Star formation around rotating black holes?

The accretion of material onto (into) black holes (and neutron stars) provides environments that are both very hot and (relatively) dense. Under these circumstances it is possible for nuclear fusion to occur, the question is whether this is significant, both energetically or as a means of producing new chemical elements (nucleosynthesis).

The answer to the first of these questions is relatively straightforward. As material falls towards the black hole, its angular momentum forces it to form an accretion disk. Viscous processes heat the disk and provide torques, cause the material to lose energy and angular momentum and eventually allow it to fall into the black hole. Much of the gravitational potential energy (GPE) gained as the material falls towards the black hole does end up heating the material.

The innermost stable circular orbit of a black hole is at 3 Schwarzschild radii $=6GM/c^2$, where $M$ is the black hole mass.  The GPE released for material of mass $m$ falling to this radius is $\sim GMmc^2/6GM = mc^2/6$.
i.e. fully one sixth of the rest mass energy of the material could be released as heat.

Compare this with nuclear fusion. The fusion of hydrogen into helium only releases 0.7% of the rest mass as energy that can heat the accretion disk.

So from the energetic point of view, fusion reactions are negligible, unless they can occur much further out in the disk

The question about nucleosynthesis yields is more complex. The more massive a black hole and the higher the accretion rate, then in general the higher the disk temperature and density and the higher the fusion rate. But it also depends on the details of the cooling processes that are possible and how much material is advected into the black hole. Hu & Peng (2008) present some models of accretion onto a 10 solar-mass black hole and suggest that it may be possible to produce certain rare isotopes by this mechanism.  Stellar-sized black holes probably need very substantially super-Eddington accretion rates to achieve the necessary temperatures to sustain nuclear fusion (i.e. much greater accretion rates than are possible by radiation-pressure opposed spherical accretion flows), according to Frankel (2016). Such rates are likely only in the cases where black holes disrupt a binary companion, rather than through a steady accretion flow.


How far out can the sun keep celestial objects revolving?

The suns gravitational pull extends unlimitedly far and, if the universe were otherwise empty, a body could orbit it (very slowly) at unlimited distance.

However, the universe is NOT otherwise empty. The nearest other stars are about 4 light years from the Sun at the moment, so a body more than two lightyears away will typically feel equally strong attraction from other stars as from the Sun, and so will not really orbit the sun.

That's a bit of an over-simplification, because (a) the sun is not surrounded by other stars 4 l.y. away in all directions (here is a chart of nearby stars) so a body might orbit in such a way that it didn't pass between the sun and any of its closest neighbours; (b) not all stars have the same mass, many are smaller than the Sun but some (eg Sirius) much bigger, so their gravitational influences vary and (c) the stars are all moving, so that from time to time one comes quite a bit closer, disrupting orbits.


Do the Bullet Cluster remnants prove that dark matter consists out of particles?

TLDR: Not as such, but it does give us some constraints on its properties.
Background:
A galaxy cluster consists of three major components, from smallest to largest proportion of mass: galaxies (visible with optical telescopes), intracluster gas (visible with X-ray and radio telescopes) and dark matter (not directly observable). Dark matter makes up about 80% of the total mass; of the remaining mass, 90% is the hot gas between the galaxies (not exact numbers, but order of magnitude is sufficient here).
The bullet cluster image:
In the image, we see two accumulations of galaxies centred roughly where the blue colouring is. Overlaid in red is the X-ray emission from hot gas. Note the conical shock front to the right. The blue colour, finally, shows the mass distribution as measured through gravitational lensing.
What does that mean?
We see that gas and galaxies are separated, in contrast to what we see in regular galaxy clusters. The implication is that two clusters have collided. While stars and hence galaxies are essentially collisionless and pass right through each other (there is a lot of space between stars within a galaxy so collisions happen rarely if at all), the same cannot be said for the hot cluster gas. So the gas remained in the centre of the collision while the galaxies passed through.
Now we know that the mass ratio of gas to galaxies is around 10:1. But gravitational lensing (which doesn't care what kind of mass is present) shows us that the majority of the mass in this system is where the galaxies are.
And the dark matter?
We see that the dominant component of the cluster mass is collisionless (it is where the galaxies are). It also does not interact with normal matter except through gravity (else we'd see it).
An exotic elementary particle would fit this description well, but so might primordial black holes (hypothetical low-mass black holes formed shortly after the big bang).
From observing the bullet cluster alone, the dark component might also be made up of quite massive black holes or other massvie dark compact objects -- these can be ruled out by observing microlensing events around our own galaxy.
What about modified gravity?
Mond and similar theories would have to explain why gas of a given mass would affect space-time differently than the same mass clumped together into a star.
Note, however, that there is a known component of dark matter: shortly after the big bang, neutrons that were not bound into helium (or deuterium etc.) decayed into a proton, an electron and an electron neutrino. This cosmic neutrino background is still around, but measurements constraining the upper limit of neutrino rest mass make it highly unlikely that modified gravity theories could use them to explain the bullet cluster. Using our standard theory of gravity, its influence on cluster physics is utterly negligible.
Conclusion
The observation of the bullet cluster provides a big problem for theories that try to explain how our universe works without resorting to some form of collisionless, weakly interacting matter as the major component of the matter in our universe.
It puts some constraints on the nature of dark matter but does not by itself rule out a form of dark matter that is much more massive than an elementary particle.
Further reading
Galaxy cluster composition
Neutrino background
Gravitational lensing
Observing intracluster gas with radio telescopes


What is the connection between moons, rings, and gaps?

It's mainly by the clear definition of the rings, and their mere existence.

Without a moon the rings would be short-lived, hence unlikely to be detected just in time shortly after they've formed. And they would tend to wash out to broader rings. The space between the rings seems to be empty. That's easier to explain by one or more moons keeping the gap clean, and keeping the rings on their track (shepherd moons).

A way of thinking of the shepherd moons:
Our Earth's moon slows down Earth's rotation by tidal forces and friction.
Think of a ring a little inside the orbit of a small moon as an analogon of Earth inside the orbit of Moon: It's slowed down by tidal forces and friction (non-elastic collisions between ring particles). Slowing down ring particles results in an orbit closer to the planet/asteroid. This way a small moon can act repulsive to a nearby ring, confining it.

More details about the dynamics of planetary rings, including confinement by shepherd moons, see this paper, p. 491 ff.


Why does the cloud of Orion Nebula appear more distinct when I am not directly looking at it?

This is called the Averted vision. It involves not looking directly at the object, but looking a little off to the side.

How does it works?

Because the retina is less sensitive at the center, and this is very remarkable on the very faint objects (like nebulas or galaxies)

This is a common technique used by astronomers.

More info here

How to make the most of Averted vision?

To maximise the effects of your Averted Vision, you first need to know which eye is your dominant eye. To do so:

Point with your finger a distant object
Then blink an eye, then the other eye.
The eye in which your finger is perfectly aligned with the object is your dominant eye.

Then, for the best Averted vision:

Look at the top-right of the object if your dominant eye is the right one
Look at the top-left of the object if your dominant eye is the left one


You might want to look at noaa.gov (https://www.nnvl.noaa.gov/view/globaldata.html in particular)

and similar sites for weather for other countries. In other words, the satellites in geostationary orbits (GEO) that take images are mostly weather satellite. I do not have a breakdown by type, but I would guess that a vast majority of the satellites in GEO are communication satellites, and they have no reason to take images.

I suspect that satellites in high earth orbit are there for astronomical purposes (to "get away" from the Earth, such as Chandra) or communication (such as the Russian Molniya satellites), so they are not imaging the Earth.


What would theoretically be dissolved in the interior oceans of icy moons and planets?

There is no guarantee of solutions for life (known), but basically you need solutions of anions and cations, as phosphates for building long chains as well as maintaining neutrality (acid and electronic equilibrium) weather, or any CHONSP (carbon, hydrogen, oxygen, nitrogen, sulfur, and phosphorus) with neutral acid (water) solution but this only applies for well known carbon life.


Starlight: are the rays that we see here on Earth parallel?

Light rays travel along things called geodesics in spacetime (specifically they travel along a certain class of geodesics called 'null geodesics').
If spacetime is flat, which will be the case in the absence of gravity, then those geodesics are just straight lines as we normally think of them, and so yes, light rays travel along straight lines in that case.
But there is gravity and spacetime is therefore not flat.  This means there there is really no definition of 'straight line' any more: geodesics are the best thing we have (that's why they're called geodesics rather than straight lines).
But in very many cases the deviation caused by spacetime not being flat is pretty small: for instance it's possible to see the difference in apparent position of stars if the light from them passes close to the Sun, but it's a small deviation.
In some cases – if the light passes close to very massive objects, the deviation can be very large (and in fact if the object is massive enough, it can be as large as you like).
The general effect of light travelling along geodesics which are not well-approximated by straight lines is called 'gravitational lensing' and it's now a well-confirmed thing which is used in practice to detect things like exoplanets (amazingly, in this case it's the lensing of the light from stars by the planet which is observed, which is just astonishing.)
So a brief answer is yes, almost always, but occasionally not at all!


How to see other celestial things using stars for direction?

Use a constellation map! That will provide much of the information you need.

Here are however some things you can try to spot:

On the lower left hand side of Orion, you can spot a very bright star, Sirius, in fact the brightest one in the sky. It is part of the constellation Canis Major (great dog).
A bit further left and up you probably see a star nearly as bright as Sirius. That is Procyon in Canis minor (lesser dog).
Up, and a little bit to the right from Orion you can see a very red star. That is Aldebaran in the constellation of Taurus (the bull). In the same constellation you can see the Pleiades, a bunch of stars in a cluster.

Taurus with Aldebaran and the Pleiades:


Is the sun shrinking currently?

It's the other way around: The Sun is slowly growing hotter and thus is slowly expanding. The Sun accumulates ever more helium in its core as it ages. This growing amount of fusion ash results in the core getting hotter, which in turn results in the Sun expanding in volume. This completely overwhelms what one would expect from looking only at the mass loss due to fusion and solar winds.
The solar mass loss rate currently is on the order of $10^{-13}$ solar masses per year. At that rate, it would take on the order of ten trillion years for the Sun to consume or expel all of its mass. That's a much longer time span than the Sun's much faster aging process. The Sun has five to seven billion years or so left on the main sequence. Shortly before the time the Sun leaves the main sequence, it will be about twice as luminous and considerably larger than it is now. Shortly after the the Sun leaves the main sequence, it will be much more luminous and even larger than it was just before it left the main sequence.


Which binary star system has a greatest angular distance?

Wide binary stars it turns out is an active research topic, with groups using statistical searches of catalogs (Tycho Gaia ...) to find in 3D space pairs of stars that are moving together taking into account possible orbital motion. Here people are looking at the physical separation not the angular separation. On the theory side it is suggested that nearly all stars were formed as part of a binary. So somewhere out there is sister of our Sun. One paper looked at nearby stars. They find several very wide binary pairs visible to the eye plus more fainter. In particular they mention Capella and 50 Peg which have an angular separation of 20 deg. This they class as an escaped binary : so no longer in orbit (thanks Connor for pointing this out).
There is however Fomalhaut; a faint (mag 12.7) 3rd member of the system was discovered in 2013. It is 5.7 deg away from the primary. The other companion is  2deg away in another direction. The whole system spans 8 deg. The Sky and Telescope article claims this as the widest (on sky) binary, beating Alpha AB Proxima system.


What makes the Earth infinitely rotating around its axis

As for the second case, that should break our physical laws

That is precisely the wrong statement. The fact that the Earth is still spinning matches our physical laws exactly. The physical law I'm specifically referring to is known as Newton's First Law of Motion, which is loosely paraphrased as:

  An object at rest remains at rest unless acted upon by a net, outside force. Conversely, an object in motion remains in motion unless acted upon by a net, outside force.

What this means is that once something starts moving (like the Earth spinning) it will keep moving until something else stops it. The Earth long ago started spinning because of the initial angular momentum of the collapsing cloud which eventually made up the Earth. There is nothing in space to stop the Earth from spinning so it is just going to go on spinning1.

You appear to believe that if you start an object moving by applying some force, it will eventually run down and stop moving of its own accord. This is a common thought, based on everyday experiences. In fact, that was what Aristotle thought (and much of the world up until Isaac Newton's time). But what you're not taking into account is that there's always a force being applied to slow things down. Consider a car rolling along the highway. If you take your foot off the gas pedal, the car starts slowing down. But it isn't because some physical law says it should, it's because a net force is acting on the car to slow it down $-$ that net force being wind resistance on the car. So you see, even common, everyday experiences hold up to Newton's first law.

1Technically there are some small forces acting on the Earth to slow it down but those are nearly negligible and mostly from the Moon. These forces would take billions of years or longer to make the Earth come to a complete stop (if they ever even could).


Relativistic Mercury orbit around the Sun

General relativity is only one cause of the precession of Mercury's orbit, and not the largest factor. Gravitational perturbation by the other planets and the non-spherical sun also cause precession. Newtonian gravity predicts precession of 532 arcseconds per century, but 575 arcseconds is observed. The discrepancy can be explained if one uses General Relativity as the model of gravity.

The extra precession of 43 arcseconds caused by GR would be the same, the perturbations due to the other planets would not be the same, unless they were also orbiting the sun in the opposite direction.


Timing based on upper or lower part of sun/moon

It is the first one, i.e. as soon as the heavenly body can be first seen. Here a definition from http://www.ga.gov.au/scientific-topics/astronomical/astronomical-definitions

Sunrise is defined as the instant in the morning under ideal meteorological conditions, with standard refraction of the Sun's rays, when the upper edge of the sun's disk is coincident with an ideal horizon.


Beamsize in radio/microwave observations

This is a fundamental problem in modern multi-color/frequency astronomy. You can not simply add two spectrally significant different channels because they contain information about different solid angles on the sky.

If you have two maps with different inherent beam sizes one common approach is to resample the map with the higher spatial resolution to the lower resolution of the other map by convolving with a Gaussian beam of size:

$\theta_{Gauss-kernel}=\sqrt{\theta_{low}^2-\theta_{high}^2}$

For example, if you have two maps with resolution (FWHM) 5" and 15" you need to smooth the 5" map with a Guassian kernel of size

$\theta_{Gauss-kernel}=\sqrt{15^2-5^2}=14.14"$

to obtain a comparable map.

Of course the beams are never true Gaussians and it depends whether you observe point sources or extended (w.r.t. the beam) source.

As an example you can take a look at the SPIRE Observers manual. SPIRE was an instrument onboard the Herschel Space Observatory. They discuss the involved problems in somewhat more detail.


Magnetic fields are generated by currents - i.e. by the motion of charged particles. As you say, the Sun is full of freely moving charged particles, and these generate currents which in turn generate magnetic fields. No metals required.

Most of the magnetic field generation is thought to occur at the interface between the radiative interior of the Sun and an outer convective envelope. This region, called the tachocline, is subject to large shearing motions which are able to take small magnetic fields and amplify them. The stronger magnetic fields are then buoyant, because magnetic fields in a plasma exert a pressure. They therefore emerge at the solar surface in the form of loops of magnetic field.

These loops are strongly coupled to the plasma in the Sun. As the plasma moves around turbulently, driven by convective motions and differential rotation on the Sun's surface, the footpoints of the loops are sheared and twisted. There comes a point where the loop will snap back into a configuration with lower magnetic potential energy through a process called reconnection. The reconnection process has a side effect of accelerating particles within the magnetic loops. These relativistic particles smash into the photosphere (actually, they are mainly stopped in the chromosphere which sits above the photosphere), where they release their energy, heating plasma to millions of degrees, which evaporates into the corona. This is a solar flare.


Has the permittivity or the permeability of space been measured in a gravitational well?

But you are asking about the speed of light, since the product $\mu_0 \epsilon_0$ determines the speed of light in vacuum, which is fixed.
Before 2019, both the speed of light and $\mu_0$ were defined as fixed numbers in SI units and therefore by definition, $\epsilon_0$ was fixed.
With the revision of SI units, the speed of light is still a defined number but both $\mu_0$ and $\epsilon_0$ are determined by experiment (though their product must still be a defined constant). They are experimentally determined to about 1 part in $10^{10}$ via the fine structure constant $\alpha$. There is no very strong evidence that they or any other physical constants vary in different parts of the universe.
See https://en.wikipedia.org/wiki/Vacuum_permittivity#Redefinition_of_the_SI_units
More fundamentally, it isn't that meaningful to ask whether physical constants with units vary, either in space or time, since they are measured in terms of other things that could potentially vary instead. The only really meaningful question is to ask whether dimensionless combinations of physical constants, such as the fine structure constant $\alpha$, vary.
There have been suggestions that $\alpha$ might vary with time or direction. Since the fine structure constant is
$$\alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c} \ ,$$
then any variation in $\alpha$ could now be interpreted as a variation $\epsilon_0$, since the other constants are fixed in SI units. Or you could rewrite $\epsilon_0$ as $1/\mu_0 c^2$ and interpret it as a variation in the permeability of the vacuum, or some combination of the two.
The results of investigations into high redshift gas clouds along the lines of sight of quasars have proved some of the best probes of this effect. Th emost recent work I have seen (Wilczynska et al. 2020) suggests no variation with time, but there might be a change in direction at the level of 1 part in $10^5$. The present rate of change (if any) must be less than  part in $10^{16}$ per year.


What does "unremarkable transverse peculiar velocity" mean exactly, and how is it calculated here?

"Peculiar velocity" is a fixed term and describes the velocity of an object relative to a defined rest frame.
Astronomy has the problem that you need different methods to measure the 3D motion of an object. Therefor one often only gives the velocity within line-of-sight (from spectrographic data) or the perpendicular velocity as measured from astrometry, thus positional data on a sphere; that is the perpendicular velocity (to the line-of-sight).
As such the "perpendicular peculiar velocity" is the velocity of the magnetar as determined from astrometry - as also stated in your quote.
The reference frame of rest in this context is the expected motion of the stars around the galactic centre - at the distance of the observed magnetar. Galaxies don't have a simple Keplerian velocity where orbital speed reduces with distance to a central heavy central object like in our solar system. Dark matter changes the relationship between orbital distance and orbital speed in a way which is subject to investigation. With a "flat rotation curve between J810 and the Sun" they mean that orbital velocities don't differ, thus they indicate directly their normalization, their reference rest frame.
The peculiar velocity of a star within our Galaxy thus refers to their motion relative to their rest frame which orbits the Galactic center in the way expected for their distance from the center of the Galaxy.
So how was it measured? They say "from annual parallax measurements". That means they measure the positon repeatedly and very exactly. This allows to measure the parallax in reference to the background. But at the same time you get the proper motion, thus the change of the positon when you removed the expected apparent change due to parallax. A typical astrometry series looks like  which I took from http://spiff.rit.edu/classes/phys240/lectures/parallax/parallax.html#nottrivial You nicely see the annual variation (the parallax) and the proper motion which remains when the circular annular motion is subtracted. This velocity is the perpendicular or transverse velocity.
Convert this measured velocity in the image plane to a real velocity by taking into account the distance to the object, reduce it by what you expect it to be for the Galactic distance of the object, and find that it is pretty normal to what you expect and you have an "unremarkable transverse peculiar velocity"


How does the black hole at the center of M87 look like in the visible spectrum?

Disclaimer: this is speculative, since no images of a black hole have been taken with enough resolution in visible light.

The black hole represented in the movie Interstellar is moderately realistic with the knowledge we have at present. In fact, the movie makers asked astrophysicists to contribute.

From the article Gravitational lensing by spinning black holes in astrophysics, and in the movie Interstellar, this is a moderately realistic view of a black hole.

The light is not the black hole itself, but the accretion disk around it. The circle of light is the part of the accretion disk that is behind the black hole, the light of which is lensed by the black hole.

Now for a black hole spinning on itself, then it would not look as symmetrical. Doppler shifting would make one side appear more red, and the other more blue.


What would be the difference in distance between earth and moon on supermoon night (14 November 2016)

Based on some information from this article

http://www.space.com/34515-supermoon-guide.html

The Nov 14 2016 supermooon's expected peak of full phase
is on the morning ov Nov 14 at 8:52 AM EST

According to some quick calculations that I performed
using PyEphem

and assuming the moon would be viewed by an observer in NYC at 08:52AM EST

0.00236372323707 AU from observer
An AU is 1.5 * 10^km

Edit: I needed a better precision for km to AU

http://www.iau.org/static/resolutions/IAU2012_English.pdf

According to this, the AU can be more precisely defined as
149 597 870 700  meters +/- 3 meters

0.00236372323707 * 149 597 870 700 =

353 607 963.19 meters or

~ 353,608 km from earth

here is a quick run through of the inputs I used for the program

>>> moon = ephem.Moon()
>>> nyc = ephem.Observer()
>>> nyc.long, nyc.lat = '-74.0059', '40.7127'
>>> nyc.date = '2016/11/14 08:52:00'
>>> moon.compute(nyc)
>>> print moon.earth_distance

    0.00236372323707

Learn more about PyEphem package here
http://rhodesmill.org/pyephem/index.html

Note I'm just an amateur and these calculations might not take in certain critical factors. I'm uncertain if PyEphems earth_distance property calculates distance to the moon or the moon's center.

If it does calculate distance to the center of the moon this number could be about ~1700 km smaller.

Given your figure of 384,400 km average, this would put the moon just under

~30 792 km closer to the earth or about 91.98 % of it's normal distance


Could an orphan/rogue planet have a moon?

Could an orphan planet have a moon?

Yes.

s there some reason why an orphan or rogue planet wouldn't have a moon?

Yes, this answer lists some possibilities.
What you didn't ask was:

Is there some reason why no orphan nor rogue planet could ever have a moon?

and based on the previous two, No.
Any non-contact gravitational acceleration that could pull a planet away from its star has some chance of accelerating its moons sufficiently to stay with the planet.
If the perturber were a few times the Moon's orbit farther from the planet, staying with the planet might even be likely.
Looking at this obliquely from a delta-v perspective, Jupiter orbits the Sun at 13.1 km/s, and its inner Galilean moon Io orbits Jupiter at 17.3 km/s.
Without doing the math, I think it's possible that you could give Jupiter a 13.1 km/s kick impulsively in an optimal direction and Io would stay bound to it, but now in a much more elliptical orbit.


Is most hydrogen in the universe in the form of plasma, atomic neutral hydrogen, ionized hydrogen, or molecular?

@HurtHikes, I hope this answers your question,
According to Wikipedia, most Hydrogen is in the form of Atomic and Plasma states.

Throughout the universe, hydrogen is mostly found in the atomic and plasma states, with properties quite distinct from those of molecular hydrogen.

Intuition says that because, as this link says, 99.9% matter in the universe is plasma, most Hydrogen in the universe should also be present in Plasma state.

"99.9 percent of the Universe is made up of plasma," says Dr. Dennis Gallagher, a plasma physicist at NASA's Marshall Space Flight Center. "Very little material in space is made of rock like the Earth."

This link has the calculations to determine the amount of Hydrogen in Plasma in relation to the total amount of Plasma (if anyone is interested...)

Stars are made of relatively simple stuff.  By mass, our Sun is 73% hydrogen, 26% helium, and only 1% of higher Z (atomic number) atoms.

X = mHnH / r = density of hydrogen / total density

Though I am not sure about the actual answer about the same, I will be grateful if anyone can point out to the right answer. And although I have made every effort into the removal and rectification of errors, If there are any errors in my answer, I will be grateful if anyone can point them out.


DecRA is the decimal right ascension
RAh, RAm, RAs are the hms form

$${\rm DecRA} = {\rm RAh}\times 15.0 + {\rm RAm}/4.0 + {\rm RAs}/240.0$$

$$ {\rm RAh} = {\rm INT}({\rm DecRA}/15.0) $$

$${\rm RAm} = {\rm INT}(({\rm DecRA}-{\rm RAh}\times 15.0)\times 4.0)$$

$${\rm RAs} = ({\rm DecRA}-{\rm Rah}\times 15.0 - {\rm RAm}/4.0)\times 240.0$$

where INT is the operation that truncates to an integer

e.g. DecRA=24.35417

$${\rm RAh} = {\rm INT}(24.35417/15.0) = 1$$

$${\rm RAm} = {\rm INT}((24.35417 - 1\times 15.0)\times 4.0) = 37$$

$${\rm RAs} = (24.35417-1\times 15.0 - 37/4.0)\times 240.0 = 25.00$$

DecDE is the decimal declination

DEd, DEm, DEs are the dms form

posneg is -1.0 for a position below the celestial equator (negative declination) and +1.0 for above the equator

$${\rm DecDE} = {\rm DEd} + {\rm posneg}\times {\rm DEm}/60.0 + {\rm posneg}\times {\rm DEs}/3600.0$$

$${\rm DEd} = {\rm INT}({\rm DecDE})$$

$${\rm DEm} = {\rm INT}(({\rm DecDE} - {\rm posneg}\times {\rm DEd})\times 60.0\times {\rm posneg})$$

$${\rm DEs} = ({\rm DecDE} - {\rm posneg}\times {\rm DEd} - {\rm posneg}\times {\rm DEm}/60.0)\times 3600.0\times {\rm posneg}$$

e.g. DecDE = -45.67778   posneg=-1.0

$${\rm DEd} = {\rm INT}(-45.67778) = -45$$

$${\rm DEm} = {\rm INT}((-45.67778 - (-1.0)\times 45)\times 60.0\times (-1.0)) = 40$$

$${\rm DEs} = (-45.67778 - (-1.0)\times 45 - (-1.0)\times 40/60.0)\times 3600\times (-1.0) = 40.0$$


If we've only observed Oumuamua as a single pixel, how can we know its shape?

Nobody "knows" the shape of Oumuamua. However there are standard ways of deducing the objects likely shape based on assumptions, and this methodology has been tested agains asteroids for which their shape has been checked by other methods, including delay-Doppler radar (see Is delay-doppler radar imaging of NEO asteroids possible only if it spins fast enough? for example) and measurements from deep-space spacecraft that have approached and even orbited individual asteroids.

This answer to Could asteroid 'Oumuamua actually be round? shows a preliminary light curve and a simulation based on some assumptions of albedo and rigid-body rotation dynamics. It's not likely we will ever know its shape for sure, at least not with current capabilities of space travel, so this kind of analysis is the best that can be done.

For a more thorough analysis and modeling of Oumuamua's light curve see for example Modeling the light curve of `Oumuamua: evidence for torque and disc-like shape.


Which astronomer set his beard on fire because he left the cap off his telescopes finder?

From the book Turn Left at Orion by Guy Consolmagno and Dan M. Davis, Galileo might have set his beard on fire with a telescope pointed at the Sun.

  And you must keep your eyes -- and your clothing -- out of the path of the projected light. It is said that Galileo set fire to his beard doing this!

In his defense, he was the first person to look at the Sun with a telescope, so you can't blame him for not knowing about the associated risks!

But this story might also be apocryphal. Many people set various things on fire with a telescope (some even doing it on purpose).


Does time dilate during night time when earth orbits around the sun?

There's no gliding and no significant time dilation.
Your speed relative to the sun is not the same at midnight as it is at midday, since you must combine the rotation of the Earth with its orbital motion around the sun.
But for any point on Earth there will be one midnight and one midday each day. There aren't "more midnights".


What is the direction of the movement of the solar system in relation to the galaxy's plane

I think your question is answered by the duplicate mentioned: but here are the relevant highlights.

Humphreys & Larsen (1995) suggest, using star count information, a distance of $20.5 \pm 3.5$ pc above the Galactic plane; consistent with, but more precise than the Bahcall paper referred to by Schleis. Joshi (2007) is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between 13 and 20 pc above the plane.

The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this "peculiar velocity" is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (up and out of the plane).  (Dehnen & Binney 1998). Different authors arrive at velocities that differ by $\sim 1-2$ km/s from these values and so this would probably be a more conservative estimate of the uncertainties.

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from http://www.visioninconsciousness.org/Science_B08.htm to show this oscillatory motion. The oscillations will not be exactly sinusoidal because the restoring force towards the plane does not very linearly with height above the plane.
As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.

EDIT: Actually I'm glad I revisited this question because I think the picture is not very good at all. As the Sun takes ~230 million years to go around the Gaaxy, it should only execute 3 complete vertical oscillations in our Galactic year, whereas the picture implies many more. Secondly, the Sun executes a radial oscillation with a period of around 160 million years, which is not even indicated!


Are new stars less pure as generations go by?

The chemical enrichment of the Universe over time is indeed a thing. The plot below (source) shows observational measurements of the cosmic density of ionised Carbon in the Universe against redshift (higher redshift -> further back in time). The abundance of other heavy elements over time shows a similar trend.

Stars are explicitly classified based on their metallicity, and its direct relation to which generation of stars they are part of. This is known as the Baade chemical classification. The first stars to form from primordial gas are confusingly known as Population III. These stars have yet to be observed, but it's hoped that the James Webb Space Telescope may glimpse them. They are theorised to be very massive and short lived, hence the difficulty in detecting them. The next generation, formed from the enriched material from this first generation of stars are known as Population II. And the subsequent generation, of which our own Sun is a member, are Population I stars. Population I stars have the highest metallicity, due to the fact they formed from material that has already participated in Population III and II star formation.
The metallicity of a star is almost entirely determined by the metallicity of its birth cloud.  Stars do swallow up small amounts of material over their lifetimes, but this does not significantly affect the overall metallicity of a star. A stars lifetime is mostly set by it's mass, and whilst metallicity does have a small effect, it doesn't change significantly due to outside effects over a stars lifetime.


It is highly likely that what you have seen is an Iridium communications satellite. Because of the material covered, it can reflect sun light and appear very bright and sometimes flashing.


How can the oblateness of Earth impart torque to change the longitude of ascending node of a satellite?

I'd like to start by quoting from the Wiki page on Nodal Precession you mentioned:

  A non-rotating body of planetary scale or larger would be pulled by
  gravity into a sphere. Virtually all bodies rotate, however. The
  centrifugal force deforms the body so that it has an equatorial bulge.
  Because of the bulge the gravitational force on the satellite is not
  directly toward the center of the central body, but is offset toward
  the equator. Whichever hemisphere the satellite is in it is
  preferentially pulled slightly toward the equator. This creates a
  torque on the orbit. This torque does not reduce the inclination;
  rather, it causes a torque-induced gyroscopic precession, which causes
  the orbital nodes to drift with time.

(emphasis mine).  Note that this is a small force and the precession is slow compared with orbital speeds.  Precession towards (or coasting away from) the equator thus is a change in latitude compared with what would be expected, and this "twist" changes the expected longitude at a given point in time because the satellite's vector velocity changes (I believe in both direction and magnitude, but I'm not certain about the magnitude).


Why do GPS satellites move in relation to earth's surface?

Geostationary orbits (GEO) are necessarily over the equator, and satellites in GEO are not visible at all poleward of about 80 degrees -- they are always below the horizon (There's a good explanation on physics.stackexchange.com).

This is a problem because for GPS to work, you need at least four satellites to be visible and they need to be well-separated in the sky.  Because it's rare for the whole sky to be "visible" all the time -- parts are blocked by buildings, metal vehicles, hills and suchlike -- you in fact need more than four in the sky as seen from any point on Earth to have a minimum of four actually visible all the time.

So even at lower latitudes where GEO satellites are visible, they still all are in the southern (or all in the northern) sky and thus won't give a good GPS fix.  And the military (who were paying for this) wanted to be able to use GPS in the north, not just in the equatorial regions.

Also, possible GEO satellite locations trace a gently curved line across the sky, and even if there were a dozen visible and even at low latitude, they'd give very poor accuracy perpendicular to that line.

The solution was to put the GPS satellites into many different orbits which were strongly inclined to the equator. They wanted them to be fairly slow-moving as viewed from Earth, so LEO wasn't viable.  (LEO also required too many satellites to get adequate coverage.)   They needed to be above the Van Allen belts for longevity. But they also needed them to be as low as possible to maximize signal strength. 10,000 miles up turned out to be the right engineering compromise.

Gamma ray bursts were lagniappe.


What's the relation between Einstein's Gravitational Theory and Dark Matter?

Dark matter was originally hypothesised because there is a greater degree of rotational energy in galaxies than visible matter would allow for - crudely they rotate so fast that they ought to spin apart and therefore it was hypothesised that there is an additional source of gravitational force in some form of invisible matter.

This is not as a result of a discrepancy in Einstein's General Relativity. Newtonian gravitation is a very good approximation of Einstein's GR at most ordinary scales and energies and this is true in this case also.

I am not sure, exactly, what your "what about dark energy" question means, but assuming you mean, does this imply an incompleteness in GR, then again the answer is no, or at least not necessarily, but it's a more subtle point.

Dark energy is the hypothesised source of the energy that causes the universe to expand at an accelerating rate. It can be inserted as a term into the field equations of GR (other explanations of DE exist though) - but that is just a mathematical term, rather than an explanation of the physicality of it.

Einstein originally inserted such a term into his solutions - to predict an essentially static universe. When it was shown that the universe was expanding he described this as his "greatest mistake".

The evidence for the accelerating rate of expansion is relatively recent and came long after Einstein's death so he never lived to see the essential idea - of a "cosmological constant" in his equations - revived.

This explains more - https://en.wikipedia.org/wiki/Cosmological_constant


According to this, the Hubble constant for redshift z is $H_0E(z)$. Meaning we need to prove that $$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$
Take the first derivative of both sides of your second equation to obtain, by the chain rule, and the equation (this is the definition of the Hubble constant) $H = \frac{R(t)'}{R(t)}$ the following:
$$\frac{dz}{dt} = -\frac{1}{R(t)^2}*R'(t) = -\frac{H(t)}{R(t)}$$
That the redshift is zero for the current time, and some value greater than zero for some time before the current time should explain the minus sign. After multiplying with $c$ and placing the small changes on both sides this becomes $$\frac{cdz}{H(z)}=-\frac{cdt}{R(t)}$$, which looks like it can be integrated to obtain what we need to prove (the integral from z to 0 is the one from 0 to t).This proof is not very formal, but it's the best I can do, so I hope it is somewhat correct and that someone will give a more detailed answer if this is not good enough.


Of the stars in the Gaia / Hipparchos catalogs, which one is the most isolated?

From a comment by barrycarter:

  In our galaxy, it might be CX330, which is 1000 light years from any star-forming region, much less any star. However, the article describes it as "Loneliest Young Star", so there may be older stars that are even more isolated. This source contains the phrase "CX330 [...] has been described as the galaxy's loneliest star.", but this may just be bad reporting.


Understanding Scatter in Data collected for HR Plot

So, scatter in observational measurements of clusters in the HR diagram is mainly, in my experience, caused by 3 things: observational errors, binaries and contamination.

As you mentioned, there are observational errors, which are contributed to by the instrument, the data reduction, the weather conditions, etc. This error usually gets worse as you get fainter toward the limits of the telescope and looks like a "widening" of the CMD (seen here notably in magnitudes > 19).

The next problem is contamination. When you point your telescope at an object like M44 or the Hyades, it's not obvious which stars belong to the cluster and which stars just happen to be in the field of view, so you will get contamination from normal milky way stars (which will not follow that isochronic relationship that cluster members will). This looks something like this.

The final problem that I can think of is a binary population. A "photometric binary" star is what happens when two stars overlap in an observation. This happens quite often in clusters, since there are "intrinsic binaries" which are stars that are very close to, and orbit each other. They are usually different masses, and hence different colors. This is usually seen as another line, next to the main isochrone line, that is a little redder than all of the other stars and would look something like this image.

Now, this scatter looks far too big to be measurement error alone, modern telescopes will have magnitude errors of much much less than a magnitude (think milli-magnitudes, depending on specifics -- I'm unfamiliar with Google Sky, but any telescope is going to do better than a magnitude error, for reference the magnitude system is based on eyesight "just-noticable-difference", so your eyes have an approximate error of one magnitude).

And since the scatter doesn't follow some other systematic relationship, I would discard the binaries idea.

The most reasonable explanation to me is, like you said, foreground contamination.

Figures:

https://www.researchgate.net/figure/231129963_fig2_Fig-4-A-B-vs-B-V-color-magnitude-diagram-of-all-9507-stars-in-the-WFPC2-image-of-M30

http://home.strw.leidenuniv.nl/~piladiez/

https://inspirehep.net/record/1217369/plots


HD 51684 is a magnetic Ap star. It has an entry in the recent catalogue of fundamental parameters for magnetic stars by Glagolevskij (2019). This catalogue lists it as having $T_{\rm eff}=7800$ K, and a mass of $2.3 M_\odot$.
These parameters are determined from the photometric properties of the star - i.e. they are not fundamental measurements of the temperature or mass.
The mass quoted by Kervella et al. (2019), in a study of binarity using Gaia data (HD 51684 is a known binary) is $2.195M_\odot$. However this is spurious precision; this mass comes from matching the inferred luminosity and temperature to stellar evolutionary models, not from a measurement of the binary motion.
The distance has been most precisely and accurately determined by Gaia EDR3
The parallax is $4.51 \pm 0.24$ arcsec. Roughly speaking, you can invert that to give a distance of $222 \pm 12$ pc.


What is the optimal escape trajectory from near a black hole?

I'd like to know a little more about the geometry of the ship's trajectory. I would be asking for clarification in comments but I don't know how to put images in comments.

A good distance away the ship is moving nearly a straight line with regard to the large mass. As the ship gets closer the path gradually bends towards the large mass. If you're still a fair distance from the star, the path can be fairly well modeled with Newtonian mechanics and the curving path can be modeled as a hyperbola. The straight line the hyperbola is gradually deviating from is called an asymptote:

This illustration (page 36 of my coloring book) is a hyperbola about the earth, but it could also be a hyperbola about a larger mass.

Escape velocity climbs as you get closer to the mass. The hyperbola's speed is sqrt(Vescape^2 + Vinf^2). I use this right triangle and the Pythagorean theorem as a memory device:

If you did a burn in the opposite direction from your Vinf vector, it could reduce your Vinf and drop you even closer to the mass.

If your burn vector is at right angles to your velocity vector, it would increase your vinf (and thus raise point of closest approach aka periapsis). It would also change the direction of the hyperbola's asymptote.

To better answer your question, I'd need to know more about the geometry of this scenario, what Vesc and Vinf is and how much delta V the ship was capable of. It would be helpful to know at what distance our heroes discover they're in trouble.

If the ship has already fallen close enough that it's traveling an appreciable fraction of c, the above doesn't apply. Conics from Newtonian mechanics is a good approximation until you get too close to the black hole. Then you'd need general relativity to model the trajectory -- and that's above my pay grade.


What evidence is there of Earth-Like internal features of Europa?

This Image, Originally sourced from NASA shows the theorized interior of Europa:

According to this geology site Earth's core is predominantly iron and nickel.

and this space site says that Earth's mantle is predominantly silicates

So the internal similarities so far is that they both possess iron cores and silicate rock mantles. However the core of Europa is mixed with sulphur rather than Earth's nickel.

Another similarity between Europa and Earth could be the composition of the oceans. NASA speculates that Europa's oceans are also briny with salts and minerals dissolved in it. This is mostly theorised due to what we know about planet formation.

However this also comes with many differences, the Galileo spacecraft did not detect a magnetic field from Europa, which leads to speculation that, whilst Europa may have an Iron compound core, it is probable not molten any more.

Conclusion

Interior Similarities:

Iron compound core
Silicate rock mantle
Salty, briny, Oceans

Differences:

Iron core mixed with sulphur not nickel
Core likely to be solid, not molten
Lack of internally generated magnetic field


What is the difference between a neutron star and a white dwarf?

In a neutron star, the force of gravity is strong enough to press the protons and electrons together to form neutrons {1}, White dwarfs are only very compact. With even more mass, you get a black hole.
All three types are outcomes of star death, when the failing fusion in the middle of a star is no longer able to counteract gravity. What a star become when it collapses is depending on its mass.


Something to consider might be a large very dark (low albedo) asteroid.  I would say it is quite possible a large dark asteroid (e.g. P-type) "could" be headed for us, and we would not know it until it hit us (e.g. distance 0).  I think you are asking for the closest distance.  It appears the main method of detecting asteriods is with telescopes, but we are starting to scan the skies in infrared too.  Hopefully infrared will reveal these darker asteriods better.  Sources for further research include:

http://en.wikipedia.org/wiki/P-type_asteroid

http://www.space.com/3126-asteroids-data-sheet.html

http://www.nbcnews.com/science/space/how-we-scan-solar-system-find-dangerous-asteroids-f8C11480423

For the furthest distance, that question is trickier.  It would largely depend on what is considered an asteroid, for example, if we consider Pluto to be an asteroid, we can at least detect very large asteroids out to the distance of Pluto, and we could track its path and see if it were on a collision course with earth.

On average, it is said that we have already detected most of the globally threatening asteroids and it does not appear we are in immediate danger.  Still, I'd say we need to stay on the lookout, and especially look out in infrared to detect any dark asteroids we may have missed :-).

Summary:

Closest distance before large threatening asteriod detection: 0 (e.g. upon impact)
Furthest distance before large threatening asteroid detection: Beyond the orbit of Pluto (and most asteroids tracked are in the asteroid belt between Mars and Jupiter)


Are there any stars that one day we could collide with?

Statistically speaking, the Sun will probably not crash into any other stars in the near future. Currently, we do not believe the Sun is on a collision course with any other star either, but we can only predict so far due to Chaos Theory and what I believe is called Lyapunov Time.

The reason for this is that solar systemic distances are minuscule compared to interstellar distances. The distance from the Earth to the Sun is 1 AU, or 150,000,000 km. Our sun is 0.004 AU in radius. The nearest star (which isn't on a collision course) is approximately 275,000 AU from the Sun. As you can see stars are generally far apart, and thus they rarely interact.

The locations of the stars do, in fact, change over hundreds of thousands of years, but again, stars generally stay far apart. That is not to say that stars never collide. However, it only happens in very dense globular clusters, young star systems (where they may form close), and in contact stars, as well as the occasional exception, which may be 1-2 in the entire galaxy when Andromeda collides with our Milky Way Galaxy in a few billion years (the Sun might already be dead though).


How can the Sun block an X-ray telescope from observing a QPO galaxy "for several years"?

It is not that it is blocked by the Sun, but that the duration of time for which it can be continuously observed was too small to be useful for the investigations performed in the paper.

The object was observed by XMM-Newton, which has a highly eccentric ($e \sim 0.8$), 48-hour orbit arond the Earth. This is designed so that the satellite can stare for relatively long periods at the same object, whilst being beyond a lot of the geocoronal background that interferes with its detectors. Whilst doing so it has to avoid looking (i) close to the Sun; (ii) close to the Earth; (iii) at the Moon.

The "visibility" of a target is defined as the longest continuous time it can be observed before the satellite has to break-off and point somewhere else. This can be as long as 140,000 s (there's a deadspot where they switch the detectors off when near the Earth), but can often be much shorter (or zero) at any paticular time. For efficiency reasons, the satellite operations team don't want to schedule too many short-visibility observations, since this wastes valuable observing time and spacecraft resources, and scientifically you often don't want to use short-visibility windows because you want to observe some sort of variability over a certain duration (e.g. observing over a full eclipsing binary orbit, or looking for variablity on a variety of timescales and requiring a long continuous time-series to get enough signal-to-noise in the Fourier spectrum).

Since the position of the Earth, Moon and Sun change with respect to each other, the visibility of an object will change from orbit-to-orbit of the satellite. Furthermore, because the orbit of XMM-Newton changes with time in quite a complex way (it precesses and the eccentricity has changed a little bit over the years), then the exact visibility of an object will not repeat from year-to-year and some objects will "come into season" during different years. In some cases sources can go through epochs where they are not continuously visible for long periods of time. For this particular position in the sky there appears to have been an unfortunate gap of about 7 years (that is actually the gap between observations, but I haven't checked whether that was completely determined by lack of "visibility" or what the maximum "visibility" was during those 7 years) when the object was not continuously observable for the $\sim 70,000$ seconds required by the science.

I managed to find this old plot, produced not long after XMM-Newton was launched. It shows the maximum visibility versus position on the sky for orbits 3-432 (about the first 2.5 years of operation). This shows that there is a fairly substantial patch of sky which was not continuously observable for more than about 50,000 s for the whole of that period.

There is a more modern interactive map here. This was for "AO-19" (May 2020 - May 2021), and is used by astronomers proposing observations to check how long an object is continuously visible for (as a fraction of the maximum possible time of around 140,000 s; actually there is a detailed web calculator to do it for any position). You can see that in this 12 month period there are big areas that cannot be observed for very long.

NB: There's nothing special about the object coordinates. Ecliptic coords 144, +28, though I would think that high ecliptic latitudes are less likely to be afflicted by this problem, since you would always be pointing at a large angle from the Sun.


Why can't different regions of the universe be the same as they have the same origin?

We don't really know what happened at the time $t=0$. It doesn't really make sense to say that everything "touched each other" at $t=0$, particularly if the Universe is infinite. But we can calculate the distance that a photon — and hence the maximum distance that any information — can travel in the expanding Universe in a given time. This calculations depends only on the Hubble parameter, and the densities of the various components of the Universe, and whether you start your calculation at $t=0$ or $t=10^{-12}\,\mathrm{s}$ makes little difference after a second.

The answer is related to your previous question about the cosmic event horizon which was also about the particle horizon, i.e. the distance that light has been able to travel since the Big Bang. If you were present when the CMB was emitted, you could also calculate your particle horizon. It would of course be much smaller than today, since light at this time had only traveled for 380,000 years. If the Universe hadn't been expanding, this horizon would of course be 380,000 lightyears, but due to the expansion, it is quite a lot larger, roughly 850,000 lightyears.

This means that, at the time of the CMB, regions farther apart than 850 kly had not had the chance to be in causal contact.

We can also calculate how large an angle 850 kly at the CMB would suspend if observed today. It turns out to be 1.7º. That is, if the Universe had just been expanding like we observe its expansion today — i.e. depending only on the densities of the known components — patches on the CMB map separated by an angle $\theta > 1.7^\circ$ shouldn't look the same.

But they do.

Inflation solves this problem by saying that the Universe initially went through a much, much faster expansion, such that regions much, much farther apart than the 850 kly have been in causal contact. Still, even with inflation, the particle horizon is not infinite. So on scales much, much larger than the observable Universe, it may be inhomogeneous.


Look up current positions and velocities of solar system objects

Is there any resource that provides these current values?

Yes. The JPL HORIZONS on-line solar system data and ephemeris computation service provides these values, and much more.


CCD in telescopes: Observation and Astro-photography

You will typically find that CCD cameras are sold separately to the telescopes. That way the telescope can be used visually or for astrophotography, by either fitting an eyepiece, or a camera. Most telescopes do ship with eyepieces, as they are much less expensive than a CCD and it is sort of expected, but most observers will typically replace this eyepiece with a better one anyway, so to all intents and purposes, when you buy a telescope you are just buying the Optical Tube Assembly and maybe a mount, and you add to that all the accessories you need to accomplish your objectives. Different CCDs will be used for different purposes, for instance planetary observations or deep space objects.

That telescope has a fork mount. Note that this constrains how much equipment you can put on the end of the telescope  without getting in the way of the mount. It also is an Alt-Az mount, which means it is not polar aligned; stars will rotate in the field of view, requiring a field de-rotator in addition to a CCD.

It looks like a nice enough scope for visual observations, but if you are interested in astrophotography, you might want to hold off for a bit, maybe get a scope for visual use first, and maybe make use of well-set-up robotic astrographs that you can hire over the internet, like slooh.com and itelescope.net - get some experience using premium gear that you don't have to worry about setting up or maintaining, and while you do that, work out what gear you will be happy about getting for yourself. The fact you are asking this question about CCDs suggests you aren't yet ready to spend money on astrophotography gear...

Getting set up well for visual stuff can set you back hundreds... Getting set up for astrophotography can set you back many thousands, so you want to be well equipped to make good purchasing decisions. And you can rent a lot of time on some pretty good telescopes for that same $...


The Hinode satellite has an active x-ray telescope that is routinely used to image the sun.  And the NuSTAR extreme x-ray telescope can also be used for sun imaging (although its main job is looking for black holes)

Here is the latest XRT image of Hinode

There are also Extreme Ultraviolet telescopes (at 171Å that is less than an octave away from the conventional lower end of X-rays at 100Å) SDO. Stereo (Ahead) and SOHO all have EUV telescopes. The Atmospheric Imaging Assembly on the SDO satellite has sensors for wavelengths as short as 94Å, conventionally classed as X-rays

https://www.nasa.gov/jpl/nustar/searing-sun-seen-in-x-rays


How much overlap will the Andromeda galaxy and the Milky Way have when they collide?

This page, from June 2012, contains a fairly detailed summary of three papers that considered the tangential velocity and collision problem.  I'll quote a few choice bits from that page:

Andromeda (M31) and the Milky Way are the two largest galaxies in the small group of galaxies called the Local Group. At the moment, M31 is about 770 kpc away from our galaxy, but Doppler measurements of the line-of-sight velocites of its stars have long shown that it is inexorably falling toward us. The exact nature of the collision, however, has so far remained unkwnon. Will it be a miss, glancing blow, or head-on smashup? This depends on the proper motion (sideways motion) of Andromeda in the sky, which is extremely hard to measure.

At last, the authors of this study manage to collect extraordinarily accurate Hubble Space Telescope observations of the tangential motion of M31 over a five-to seven-year period, and these data remove any doubt that Andromeda is destined to strike and merge with our Milky Way in four billion years. The Triangulum galaxy M33, the third most massive galaxy of the Local group, which is bound to M31, will likely join in the cosmic boom (with a small chance that M33 will hit the Milky Way first).

The final result is that M31 has a radial velocity with respect to the Milky Way of about  -109 km/s, and a tangential velocity of about 17.0 km/s (<34.3 km/s at 1-sigma confidence). These numbers clearly imply that the velocity vector of Andromeda Galaxy is statistically consistent with a radial (head-on collision) orbit towards the Milky Way. This means that the Milky Way-M31 system is bound, and that the two galaxies will merge, as investigated further in Paper III.

The simulations show that the Milky Way and the Andromeda galaxy will merge, consistently with the radial (head-on collision) orbit deduced in Paper II for M31, and  that the first “pericenter” (closest approach) will occur at around 4 Gyr from now. For our Milky Way,  the encounter has 72.2% probability of being prograde (the galaxy spins in the same direction as the flyby). In 41.0% of the Monte-Carlo orbits M31 makes a direct hit with our Milky Way, where the authors define a “direct hit” as an encounter with a first pericenter distance less than 25 kpc. The two galaxies will eventually merge after 5.9 Gyr, and the radial mass profile of the merger remnant will be significantly more extended than the original individual profiles. Roughly speaking, this profile will follow the $R^{1/4}$ law characteristic of elliptical galaxies, in agreement with the predictions from the numerical simulations of major mergers of spiral galaxies.

I'll note that $25 \text{ kpc} = 25,000 \text{ pc} \cong 81,540 \text{ ly}$, and that the Milky Way is only about 100,000 ly in diameter, so the "direct hit" definition is quite a substantial distance here.


How can a low-mass star increase its mass to 1.4 Msun?

You were taught wrong. Stars of up to about 8 solar masses will end up as white dwarfs. But it is only their cores that become degenerate and end up as the white dwarf. The rest of the envelope is lost during the giant phase due to a dense wind.

There is a non-linear, but probably monotonic, relation between the initial progenitor and final white dwarf masses (see below, from Kalirai 2013) - the Sun will likely end as a 0.5 solar mass white dwarf, but in normal stellar evolution, degenerate white dwarf stars can only be produced up to about 1.25 solar masses by the most massive progenitors. Any more massive than this and it is likely that the core does not become degenerate before igniting and burning through the heavier nuclear fuels. The most massive, probably single, white dwarf known is "WD 33" in the cluster NGC 2099 and has a mass of $1.28^{+0.05}_{-0.08}\ M_{\odot}$, is likely made of an O/Ne mixture, and had an estimated progenitor mass on the main sequence of $>3.5\ M_{\odot}$ (Cummings et al. (2016).

In order to get a more massive white dwarf, up to the Chandrasekhar mass (about 1.38 solar masses for a C/O or O/Ne white dwarf in general relativity), it almost certainly needs to accrete mass from a close binary companion or be the result of some sort of merger.

This is the leading candidate to explain type Ia supernova.


Brightness, altitude & azimuth for satellites?

Altitude is measured in degrees above the horizon, from 0° on the horizon to 90° directly overhead.
Your fist appears about 10° wide at arm's length away from your eye.

Azimuth is measured along the horizon, usually with north 0°, east 90°, south 180° west 270°, and other directions (e.g. NNE, ENE) interpolated as in the compass image.

Magnitude is a logarithmic scale in which the brightest stars are 0 or 1, and the faintest stars visible to the unaided eye are 4 or 5 (maybe 2 or 3 in a city, or 6 in a remote area).

Each row of Heavens-Above's table links to a sky chart showing their predicted path for that satellite pass.
The brightness listed in the main table is a maximum; the table under the sky chart shows a brightness for other points on the path.


Correct units to use while plotting mollweide projection of sky chart with RA and Dec values, matplotlib

With stars extending to ±90° on the horizontal axis and rather sparse near those limits, I suspect that RA and Dec are swapped.
Try this instead:

plt.scatter(df['rarad'], df['decrad'], s=1, color='red')

Also if you want the plot to resemble the night sky, filter the star list by apparent magnitude, and reverse the horizontal axis so it increases from right to left.


the "mirror" cover on top of the glass had chipped away all over,
  giving the result of a patchy look.

Optics can take a huge amount of damage (chipping, scratching) before it really starts to affect performance. You'll be surprised at the amount of abuse a telescope can take like that. As long as it's not properly cracked in two, don't lose any sleep over it.

  The stars just move like a liquid in the center

The moving around is normal. It is due to atmospheric turbulence, which happens all the time. Unless you're in a vacuum, the images will always shimmer a little. Sometimes they will shimmer a lot. Other times not so much.

In astronomy we have a word for this - it's called seeing:

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

  Also, as the scope is a reflector it has the "three lines" on the
  front, however this interferes with the object I am looking at as i
  can see the lines on the star.

You are way, WAY, WAY out of focus. The scope might also be completely miscollimated. The star must not appear like a big blob, with the "lines" and the image of the secondary mirror visible in the middle, like in your photo.

The "lines" are called spider vanes - they are parts of the assembly called spider, which holds the secondary mirror in place. The vanes, or the secondary mirror itself, must never be visible in the image.

First, do a rough, purely visual collimation. Do this in a place with plenty of light (inside the house). Take out the eyepiece. Look into the focuser. You'll see the secondary mirror. Reflected in it, you'll see the primary mirror from the bottom of the tube. Reflected again in the primary, you'll see the focuser, with your eye in the middle.

If things are more or less centered, like in the third diagram above, leave it alone. If they are way out of center, like in the first diagram, use the adjustment screws on the mirrors and center everything. Collimation is very important to performance. A miscollimated scope will perform miserably. This is FAR more important than all the scratches on the mirror.

Again, this is purely visual, no-tools collimation. This is all you need for now. You can get a laser collimator later, after all your problems are solved. This is the article where I copied that image from, if you need more info:

http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/

Also see this document:

http://www.etnaastros.com/documents/No_Tools_Collimation.pdf

Once collimation is done, you need to find true focus. Go outside at night. Point the scope at a medium-bright star. Polaris, if you can find it, is great because it doesn't move.

Now moving the focuser back and forth, try and determine which way you can make the image of the star appear smaller. Now keep adjusting the focuser until the image of the star is super tiny, then it starts to grow again. Then back off until you make it as tiny as possible. That's when it's in perfect focus.

Now point the scope at a very bright star. If your scope has 3 spider vanes on the secondary mirror, you may see 6 spikes coming out of the star. Like this:

If your scope has 4 vanes, then you should see 4 spikes, like this (it's really 8 spikes, 2 times the number of vanes, but in this case they are superimposed 2 by 2):

The diffraction spikes are easily visible on very bright stars. On medium-size and small stars, they are not easily visible, or not visible at all. This is normal.

Unless the stars are super-tiny dots like in the images above, your scope is not ready for use. Do your best to collimate it and find true focus. Make the stars super-tiny.

Good luck and let me know if you need more advice. It would help a lot if you could find an experienced astronomer nearby to help you fix the scope.


If an exoplanet transit we are seeing is 13000 light years away, are we seeing a 13000-year-old orbit?

In one word, yes. Anything and everything we see, we see the way it was a certain time ago—about 1.3 seconds for the Moon, about 13,000 years for your hypothetical planet. Like @Richyt pointed out, there is no way for information to travel faster than the speed of light.
Because any and all celestial objects gravitationally affect any and all others, orbits are not frozen in time and do “evolve.” If this planet is the only “major” thing gravitating around its star, its orbit will change very slowly, but there is still a change, which may or may not be noticeable in 13,000 years, depending on many factors such as the proximity of the planet to its star (the closer it is, the stronger relativistic orbital precession would be—e.g. it is 43″/century for Mercury, 8.6″/century for Venus, and 3.8″/century for the Earth [source for these latter two: https://arxiv.org/pdf/0802.0176.pdf).


How high does light pollution reach into the sky?

Light pollution occurs because light from the ground refects off atoms in the atmosphere. So you can reduce light pollution either by getting away from  light, or getting above the atmosphere. 50% of the atmosphere lies below 5500m,  if you can get 5500 m high, you half light pollution.  Getting high also improves clarity, and reduces the disturbance caused by atmospheric turbulence, which is why most major telescopes are situated on mountain tops.


For locating objects in the sky, the horizontal and equatorial coordinate systems are commonly used. These systems describe the position of some object in the sky very well, but do not explain the position of the object in space (if you know the distance you know "where" the object is, but this is relative to the equatorial/horizontal plane and is tricky to wrap your head around).
If you want to precisely locate the position of some object in space relative to the ecliptic, the ecliptic coordinate system exists for that, and for locating deep sky objects, the galactic coordinate system is normally used, as it nicely describes positions of objects relative to our galactic disc.
The space with the fourth coordinate is used for relativistic models (it fits well into special relativity), but not for describing positions of objects (to see more about that, look up Minkowski space).


How significant and accessible is modern Chinese astronomy?

China is pretty well integrated into modern astronomy -- there are regular international conferences held in China (the most recent General Assembly of the International Astronomical Union was held in Beijing in 2012), lots of Chinese astronomers at other international conferences, Chinese participation in international observatory projects (the Joint Institute for VLBI, the Thirty Meter Telescope, the Square Kilometer Array, etc.), and so forth. I'm not sure I'd say they have any really significant facilities yet (with the possible exception of LAMOST), but they have very ambitious plans, as you've noted.

(ITAR has very little to do with astronomy except when it comes to a few things like nuclear physics codes produced by places like Los Alamos.)

I don't know how much international participation is anticipated for the FAST project, though I gather the instrument development has some involvement from British and Australian institutes, so they'd likely get some time.

Yes, results from the Chang’E-3 mission have been published in scientific papers, e.g.
http://www.sciencemag.org/content/347/6227/1226.abstract

There are several Chinese astronomy journals (e.g., Acta Astronomica Sinica; Research in Astronomy and Astrophysics), but like most "national" journals they aren't that important internationally.


Function for current distance between in light-seconds between Mars and Earth

There is no simple function that is accurate. Both Earth and Mars are orbiting in ellipses (Kepler's first law). They are travelling at varying speeds (Kepler's second). To find the current distance between them you have to calculate their positions in three dimensions.

In other words you have to do orbital kinematics. How do I calculate the positions of objects in orbit?

If accuracy is not important. You can approximate the orbits with circles in the same plane. The Earths position is then
$$(\cos(2\pi t + \alpha), \sin(2\pi t +\alpha))$$
and Mars is
$$(1.5\cos(2\pi t/1.882 +\beta), 1.5\sin(2\pi t/1.882 + \beta))$$

The distance between them is calculated by pythagoras

The distances are in AU and the time is in years (converting to lightseconds is then easy). $\alpha$ and $\beta$ are constants chosen to fit the positions of the planets at time 0. The values 1.5 and 1.882 are chosen because Mars's orbit is about 1.5 times bigger than Earth, and a Mars year is 1.882 times longer than Earth.

The circular approximation however is poor for Mars, and this ignores the fact that Mars isn't in exactly the same orbital plane as the earth, so the values will only be rather approximate.


Is there, in fact, any close-up photography of Saturn's rings, showing individual pebbles/rocks?

The rings are about 3% solid in the densest parts, but this translates to a separation between 30cm particles of about 1metre.

There are no images Because approaching the dense part of the rings would be extremely difficult. A probe would be travelling at orbital speeds, and even a small percentage difference in speed would translate to a large absolute difference. As the probe would also be in orbit, it would be extremely difficult to find an orbit that doesn't collide catastrophically with ring particles.

Cassini did travel through one of the gaps in the ring system during Saturn orbital insertion. It survived 100000 dust impacts in 5 minutes.

Nature has some artist's impressions of the ring particles.


Equation for solar radiation at a given latitude on a given exoplanet?

The rotational period ( or whether a body is tidal locked ) is a factor here.  A tidal locked body has a hot spot on one side and a cold spot on another.  Rotation may also affects average temperatures.

Season depends on orbital factors.  Simply saying it's one season or another does not tell you the distance you are from the star, which is the single most important factor ( combined with the power of the star ).

However in terms of making a very basic calculation you can use a black body radiation calculation and thermal equilibrium to get something.  You can see more on this at these links :

http://claesjohnson.blogspot.ie/2010/04/black-body-temperatures-of-planets.html

http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html

It's a pretty good predictor considering it's simplicity.  Note the impact of atmosphere and ocean.  Note also that significant energy is being radiated by some bodies.

This slide show has several equations that may help you.  It also has an interesting model on the effect of an ice cap, which I'd not seen before.

http://www.math.umn.edu/~mcgehee/Seminars/ClimateChange/presentations/2011-2Fall/20111102Handouts.pdf


Have we observed any rogue/wandering stars?

Different types of observations have been made:

hypervelocity stars: there are stars in the halo of our Galaxy (it is kind of the suburbs of the Milky Way) with such velocity that they are escaping it, and aiming to the intergalactic medium (IGM). There not yet fully in the IGM, but the're on their way. (Reference)
molecular clouds in the IGM: there are molecular clouds detected in the IGM, that are forming stars. Stars are not directly observed, but these star forming molecular clouds are proof of their existence. (Reference)

It's hard to do better, because it is hard to observe directly stars outside of the Milky Way.


How much does cloud cover affect viewing the total eclipse

I don't know if there is a list but I'me sure your local weather surface will give you some idea. Cloud cover could have a big effect of course. But it will be the same as it is for any day or night. Total overcast will allow you to notice the darkness only. Just use your common sense on what cloud cover you deem too risky for travel. There is an internet site: https://clearoutside.com/forecast which is supposedly developed by astronomers to help determine viewing conditions. Just input a latitude and longitude and it gives you info related to viewing conditions. Different days are on left after local forecast. Click the triangle in the lower right of the date and it gives you a detailed forecast for the location you entered. This includes predicted cloud cover by the hour.


Any possibility for life on moons of planets in our Solar System

If you check 2001 A Space Odyssey saga, Europa is actually a satellite with Life.

Coming back to real space, the "comfort zone" for life is approximately from Venus' orbit to Mars' orbit. There, only Moon, Phobos and Deimos are satellites, and all of them are too dry to have any kind of life.

Beyond the comfort zone you need some other way for water not to frozen (a commonly referred prerequisite of Life), so you'll need a volcanic or tidal-heatened satellite. There are a few of these. Europa and Io come to my mind just now. Life there could be possible.


Are there any hot jupiters orbiting red dwarfs?

I found one "hot Jupiter" in the Kepler data (Kepler 45b). The star is a M dwarf with an effective temperature of 3820K. The planet has an estimated mass of 160.5 M(Earth) and radius of 10.76 R(Earth). This gives a density of about 0.8 g/cm2 which is consistent. The planet is located at approximately 0.03 AU from the star with an orbital eccentricity of 0.11. Also, it may be interesting that the star has a very high metallicity of Fe/H = 0.28. (This data is from http://kepler.nasa.gov/Mission/discoveries/ ).

It must be said, however, that M dwarfs only made up about 4.3% of the initial Kepler study which found 82 "Jupiter size" planets (as of my tally in April of 2015). Thus, there is an indication that there are less "hot Jupiters" around M dwarfs, but sample size is way too small. There does seem to be a correlation between stars hosting "hot Jupiters" and their metallicity (more common around higher Fe/H stars).

My gut feeling is that "hot Jupiters" in general are uncommon. They are just the easiest thing to find.


Is the ratio between earth's distance from the sun and the speed of light just a coincidence?

It is a co-incidence. The units are what you need to look at, in both the speed of light and the distance of the Sun the units of length are the same (m) (your ratio is $\approx 0.002 s^{-1}$ when you use metres as the unit for the distance to the Sun). The definition of the second is in a sense arbitrary and is a consequence of how we have chosen to sub-divide the day into smaller units of time (there is nothing cosmic in dividing the day into 24 units each subdivided into sixtieths which are again subdivided into sixtieths).

If we had chosen to take as our basic unit of time $1/100000$ of a day rather than $1/86400$ the nice round answer vanishes (if my calculations are right the $\approx 0.002$ becomes $\approx0.0023$)


Are the planets in the TRAPPIST-1 system so close that inhabitants on one planet could see surface details on the other planets?

Yes!

How big would each planet appear from each of the other planets during conjunction or opposition? Bigger than Earth's moon?

For some combinations, definitely!
Here's a diagram of the size of each of the planets as seen from each of the other planets. I've taken the semi-major axes and planet radii from a table in Wikipedia and used the Python script below to show the apparent sizes at conjunction or opposition. The axes are labeled in degrees and the largest case is 1.28 degrees or 2.5 times the apparent size of the Moon from Earth!
   a (Gm)        R (km)
b:  1.73          7150
c:  2.37          6984
d:  3.33          5000
e:  4.38          5804
f:  5.76          6671
g:  7.01          7322
h:  9.27          4930

For example, the top row shows the largest possible size of planets c through h from planet b. The black dots indicate the planet with itself.

import numpy as np
import matplotlib.pyplot as plt

data = (('b', 1.73, 7150), ('c', 2.37, 6984), ('d', 3.33, 5000),
        ('e', 4.38, 5804), ('f', 5.76, 6671), ('g', 7.01, 7322),
        ('h', 9.27, 4930)) # https://en.wikipedia.org/wiki/TRAPPIST-1#Planetary_system_data_charts

class Planet(object):
    def __init__(self, name, a_Gm, r_km):
        self.name = name
        self.a    = 1E+09 * a_Gm
        self.r    = 1E+03 * r_km

halfpi, degs = 0.5*np.pi, 180/np.pi

planets = []
for thing in data:
    planets.append(Planet(*thing))

arc      = np.vstack([f(np.linspace(halfpi, -halfpi, 201)) for f in (np.cos, np.sin)])
f        = np.array([-1.0, 1.0])[:, None]
c        = np.array([+0.8, 1.0])[:, None]
full     = np.hstack((arc, f*arc[:, -2::-1]))
crescent = np.hstack((arc, c*arc[:, -2::-1]))

all_sizes = []
for p in planets:
    p.arcs = []
    for pp in planets:
        if pp == p:
            p.arcs.append(None)
        else:
            size = 2*pp.r/(pp.a-p.a)
            all_sizes.append(abs(size))
            if size > 0:
                p.arcs.append(size*full)
            else:
                p.arcs.append(size*crescent)

max_size = max(all_sizes)
print('max_size: ', max_size)

if True:
    plt.figure()
    for (i, p) in enumerate(planets):
        y0 = -max_size * degs * i
        for (j, arc) in enumerate(p.arcs):
            x0 = max_size * degs * j
            if type(arc) is type(None):
                plt.plot([x0], [y0], 'ok', markersize=6)
            else:
                x, y = degs*arc
                plt.plot(x+x0, y+y0)
    plt.gca().set_aspect('equal')
    plt.show()


The thing about time travel is it's hard to define. Technically we're constantly traveling into the future.
Now, if you went at near-light speed for a long time and then stopped and came back, you would have the illusion of having time travelled into the future because time essentially went on without you. This is entirely physically possible, though not feasible of course.
But past time travel, what does that mean? Does it mean rewinding all time for everything? Or one entity aging backwards? Or time goes backwards while you remain the same age, like the reverse of the future travel?
In any case, unlike the future travel, there is no way, with our current model of physics, to do any of them.
And I must reiterate, future travel by the means I described is only an illusion. You didn't move though time, time just went on without you. Obviously, this won't happen the other way.


Is the delay to correlate the signal from one dish to another on a VLA radio telescope longer for red shifted objects?

...is the delay to correlate the signal from the closest dishes to the further ones longer for the red shifted object?

Good question!

The short answer is no. The path difference between the closest and farthest dish in the array comes from local geometry, a big triangle in the air right above the VLA dishes. Here the speed of light is essentially constant (except for local temperature and humidity gradients in the air above the array, or maybe wind sheer) and so wouldn't vary between the to signals very much for reasons related to the sources' Doppler shifts.

The interference path length difference is divided by the local speed of light (in the air) to get the time difference needed for best correlation of those two dishes, and it doesn't make any difference what the source of the signal is or the speed that it was moving.

Doppler shifted photons don't look or act any different than non-shifted photons, there's no way to tell the difference in an experiment, unless you have some information about the original frequency before the shift.

The only teeny tiny difference might be if the air has a different index of refraction at the shifted frequency than the unshifted frequency. That's certainly taken into account during normal analysis, and I don't think it rises to the level and intent of your question.


How do elements with only one shell emit light?

Actually, hydrogen can have as many electron levels as you want. However, they only exist at specific energy levels, with the potential energy given as:

$$E=-\frac{R_H}{n^2}$$

Where $R_h$ is the Rydberg constant, and $n$ is any integer. When electrons are exited to a higher energy lever, they will fall back after some time and emit photons (not protons, by the way). So how many levels do they drop? Well, in the case of hydrogen, it will eventually go back to the basic energy level, but this is not guaranteed to happen in one jump. This may or may not take multiple jumps, that depends on probability. For a simple atom like hydrogen, those probabilities may be possible to calculate analytically. For more complex atoms however, this becomes an extremely difficult task, usually approached by simulations.


Position and velocity vectors of nearby stars?

You may use the XHIP star catalog, using the fields X,Y,Z for the heliocentric position and U,V,W for the heliocentric velocity vector.


Gear teeth usually have a special shape which is either an epicycloid or involute curve. Making worm gears is even more complicated because the teeth have a helical curve in addition to having an involute profile.

Making good-quality gears requires specialized tools, knowledge and equipment.


How exactly does Hawking radiation decrease the mass of black holes?

I'm going to give you an intuitive answer. Keep in mind, this is not the "actual" answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.

  I don't see how this event contributes to evaporation of the black
  hole (, since the particles do not originate from the black hole).

You're missing a key point here.

When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.

Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual "existence", per the Heisenberg equation. In many ways they are just a mathematical trick.

Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.

What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.

(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)

But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.

How did that particle become real?

This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.

It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.

What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.

Again, this is not the actual process. The actual process is more complex. This is just a pop-sci fairy tale.

EDIT: To hit closer to home, Hawking radiation is more like a close relative to the Unruh effect. Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.

Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.

http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html

EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.


Power spectrum of matter in the Universe

Power spectrum

The distribution of matter in the Universe is not uniform$^\dagger$. Rather, it has clumped together by gravity to form the structure we see, from planets and stars, to galaxies, to groups and clusters of galaxies. Exactly how much the Universe clumps on the various scales can be described statistically by the two-point correlation function $\xi(r)$, which gives the excess probability of finding a clump of matter at a certain distance $r$ from another clump, relative to a random, Poisson-distributed matter.

The Fourier transform of $\xi(\mathbf{r})$ is called the matter power spectrum $P(k)$, where $k = 2\pi/r$ is the wavenumber. The power spectrum thus shows readily how much structure there is on various physical scales, i.e. a high value means much structure. It evolves (increases in amplitude) with time or, correspondingly, redshift $z$, and so may be written $P(k,z)$.

The power spectrum can be written as the product of the primordial power spectrum $P_*(k) \propto k^{n_s-1}$ originating from cosmic inflation, and a transfer function $T(k,z)$ (squared) describing what happens at a later epoch. A good description is given by Peebles (1980).

The figure

The figure you show is taken from Tegmark & Zaldarriaga (2002). It shows the power spectrum as a function of wavenumber, but on the top $y$ axis you see the corresponding spatial scale. This scale is expressed in comoving megaparsecs (Mpc) (where 1 parsec equals 3.26 light-years). This factors out the size of the expanding Universe — or, equivalently, the epoch at which an observation is made — such that observations at different epochs in the history of the Universe can be readily compared. A distance $r$ at redshift $z$ will expand to $r(1+z)$ at redshift $z=0$. Using comoving coordinates, no matter at what value of $z$ we observe $P(k,z)$, we "extrapolate" to the current power spectrum $P(k)$.

Moreover, the distance is divided by the dimensionless Hubble constant, $h \equiv H_0\,/\,100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1} \simeq 0.7$. The reason for this is that different observations may have made different assumptions about the exact value of $H_0$; in general inferred distances will scale as $H_0$, so measuring distances in $h^{-1}\mathrm{Mpc}$ allows for direct comparison.

In other words, $1 h^{-1}\mathrm{Mpc} \simeq 4.7\times10^6$ light-years at redshift $z=0$, but at redshift, say, $z=4$ — when the Universe was five times smaller in all directions — we had that $1 h^{-1}\mathrm{Mpc} \simeq 9.3\times10^5$ light-years.

Observational probes

The power spectrum depends on various cosmological parameters, mostly on the matter density parameter $\Omega_\mathrm{m}$ and the expension rate $H_0$, but also the baryons ($\Omega_\mathrm{b}$) and dark energy ($\Omega_\Lambda$).
Given these cosmological parameters, the transfer function and the resulting matter power spectrum can be calculated. In the plot, the red line shows $P(k)$ as calculated theoretically from linear perturbation theory. For details on the calculations, see Tegmark & Zaldarriaga (2002). It is compared to constraints from various observational probes of the cosmological parameters:

Black: Anisotropies in the CMB shows how matter clumped in the very early Universe, and its polarization pattern, caused by subsequent scattering, gives constraints on matter distribution in later epochs, on the largest scales.
Green: Clustering of galaxies (here from the 2dF survey) shows how matter has formed structure on (comoving) scales roughly from a few tens of Mpc (a bit larger than the typical distance between galaxies) to several hundred Mpc (above which the Universe is homogeneous).
Blue: Abundances of galaxy clusters. Here, the error bars reflect the spread in the literature.
Magenta: Weak lensing measures photons that have traveled through fluctuations in the metric caused by intervening matter; this technique probes the power spectrum on $\sim10\,\mathrm{Mpc}$ scales. These data are from Hoekstra et al. (2002).
Red: The smallest scales (largest wavenumbers) are probed by the Lyman $\alpha$ forest (LAF). Whereas all the previous techniques measure light, the LAF is the absence of light, namely absorption lines in a bright background source, caused by the neutral hydrogen in all the diffuse clouds and filaments of gas lying between the source and us. The background source is usually a quasar, but may also be, e.g., a gamma-ray burst. As light blueward of the quasar's own Ly$\alpha$ emission line travels through the Universe, it is gradually redshifted toward the rest wavelength of Ly$\alpha$ (1216 Å), so whenever a parcel of hydrogen gas is present along the line of sight, it will cause an absorption line at the corresponding wavelength, creating a "forest" of absorption lines in the blue part of the quasar's spectrum. These data are (primarily) from Croft et al. (2002).

$^\dagger$This is fortunate, since otherwise you wouldn't be here to ask the question.


You can never tell the luminosity of a star from knowledge of its color X-Y alone, that only tells you something about the surface temperature.  The problem is not solved by knowing the distance modulus, that only tells you how far the star is, but not how big it is.  If you want luminosity from those things, you need to know how bright the star looks, not just its color and distance.  Or you could look at the width of the spectral lines to infer luminosity class, but that brings in line resolution.  But if you know the distance modulus and Lx and Ly, not just Lx/Ly, then you can get the luminosity because you know its color, how bright it looks, and its distance.


What is the underlying nature of the dark spot found on Uranus?

There is a science-based approach to explaining the dark spot on Uranus. In 2009 (the image and sighting of the spot in question are from 2006) a paper was published titled: The Dark Spot in the atmosphere of Uranus in 2006: Discovery, description, and dynamical simulations by H.B. Hammel a, L.A. Sromovsky et. al..
In this work they say that the dark spot is a wind vortex. Quoting the paper:

Regardless of how it is computed, the dark features's zonal velocity is higher than all other velocities measured to date near that latitude.

Using computer simulations they are able to predict possible initial velocities, lifespan of the Vortex, center latitude of the initial and center latitude when it reached the end of its lifespan. They go further and state that the mere existence of the vortex shows the dynamic and evolving seasonal nature of Uranus's atmosphere.

So I'd like to know if there is any science-based thinking on the underlying nature of the dark spot on Uranus beyond "it looks dark, like the ones on Neptune".

The abstract explains that the spot has a markedly different spectral character than the one on Neptune:

The dark feature's contrast and extent varied as a function of wavelength, with largest negative contrast occurring at a surprisingly long wavelength when compared with Neptune's dark features: the Uranus feature was detected out to 1.6 μm with a contrast of −0.07, but it was undetectable at 0.467 μm; the Neptune GDS seen by Voyager exhibited its most prominent contrast of −0.12 at 0.48 μm, and was undetectable longward of 0.7 μm. Computational fluid dynamic simulations of the dark feature on Uranus suggest that structure in the zonal wind profile may be a critical factor in the emergence of large sustained