravity)?

The technical content of NASA's web pages have sadly declined over the years. The planet fact sheets are exemplary of that decline. The December 8, 2002 version of that page provides only one value instead of two, titled "surface gravity". The values are 3.69 m/s2 for Mars and 9.78 m/s2 for Earth.
The gravitational acceleration at the surface of the Earth (or Mars) is not constant. It varies with latitude and altitude. The value of 9.78 m/s2 for the Earth represents the gravitational acceleration, including centrifugal acceleration, at Earth sea level at the Earth's equator. The gravitational acceleration at the Earth's North Pole is a bit greater than this, about 9.834 m/s2. There is no centrifugal acceleration at the Earth's poles, and the Earth's poles are a bit closer to the center of the Earth than is the equator thanks to the Earth's rotation. The same concepts apply to Mars.
The value of 9.80 m/s2 added to the cited page a bit later is arguably erroneous; it should be 9.81 m/s2 (or even better, 9.80665 m/s2) rather than 9.80 m/s2. As I've said elsewhere regarding Wikipedia, "that's Wikipedia for you." Unfortunately, the same applies to NASA web pages, in spades.


The Chandrasekhar limit applies only for white dwarfs. Stars on the main sequence (or even off the main sequence) can easily surpass it, but if a white dwarf's mass is greater than the Chandrasekhar limit ($1.39 M_{\odot}$), it will undergo some sort of collapse.

First, though, in response to

  Or has it already undergone supernova explosion?

White dwarfs are not formed from supernovae, but from stars that (relatively) gently shed their outer layers to form planetary nebulae. So no, the stellar remnant will not have gone supernova already.

Back to the main question. If a white dwarf exists, its mass is less than the Chandrasekhar limit. To surpass that, it needs to gain mass. This can happen in binary systems where the white dwarf accretes matter from the more massive star (generally a giant). The white dwarf can then undergo a Type Ia supernova and explode. The light curve will look a bit like this:

A second (but far less likely) scenario is that two white dwarfs merge. The result? A Type Ia supernova and, possibly, a neutron star.


Star formation: When a nebula collapses, is it only the gasses that form the star?

The very first stars to form indeed consisted essentially only of hydrogen and helium. When stars die, they leave behind them more massive atoms, as you say. These heavier elements are too incorporated in newer stars when they form. This results in stars which start out with a lower portion of hydrogen and helium, thus making them somewhat less effective; see for example this question and answer. The Sun, for instance, is a so-called third generation star, which means that it is made up of materials which have already been used at least twice since the Big Bang, and thus naturally contains heavier elements it has not produced itself.


How do you calculate length of day in a multiple-star system?

Are you asking how to measure a day for an exoplanet?  Because that is pretty hard given that they are far away and hard to image their surfaces.  But, if you are asking how to define the day, say for simulations, then that is pretty easy, actually.
From Wikipedi's entry on Day.

For a given planet, there are three types of day defined in astronomy:
stellar day - an entire rotation of a planet with respect to the distant stars
sidereal day - a single rotation of a planet with respect to the vernal equinox
mean solar day - average time of a single rotation of a planet with respect to the sun as the central star

The first two have to do with the spin of the planet with respect to the remote stars and therefore the complexity of the stellar system does not enter, to first approximation.  The 'stellar day' depends on how much time it takes for some remote galaxy, for instance, to come back to the same point in the sky as seen by an observer on the planet.  The 'sidereal day' depends on how much time it takes for the direction that is at the extension of the equator of the planet with the plane of the orbit (pick 1 of the 2 points) to come back to the same point in the sky. So presuming that the spin axis and orbital plane are not changing much (and conservation of angular momentum law helps here), this is all well defined.  If the planet's orbital plane is changing rapidly then there simply is no well defined sidereal day, but still there would be a valid stellar day.
The 'mean solar day' is the average time it takes for the nearest star to return to the same spot in the sky.  Generally, even in a binary or triple system, the planet is close enough to one that this is well defined.  This is because, if  not, the planet is probably not in a stable orbit and will quickly leave the system.  However, if the stars are in a tight configuration  and the planet's orbit is wide and far enough away from them, it could be stable.  Then the simplest definition for 'mean solar day' would use the center of mass of the stellar system instead of the nearest star (which would change over time).
There are a few stable complicated orbits, like a figure 8 orbit, where the planet can switch back and forth between two stars.  Unlikely, but theoretically possible.  In such a case, the mean solar day is ill-defined, but the stellar and sidereal day are well defined.


How is the Dating of Samples of Cosmic Objects Performed, Including Pre-Solar Dust?

The BBC article pointed out by @uhoh in a comment answers the question nicely.

  To work out how old the grains were, the researchers measured how long
  they had been exposed to cosmic rays in space. These rays are
  high-energy particles that travel through our galaxy and penetrate
  solid matter.

  Some of these rays interact with the matter they encounter and
  form new elements. The longer they are exposed, the more of these
  elements form. The researchers used a particular form (isotope) of the
  element neon - Ne-21 - to date the grains.

  "I compare this with putting out a bucket in a rainstorm. Assuming the
  rainfall is constant, the amount of water that accumulates in the
  bucket tells you how long it was exposed," said Dr Heck.

  Measuring how many of the new elements are present tells scientists
  how long the grain was exposed to cosmic rays. This in turn informs
  them how old it is.

  Some of the pre-solar grains turned out to be the oldest ever
  discovered.

  Based on how many cosmic rays had interacted with the grains, most had
  to be 4.6-4.9 billion years old. For comparison, the Sun is 4.6
  billion years old and the Earth is 4.5 billion.

  However, the oldest yielded a date of around 7.5 billion years old.


Is there a connection between the solar wind and the cosmic radiation?

Cosmic radiation (which consists of high-energy protons and atomic nuclei) comes from the sun, other stars, even other galaxies.
Solar wind is a part of that - the part coming from the sun, which mostly consists of electrons, protons and alpha particles with kinetic energy between 0.5 and 10 keV


Solar flares are observed at wavelengths right across the electromagnetic spectrum, not just H alpha.

The basic model for a solar flare starts with the magnetic field in the corona. You can think of the topology of the magnetic field to consist of loops that poke up out of the photosphere and extend into the corona. However, the photosphere of the Sun is turbulent and constantly in motion due to convection and differential rotation. Whilst a loop may be formed in a minimum energy state, it can get twisted and stressed by these motions.

At some point an instability is reached and the magnetic field can undergo a "reconnection" event, to flip back to a lower energy configuration. During this event, charged particles are accelerated and travel down the magnetic field lines towards the photosphere.

Before they get there, they encounter the chromosphere, which is where the bulk of the particle kinetic energy is deposited. i.e. the density increases as you go down towards the photosphere and once a certain column density is reached, the accelerated electrons are stopped and deposit their kinetic energy. This results in heating and excess H alpha emission from material at about 10 thousand kelvin in the flare footpoints. Any hotter than this and all the hydrogen is ionised. The H alpha is in emission because the only material above it is optically thin to the H alpha radiation. There is hotter, ionised material produced too, and much of this is evaporated such that it fills the magnetic loops with X-ray emitting plasma at temperatures of more than a million kelvin. Some of the flare energy may also be used to accelerate material away from the Sun in "coronal mass ejections".


How does the earth's plane of orbit (ecliptic) vary over time?

The position of the ecliptic does change, due to perturbation of other planets, but only very slowly.

The invariant plane is the plane of angular momentum of the whole solar system. This is dominated by the orbital momentum of the planets, particularly Jupiter, and since other stars are much too far away to perturb the solar system, it is invariant with respect to distant stars.

The ecliptic does vary. It is currently decreasing by about 1.4 degrees per 10000 years, as indicated by this graph, (from Wikipedia). This variation is one hundredth of the variation in the Earth's rotational axis.


How can I validate star names and designations?

As far as I know, there is no standard that defines valid characters for a stellar designation. Sometimes people write using Greek letters (e.g. "α Centauri") and there are many odd catalogue notations. I would not trust that star designations could not contain commas or pipes (aka vertical bar, |), although I think it would be rare.
The IAU has stated that star names are separate from the catalogue designations, and in regard to designations "it is standard practice to identify them with an alphanumeric designation. " But standard practice does not guarantee that somebody will not disregard it, and there are clearly plenty of designations already with non-alphanumeric symbols.
So, one should probably not assume too much about what characters are in the string. Presumably for safety and generality one needs to assume a Unicode string and some robust way of escaping entries.


Why is [O III] a good density probe in interstellar medium?

The key to understanding this is the concept of forbidden lines and forbidden transitions.

A forbidden transition is one that cannot occur radiatively via an electric dipole interaction. Instead, it must proceed either through magnetic dipole or electric quadrupole emission, with much lower probability, or the transition may be accomplished through collisional (de)excitation.

A forbidden line is the radiation at a wavelength corresponding to a forbidden transition. In laboratory plasmas these are usually not seen - hence the term forbidden lines - because the transitions are normally accomplished through collisional de-excitation on a timescale much shorter than the radiative lifetime through magnetic dipole/electric quadrupole emission. However, in astrophysical plasmas the densities can be many orders of magnitude lower than even the best laboratory vacuums. In this case, forbidden lines are seen and indeed are often an important means of radiative cooling.

The property of forbidden lines which makes them an excellent density-sensitive tool is that they can be "quenched" if the density becomes large enough to make collisional de-excitation more probable than radiative de-excitation. The strength of the forbidden line is thus sensitive to the electron density (which is what dominates the collisions) between densities where "quenching" starts to become effective and higher densities where the line essentially becomes unobservable.

Taking the specific example of OIII (doubly ionised oxygen) optical line emission. There are three forbidden transitions of interest between the $^1D_2 \rightarrow\ ^3P_2$ (501 nm), $^1D_2 \rightarrow\ ^3P_1$ (496 nm) and $^1S_0 \rightarrow\ ^1D_2$ (436 nm) states.
These transitions produce optical forbidden lines which are quenched at different characteristic densities. Measuring a line strength ratio is important because the ratio will be independent of the abundance of OIII. The ratio will depend on electron density, providing the electron density is $>10^{5}\ cm^{-3}$, and temperature (at lower densities, the ratio is just temperature dependent).

To get the density in this case requires more information - usually supplied by a similar line ratio for something like NII, which has a similar set of forbidden lines, but with differing density and temperature dependence.

A more straightforward route is to use the line ratio for forbidden lines of OII or SII where there are a closely spaced pair of energy levels, both undergoing forbidden transitions to the same lower level. In this case the ratios are density sensitive, but not temperature sensitive, again over a range of densities where one or other of the transitions is in the process of being quenched (e.g. for OII 372.6/372.8nm, $10^2 <n_e < 10^{6}\ cm^{-3}$.)


Someone suggested this object to be 31 Ori. After further research, equipped with the name, I confirmed this object to be 31 Ori. Source of confirmation: http://www.astrostudio.org/xhip.php?hip=25737


Do solar systems have to evolve in a galaxy?

Probably not. You'd need a could of gas dense enough to contract under its own gravity. In the space between galaxies, matter is spread too thin for a solar-system sized cloud to exist. A much larger cloud could be more diffuse and still collapse, but being much larger, it would collapse into a galaxy.

Even within a galaxy, stars seem to form when shock waves from nova or some other event initiates the gravitational collapse.


What is the star whose western traditional name is the longest?

According to Google, the longest name for any star is “Shurnarkabtishashutu”, the Arabic for “under the southern horn of the bull”. The star is known as Zeta Tauri. You could also scroll thru Wikipedia's list of star names for different countries.


Why is the eccentricity of Triton so low?

Tidal circularization is indeed a plausible mechanism for why Triton's orbital eccentricity is so low, and I believe it is in fact the prevailing theory, at the moment. In short, tidal forces from Neptune dissipate energy in Triton's interior, essentially "squeezing" the moon. This results in a loss of orbital kinetic energy. The exact calculations that lead to the decrease in eccentricity are a little complicated - and described in the paper you linked to - but in a nutshell, Triton's orbit could have circularized on a timescale of about $\sim10^7$ to $\sim10^8$ years (Ross & Schubert (1990), for instance, computed the latter order-of-magnitude timescale).

Tidal circularization is not the only possible explanation for Triton's low eccentricity. Cuk & Gladman (2005) hypothesized that there was once a debris disk around Neptune, possibly caused by the collision of two or more other satellites after gravitational interactions with Triton. Interactions with the disk then contributed to energy loss from the moon in the same way that tidal circularization might have, thus leading to a smaller eccentricity. Interestingly enough, though, this could have happened within only $\sim10^5$ years! The authors have suggested that a hybrid of the two models may be the best explanation.


Was this an astronomical phenomenon observed in 1689?

This is a really interesting question!  One thing to note is that since it appeared in the northeast at sunset, it is nowhere near the sun - in fact, it is pretty much in the opposite direction as the setting sun.  In the winter in the north, the sun would set in the southwest.  And regarding your comment about the text describing the event as "around sunset" when it would have been 45 minutes later, I think that's not surprising - at northern latitudes in the winter, around the winter solstice, the path of the sun is at a very oblique angle to the horizon, so twilight lasts quite a long time.  So it could easily feel like "sunset" for an hour or more.  Though the sun set at 3:15 PM as you note, at 4:00 PM the sun would still only be 5.3 degrees below the horizon, so civil twilight had not yet ended. Plus, no one in those days was wearing a wristwatch, so a given observer's estimate of time of a particular event may have been less precise than we would be accustomed to now.
So, my best guess is that this might have been related to anticrepuscular rays.  Pros and cons for this hypothesis:
Pros:

Appears at the right time of day, in the right location (opposite the setting sun).
Would give a vertical ray.
Doesn't need clouds in the direction that the image is seen.  (But see my comment below about the crossbar.)

Cons:

Most of the cases I can find pictured have multiple rays, which isn't what is described here.
There is no obvious explanation for the horizontal part of the cross.
I'm not sure if this would persist without obvious changes for as long as described.

Another thought I had, which looks very similar in appearance, but is in completely the wrong direction, is a sun pillar, combined with a stratus cloud, like this (from here):

The appearance is striking, but (a) it would have to be in the sunset direction, and (b) there would have to be clouds.  So it doesn't seem to quite work.
I also thought of the moon, and whether that could give something similar, e.g. if it was just below the horizon.  But that doesn't quite work, either.  I did some calculations, and the moon was 38% full that day (waxing crescent, one day short of first quarter).  At 4:00 PM (assuming that the location is UTC+1 hour), the moon would have been 23 degrees above the horizon, at an azimuth of 161 degrees, i.e. low in the south-southeast.  So that doesn't seem like it would contribute much to something in the northeast.
So clearly I don't have a definitive answer, but maybe something here will be helpful or give others some ideas.


If you know distance from Earth to both objects and the angle between them viewed from Earth, it is just a matter of trigonometry.


What implications does younger volcanism have on the Moon's thermal evolution?

A younger lunar thermal history implies that the Moon was still undergoing a contraction and cooling process, which was initiated when it began accreting material together. Evidence has shown that the moon has no tectonic activity, so all geologic features are volcanic.

The mare are a more recent example of this activity, and have been shown to have fairly young ages, as you indicated. The moon still has some internal heat from its formation that may be leaking out through convection and cooling.

Even though the moon has a relatively small mass, it didn't lose its internal heat as quickly as the earth because there was no tectonic activity to facilitate it fast enough, and convection is a slow process.

As the moon cools, it contracts, creating sharp ridges and cracks in the surface (see features known as 'scarps', or 'garden'). These contractions contribute to the "deposit" of material formed in the images that paper discussed. As the cooling surface shrinks, it breaks the crust through a thrust fault, which pushes some material higher and shoves other material lower.

What may appear to be river channels on the moon are also lava flows. These were initially underground lava channels whose roofs collapsed to expose these subterranean features. More recent incidents of molten material rising to the surface and cooling can contribute to some of the younger ages recorded in the basaltic rocks.

Tldr; Evidence of younger volcanic activity confirms existing theories for the contraction and cooling process of the moon, just that it didn't end as soon as we initially thought


I think that there isn't a strict answer to this question. However, I believe the answer is that there's a difference between the core of a hydrogen-burning star and the core of a protostar or star-forming, gas cloud.

For a hydrogen-burning star, the core, as you say, is the region of the star where fusion is taking place. This is surrounded by the radiative and convective zones (which one comes first depends on the mass of the star). For true stars, the concept of the core is well defined.

For things like protostars and star-forming gas clouds, the concept of the core is less well defined and more a nomenclature for a region of the object rather than a strict definition. I don't think it is correct to assume that because one is talking about the core of a protostar, that they are referring to the core as it is defined for a hydrogen-burning star.

So what does it mean to be a core in a protostar or star-forming gas cloud? I will first point out that protostars can still be fusing. In the pre-main sequence stage, most stars will be fusing deuterium and we can consider the core to be the region where this is occurring. A protostar does not become a fully fledged star until it begins hydrogen burning, but deuterium fusion requires lower temperatures and pressures than the full proton-proton chain of hydrogen fusion (of which deuterium fusion is a part) and so can occur in protostars.

We might also distinguish a core from outer regions based on a density or compositional change. If we plot up something like the density (or really any property that changes with radius), we are likely to see a drastic jump signifying some sort of boundary. Cores for star-forming regions may be defined by highly dense regions, high temperature regions, or regions "where the physics changes" because we've passed some critical threshold at some radius. These cores are less well defined and can't be likened to stellar cores very well.


Even rocky planets would explode. I think there are two ways to see this.

From the perspective of forces, the earth is in equilibrium between the compressive force of gravity and the elastic resistance to compression of the materials that make it up. By Newton's third law, the mantle is pressing upwards on the crust with a force equal to the weight of the crust. If gravity disappears, you will still have that upward force with no downward force to balance it, so you will get upward acceleration.

You may think that the tensile strength of the crust will hold the planet together. But the upward force is the same as the weight of the crust. We know that very large stone structures cannot support their weight under tension.

You may also think that iron and rock aren't very compressible, so the mantle wouldn't expand very far. But the pressures are very great. The core has a density at least 25% greater than iron does under normal pressures. Plus, the temperature is around 6000K, equal to the surface of the sun and much higher than the boiling point of iron at atmospheric pressure. So the expansion won't just be a matter of some cracks forming.

From an energetic point of view, the potential energy of all the parts of the earth spread out across the protoplanetary disk was converted to heat and elastic potential energy when the earth formed. At least one study suggests that about half that heat remains, and in any case heat from radioactive decay has been added over time. If all the energy remained, it would be enough to disperse the earth with the same velocity that parts came together on average when it formed. Given that the energy is at least on the same scale, I think it's reasonable to expect an explosive breakup.


I live in an area with a lot of light pollution, how can I view the stars without building an observatory?

telescope.com has a quick paragraph on astronomy. Below I summarise the important points with a few of my own suggestions thrown in.

Dark & dirty places

Set up on grass or dirt, pavements and buildings radiate the heat again at night and the air flow created by this can distort your image. If possible you might be able to make use of a public park.

Failing the above you might be better off going to the outskirts of the city and finding a nice dark hill and avoid as much of the light pollution as possible, but obviously this depends on your situation.

Telescope

This doesn't require a special telescope, you should be able to manage just fine with what your budget allows for.

Though a computerised mount can aid you plenty in locating objects, as you have fewer stars to use as way-points in light polluted areas.

Here is an example I found earlier - yes they are quite pricey.

Note: This picture and the mount in the link are NOT the same mount.

Try to set up out of direct view of a source of light, i.e street-light or porch light. This is because you will suffer from a lack of contrast, making it harder to view the objects. So dark spots are good!

Light Filters

A good, fairly cheap method*, for negating light pollution that I have found is to use a light filter, these usually just screw on over the telescope and filter out light of that wavelength.

However, a very important point to note is that light filters will also filter out light from stars, galaxies and star clusters, making them appear duller.

Light filters actually decrease 'sky glow' which is, in part, due to the upper atmosphere.

What these filters are best for is looking at nebulas

Nebulas emit light of a different wavelength that are not dulled by the filter and they pass straight through.

If you are not hoping to look for nebulas then I do not suggest a light filter, as this will also dull the stars.

Examples of which can be found here

*cheap relative to buying a go-to mount that is


Calculate Atmospheric Extinction Using Source Altitude Angle

You seem to be referring to the computation of $m$ in equation (15) for the extinction as the light from the source passes through the atmosphere. The author intends to say that $m\approx \frac{1}{\sin(\alpha_s)}$, where $\alpha_s$ is the altitude of the source (in radian), so restricted to the range $0$ to $ \pi/2$. (Note this is not the inverse of the sine function as the author says but the reciprocal of the sine function)  The altitude is the angle measured from the horizon to the source, which is $=\pi/2-\theta$ where $\theta$ is as in figure 2. Note this blows up when $\alpha_s=0$ ($\theta=\pi/2$) because in the simple approximation being used the path length through the atmosphere in infinite.

The approximation being used here is that of a flat Earth, and $m$ is the number of equivalent atmosphere heights that the light path travellels through


Need help with the calculations/conversion of a celestial object

I couldn’t really figure out the way to get RA Dec from the data mentioned. I ended up using a catalog that already has RA Dec of objects.


Well according to the article "Have we seen a white hole?" (Byrd, 2011) there are some claims that scientists (specifically "The Revival of White Holes as Small Bangs" (Retter and Heller)) may have detected a white hole; however, this claim seems to be shrouded in controversy, as this observation may actually have a more stellar explanation.

However, the article from Retter and Heller offer a hypothesis as to what white holes are, from the article linked before:

  We thus suggest that the emergence of a white hole, which we name a ‘Small Bang’, is spontaneous – all the matter is ejected at a single pulse. Unlike black holes, white holes cannot be continuously observed rather their effect can only be detected around the event itself.

Which, if the authors are correct, would explain why we don't see/detect them and white holes (unlike black holes) are unique because they supposedly spew matter, rather than extract.

One major caveat is that it is not clear as to where this matter that spews out comes from.


What is meant by the notation $A^{\times{B}}_{\div{C}}$; where A,B,C are real numbers?

A confirmation of the explanation is given slightly further on where it talks about how the uncertainty is derived:

In equation (4) we use the 0.6 dex scatter of $\log L_{\rm UV}$ around
the $L_{\rm UV}$ versus $L_{\rm [O\ III]}$ relation, as an estimate for the uncertainty
in deriving $L_{\rm bol}$ from $L_{\rm [O\ III]}$.

The term "dex" indicates a base-10 logarithm of a unit, which translates into a factor of $10^{0.6} \approx 4$. So translating the error expressed in the logarithm of the quantity into an error expressed on the quantity itself, this becomes "multiply-or-divide by 4", which here is rendered $4000^{\times 4}_{\div 4}$.
Pela's excellent answer explicitly goes into detail about what this implies for the probability distribution. I usually prefer notation like $10^{x \pm \sigma}$ which sticks to more familiar notation and explicitly implies something vaguely log-normal rather than normal, although perhaps it gets a bit cramped when asymmetric error bars come into play.


Software to convert RA and DEC into ALT and AZ

Based on stellarium, I think you're not accounting for precession (more after the picture):

As you can see, Rigel's J2000 coordinates and Rigel's current coordinates are different by about the same amount as your calculation.


Does terrestrial planets conduct heat from near to far side?

The transfer of heat would be insignificant. Consider the Earth, the core is as hot as the surface of the sun. Some of this heat does get transferred to the surface, but it is only 0.03% of the energy received from the sun. Rock is a good insulator, most of the heat is transferred by convection.

The far side of a tidally locked planet would be very cold. It would receive very little heating from the planet's interior or the near side.


Why does moon moves to side if I change my position directly into it?

I can approximate the scenario in question at 2020-03-15 04:23 UT.
At location 1, the Moon is at alt=34.6°, az=90.0°.
At location 2, the Moon is at alt=41.2°, az=85.1°.

We see a shift in azimuth because the Earth's surface is not cylindrical.
You can get from location 1 to 2 by moving 721 km on a constant heading of 90°.
However, the 36°S parallel curves almost 5° to the right in that distance.

Moving the same distance constantly toward the moon, along a great circle, would put you about 29 km north of location 2 with final heading 85.3°.
Continuing around the world on the same great circle, you'd cross the equator at 39.6°E, heading 54.1°.

A view of the zenith showing the equatorial grid (blue) may help to illustrate the issue.
The left half of the prime vertical (green) is due east.
The longitude difference between locations 1 and 2 is equivalent to 0.53 hour of right ascension.


Exposure details for this now (in)famous image from Lowell Observatory?

I was the one operating the telescope at the time! I was using our 11 inch telescope fitted with a 1 MP MallinCam camera. The image was taken at 9:21 pm with a 25 second exposure.


Supernova in Hertzsprung–Russell diagram

It would be located way off the diagram to the top and left  i.e. moderately hot and very, very luminous. Of course it then changes with time as the supernova fades and cools.

This paper by Faran et al. (2017) shows plots of luminosity and temperature versus time for type II supernovae. They start at 15,000K and luminosities of about a billion suns and then fade to about 5000K and by an order of magnitude in luminosity after 50 days.


What's going on here is that you have been misled into thinking the ring-like structure has anything to do with the accretion disk. It doesn't, or at least only indirectly.

The disk is referred to as geometrically thick, but optically thin (see sections 1 and 2 of paper V issued by the Event Horizon Telescope collaboration on 10 April 2019). This is actually the opposite of the disk visualised in the film "Interstellar".

Because the disk is geometrically thick, it covers the whole picture. Because it is optically thin, we can see through it to the black hole. That is the basic answer to your question.

In an optically thin plasma, the brightness you will see is proportional to the optical path length (physical length multiplied by an absorption coefficient) along that sightline. The photon ring marks radiation travelling towards us that has been bent around the black hole, or has even orbited it several times. Hence those sight lines have larger optical path lengths and that is why we see it as a bright ring.

Radiation travelling towards us from plasma in front of the black hole, has a small optical path length and is not very bright. In addition, the sight lines to plasma behind the black hole cannot travel through the black hole or even close to the event horizon. Hence the circular "shadow" inside the photon ring.


How do we know how the Sun (or any star, for that matter) will evolve?

The thing you are not taking into account is that our predictions for the future evolution of the Sun are not based on our understanding and observations of the Sun alone. They're based on our understanding and observations of all the stars we've ever looked at. The thing is, stars of the same mass and composition must necessarily evolve in the same way. So we can look out at other stars and find some which are more or less the same as our own, but might be farther along the evolutionary track. We can study these stars and see what is happening to them. By conglomerating a lot of data about many stars at many different points in their evolutionary life cycle, we can put a coherent picture together for our own Sun.

Scientists have spent a lot of effort to create stellar evolutionary models. To do this, they put together the multitudinous information gleaned from the tens of millions of stars we've observed, add in a lot of math and physics representing the physical processes occurring in these stars and what you find is that the "predictions just inevitably fall out". There are many stellar evolutionary code bases out there you can find. For example, Pols et al. 1998 applied stellar evolutionary code to clusters of stars. A figure pulled from their paper is shown below. It shows a Color-Magnitude diagram of the M67 open cluster of stars, along with predictions of the evolutionary track of those stars. There's of course some variance as every star is not exactly the same, but overall, you can see that the stars follow the predicted evolutionary track pretty well.

Once you have a well calibrated stellar evolutionary program, it's not too difficult to apply that program to our Sun. There will be some unknowns and some variance, e.g., we might not know the physical extent of the red giant phases' expansion, but we know it will happen.


Is the Sun visible from Proxima Centauri to human eyes?

Well, there's two things we'll need for this: apparent magnitude (the brightness that an object appears to have) and absolute magnitude (the actual brightness an object has). Both of these scales are logarithmic, with brighter objects being lower and dimmer objects being higher. Astronomers have determined that the Sun's absolute magnitude is 4.83. Knowing this, we can find the apparent magnitude of the Sun from Proxima Centauri's location. Apparent and absolute magnitudes are related by the equation:

$$M = m - 5 (\log_{10}{d}-1)$$

Where $M$ is the absolute magnitude, $m$ is the apparent magnitude, and $d$ is the distance, in parsecs. Astronomers have determined that Proxima Centauri is 1.3 parsecs from us. So the apparent magnitude can be determined as:

$$m = 4.83 + 5(\log_{10}{1.3}-1) ≈ +0.4$$

As made clear in this paper, most humans can see objects with apparent magnitudes as dim as $5$ without using tools. So yes, it is certainly visible, and would be quite bright. It is between Procyon and Achernar, the 9th and 10th brightest star on Earth's night sky.

For a comparison, Proxima Centauri has an apparent magnitude of $+11.13$ from Earth's perspective. If we want to compare the two, we could use this formula:

$$v_b = 10^{0.4 x} = 10^{0.4×(11.13-0.4)} \approx 19588$$

So the Sun would appear almost 20,000 times brighter from Proxima Centuari than PC appears from the Sun.


Why do astronomers not use past tense when discussing observations of the universe

You're being a pedant and are not necessarily correct!

The present tense could be clearly defined as what is experienced at the time of observation. To write that it had a black hole 70 million years ago would (pedantically) only be correct at one instant of time, not correct when the reader read the sentence and even ignoring that pedantry, not necessarily true for gravitationally lensed light where the light has not travelled along a straight path!


Intuitive explanation for why the Doppler effect (and red-shift) happens?

Your intuition is correct - a moving source emitting wavefronts periodically will be closer to the previously emitted wave in the direction of motion, and farther from the previously emitted wave in the opposite direction - see the simulation here.
You are also correct that the size of the effect depends on the speed of the observer relative to the speed of the wave.  That is why we can easily experience the Doppler shift for sound in everyday life (e.g. the sound of a car when it is approaching vs. receding) but we never notice the Doppler shift for light.  The speed of sound is a lot slower, and thus a lot closer to the velocities of everyday life, so the effect is larger for sound.
While the shift for light is too small for our eyes to perceive (e.g. not enough to change an object’s perceived color), sensitive instruments can measure extremely small changes in wavelength.  For example, the spectrographs that astronomers use to make radial velocity measurements of stars for the purposes of detecting extrasolar planets (via the star’s motion as the planet orbits) can now routinely measure velocities with a precision of 1 m/s - a leisurely walking speed for a human!


How does gradual crossing over of the Roche limit transform a planet or moon?

A large moon (the size of a planet) would be modeled by the fluid model, meaning friction and tensile strength are too weak to modify the shape of the moon significantly, shape is determined by rotation, vortices, self-gravitation, and tidal forces.
Smaller moons, like Phobos would likely follow a rubble pile model.
On circular orbits tidal heating doesn't play a major role.

The fluid model has been studied extensively. The occuring shapes are known as equilibrium figures.
For a full understanding, the parent body should be considered together with the moons, as one system. (Fluid) moons getting too close to the Roche limit gradually deform (tidal heating depends on inner friction), and start to shed mass at some point. There exist solutions with vortices, and vortex-free solutions, depending on the properties of the fluid. The lost mass may either form an n-body system outside the Roche limit, or form a ring system, mostly inside the Roche limit.
Closer to the planet, it may either impact, and fuse with the planet to a the overall spheroid of the planet, or - if the planet is already rotating fast, and the mass of the moon was sufficiently high - increase the rotation of the planet, such that it may undergo e.g. the Maclaurin - Jacobi series, up to the point, that it cannot accrete more mass. There are several more possibilities in detail.

Rubble piles with tensile strength are more likely to undergo sudden shape changes or disruptions than rubble piles without tensile strength. Behaviour of rubble piles without tensile strength depends on friction. The less friction, the closer to the fluid model.
The more tensile strength of the rubble pile, the more it behaves like the ridgid model. Ridgid objects may survive well below the Roche limit for fluid objects, but disrupt suddenly when tidal forces get too strong.


Do we finally have a decent set of parameters of 162173 Ryugu?

A science paper from 2019 by K. Kitazato et al. shows a shape model of the surface in Fig. 2, so  we now indeed know the exact size and  the exact mass, even the surface temperature distribution:

The quoted paper refers to S. Watanabe et al., Science, 2019 which has all the details

Ryugu has an oblate body, with an equatorial radius of 502 ± 2 m and polar-to-equatorial axis ratio of 0.872 ± 0.007. The total volume obtained from the SPC-based shape model is $0.377 {\rm km}^3$, with an uncertainty of 1.3%. We conducted a gravity measurement during a spacecraft ballistic descent down to 0.85 km from the asteroid surface and a subsequent ballistic ascent up to 5.4 km. The estimated mass is $4.50\times 10^{11} {\rm kg}$, with an uncertainty of 1.3%, mainly because of uncertainties in the solar radiation pressure on the spacecraft.

The highlighting in the quote is from me.


Is it possible that in the Local Interstellar Cloud there are undetected stars?

It is unlikely, unless they are extremely faint. The reason is that the cloud obviously does not absorb enough light to keep us from detecting remote red and brown dwarfs, and hence a closer star would also be detectable.


Identification of Lunar mare (maria) from a taken picture

So your image looks more like this, with the red line being the terminator. The illuminated side in your image is the left side of the reference image, turned upside down.  I think you've gotten it backwards, the mare that you've labeled "1" is the Sea of Tranquility, and the one you've labeled "2" is the Sea of Serenity.


If the speed at which the universe contracts in reverse time decreases over time, then how did scientists arrive at a date for "The Big Bang"?

Under the assumptions of large-scale homogeneity and isotropy, cosmic evolution can be described by the scale factor $a(t)$ that describes by what ratio the large-scale structure of the universe has expanded or contracted. The scale factor is unitless, and by convention taken to be $1$ in the present epoch, and the rate of change of $a$ can be denoted $\dot{a}$, so the the universe is expanding whenever $\dot{a}>0$ and contracting whenever $\dot{a}<0$. The existence of a past singularity is then the claim that $a\to 0$ for some finite amount of time in the past.

The scale factor is related to the Hubble parameter by $H \equiv \dot{a}/a$. The Hubble parameter is very frequently and erroneously also called the Hubble constant, which is probably the source of your confusion. It is true that if $H$ is taken to be constant, then trivially $a = a_0e^{Ht}$ is the only possible solution, which only asymptotically tends to $0$ in the infinite past. If spatially flat, as our universe is, this is the de Sitter solution describing a universe dominated by a cosmological constant and negligible matter or radiation.

But that's simply not our universe. In the past, matter was dominant (and prior to that, radiation was dominant), and the de Sitter model simply isn't applicable. Simply put, the Hubble constant is not constant, and so should properly be called Hubble parameter.

In general, the evolution of the scale factor or the Hubble parameter is described by the Friedmann equations:
$$\begin{eqnarray*}
H^2\equiv\left(\frac{\dot{a}}{a}\right)^2 &=& \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}\text{,}\\
\dot{H}+H^2\equiv\frac{\ddot{a}}{a} &=& -\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)\text{,}
\end{eqnarray*}$$
where $\ddot{a}$ denotes the second time-derivative of $a$, the cosmological constant (dark energy) is included in the density $\rho$ and pressure $p$ of the universe, and for our spatially flat universe, $k=0$.

You may also wish to look up the deceleration parameter $q\equiv -\ddot{a}a/\dot{a}^2$ (so defined because the universe was once thought to be decelerating). But the main point is that $H$ is in fact not a constant, so it is consistent for the scale factor to become $0$ a finite amount of time in the past. How much time depends on the changing density $\rho$ and pressure $p$, which is studied in detail in the standard ΛCDM model.


No.

No such effect has been discovered. Gravity seems to be always attractive, never repulsive. Nor has a particle with negative mass ever been discovered. In particular, antimatter appears to have positive mass.

It is not theoretically impossible, but a particle with negative mass would be very odd, as the faster it moved, the less momentum and energy it would have.


Is there a closest free-return trajectory to a black hole?

The trajectory of a ballistic body, whether in Newtonian or Relativistic physics depends on the initial energy and angular momentum.
The difference is that in Newtonian physics, if the mass is compact enough, the infalling object (if given an initial kinetic energy) will never hit the central object, unless it is a direct hit, and will scatter off to infinity.
This can also happen in GR, but only if the specific angular momentum exceeds a threshold that increases with the initial energy of the particle, but can never be less $4GM/c$, and only as small as this for an object that falls from infinity with no initial kinetic energy. If the specific angular momentum is less than this then the infalling object will always end up in the black hole (for a freefalling trajectory).
For the limiting case, the closest approach to the black hole is $2r_s$, which is the local maximum of the effective potential.
However, the body can get closer if it has more angular momentum and more initial energy. If we allow the angular momentum and energy to get arbitrarily large then the body can get to $r=1.5 r_s$.
So I think this is the answer you are looking for. This is the value for a Schwarzschild black hole. For a spinning black hole, things are more complicated.
EDIT: Some details
The effective potential in the Schwarzschild metric is given by
$$ \frac{V_{\rm eff}}{mc^2} = -\frac{r_s}{2r} + \frac{L^2}{2m^2 r^2 c^2} - \frac{r_sL^2}{2m^2r^3c^2}\ , $$
where $r_s$ is the Schwarzschild radius and $L/m$ is the specific angular momentum of the free body.
The radial equation of motion in terms of the proper time $\tau$ of the body is
$$\frac{1}{2c^2}\left( \frac{dr}{d\tau}\right)^2 = \frac{1}{2}\left( \frac{E^2}{m^2 c^4} -1 \right) - \frac{V_{\rm eff}}{mc^2} \ .$$
To find the local maximum of the effective potential, differentiate with respect to $r$ and equate to zero. This gives you that the specific angular momentum at the maximum is
$$ \frac{L^2}{m^2} = \frac{c^2 r_s r^2}{2r - 3r_s}\ . $$
Allowing the energy of the body to be $E = \gamma mc^2$ (where $\gamma$ is the Lorentz factor) one can set $dr/d\tau = 0$ at the scattering minimum radius and then subsititute in the value for $L^2/m^2$ at the maximum of the effective potential. The rationale here is that the body will get to its closest point when it has an energy corresponding to the maximum in the effective potential. If it has any more energy than this it will just plunge into the black hole. Any less and it will scatter at a larger radius.
This condition gives you a quadratic equation in $r$ which can be solved in terms of $\gamma$. If $\gamma=1$ then the quadratic becomes a linear equation with $r=2r_s$. This corresponds to the case of a particle falling in from rest at infinity and meeting a local maximum of the potential of $V_{\rm eff}=0$. The body then scatters off to (just) infinity (i.e. the marginally bound case). This is illustrated below, where the horizontal purple line is a line corresponding to an energy $E/mc^2=1$. The peak at B is at $r=2r_s$ where $V_{\rm eff}=0$ (I've also shown the equivalent Newtonian potential for contrast).

If you allow $\gamma$ to be larger then the the viable solutions for the minimum $r$ lie in the bounds $1.5 r_s \leq r \leq 2r_s$. If $\gamma \rightarrow \infty$ then the scattering minimum radius $r \rightarrow 1.5r_s$, which also requires $L/m \rightarrow \infty$. The plot below shows the potential for $L/m = 10GM/c$, with a purple line representing a body with $E/mc^2=2$ which would scatter to infinity. The peak at B is just above $r=1.5r_s$.

If you want to play with this (my) Geogebra widget, you can find it at https://www.geogebra.org/m/u7y6huer


Do large telescopes, especially plans for the LSST, avoid saturation artifacts from the brightest stars and planets? If so, how?

Large surveys don't usually bother to do anything about bright stars saturating the images.  It is more trouble than it is worth to deal with, and these kinds of surveys are designed to study much fainter things.  The data from bright stars are just unusable.  If an astronomer really wants data on a bright star, they can go to a smaller telescope and/or take their own observations with filters/apertures to reduce the light from the star going through the optics.

Here you can see one of the many examples of bright stars saturating the detectors on the SDSS survey: http://skyserver.sdss.org/dr15/en/tools/chart/navi.aspx?ra=81.5049695271329&dec=-5.52153697792986


Angle formed between a satellite, a ground station, and then another Satellite

This figure from
Wikipedia:Spherical law of cosines
applies if the ground station is at the center of the sphere,
u is the ground station's zenith, and v and w are the apparent positions of the satellites as seen from the ground station.

Then a and b are the satellites' zenith angles, i.e. 90° minus their elevation angles.
C is the difference between the satellites' azimuths, and c is the angle you want.

Using the cosine rule for sides
$$\cos c = \cos a \cos b  + \sin a  \sin b  \cos C$$
and the rules for sines and cosines of complementary angles,
the expression in the question looks correct if the absolute value operations are omitted.


The loss of the Martian atmosphere can be mostly attributed to its mass. The reason why Earth still has an atmosphere made of lighter elements is because with larger mass comes larger escape velocity, which is the speed at which an atom's kinetic energy overcomes the gravitational potential energy of its planet.

The distribution of speeds of most gasses can be described by the Maxwell–Boltzmann distribution.

This curve represents the probability of finding a particle with a given speed at a given temperature. Holding temperature constant, the above chart illustrates that lighter molecules will have higher probabilities of being found with higher speeds.
$$P(\epsilon) = \frac{2}{\sqrt{\pi}k_{B}T} \left( \frac{\epsilon}{k_{B}T}  \right)^{1/2} \exp{-\left(\frac{\epsilon}{k_{B}T}\right)} $$
where $\epsilon$ is the energy of the particle, $\frac{p^{2}}{2m}$ in the absence of a gravitational field (see this lecture for information regarding how to include the gravitational term).

Integrated over time, the lighter gasses tend to exceed the escape velocity more often than their heavier counterparts. This is why the larger planets like Jupiter and Saturn still have atmospheres dominated by hydrogen and helium.


How long does it take the center of the Earth to complete a full orbit?

It's not exactly one year because of the 28/29 February thing.

The exact period is 365 days 6 hours 9 minutes 9 seconds [Random House Encyclopedia. New York: Random House: 174.]


What is the relative time difference between us and a star system in outer layer of our galaxy?

Your third question has a very well known answer here:

What means, that although a big time difference had to exist, there is a much smaller one, and in the opposite direction. This is considered the effect of the dark matter.

Our Sun is around 30000 ly from the galactic center, the edge of the galaxy is around 50000 ly, so we can read on the diagram, that the actual speed difference is around 15 km/s, which is 1/20000 of the speed of light. This causes a (not really big) time dilation due to the special relativity.

There is another source of the time dilation, which is caused by the gravity of the galaxy, which differs in the case of us and in the case of somebody living on the outer edge. On such weak gravitation (i.e. for from any black hole, neutron star, etc) it practically depends on the newtonian gravitational acceleration.

But this acceleration is very small - it takes a half galactic year (some hundred millions of year) to revert the orbital speed of the Sun around the galactic core! So, we can consider that actually negligible.

Thus only the special relativistic effect remains. 15 km/s is around 1/20000 of the speed of the light. In cases of speed which are much smaller as the light speed, we can use the speed dilation formula 1/(2*(v/c)^2). Substituting 1/20000 into this formula, we get 1:800 000 000 .

For laymans, we can say, it takes around 25 years for the clocks of this people living on the outer edge of our galaxy to dilate 1 second.


I refer to the book : Astronomical Algorithms (Second Edition) by Jean Meeus.

To calculate the values of position angle of the bright limb and position angle you will need to calculate  the apparent RA and DEC of the Sun and the Moon.

Code the equations for the Sun RA and DEC using Chapter 25. Please note that L0 (equ. 28.2)  is in error To calculate T (Julian century you will need to compute Julian day (Chapter 7).

Code equations for the Moon RA and Dec from the equations given in Chapter 47.

For the calculation of Apparent RA and Dec you will need to compute Nutation and Obliquity - Equations in Chapter 22.

BTW. For all my calculations I have "standardized" on using the equations on page 144 for the variables : D, M, Mdash, F, and Omega.
Fyi compare the equations for M given in this book ie. equations 25.2 and 47.3

Chapter 48 provides a formula (48.1) for the illuminated fraction of the moons disc.

5.1 I used 48.2 and 48.3
Addition:
5.2 To use equation 48.5 you will need to obtain or write the atan code yourself to resolve the angle X in the correct quadrant - I wrote a routine: ATAN2(sin_tl,cos_bl).
I have coded all the above to run on my HP Prime graphing calculator and they have proved to be very, very accurate.
For the purpose of accurate determination of the moon phase angle you will need to calculate UTC(date and time) using your zone longitude or better still your local longitude.


How do we know that our galaxy is a spiral galaxy?

To elaborate on Conrad Turner's excellent list:

The disk structure is even more easily seen in the infrared, where dust extinction is much less than in the optical. See for instance this image from the 2MASS infrared survey:

Stars are not really (as Milan van Dijck suggests) used for mapping the spiral arms. Rather the gas clouds of gas in between stars are used, more specifically neutral hydrogen observed through its  21 cm observation. The figure below (from Oort 1959) shows the resulting map. As you can see, its not really as well-defined as seen in many artist's-conception pictures.

Neutral hydrogen is fuel for stars. Massive stars ionize the gas around them, and since star formation mainly takes place in the spiral arms, the gas clouds are accompanied by pockets of ionized hydrogen, as seen in this map from Georgelin & Georgelin (1976):

The physical properties of the Milky Way (stellar population, gas fraction, color, kinematics, etc.) are also similar to other spirals.


Is our universe included inside a black hole?

There's a lot to pick apart in everything you try to propose, as it includes a lot of far fetched (or at least rather non-standard) claims.  I am frankly not up to attempting to address every one of them, if for no other reason than that it makes the question as a whole rather too broad for my taste (and perhaps more in the territory of Physics.SE, which has quite a lot of answered questions concerning black holes).

There is, however, the following simple and amusing observation: current estimates of the mass-energy of the observable universe tell us that it is too dense to be a black hole.  That might sound a little weird if you're not familiar with black holes.  In fact, the density of a (non-rotating, Schwarzchild) blackhole is inversely proportional to the square of its mass, and the radius is directly proportional to the mass.  More explicitly:
$$r=\frac{2 G M}{c^2},$$
$$\rho(M) = \frac{3 c^6}{32 \pi G^3}\cdot \frac{1}{M^2},$$
where $r$ is the radius, $\rho$ is the density, $M$ is the mass, $c$ is the speed of light, and $G$ is the gravity constant.

The order of magnitude estimate for $M$ is $10^{54}$ (and $M\geq 10^{54}$ in particular), which makes the universe ~3 times too small at least.

A key fact here is that the universe is not static with respect to itself. See this Physics.SE Q&A in particular.  I'll quote the end of Lubos Motl's answer, in particular:

Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the exterior of a normal black hole, and the analogy of the interior of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.

In other words, there are lots of important and sometimes subtle issues with the whole "the universe is a black hole" concept.


The precession of the Earth's spin axis orientation takes place on a 25,700 year cycle. But this isn't connected with the Sun's movement through space.

The Sun orbits the Galaxy about once every 220-240 million years. It also oscillates about the Galactic plane with a period of about 70 million years.


What would happen if a star passed through the barycenter of two binary stars?

If we assume the binary pair have equal masses, then by symmetry a star passing through their barycenter at right angles to their rotation plane will not change their symmetry.

I did some simple simulations where a star with mass 0.1 flies through a binary in an originally circular orbit.

The first case is with an interloper moving at speed 20. It passes through, losing a bit of momentum to the binary, that now moves slowly in the same direction.

The second is an interloper moving more slowly, with speed 1. Now the binary moves along more, and the interaction causes a quasiperiodic oscillation in seimajor axis.

Just dropping the interloper with zero velocity makes it bound to the binary, oscillating back and forth, driving complex quasiperiodic or chaotic motion.

It is possible to make the pair scatter by throwing in a sufficiently heavy interloper with enough momentum that first drags the binaries close, transfers enough momentum to unbind the system, and then passes along.

Of course, if the direction is not exactly symmetric things can turn messy. Note that there are non-chaotic solutions too.

To sum up, interlopers can deposit energy and momentum, split binaries, or create chaotic tangles. But it is not the barycenter that is the important part.


Does the mass of the observable universe ever change?

Even if you're only referring the "ordinary" matter (such as stars, gas, and bicycles) and dark matter, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we can see stuff that today is too far away for the light to have reached us, so its radius has increased. Since the mass $M$ equals density $\rho_\mathrm{M}$ times volume $V$, $M$ increases.

As called2voyage mentions, we have several ways of measuring the density, and we know it's close to $3\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$. The radius is $R = 4.6\times10^{28}\,\mathrm{cm}$, so the mass is
$$M = \rho_\mathrm{M} \frac{4\pi}{3}R^3 \simeq 10^{57}\,\mathrm{g},$$
or $5\times10^{23}M_\odot$ (Solar masses).

However, another factor contributes to the mass increase, namely the so-called dark energy, which is a form of energy attributed to empty space. Since the Universe expands, dark energy is being created all the time.


Well the spin is easy. Your system has zero angular momentum, so the spin will be zero.

I think there will also be no gravitational radiation. for the slightly technical reason that gravitational waves only come from changes to the quadrupole moment of a system and in your system the quadrupole moment is always zero.

So everything ends up in the final black hole which will have mass about 4.6 billion solar masses, coming from the mass of the two colliding holes, and their kinetic energy.

I'm not sure exactly how long the merger would last, but it will be comparable to the Schwarzschild radii involved, divided by $c$, so around three hours (as seen by a distant observer).


Would Pluto keep an orbit without its moon?

The orbit of a planet does not in any way depend in its moons, mass or own gravity. The orbit is the same for a grain of dust as for a giant planet. It only depends on the mass on the Sun and the distance to the Sun. The eccentricity of the orbit is then given by the initial angular momentum of the orbiting object. This might explain it a bit better.

This was a really BIG breakthrough in physics, actually the foundation of physics, about 400 years ago. Big because it is irrefutable although it is counter-intuitive. It is really difficult to believe unless one tries to accept the actual observations and logic.


Why does the angular size of the Moon change?

The moon's orbit is elliptical, and it moves faster when it is closest to the Earth.  This is the case for all elliptical orbits, and was discovered by Kepler, who gave a neat rule for predicting how fast an orbiting body moves. (https://kids.britannica.com/students/assembly/view/90830)

For the moon, this means that when it is largest, it is also moving fastest. So the peaks in the graph (where the moon is large) are narrow, as the moon quickly moves away from periapse.  When the moon is smallest, it is moving slowly, so the troughs in the graph are wide. It changes size slowly when it is far from the Earth.
The bumps are not all the same size because the orbit of the moon is perturbed by the sun.  This changes the amount that the moon changes size. https://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html
When the eccentricity is low, there is less difference in size.  When it is large there is a big difference.
This is the cause of "supermoons", when the full moon occurs at the same time as periapse.  When the eccentricity is at a maximum, the supermoon will appear especially large and bright.

The Moon's orbit is one of the most complex orbits in the solar system!


How do we know how fast the surface of a neutron star is rotating?

How do we know how fast surface of neutron star is rotating?

I suppose we don't know the answer with total confidence, but I'd say we think we know with good confidence. Because we have robust evidence along with a model that matches that evidence.

  I understand that we can count pulses in time, but how do we know that magnetosphere is rotating along with surface of neutron star?

I suppose we don't actually know for sure. But I can't think of any process whereby a stationary spherical body can emit regular pulses of radiation.

  Can we measure rotation of a neutron star in any other way like red-blue shifts? It should be pretty difficult due to small radius and big distance but is it possible? Is it already done?

No, and no. At least as far as I know. But most of what I know came from Frozen Star by George Greenstein. The first half of the book is all about neutron stars. It dates from 1984 and it could be a little out of date. But as far as I know there's not much wrong with it, and it's a good read regardless. Note that you can buy a second hand copy fairly cheaply.


Can expansion of space cause the motion of galaxies, does expansion of space do effect motion of our sun around center and earth around Sun?

No. The solar system is a gravitationally bound system so it is unaffected by the large scale expansion of the universe - see this Wikipedia article for further explanation. Similarly, galaxies like the Milky Way, and even groups of galaxies such as the Local Group, which includes the Milky Way and Andromeda, are gravitationally bound systems and are also unaffected.


Which camera body is best introduction for beginner amateur astrophotographer?

When I first started out, I used my Olympus pen e-pl1, so you're doing great with the e-pl3. It takes higher resolution photos (330 vs 300 dpi) than my wife's Nikon D5300, so your camera should work just fine. I still whip this one out every now and again just because of its simplicity.

The one on the right is the adapter that I use regularly. I mounts up easily to each of my telescopes. Works great and has held up nicely the last (almost) decade.

To avoid the shake, set as long of a timer as you can! I'm not sure if the newer pen cameras have the option for the tethered remote, but mine definitely does not.

As for the software, I wrote some python codes that I use for image processing, and DS9 is the interface that I use regularly, however good old photoshop (pre-everythingisstupidandcloudbased) really works wonders if you're on a budget and have or "find" a copy laying about.

p.s. the adapter is from telescopeadapters.com


Theoretically, if you're talking crater to planet size, I (think) a crater that covers half or a bit more than half of the planet is possible if you have a dense planet (Say, Mercury) and a lighter, more icy object (say, Titan).   Mercury would break Titan apart like a hammer hitting glass (er, maybe), but Titan, which is slightly larger than Mercury but weights only 41% as much, with a lower velocity impact, might not make the entire surface moulton and could leave an impact crater over 50% of the surface with debris scattered unevenly leaving something like an impact crater.   That's, granted, pretty speculative, but I think that type of impact, a rocky world and an icy one of similar size might provide a collision that could create an impact crater a bit larger than 50% of the planet's surface.   Such an impact would be interesting to observe, but . . . I probably won't see one in my life time, unless they get some very good telescopes out there and get lucky.

Theoretically if you count surface melt, it's arguably possible to have an impact that would melt 75% or more of the surface but leave 25% or less intact, but I don't consider surface melt the same thing as an impact crater, so I wouldn't count that, but that's optional.

As far as largest known craters, if we measure crater diameter against planet or moon radius (I'm going to ignore depth cause depth is easier on lower gravity bodies), but just diameter to radius.

Hershal on Mimas (the death star crater)  (Props to Andy)

Diameter of Hershal, 139 km.    Radius of Mimas, 198 km.   Ratio, 70%.

But there's an even bigger one.

The South Pole–Aitken basin on the moon.

Article here

It went undetected until relatively recently because it's large enough that it no longer looks like a crater, but it is, and it affected the Moon's density where one half of the moon has a measurably thicker crust than the other half. (The crater half, I presume, the crust is thinner).   This makes orbits around the Moon weird cause the moon is gravitationally quite lumpy.  (Proximity to the Earth also makes orbits around the Moon weird, but that's another story.

This ancient crater has a diameter of about 2,500 km compared to the radius of the moon of 1,737 km a ratio of about 144%, over twice the relative diameter and 4 times the relative size relative to the crater on Mimas.

and Mars might have an even bigger one.   See here.  It's size isn't listed, but Mars (like the Moon) has a "split personality" where one side has a measurably thinner crust than the other side and one possible, perhaps likely cause of that is a giant impact.   This "Crater" if we want to call it a crater covers about 1/2 the planet.   This is different than the theorized Earth-Theia impact, which liquefied the surface and left no visible crater, but it may have liquefied half the surface and not left a very visible impact.     One problem with measuring craters as they get this large is they look less like craters.


How do we know that the Universe is still expanding now?

You're right that in principle we cannot know if the distant galaxies are still receding from us. When, however, we think that they indeed are, it hinges on the cosmological principle, viz. that the Universe is homogeneous and isotropic, i.e. "looks the same everywhere and in all directions". This is a philosophically motivated statement, which is continuously tested observationally. So far, there has been no significant experimental evidence that this is not correct.

When we observe the velocities of galaxies throughout the Universe, we see a smooth velocity as a function of time, or distance, with some intrinsic scatter which is due to the gaalxies' peculiar motion, i.e. their velocities through space, which is usually of the order of 100–1000 km/s. That is, distant galaxies move away from us fast, and nearby galaxies move away from us slowly. As you say, the expansion rate decreased with time, but only until a certain point, after which it started increasing again (due to dark energy).

The fact that we see that nearby galaxies are still receding, together with the cosmological principle, implies that the distant galaxies should also still be receding. We know of no reason that they shouldn't. Of course this is no proof, but in physics, there are no proofs, only verifications and falsifications. To postulate that the distant galaxies no longer recede, and to be taken seriously, you would need to come up with an experiment that could verify this hypothesis, preferably along with a mechanism that would be able to cause this phenomenon (this looks a bit condescending when I type it; this is not my intention).

The expansion rate is currently — and hence locally — around $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. Due to the aforementioned velocity scatter, we cannot probe expansion below ~1 Mpc, but at this scale, space doesn't really expand anyway due to the mutual gravitational attraction of the Local Group. This means that in principle, the expansion of the Universe could have come to a halt within the last few million years, and still be in accord with the cosmological priciple. However, this would mean that the evolution of expansion velocity as a function of time would have a sharp kink. Again, this is not impossible, but we know of no physical mechanism which would be able to cause this.


What is the density of the solar system per cubic light second outside the Jupiter ecliptic?

I assume you want density expressed in units of average mass in kg per cubic light second.
An ecliptic plane is infinitely thin, so we might assume this is irrelevant to the question.  Perhaps you meant "not including matter within a Jupiter radii of the Jupiter ecliptic," but this isn't very clear from the question.
The Solar System is defined as all objects gravitationally bound to the Sun, so it doesn't have a volume per se.  For example, when Oumuamua cruised by the Sun, it was never part of the Solar System because it always had greater than escape velocity.  But the interplanetary medium is not a perfect vacuum, but is rather about 5 particles per cubic centimeter. Even the interstellar medium isn't quite a vacuum.
If we assume the edge of the Solar System is at the edge of the Oort cloud at 50,000 AU, which is given as an answer to Where does the Solar System end? Then the volume of the Solar System is $4/3\pi r^3$ or $5.24\times 10^{14}$ cubic AU, or $6.5\times 10^{22}$ cubic light seconds.  The mass of the Solar System is well approximated by the mass of the Sun, or about $2\times 10^{30}$ kg.  So the Solar System average density is about $3.07\times 10^{7}$ kg per cubic light second.


The SM developers seem to have anticipated such issues.
In your nm output sample,
T sm_ylabel is the native C function, and
T sm_ylabel_ is its wrapper to be called from Fortran.
gfortran usually appends single underscores to external symbol names.
If the MOOG object files were in fact built with the gfortran options in the question, the symbol references should match the wrappers in the SM library.
The double underscore suffixes may be left over from a prior build attempt with -ff2c.
When you change compiler options, remember to make clean before rebuilding.
If that doesn't lead to a solution, and nothing else is using SM,
the SM manual suggests
that you could rebuild it with this in its options.h:
#define FORTRAN_APPEND __


The most feasible hypothesis about Theia

According to new simulations, a better model seems to be one where the Earth had a magma ocean at the time of impact, see here:

We perform density-independent smoothed particle hydrodynamic collision simulations with an equation of state appropriate for molten silicates. These calculations demonstrate that, because of the large difference in shock heating between silicate melts and solids (rocks), a substantial fraction of the ejected, Moon-forming material is derived from the magma ocean, even in a highly oblique collision. We show that this model reconciles the compositional similarities and differences between the Moon and Earth while satisfying the angular momentum constraint.


Throwing a baseball from orbit to the earth

Orbits are all ellipses, so when you throw a baseball you change the orbit of the base ball from your circular ellipse to another ellipse.
If you throw it prograde (in the velocity direction), then you baseball will have an elliptical orbit. It will be closest to the Earth at the point at which you threw it, and furthest from the Earth (and higher than the circular orbit) at 180 degrees. So throwing prograde will never make the baseball deorbit.
The speed of the baseball will now not be constant.  It will be faster than the circular orbit speed at the lowest point in the orbit, but slower than the circular orbit speed at the highest point.
This is your misconception: Adding velocity doesn't make the ball to a lower orbit.  Instead, adding velocity makes the ball go to a higher orbit, which in turn reduces the speed of the ball.
Throwing retrograde (wake direction) has the opposite effect. The baseball will enter an elliptical orbit that is close to the Earth at 180 degree from the point at which you threw and is, on average at a higher speed.  If you throw retrograde hard enough you can reduce the height at 180 degrees to less than 0, ie the baseball hits the ground.  But you have to throw far far harder than is physically possible, because you and the baseball start out at orbital speeds, about 7km/s so throwing at 10m/s won't be enough to deorbit.
Throwing at the nadir would also just change the baseball's orbit into an elliptical orbit. The baseball would initially approach the Earth, but as it did it would gain speed and then move away from the Earth in an ellipse.  The closest point would be at about 90 degrees ahead of the throwing point. This is actually not a very effective way of deorbiting. It takes a lot harder throw to reduce the height by the same amount that throwing retrograde would. As with the other throws, the ball will pass through the throwing point after going all around the Earth.  But you won't be at the same point again because your orbital period will be different to that of the ball.
Throwing anti-nadir has a similar effect, except now the furthest point is at about 90 degrees, and the ball falls to the closest point at 270 degrees.  It is equally ineffective at de-orbiting or gaining height. This is why nearly all spacecraft manoeuvres happen either in the prograde direction (to add height) or retrograde (to reduce height)
"Bouncing off the atmosphere" is a practical problem not directly connected to orbits. If you enter the atmosphere at the wrong angle, you can actually start to fly and so gain altitude. That's not really connected to the orbital manoeuvres here


I like sharing the following quote from Wikipedia on the Star of Bethlehem

Astronomers have made several attempts to link the star to unusual celestial events, such as a conjunction of Jupiter and Saturn or Jupiter and Venus, a comet, or a supernova.
Some modern scholars do not consider the story to be describing a historical event but a pious fiction created by the author of the Gospel of Matthew.

This said, luckily you specify that you are after a conjunction of Jupiter and Venus. WolframAlpha can determine the conjunction of Venus and Jupiter closest to the year 0:

8:07:26 pm LMT  |  Thursday, October 14, 2 BC (extrapolated Gregorian calendar)

I have never used Stellarium, but other software, where entering that date and searching for Jupiter did the job.


Why doesn't the night sky keep on getting brighter and brighter every day?

This assumes that the stars formed at the same time as the Earth.  This is not the case.  Stars had been forming for billions of years since the Earth formed.
The Earth didn't form in a single day, but for convenience let's pick one Early day in the Earth's formation as "day 0".
On day 0 light from many stars that had been travelling for years reaches Earth. The night sky is full of stars.
On day 1 light from those same stars that was emitted one day later reaches Earth, the sky brightness doesn't change.
On day 2 the same and so on.
Over time some stars will have moved nearer, some will have moved away, some stars will have reached the end of their life and some will have been born. But, on average there is no change.


Is it possible that a satellite creates a forever eclipse?

The only potentially valid way to cause a long-lasting eclipse that I can think of is to have your moon orbit at the planet's L1 Lagrange point between the planet and the star (and the moon has to be the right size). If it had just the right orbit and was the right size, it could then cause a long-lasting eclipse. Note I'm not saying permanent because such a system wouldn't be permanent. I'm not going to run through the math$^1$ to determine if such a system is stable or for how long. I'd guess it may be stable for a few orbits at least.

The main problem is getting your moon there in the first place. One could conceivable say that by some freak chance a third fairly massive body came close to your planet/moon (either from outside the solar system or near the edge) and gravitationally perturbed your Moon's orbit such that it ended up in the L1 point. But the chances of such an event are so slim that it would be incredulous if it actually happened.

And just to add a technicality, once your moon moved to the L1 point, it would now be orbiting the Sun and wouldn't be a moon anymore according to the definition of a moon.

$^1$ Out of curiosity, I ran through some of the math very loosely and found that if you put the Moon in Earth's L1 position, the Moon survives for about 3-5 months before perturbations knock it out of the L1 position. Of course our Moon isn't the right size to cause an eclipse at that distance, but this gives a very rough estimate for how "permanent" the scenario I described above really is.