Why can't we see a solar eclipse every month?

The Moon orbits in a plane that is inclined with respect to the plane in which the Earth orbits. The intersection between these two planes is a straight line. For an eclipse to occur, this straight line needs to be roughly aligned with the line that goes between the Earth and the Sun.

From this blog post

As the Earth goes around the Sun, the plane in which the Moon's orbit keeps the same inclination compared to the stars (well actually, not quite, because of the precession of the orbial plane, but this gives you the general idea), so it turns with respect to the Earth-Sun system. This is why eclipses occur (roughly) every six months, and not every single month.

The frequency with which you can actually see an eclipse from a given location will be even less than once every six months. This is because to see a lunar eclipse, you need to be on the night side of the Earth while the eclipse is happening. And to see a solar eclipse, you need to be under the track of the Moon's shadow. This is a band a only few hundred kilometers wide. Here are the tracks of a few past eclipses.


Correlation between stellar mass and galactocentric distance

In terms of the distribution of stellar masses when stars form -- the Initial Mass Function (IMF) -- there doesn't seem to be much variation, at least for spirals. (There is evidence that massive elliptical galaxies may have different IMFs as a function of radius, with relatively more low-mass stars formed near their centers. The cause of this is unclear; it might have to do with different conditions for star formation, with very dense, high-pressure gas in what would become the centers of massive elliptical galaxies possibly producing different distributions of stellar masses.)

The current distribution of stellar masses at a given radius in a spiral galaxy will depend almost entirely on the local star-formation history. Since stellar lifetime depends on stellar mass, regions with recent star formation will have more stars that are massive; regions where star formation ceased a long time ago will be dominated by low-mass stars, since all the higher-mass stars there have died.

In most spiral galaxies, the mean age of stars tends to decrease as you move further out in radius (as indicated by optical colors: reddish central regions are mostly old populations dominated by red giants, while the disk further out is blue because there are younger populations dominated by hot, massive stars). This is an example of what's called "inside-out" star formation.

However, a lot of lower-mass spirals seem to show a reversal of this trend at large radii, with redder (older, mostly lower-mass) stars beyond a certain radius. Since this is the same radius at which the density of stars in the disk stars falling off more rapidly (the "break" or "truncation" radius), it's usually thought this is the result of a combination of two effects: 1) A radial threshold/cutoff in star formation, with very few stars being formed beyond a certain radius (possibly because the gas density becomes too low); and 2) Radial scattering of stars over time by spiral arms, which means that older stars tend to be scattered to larger (or smaller) radii simply because they've had more encounters with spiral arms. In this case, the relative fraction of more massive stars would increase with radius out to the truncation radius, and then decrease beyond.


Exactly how elliptical is Mercury's orbit, visually, without exaggeration?

Some things we know about Mercury's orbit:

Semi-major axis: 0.387 AU, about 57.9 million km
Eccentricity: 0.205

We can calculate the semi-minor axis, $b$ from the equation
$$e=\sqrt{1-\frac{b^2}{a^2}}\to b=0.379\text{ AU}$$
We can also calculate the distance to the focus from the center of the ellipse, $f$, as
$$f=ae=0.078\text{ AU}$$
I used Mathematica's ParametricPlot to plot Mercury's orbit using these parameters, along with Earth's. Earth's semi-major axis is 1 AU, and with an eccentricity of 0.016, its semi-minor axis is about 1 to three significant figures; I've treated it as a circle. Mathematica gives

ParametricPlot[{{0.387*Sin[x] - 0.0793, 0.379*Cos[x]}, {Sin[x],
  Cos[x]}}, {x, 0, 2*Pi}, AxesLabel -> {AU, AU}, PlotLegends ->
  Placed[LineLegend[{Red, Blue}, {"Mercury", "Earth"}], {1, 0.25}],
  PlotStyle -> {Red, Blue}]

I'd consider the deviation from a perfect circle to be noticeable insofar as the focus is shifted from the point $(0, 0)$, but it's not incredible.


If oxygen were pumped into space where would it go?

Gases tries to occupy all the volume of the container in which they are encased in equilibrated pressure. This happens because the gas molecules are always colliding and kicking each other in a 3D high velocity brownian motion. The result is that the gas molecules occupies all the container's volume homogeneously.

So, if the sealed container is opened up in space, the oxygen would quickly leak out, because the molecules near the vacuum edge would be instantaneosuly kicked out by the gas innermost molecules. This would happen rather quickly and in a chained manner, but is dependent on the size of the open hole. Macroscopically, the result would be that all the gas would be running out in a form of a wind/jet out of the open hole into outer space. This would only stop when all the molecules becomes too far apart from each other to have any further intermolecular interaction.

This means that the oxygen (or any other gas) would be dissipated into outer space. The gas molecules would initially be spread out by the intermolecular collisions and then would go outwardly due to the momentum acquired from each individual gas molecule.

When freed in outer space, each molecule would eventually reach something else to interact, so they could be attracted and retained by some gravity well (most likely from a planet or moon), or interact with some magnetosphere (if near a planet) or interact with some other molecule wandering around, or with a cosmic ray or be catched by the solar wind particles (or any of the equivalent if this happens near some other star).

If this happens in intergalactic space where there is no body with appreciable gravity or magnetic field nearby, no stellar wind and the cosmic rays are very sparse and rare and the very few that exists are randomly directed, then the molecules would just spread out in virtually linear trajectories and travel the intergalactic space lonely, quietly and virtually undisturbed for some billions or trillions of years. In their journey, each molecule would only very occasionally interact with some other molecule (likely to be hydrogen) or some other lost particle wandering around.


Did Gaia actually generate complete light curves for 212 Cepheids in other galaxies?

Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations?

No, the more I read about it, the more difficult it seems to be. In principle, if you sample regularly and often, you should eventually get a light curve, unless the period is the same as or a multiple of your sampling period.

It's not easy to find how often GAIA observes a given patch of sky, but here it says that after one year of observing

  Gaia has made an average of roughly 14 measurements of each star on the sky thus far, [...]

so they seem to have repeat observations a bit more often than once a month. Cepheids have a period of typically 2 to 45 days (says german Wikipedia - didn't find it in the english article). If your sampling is regular, you should be able to get a decent light curve for at least some Cepheids.

Complete guesswork on my part: Apparently, longer period Cepheids are redder at the minimum. Perhaps they use that to distinguish between the actual period and higher harmonics?

  And could someone help me understand - what is a TGAS parallax?

In the introduction it says

  Tycho-Gaia Astrometric Solution (hereafter TGAS; Michalik et al. 2015)

From the abstract of that paper, it seems to describe their method (they practised with mock GAIA data). Seems like they use the previous Tycho catalogue to improve astrometry and parallax measurements from GAIA data (they mention using Hipparcos parallaxes as a consistency check).

So, figure 1 seems to be Cepheid distance vs. Parallax distance.


1MW of transmitted power is very modest by RADAR terms, to give one example. Thus, you probably need to establish what exactly your professor means by "brightest" (may it be a specific wavelength or point in time, or some other definition). Otherwise, there's no reason to believe that Arecibo transmission was any special or outstanding from radio engineering stand point, while there were such nice toys operating at the very same time:

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


What is the orbital period of our sun around its barycenter?

As you can see from the diagrams below of the motion of the Solar System's barycenter relative to the Sun (courtesy of Wikipedia), the Sun's motion with respect to the system's barycenter is not a simple closed curve, and it doesn't have a simple period.

(Click the above image for a SVG version).

It is primarily dominated by Jupiter's period (11.862 years) because Jupiter has most of the non-solar mass of the Solar System. (On a related note, Jupiter has most of the angular momentum of the solar system). Jupiter's mass is approximately 0.0009543 solar masses, so the Sun is almost 1048 times more massive than Jupiter.
According to Wikipedia

To calculate the actual motion of the Sun, you only need to consider
the motions of the four giant planets (Jupiter, Saturn, Uranus,
Neptune). The contributions of all other planets, dwarf planets, etc.
are negligible.

Here is a Sage / Python script which displays an interactive 3D plot of the above diagrams, using data from
the JPL Horizons on-line solar system data and ephemeris computation service. Horizons can produce data covering a wide range of dates, please see the Horizons documentation for details.


Which structure is represented in this video (if any)?

I think it is supposed to be a representation of the Centaurus A/M83 Galaxy group.

https://en.m.wikipedia.org/wiki/Centaurus_A/M83_Group

These are a bunch of galaxies between 12 million and 15 million light years away.

The big central dust lane in Cen A is very recognisable.

I guess the sequence then continues through the Virgo cluster and supercluster which are roughly in the same direction.


Calculating star radius from temperature and absolute magnitude - what am I missing?

The page you quote is guilty of a gross simplification (or error). The luminosity ratio cannot be estimated from the difference in visual magnitudes for stars of different spectral type. You need to use the difference in bolometric magnitude, which indicates the luminosity integrated over all wavelengths. This will involve making bolometric corrections to the visual absolute magnitudes.

A star like Proxima Cen radiates much of its light outside the visible band and has a much different and larger bolometric correction than the Sun or Sirius. This accounts for your problem.

Example:

The correct calculation uses a table of bolometric corrections vs temperature. This rapidly changes for M-dwarfs in the visual band (you would be more precise using K magnitude), but is about 2.6 mag for a star at 3100 K (see Zombeck 1992). That is, the bolometric magnitude is smaller than $M_V$ by 2.6 mag. The equivalent correction for the Sun is only 0.1 mag.
$$ M_{\rm bol}= M_V - BC$$

Thus there is a 2.5 mag less difference in the bolometric magnitudes in your final equation than the difference in absolute visual magnitudes, and I get a radius for Barnard's star of $0.23\ R_{\odot}$.

The probable reason for the discrepancy between this and the radius quoted in wikipedia (actually it is the radius determined by an interferometric angular diameter measurement plus the parallax by Lane et al. 2001; a more recent measurement by Boyajian et al. 2012 gives $0.187\pm 0.001\ R_{\odot}$) is that (a) the assumed temperature is too low or (b) that the bolometric correction scale for V-band magnitudes is not very well defined for very cool stars. Both (a) and (b) are possible, especially as Barnard' star probably has a significantly subsolar metallicity.


What part of the EM spectrum was used in the black hole image?

According to the EHT website The observations were done using radio telescopes observing at a wavelength of 1.3mm. Visible light or infrared from the accretion disk would be nearly impossible to detect at such large distances. The radio measurements from all the synchronized telescopes were then combined and converted into an image.


What is the naming convention for newly discovered objects?

The International Astronomical Union's Naming Astronomical Objects webpage is a comprehensive guide to how astronomical objects are named.

For example, for "Specifications concerning designations for astronomical radiation sources outside the solar system", the IAU states that the designation (or sequence of letters and numbers) follow a general form:

  Acronym  ^ Sequence ^ (Specifier)

Where the Acronym

  is a code (i.e., alphanumerical string of characters) that specifies the catalog or collection of sources.

The sequence

  The sequence (or numbering) is an alphanumeric string of characters, normally only numerical, that uniquely determines the source within a catalog or collection. It may be a sequence number within a catalog (e. g., HD^224801), a combination of fields, or it may be based on coordinates.

and finally, the Specifier

  The specifier is optional and allows one to indicate other source parameters. However, they are not required syntax and are enclosed in parentheses.


Why is sodium such a common ion for in ion tails?

From http://science.nasa.gov/science-news/science-at-nasa/2000/ast26oct_1/:

  "When a Leonid meteoroid hits the Moon it vaporizes some dust and
  rock," explains Jody Wilson of the Boston University Imaging Science
  Team. "Some of those vapors will contain sodium (a constituent of Moon
  rocks) which does a good job scattering sunlight. If any of the impact
  vapors drift over the lunar limb, we may be able to see them by means
  of resonant scattering. They will glow like a faint low-pressure
  sodium street lamp."

Also, relative to other elements, sodium has a low ionization energy.


There are two things to discuss here: (a) why the Sun does not explode; and (b) why the Sun will not explode.

(a) An explosion occurs when the timescale for the energy release by some process is much shorter than the timescale on which a system can adjust to damp the energy release process. In the present day Sun, nuclear fusion is a very slow process: on average it takes many billion years for a proton to fuse with another. This timescale is quite temperature dependent, so you might have thought the centre of the Sun might heat up quickly, leading to a runaway "explosion". However, an increase in temperature leads to an increase in pressure that would expand the Sun, reducing the core density and temperature and decreasing the rate of nuclear fusion again. The timescale for the Sun to react in this way is just millions of years, so this acts like a thermostat that keeps the reactions under control.

(b) Stars more massive than the Sun burn through their hydrogen and other heavier fuels, and end up with an inert iron core from which no further energy can be extracted. The core subsequently collapses and the gravitational potential energy released by the core collapse powers a supernova explosion. The reason that the core does not quietly collapse into a black hole (at least in the supernovae we see!) is that the core "bounces" when neutrons in the core are squeezed very close together. This provides a quantum mechanical effect called degeneracy pressure that resists a complete collapse.

Ironically, it is degeneracy pressure in the solar core that prevents our Sun reaching the supernova stage. After burning through its hydrogen and helium the core of the Sun would consist of carbon and oxygen; but it will never become hot enough to commence fusion of these elements, because electron degeneracy pressure will support it from contracting and getting any hotter. Only more massive stars will ever attain the core temperatures required to ensure progress towards a terminal iron core.


Can Space-Time Itself Have Energy Qualities Like Momentum?

I'm going to build off of chris's original comment. I haven't had a response from him during the 24 hours since I asked the question, and there doesn't appear to have been any activity from him recently (Note: don't confuse him with another user who goes by Chris), so I might as well expand upon what he said.

Gravitational waves do appear to be what you're looking for. They are emitted by systems with varying quadrupole moments (see https://en.wikipedia.org/wiki/Gravitational_wave and https://en.wikipedia.org/wiki/Quadrupole for more information); commonly cited examples are binary neutron stars. In fact, one such binary, the Hulse-Taylor binary, was the first discovered system to emit gravitational waves.

Gravitational waves carry energy away from the system, at a rate of
$$\frac{dE}{dt}=-\frac{32}{5}\frac{G^4}{c^5}\frac{(m_1m_2)^2(m_1+m_2)}{r^5}$$
where $E$ is eneryg, $t$ is time, $m_1$ and $m_2$ are the masses of the objects in the system, $r$ is the distance between them, and $G$ and $c$ are the constants, the universal gravitational constant and the speed of light. I invite you to do the calculations for a given system, if you please. I can assure you that it's one of the easier calculations in general relativity! This release of energy causes the orbits of the two neutron stars to gradually decay, and it is thought that eventually the two will merge.

The answer boils down to this: Yes, gravitational waves can carry [angular] momentum, just like many other types of waves. They also have frequency, amplitude, wavelength, and speed, just like the "normal" waves we are familiar with.

I hope this helps.


Calculate the time when Arcturus passed through the meridian

Find the time that Arcturus passed the meridian in 29 April.

With respect to the Sun, the Earth rotates once every 24 hours.  But because the Sun's apparent position changes in the sky due to Earth's revolution around the Sun, the Earth rotates once every 23 hours, 56 minutes, 4 seconds with respect to the stars.  Because of this 3 minutes, 56 seconds difference, all stars rise and set 3 minutes, 56 seconds earlier each succeeding day.  This applies to the time of a star passing through the meridian as well.

So, on the next day, April 29, Arcturus will pass the meridian at 22:56:04.

  And what day does the Arcturus will pass the meridian about 01:00:00?

Over the course of an entire year, about 365 days and 6 hours, these 3 minutes and 56 seconds adds up to an entire day, so that exactly one year later, the same star will pass through the meridian at the same time of day.

For Arcturus to pass the meridian exactly 22 hours earlier, it should be 22/24 of a year before then, or about 335 days later, or about 30 days earlier, or March 29.

  solar time equal sidereal time minus 12hr

Generally not correct.  Solar time is based on the average time it takes for the sun to cross the meridian (24 hours), and sidereal time is based on the time it takes for a star to cross the meridian.  Sidereal time counts 24 "hours" in 23 hours, 56 minutes, and 4 seconds, meaning that it runs slightly faster than our normal solar time.

There is one day in which solar and sidereal time are 12 hours off, on the northward equinox in March.  The Sun is on the northward equinox, but solar time is measured since midnight where sidereal time is measured since passing the meridian (local noon).

There is one day in which solar and sidereal time coincide, on the southward equinox in September.  The Sun is on the southward equinox, 180 degrees in the sky from the northward equinox, so the 12 hours off from midnight and the Sun being 12 hours off from the northward equinox cancel.


How dependent is Earth is with the other Planets' gravity?

The orbital speed of the Moon around the Earth is 1022 m/s. The Earth is 81 times more massive, than the Moon. Thus, a sudden disappearance of the Moon would cause an $\frac{1022}{81}=12 \frac{m}{s}$ change of the Earth's orbit, which would be negligible.

The disappearance of the Mars or Jupiter would have yet lesser effect.


If wormholes exist and behave like we think they do, can we cross them without "negatively charged matter "?

The best current answer is "We can't".  We do not presently know of any way to stabilize wormholes without large negative energy densities. (Some argue that even that isn't enough, but I don't believe that the matter is settled, even on a theoretical level.)

Note that:

(a) No wormhole has ever been observed, nor have we observed anything which is a predicted effect of a wormhole: Wormholes are entirely theoretical.

(b) Having said that, Wormholes are a prediction of General Relativity, which is a very well-tested and well-established theory, so it's unwise to bet against them existing.

(c) Macroscopic wormholes are a purely GR phenomenon and while the fields involved are strong, they are not so strong that they are in the realm where we expect the eventual unified theory of quantum gravity to give markedly different results.

(d) And by the way, we haven't a clue how to create large negative energy densities.


How to convert quasar flux time series from observed frame to rest frame?

This is not an answer, but it is too long to put in the comment.

I am not familiar with quasars. In general, converting fluxes from an observed frame to a rest frame requires following consideration: extinction correction, cosmological expansion, k-correction. I think the simplest form is $f_{rest} = (1 + z)^n f_{obs}$ where $z$ is redshift and $n$ is some integer depending on how to express the unit of fluxes. This considers only the cosmological expansion. But, I am not sure if this transformation is valid with your quasars because they might be too far that $(1+z)^n$ is no longer a good approximation.


Plot of best available resolution vs wavelength - radio through gamma rays?

After a bit of searching, I found this blog page, which has several charts about various observatories, including this one:

Image courtesy of Olaf Frohn under the Creative Commons Attribution-Share Alike 4.0 License.

The majority are space-based, although the radio telescopes are largely land-based. They cover existing and future telescopes, at energies from the gamma-ray spectrum to radio waves. You are correct, too, in assuming that adaptive optics can cause dramatic increases in angular resolution; CHARA and the European Extremely Large Telescope both use adaptive optics, and actually can have better angular resolutions than some space-based telescopes.

I annotated the graph to cover in green the smallest angular resolution at various wavelengths:

Notice that most of the lines in the radio, microwave, and infrared part of the spectrum are diagonal, with roughly the same slope. This is because they are limited by diffraction. In the case of radio waves, this is because the atmosphere has little impact. In the case of infrared- and visible- wavelength telescopes in space - and in space-based telescopes in general, the main thing that stops them is the diffraction limit.

The diffraction limit is
$$d=\frac{\lambda}{2n\sin\theta}$$
where $\lambda$ is wavelength and $n\sin\theta$ is the numerical aperture. On a log-log plot, such as the one above, we have
$$\log d=\log\lambda-\log(2n\sin\theta)$$
and
$$\frac{\mathrm{d}\log d}{\mathrm{d}\log\lambda}=1$$
for all telescopes limited by the equation. Thus, telescopes restricted by this limit should be described by a diagonal line with a slope of 1 (-1 on this graph).


How did scientists determine an estimate of the number of planets greater than Sedna's size to exist in the Inner Oort Cloud?

Unfortunately, the paper is not available on ArXiv (oh, what hardships we must overcome!), but I have found it here. In it, where the "900" figure is mentioned (2nd page, I believe), the authors (Trujillo and Sheppard) say that they ran simulations with the data already found and their additional findings, and found that 900$^{+800}_{-500}$ bodies larger than 1,000 km in diameter could exist. They also found that 430$^{+400}_{-240}$ relatively "bright" bodies could exist, as well. Notice that there is quite a lot of uncertainty there; they are rather cautious (and rightfully so) when announcing the results of their simulations.

Interestingly enough, they also mention a paper in that section, written partly by none other than Mike Brown (!) that also used simulation methods. The authors, Schwamb et al, concluded (in 5.4, as well as the conclusion) that 393$^{+1286}_{-264}$ and 74$^{+249}_{-47}$ objects greater than or equal to Sedna in brightness, not size, could exist (393 is for the "hot" population, while 74 is for the "cold" population). The reason Trujillo and Sheppard cited this paper is that their simulation results fit the results of the simulations by Schwamb et al., which is encouraging.

Summary: Trujillo and Sheppard arrived at 900 via simulations based on prior observations. There is a large amount of uncertainty in this number, but it does agree with prior simulations by Schwamb et al. While zibadawatimmy 's comment is perhaps the funniest I have seen, it is, fortunately, inaccurate. Those statisticians must be relieved.

I hope this helps.


Could the trajectories of non-periodic comets be used to infer properties of the 'ninth planet'?

The idea is basically to look for comets coming from the same direction. Look for a pair of comets that at some time earlier in their orbit was at the same place at the same time. That is a data point, but an extremely inaccurate one. It does not have to mean anything, and you are going to collect a large amount of statistical noise. For this purpose, normal periodic comets are close to useless, as they are in the same region for decades or even centuries as they fall slowly towards the sun. Hyperbolic (non-periodic) comets on the other hand, have a shorter stay in each region, allowing us to better pin-point a possible encounter. The problem is, there are not actually a lot of them observed, and the length of the time horizon for accurate enough observations is not that long.

The properties you can obtain about the ninth planet, if it exists, from large amount of data on non-periodic comets are:

Current distance within a reasonably large error margin. (+-10 AU)
Orbital period; not so accurate, but you may get an OK estimate.
Inclination; not accurate, but together with the orbital period, estimates will grow more accurate over a longer time horizon of observations.


Are high-speed galactic collisions survivable?

Interesting question.   I did a bit of reading and gave it some thought and and I can sort of give an answer, though I invite corrections and input from anyone more knowledgeable than me.

First, those 2 galaxies are enormous.  I didn't see specific sizes listed, but per this paper:

  1.2-1.5 x 10^14 solar mass fossil group

That's about 100 Andromeda galaxies.  Also, since what we're seeing is 4.6 billion years old, it's reasonable to assume that those 2 galaxies probably have more free dust and gas and, being larger, the gravitational collision is happening at higher speed.

That's really the question with how much background x-ray radiation you get when 2 galaxies collide.   how much free gas and dust and how fast the galaxies move into each other.   My guess is that being larger, and younger, those 2 galaxies are glowing much more brightly than the Milky-way and Andromeda will 4 billion years from now.

As to how intense the glow is, my hunch agrees with yours, it's probably not very bright if you're there.   The faint glow and the 2 bright eyes you see in the picture are long exposures.   This site (scroll to the bottom) says it's 19 hours and 30 minutes of exposure.  Andromeda, for example is barely visible to the naked eye, but when photographed it's made many times brighter.   I'm not smart enough to say for 100% sure, but my guess is that the faint purple glow would be barely visible, if visible at all, assuming it even reached into the visible spectrum, which X-ray's don't.  Chandra might be setting the color to a visible spectrum in the picture.

How dangerous is it?   Again, I'm guessing, but I think in the Milky-way-Andromeda collision, we'd probably get more UV from the sun than we would from high speed gas and dust collisions, but that's just a guess.   It would be kind of fun if the entire night sky glowed a little bit though.

As far as the Milky-way-Andromeda collision, Wikipedia says that there won't be much free gas in the collision, here, so it's a safe bet there won't be any visible glow in the sky and not even that much of an acceleration of new star formation, though there should be some.   The Milky-way currently forms about 7 new stars every year, and that I imagine will gradually slow down over the next 4 billion years, then speed up significantly as the collision gets underway, but from the point of view of Earth, I don't think there's any guarantee we'd get a close look at any new star formations.

Here's a fun article/interview on the collision between the Milky Way and Andromeda, though I think Roeland van der Marel might be exaggerating the new star formation a bit if we're to believe Wikipedia where it says there won't be much free gas.

As for the speed of our collision, Andromeda is currently heading towards us (or, we're heading towards it, whichever you prefer) at about 110 km/s source.  and it's currently about 2.5 million light years away.  Source.

Covering 2.5 million light years (about 24 million trillion km) in 4 billion years (126 million billion seconds), works out to an average speed of 190 km/s, so we can roughly estimate that the speed of impact between our two galaxies will reach a peak somewhere around 270 or so km/s.   But (see video you also have to take into account the rotation speed and it appears (video above) the rotations will be going in opposite directions when the galaxies collide.    We orbit the Milky way at about 250 km/s and Andromeda, being larger, probably a bit faster orbit speed.   Adding the 2 velocities together, in our neck of the woods, we might see some relative velocity and some gas & dust collisions at (guessing here, cause I don't know if the rotations will slow down as the galaxies approach), but lets say 500 to 600 km/s or so.   For any close bright stars from Andromeda, that could be fast enough for visible changes to a few near-by Andromeda stars in constellations over a single human lifetime.

Perhaps, we'll pass through an oort cloud of another star every few thousand years or so, perhaps even the occasional kuiper belt equivalent, every ten or 50 million years - er, maybe.   We could see some very impressive meteor showers and perhaps an occasional, slightly more frequent, dinosaur killing level comet or meteor impact - but I'm just speculating.  There might be lots of interesting things to see between 4 and 5 billion years from now.

When stars form they can be (per this website) 100 to 200 times brighter than they are during their main sequence, so if our solar-system finds itself nearish to a colliding gas cloud and some new star formation, that could be interesting.  Likewise if we pass near the core of Andromeda we might, for a while, have a very bright night sky.


What is the difference between a depleted comet and an asteroid?

Extinct comets are a sub-category of asteroids.
Minor solar system bodies are classed as "Comets" if they show a coma, or "Asteroids" if they don't.  A depleted comet that has never shown any coma would be categorised as an asteroid.
It is hard to determine if an asteroid is an extinct comet. Some may be distinguished from other asteroids by its spectrum. F and B type Asteroids with low albedo may be extinct comets;  comets are very dark.  Asteroids with eccentric orbits at a high inclination are also often suspected of being extinct comets.
There is another meaning of "depleted", it means that the comet (with a coma, and perhaps a tail) is low in one or more species of molecule, which suggests it formed in an atypical location, or has been modified by some process since its formation.  For example, C/2016 R2 is rich in CO and N2, but depleted in HCN. (Which suggests that it formed in a region of space at ~50K, so much closer to the sun than the outer Oort cloud.


What's the actual speed of electromagnetic radiation in space?

I believe this question was addressed in the paper , Abbot, et. al., Gravitational Waves and Gamma-Rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A , ApJ. Lett., 848:L13, 2017 October 16.

  "We use the observed time delay of ($+1.74 \pm 0.05$) s between GRB 170817A and GW170817 to: (i) constrain the difference between the speed of gravity and the speed of light to be between $-3 \times  10^{-15}$ and $+7 \times 10^{-16}$ times the speed of light, ..."

In Section 2.2 of the paper they discuss the uncertainty of delay between GW and GRB emission, predicted to be a few seconds. In Section 4 of the paper they compute the range of speed uncertainty quoted above. They conservatively use 26 MPc, their lower bound on the 90% credible distance interval, and for the upper bound on speed difference they assume the GW and GRB were emitted simultanously, and use the 1.74 s $\Delta t$. For a lower bound on speed of gravity, they assume the GRB were emitted 10 s after the GW, and a faster EM signal made up some of the difference.

I believe that X-rays were also detected several (~8-9) days later. This delay was attributed to time needed for the shock wave from the merger to interact with surrounding matter. I was unable to find any reference to signal arrival 27 minutes after GW detection in the LIGO papers.


What would happen if asteroid 99942 Apophis collides with the earth?

If it hits the deep ocean, not much relatively speaking. In the recent past, people wildly exaggerated the capability of an asteroid (at least ones in the size range of Apophis) to cause massive tsunamis; in reality, the ocean efficiently and quickly absorbs the impact. That said, it'd still be terrifying to see such an impact in deep water.

If it hits a city, it would cause violent devastation, immediately killing 10's or 100's of thousands with its initial impact, and millions with the resulting shock wave and debris.


Are there recommendations to use "Hubble-Lemaître constant" instead of "Hubble constant"?

No, only the Hubble law was recommended to have its name changed (I'm a member of the IAU, so hopefully I'd have known if there were more votings).

However, several astronomers (including myself) found the voting a bit… weird; while acknowledging the work of George Lemaître is admirable, many more people than him and Edwin Hubble contributed to the understanding that the Universe is expanding. So, some were joking that we should rename the Law to the Slipher-Friedmann-Wirtz-Lundmark-Lemaître-Robertson-Hubble-de Sitter-Humason law, or something like that.

Others joked that if the Hubble law should be renamed, we should also start referring to "the Hubble-Lemaître Space Telescope", "the Hubble-Lemaître Ultra Deep Field", "the Hubble-Lemaître Bubble", and, as you have seen, "the Hubble-Lemaître constant". Others probably also did this non-jokingly.

Btw., astronomer Virginia Trimble wrote a paper six years ago (Trimble 2013), arguing that we should name the Law after anybody but Hubble…


It is a very good place for Earth Life, but that is because we have evolved and adapted to this planet. For example, we like to have an atmospheric pressure of about 1000 mBar, and a partial O2 pressure of about 200 mBar, which is exactly what the atmosphere provides.

Earth would be a very bad place for a species to which oxygen was toxic. A species that hadn't evolved here would probably find conditions challenging in one way or another.


The life cycle is just one of gradual cooling and settling. The heavier elements will sink towards the centre; the planet will get slightly more dense - probably reaching about 90% of its current radius. It will essentially become a cold mini-white dwarf made mostly of hydrogen and helium and supported by cold electron degeneracy pressure. The density is probably not high enough for it to crystallise before differentiation of the chemical elements has completed.
The density profile will not change very much from what it is now. If anything the desnsity gradient will become slightly shallower become because a degenerate gas of heavier elements has more electrons per unit mass and so exerts a higher pressure for the same density. Since the thermal conductivity is extremely high then a temperature gradient will gradually be eliminated in the inner regions.
These processes take place on tens of billions of year timescales. Jupiter will be quite similar to the way it is now by the time the Sun becomes a white dwarf.


Why doesn't the Moonrise appear red when viewed from orbit?

In this video there is blue diffused sunlight present in the atmosphere.  At any point where you might expect blue light to be diffused from the Moon, there is also blue light diffused from the Sun being added back into what you see. To know for sure, you would need to have a recording from night.

The density of air could also be a consideration. The perspective in the video shows the moon from above cloud clover passing below cloud cover. The further the light passes from the Earth's surface doesn't only result in less distance spent in the atmosphere, but also the density of air throughout the entire passage.


How to calculate magnitude of a star in a triple star system?

Basically you need to convert between luminosities (which you can add) and magnitudes using
$$M-M_\odot=-2.5\log_{10}(L/L_\odot)$$
Let's call the total luminosity $L_0$ and magnitude $M_0$ and the individual luminosities and magnitudes $L_1$, $L_2$ and $L_3$ and $M_1$, $M_2$ and $M_3$.
Then, you have the total luminosity of the system, directly
$$L_0/L_\odot=10^{-0.4(M_0-M_\odot)}$$
and as the sum of the components
$$L_0/L_\odot=(L_1+L_2+L_3)/L_\odot=10^{-0.4(M_1-M_\odot)}+10^{-0.4(M_2-M_\odot)}+10^{-0.4(M_3-M_\odot)}$$
Solving these equations for $M_3$ gives
$$M_3-M_\odot=-2.5\log_{10}\left(10^{-0.4(M_0-M_\odot)}-10^{-0.4(M_1-M_\odot)}-10^{-0.4(M_2-M_\odot)}\right)$$
I'm assuming you have absolute magnitudes, but you can rewrite the formulae in terms of apparent magnitudes using
$$M=m+5(1-\log_{10}d)$$
but I think the result then also depends on the distance.


What is the reason for high lithium concentrations in Sun-like stars?

I could offer a couple of speculations - but there is no "theory". Incidentally, I was looking at the refereed journal paper by Baumann et al. (2010).

The atmospheric parameters of these stars are not correct. Given that the ages are derived from the gravities and temperatures of the stars and that the lithium abundances are also dependent on the gravities and particularly the temperatures, there will be correlated errors. I'm actually struggling to understand how you get stars which are "solar analogues", all of which have similar gravities, yet seem to have a spread of ages (estimated from the Teff vs log g plot) from 1-10Gyr? It is notable that there appears to be just 1/117 in Baumann's sample, compared with 10/82 in Israelian's work...
Maybe they are actually PMS stars, so the estimated ages are completely spurious since they assume that the stars are on the main sequence. I would however have expected the abundances to be 3 or higher.

Neither of these explanations seems completely satisfactory.

I mention in passing that the question of planets and lithium is far from settled. The latest salvo in the war was fired by Figueira et al. (2014) who claim there is statistical evidence for enhanced Li depletion in solar analogues with exoplanets.


If this is really Mirach, then I would say the point-like object above and a little to the left of it is the star mu And (another designation: 37 And) in the same constellation of Andromeda. Then the next point-like object slightly further above and to the left from mu And would be the star nu And (35 And). And a barely visible fuzzy object above nu And would be the Andromeda galaxy (M31). Then in the left lower corner we see another star Nembus (51 And), and above it the star phi And (42 And). And at the same height with nu And farther to the right from it at the edge of the building there is the star pi And (29 And).

I would strongly suggest you to use Stellarium (www.stellarium.org) for identification of nearly any night sky object visible at your or any other location on Earth at any time.


"How the Universe Works": is it scientifically accurate?

I don't think that the Discovery Channel would produce science fiction and call it a documentary. The episodes are bases on factual information at the time of production.


How can the next supermoon be analytically predicted?

A slightly more useful answer than my comment above:

Using http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation "Angular
Separation Finder" feature, we can determine when the angle between
the Sun and the Moon is at a local maximum. It turns out this isn't
quite the definition of Full Moon, but it's very close. I couldn't
find a way to make webgeocalc compute full moon times exactly, though
I'm sure there is a way, and, of course, there are plenty of tables of
full moons online.

Starting at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder
we fill out the form as so:

and click the "Calculate" button.

This yields a list of all "full" moons. Scroll down and click on the
"Save All Intervals" button:

We now want to find the moon's distance at all of the times we just
saved (ie, the times of the "full" moons). To do this, start at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#StateVector and fill out as
follows:

NOTE: to get the list of intervals above, drag from the saved
intervals result window into the list of intervals window.

The results, which you can also download, tell you the moon's distance from the Earth at the times of the "full" moons we computed earlier:

I downloaded the results in Excel format, imported them into gnumeric
and sorted by radius (distance from Earth to Moon) and did a little
additional cleanup to get:

(the full spreadsheet is available at https://github.com/barrycarter/bcapps/blob/master/STACK/StateVectorResults.xls)

So, yes, tonight's full moon (yellow) is the closest we'll have until
2034-11-25 (red). It's also the 6th closest moon we'll have this
century.

To see other examples of geocalc in action, see my:

What is the format of the data from the JPL's HORIZONS system?
https://space.stackexchange.com/questions/18718/calculating-positions-of-moons-ascending-and-descending-nodes-from-ephemeris-fi

For more about how NASA calculates these values in the first place: Where can I find a set of data of the initial conditions of our solar system? which is also now linked to the general "how to compute positions" wiki answer: Where can I find/visualize planets/stars/moons/etc positions?


Does Eddington's variable polytropic index better fit data from the Standard Solar Model?

The definition of a polytrope is one for which the pressure and density are related by
$$ P = K \rho^{(n+1)/n}$$
The pressure is thus assumed to be independent of temperature and this simplifies the equations of stellar structure so that certain analytic solutions and approximations are possible.

It is however a  big assumption and is generally only accurate for stars/planets supported by degeneracy pressure, where $1.5<n<3$ depending on whether the degeneracy is non-relativistic or relativistic. The other instance is where the energy transport is by adiabatic convection, which gives an independent relationship between $P$ and $T$ and means that for a completely ionized gas that $P \propto \rho^{5/3}$ and $n=1.5$.

A further polytropic approximation can also be made where the energy transport is predominantly radiative and where it is assumed that the radiation pressure is a fixed proportion of the total pressure. In these circumstances $P \propto \rho^{4/3}$ and $n \simeq 3$.

I found a bit of a primer here, but most stellar astrophysics textbooks will go through this.

The Eddington standard model assumes a radiative ($n=3$) polytrope. The vast majority of the Sun, by mass, is indeed radiative and the polytropic approximation works reasonably well. However the outer parts of the Sun (beyond about 70% of the radius) are convective and so a single polytropic model does not fare so well. So what you can do is have a "mixed polytropic model", where you divide the star up into polytropic shells with different polytropic indices, each forming the boundary conditions for the ones above and below.
A double polytropic model for the Sun, with two zones that have $n_1\simeq 3$ in the interior and $n_2\simeq 1.5$ in the outer convection zone should be much more accurate.

But modern astrophysical calculations do not rely on the analytic approximations of polytropes and the Lane Emden equation. They numerically solve the coupled differential equations of stellar structure, subject to the boundary conditions imposed by an atmosphere on the top of the star.

EDIT: Unashamedly tempted by the bounty, I did a little bit more research. I came across some sort of high-level set of lecture notes by
Robert French of Swinburne University. He discusses modelling the Sun with composite polytropes at some length. On p.8 he describes how the matching criteria at the boundaries between polytropic zones is actually quite tricky and you have to let the matching radius and central density float in order to get a smooth join.

The conclusion he arrives at is that a combination of a $n_1=3$ and $n_2=1.5$ actually appears to do a worse job of fitting the entire Sun than modelling with a single average polytrope (although with $n$ varied depending on which physical quantity you want on the y-axis - e.g. $\bar{n}=3.11$ for density vs radius). If on the other hand you allow the polytropic indices and the interface radius be free parameters in order to get a good match to the standard solar model, you can get a really good fit with $n_1=3.94$ in the radiative zone, $n_2=1.58$ in the convection zone and an interface at 68% of the solar radius. (Plot reproduced below - compare these with the plots in your question!).


Is the Bondi Radius the Same as the Schwarzschild radius?

The Bondi radius is $2MG/s^2$ where M is the mass, G is the gravitational constant and $s$ is the speed of sound in the medium of the accretion disk.  It represents the boundary between sub-sonic and super-sonic infall of matter into the black hole.

The Schwartzchild radius is $2MG/c^2$ where $c$ is the speed of light and is the radius of the event horizon.


Just going off the Wikipedia article you posted, it says the hot jupiter superflare theory was abandoned.

  The flares were initially explained by postulating giant planets in
  very close orbits, such that the magnetic fields of the star and
  planet were linked. The orbit of the planet would warp the field lines
  until the instability released magnetic field energy as a flare.
  However, no such planet has showed up as a Kepler transit and this
  theory has been abandoned.


which pulsar has the longest spin period so far?

Vela X-1 is an example of an accretion-powered pulsar. These emit pulsed emission because they are accreting material onto their magnetic poles. If the magnetic poles and rotation axis are misaligned this results in pulsed X-ray emission. The power for the pulsar comes from the infalling material.

The classic P-Pdot diagram you show is for radio pulsars, or rather rotationally-powered pulsars. These are objects that are pulsators by virtue of accelerating charged particles from their magnetic poles out along the field lines. These accelerated relativistic particles emit synchrotron and curvature radiation that is beamed and intensified. The energy ultimately comes from the rotational kinetic energy of the pulsar.

In other words, other than the fact that they both involve neutron stars, these are completely different phenomena.

In terms of radio pulsars, your diagram looks reasonably up to date. I think the longest period object on your diagram is PSR J2144-3933, which has a period of 8.51 seconds (Young, Manchester & Johnston 1999).

Your diagram has a line marked as "graveyard". I believe this is a locus defined by Bhattacharya et al. (1992) and has the form
$$P = 2.42\times 10^{-6} B^{1/2}\ s,$$
where the pulsar magnetic field $B$ is in units of Gauss (typical pulsar values would be $B=10^{10}-10^{13}$ Gauss (as also marked on your diagram). The theory behind this "death line" is discussed by Ruderman & Sutherland (1975). Briefly it arises from the requirement of a minimum potential difference to be generated such that accelerated particles produce energetic enough radiation to stimulate the production of further electron/positron pairs. If the magnetic field strength falls or the rotation period gets too long, then this mechanism fails and the pulsar is quenched.

PSR J2144-3933 appears to lie beyond this death line, but there are other ideas and models of how this death line may arise (of which I am not very familiar, but see for example Zhang et al. 2000).

Note that millisecond pulsars are thought to recycled accretion-powered pulsars. That is that they gain sufficient angular momentum from an accretion process that they spin up to become radio pulsars again.

Radio-quiet pulsars are thought to be rotationally powered but where the radio beam is narrower than say the beaming of gamma rays in which they are detected.

Magnetars are something different again. Their power comes from the decay of extraordinarily strong magnetic fields.
The soft gamma ray repeaters and anomalous X-ray pulsars fall in this category.

So you can see that the word "pulsar" might include all sorts of different types of objects and physics.


Why don't the inner moons of Jupiter have tidally-induced volcanism?

It’s because they are much smaller than Io.  Tidal forces are differential forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other.  When an object is small, the difference in distance to the two sides of it is necessarily small as well.
According to Wikipedia, Amalthea, the largest of those four innermost moons has a long axis that is only 250 km, and the others are smaller yet. The strength of the tidal heating scales as the body’s radius to the 5th power, quite a strong dependence.   The small mass plays a role, too, but the dependence on radius is stronger.
For this same reason, small objects on Earth don’t feel tidal stretching from our Moon, yet the Earth as a whole does.
On a related note, the inner three of those four satellites are inside Jupiter’s Roche limit, the orbital distance at which a purely self-gravitating body should be pulled apart by tidal forces.  But they survive because most small bodies aren’t held together primarily by gravity (just as your own body isn’t) - the internal electromagnetic forces between atoms (which manifests macroscopically as the rigidity of the rock) is the main source of the bodies’ structural strength.


I don't know how to calculate the minimum possible forbidden region for a planet of a specific mass orbiting a star of a specific mass at a specific semi-major axis. So I can't say what the absolute theoretical minimum separation between the orbits of two planets in the same planetary system can be.
A reductio ad absurdum calculation suggests that the minimum possible separation between planetary orbits could be about 2,500 to 5,000 kilometers.  Probably astrophysicists should be able to calculate a minimum possible separation of orbits which would be many, many times that.
There are also some theoretical configurations of planets which would result in two or more planets sharing the same orbit,  in which case one could be facetious and claim there is a zero separation between their orbits.
More Detailed Answer:
Part One of Six: Some "Dole ful" Calculations.
I do know that there have been formulas to calculate the minimum possible spacing between planets in a star system for decades.
Habitable Planets for Man, 1964. 2007, by Stephen H. Dole, is a scientific discussion of the parameters of planets which are habitable for humans.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf
Chapter Three: introduction to General Planetology, has a section "Spacing of Planets in the Solar System", on pages 49 to 52, discusses the spacing of planets in our solar system.
Dole bases his discussion of forbidden regions on his own paper:
"Limits for stable Near-circular Planetary or Satellite Orbits in the Restricted Three-Body Problem". ARS J, 31, no. 2 (February, 1961), pp. 214-219.
Dole calculated the limits of the "forbidden" regions of all the planets in the solar system, based on the masses of each planet and the Sun, the eccentricity of each planet's orbit, and the semi-major axis of each planet's orbit around the Sun.
The "forbidden" region is a ring around the Sun that a planet orbits within and which no other planet should be able to have a stable orbit within.
According to Dole's calculations, our solar system out to Neptune or Pluto is approximately half full of forbidden regions around planetary orbits:

This pattern of regularity should also be found in other planetary systems. Forbidden regions take up about 50 per cent of our solar system, and if this is true of other planetary systems (or multiple star systems); then it would be a simple matter to design any number of stable planetary systems by random mechanical processes.

So, according to Dole's calculations, a star like the Sun might have the same number and masses of planets as the Sun does, if its planetary system was about 50 per cent filled by planets and their forbidden regions.  If planets could be spaced close enough together that the limits of their forbidden regions touched, then our solar system could have perhaps twice as many planets with the same masses, about 16 to 18.
More importantly to me, if there are one, two, or three planets within the circumstellar habitable zone of the Sun, in the planetary system of a star like the Sun with the forbidden zones of the planets just touching, there could be only two to six planetary orbits within the circumstellar habitable zone of that star.  As a kid who liked science fiction stories where many planets in a single solar system were habitable, I found that idea rather "Dole ful".
Part Two: Some Early Ideas to Increase the Number of Habitable Planets in a system.
A number of science fiction stories have habitable planets in Trojan orbits, in which a planet orbits around a star in the L4 or L5 Lagrange point of another planet or another star, 60 degrees ahead of or behind the other astronomical object.
Such Trojan orbits are known to work when the primary, secondary, and tertiary objects differ in mass by hundreds or thousands of times.  In our solar system, the Sun is thousands of times as massive as planets which are thousands of times as massive as the asteroids in their Trojan positions.  And Saturn is thousands of times as massive as its moons Tethys and Dione, which in turn are thousands of times as massive as their Trojan moons.
But could two planets in the size range to be habitable share the same orbit, one planet being in the L4 point of the second one, which would be in the L5 point of the first one? I don't know.
But if that was possible, and if there can be two to six planetary orbits within the circumstellar habitable zone of a star like the Sun, then if there are two Trojan planets in each orbit that solar system could have four to twelve potentially habitable planets in its circumstellar habitable zone.
Another way to increase the number of habitable planets within the circumstellar habitable zone would be to replace each Earth mass habitable planet with a double planet having the same total mass as the Earth.  According to Dole, the minimum mass for a habitable planet would be about 0.4 Earth mass, so a double planet with twin planets of about 0.42 Earth mass would have a total mass of 0.84 Earth, and thus a slightly smaller forbidden region than Earth.
So theoretically about four to twelve habitable planets could orbit in the circumstellar habitable zone of a star identical to the Sun, if arranged as two to six sets of twin planets.
Part Three: Co-Orbital Planets?

In astronomy, a co-orbital configuration is a configuration of two or more astronomical objects (such as asteroids, moons, or planets) orbiting at the same, or very similar, distance from their primary, i.e. they are in a 1:1 mean-motion resonance. (or 1:−1 if orbiting in opposite directions).1
There are several classes of co-orbital objects, depending on their point of libration. The most common and best-known class is the trojan, which librates around one of the two stable Lagrangian points (Trojan points), L4 and L5, 60° ahead of and behind the larger body respectively. Another class is the horseshoe orbit, in which objects librate around 180° from the larger body. Objects librating around 0° are called quasi-satellites.
An exchange orbit occurs when two co-orbital objects are of similar masses and thus exert a non-negligible influence on each other. The objects can exchange semi-major axes or eccentricities when they approach each other.

https://en.wikipedia.org/wiki/Co-orbital_configuration
The space probes Pioneer 11, Voyager 1, and Voyager 2 discovered several new moons of Saturn when they passed it in 1979, 1980, and 1981.  This included the discovery of the tiny moons Epimetheus and Janus, which are co-orbital.

The Saturnian moons Janus and Epimetheus share their orbits, the difference in semi-major axes being less than either's mean diameter. This means the moon with the smaller semi-major axis will slowly catch up with the other. As it does this, the moons gravitationally tug at each other, increasing the semi-major axis of the moon that has caught up and decreasing that of the other. This reverses their relative positions proportionally to their masses and causes this process to begin anew with the moons' roles reversed. In other words, they effectively swap orbits, ultimately oscillating both about their mass-weighted mean orbit.

So it would potentially be possible for two habitable planets to share the same orbit in an exchange orbit like that of Epimetheus and Janus, thus having two habitable planets in the orbit and forbidden region.  That could be used to double the number of planets orbiting within the habitable zone of a star.
I can't help thinking that inhabitants of a planet in an exchange orbit with another planet would find the exchange process terrifying until and unless they could calculate that the two planets were not going to collide.
Part Four: Exoplanet Discoveries.
IN the last generation thousands of exoplanets have been discovered orbiting other stars.  And planetary systems with two or more planets orbiting the same star have also been discovered.  Because of the great difficulty in discovering exoplanets, it is reasonable to assume that in most cases there are more planets in a system than have been discovered yet, perhaps more planets than can be discovered until decades, centuries, or even millennia of future scientific progress is made.
Even though Dole wrote in 1964 that:

This pattern of regularity should also be found in other planetary systems.

The majority of planetary systems discovered around other stars have been significantly different in one or more major ways from our solar system.  Thus there must be a great variation in the processes which form and shape planetary systems.
According to Wikipedia's list of exoplanet extremes, the smallest distance between the semi-major axis of the orbits of two consecutive planets is between Kepler-70b and Kepler-70c, about 0.0016 AU, or about 240,000 kilometers, or about 149,129 miles, closer than the distance between the Earth and the Moon.
https://en.wikipedia.org/wiki/List_of_exoplanet_extremes
If each consecutive planetary orbit around the Sun was only 0.0016 AU farther out than the previous one, 262 planetary orbits could fit within Kasting's conservative habitable zone for the Sun, and 518 in his optimistic habitable zone.
However, the article on Kepler-70 indicates that it is now believed the planets do not exist and their detection was probably an error.
https://en.wikipedia.org/wiki/Kepler-70
I do not know what is the smallest known difference between the orbits of two confirmed exoplanets.  But the two exoplanets with the smallest ratio between their orbits are Kepler-37b & Kepler-37c.  Both planets are several times as massive as Earth but orbit very close to their star Kepler-37 and thus to each other.
The orbit of Kepler-37b has a semi-major axis of 0.1153 AU, or 17,248,634 kilometers, or 10,747,804.57 miles, and the orbit of Kepler-37c has a semi-major axis of 0.1283 AU, or 19,193,407 kilometers, or 11,926,230 miles, a difference of 0.013 AU, or 1,944,772.3 kilometers, or 1,208,425.49 miles.
If the planetary orbits around the Sun could have an average separation of 0.013 AU, 32 planetary orbits could fit within Kasting's conservative habitable zone and 64 planetary orbits could fit within Kasting's optimistic habitable zone.
0.1283 AU is 1.1127 times 0.1153 AU, and is the smallest known ratio between consecutive planetary orbits.  If I remember correctly I calculated that 4 planetary orbits with that ratio could fit within Kasting's conservative habitable zone and 6 planetary orbits could fit within Kastings optimistic habitable zone.
So should the distance of 0.013 AU between the orbits of Kepler-37b and Kepler-37b be considered the minimum possible spacing between consecutive planetary orbits?
Or should the ratio of 1.1127 between their semi-major axis be considered to the minimum possible ratio between consecutive planetary orbits?
If it is the ratio between orbits which determines the minimum spacing of planetary orbits, if a planet orbits very close to its star the minimum possible distance to the next planet's orbit could be less than the 0.013 AU between the Kepler-37 planets. For example, if the inner planet orbits at 0.01 AU, the next planet could orbit at a distance of only 0.011127 AU, a difference of only 0.001127 AU, a little less than the distance between the orbits of the alleged Kepler-70 planets.
If it is the ratio between orbits which determines the minimum spacing of planetary orbits, if a planet orbits very far from its star the minimum possible distance to the next planet's orbit could be many times the 0.013 AU between the Kepler-37 planets.  For example, if a planet orbits 100 AU from its star, and the minimum separation is 1.1127 times the inner orbit, the next planet out would have to orbit at a distance of at least 111.27 AU.
Par Five: Planets in Rings.
Astrophysicist Sean Raymond, in his PlanetPlanet blog, has a section called Ultimate Solar System designing planetary systems with as many planets in the habitable zone as he can fit in.
https://planetplanet.net/the-ultimate-solar-system/
In the Ultimate Engineered Solar system Raymond designs a planetary system with 416 planets in the habitable zone, using rings of planets sharing the same orbit.
That is based on this paper by Smith and Lissauer:
https://ui.adsabs.harvard.edu/abs/2010CeMDA.107..487S/abstract
Smith and Lissauer calculate that seven to forty two planets can share the same orbit around their star, if the planets have equal mass and are equally spaced around the star.
The planets would be separated by 8.57 degrees if there are forty two planets in the ring, increasing as the number of planets decreases, so that seven planets would have gaps of 54.42 degrees between them.  The size of those gaps in AU, kilometers, or miles would have to be calculated from the semi-major axis, and thus the circumference, of the orbit.
Raymond also states that planetary orbits should be separated by 5 to 10 times their Hill radius to be stable.  For example, a planet with the mass of Earth 1 AU from a star with the mass of the Sun would have a Hill radius of about 1,500,000 kilometers or 0.01 AU.
So planetary orbits of Earth-like planets orbiting Sun-like stars at distances of about 1 AU should be separated by at least 0.05 to 0.10 AU or 7,500,000 to 15,000,000 kilometers.
Part Six: Conclusion.
The minimum possible distance between the semi-major axis of the orbit of a planet and the semi-major axis of the the orbit of another planet orbiting the same star can be calculated from factors such as the mass of the planet and the mass of the star, and the semi-major axis of the planet's orbit.
Known examples of the separation of the semi-major axis of consecutive planets in a planetary system range from as much as about 662 AU between CVSO 30 b & CVSO 30 c down to as little as 0.013 AU between Kepler-37b & Kepler-37c.
Known examples of the ratios of the semi-major axis of consecutive planets in a planetary system range from as much as about 78,998 times the distance in the case  of CVSO 30 b & CVSO 30 c down to as little as 1.1127 times the distance in the case of Kepler-37b & Kepler-37c.
Of course if additional planets are discovered between the orbits of the known planets in those systems the records could change.
Astronomers are fairly certain that any planetary orbital separation within those ranges is possible.   I don't know the theoretical limits to how much larger or smaller the separation between the semi-major axis of two consecutive planetary orbits could be.
For minimum spacing I guess I can do a reductio ad absurdum calculation.
According to Sean Raymond here
https://planetplanet.net/2017/05/03/the-ultimate-engineered-solar-system/
The minimum possible separation between two planetary orbits should be five to ten times the Hill sphere radius of one of the planets - the planet with the larger Hill sphere radius.
The absolutely smallest the Hill radius of a planet orbiting very close to a very massive star could get would be at the surface of that planet.  If the surface of the planet extended above the Hill sphere radius the surface material would be ripped from the planet by the gravity of the more massive object until the planet was stripped down to to the radius of the Hill Sphere.
If a astronomical object has to be pulled by its gravity into a spheroidal shape to be considered a planet, that establishes a minimum size for a planet or a planemo (planetary mass object), and thus for its smallest possible Hill sphere.  The minimum radius necessary for astronomical object to be spheroidal is not known precisely.
More or less arbitarily making a radius of 500 kilometers the lower limit for a planetary mass object, and assuming that such an object could orbit close enough to its star to have it's hill sphere at its surface, the minimum possible separation between its orbit and that of the next planet in the system would be about 2,500 to 5,000 kilometers.
And of course if two planets shared the same orbit, such as Trojan planets, planets in an exchange orbit, or planets in a ring, someone could facetiously say there would be approximately zero difference between the semi-major axis of their orbits, since their orbits could be the same orbit.


What to observe with a remote telescope close to sunrise time?

I have been an amateur astronomer for over 60 years. In all that time I have not had more that a passing interest in variable star observation or spectroscopic measurements. What makes you think a bunch of high school kids will?

As I recall the best teachers I had were the ones that could make a subject interesting. Kid want candy. Give it to them. Point that monster at the most visually interesting objects. If you want to start a fire you use use a bit of kindling not a pail of cold water.

Put yourself in their position. A once in a lifetime opportunity and you want to turn it into a lecture on page 347 of the Science textbook.


From whence is the Event Horizon Telescope black hole data available for amateur reconstruction?
So I'm wondering what kind of data is available that would allow someone to reconstruct their own EHT image?

In the CalTech video Imaging the Unseen: Taking the First Picture of a Black Hole - Katie Bouman - 6/7/2019  which I just found in this question, after about 25:45 Dr. Katie Bouman says (edited from closed captions):

So we collected five petabytes of data during that time... and a little less than a petabyte was for M87... but actually what we use for imaging; we don’t work with the petabytes of data.
This is reduced down through a process of correlation and a calibration stage that initially it’s at petabytes, then it goes to terabytes, then eventually what we use for imaging is in the megabytes range, where you can just email the file. And you can actually download it online now. We have all of the data online that we use for imaging and you can try to make your own image...

It seems that the MB sized raw data is available here, found on the EHT website here per @PeterErwin's comment.
More information is available at https://github.com/eventhorizontelescope/2019-D01-02
One can only speculate on how the tweeted image was generated, but luckily that wasn't the question.
A sample of a random selection of the text files begins with:
# SRC:M87 DATE(MJD):57849 FREQ:227.0707GHz
# time(UTC)      T1     T2   U(lambda)        V(lambda)         Iamp(Jy)    Iphase(d)   Isigma(Jy)
00.86805555      AA     PV  -4396973056.0000  -4510950912.0000  0.13778746    178.1494  0.00382148
00.87083334      AA     PV  -4397497344.0000  -4511639040.0000  0.13610664   -179.6852  0.00413867
00.87361112      AA     PV  -4398009344.0000  -4512314368.0000  0.13498885    179.0204  0.00400464
00.87638891      AA     PV  -4398538752.0000  -4513015296.0000  0.13268795    179.9253  0.00369624
[...]


So is there nothing like satellite anymore?

This is really a question of Language rather than Astronomy. The meaning of words is defined by their use. The IAU has proposed one way of using the word "planet", this group of scientists have proposed another.

Anyone can propose a definition of any word. This group proposes that a "planet" is any astronomical body which 1) Does not have, and never has had sustained nuclear fusion in its core, and 2) Is large enough for gravity to pull it into an ellipsoidal shape.

The intention of this definition is to make a planet defined entirely by "what" a body is, not by "where" it is. Under this definition the classic 8 planets, Ceres, Pluto, Charon, the Moon, the larger moons of Jupiter, Saturn Uranus and Neptune, Eris, Sedna, and the larger Kuiper belt objects would all be "planets" - about 100 known in total. The exoplanets would also be "planets"

Asteroids (except Ceres), comets and other small solar system bodies are too small, and so would not be planets.

The advantage of this definition is that the usually where something is located is not part of the definition of what it is. However, it contradicts several hundred years of use by defining the larger moons as planets.

Unlike the 8 planet definition, this does not have the support of a large international body of astronomers (the IAU). If you wish to use this definition you are free to. I doubt that without the support of the IAU this definition is ever going to be widely used. Most people will either use the 8 planet definition or the 8+Pluto definition.


Working with high-magnification eye-pieces

Try the moon first.

If you see nothing but black, assuming that you don't have a lens cap on or something, then most likely, you are zoomed in on, well, relative blackness. The star you were viewing is probably off to the side now. Or, you may just be looking through the eye-piece at the wrong angle or something. The moon is too big a target to miss, and you have nice rough features to use to tune in your focus. And if you see black when you are looking at the moon, it's much easier to troubleshoot. You should definitely see something!

Now, secondarily the reason you expect to see something, but instead see nothing may be a viewfinder problem. If you have a small, low magnification viewfinder scope attached to your telescope, you usually can't just trust that what you see in the center of that view is exactly what your telescope will be aimed at. You may have to align it. Search "align telescope viewfinder" for help with this. From your description, it sounds like you are not using a viewfinder, but this seems to be a common problem, so I thought I should mention it.


Effect of the absence of atmospheric refraction to sun rise and sunset times

The simple answer is that atmospheric refraction makes the disk of the sun visible at sunrise or sunset, when it's actually half a degree or so below the horizon:

In reality, both the extent of refraction of the sun's disk and the refraction of the apparent horizon come into play. Both depend on the atmosphere's density and temperature profile; which changes irregularly. That makes it impossible to predict exact sunrise or sunset times when there is an atmosphere.


What contributes the most to the seasonal temperature variation?

[ The earth is actually closer to the sun in the Northern hemisphere's winter. The seasonal temperature variation is predominantly due to the angle the earth makes with the sun. In the northern hemisphere in winter the angle is such that the earth is tilted with north pole away from the sun and the sunlight hitting the earth is spread over a much larger area than if it was pointed towards the sun. Due to this tilt the sun is also lower in the sky and has the effect shown in the 2nd figure.


How does glass affect taking solar spectra?

you'll need to find the spectral transmissivity of each piece of glass.  Most likely they all block UV and large sections of the IR bands.  Further,  unless you've removed the internal IR filter, your camera won't record any IR.  CCDs also cut off in the blue, so no UV will be recorded.

BTW, how are you separating the wavelengths at the camera?  Typically one uses a (expensive) grating spectrometer.


Van-der-Waals forces and gravitation are the main answer.
The main property keeping together bodies smaller than a few $100 {\rm km}$ are their constituents surface forces between the grains (van-der-Waals) and chemical/ molecular bonding within the grains themselves. It's the same mechanic which makes the dust and grease in your flat stick to surfaces and the fluffy dirt clumps under the bed stick together, thus cohesion/ adhesion.
The typical internal strength of such dirty iceball as comets are, or rubble piles as some minor asteroids like Ruygu seem to be (e.g. Wada et al, 2018), does not need to be large - there is nothing in deep space which pulls it apart. And even the few µg of self-gravitation help to keep smaller fragments and ejecta to return to the parent body when hit by the inevitable micro-meteorites after their intial formation (Blum et al, 2017) (so yes, there is still self-gravitation involved which helps form them; they are just too small to achieve hydrostatic equilibrium and differentiation).
67P Churyumov-Gerasimenke provided a nice cometary physics lab which allowed in-situ investigation. The strength of the material could be analysed directy (Heinisch et al, 2018), (Grousin et al, 2018) with the lander Philae, even though it landed somewhat precariously (see also this overview by Blum et al (2018) focussing on planet formation and lessons learned from 67P in particular).
In summary: most measurement, both in-situ and lab experiments, indicate that the internal strength of comets is within the range of $1 {\rm Pa} \ldots  ... 10 {\rm kPa}$ and most likely at the lower end of this range. Their strength is thus determined by their small gravity and the inter-particle forces, thus van-der-Waals which in part is possibly enhanced a bit by sintering together the different constituents grains.


How was precision astrometry done before digital imaging?

Prior to digital imaging then photographic plate negatives were analysed with scanning microdensitometers to produce astrometric catalogues. Many of these catalogues are still in use today, they are valuable sources of early epoch positions that enable proper motion measurements.

For more details you could look at the descriptions of the SuperCosmos project http://ssa.roe.ac.uk// (which is based on Schmidt plates) or the UCAC4 catalogue, http://ad.usno.navy.mil/ucac/readme_u4v5, which uses plates to get positions and proper motions for faint stars.

In the good old days, before even these catalogues existed, astronomy groups would have copies of the whole Schmidt and Palomar sky surveys. You would put the relevant plate on a massive, concrete-based, hydraulic X,Y measuring machine, with a binocular microscope. You would measure sets of Standard stars, get a 6-coefficient fit to convert x and y into RA and Dec,then find your objects on the plate, measure x,y, calculate RA and Dec. Then snap a polaroid to use as a finder chart at the telescope.

I was doing this as late as 1995 before the advent of the Digitised Sky Survey.


What was the bright orange glow in the sky?

This is low on the horizon and there is glow from distant lights.
I have seen stars or planets suddenly disappear when fairly low, and it happens when clouds that were not visible at night move in between the object and myself.
If there is some hazy glow already in the sky, and the cloud is darker, it could be completely invisible and undetectable until it blocks something.
If it move just a tiny bit, it could be that the viewer moved and the the movement relative to the branches is what was actually seen. Assuming its a star or planet it would not have moved quickly in the sky, but relative to foreground objects for reference if one's head moves slightly it could have seemed to move.


Why does the outer layer of the sun have differential rotation?

I think they're not taking into account the Coriolis force in the frame of reference of the part of the outer layer of the sun at each latitude. I can't quite think of a proof that the outer layer of the sun would be predicted to have differential rotation. However, when you realize about the Coriolis force acting on the convection currents, you see that it does not give you a proof that there shouldn't be differential rotation. Maybe the part at one latitude then exerts a force on the part at an adjacent latitude until the rates of the two layers satisfy a certain equation.


Can you see the starting and the ending of a light beam passing in the distance?

Nothing forbids this, and it is actually observed astronomically. You need a very bright source of light: such as a supernova (which isn't a beam, but radiates in all directions) and very large distances. The flash of light can be seen spreading out from supernova in a circle, as it illuminates the dust and gas ejected by the supernova progenitor star in the years before

The effect is known as a light echo. Over time the "beams" of light move further out, and so the echo widens, as shown in this series of images (from an Australian telescope so you have to read from the bottom up, 507 days means 507 days after the supernova in 1987.) For scale, 5 arcmin is quite small: The moon is about 30 arcmin. source


Mountains are higher than the atmosphere?

This is a bit of a gray area, as an atmosphere doesn't have a clear boundary. That being said, Olympus Mons on Mars is so tall, the atmospheric pressure on top of it is only 12% the average pressure on the surface of Mars. That's near vacuum by terrestrial standards.

https://en.wikipedia.org/wiki/Olympus_Mons#Description

In general, for this to happen you need:

a pretty thin atmosphere to begin with
some exceptional geology that ends up producing very tall anomalies like Olympus

It's not a very likely combination, but it can happen, as seen on Mars.


What would be the dynamics of a double-planet system, similar to Earth / Moon, but with both bodies nearly Earth-sized?

This is a duplicated question but I think it fits better into astronomy than into physics. I'll try to answer all aspects of this question including the consequences.
First, is it possible? As already pointed out it is, like Pluto and Charon. But is it possible for bigger planets, with the size of Earth? Well we didn't find this planet system yet, but the universe is big. There is no physical law that would forbid this system, it might be unlikely and very rare but it would be possible.
Let's assume somewhere is a planet system with two Earth-sized twins. What would be the dynamics of this systems on the consequences for possible living beings on one of this planets?
One consequence is that the gravity center of this system would be between both planets, if they exactly are the same size (same mass) the gravity center would be in the exact middle of both. That means both planets are rotating around this point (while they are rotating around a sun). This would lead to some effects like in a centrifuge and lead to some weird appearances. The gravity point of our Earth – Moon system is already shifted from the geometrical center point of our earth. And this leads into tides. The tides on our double Earth system would be vast (assuming at least one of this planets is just like our earth and has water on it). Complete areas would be periodically covered with water. Most living beings would probably be amphibians. This of course means the planet would rotate. In the beginning, the planets would rotate but they would slow each other down until they show the same side to each other. Just like our Earth did to the Moon. The Moon continually slows down the rotation of our Earth, but with his minor mass this takes a while. This processes would be much quicker on the double earth system. But even this would take some time. So it would be a matter of time when you look at this system. If it's a young system they would still rotate. If it's an older system they would show the same side each other.
The last point would fix the tides problem, because the water level would be stationary. But this would lead into new problems. The day length would be as long as they need to rotated around their common gravity center. I'll be honest. I am not sure how long this would be. For a stable double Earth system the distance between both planets would be further then the Earth-Moon distance. Let's just assume a day in that system would take 2 Earth months for simplification. This would mean on daylight it would be very hot and at night it would be very cold. One side of the planet would be heat up for hundreds of hours (our earth hours) and the other half would cool down for the same time. Temperatures over a period would jump up and down in big steps. There is no living being known on earth that can handle both extreme hot and extreme cold in such a short period of time. Also this would lead into very heavy and constant storms in the area where the sun sets or rises, making living there even harder.
Also your weight (stepping on the bath scale) would vary depend on where on the planet you are. If you on the side facing the other planet you would be heavier than on the other side of the planet.
Another point that would be possible, but I will point out I am not completely sure about this. The planets would struggle keeping their atmosphere for a long period of time because of the rotation. The side facing away from each other would have thicker atmosphere than on the sides (see my sketch).

(source: directupload.net)
At this point the gravitational pull of the planet is also lower then everywhere else. This could help the molecules to escape the Earth, in a long period of time.


The altazimuth system is directly linked to your position on the Earth. It is the “left-right/up-down” coordinates system; the azimuth being the “left-right” (with 0° for North, 90° East, etc.), and the altitude is the “up-down” (with 0° being the horizon, 90° being overhead, also called zenith). Of course, a star’s (or planet’s, comet’s, galaxy’s, etc.) position will change during the night and over the year.
The equatorial system is independent of any location on the Earth. It is akin to the longitude and latitude of the Earth, but for the sky. As you can guess from the name, its reference is the celestial equator, which is the projection of Earth’s equator on the celestial sphere. There’s an imaginary point along it, where the Sun crosses it from south to north (near March 20), that’s the reference point, to be more specific; it’s called the vernal point. Like longitude, right ascension is measured (in hours, minutes, and seconds of time) eastward from that point, from 0 h at this point to 12 h opposite it to 24 h back at this point. Like latitude, declination is measured north (+) or south (−) from the equator to the poles (+90° North, −90° South). Celestial bodies virtually keep the same right ascension and declination over a human lifetime, except for planets, asteroids, and comets, which keep moving.
Finally, the ecliptic system is used mostly for bodies of the Solar System. As you can guess by the name, it’s measured along the ecliptic instead of the equator. The ecliptic is the apparent path of the Sun in the sky over the year—it corresponds to the plane of the Earth’s orbit around the Sun. Coordinates in the ecliptic system are ecliptic longitude, measured eastward from the vernal point mentioned in the previous paragraph, from 0° at it to 180° opposite it to 360° back at it; and the ecliptic latitude, measured perpendicularly to the ecliptic, from 0° along it to +90° at the ecliptic north pole and −90° at the ecliptic south pole.
There are indeed formulas for converting from one system to the other. Between the equatorial and ecliptic system, it’s a fixed conversion, but between any of these and the altazimuth system, there’s a time factor, so you need to account for it. And as time zones vary from country to country (and sometimes within the same country!), we don’t rely on them but instead on true local sidereal time, which is basically the right ascension of the point in the sky that is due south (or north, in the southern hemisphere) of your location.
You also mentioned, in the beginning of your question, orbital elements, but these are completely independent of the coordinates system being used, although it is standard to consider them in the ecliptic reference system.


Is it possible to have a part of space devoid of matter?

OK, so I don't think the article is claiming that it is a place "without matter", but that it is a large region of space where the density of matter is lower.

The original research, published in 2007, suggests that a lack of radio sources seen in a particular direction coincides with a "cold-spot" that is seen in the cosmic microwave background by WMAP (which has since been confirmed by Planck). They calculate that if the void were completely empty it would need to be about 280 Mpc across and at redshifts $z\leq 1$.

Galaxies certainly could be seen on the other size of this void. Many of the galaxies in their survey must have considerably higher redshifts and are more distant.

The cold spot could be caused by such a void, because light is redshifted as it travels through gravitational fields. The Sachs-Wolfe effect, as it is known, causes large scale perturbations in the cosmic microwave background as light travels from the distant past, when the CMD was generated, to us in the present day. It is argued that a large void that is comparatively empty of both visible and dark matter, but which has dark energy. The way it works is that photons works against gravity to enter a void, loses energy, but then gain back some of that energy as it leaves the void. However, the expansion of the universe means that in the time taken to cross the void the amount that is regained is not as much as was lost in the first place.

"What if a black hole were to appear"? Not clear what you mean by this. Black holes are black, you wouldn't necessarily see them, however if the "void" or lack of radio galaxies was compensated for by an equivalent amount of mass that was for some reason invisible then it would not explain the CMB cold spot.

The idea of a "supervoid" causing the CMB cold spot has also been supported by other surveys of extragalactic sources (Szapudi et al. 2015), but you are correct that the size of this void is difficult to understand in terms of current cosmological assumptions.


Can matter fly directly into a black hole and avoid the accretion disc?

In your analogy: A satellite stays in orbit because it has angular momentum. This always balances the force of gravity, thus when being in the satellite (or in the ISS) it seems as if you're weightless, because force balance. This is very important to understand, because it is the reason that in space all orbital motions can -in principle- go on forever. Because angular momentum is conserved.

Now in order to crash into the Earth/BH you need to have very low angular momentum (which is equivalent to having only linear momentum straight down into the BH) or you have to loose the angular momentum somehow.
Near Earth, this happens to the ISS, because it has a low-earth-orbit between 350-450 km and at this height there are still enough atmospheric particles around to produce drag.

Around black holes it is a very similar thing.
I will define the Schwarzschild-radius $R_s = \frac{GM}{c^2}$ as a relevant length-scale. Then, as long as we are outside $3 R_s$ distance from the BH, we need, as on Earth, friction to dissipate our angular momentum. In the case of accretion discs that form around BH's this can be accomplished through turbulent dissipation as laid out in the famous paper by Shakura and Sunyaev.

Then, only very close to the BH, namely at a distance $< 3R_s$, space-time distortions help us to fall into the BH without the need for any additional source of friction. As you see, this is a region very close to the Black Hole and it is misrepresented in popular accounts of BHs that all matter orbiting it will magically fall into it.


How large could gravitational waves get and what effect could they have on us?

Wave strain scales as $1/r$; the waves that were detected came from a source about a Gpc away and had a strain of $10^{-20}$.

To get a strain that would affect things at the 1% level, you would need to be $10^{18}$ times closer, or at about 30,000 km from the merging black holes.

The tidal force due to the black holes is roughly $GMmh/4r^3$, where $h$ is your height, $m$ your mass and $M\sim 60M_{\odot}$ the combined mass of the black holes. Assuming 1.8m height and 80kg mass, the tidal force stretching you is around 50N. So I think you would notice this more than the gravitational wave, and even more so if you were to get any closer...


I personally use Astropy, specifically astropy.io.fits, although I'm not a seasoned user of FITS files and I don't really know their layout.  As an example snippet of code, I often load data from FITS files using

from astropy.io import fits
data = fits.open('data_file.fits')[0].data

You'll find more information in the documentation on the FITS module.


How can ionized emission line flux decrease as a function of increasing metallicity or abundance?

Metallicity and abundance

Metallicity

Without specifying a given metal, the term "metallicity" — abbreviated $Z$ — usually refers to the total metallicity of all elements, i.e. the mass fraction of all metals to the total mass of some ensemble of elements, e.g. a star, a cloud of gas, a galaxy, etc. (as usual, the term "metal" refers to all elements that are not hydrogen or helium). For instance, the mass of all metals in the Sun, divided by the Sun's mass, is 0.02:
$$
Z_\odot \equiv \frac{M_\mathrm{C} + M_\mathrm{N} + M_\mathrm{O} + \ldots}{M_\odot} = 0.02.
$$

Sometimes we talks about the metallicity of a given element, e.g. oxygen. The mass fraction of oxygen in the Sun is 0.005 (i.e. oxygen comprises 1/4 of all metals by mass), so we say $Z_\mathrm{O} = 0.005$.

Unfortunately it is not uncommon to implicitly talk about the metallicity of an object, divided by Solar metallicity, such that a galaxy which has one-tenth of Sun's metallicity is said to have $Z=0.1$, rather than $Z=0.002$.

Abundance

The term "abundance" is only used for a single element. It basically expresses the same thing as metallicity, and is often used interchangeably, but is expressed in terms of the number $N$ of element nuclei, and as the ratio not to all nuclei but to hydrogen nuclei. For wacky historical reasons, we also take the logarithm and add a factor of 12. Taking again oxygen as an example, the mass fraction of 0.005 corresponds to a nuclei fraction of roughly $5\times10^{-4}$, so we say that the abundance of oxygen is (e.g. Grevesse (2009))
$$
A(\mathrm{O}) \equiv \log \left( \frac{N_\mathrm{O}}{N_\mathrm{H}} \right) + 12 = 8.7.
$$

Metallicity of a given species vs. total metallicity

In general, the ratio of a given element to all metals is roughly constant. That is, various elements are produced by stars approximately by the same amount. But various processes may cause elements to exist in various forms. For instance, metals deplete to dust, but some elements tend not to form dust, e.g. Zn. For this reason, Zn is a better proxy of the total metallicity than, e.g. Mg, since half of the Mg may be locked up in dust.

Metals increases cooling

Elements also appear in various excitation states, which depend on various processes. The lines you mention, [O II] and [O III], arise from collisionally ionized oxygen, which subsequently recombines (in my first answer I wrote, wrongly, that it was excited), and thus depend on the temperature of the gas. As the metallicity of the gas in a galaxy increases, the ratio of the intensity of these lines to that of hydrogen lines (e.g. H$\beta$) first increases, as expected. However, the increased metallicity also allows the gas to cool more efficiently. The reason is that metals have many levels through which electron can
"cascade" down.
If the electron recombines to the level where it was before, a photon of the same energy will be emitted, which itself may radiatively ionize another atom. But the many levels in metals makes de-excitation to intermediate level more probable, such that the electron cascades down, emitting several low-energy (infrared) photons, which are incapable of ionizing atoms and thus escape. The result is that energy leaves the system, i.e. the system is cooled.

This in turn means that, above a certain metallicity threshold — which is specific to a given species — the abundance of the collisionally excited lines begin to decrease. The following figure is taken from Stasińska (2002), and shows the turnover for the two oxygen lines:

This means that measuring the metallicity of a single species in general gives two solutions for the total metallicity. Luckily, as the turnover is different for different elements, measuring the metallicity for several species can constrain the total metallicity.