To expand a bit on FJC's answer: galactic disks are not rigid bodies, and so we should not expect them to rigidly precess.

Most disk galaxies do show warps at large radii (including our own), in both the interstellar gas and the stars. From observations of warps in gas disks, we know that the azimuthal angle of the warp often changes with radius: i.e., the tilt of the disk changes as a function of radius.[*] Since the angular rotation speed changes as a function of radius,[**] this suggests the tilt could precess at different rates for different radii.

Unfortunately, since there isn't really a consensus on what causes or maintains warps, it's very hard to go from observing a warp to predicting what the underlying precession rates might or might not be. (Although one can argue that the precession rate can't vary too much with radius, otherwise the warp would rapidly "wind up" and lose its coherent warped shape, and we wouldn't see warps as often as we do.)

See here for some discussion of the theoretical issues.

[*] This is a kind of precession, although it's a change in the orientation of the rotational axis with radius rather than a change with time (as in the Earth's precession).

[**] Even if the circular velocity is approximately constant, as it often is in the outer regions of galaxies, the increase in the circumference as a function of increasing radius means the angular velocity goes down.


Why does the concentration-mass relation of galaxy clusters have a negative slope?

The hierarchical picture of cosmology posits that the smallest structures formed first, early in the evolution of the Universe, and that more massive systems assemble from these initial structures.  The smallest structures therefore have the highest concentrations according to the picture you present, as the Universe had a higher mean density at the time. As the Universe evolves, the average density decreases, and the largest systems form from the smaller structures. Therefore, the higher the mass of the system, the smaller the concentration, whereas the smallest structures keep their initial concentration from their formation time.

The key here is that smaller structures are older than larger structures, therefore have a higher concentration.


Spinning black hole vs non spinning black hole

It makes a difference, but not to the strength of the gravity.

Around a spinning black hole (or any other spinning mass) space time is dragged. This effect has been measured around the (spinning) Earth by Gravity Probe B. It would cause a plumb line not to point directly at the centre of the Earth, but slightly forward.

For a rotating black hole, there is a region close to the event horizon, but outside it, in which the gravity (a combination of downwards and sideways accelerations) is so strong that it is not possible to remain stationary, relative to a distant observer, for to do so one would have to travel faster than the speed of light (relative to the local spacetime)

A similar situation exists in a non-rotating black hole, but only behind the event horizon.

So a rotating black hole does not have stronger gravity, but it is different: It is sideways.


When will the Sun end all human life on Earth?

The Sun is gradually getting larger and brighter. In fact, as called2voyage pointed out, its brightness is increasing by 1% every 100 million years. You can see how the Sun will change in the future from this graph: (Source)

According to this paper, within 1 billion (short scale) years from now, the ever-increasing luminosity will have made Earth nearly uninhabitable. The average temperature will have reached 47°C, compared to its current 15°C. Essentially no water will be left either, except at the poles. This may allow for simple life to survive for a while.

By 3.5 billion years, Earth will no longer resemble its current self. Its oceans, magnetic field and ozone layer and plate tectonics will be no more. Its surface temperature will skyrocket to roughly 1,330°C, hot enough to melt surface rock. No longer will our planet resemble a pale blue dot, and it will be more like Venus. Our planet is officially dead, along with all life on it. (Source)

~4.5 billion years from now, the Sun will become a red giant and possibly consume Earth. However, according to this paper, it may heat up potentially habitable bodies like Triton, to the point where they would support life. Unfortunately, the Sun won't remain in this stage for long enough — life usually takes billions of years to develop.


As I have discovered afterwards, the described bubble exists and is in fact a known phenomenon. For those, who want to read more, here is a wiki link http://en.wikipedia.org/wiki/Local_Bubble and the links therein.

Otherwise, it is an underdensity in ISM (not only dust) in the region of Orion Arm (which is a minor spiral arm). The underdensity is up to ten times comparing to normal density in the Milky way, depending on the component of ISM in consideration. The shape of the cavity is complex, typical sizes being of order $100 \textrm{pc}$, but when it comes to practical observations, one should be wary when neglecting dust extinction at distances over $50 \textrm{pc}$.


A vertical stick's shadow at solar noon should be straight north/south, right?

Your understanding is correct, the shadow of a vertical stick on a level plane always points north or south at solar noon, except in the Tropics where sometimes it will not have any shadow due to the Sun being at the zenith, and at the Poles, where solar noon is undefined. The sun can also be below the horizon at solar noon: places inside a polar circle can experience polar night, when the Sun does not rise for at least an entire day. On those days, even at solar noon, the Sun remains below the horizon.

This erroneous passage seems to originate from edits on 13 April 2012. The original text made more sense:

  On the Equator, the sun will be straight overhead and a vertical stick
  will cast no shadow at noon (solar time) on March 21 and September 23,
  the equinox. 23.5 degrees north of the equator on the Tropic of
  Cancer, a vertical stick will cast no shadow on June 21, the summer
  solstice for the northern hemisphere. The rest of the year, the noon
  shadow will point to the North pole. 23.5 degrees south of the equator
  on the Tropic of Capricorn, a vertical stick will cast no shadow on
  December 21, the summer solstice for the southern hemisphere, and the
  rest of the year its noon shadow will point to the South pole. North
  of the Tropic of Cancer, the noon shadow will always point north, and
  conversely, south of the Tropic of Capricorn, the noon shadow will
  always point south.


Why doesn't the vertical light beam get out of a black hole?

There is no "up" direction within the event horizon.

Most people get fixated on the speed of light, or energy or whatever. They're like, if light was faster, could it escape the black hole? If my rocket had bigger engines, could I escape? The problem is, all these questions make no sense. You can't get out because there is no way out.

A black hole is formed when gravity is so strong, it ties spacetime into a giant knot. Space is tied into itself. It's not just a little bent; it gets curved until it closes onto itself.

Inside the event horizon, all paths lead to the center. No matter how you're turning, which direction you're looking, you're actually facing the center. It's hard to visualize, but that's how it is. This is not normal spacetime, it's something unlike anything you've thought about.

Starting from point A inside the event horizon, there is no path you could draw that leads to point B outside. All paths lead to the center. Spacetime is really sick and broken.

This is the real reason why nothing gets out of a black hole.


Is it possible to see a spacecraft orbiting the moon with a pair of binoculars?

No. There are currently 5 spacecraft orbit the Moon (Or the E-M L2 point). These are LRO, Artemis P1 and P2, Chang'e 5 T1, and the Queqiao relay satellite. Of these, the largest is probably LRO. The solar panels, which are the largest part of most spacecraft, "he three panels together are 168 inches wide and extend out from the spacecraft 126 inches". Converted to meters, that is 4.25 x 3.2m. Just to give an idea of how visible that would be, Elon's Tesla Roadster, according to JPL Horizons, when it was at the distance of the Moon, had an apparent brightness of about 15.27. Using this chart, an aperture of about 50 cm is required. LRO should be a bit brighter, but not enough to make a huge difference. And none of this takes in to account the glare of the Moon itself. Whatever you saw, it probably wasn't a satellite.

I suspect it was a star of some kind. If you can provide the date and time we might be able to figure out a candidate star.


Can moons exist inside a planet's roche limit?

You've answered you own question. If there are cohesive forces beyond that of simple self-gravitation, then objects can survive intact inside the self-gravitating Roche limit - as does every solid item on the surface of the Earth for example.
Saturn's rings are made of ice, not rocks.
The tensile strength of ice is about $10^6$ N m$^{-2}$.
For a self-gravitating body, the Roche limiting radius is about
$$ R_L \simeq 1.4R_p\left(\frac{\rho_p}{\rho_s}\right)^{1/3},$$
where $R_p$ is the radius of the planet and $\rho_p$ and $\rho_s$ are the densities of the planet and satellite respectively. This is for a rigid object. The leading constant would be bit higher at about 2.4, for a fluid.
Thus the self-gravitating Roche limit does not depend on the size of the satellite, it just depends weakly on its density.
However this limit becomes much larger if the cohesive forces holding a body together exceed the gravitational forces. This becomes true for smaller objects because the gravity at the surface of an object does depend on its size, whereas the cohesive forces do not.
A critical size will depend on the geometry, density and tensile strength. Roughly speaking, gravity and the cohesive forces will be equivalent for an satellite of radius
$$ r \sim \frac{T}{ \rho_s  g},$$
where $T$ is the tensile strength and $g$ the surface gravity. But
$$ g \simeq GM_s/r^2 = \frac{4\pi G}{3}\rho_s r$$
Thus
$$r = \left(\frac{3T}{4\pi G}\right)^{1/2}\rho_s^{-1}$$
For ice of density $1000$ kg m$^{-3}$ and $T\sim 10^6$ N m$^{-2}$, then $r = 60$ km.
This rough and ready analysis suggests that ice objects smaller than 60 km could survive inside the self-gravitating Roche limit.
FYI, the innermost named moon of Saturn, Pan, has a radius of about 15 km, an icy composition and is inside the Roche limit. There are other satellites in the ring system (Epithemius, Pandora) with radii of 30-60 km. Whereas the largest inner moon, Mimas with a radius of 200 km, is in an orbit at just over 3 Saturnian radii and outside the Roche limit.


What is a debris field called before it becomes a planetary ring?

I don't think such an intermediate state exists for very long. Once you shred the moon, it's shredded and the particles orbit on locally Keplerian orbits.
This would look like Saturns rings, which are also well inside the planet's Roche limit.

The intermediate state you propose to name would have to form either
a) a steaty state of in some dynamic variable or
b) have a sufficiently long lifetime (which would be your job to prove)
in order to earn its own name.

Edit: As OP has asked for material on ring spreading, I can point to A. Crida's work who had a bit of a controversial idea about the origin of the gas giant moons. However his calculations involve also classical calculations of collisional disc spreading.


An adiabatic process in physics is at first a process where a system of limited size does not exchange any heat with the outside world, or where the energy exchange timescale is very long compared to the interesting timescale under which the adiabatic process occurs.

It is also loosely used as a term meaning 'constant in the framework of a certain approximation'. One sees this often used in quantum mechanics where it essentially means 'quasi-static', or in plasma physics where particles can gyrate in a field gradient. If the field gradient is much smaller than the gyration radius, such that every individual particle gyration experiences a near uniform field, then the particle experiences an adiabatic change of the underlying field.

Coming back to planetary/stellar atmospheres, here the two concepts are used interchangeably. Should a piece of atmosphere experience a super-adiabatic temperature gradient, then this situation is usually unstable towards buoyant instabilities. As soon as convective motion sets in, the temperature gradient will be brought down towards the adiabatic one.
This is a very general process and happens in stars as well as planets, if no other physics is present. Thus it is used interchangeably, while people using it are aware that this is a bit sloppy.


The structure (mass versus radius and density profile) is influenced by its rotation rate, but not by as much as you might think.

Even in Newtonian physics you can think of a mass element $m$ at the surface of a star of mass $M$ and radius $R$, rotating with angular velocity $\omega$.

A condition for stability would be that the surface gravity is strong enough to provide the centripetal acceleration of the test mass.
$$ \frac{GMm}{R^2} > m R \omega^2$$
If this is not satisfied then the object might break up (it is more complicated than this because the object will not stay spherical and the radius at the equator will increase etc., but these are small numerical factors).

Thus
$$ \omega < \left(\frac{GM}{R^3}\right)^{1/2}$$
or in terms of rotation period $P = 2\pi/\omega$ and so
$$ P > 2\pi \left(\frac{GM}{R^3}\right)^{-1/2},$$
is the condition for stability.

For a typical $1.4M_{\odot}$ neutron star with radius 10 km, then $P>0.46$ milliseconds. A proper General Relativistic calculation of this limit would give a similar result, but depends to some extent on the equation of state of the neutron star.

Happily, this is easily satisfied for all observed neutron stars - they can spin extremely fast because of their enormous surface gravities and all are well below the instability limit. I believe the fastest known rotating pulsar has a period of 1.4 milliseconds.

You also ask how pulsars can attain these speeds. There are two classes of explanation for the two classes of pulsars.

Most pulsars are thought (at least initially) to be the product of a core-collapse supernova. The core collapses from something a little smaller than the radius of the Earth, to about 10km radius in a fraction of a second. Conservation of angular momentum demands that the rotation rate increases as the inverse of the radius squared. i.e. The spin rate increases by factors of a million or so.

Pulsars spin down with age because they turn their rotational kinetic energy into magnetic dipole radiation. However, the fastest rotating pulsars - the "millisecond pulsars" are "reborn", by accreting material from a binary companion. The accreted material has angular momentum and the accretion of this angular momentum is able to spin the neutron star up to very high rates because it has a relatively (for a stellar-mass object) small moment of inertia.


67P/Churyumov–Gerasimenko is actually not really warm when compared to these bodies.
You quoted figures for the minimum temperatures for Mercury and the Moon. Those are accurate but misleading. Both are for nighttime temperatures, taken from locations on the bodies that are facing away from the Sun. Daytime temperatures can skyrocket, to 427 degrees Celsius during the day on Mercury. The Moon can also get very hot during the day, although nowhere near as hot as on Mercury,
Looking at it like this, 67P/Churyumov–Gerasimenko's surface temperatures are pretty moderate, and nowhere near extreme. Rubin et al. (2015) suggest that it formed in extremely cold conditions - albeit ones not too extreme, for a comet.
The ESA has something related to say on the matter:

At these distances, the comet covered only a few pixels in the field of view and so it was not possible to determine the temperatures of individual features. But, using the sensor to collect infrared light emitted by the whole comet, scientists determined that its average surface temperature is about –70° Celsius.
The comet was roughly 555 million km from the Sun at the time — more than three times further away than Earth, meaning that sunlight is only about a tenth as bright.
Although –70° C may seem rather cold, it is some 20–30° C warmer than predicted for a comet at that distance covered exclusively in ice.

That said, evidence from the mission has made scientists think that an surface with a lot of dust may be the norm for comets.
67P/Churyumov–Gerasimenko isn't warmer or colder than the other bodies at this distance from the Sun. It's simply milder.

I would wager that the reason Mercury and the Moon can get so cold is because they don't really have atmospheres. Over the course of a day, Earth's atmosphere heats up a bit, and over night, it retains that heat, meaning that the surface can stay relatively warm. Mercury's atmosphere is extremely tenuous. The same goes for the Moon.
This might not be an issue, but one day on Mercury lasts 59 Earth days. One day on the Moon lasts for about one Earth month. This means that while there's time to trap some heat, there's also the same amount of time to lose it, and without an atmosphere, the heat gained is soon be lost.
67P/Churyumov–Gerasimenko also doesn't have an atmosphere to speak of, but it rotates quickly - rotating once every ~12.5 hours. This isn't necessarily perpendicular to the orbital plane, so not every part of it will face the Sun during a rotation, but it does mean that there's not as much time to lose heat.


Why aren't all planets in the same plane?

Your reasoning is correct: if Mercury orbited in the same plane as Earth, we'd see it transit the Sun every 4 months or so. In fact these orbital planes are inclined 7 degrees to each other, and the other major planets' orbits are inclined 1 to 3 degrees relative to Earth's.

The planets perturb each other's orbits slightly, so no planet's orbit is perfectly planar.
However, the Solar system average plane of all orbits and rotations is
invariable,
and most individual planets' orbits will remain near it for millions of years.


What is the most distant observable White Dwarf known?

There is the unusual white dwarf RX J0439.8−6809.  It seems to be a carbon-oxygen white dwarf with a temperature of around 250,000 K!  It is 9200 pc away.  You might want to look for other PG 1159 stars to see how far they can be detected.


How would a planetary defence against comets work?

They used several Mars and earth gravity assists to raise Rosetta's aphelion to around 5 AU. The comet's aphelion is also in that ball park. The comets from Jupiter's neighborhood (about 5 AU) are moving about 40 km/s when they're in our neck of the woods (earth moves about 30 km/s). A comet moving 40 km/s wrt the sun at 1 AU can be moving anywhere from 70 to 10 km/s wrt to the earth, depending on inclination. A head on collision would be 30 + 40. If the comet was moving the same direction as the earth, it'd be 40 - 30.

A comet from the Kuiper Belt would be moving just a hair under escape velocity when in our neighborhood. Escape is sqrt(2) * circular velocity. So a comet from the outer system would be moving 42 km/s wrt to the sun.  The velocities wrt to the earth could range from 72 to 12 km/s. The Kuiper Belt is 30 to 50 AU from the sun.

A comet from the Oort Cloud would also be moving a hair under escape velocity, 42 km/s in our neighborhood. But Aphelion would be much higher, up to 50,000 AU. Presumably many of the non-periodic comets are from the Oort. Click the comet links in this list and you can compare discovery date with last perihelion. For example Comet C/2011 W3 (Lovejoy) was discovered November 27, 2011 and it's .55 AU perihelion occurred December 16, 2011.

In JPL's description of the LINEAR NEO search program, they say "… with most of the efforts going into searching along the ecliptic plane where most NEOs would be expected. " But objects from the Oort don't tend to the ecliptic plane. In my opinion a high inclination comet from the Oort with our name might escape undetected until it was way too late to do anything about it. But an Oort comet with our name on it is very unlikely.


Planet solidifying.

No molten core iron -> no magnetosphere -> solar wind strips atmosphere.
Without atmosphere to cycle in exposed water also leaves.

Why did Mars freeze solid & Earth has not (yet).

is further from sun
smaller means less volume to hold heat
Earth may have the iron cores from two planets (moon forming impact)
Radioactive decay?

On oil, possibly.
If there was life there and it colonized the land and it found it needed to create a scaffolding structure to compete for photons such as ligand as happened here, and there were several hundred millions years until other life there figured out how to digest those ligands then there would be a layer there like our Carboniferous era in which a large amount of carbon is sequestered underground.

But chances are vanishingly slim.
If there was life there it most likely

recycled carbon just as we have here for all time except that little blip* after trees showed up and before bacteria figured out how to digest them.
did not survive long enough to need to compete for photons by growing very tall

(*)  300,000,000 / 4,300,000,000  is about 1/14th of life on earth's history.


Creating a 2D matrix from 2 FITS images of unequal dimensions for Radio/FIR correlation

If you have the WCS for both images, you can use the astropy reproject module. Once you have projected image 1 onto image 2, the (x,y)th pixel from the output will correspond to the (x,y)th pixel from image 2.

Then you can flatten both images with numpy, and assemble them into a 2-column array.


Above I have plotted few Planck's radiation contribution:
The violet is due to CMBR, Green is due to Milky way Galaxy, and Blue is due to contribution from both.

One way to resolve out CMBR is by modeling milky way from higher frequency of the graph since CMBR contribution is negligible at higher frequency. At particular direction one can associate temperature due to Milky way by setting a high pass frequency. Then one can subtract this temperature contribution at lower frequencies where CMBR contribution is significant. This is the one of the way probably not the best, but doable.

Note1: The graph is not from actual data. It is just for explaining how to go about it.

Note2: What should be high pass frequency will be a good question. With good arguments one can decide what should be the high pass frequency. And there's always way about it.


Converting comment to community wiki

Possibilities:

Aircraft lights (but only if non-blinking);
genuine satellites/old launch stages not in the database;
random ("sporadic") meteors not part of an established shower;
birds (yes, sometimes light-coloured migrating birds reflect city light from their bellies).


How to Interpret Lomb Scargle periodogram

The x-axis of a periodogram is usually a frequency, measured in inverse time units. It is possible to plot the inverse of frequency, in which case the axis would be non-linear.
In your case, the x-axis looks linear, and the the code you present appears to be using something labelled as a frequency on the x-axis. Therefore I would assume (though I would read the documentation for the functions you are using) that the x-axis is frequency. The units will be based on whatever units are in your input. i.e. It is just the inverse of whatever you are giving it for the times - in this case, $d^{-1}$.
If your sampling rate is once per day, then you are getting no reliable information at all about frequencies $>1$ d$^{-1}$. The stuff you should be concentrating on lies between $0\leq f < 1$ d$^{-1}$. Even in this range you won't have good fidelity on signals with $0.5<f<1$ d$^{-1}$ because they are undersampled. You can also expect to get aliasing signals in the range $0\leq f <1$ that are caused by true signals (including those with higher frequencies) aliasing with the 1 d$^{-1}$ frequency of your sampling.
It looks to me, by eye, like you might have a genuine signal at $1/7$ d$^{-1}$, corresponding to a weekly variation. There may be a weaker signal at much lower frequencies that might correspond to a monthly signal. The 2 signals either side of 1 day look like aliases of the weekly signal with the sampling frequency.
I would highly recommend reading VanderPlas 2018 for a thorough discussion of the Lomb-Scargle periodogram and its interpretation.


According to the NASA page about news and events related to the moon, the start of October, 2016 saw the moon enter a New Moon phase - where it is not visible, as explained on the site:

  the new moon isn't visible, because it's between Earth and the sun, and the unlit side is facing Earth.

The 'unlit side' is often referred to as being the 'dark side' of the moon.

and as shown in the terribly-but-necessarily-not-to-scale diagram below:

Image source and further information: SEMS: Moon Phases

The lunar cycle is roughly 29 days (the approximate time between consecutive new moon cycles as an example).  There are many online 'lunar calendars' are available, such as from Moon Connection.

However, it is just as easy to observe the lunar cycles throughout the nights (weather permitting) - when the lunar cycles are heading towards a 'new moon', then it will appear to be 'waning', 'waxing' refers to the image of the moon 'growing' towards a full moon.  Examples of how the moon would appear is shown in the NASA image below:

(Image from the NASA page linked above) - the cycle progresses anti-clockwise in the diagram.


Yes, it is correct. For better precision:
Declination =23.5*sin(longitude)
This is true only if you don't take into account the eccentricity of Earth's orbit and other movement. It isn't easy to understand why, but I'll try ti explain it.
The different declination of the sun during the year is generated by the different angle of inclination of the Earth (the north pole point always in the same direction, so during the earth revolution the poles don't point always in the sun direction and the protection of the inclination on the sun-earth plane change during the year)
So, what wee see of the sun is the projection of the circular movement of the earth on a line of sight, so an harmonic motion.


How long was a lunar month in ancient times?

'Day' is the worst possible unit to measure the revolution of the Moon around Earth as it is the temporal unit which is directly influenced by the tidal friction and deceleration. We can reasonably assume that the length of the year is not impacted by lunar tides and subject to much less variation.
This is only a partial answer as I will have to leave the math to get an actual number for time reasons.
The only measurable quantity for the past is the length-of-day of Earth and the tidal rythm by means of their impact on biorythms and in turn its impact on sediment stratigraphy and growth markers e.g. in fossil shells and similar.
One summary with some values and further references on how the past rotation can be estimated is this paper by Deines & Williams. If you know the length-of-day at a time, you can calculate the yearly angular deceleration of Earth. That knowledge can be employed, using the conservation of angular momentum and calculate the orbit of the moon and thus length of month (thus one full orbit around Earth).


Can life on earth be supported by only a Red Dwarf star or a White Dwarf(state still hot not cooled enough yet)?(Provided it is near to the star)

Certainly a red-dwarf star can have enough energy for a planet around it to be in the goldilocks zone.   There are some difficulties with red-dwarf stars and Earth like planets.   The planet would need to be very close to the star and as a result, tidally locked.   The orbital period would be quite short, so there would be no seasons and one side of the planet would be in permanent sun the other side in permanent shade.

Red-Dwarf stars, unlike our sun, grow colder with time and they can be quite magnetically active, sometimes called flare stars, though not all red-dwarf stars are flare stars, many start out that way.   The proximity to that much solar-ejection could strip planets of their atmosphere early in the planet's life.

None of that makes it impossible.  It's certainly possible for a Red-dwarf star to host an Earth-like planet, but there's some difficulty.  Orange or Yellow stars are, on average, considered better candidates.

A white dwarf, however, likely not possible.   A white dwarf follows the red-giant stage of a star.  It would be too small and give off too little energy and any planets that survived the red-giant phase would likely be very distant.   Life and/or an Earth-like planet around a white dwarf is virtually impossible.


Why do J2000 coordinates change with time in Stellarium?

Apparently, Stellarium's "J2000.0" reports the coordinates of stars in J2000 frame, but at the epoch of the date you specify, instead of reporting the coordinates of the star at epoch 2000.0, which can be misleading to say the least. So the difference in J2000 coordinates that you noticed corresponds to the proper motion of the star. And the star you picked, 61 Cygni, actually has one of the largest proper motion in all of the catalogued stars, with about 4.1"/year in RA and 3.2"/year in declination. Looking at the differences that Stellarium reports in your example, we can notice that it matches perfectly with this proper motion : 31.3" - 28.2" = 3.1" ± 0.1" in declination, for RA: (55.27s - 54.92s) x 15 x cos(declination) = 4.1" ± 0.1".

If you pick a star with a smaller proper motion in Stellarium, like Deneb (only 2 mas/year for each axis), the J2000 coordinates change much more slowly.

As for the difference between J2000 RA/Dec and apparent (on date) RA/Dec, this is caused by the fact that the J2000 frame's equator/equinox was fixed at the mean position of the equator/equinox in 2000 (mean position means that the small periodic effects of nutation are averaged out). So the apparent coordinates on Jan 1 2000 are slightly different because apparent coordinates take into account nutation N of the rotational axis R around the mean P (precession):


Yes, you can have more than one ring. Saturn is an example of a planet with multiple rings on the same axis separated by divisions (gaps between each ring).

About rings that are not on the same axis (similar to your example picture) is addressed in this previous question.


Can you identify a star's 'signature' from its spectroscopy?

Realistically, no - there are too many stars with the same spectrum.

The pulsar idea is a good one, though. There is a reason our location relative to some bright pulsars was sent out on the Voyager plaques: we are confident that a civilisation finding one could then find our sun, and us, as the pulsar rates are unique enough.


How do I plot galactic coordinates using matplotlib and astropy in python?

This code reads coordinates as equatorial (ra, dec) and transforms them to galactic (l, b):

eq = SkyCoord(xarr[:], yarr[:], unit=u.deg)
gal = eq.galactic

The contents of 'galacticwperiod.csv' are already in galactic coordinates and should not be transformed.
Something like this may give better results:

gal = SkyCoord(xarr[:], yarr[:], frame='galactic', unit=u.deg)

The other issue is that pyplot's geographic projections seem to expect angles in radians.
This code:

plt.subplot(111, projection='aitoff')
plt.grid(True)
plt.scatter(gal.l.wrap_at('180d').radian, gal.b.radian)

produces this plot:


Fact Check: How bright will the Perseids be in 2017?

There is variation in meteor showers from year to year, but for most showers, the big difference is how bright the moon is. If the moon is up, then it's hard to see many meteors.

The Perseids of 2017 will be rather ruined by the moon. There will be a waning gibbous moon, rising shortly after sunset and staying up for the rest of the night. It will make the sky too bright to see many meteors

2018, with a new moon on the 11th, will be much better.


The right ascension of the point on the horizon and due west is LST - 6 hours, where LST is the Local Sidereal Time. The LST can be calculated from knowing the date, time, and longitude. The declination of the west point is 0 degrees. So the angle between that right ascension and 0 hours right ascension (the vernal equinox) is just the difference between the two.


Electron Degeneracy Pressure and the Pauli Exclusion Principle

Basically, the Pauli exclusion principle says that two fermions (in this case, electrons) can't be in the same quantum state. To expand: No two electrons in an atom can share the same numbers for their four quantum numbers, properties that help describe the state of a particle. What are quantum numbers? The important consequence here is that no two electrons can have the same spin and energy level.

In a white dwarf or a neutron star, the fermions are packed very close together, and there's quite a lot of force due to gravity. However, the exclusion principle triumphs. Fermions near each other must have different energy levels; this leads to energy differences and degeneracy pressure pressure, which counteracts the force of gravity. Above a certain mass limit (the Chandrasekhar limit, roughly $\sim1.40M_{\odot}$), electron degeneracy pressure is no longer sufficient; the white dwarf collapses to a neutron star. There appears to be a similar limit for neutron stars, where neutron degeneracy pressure cannot support the remnant against gravity, and it collapses into a black hole.


As Wikipedia explains in the article on the Zodiac, the Zodiac was originally developed with the assumption of fixed equinoxes. It is assumed this is because the Babylonians who developed the Zodiac were unaware of the precession of the equinoxes. At that time, about 2500 years ago, the vernal equinox did in fact lie in Aries. For this reason, in astrology, that time is referred to as "the Age of Aries".

Of course, in modern times these astrological concepts are scarcely important to astronomy, and in the few cases where the astronomical symbol for the vernal equinox is used, it is more practical to have a fixed symbol than a changing one.


Can a galaxy splits into multiple galaxies?

Small galaxies (say of masses $\sim10^7M_\odot$) can be torn apart by stellar feedback, i.e. the energy provided by exploding stars in a starburst. Whether you want to call the debris galaxies is a matter of opinion, I suppose, but since these guys typically are satellites of larger galaxies, it will probably be torn completely apart by tidal forces, eventually accreting onto the larger one.

Large galaxies (say of masses $\sim10^{11}M_\odot$) are too massive to be split by feedback, but may lose matter (mostly gas, rarely stars) through large-scale outflows created by stellar feedback (or a quasar jet). However, when two large galaxies collide and merge, sometimes large chunk of gas, stars, and dark matter are slung out, which arguably could be called individual galaxies. Most of the matter of the two original galaxies become one, though.


Green objects in Hubble Ultra Deep Field

what are the green objects that you see scattered in the Hubble Ultra Deep Field image

The Hubble Ultra Deep Field (HUDF) has approximately 10,000 objects in it. Between 25 and 50 of them are identified as extremely faint stars. That means the remaining ~10,000 objects are galaxies. To answer your question literally then, the green objects are galaxies.

However, I believe what you probably really mean to ask is, why are these galaxies green? To answer that, there are actually a number of points to address.

You seem to be under the belief that the concept of redshifting causes objects to appear red. While this is certainly a true statement, it is not exclusively true. Redshifting simply makes wavelengths longer (and thus shifted red-ward, towards the red end of the spectrum). It is certainly possible that an object which was once blue could be redshifted into the red and thus appear more red than it naturally is. However, an object that emitted most strongly in the UV could redshift into the green and thus appear more green than it naturally is. Some such effect may be going on in this image, but in actuality, I think there's other stuff going on.
You have to account for image processing. It's very difficult to take a "real color" image of a galaxy. Very often (as was the case with the HUDF), images are taken in a few different color bands and combined together with some sort of image processing program. Its up to the discretion of the image processor to balance the colors correctly, get the right saturation, hues, luminosities, etc. There are actually a variety of different forms of the HUDF that include different bands and different forms of image processing where the images that appear green in your image don't appear green in those. An example is shown below of a rendition of the HUDF, taken directly from the same NASA/STScI Hubble website you found your image on. The coloration is significantly different than yours and many of the greenish galaxies from your image appear more blue.

I'm not saying this is the case with the galaxies in the HUDF, but some galaxies actually are green! A few examples include the Pea Galaxy or the Green Bean Galaxy. When you plot up galaxies based on color (note that this is the astronomical definition of color which doesn't necessarily have a 1-1 correspondence with visual color) you get a Color-Magnitude diagram (CMD) wherein most galaxies are clumped into a red portion or a blue portion. However, there are a small group of galaxies in what is known as the Green Valley which do appear green for one reason or another. What actually causes their greenness, I think, is beyond the scope of this question and would be better suited for a new question.


In km/h, what actually is the "speed" of Andromeda away from us: cosmologically?

The rate of expansion, measured in the customary units of (km/s)/Megaparsec
is not known with great accuracy. Recent measurements include 67.6 (SDSS-III), 73(HST) 67.8 (Plank) 69.3 (WMAP) [wikipedia]

The Andromeda galaxy is 0.78 Mpc from us, so taking the Hubble constant to be about 70, gives a recession of about 55 km/s. This is not a very great speed: compare with the orbital velocity of the sun around the galaxy at over 200 km/s, or the escape velocity of the galaxy (over 500km/s)

As you note, this is pretty much swamped by the proper relative motion of our galaxies. Its blueshift indicates that Andromeda is approaching us at over 100 km/s. For galaxies outside of the the Local group, the Hubble flow dominates.

Now the value of 55km/s assumes that space is smooth and homogeneous. This is approximately true on a universal scale, but it is not true on the scale of a galaxy cluster, where local gravitational effects dominate the curvature of spacetime. The general expansion of spacetime has very little effect on the motion of galaxies in the local group, as discussed by Iorio's paper on the motion of a gravitationally bound binary system


How can we tell the age of a rogue planet?

At the moment there is basically only one way. That is to associate the planetary-sized object with a cluster of stars or moving group of stars of known age.

That's basically it. If the planetary-sized object really can't be associated with another object, then only limits can be placed on its age by comparing it's luminosity to theoretical planet cooling models. But to use these models you need to know the mass!

So at the moment, the only "free-floating" planetary mass objects known (they could all be low-mass brown dwarfs actually) are those for which an age is estimated by association and therefore a (planetary) mass has been deduced from cooling models.


Emission measure is (usually) used in X-ray and EUV astronomy, though I suppose also in cases of optically thin radio emission. It is defined as the square of the number density of free electrons integrated over the volume of the plasma.

$${\rm EM} = \int n_e^2 \ dV$$

The flux of optically thin emission from a plasma (e.g. thermal bremsstrahlung) is then directly proportional to the emission measure of the plasma multiplied by a temperature dependent cooling loss law.

In other words, when you measure the flux of X-rays from an unresolved optically thin emitter, there is a degeneracy between the electron number density (squared) and the overall plasma volume.

When you fit an X-ray spectrum with an optically thin model, the emission measure (divided by $4\pi d^2$, where $d$ is the distance to the object), is a multiplicative free parameter.

Your question about calculation is extremely difficult to answer. Suppose I measure a count-rate of $N$ X-ray counts per second using some X-ray telescope (I can only assume that's what you mean by a "X-ray light curve".).

The count-rate received at the telescope depends on: the emission measure (as defined above) multiplied by a term that depends on the temperature (or temperatures) of the source, the chemical composition of the source and the adopted emission process (is it free-free thermal bremsstrahlung, a thermal plasma or something else). It is then attenuated by any intrinsic absorption in the source and any absorption between us and the source and by the distance to the source (assuming the radiation is isotropic). Finally what is detected is determined by the response of the X-ray detector to X-ray photons as a function of energy.


What would happen to the solar system if Earth's mass was removed?

There would be three effects to consider, one very short-term, one medium term and one long-term.

The medium-term effect is that the Moon has an orbital velocity around the Earth of about 1 km/s.  If the Earth suddenly disappeared, the Moon would just keep on going in a straight line at 1 km/s.  This would have the effect of putting it into a Solar orbit quite different from the Earth's. It would have a different inclination to the ecliptic (the Moon's orbit is inclined to the ecliptic by 5 degrees), different period and different eccentricity. (How different depends on exactly where in the Moon's monthly orbit the Earth disappeared.)

The effect wouldn't be enough to cause the orbit to intersect that of Venus or Mars right away, but perturbations on the new Planet Luna would potentially be very different from those on Earth and it might evolve fairly quickly (centuries) to a Mars-crossing or Venus-crossing orbit with an eventual collision.

It is highly unlikely that it would be very near the spot where Earth reappears.

The long-term effect is smaller: The Earth is part of the generally stable Solar System and contributes small perturbations to all the planets. It is possible (but I suspect unlikely) that the disappearance of the Earth and the subsequent lack of the Earth's perturbations would result in significant changes to the Solar System over many centuries to tens of thousands of years.

The very short-term effect is that the disappearance of the Earth's mass would result in a speed-of-light propagation of an abrupt change in the curvature of spacetime.  I.e., gravitational waves.  I don't know enough to calculate if these waves would have any noticeable effect within the Solar System. My inexpert guess is that it's plausible that they would.


If an asteroid twice the size of Earth passed super close would half of the Earth be pulled towards it?

In a close approach, Earth would feel tidal forces: the near side to the approach would be affected by the vistor's gravity more than the far side, because the force of gravity decreases with distance from a mass. If you were standing on the far side, though, your effective weight wouldn't increase -- it would decrease! This is because the visitor would be pulling the center of the Earth more strongly than it would be pulling on you, again because of the gravitational falloff with distance.

In a very close approach, closer than the Roche limit, the tidal forces would overcome the Earth's gravity completely, and Earth would begin to break apart. So it matters very much how close an approach we're talking about. Assuming the visitor has the same density as the Earth, the critical distance is about 10 Earth radii, or 64000km -- about a sixth of the distance from the Earth to the moon.


Is the mechanism of solar flares on red dwarfs and brown dwarfs the same as that on the Sun?

Yes, the basic mechanism is thought to be the same on red dwarfs and at least the hotter brown dwarfs, but the details can be different.

As you say, magnetic reconnection in the corona is the starting point. Well, actually it is fluid motions at the magnetic loop footpoints that is the starting point. The B-field and partially ionised plasma are coupled and photospheric motions put (magnetic) potential energy into the B-field structures.

This potential energy can be released suddenly in reconnection events. These can drive coronal mass ejections or accelerate charged particles along the loop field lines.

A flare occurs when a significant amount of energy goes into accelerating charged particles down the field lines towards the loop footpoints. These charge particles emit radio waves and then non-thermal hard X-rays as they impact the thicker chromosphere/photosphere. Their energy is then thermalised, heating the chromosphere and possibly causing hot ($>10^{6}$ K) material to evaporate into the coronal loops. Here it can cool by radiation and conduction before falling back, or perhaps forming cool prominences.

Similar things must be going on in low-mass stars. The X-ray and optical light curves of their flares do resemble what is seen on the Sun as do the relationships between hard- and soft X-rays and the evolution of plasma temperatures. The details may be different because the temperature and density structure of their photospheres, chromospheres and coronae are a bit different to the Sun's, and there are some indications that their coronae can be denser or sometimes that flares occur in much larger structures than are seen on the Sun. "White light" flares are also more common in M-dwarfs.

Why are flares on red dwarfs so powerful? Partly it is contrast - you are comparing the flare emission with something that is intrinsically less luminous. In absolute terms the flares on the Sun and flares on red dwarfs are not hugely different. What is different is that the flare power as a fraction of the bolometric luminosity and the frequency of large flares can be much higher on M-dwarfs.

The underlying reasons are likely to do with the magnetic field strengths and structures on red dwarfs. Magnetic activity is empirically connected to rotation and convection. Magnetic activity is higher on rapidly rotating stars and those with deep convection zones. M-dwarfs have very deep convection zones (or are even fully convective). They also tend to rotate much more rapidly than the Sun, since their spin-down timescales appear to be much longer than G and K dwarfs. They are thus more magnetically active relative to their bolometric luminosities. It appears that active M dwarfs have very strong magnetic fields (like those in sunspots) covering a very large fraction of their surfaces and this along with the convective turbulence where the magnetic loop footpoints are anchored likely leads to their strong flaring activity.

Brown dwarfs are a bit more tricky. The younger, hotter ones probably behave much like low-mass M-dwarfs (in fact they are M-type objects). The magnetic activity on cooler/older L- and T-dwarfs is much more mysterious. I guess a few flares have been seen, but it is not clear this is related to similar mechanisms as in higher mass and hotter stars. These cool brown dwarfs have neutral atmospheres and the magnetic field is not frozen into the plasma like it is in the partially ionised photospheres of hotter stars. This means that the loop footpoints may not be stressed by photospheric motions in the same way. It is not even clear that brown dwarfs generate a magnetic field in the same way as more massive stars, though it is clear that at least some of them do have magnetic fields.


Would a tablespoon of a neutron star remain intact?

If we take neutron star material and somehow transport it somewhere for examination (say the Earth!), the results would be catastrophic. At say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have a number density of $\sim 6\times 10^{43}$ m$^{-3}$ an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$ (calculated using the relevant equations for an ideal gas of degenerate neutrons at this density). So even in a tablespoonful (say 20ml, which would have a mass of 2 billion tonnes!), there is $6\times10^{27}$ J of kinetic energy (15 times more than the Sun emits in a second, or a few billion atom bombs) and this will be released instantaneously.

The energy is in the form of around $10^{39}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about a billion tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 50km radius near-earth asteroid hitting the Earth at 30 km/s.

The neutrons in a dense neutron star gas are relatively stable (beta decay is blocked by electron degeneracy). The expansion described above would allow beta decay into protons and electrons, but as this happens on timescales of 10 minutes, it is hardly relevant to the initial destruction. However, you would end up after a few tens of minutes with an expanding cloud of ionised hydrogen a few light minutes across.

The minimum possible size to gravitationally bind neutron star material is thought to be around $0.15 M_{\odot}$ (see here). The equilibrium electron density (there are always some electrons and protons present in neutron star material) for lower masses is too low to block neutron beta-decay.


If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?

TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.

thanks to PeterCordes

Jupiter's gravity will pull on the Earth itself, as well as everything on it.

It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger.

What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't "feel the tug" anywhere near as strongly as one might suspect.

Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.

Since people nearer to Jupiter will feel a slightly stronger acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute.

But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.

This kind of force is called a tidal force and this is the picture that's often used with the concept:

 Source Replace "Satellite" with "Jupiter"

The acceleration we feel due to gravity is expressed as

$$a_G = \frac{GM}{r^2}$$

where $G$ is the gravitational constant and equal to about $6.674 \times 10^{-11}$ m^3/kg s^2 and M is each mass that's pulling on you.

If you put in 6378137 meters and the mass of the Earth ($5.972 \times 10^{+24}$ kg) you get the familiar 9.8 m/s^2.

If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.

If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. But that wouldn't last very long!!! Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2.

On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.

When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.

However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure.

It would certainly be fun, but it wouldn't last long!


Telescope light gathering power and resolution

The collecting area of a telescope is roughly proportional to the square of its radius (for a circular aperture). I say roughly because you also have to factor in the small fraction of light blocked by the secondary if we are talking about reflecting telescopes.

In principle the angular resolution decreases as the reciprocal of the radius (that is, the smallest resolvable angle gets smaller). In practice, this may not be the case for ground-based telescopes without adaptive optics capabilities, where the smallest resolvable angle could be determined (for large telescopes at least) by turbulence in the atmosphere (a.k.a. the seeing).


Visible satellites real time position API

I've already used the site N2YO.COM and the API. There is a lot of satellites on it and the most important ones.

https://www.n2yo.com/api/


What spectral feature was used to confirm that S0-2 is "single and available"? What is Brγ?

Brγ is a line in the emission spectrum of Hydrogen, or single-electron (and therefore) hydrogen-like Rydberg atoms in general. In particular, it is the emission due to an electron moving from the 7th to the 4th energy level.

Br is short for Brackett, after Frederik Brackett, and gamma because

$\alpha: \ \ 5 \rightarrow 4,$
$\beta:  \ \ 6 \rightarrow 4,$
$\gamma: \ \ 7 \rightarrow 4,$

...and so on.

Brγ has a wavelength of 2166nm, as you would get from the Rydberg formula, which is well in the infra-red part of the spectrum. It is probably chosen as it has low absorption by the dust towards the centre of the galaxy.

Screenshot from here.


Is there a better explanation of Hawking radiation?

Andy Gould proposed a classical derivation of Hawking radiation in a somewhat obscure paper from 1987.  The essential argument is that a black hole must have a finite, non-zero entropy (otherwise you could violate the second law of thermodynamics with a black hole).  Moreover, the entropy of the black hole must depend only on its area (otherwise you could change the area of a black hole via the Penrose process and lower its entropy and make a perpetual motion machine).  If a black hole has an entropy and a mass, then it has a temperature.  If it has a temperature, then it must radiate thermally (otherwise you could again violate the second law of thermodynamics).

Of course, if you look at the Hawking radiation temperature, there's a Planck's constant in there, so it has to know something about quantum mechanics, right?  But it turns out that it's actually thermodynamics in general that knows about quantum mechanics, not general relativity --- Planck's constant is only needed to keep entropies finite (and therefore temperatures non-zero).  This is true of black holes and blackbodies alike.


Has anyone ever tried to make a simple telescope using ice?

I'm not sure it counts as "simple" but there is the ice cube neutrino observatory whose detector consists of a cubic kilometer of very clear ice a mile or so down in the antarctic icecap.


Why Jupiter's Great Red Spot is Reddish?

It's red, because it's a 'sunburn'. The clouds in the red spot reach to higher altitudes than the surrounding and are more exposed to Solar UV radiation, which in turn changes the structure of some of the organic molecules etc. This is at least the explanation suggested by recent data from NASA's Cassini mission, see this 5 day old press release.


What apparent sizes of the sun an Earth like planet can have?

I'll make a very simple dimensional calculation here.

Total power from of a star with radius $R$ and temperature $T$:

$$ P_{tot} \ = \ R^2 \ T^4$$

Flux $I$ (power per unit area) at an orbital distance $r$:

$$I \ = \ \frac{R^2 \ T^4}{r^2} \ = \ T^4 \ \left( \frac{R}{r} \right)^2$$

Apparent size (radians, small angle approx):

$$\theta \ \approx \frac{R}{r}$$

So:

$$\theta \ \approx \ \frac{\sqrt{I}}{T^2}$$

Surprisingly, for a fixed incident flux of light (ignoring different planetary warming effects of different wavelengths) the apparent size of a sun viewed from a star varies roughly as the inverse square of its temperature, and does not depend on the actual size of the star.

Your milage may vary.

Using Wein's dispacement law which relates the wavelength of maximum spectral radiance of a blackbody $\lambda_{max}$ with its temperature $T$ using Wien's displacement constant $b$:

$$\lambda_{max} \ = \ \frac{b}{T}$$

The apparent size scales with the square of the peak wavelength of a star's blackbody spectrum.

$$\theta \ \approx \ \lambda_{max}^2 \ \sqrt{I}$$

Yay! Reddish stars will appear much larger than bluish stars - the result we were all expecting.

And if you'd like do some landscaping, the 2008 Scientific American article The Color of Plants on Other Worlds is a good read in stellar spectra, atmospheres, and photosynthesis.


The point you appear to refer to, is called the Lagrangian point $L_1$.
This point is a saddle in the field of gravity, hence not to be considered to be stable in the strict sense.
Two other Lagrangian points, called $L_4$ and $L_5$, can be stable, provided the considered orbiting objects are of small mass in comparison to the two main bodies of the system, and if the masses of the binary components are sufficiently different.

According to theorem 4.1 of this paper, $L_4$ and $L_5$ are stable in all directions, if and only if the mass ratio of the two main binary components $\frac{m_1}{m_2}\geq\frac{25+3\sqrt{69}}{2}\approx 24.9599$.
According to theorem 3.1 of the same paper all Lagrangian points are stable in z-direction, which is the direction perpendicular to the orbital plane of the binary system.
(Credits for this corrected version go to user DylanSp.)


Why do spectroscopic binaries have approximate circular orbits?

Spectroscopic binaries are not necessarily close and not necessarily in circular orbits. If the orbit is circular, then a time series of radial velocity measurements for one or both stars would be well modelled with a simple sine wave.

However, eclipsing binaries are almost certainly close, since eclipses in a system with widely separated components are improbable (but not impossible). If you have the eclipsing light curve, then the primary and secondary eclipses will be equi-spaced in time if the orbit is circular.

The key factor for circularisation is the orbital period. If it is less than about 7 days (for solar-type stars), then circularisation due to tidal forces is almost certain.


Which moon has a scar as if it was torn in half and put back together?

I believe the moon in question is Miranda, as James K suggested. It is hypothesized that it suffered a collision with another large body sometimes in the past. There are several large features that suggest this:

Three large grooves, each > 200 km long, cover parts of the surface (the "scar" you talked about). These are the coronae.
Observations indicate that some parts of the surface appear less than 100 million years old.
Verona Rupes, a 10 km-high cliff, is the largest in the Solar System. Other, smaller, rupes exist elsewhere on Miranda and across the Solar System, but not quite as high as this one.

Alternate explanations include heavy crustal activity in the past. Tidal effects from Uranus may also have played a role (either at different times than the putative impact, or instead of it entirely). Gas giants strongly interact with their moons through tidal forces; cases include Io's volcanism, driven by Jupiter, and the covered oceans of Europa and Enceladus.

Here is a partial map of Miranda, displaying the coronae and several other features:


Do the number of moons a planet has affect the speed of the planet's rotation?

Any moon exerts a tidal force on its primary, tending to cause the primary to bulge outward along the axis passing the moon and to narrow slightly in the other two directions. This effect is greater if the moon in question is larger and/or closer to the primary, but a solid primary may largely resist it, so that it changes shape little if at all. In the case of the Earth and its Moon, the solid Earth moves relatively little, while the water of the oceans is moved quite a bit. As a result, water has to flow from one place to another over the solid crust. That flow causes friction, which causes the "bulge" of water to lag behind the Moon and pull back on it by gravity. This process gradually slows the rotation of the Earth and drives the Moon further away.

This process acts, in general to bring the rotation of the planet and the revolution of the Moon to the same rate, so that the planet keeps one face to the Moon. Usually this means slowing the planet's rotation, but a few moons are so close to the planet that it would be the opposite. This process will be less pronounced on a planet which  has a deeper liquid layer, since there will be less friction, or on a planet which has a stiff solid structure, since there will be less movement.

The number of moons is not directly relevant. To a reasonable approximation they all work independently.


Calculate the absolute magnitude for a multi-star system

Yes, you are correct. The luminosities can be added. Luminosity is the amount of electromagnetic energy emitted per unit of time (measured in $ \textrm{J} \cdot \textrm{s}^{-1} $ or $\textrm{W}$). So if you have a multiple star system, the total amount of energy emitted (i.e. the total luminosity) is simply the sum of the energy emitted by each of its components.
This does not take into account any absorption of the radiation of one of the components by one of the others.
In the book by Jean Meeus [1], the formula for adding stellar magnitudes can be found in Chapter 56:
$$ m = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 m_i} $$
This formula is for the apparent magnitude, but would also apply for the absolute magnitude, which, after all, is only the apparent magnitude at $10\,\textrm{pc}$ distance.
$$ M = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 M_i} $$
[1] Jean Meeus, 1998, Chapter 56: Stellar Magnitudes, from Astronomical Algorithms, Willmann-Bell Inc. Second Edition.


Could supermassive black holes form in dwarf galaxies?

The prediction was that dwarf galaxies have SMBHs, but with smaller masses than regular galaxies.  This was based on the idea that regular galaxies were built from dwarf galaxies and the BHs need to have started growing early on.  Also the correlation between bulge mass and BH mass requires the BHs to come from the building blocks.  A recent paper at arxiv detects 28 AGN in dwarf galaxies with BHs in the $10^3 - 10^4 M_{\odot}$ mass range.


Can you translate a GPS coordinate to a Galactic coordinate?

Sure you can! Galactic coordinates have the same origin as other J2000.0 systems; the solar system barycenter (center of mass). This is very close to the Sun, usually but not always inside the Sun, because the larger planets, especially Jupiter, pull it around a little bit. You can read a little more here for example, and also read @zephyr's excellent answer.

At first you might ask why the origin of the galactic coordinates isn't the center of the galaxy. I'm pretty sure that the answer is that we don't know where that is! We'd have to know the masses and locations of everything, and of course since most of the galaxy's mass is dark matter we're not going to know where the center of mass is any time soon.

However, the XY plane of the galactic coordinates has been chosen for now, based on an estimate of the Galaxy's apparent equator. Since it's a different plane than our solar system's plane, otherwise known as the ecliptic, the coordinates will be different even though the origin is the same.

Since galactic coordinates are centered near the Sun, the distance of our position from the origin will still be about 1 AU (150,000,000 km).

Below I wrote a little script in Python using the easy-to-use Skyfield python package. At the moment that I've run the program , the coordinates are:

time (JD):   2458099.18846
time (UTC):  (2017, 12, 11, 16, 31, 23.049599826335907)
latitude (degs):    41.42
longitude (degs):   2.15
galactic (km):     [ -1.46347711e+08  -2.89156773e+06  -2.34254700e+07]
barycentric (km):  [  2.69276456e+07   1.33751808e+08   5.79665590e+07]
and just for fun...
galactic latlon (degs):  [-9.0922867698877123, 181.13191434418721]

Here is the Python script:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

load    = Loader('~/Documents/SkyData')
planets = load('de421.bsp')
earth, sun = planets['earth'], planets['sun']

ts = load.timescale()
tnow    = ts.now()
tmonth = ts.utc(2017, 12, range(1,31))
topo = Topos(latitude_degrees=41.42, longitude_degrees=2.15)

position = earth + topo

print "time (JD):  ", tnow.tt
print "time (UTC): ", tnow.tt_calendar()
print "latitude (degs):   ", topo.latitude.degrees
print "longitude (degs):  ", topo.longitude.degrees
print "galactic (km):    ", position.at(tnow).galactic_position().km
print "barycentric (km): ", position.at(tnow).position.km
print "and just for fun..."
print "galactic latlon (degs): ", [x.degrees for x in position.at(tnow).galactic_latlon()[:2]]


Radioactive decay is the main source of the earth's interior heat: fact or theory?

There are really only six ways you could heat the interior of a planet. Gravitational collapse and compression, fission, fusion, crystallization, a very large impactor, and tidal flexing.

Fusion is out because Earth would be a rock-star. (Been waiting to use that one.)

Of the remaining possibilities, nuclear fission explains the majority of Earth's current temperature gradient.

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


Find a planet with incomplete information of a star chart

I don't know if this will be closed as too general, but I think there's a fun question in here about galaxy mapping.

The trick is to have a good map of the galaxy.   We can't see large sections of the Milky way, so we might have a better map of Andromeda than the Milky way overall from our perspective.

And yes, there are landmarks.    Pioneer and Voyager used Pulsars as landmarks in a diagram on their plaques.

There are by this estimate, at least 200,000 pulsars in the milky-way.  A map of 14 of them won't do someone on the other side of the galaxy if our map has some errors (which Voyager's might) or if they don't have a good chart on the other side of the galaxy.  Pulsars are very precise and they can make  very landmarks if you have enough of them and properly spaced.

Also, stars move, so maps age and lose accuracy over time, but in your star-wars example, a human lifetime wouldn't change things much.

Other possible landmarks include X-ray binaries,  very large stars (less common than smaller ones), black holes, neutron stars, the age of stars could be used, contact binaries, nebulas (or is that nebulae), stellar nurseries.   Identifying an individual planet would be easy given good data and a complete map.   A typical modern computer could do it, but the trick is having a good map.   There's about 100 billion stars in the Milky way, so it comes down to how good the map is, but you're correct, if the small map is good, R2's larger galaxy map is irrelevant.   To be relevant, the smaller map should have been maybe 100-500 stars and R2's map nearly the entire galaxy, and there would be ways to make the map useless if it fell into enemy hands, by not including pulsars or star size, making it very very difficult to identify which part of the galaxy by just 100 stars, none of them distinct.   The principal of needing both maps is valid and could be done, but you're right that the size of the 2nd map was ridiculous, also, why would they need to track his journey like in the picture?.    Movies often ignore the finer points like that and just focus on what looks cool.


If you take a slowly rotating asteroid with a large basin of trapped water, some solved minerals, and may be some gases, close to the asteroid's surface, to get an environment similar to a black smoker by the heat of the sun, I wouldn't call it totally impossible.

The formation of such an asteroid is close to the limits of my imagination, might be as a remnant of a comet with a crust of dust baked together to a water-proof surface layer.

But that's pure speculation at the moment. May be we can learn more this year with Rosetta.

Some biochemistry details: By the rotating asteroid part of the water undergoes a heating/cooling cycle. Provided there are enough prebiotic molecules, by this heating/cooling cycle kind of polymerase chain reaction (PCR) could multiply short occasional RNA or DNA fragments (nucleotide sequences). The difficult part is providing polymerase. It has originally been isolated from a bacterium living in hot springs, not quite dissimilar to black smokers. Information storage and replication (e.g. in RNA/DNA) is one important ingredient for life.

Polymerase is an enzyme, mainly consisting of amino acids. Glycine is the first amino acid found in cometary dust in 2009. That's a first step, but there is a big gap. As an enzyme polymerase is a protein, hence can be encoded by RNA or DNA.

To decode RNA to a proteine, kind of ribosome is needed. A ribosome itself consists of RNA and protein (chain of amino acids). That's the basics of the replication mechanism.

Now we need a containment.
The simplest form of a cell-like structure is a micelle. Organic molecules with a polar group can form micelles.
This paper contains a list of organic molecules found in interstellar medium.
Some of them could be used to synthesize appropriate molecules, but it's also quite a gap in this case.

A long way from what we know of extraterrestrial organic molecules to a bacterium.


how long a light year on earth will seem

You can never reach the speed of light but can get very close to it. Thus, when you travel with, say, 99.99% speed of light it will take you a little more than 31 years to reach the star 31 light years away. This is, however, the time it takes seen from earth. In the spaceship, the time taken will be dilated by
$$\gamma ={1\over \sqrt {1-{v^2\over c^2}}}$$
Which is roughly a factor of 70.7. This means in the space ship it is only 0.44 years or 160 days.


Mathematically calculate if a Planet is in Retrograde

Here is a quick answer. Apparent retrograde motion of a planet as seen from Earth means that its ecliptic longitude is decreasing. (Thanks to @barrycarter for correcting my previous statement.) You have to calculate the apparent angular position of the planet from the Earth in ecliptic coordinates (as explained here and here) and find places where the ecliptic longitude is decreasing.

You can subtract the positions of Earth from the position of the planet in either coordinates. Then you express that vector in spherical coordinates. It doesn't matter if it's heliocentric or geocentric. What matters is the direction that the axes of the coordinate system is pointing, not where the center is.

If you want RA and dec, use ICRF coordinates, and these can be transformed to ecliptic coordinates as well. An easy way to do this is to use the python package Skyfield.

Here's a sample script and a plot. This is Mars from 2010.0 to 2020.0 and the thick red line represents retrograde motion.

import numpy as np
import matplotlib.pyplot as plt

from skyfield.api import load

planets = load('de421.bsp')

earth   = planets['earth']
mars    = planets['mars']

year_zero = 2010
days = np.linspace(1, 3650, 10000)
years = year_zero + days / 365.2564
ts = load.timescale()
t = ts.utc(year_zero, 1, days)

# thanks to @barrycarter's comment, do it the right way!
eclat, eclon, ecd = earth.at(t).observe(mars).ecliptic_latlon()

eclondgs      = (180./np.pi) * eclon.radians
eclondel      = eclondgs[1:] - eclondgs[:-1]

eclondel[eclondel < -300] += 360. # this is a fudge for now
eclondel[eclondel > +300] -= 360. # this is a fudge for now

prograde   = eclondel > 0.

eclon_prograde   = eclondgs.copy()[:-1]
eclon_retrograde = eclondgs.copy()[:-1]

eclon_prograde[-prograde]  = np.nan
eclon_retrograde[prograde] = np.nan

plt.figure()
plt.plot(years[:-1], eclon_prograde,   '-g', linewidth=1)
plt.plot(years[:-1], eclon_retrograde, '-r', linewidth=3)
plt.ylim(0, 360)
plt.title('Mars Ecliptic longitude (degrees) AD 2010.0 to 2020.0',
          fontsize=16)
plt.show()


Why are rocky and small planets nearer to the Sun whereas big, gas giants are farther?

The planets all have differences based on where they are and their compositions. The outer gas giants are so large today because of how quick they were to accrete material. In a protoplanetary disk, once an object gets massive enough, it can attract others through gravity, making the process much quicker than granules sticking together through electromagnetic forces or random collisions. The reason the outer planets were able to get massive much quicker than the inner planets is that there is more material with larger circumferences and that material was able to stay a solid rather than become gaseous.

With a larger mass, the outer planets attracted material quicker and the process was self-perpetuating. In addition, these gas giants had the opportunity to collect more gas that the sun did not end up taking as they are farther out. Hot Jupiters can confuse scientist because of how close they are to their host star when they could not have formed there.

The Kuiper Belt and Oort cloud are also two interesting topics in planetary formation and are most likely relics of planetary formation. This would result from the material that was created around that area, but also material that was flung out by gas giants to larger orbits. The Oort Cloud could also have been formed from interactions of the sun with neighboring stars but we do not have direct observation of this cloud yet.


If I were to point into the sky, how many galaxies would be in this line?

Take a look at the "Deep field" image from Hubble: http://www.spacetelescope.org/science/deep_fields/

You can see in this tiny part of the sky, hundreds of galaxies, but between there is lots of dark sky. Most lines of sight don't end on a galaxy. I'd say dad was right. I would estimate, from the deep field, that you have about a 1/10 chance of pointing at a galaxy.

Even if your finger does point at a galaxy, the distances between stars are vast. Think of two footballs separated by the distance between Europe and China as a model for the distances between stars. A finger pointing at a galaxy would be unlikely to point exactly at one of the stars.

This is probably a Good Thing. Olber's paradox supposed that if each line of sight ended on a star, the night sky would be a uniform blaze of light.


Are there any galaxies which fell out of sight horizon due to cosmic expansion?

No. In fact the opposite is the case.

(See the last paragraph for an intuitive explanation.)

It is a common misbelief that galaxies receding faster than the speed of light are not visible to us. This is not the case; we easily see galaxies moving at superluminal velocities. This does not — as I think most people would think — contradict the theory of relativity, which says that nothing can travel through space faster than $c$. Galaxies do not travel through space (except with small velocities of 100-1000 km/s); rather, space itself is expanding, causing distances between the galaxies to increase.

We see "super-luminal" galaxies

The recession velocity $v_\mathrm{rec}$ of a galaxy is given by Hubble's Law:
$$
v_\mathrm{rec} = H_0 \, d,
$$
where $H_0 \simeq 67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ is the Hubble constant (Planck Collaboration et al. 2016). This law implies that galaxies farther away than
$$
r_\mathrm{HS} \equiv \frac{c}{H_0} \simeq 4400\,\mathrm{Mpc} \simeq 14.4 \, \mathrm{Gly}\,\,\mathrm{("\!\!Giga\mbox{-}lightyears\!\!")}
$$
recede faster than $c$. Here, the subscript "HS" is chosen because the ragion within which galaxies recede slower than $c$ is called the "Hubble sphere". Objects at a distance of $r_\mathrm{HS}$ have a redshift of $z\simeq1.6$.

Consider a photon emitted from a distant galaxy (say, GN-z11 at redshift $z=11.1$) in the past, in the direction of the Milky Way (MW). What special relativity tells us is that locally, the photon always travels through space at $v=c$. Initially, the photon thus increases it distance from GN-z11 at velocity $c$. However, even though the photon travels toward us, its distance to MW increases, due to the expansion of the Universe. As the photon increases its distance to GN-z11, the same expansion causes it to recede from GN-z11 at an ever-increasing velocity. Moreover, as it travels toward MW, it will slowly "overcome" the expansion until it reaches the point where $v_\mathrm{rec} = c$. For an infinitesimally small period, it will stand will wrt. MW, after which it will begin to travel faster and faster as measured from MW. Eventually, its velocity — still in MW's reference frame — will reach $c$, at which point it will have reached MW.

Thus, even though GN-z11 and MW recede from each other at $v_\mathrm{rec} = 2.2c$, we are still able to see it. What is perhaps even more counterintuitive is that when GN-z11 emitted the light we see today, it receded even faster, at $v_\mathrm{rec} \sim 4c$.

We see more and more distant galaxies

There is, however, a limit to how fast a galaxy visible to us can recede, given by the distance $r_\mathrm{PH}$ that light has had the time to travel since the Universe was created. Light comes to us from all directions, so we're situated in the center of a sphere of radius $r_\mathrm{PH}$. This sphere is called "the observable Universe", and its surface (which is not a physical thing) is called the particle horizon (hence the subscript "PH"). Galaxies at the particle horizon are receding at $v_\mathrm{rec}\simeq3.3c$.

As time goes by, light from ever-more-distant galaxies$^\dagger$ will reach us; that is $r_\mathrm{PH}$ increases. In other words, the observable Universe always increases in size, and no galaxy visible today will ever leave the observable Universe, no matter its speed.

However, since future observable galaxies will be more and more redshifted, their light will eventually shift out of the visible range and into longer and longer radiowaves. Furthermore, the time between each detected photon will increase, so they will be dimmer and dimmer, and thus in practice, they will disappear.

Intuitive explanation

A good analogy for better understanding why light can reach us from a galaxy that recedes faster than light, is the "worm on a rubber band": Attach an (infinite stretchable) rubber band (of length, say, 10 cm) to a wall and walk away at any constant speed you choose, e.g. 1 m/s. Before you start, put your pet worm at the end near the wall. It wants to get back to you, and starts crawling at 1 cm/s, i.e. 100× slower than you. Will it ever reach you? If you look at it from the perspective of the wall, both you and the worm move away, but whereas you recede at a constant speed, the worm, although slower in the beginning, accelerates because it moves on the rubber band, but the part of the rubber band between the worm and the wall increases in size. The rest of the rubber band of course also increases in size, but that doesn't matter — as long as you have a constant speed, and the worm accelerates, it will reach you (although in this example, it will take the worm $10^{26}$ billion years, at which point it may have lost its patience. But if you walk at only 10 cm/s, it will take just 6 hours).

In this analogy, you're the MW, the wall is GN-z11, and the worm is a photon. Now if you don't walk at a constant speed, but also accelerate (this is an analogy of the effect of dark energy), the worm may or may not reach you, depending on your speeds. Just like there is a limit to how distant galaxies we will ever be able to see.

$^\dagger$Note that since large distances also means looking back in time (since the light has spent a long time traveling), we actually don't see galaxies this far away, as they hadn't formed this early in history. We do however see the gas from which the galaxies were born, as far back as 380,000 years after Big Bang.


Are magnetic field lines Spiral for Spiral Galaxies? If no, then how it look like topologically?

The short answer is: yes, magnetic fields in spiral galaxies are generally spiral in orientation. (But it probably isn't directly correlated with your solar model.)

For example, this is a map of radio polarization (a reflection of magnetic field direction) in the galaxy IC 342 by Beck et al. (2015):

(The short yellow lines are indicators of polarization, and thus magnetic field, direction. See here for details of the observation)

A very good, if somewhat technical, reference is Rainer Beck's recent (2015) review article "Magnetic Fields in Spiral Galaxies"; this discusses both observations of magnetic fields in nearby galaxies (and the Milky Way) and the models for how they might be generated.

He also discusses, for example, evidence for magnetica fields in the halos of galaxies; these are (seen in projection) sometimes radial and sometimes more X-shaped, as in the case of NGC 891:

(From Krause 2009; the short white lines indicate magnetic field line directions, based on radio polarization.)

The evidence for the Milky Way is somewhat ambiguous, for the usual reasons that it's hard to get a global picture of something you're inside of. One intriguing finding is that the magnetic field sometimes reverses direction as a function of radius; this hasn't been seen in other galaxies (though this may just be a function of resolution or the fact that we are measuring the Milky Way's magnetic fields primarily in a narrow layer of gas near the Galactic plane).


How bright would it be on the "face" of Titan

It's not hard to calculated brightness to distance, but that doesn't take into account cloud cover, which is important for Titan.

Just looking at distance first, Titan (based on Saturn's distance) averages 9.6 AU from the sun, or about 1/91st as bright as the sun appears from Earth, assuming equal atmospheres.  1/91st isn't a bright sunny day, but it's probably bright enough to read by.    The sun has an apparent magnitude of about -26.7 from Earth.   91 times less bright works out to -21.8 from Titan.

Saturn is a little more tricky because from Earth, we often see Saturn's rings but from Titan, you mostly don't.   Titan orbits almost entirely in line with Saturn's rings (just 0.34854° inclination), which would make the rings appear like a thin band.   Visible but probably not very reflective.    You could also probably see ring shade and ring shine reflecting off the surface of Saturn from Titan.

How bright would Saturn be?

From Earth, Saturn's brightness varies from (mostly) between 0 and 1 apparent magnitude.  That's a range of 2.5 depending on distance and angle, which is kind of a lot.    Titan would probably receive less variation due to no ring reflection (see Saturn as viewed from Earth).

There would still be variation based on the angle of the sun reflecting off Saturn's surface.   Perhaps an estimate of between 0.2 and 1 is in the ballpark (as a guess) at Earth distance but Titan's viewpoint, almost entirely above the ring.

Using an an average distance between Earth and Titan of about 9.6 AU, and Titan's distance from Saturn, about 1/122 of 1 AU.  Titan is some 1170 times closer to Saturn than the Earth on average, which makes Saturn about 1.38 million times brighter from Titan than from the Earth.   which works out to between 15.3 and 15.4 apparent magnitude units.

Using my rough estimate, Saturn, from Titan would be in the range of -15.2 to -14.3 apparent magnitude, which would be some 400-800 times less bright than the Sun from Titan (it's not possible for a planet to outshine a star from the point of view of the moon), but several times brighter than the full moon from Earth.   You could probably see, well enough on Titan under a "full Saturn" - being some 5-10 times as bright as the full moon (peak -12.9/average -12.6).  It would also be much larger, so it wouldn't appear bright in the sky, but it would provide enough light to see well enough, but maybe not enough light for reading.   Enough to see though.

Permanent Clouds

The numbers above, however, doesn't take into account Titan's cloud cover and it has nearly constant clouds.   Here is an article with a picture of a rare situation of sunlight reflecting off the surface of titan.   Mostly it's under a permanent haze.   I don't know how effective that haze is in blocking sunlight, maybe 90%, maybe even 99%, but it would certainly block a fair percentage, so, while you could probably see under sunlight, I think reading is no longer possible under those clouds.

Here's an interesting article that says Titan is brightest at twilight, not daylight, because it's atmosphere is very effective at scattering light.   It's atmosphere is thicker and much higher than Earths, and also more massive overall.     It's likely for any real work, Astronauts would probably want a permanent artificial light source and not rely much on light from the tiny sun though all those clouds.    But they might be able to see well enough to move around and explore without flashlights a little bit, but flashlights would help.

Night vision goggles would do nothing.   They see heat or longer wavelength IR light outside the visible wavelengths.   They could spot heat sources, like other astronauts, or a rip in a space suit or hut - so they'd have uses seeing where the heat is escaping for example, but they wouldn't help you see the on surface of Titan.

Titan takes nearly 16 days to orbit Saturn anyway, and nobody would want to put up with 8 days of darkness.  Artificial light would be essential which, I would think, makes choosing a location based on sunlight not a key factor.  Perhaps someplace where there's less methane rain would be a priority, or less chance of solid ground becoming liquid under a warm spell like a nice "dry" water ice Island.


What is a "semester series" for eclipse prediction? - Solar saros series 147 and 152

Any eclipse is a coincidence of a new or full moon (repeating every 29.53 day lunation or synodic month)
and a lunar node (moon crossing the ecliptic, repeating every 27.21 day draconic month).
A semester
is 6 synodic months, close enough to 6.5 draconic months that eclipses can occur at that interval in a series 3 or 4 years long.

The interval between any two solar or lunar eclipses can be expressed as $aI + bS$,
where $S$ is the saros interval of 223 lunations,
$I$ is the inex interval of 358 lunations between adjacent saros series,
and $a$ and $b$ are integers.
The semester interval of 6 lunations is $5I - 8S$; for each eclipse in saros $n$, there may be an eclipse in saros $n-5$ one semester earlier, an eclipse in saros $n+5$ one semester later, or both.
As the eclipses in a given saros series are about 18 years apart, each is in a different semester series.

For example, one semester series runs from 2015 to 2018 and includes one solar eclipse in each of saros series 120, 125, 130, 135, 140, 145, 150, and 155.
Another runs from 2018 to 2021 and includes one solar eclipse in each of saros series 117, 122, 127, 132, 137, 142, 147, and 152.


how to define the universe itself is expanding greater than than the speed of light?

You don't see anything in real time.

When you check your phone, you look a nanosecond back in time. When you look at the Sun,$\!^\dagger$ you look 8 minutes back in time. How do you know your phone and the Sun are still there? Because you have a model of how the world works, and that model tells you that, to the best of your knowledge, they won't just disappear.

Likewise, we have a model of how the Universe works, based on theories and observational evidence for these theories. This model predicts that the Universe expands in such a way that two objects recede from one another at a rate proportional to their distance (the so-called Hubble-Lemaître law). Hence, for large enough distances, this rate exceed the speed of light. This is described by general relativity, and is not in conflict with the speed of light being a fundamental "speed limit", since that is a prediction of special relativity.

In physics, we can never be 100% sure of anything, but we can have more or less faith in our models. If you're asking more specifically how can we trust the models, then that would become a longer description of the assumptions entering the hypothesis, the theoretical framework, the equations, the observational evidence, etc.

Actually, the fact that we look farther back in time, the more distant an object is, has helped us developing our models, since this allows us to observe the Universe not only today, but throughout its entire history. This is one of the coolest differences between astronomy and other sciences, I think.

The redshifting of light is also sometimes helpful. The earliest galaxies (i.e. the most distant ones) are often only visible due to a certain type of light called Lyman $\alpha$. This light is ultraviolet, and hence doesn't easily penetrate our atmosphere. But when observing distant galaxies, the light has been redshifted into the optical or even infrared, and so makes it possible to observe from ground, rather than form space-based telescopes (which is much more expensive).

$^\dagger$Don't.


How many stars can be seen naked eye from outside the atmosphere? How dense is the "sky" from there?

Depending on your eyesight, the faintest stars you can see in a perfect environment have an apparent magnitude (brightness) of 6-6.5 (though some people have managed to see magnitude-7). According to this site, that translates into about 10,000 stars.

As for how it would look, I have seen the sky from the stratosphere (about 3 miles up, in a plane, where there is a lot less atmosphere to worry about, and no light pollution), and it is a surreal sight - with the sky absolutely littered with stars. However, it doesn't  quite look like a Hubble image, though - there's very little colour in these fainter stars, and Hubble tends to focus on narrower fields of view, at higher zoom.


Why are these objects moving at Vastly Different Speeds along the same orbit?

Looking at the simulation, and the gifs attached here, some things are clear. The dots cannot represent actual meteoroids in their orbits, as the 6 orbital elements fully determine the state vectors of a body. You can't be in the same orbit, at the same place, at the same time and have very different velocity. Thus the dots don't obey Kepler's third law.

It is not clear that they obey the second law either (that equal areas are swept out in equal times) They do seem to move faster at perihelion but perhaps not fast enough.

The graphic doesn't seem to show the "lumpiness" of some meteor streams. Notably the Leonids, which have a very intense storm, roughly every 33 years. Other streams are also lumpy which doesn't seem to be apparent in this graphic.

What it does show is the variation in orbit of the meteoroids that form a meteor shower. This is where (I guess) Nasa was involved. From the observation of a meteor (ideally several observations of the same meteor to triangulate its position) it should be possible to work out its orbit before it fell into Earth's gravitational field and hit the atmosphere. Observe enough meteors and you can get an idea of the distribution of orbits: the mean and standard deviation in the inclination, eccentricity, semimajor axis, ascending node and angle of perihelion. From these, you can get a distribution of elliptical orbits. There may be some modelling work do be done: we can only observe those meteorids that hit the Earth, but we can suppose the they are distributed all around the parent body. But from our sample of meteoroids that hit the Earth, we can model the distribution of those that never will.

The dots are drawn on these Keplerian orbits, but the speed at which the move on the orbit doesn't seem to represent the actual speed of meteoroids in space. Rather it is a way of illustrating the variation in elliptical orbits, without drawing a solid ring for each orbit.


Ratios of magnitudes instead of differences

Consider for example these bright stars (V and B-V values from Yale Bright Star Catalog).
Near V=0, B/V is ill-behaved, and B*V is near 0.

Name       B     V     B-V    B/V   B*V
Arcturus  1.19 -0.04  +1.23 -29.7  -0.05
Vega      0.03  0.03   0.00   1.00  0.00
Capella   0.88  0.08  +0.80  11.0   0.07
Rigel     0.09  0.12  -0.03   0.75  0.01

Now imagine that each of those stars were 10 times as far away.
Neglecting interstellar extinction, this would decrease the apparent brightness by a factor of 100 and increase the apparent magnitude by 5.0.
As B and V increase together, B/V approaches 1 and B*V approaches V^2, telling us nothing meaningful about a given star.

Name       B     V     B-V   B/V    B*V
Arcturus  6.19  4.96  +1.23  1.25  30.7
Vega      5.03  5.03   0.00  1.00  25.3
Capella   5.88  5.08  +0.80  1.16  29.9
Rigel     5.09  5.12  -0.03  0.99  26.1

B-V, on the other hand, has stood the test of time as a distance-independent metric of a star's color and surface temperature.