What was the age of the universe when the average density was one atmosphere? How long did it take after the Big Bang for the average density of the universe to reach one Earth atmosphere at sea level? How about the density of water? Is there a chart of densities that relate to everyday objects (so easier to understand)?

The critical density of the universe is
$$\rho_c=\frac{3H_0^2}{8\pi G},$$
with $H_0=67.8 \frac{\mbox{km}/\mbox{s}}{\mbox{Mpc}}$ the Hubble constant and $G=6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg }\mbox{s}^2}$ Newton's gravitational constant.
Hence with one parsec $1\mbox{pc}=3.0857×10^{16} \mbox{m}$,
$$\rho_c=\frac{3\cdot (67.8\cdot 10^3\cdot \frac{\mbox{m}/\mbox{s}}{3.0857×10^{22} \mbox{m}})^2}{8\pi\cdot 6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg}\mbox{s}^2}}=8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}.$$

Ordinary matter content is $4.9\%$, the universe has total density close to critical density, hence ordinary matter density is
$$0.049\cdot 8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}=4.23 \cdot 10^{-28}\frac{\mbox{kg}}{\mbox{m}^3}.$$
(The Hubble constant is not precisely known, so the calculated density depends a bit on the precise value.)

Density of water:
The density of water is about $10^3 \mbox{kg}/\mbox{m}^3$.
That's a factor $4.23 \cdot 10^{31}$ denser than the average universe.
The space needs to have been reduced by a factor $\sqrt[3]{4.23 \cdot 10^{31}}=3.48\cdot 10^{10}$ in each if the three spatial dimensions. That's the inverse of the scale factor, hence corresponds to a redshift of $z=3.48\cdot 10^{10}$.

Using the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc (slightly different to the above value),
$\Omega_{\Lambda} = 0.6825$ provided by the Planck project, and setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$ the age of the universe was 0.022 seconds at the redshift $z=3.48\cdot 10^{10}$, hence when it was of the average density of water.

Density of air at sea level:
The density of air at sea level is $1.225 \frac{\mbox{kg}}{\mbox{m}^3}$. The corresponding redshift is about $z=1.423\cdot 10^9$ yielding an age of the universe of 12.68 seconds.


Black dwarf binary systems I know black dwarf stars do not exist yet, and that they are what white dwarfs end up after a long time. The question is really simple: would it be possible to have a binary system in which one of the two is a black dwarf, and the other companion star is dumping mass on the black dwarf, causing the black dwarf to supernova? I guess based on the timescales, the maximal initial mass of the second object could be calculated to see if it is possible that one object is at the black dwarf stage and the other still a giant. Or is the timescale on which the star goes from white dwarf to black dwarf big enough that it will never reach the black dwarf stage while the other one is still an AGB star?

Second part: If not possible that one is black dwarf and the other AGB star, is it possible if the companion is a brown dwarf, that they are close binary, and that a brown dwarf would dump mass on a black dwarf? This would need the binary system to be a very close binary system at this stage of the system.

Third part of the question: would the spectrum of the supernova be any different from the normal one with a white dwarf and a companion AGB star? I guess the spectral lines would be different, since the companion would have to be a very low-mass AGB star, since that companion would be about 20billion years or so, when the black dwarf star becomes a black dwarf.

I see two real questions here. First, whether it's possible to have a black dwarf with a companion object. For a given black dwarf, this is unlikely, since the orbits would likely be unstable at the time scale required to produce a black dwarf. Given the size of the universe, however, it's not out of the question. A black dwarf could even capture a companion star after cooling down, although this would be extremely rare.
Assuming it's possible, the next question would be whether the supernova produced would be the same as a normal type Ia supernova. The supernova is generally triggered based on the critical (Chandrasekhar) mass which triggers carbon fusion. Since it's dependent on mass, it would likely be a "normal" type 1A supernova. Keep in mind not all 1A supernovae are identical to begin with, although they can still be "normalized" to function as a fairly reliable standard candle.
One thing to note is, the accretion process would cause the black dwarf to heat up, meaning it would not remain black.


If GRB 080319B happened next to nearest star: Proxima Centauri, what effect would that have had on us? From a Reddit post

GRB 080319B was a gamma-ray burst (GRB) detected by
the Swift satellite at 06:12 UTC on March 19, 2008. The burst set a
new record for the farthest object that was observable with the naked
eye: it had a peak visual apparent magnitude of 5.7 and remained
visible to human eyes for approximately 30 seconds. The magnitude was
brighter than 9.0 for approximately 60 seconds. If viewed from
1 AU away, it would have had a peak apparent magnitude of −67.57 (21
quadrillion times brighter than the Sun seen from Earth).
Wiki

Someone asked what would happen if that GRB happened from the distance of  Proxima Centauri, what effect would that have had on us?

Faster than light's answer is fine if the gamma ray burst is "pointing" towards us. The emission from rapidly moving material in a GRB is thought to be beamed in the direction of that motion, along the rotation axis of the progenitor object in fact.
The opening angle of the beam may only be a few degrees (e.g. Frail et al. 2001), which would mean that for random orientations of the GRB source there would only be about a 1 in 500 chance of the solar system being "in the beam". Thus the highly destructive conclusion of Faster than Light's answer is a remote possibility.
As the fireball phase of the GRB develops then it is thought that the beam opening angle will widen over days and weeks, but of course the source brightness also fades drastically on these timescales too. It should also be noted that most of the highly destructive X-ray and Gamma ray emission arises during the tightly beamed phase within minutes or hours of the burst.
Thus although I'm not saying we're not "screwed", I am saying that the issue is more complex than scaling the apparent magnitude of a GRB and comparing it to the Sun. I suspect that in most cases the effects would be similar to the explosion of a supernova at the distance of Proxima Cen.
The conclusions I can garner from such as https://physics.stackexchange.com/questions/26578/near-earth-supernova?rq=1 , suggest that would still likely remove the ozone layer for some time and therefore could have serious and deadly consequences for life on Earth.


Convert orbital elements to positions and velocities I've written a quick and dirty n-body simulator, but currently it requires initial positions and velocities. I've been looking for a way to convert Keplerian orbital elements to positions and velocities. I've searched the web, but I only seem to get conversion from orbital elements to positions and velocities in ECI coordinates, or to a system that orbits the Earth. Ideally, I am hoping that there is an easy conversion between orbital elements and position and velocities for an observer on Earth looking at this n-body system somewhere out in space. I'd appreciate any help in determining this conversion.

Is this what you're looking for? When you say you want something defined for an observer on the earth looking at the system, do you mean you wish to set the origin of the system at the earth? This can be done, but I don't really see why you would do it unless the earth was intrinsic to the system.


Why doesn't a singularity spin infinitely fast? A question for a random reader passing through: because of the nature of things spinning faster when it gets smaller, wouldn't a singularity spin infinitely fast as it's infinitely small, if it has a spinn from the beginning that is? Btw I know what a "ringularity" but does it automaticly become one when rotational velocity is added? Is it just "fake news"?

Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a "theory of quantum gravity" which would take over where GR breaks down (at the physical singularity).

For a non-rotating (Schwarzschild) BH, the physical singularity is at $r = 0$, but the coordinate singularity is at the event horizon, $r = 2M$.

For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is the ringularity. We literally don't know what should happen here (need "quantum gravity"), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).

OKay, with all that out of the way, we can still make a similar comment like what you want to within the framework of GR:

Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin $a$ whose symmetry axis is along the $z$, so that the ringularity lies in the $x$-$y$ plane. One can show that at $z = 0$ and at $r = 0$ (where $r$ is not the physical radial distance), the equation of the ringularity is

$$ x^2 + y^2 = a^2$$

and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.

  Why doesn't a singularity spin infinitely fast?

the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of "spin" of the ringularity, but it wouldn't make much physical sense.

Now, if the BH was spinning infinitely fast, i.e. $a \rightarrow \infty$, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the naked singularity hypothesis. You see, if BH's could have a spin greater than 1, then the ringularity would extend "outside" the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)


Why is constant velocity drift not an accepted explanation of Hubble expansion? Edward A Milne at the Oxford University proposed that what Hubble was
  observing was simply a natural sorting of random galactic motions. If
  a group of galaxies formed together moving at various speeds, it was
  natural that the fastest ones would now be the most distant, while the
  slow-moving ones would remain nearby.

Source: https://aeon.co/essays/how-they-pinned-a-single-momentous-number-on-the-universe

Why is this incorrect?

I understand that gravity wouldn't normally allow expansion at constant velocity, but we have already conjectured dark energy to account for that.

The simplest answer is that Milne's explanation requires a non-uniform cosmos, with a formerly crowded "starting area" that expands into empty space.  Observations indicate that the cosmos is nearly uniform at very large scales.  With Milne's model of a non-expanding space and uniform cosmos, galaxies that started near to us with high velocities that ended up far away would be replaced by galaxies that started far away and ended up near, so there would be no observable relationship between distance and velocity.

There is a slew of other observational evidence against the naivete' of Milne's model, of course.  The cosmic microwave background shows that space use to be dense and hot; galaxies evolve, interact, and combine over time; there are intergalactic clouds that influence galaxies' trajectories.  Only when you can account for all that does incorporating elements like dark energy into your model make sense.


Are there stars that don't emit visible light? Are there stars that do not emit any light in the visible part of the EM spectrum?

There are two possible reasons why a star would not be detectable in the visible part of the spectrum (even with the most powerful telescopes not yet invented), apart from the trivial (too far away, hidden behind screens of dust).

It has too low a temperature. To not emit much in the visible part of the spectrum, a star has to be very cold indeed, at most a few 100K, when radiation is mostly in the infrared (which is still detectable). By definition, a star is an object that undergoes (or has undergone) hydrogen burning (H$\to$He fusion). This sets an lower mass limit of 0.08M$_\odot$. All objects of that mass (or more) ever formed in the universe are still much hotter than 100K. However, in the very distant future, some 'dead' stars (without remaining fusion energy source) will cool down to such temperatures, this includes white dwarves.
It has too high a gravitational redshift for any light to appear in the visible. In fact, the so-called stellar-mass black holes (remnants of supernova explosions of massive stars) may in fact be such objects: strange stars denser than a nucleus (consisting of a quark-gluon plasma as dense as a neutron) but with such a small size and high mass that any radiation emitted from their surface is redshifted by a factor 1000 or larger.

So if you consider those (still hypothetical) strange stars, then these are possible candidates.


What satellite is this? While out viewing the Perseid shower last night (8/11/20), we saw an interesting object that I'm trying to identify. We're in the vicinity of Portland, Oregon. It appeared to be traveling west->east and would be almost invisible to the naked eye for 10 or so seconds, then flare up to be the brightest object in the sky for ~2 seconds. We saw two passes (or two separate objects) over the course of an hour, the first was a bit to the south (say 1 o'clock), the second was farther north (11 o'clock), it might have been the same object.
I was thinking it might be a Starlink satellite, but I would have expected to see more than just one. Also, I wouldn't think they would be that bright, being as small as they are.
I've not seen anything like that before, and it surely would have been noticeable in my past stargazing.
Anyone know what it might be?

It sounds like a tumbling satellite and by your description of the different locations in the sky during the second pass, it sounds like you saw the same one over the course of two orbits.  Maybe not, though, because it'd (most likely) be 90 mins from one orbit to the next and conditions would have to be just right for the geometry of the Earth/Sun/Satellite angle to make it visible for two orbits so brightly. But not impossible!  The Starlink satellites don't vary in brightness that way, so I can safely say it wasn't a Starlink. Tumbling sats can be wild to see, some are spinning very fast and flicker quickly (a couple of times a second) while others are doing a much slower tumble.  Sadly, there is a lot of space junk up there launched without a plan to decommission it (bring it down over the Pacific to burn up harmlessly).


What is the resolution in megapixels of the Hubble Telescope? What is the resolution in megapixels of the Hubble Telescope?

The HST cannot see the planet Pluto clearly. Why is that?

The size of the cameras that took the Pluto pictures is easy enough to find: the ACS Wide Field Camera has two 2K by 4K (hence 8 Mega pixel) CCDs, with a field of view of 202×202 arcsec. The high resolution HRC had a square 1 Megapixel camera with a 26×29 arcsec field of view.

The reason that the images of Pluto are relatively poor, is that Pluto is small! The size of the camera is irrelevant; it is the size of the pixels and angular resolution that are vital.

The pixel size of the HRC (which I think is what the pictures are mainly due to) is 0.025 arcsec, which matches the angular resolution of 0.05 arcsec. The angular diameter of Pluto is about 0.1 arcsec. Hence you effectively have an image that is only made of 4×4 pixels!  I'm surprised it looks as good as it does - but that is because these are not the raw images (see below).

Edit: In response to comments I had a look at some details from the original paper describing these images, by Buie et al. (2010).  http://iopscience.iop.org/article/10.1088/0004-6256/139/3/1128/meta  It seems that what is shown is not an actual image from the ACS, but is a best-fitting map of what Pluto looks like, obtained by modelling all the images available.

The model image is described by a much finer pixel scale than the original images, but then blurred with a Gaussian filter. However, to quote from the paper, "The width of the Gaussian filter was chosen to impose a limit on the spatial frequencies in the output image comparable to the diffraction limit of HST. "

Thus the angular resolution of the final "reconstructed images" is still similar to and limited by the angular resolution of the HST.

Just to hammer home the message. It really doesn't matter here how many pixels the HST cameras have, so long as they have enough to sample the angular resolution properly. At a 0.1 arcsec angular diameter as seen from Earth, Pluto has an angular size equivalent to trying to take a photo of a quarter from 50 km. It really isn't going to help you if you have the latest camera phone with a gazillion pixels unless you also connect it to a 2.4m telescope like HST with an angular resolution better than this.


Why doesn't the equipartition theorem disallow spinning dust? This excellent answer to the question What is the physics of the “spinning dust” contribution to Cosmic Microwave Background measurements? Is well sourced. The abstract of one of the papers there, Purcell 1975 Interstellar grains as pinwheels begins with:

  Contrary to the assumption usually made in theories of grain alignment, the rotational energy of an interstellar grain is likely to be very much greater than 3/s kT, where T is the gas temperature, or the grain temperature, or any other temperature in the system. Any more or less permanent irregularity of the grain’s surface with respect to accommodation coefficient, distribution of H-H recombination sites, or photoelectric emissivity will result in an unbalanced torque capable of spinning the grain up to high angular velocity. Such a grain is, in effect, the rotor of a heat engine.

In the body of the paper, the first example given is of a sort-of gedankenexperiment involving a hypothetical Crookes radiometer with the gas removed. The author explains that it would spin up to high energy (without saying how):

  [...] The kinetic energy of the rotating paddle wheel is now enormously greater than it would be if it were immersed in, and en equilibrium with, a 6000 K radiation field, for then it would exhibit only a residual “Brownian rotation” appropriate to that temperature. So the wheel is really a heat engine, depending on the difference between the temperature of the vanes and that of the radiation field to which they are exposed.

Without the gas to cool them, I suppose that after the whole thing reaches equilibrium temperature the black sides radiate more strongly than the white sides and so there should always be a torque, and the final speed and kinetic energy would be limited only by friction and other practical losses.

This excellent answer to the question What is the difference between gas and dust in astronomy? points out that there is no real distinction between the two, but that they are limits in size, gas particles being small and dust particles being large(r).

So if dust can become pinwheels, then I have a hunch that very very large molecules can do this too, and if so, then so can smaller molecules. All you need is some mechanism for torque. Molecules can certainly have a non-uniform distribution of  H-H recombination probability just as dust can, and they can have some regions that are concave and some that aren't.

This makes me wonder if these gas molecules no longer obey the Equipartition theorem of thermodynamics. The spinning dust certainly doesn't.

Question: How can spinning dust and by induction, my hypothetical spinning molecules avoid conforming to the equipartition theorem? Why doesn't the equipartition theorem disallow spinning dust?

Source borrowed from here

Bonus points: Is that little arrow ($\omega_{rot}$) pointing the wrong way?

The first relevant point is that the equipartition theorem only formally applies in thermodynamic equilibrium, which requires that all temperatures be the same.  Since the spinning dust only happens when there are temperature differences, holding to the analogy with a heat engine mentioned in the source, we should not expect that theorem to be the whole story.  Granted, we often apply that theorem even when there are temperature differences, so the issue boils down to, when is the behavior more or less the same when all temperatures are the same, and when is the behavior fundamentally a heat engine?  I will assume the source is correct that this is heat engine-like behavior for spinning grains.

Once you have a heat engine in operation, the sky is the limit on how much energy you can put in any given mode (as long as you can maintain the temperature difference), because a heat engine does work, and work energy never needs to be equipartitioned, you can partition it any way you like by setting up an appropriate apparatus.  The dust particle is therefore a kind of apparatus for doing work, in the presence of temperature differences, and channeling the work energy into rotation.

How and why that heat engine operates and picks rotation to channel the work is a much more subtle issue.  For example, do we need temperature differences across the dust grain itself, or just between the grain and the surroundings?  If the "hypothetical" Crooke's radiometer (meaning, one that works quite differently from actual Crooke's radiometers and actually spins the opposite direction) is a good analogy, then the spin is from temperature differences across the dust grain itself, which warm the gas in the surroundings of the dust grain, producing gas flows that maintain constant pressure.  Those flows require the gas to receive torques, and conservation of angular momentum requires an opposite torque on the dust grain.

To get torque like that, it seems that you not only need a temperature difference across the dust grain, but it also has to have a kind of "quadrupole" character-- hot-cold on one side of the grain, cold-hot on the other, producing a "handedness" to the temperature structure in the grain.  So we immediately see that one cannot get spin until one has a large enough system to be able to support the concept of "temperature difference" across the object, and molecules are generally not thought of that way-- though maybe in the case of very large molecules, they could be.  So the question then comes down to, how large does a molecule need to get before it can support a concept of spatially nonuniform temperature? I don't know how big that needs to be, but my guess is, very big indeed-- there are certainly polymers that can get very large, but the molecules talked about in astronomy are generally not polymers.

As for the sign of the rotation, that can be very difficult to figure out!  It seems to depend on how the temperature differences get created in response to absorption of a radiation field (or other interactions with the environment, perhaps thermal contact with warmer gas).  Generally speaking, concave sides warm up more quickly than convex, so cause the gas in the concave part to be warmer than in the convex, so a flow sets up from concave to convex.  That acts like a jet engine that pushes the concave side forward.  That this is subtle can be seen from the fact that the higher pressure gas would seem to push the other way on the dust, but it's like the way a sail on a sailboat works-- follow the deflection of the air and conclude that the boat deflects the opposite way.

If this is all correct (see the Wiki on Crooke's radiometers for background information), then I would say the shape of the dust grain you drew produces spin in the direction that you indicated, but that no astrophysical molecules are large enough to produce the same effect.


Have Pluto or any of its moons been mapped? I was looking over some of the data for the terrestrial planets, and noticed that Venus, despite its cloud covering, had been mapped by the Arecibo Observatory. Also that other planets such as mars, and some of the Jovian moons were mapped by the Voyagers. So, do any maps exist for Pluto or its moons?

Yes, but they're not very good and they're amazing.

The Wikipedia article for Pluto shows a low-resolution map of the surface, generated from Hubble images:

And the Wikipedia article for Pluto's largest moon Charon shows a low-resolution map of the Pluto-facing side of Charon (not to scale):

Larger image here.

Only the Pluto-facing side is shown because the map was generated from brightness variations during Pluto–Charon occultations.

Maps with similar resolution would show no details of any of the other moons, which are much smaller.

Expect much better images in a couple of years, likely including moons that haven't yet been discovered.

UPDATE : Thanks to user HDE 226868 for reminding me: it's been a couple of years, and we have those much better images now. Here's one from New Horizons via Wikipedia:

Since the question was about a map of Pluto, here you go. You can see the much lower resolution on the opposite hemisphere, which New Horizons saw only from a greater distance (it was a flyby mission).

We also have some good images of Charon:

And here's a map of Charon:

and some lower resolution images of the other moons, Styx, Nix, Kerberos, and Hydra.

My guess that New Horizons would discover new moons turned out to be wrong.


Adjective for things outside our solar system I know we have "exoplanets" etc. but those are nouns for specific objects outside our solar system.

We have extraterrestrial for objects outside Earth's atmosphere, but I don't know if we have a general adjective for objects outside our solar system.

As @AtmosphericPrisonEscape noted in comments, above, "Extrasolar" is the correct term.  E.g., we refer to "extrasolar planets", not "interstellar planets."

"Interstellar" refers to things among the stars (Merriam-Webster: "located, taking place, or traveling among the stars especially of the Milky Way galaxy")

"Extrasolar" means "outside the Solar System." (Merriam-Webster: "originating or existing outside the solar system")

One distinction is that "interstellar" doesn't include all space outside the Solar System -- "Interstellar space" is usually spoken of as distinct from, and much smaller than, "Intergalactic space".

The original question asked for "a general adjective for objects outside our solar system" and that would be "extrasolar".


What will happen to life on Earth when the Andromeda and Milky Way galaxies collide? It is said that the Andromeda and Milky Way galaxies are coming close to each other with a speed of approximately 400000 km/hour. They will be together in the next 4 billion years.

What will happen to life on Earth or human beings on Earth?
If we are about to collide in the next 4 billion years then how long before we should take action for interstellar voyage?
Are scientists working on such projects for an interstellar voyage?
Will we be able to get the something like Earth to go away from this galaxy/Earth/ solar system and, considering speed of human instruments/spaceships, how long will it take to go to a safe place?

What will happen to life on earth or human beings on earth?

Assuming that human beings, or life, still exists on Earth at that time, they will have survived so much due to the ongoing death of the sun, that the gravitational pertubations due to the galactic collision will be nothing.

Keep in mind that in about 1-2 billion years, the sun will be so hot and large that all the water will have boiled off the earth into space.  About 3 billion years from now, the surface of the Earth will be so hot that metals will be melting.

Any life that has survived those events and still lives on Earth will surely take a galactic collision in stride.

I imagine, though, that most humans will have fled Earth - if not for distant star systems, then at least for planets in our own system that are going to be warming up enough for human habitation.

  If we are about to collide in next 4 billion year then how long before we should take action for inter-stellar voyage?

As soon as possible. When interstellar voyage becomes possible, we should start sending out ships to colonize other planets and star systems.  This will likely take a long time, but if we are to survive more than a billion years, it is necessary. Keep in mind that the Sun and Andromeda are events we can predict.  We don't know, and can't predict, the next cataclysmic asteroid strike, which is likely to happen in a shorter time than a billion years.  There are lots of reasons to exit the planet, we should be worried about the ones we can't predict or see, not the ones we can predict.

  Are scientists working on such projects for inter-stellar voyage?

Yes, but in small steps.  Manned missions to space, to the moon, and living aboard the ISS have provided significantly valuable information that will be used in such interstellar missions.  As we continue to push the boundaries of our ability to survive in space we eventually will be able to live in space, perhaps whole lifetimes will be spent in space.  As engine technology progresses beyond simply lifting people out of Earth's gravitational well, we will eventually be sending people on long voyages outside our solar system.

It's a very, very long way off, but each advance takes us closer to that eventual goal.


What you would see in a wormhole? Science fiction seems to depict the experience of traveling through wormholes (if possible of course) as seeing psychedelic colors or lines of light where stars once were (ie. 2001 or Contact). I wonder if theorists have any idea of what, if anything, the traveler would see.
This question may be subjective, but I know that there are theories about what an observer would see if traveling near the speed of light.

Around the worm hole would be a region with an intense gravitational field, and there would be gravitational lensing occurring, similar to that around a black hole. The difference would be that light that has passed through the worm hole would would be visible. So, a distorted view of the "other side" could be seen.

This physicsbuzz article has some images and discussion. Note that the wormhole would be a spherical region in space, and not a 2d "stargate"


Has New Horizon's data updated Charon's orbital elements? For years I've been fascinated with the mutually tide-locked bodies Pluto and Charon. In July 2012, The Astronomical Journal published an article The Orbit of Charon Is Circular by Buie, Tholen and Grundy. The authors thought that the eccentricity of Charon's orbit is very close to zero.

It seems to me their opinion is somewhat speculative given the low quality of images from Hubble. Has New Horizons data verified Charon's circular orbit? Is there an online source giving more precise orbital elements from New Horizons data?

I'm also interested in the obliquity of Pluto and Charon.

I suggest you use the JPL Horizons system. You can investigate how Charon's orbital elements change with time. The short answer is, not surprisingly, not by much. The values for the eccentricity I see hover around 0.00015 and those for the orbital inclination relative to Pluto's equatorial plane are about 0.09 degrees.

http://ssd.jpl.nasa.gov/horizons.cgi


Could a rocky rogue planet get trapped in the orbit between Earth and Mars? I know it's extremely unlikely a rogue planet passing by there, and I know the size of rocky planets can vary a lot (IIRC there is a rocky planet 5 times the size of Earth) but in any hypothetical case, could a rogue planet be trapped in any part of the orbit between Earth and Mars? If so, which are the possible effects it could have over Earth and Mars? In which cases would they be expelled from the solar system?

It is exceptionally unlikely.

Imagine that there was a planet already there. What could cause it to suddenly escape from the solar system?  It would have to be a massive event, such as a second passing rogue planet catapulting it out of orbit.

Running time backwards, for a rogue planet to be captured and neatly end up orbiting between Mars and Earth would require an unbelievably unlikely sequence of events, such as

  two rogue planets happen to pass between Earth and Mars at the same time and the interaction between the two results in a transfer of energy and momentum, which (by lucky chance) leaves one of the planets in a nearly circular orbit, (while the other escapes)".

Given how big space is, and in consequence, how rarely a rogue planet enters the inner solar system, this would make the above sequence of events practically impossible.

In this unbelievable scenario, the effects on Earth and Mars would depend on how closely the new planet passes. If we are close enough for significant effects on the planet, we are in a great deal of danger.


Optics: "Airy disc" and size of objects Jupiter has a minimum brightness that is brighter than stars. Is the reason an “airy disc”, concentric rings as a diffraction effect, is seen around stars and not Jupiter because Jupiter has an angular size that is thousands of times larger? With similar reasoning, if the field of view was scaled up a thousand fold, would the optical effects be similar to that around stars?

No.
The size of the Airy disc, and rings, is defined for a point source and depends on the system magnification, and the system aperture. It does not depend on brightness other than depending on the low-level detectability limit of your sensor.

It is true that the diffraction pattern of an extended object is roughly the superposition of Airy disks from every (geometric) point in the object's image. There will be a 'halo' around the image of width (approximately) Airy disk radius, and then there will be secondary maxima generated by the (interfering) Airy rings from all the image points.


Find distance from star to star? How do you find the distance from a star/planet/black hole to another? I know people can calculate the distance from Earth to a star, but what about from one to another?

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


Which will destroy the Earth first: the Sun or Jupiter? The sun is expected to grow in brightness over time, slowly heating the Earth until the oceans evaporate, leaving the planet uninhabitable.
Jupiter is the source of the largest perturbations to Earth's orbit. In general, the 3 body problem (and higher) is not stable. So we can expect Jupiter to, eventually, perturb Earth's orbit until Earth falls into the sun or gets kicks completely out of the solar system.
So, which do we expect to kill the Earth first: solar baking or orbital chaos?

I found an article by Ian O'Neill posted on May 2, 2008 at universetoday.com with the title Could Jupiter Wreck the Solar System? which says

But here’s the kicker: There is only a 1% chance that these gravitational instabilities of the inner Solar System are likely to cause any kind of chaos before the Sun turns into a Red Giant and swallows Mercury, Venus, Earth and Mars in 7 billion years time.

This is clearly after the time when the Sun is becoming a red giant in 4.5 – 5.5 billion years - this is the number given in an article Will Earth survive when the sun becomes a red giant? by
Matt Williams from 2016.
References
The first post is refering to a scientific article On the Dynamical Stability of the Solar System by Konstantin Batygin and Gregory Laughlin.


What planets or exoplanets orbit the Sun’s elder twin HIP 102152? Has anyone done any research on the planets that orbit the HIP102152? Since that star is similar to ours and older, I postulate that it’s likely those planets are most worthwhile searching for advanced civilizations.

No planets have been detected in orbit around HIP 102152.

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

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

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

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


How to calculate magnitude of a star in a triple star system? I want to find a star's magnitude in a triple star. I already knew the the magnitude of the triple star and the other two star, but I don't quit know how to solve it. Is there a way to find it?

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


Photographing Mars, Jupiter and Saturn I want to take pictures of planets, well at least Mars, Jupiter and Saturn with some surface details.
I have a Canon EOS 7D Mark-II and a Tameron 150 - 600mm lens (very heavy), a 1.7X extender, and a... meh... half decent tripod that can take a peak in the sky directly overhead, with some neck discomfort. So far, I have been able to only capture a fuzzy blob of Mars with aura of orangish red light around it. But that is partly because the skies were not very clear, and some clouds quickly tried to hide the sky when they saw me taking a picture.  And it is Late October. So... Any suggestions, ideas,

a 1.7X extender

By 1.7X extender I think you mean a Teleconverter; a projection lens that re-images your main lens's focal plane back on to the sensor, but magnified.

a Tameron 150 - 600mm lens (very heavy),

This says Tamron SP 150-600mm f/5-6.3."
So 600 mm  x 1.7 = 1020 mm focal length, and I'm guessing f6.3 at 600 mm so 95 mm aperture.
This combination is a f=1020 mm f/10.7 telescope!
If you magnify the planets they will get dim at f/10.7 so you'll have to expose longer, meaning atmospheric seeing will cause some trouble.
From here I see that

Canon 7D Mark II comes with a 22.4 x 15 mm CMOS sensor, which has a diagonal of 26.96 mm (1.06") and a surface area of 336.00 mm²... The sensor has a surface area of 336 mm². There are approx. 20,200,000 photosites (pixels) on this area. Pixel pitch, which is a measure of the distance between pixels, is 4.08 µm.

So with f = 1020 mm and 4.08 E-03 mm pixels, each pixel will be about 0.83 arc seconds.
You are ready to go, and you should probably skip the teleconverter at first and just gain some experience; setting up, avoiding dew, pointing at the planets, watching them drift by as the Earth rotates, then re-pointing over and over.
Then taking photos without vibrating the whole thing! DSLR's usually (always?) have mirrors that flip, you may want to delay the exposure until it stops vibrating.
Then taking many photos and learning to stack them, or sort them for lucky imaging.
Photography SE has lots of astrophotography questions and answers I recommend you go there and read a bunch before asking further so as not to ask a duplicate question.
-https://photo.stackexchange.com/questions/tagged/astrophotography
Have fun!


Calculating the orientation of the night sky How would one calculate the rotation/tilt of the earth to simulate the Night Sky in a self-written tool or app. I am trying to built an app for my telescope to show me on my phone what I am looking at. I can already simulate the exact positions of stars at the 1.1.2000 at 00:00:00 but I dont know how the "earth" should be rotated so I can simulate the sky from my current position and accurate time.

There are App and tools like stellarium that can achive this but I just cant get the angles right:

If anyone could point me in the right direction and tell me what I need to change/calculate it would help me a lot.

If you do not require arcminute precision, you can approximate Greenwich sidereal time as the Earth rotation angle $\theta(t)$.
IERS Technical Note 32 §5.4.4 gives
$$ \begin{align}
   \theta(t) &= 2 \pi (0.77905~72732~640 + 1.00273~78119~11354~48~ t) \\
      &\approx 280.46^\circ + 360.985612^\circ~ t
\end{align} $$
where $t$ is a real number of days since JD 2451545.0 (2000-01-01 12:00 TT ≈ 11:59 UTC).

For an observer at north latitude $\phi$ and east longitude $\lambda$, things should line up like this:

Local sidereal time $ = \mathrm{LST} \approx \theta(t) + \lambda$

Zenith (RA, Dec) $ = (\alpha, \delta) = (\mathrm{LST},~ \phi)$

North horizon $(\alpha, \delta) = \begin{cases}
   (\mathrm{LST + 12h},~ 90^\circ - \phi) & \mathrm{if}~\phi >= 0 \\
   (\mathrm{LST},~ 90^\circ + \phi) & \mathrm{if}~\phi < 0
\end{cases}$

East horizon $(\alpha, \delta) = (\mathrm{LST + 6h},~ 0^\circ)$

The transformation between equatorial and horizontal coordinates can be composed of two rotations, similar but not necessarily identical to those in Wikipedia: Celestial coordinate system.
For example, you could:

Start with the north horizon vector pointing at the north celestial pole, and the zenith vector pointing at (0h, 0°).
Rotate the ground counterclockwise as seen from the north horizon vector by LST.
Rotate the ground clockwise as seen from the east horizon vector by $\phi$.


Which is more rare: Lunar eclipse or Solar eclipse? Which is more rare: Lunar eclipse or Solar eclipse? Please explain why either one is more rare than the other.

Thank you in advance for your inputs.

Solar and lunar eclipses occur about equally often - between two and four times per year.

However, when you do not intentionally travel around the world chasing solar eclipses, you are more likely to observe more lunar eclipses. The reason is that solar eclipses can only be observed from a comparatively small area while lunar eclipses can be observed from anywhere were the moon is visible when they occur, which is half the planet.


What is the "lunar irradiance" received by the Earth from the full Moon? The solar irradiance received by the Earth is 1361 $W/m^2$. What is the lunar irradiance received by Earth from the full Moon in the same units? My first thought was to work it out via apparent magnitude but I don't think that would work out.

You can work in magnitudes, but the magnitude scale is logarithmic.  Instead you can use luminosity.  The moon has a luminosity that is 400,000 times less than the sun, so the maximum theoretical "lunar irradiance" is about 0.0034 watts per square metre.
In practice, the luminosity at the surface is less than this, with Wikipedia quoting "how bright is moonlight" typical values on the Earth's surface of 0.1 lux, equivalent to 0.001 watts of visible light per square metre.


Why haven't I seen the Moon? It has been some time since I've seen the Moon. What are the conditions that might affect whether or not the Moon is visible within a 24-hour period?

Please take into consideration any notable astronomical effects: moon phases, light pollution, other viewing conditions. I want this to be applicable to other cases as well as my own.

Here is my particular case:

  I live in Wales. The date today is October 3, 2016. There has been no moon visible in the sky for the last 5 days, and I don't know why.

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

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

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

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

Image source and further information: SEMS: Moon Phases

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

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

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


What would theoretically be dissolved in the interior oceans of icy moons and planets? I know we have some indication of "what's in the water" from the plumes of Enceladus, but would other places be the same. For instance, could the deep liquid water of Titan contain dissolved methane gas?
I'm mainly contemplating what possible "solutions" would be beneficial to some form of microbiology.

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.


How do astronomer measures the size of any celestial objects? What techniques and tools are available to the astronomers to measure the size of any celestial objects such as, stars or perhaps black holes that doesn't emit light nor reflects starlight?

The main tool to measure the diameter of a star is interferometry combined with a parallax-based distance measurement - a brief review by Kervella (2008) might be useful. The principles behind interferometry are described here.

Interferometry involves measuring the light from a star using two (or more) telescopes that are separated by some distance. Together, the signals from these telescopes can be combined to give an angular resolution that can be (in the best circumstances) equivalent to a telescope with a diameter equal to the telescope separation. These measurements give the angular size of the star, which must then be multiplied by their distances to get a physical diameter.

One of the most successful experiments is the Chara array, which has yielded diameters for many nearby stars. Precisions can be as good as a few percent, but more usually 10% and of order 100 (predominantly nearby) stars have had their radii measured in this way.

A second main direct technique is to use eclipsing binary systems. The measured light curve can be used in an almost model-independent way to estimate the radii of the two stars involved. Of course most eclipsing binaries are close pairs with short orbital periods and with orbital inclinations that allow us to see the eclipse. They are therefore highly prized objects. Radii can be measured with precisions of 1%. A reasonably complete catalogue of the $\sim 100$ known eclipsing binaries with precise radii can be found here.

Another technique is lunar occultation. The passage of a star behind the limb of the moon results in a changing diffraction pattern that can be used to estimate the angular size of the star. Again a distance is required to convert this into an actual diameter.

More distant stars are inaccessible - their angular diameters are simply too small. At the moment only indirect estimates of their radii are possible. For example, if we were to assume that a star radiates as a blackbody, then its luminosity ($L$), radius ($R$) and temperature ($T$) are related by Stefan's law.
$$ L = 4\pi R^2 \sigma T^4,$$
where $\sigma$ is the Stefan constant. If the star has a measured flux at the Earth and we know how far away it is, then $L$ can be estimated. If we take a spectrum and estimate its temperature, then the equation above can be rearranged to give the radius in terms of the measured luminosity and temperature. Real stars are more complicated than blackbodies, but the principle is the same.

Neither of the above techniques can work for black holes, and the sizes (event horizon or Schwarzschild radius) of black holes have not yet been directly measured. The physics of a black hole is relatively simple(!) and so there is a direct relationship between their Schwarzschild radii and their masses (modified somewhat by rotation). Basically it is 3 km multiplied by the mass in solar units. Therefore a measurement of the black hole mass gives its "radius". The masses of black holes are measured by looking at the motions of stars and gas around them and applying our knowledge of how gravity works.


What is in the center of the universe? If the universe has formed & originated by a Big Bang Explosion, then there must be empty space left in the center of the explosion site, as all the matter is travelling at tremendous speeds away from the center, and there must be more matter, stars, galaxies and dust, etc near the present periphery or circumference or horizon of the present universe. As that big explosion has taken place about 13.7 billion years back, then the outer boundaries of our universe are 13.7 billion light years away from the centre of the explosion of Big Bang.

Have our astronomers discovered hollowness or emptiness anywhere in the centre of the universe or not?

I think your question is on topic, but @RhysW has linked a very helpful post in understanding why your question is a common misconception about the Big Bang.

No Center

There is no 'center' to the universe. At any point, a local observer will claim that they are at the center of the universe by the way galaxies move away from them. How can we possibly know this? The universe appears to be both homogeneous (has the same structure everywhere) and isotropic (there is no preferred direction). If these are indeed properties of the universe, then the expansion of the universe must be the same at all other locations (See: The cosmological principle).

How the Big Bang and Explosions Differ

Additionally, the Big Bang is different from an explosion in the following ways:

1) Particles involved in an explosion slow down eventually due to frictional forces. Think of fireworks (http://www.youtube.com/watch?v=qn_tkJDFG3s). Particles are moving the fastest at the instant of explosion, and slow monotonically with time. The expansion of the early universe does not follow this trend, though sometimes people use the word 'explosion' to describe the enormous volumetric increase (an increase by a factor of $\sim10^{76}$) which occurred between $10^{-36}- 10^{-32}$ seconds after the Big Bang, which is aptly named inflation.

2) An explosion implies the existence of space. For an explosion to take place, particles (whether we're talking about matter or light) must have space to explode into. Strictly speaking the inflation of the universe is an expansion of space-time coordinates, and so the word explosion cannot really apply since there was nothing for space-time to explode into.


Why does the Sun move slower during the solstices? A comment under this answer states that the apparent angular speed of the Sun is 8% slower during the solstices. This is rather counter-intuitive, since the rotation speed of the Earth is constant (or close enough for the timescales considered).

Why does the Sun appear to move slower in the sky at the solstices?

All points on the celestial sphere execute a full circle every (sidereal) day, but the speed of a point with nonzero declination is slower than a point on the celestial equator because it's traveling on a small circle, not a great circle. This is exactly the same as how a point on the terrestrial equator travels at a higher speed than a point not on the equator. Eg, a point with a latitude or declination of 60° travels at half the speed of a point on the equator, because cos(60°) = 0.5

At the solstices, the Sun is on either the tropic of Cancer or Capricorn, so it has its maximum or minimum declination, approximately ±23.4°. So its speed is cos(23.4) $\approx$ 0.9178 relative to a point on the celestial equator, or about 8% slower, as Mike G mentioned in that comment.

Here's a diagram from an answer about latitude speed by SF. on the Space Exploration Stack Exchange:


Is a supernova powerful enough to destroy the other star in a binary system? (This question was originally posted via World Building, but I was instructed to post it here)
https://worldbuilding.stackexchange.com/questions/109457/is-a-supernova-powerful-to-destroy-the-other-star-in-a-binary-system?noredirect=1#comment332775_109457

The question is basically the same:

Can a binary star system have one of the two stars go supernova in the first place, and if so:

The first scenario would be:

Star A of the binary swells to red giant status, and then is either
touching or consumes star B: what would happen?

Scenario 2:

Star A goes supernova, with maybe 1 AU of distance between it and
Star B.

What info Id like to know:

Can it destroy the other star via supernova?
What is the effective    range of the supernova, and can it destroy
stars a light year or more    away?

The gravitational binding energy of a solar mass star of uniform density is $3GM^2/5R = 2.2774\times 10^{41}$ Joule. The energy of a typical supernova is $10^{44}$ J. So it looks at first glance like the explosion could disperse the other star...

...except that this assumes more than a thousandth of the supernova energy is absorbed and does work on the star. If the other star is more than 16 stellar radii away then it will definitely not be hit by this much energy. And even if it is close enough, converting radiation to mass lift is not going to be efficient. The nova will mainly heat up the surface layer down a few optical depths, meaning a few thousand kilometers get very hot but it is doubtful this does more than push the star away a bit and induces some wobble.


Serious alternate form of the Drake Equation, or graffiti? Where I live it is very common to see jackets and T-shirts with familiar yet casually garbled or modified content (example)
I recently spotted what looked a lot like the Drake equation on the back of a motorcycle, shown in the photo.
Here is the equation from Wikipedia (more here):
$$N \ = \ R_* \ f_p \ n_e \ f_l \ f_i \ f_c \ L$$
and here is what I can make out from the photo
$$N \ = \ R_* \ f_p \ n_e \ f_l \ X \ X \ L \ B_X$$
where $X$ represents something I can't read, and note that there is a $B$ at the end that doesn't appear in the Wikipedia version.
There are also unspecified callouts on the $N, n_e, B$ and the first "$X$".
The sleeve also contains text which is only partly visible:
...CATING
...TIONS
...ALAXY

Question: Does this representation seem to reflect a serious new or alternate version of the Drake equation, or is it better or more likely characterized as "art" or graffiti?

Given the font, it's xkcd 384, The Drake Equation.
The two $X$'s are the other $f$ terms from the original Drake equation, while the $B_S$ is "amount of bullshit you're willing to buy from Frank Drake"


length of Star lifetime In a star, the fact that not all the hydrogen is used up, not all the helium is used up etc. is a little counterintuitive.

Shouldn't it be that a star looses the hydrogen layer passively, in other words all the hydrogen eventually fuses into helium forming a helium star which then fuses all its helium into carbon forming a carbon star etc. all the way to iron? In this way you get up to the temperature required to form iron and nickel in stars with a much lower mass and in massive stars you might even have Iron being fused to form heavier metals.

Why isn't this so?

Why isn't this so?

If the mass of the star is about 1/4 the mass of our sun, then the core's (where fusion occurs) temperature and pressure will never be enough to fuse anything other than hydrogen into helium. The death of this star may be something like a fire dying. The flames disappear for a while, then you may get one or 2 popping back into existence.

Our sun will start to fuse helium into carbon before all the hydrogen in the core is used. In the description of a heavy star, it is described as a onion with its layers with the higher fusion products occurring in the lower layers. They use this model since the iron fusion requires temperature and pressures above a certain level. Neon has lower requirements, but still greater than carbon fusion, and so on. What is not mentioned, is that there is still hydrogen fusion occurring in the center, as well as all the other fusion, helium, carbon, neon, etc.


Is their any clue where gravitons come from? Everything which has mass has gravity. Some theories hypothesis that gravitons are responsible for that, but do they have any clue where they come from? Like gluons are interacting between quarks and photons between eg electrons. But does gravitons have, according to those theories, also have a kind of particle on their own with whom they are interacting with. Or is it possible that fe gravitons are also coming from quarks?

There is no fully satisfactory quantum theory of gravity. The electromagnetic field is quantized, the quantum excitations are photons. It would be reasonable to believe that the gravitational field would also be quantized, and the quanta would be particles, called gravitons. In fact it would be very hard to explain how gravity could not be quantised.

The image of a particle emitting photons is rather misleading. Instead, two particles exchange a photon in an electromagnetic interaction. And the particles are not classical, but quantum mechanical.

Photons carry the electromagnetic force between any charged particles: not just electrons, but also muons, tauons, quarks, and the W particles. By this analogy, anything that can create a gravitational field would be exchanging gravitons: that includes anything that has mass or momentum. (ie more or less anything).

However, there is neither a theoretical, nor an experimental basis for the existence of gravitons.


Is the colour of a wave from a far galaxy the same for us as for a galaxy which lies between? Because of the expansion of the universe, light from a far galaxy is redshifted. The expansion of the universe will make the wavelength of light longer. But is the colour of such light the same for us, as for an galaxy which is between us and the far galaxy.

I first thought that a further galaxy's speed was higher than ours, but I don't think that's right it is relative. But is the speed of the far galaxy from a view of the further galaxy higher than from our view? I don't think so because the universe is expanding faster the further you look. So for the further galaxy the speed of the far galaxy is less high than from our view because the far galaxy is further away from us.

So is the conclusion right that the colour of the same wave is more red for us on earth than from the further galaxy?

If I understand you right, you're asking whether or not the redshift of the photons emitted from a far-away galaxy happens the instant it leaves the galaxy.

Redshift is gradual…

If so, the answer is no. The redshifting of photons happen gradually as they travel through the expanding Universe. You can find the derivation here where you'll see that every infinitesimally small increase $da$ of the scale factor $a$ of the Universe (its "size") increases the photon's redshift by an amount $dz$, or, in terms of wavelength, by an amount $d\lambda$.

If galaxy $B$ lies at redshift $z_\mathrm{B}$, then an observer in galaxy $A$ at redshift $z_\mathrm{A}$ lying between us and $B$ (so that $z_\mathrm{A}<z_\mathrm{B}$) would measure $B$'s redshift to be
$$
z_\mathrm{B\,seen\,from\,A} = \frac{1+z_\mathrm{B}}{1+z_\mathrm{A}} - 1,
$$
which is less than $z_\mathrm{B}$.

…at least in our Universe

The redshifting is not due to the source moving away from us. If the expansion hadn't been gradual, but we instead lived in a crazy universe that were static when the distant galaxy emitted the light, and static when we observe, but somehow expanded suddenly by some factor in the meantime, then we would still observe a redshift, even though the galaxy were static both when it emitted the light and when we observed it.


What is the YORP effect exactly? Is it just the non-central component of the Yarkovsky effect? This answer to Where have all the Vulcanoids gone? links to the aptly-titled The YORP Effect Can Efficiently Destroy 100 Kilometer Planetesimals At The Inner Edge Of The Solar System which says in part:

... The YORP effect destroys Vulcanoids by spinning them up so fast that the gravitational accelerations holding components of the body together are matched by centrifugal accelerations, this causes the body to rotationally fission. i.e break apart. We calculated the timescale of this fission process for a parent Vulcanoid and for each of their subsequent generational fragments. We show that objects with radii up to 100 kilometers in size are efficiently destroyed by the YORP effect doing so in a timescale that is much younger than the age of the Solar System...

Tha answer also links to the explanation of the YROP effect in The YORP Effect and Bennu
 which says:

The YORP effect is a similar phenomenon that affects the rotation rate and pole orientation of an asteroid. YORP is an acronym that combines four scientist’s names: Yarkovsky, O’Keefe, Radzievskii, and Paddack. Building on the work of Yarkovsky, V. V. Radzievskii showed in 1954 that variations in albedo across the surface of a small body in space could increase its rotation rate. This phenomenon is essentially the Crooke’s radiometer effect. Stephen Paddack and John O’Keefe, who were separately working on the origin of tektites and interplanetary dust, went on to show that an object’s shape strongly influences the change in rotation. David Rubincam synthesized these ideas in 1999, and showed that YORP creates a thermal torque akin to a “windmill” effect on asteroids. This torque can modify the rotation rate and obliquity of an asteroid, depending on the external geometry of the body.

This is frustrating because of course the Crooke's radiometer works by interacting with the molecules of the low pressure gas intentionally included inside the glass bulb, and not by radiation, but at least we get the idea that a net torque is produced due to a non-uniformity of some optical property.
Wikipedia's Yarkovsky–O'Keefe–Radzievskii–Paddack effect works hard to try to explain but the explanation is long and I am not confident it's actually accurate.
Answers to What is the difference between the Yarkovsky effect and YORP effect? touch on this but do not answer this question.
Question: What is the YORP effect exactly? Is it just the non-central component of the Yarkovsky effect?

Yarkovsky only considers the center of mass recoil from thermal radiation of a rotating body in sunlight, is YORP just the tangential component of this recoil?
Or does it require the body to be non-uniform in order to spin-up the body?
Would a uniform sphere in fact spin-down?

From What is the difference between the Yarkovsky effect and YORP effect?, both the Yarkovsky and YORP effects describe the changes in momentum to a body in orbit due to the re-radiation of photons.  The Yarkovsky effect only describes the changes to the orbital parameters of the body:

The Yarkovsky effect describes a small but significant force that
affects the orbital motion of meteoroids and asteroids

The YORP effect only describes the changes to the rotational momentum of the body:

This rotational variant has been coined the
Yarkovsky-O’Keefe-Radzievskii-Paddack (YORP) effect

Consider a perfect uniform sphere in orbit around the Sun.  For any point on the sphere, the average direction of re-radiated photons will be normal to the local tangent plane (or directly opposite the center of gravity). In this case, we expect a Yarkovsky effect (change in orbital parameters), but no YORP effect (change in rotational momentum). Graphic from this source.

In contrast, consider an irregular body with a huge flat face offset from the center of gravity.  When photons are re-radiated from the flat face, on average they will impart some angular momentum change (YORP effect) and some orbital parameter change (Yarkovsky effect).

Yarkovsky only considers the center of mass recoil from thermal
radiation of a rotating body in sunlight, is YORP just the tangential
component of this recoil?

The forces causing the YORP and Yarkovsky effects are not tangential in general. For a recoil for some impulse on a sphere, the rotational and linear forces imparted are tangential.  However, for an irregular shape like a rod, we wouldn't expect the two forces to be tangential. Here is a little drawing (extremely exaggerated for effect):

In the above drawing, the Yarkovsky vector represents purely translational momentum change and the YORP vector represents purely rotational momentum change. The sum of the vectors that represent forces that cause the YORP and Yarkovsky effects should sum to the mass recoil vector (which is the negative of the impulse):

They YORP and Yarkovsky forces don't necessarily need to be at right angles.  In three dimensions, sorting out the forces can get extremely complicated.    It can be quite complicated for a rod even if the reaction mass vector is at the tip. It's hard for me to imagine the recoil from a "crescent moon" shaped asteroid that has a center of mass exterior to the asteroid body.

Or does it require the body to be non-uniform in order to spin-up the
body?

The body must be irregular in shape to spin it up.  A perfect sphere with a non-uniform surface would not be subject to the YORP effect.  This is as mentioned above, since on average, photon emission at a specific point will be 90 degrees from the tangent plane, effectively canceling all YORP effects from individual emissions.

Would a uniform sphere in fact spin-down?

Again, a sphere would not have any YORP effects because the average re-radiation vector would be perfectly opposite the center of mass.
Note:
When an asteroid strikes a planetary sized body in an inelastic collision, momentum is also conserved.  Some of the asteroid's momentum goes into rotational momentum of the body and some of the momentum goes into the momentum of the center of mass (equivalent to the orbital parameters).  I like this analogy for thinking about the Yarkovsky and YORP effects.


Does the Lagrange L1 point have a 'size'? Like in cubic kilometers, what is the size of L1 "area of influence"? being unstable I guess that size can vary?

From this source I get:

  The size of these islands varies. Each planet in the solar system has
  its own Lagrangian points. The islands of stability get bigger farther
  from the Sun and also for more massive planets. The ones associated
  with Earth are roughly 500,000 miles (800,000 kilometers) wide. The
  biggest zones (at least in the solar system) are Neptune’s; they are
  about 2 billion miles (3.2 billion km) across.

L1, L2, and L3 are unstable. If a Sattelite ventures too far from the Lagrangian area it will fall towards the sun. As you can see from the image below. This image came from the source linked above.


How does a planet-sized robot standing on a planet affect its gravity? I originally asked this at the sci-fi exchange but was told it was off-topic as it was about real-world science.
Here is a picture of Unicron standing on top of the planet Cybertron, as written by Simon Furman and drawn by Geoff Senior in issue #75 of the Transformers comic.
Notice how Unicron is pretty much as tall of the planet's diameter. Assuming he was able to land there without causing the entire planet and himself to explode, wouldn't this have affected the planet's gravitational forces and possibly thrown it off its orbit?

Of course it would have had some effects! Gravity obeys the mathematical function:
$F = {G\ m_1\ m_2\over D^2 }$
where G is the constant of gravity (approx. $6.67430 \times 10^{-11} N m^2 kg^{-2}$), $m_1$ and $m_2$ are the masses of each body, and $D$ is the distance between each body (center-of-mass to center-of-mass).
Supposing the robot’s mass is about as much as the planet’s mass, it would definitely have thrown it off-course as it approached the planet—but, then again, the planet would also have affected the robot’s course.
It would not, however, have “affected the planet’s gravitational forces,” as the gravitational force exerted by a body depends only on its mass.
I would presume, though, that there are,  on planet Cybertron (though we can’t see them on this image) bodies of water, or simply loose objects. These would have been affected by Unicron’s presence/approach, and would have been somewhat displaced by his presence/approach, as there was a force (basically) as strong as Cybertron’s gravity pulling them in the direction where Unicron was at that moment. Such objects on the opposite side of Cybertron may have been crushed (if they were not made very resistant), for example, and such objects on the side of Cybertron facing Unicron would have “lost their weight” (at least partly) and—no, not float—being easier to displace…
A lot is not know in the context of your question. Maybe reading the comix where this comes from would help know Unicron’s trajectory, etc., but I’ll leave that for others.
Be well!


Can a star eat a black hole? I know there are stars much larger than the sun and that we've discovered "small" black holes, so, let's say that the Pistol Star, which is at least 80x the mass of the sun, and the smallest known black hole, XTE J1650-500, around 4x the mass of the sun, collide. Would the pistol star "eat" the black hole, or is there some property that a black hole has that is immune to the star's power?

I would think that the star would turn the black hole inside out, and like explode or something, but there's also the possibility that once a black hole rips space time the way it does maybe that can't be easily undone. I'm assuming the matter within the black hole isn't going to magically pop back into existence either as it's been broken down into it's most basic parts.

You ask if

  there some property that a black hole has that is immune to the star's
  power?

The short answer is "yes" -- the property of being a black hole. Assuming the collision is at a reasonably low relative velocity, the black hole would consume most of the mass of the star, so that eventually you would be left with a much bigger black hole and a certain amount of matter and radiation heading away from the site where the collision took place. The details of how long this would take, how much matter and energy would escape and so on will vary depending on the exact structure of the star and the details of the collision.

Essentially you can't disrupt a black hole much by throwing matter into it (if the matter you throw in is VERY dense, like another black hole or a neutron star there will be some temporary disturbance to the event horizon, but it will settle down in a fraction of a second), all it does is absorb the matter and get bigger. Even if the matter is super-hot dense fusing plasma from the centre of giant star. In the course of falling in, the matter gets distorted and heated (much hotter than the interior of even a large star) and sometimes whirled around the black hole, and this can appear as a huge explosion, but in the middle of it, the black hole is sitting there absorbing matter, largely unperturbed.


Is there any thing special with an axial tilt of roughly 25°? Half of the planets in the solar system have an axial tilt of $25.5\pm 3^\circ$: earth (23.5°), mars (25.3°), saturn (26.7°) and neptune (28.3°). If random collisions in the early solar system caused the axial tilt of the planets, it seems like there should be a lot more variation in the planetary tilts. Is there some mechanism that favors tilts around 25°, or is this just a statistical fluke?

Mars (the planet with an axial tilt closest to that of the Earth) has an axial tilt that varies greatly over time. For instance, 4 million years ago, the mean obliquity was $∼35 ± 10°$. From the same article:

Earth on the other hand has had its rotational features stabilized due to the effect of our large Moon.
So it is only by chance if the axial tilt of Mars is currently similar to that of Earth.


Meteors made of differing substances. Will they all burn up? If an object enters Earth's atmosphere, what determines whether it burns up or not?

Example

A feather is released from a space-ship 30,000 km away from Earth and falls under the influence of Earth's gravity.

Will it burn up on re-entry or will it float down to the surface unharmed. If it would be unharmed, where is the cutoff point for which objects survive?

I can give a partial answer to this.   In general, larger objects are more likely to break up than smaller ones, and comets are more likely to break up than meteors.   A feather would likely burn up too, here's why:

30,000 KM above the surface, or roughly 4.7 earth radii, the escape velocity at that height would be the square root of that or roughly 2.17 times less or 46% of the escape velocity from the surface, so a feather dropped from that height would be traveling at nearly 54% of earth's escape velocity when it hit the atmosphere, or about 6 KM per second.  That's about 7 times as fast as a bullet and I'd wager, that would burn up a feather pretty quick.

Speed of space collisions is faster than free-fall velocity can reach cause anything that hits the earth from space is also in free fall and you can add to that any relative velocity between the objects.   The slowest meteors hit the earth at about 11 KM per second, not surprisingly, the Earth's escape velocity.

http://www.amsmeteors.org/fireballs/faqf/#12

Now, if you drop a feather from the highest hot air balloon, then it probably would float down to the earth, accelerating faster at first in the thinner atmosphere, then more slowly as the air got thicker.

Small point to add, angle of approach matters too.   A glancing blow and the object can effectively bounce off the atmosphere, not burn up in it.


Maximum Amplitude of a Lissajous Orbiting Object in a L4 or L5 Position I stumbled on The Lagrangian points during some wikipedia reading.  After looking at the gravity contours, I naturally come to the conclusion that the L4 & L5 should have a wave pattern and then found the Lissajous orbit page.  It states:

  Orbits about Lagrangian points L4 and L5 are dynamically stable in theory so long as the ratio of the masses of the two main objects is greater than about 25, meaning the natural dynamics keep the [third object] in the vicinity of the Lagrangian point even when slightly perturbed from equilibrium.

After reading it I started to wonder if there is a maximum possible amplitude (height of the peaks and troughs relative to the orbital plane of the second object) of the pattern?

Also, if there theoretically is none, for cases where it is extremely large, say larger than the 2 times the radius of the second orbiting object, what would the respective object weight ratios have to be to keep such a perturbed orbit stable for any realistic period of time?

FYI, I'm a computer scientist who loves reading about physical cosmology but can be kinda a noob sometimes. Please forgive me if I'm asking for the wrong parameters.

Edit: Here is an animation of 2010 TK7, Earth's first trojan asteroid, showing the wave pattern I'm referring to. Recall that my question is referring to the height of the peaks and troughs relative to Earth's orbital plane. Since the video is a top-down view, the peaks and troughs are going into and out of the screen.

Is there a ceiling for stable L4 or L5 masses?

Your link says orbits about Lagrangian points L4 and L5 are dynamically stable  so long as the ratio of the masses of the two main objects is greater than about 25.

Earth < 4% Sun = Stable L4,L5 points

http://en.wikipedia.org/wiki/Giant_impact_hypothesis says that the Lagrangian object would destabilize once above 10% of the Earth's mass.

L4 or L5 object < 10% Earth = Stable object


Telescope size to detect Ceres (Newtonian) What is the minimum size telescope required to enable clear visual detection of Ceres?

Ceres varies between mag 9.3 and mag 6.7 which are below the nominal naked eye limiting magnitude of ~5. The gain in limiting magnitude for something like 10x50 binoculars (~4.5 magnitudes) would bring Ceres even at its faintest above the limiting magnitude of about 9.5.


How to compute Ascension / Descension cycle of the Moon I am struggling to find out an information about how to determine ascending/descending cycle of the Moon.

I can find from this source and a bunch of others, that Moon begins to descend when reaches Gemini and minimum when reaches Sagittarius.

I tried to do a simple comparison using flatlib library:

from flatlib import const
from flatlib.datetime import Datetime
from flatlib.geopos import GeoPos
from flatlib.chart import Chart

d1 = Datetime("2020/1/21")
d2 = Datetime("2020/1/22")
d3 = Datetime("2020/1/23")
pos = GeoPos('50n06', '14e24') #I concentrate to a specific place on northern hemisphere
c1 = Chart(d1, pos)
c2 = Chart(d2, pos)
c3 = Chart(d3, pos)

m1 = c1.get(const.MOON)
m2 = c2.get(const.MOON)
m3 = c3.get(const.MOON)
print(m1)
print(m2)
print(m3)

Which prints:

Moon Sagittarius +14:07:19 +13:16:13
Moon Sagittarius +27:17:06 +13:03:28
Moon Capricorn +10:14:25 +12:51:13

Am I right that moon is ascending during these days (21-23rd Jan 2020) as it moves from Sagittarius to Capricorn and further to Gemini?

I am aware of this answer, but I am not exactly sure if it deals with same problematics.


Orbital path projections onto a 2D plane In ProfRob's excellent answer to Is S2 still the fastest known star in the galaxy?, he posted a figure of the orbital paths of stars passing close to the supermassive black hole near the center of the galaxy, which I am re-posting here:

This figure caused some confusion in the comments, especially concerning the fact that there doesn't seem to be a common focus for all the ellipses in the image, as might be expected by Kepler's First Law.  ProfRob reminded the other users that the figure was a projection of the orbits in 3D onto the plane of the sky.
Question: How are the geometric properties of orbits conserved (or not) when projected onto a plane?
Specifically:

Are the 2D projections of ellipses also ellipses?
Do the ellipse projections all contain the corresponding projection of the central mass?
Why don't the ellipse projections share a focus?
Under what conditions (if any) will the projections share a focus?

Yes, the 2D projection of a 3D ellipse (or circle) is always an ellipse (or circle). However, in general, the various elements of the 3D ellipse do not project to the corresponding elements of the 2D ellipse.
A simple parameterization of the 2D ellipse uses the eccentric anomaly. For an ellipse centred on the origin, let $a$ be the semimajor axis, aligned with the X axis, $b$ the semiminor axis, aligned with the Y axis, and $\theta$ the eccentric anomaly. Then
$$x = a\cos(\theta)$$
$$y = b\sin(\theta)$$
Note that $\theta$ is not the central angle of the ellipse, it's the central angle of the auxiliary circles associated with the ellipse. Here's a diagram, courtesy of Wikipedia (click the image for an SVG version):

The eccentric anomaly of point P is the angle E. The center of the ellipse is point C, and the focus is point F.

The blue circle provides the X coordinate of P, and the green circle provides its Y coordinate.
The vertices of an ellipse are the points where the major axis crosses the ellipse, the co-vertices are the points where the minor axis crosses the ellipse. The focal points lie on the major axis at a distance $f=ae$ from the origin, where $e$ is the eccentricity. Note that $a^2=b^2 + f^2$.
If we rotate the ellipse in 3D around the X axis by $\gamma$ and project it back onto the XY plane, the projected ellipse has the same semimajor axis, but its semiminor axis shrinks to $b\cos\gamma$. Similarly, if we rotate it around the Y axis by $\beta$, the semiminor axis is preserved, but the semimajor axis shrinks to $a\cos\beta$. With more complex rotations, things get complicated...
We can represent rotations using rotation matrices. This is useful because we can combine rotations by multiplying their matrices. Please see the Wikipedia article for details.
Using rotation matrices, we can produce parametric equations for the general 3D ellipse. I won't reproduce those equations here, since they're a bit messy. But using those equations we can show that the projected ellipse is, in fact, an ellipse.
Below is a Sage / Python script which creates an interactive 3D plot of a rotated ellipse and its projection onto the XY plane. I use the same conventions as that Wikipedia article: I use a right-handed coordinate system, with angles $\alpha, \beta, \gamma$ representing anticlockwise rotations (in degrees) around the Z, Y, X axes, respectively. (If you "grab" an axis with your right hand, with your thumb pointing in the positive direction, then your fingers curl in the positive angle direction).
The $z=0$ XY plane is rendered in translucent grey. I plot the vertices, co-vertices, and foci of the 3D ellipse in primary colours, and use the same colours for the projections of those elements onto the 2D ellipse. The major axis is red, the minor axis is blue, the foci and the 3D ellipse itself are green. I also plot the actual vertices, co-vertices, and foci of the 2D ellipse: magenta for the major, cyan for the minor, and pale green for the foci and the 2D ellipse. When one of these points coincide with a projected point, the graphics system "decides" which one to render.
""" Plot an ellipse in 3D & its projection onto the XY plane
    Written by PM 2Ring 2012.04.11
"""

from math import radians
t = var('t')

ps = 10
# Plot a 3D line segment & its endpoints
def pline(pts, color):
    P = point3d(pts, size=ps, color=color)
    P += line3d(pts, color=color)
    return P

# Project a 3D vector to the z=0 plane
def proj(v):
    return vector((v[0], v[1], 0))

@interact
def main(ecc=slider(0, 1, step_size=0.05, default=0.6),
  alpha=slider(-180, 180, step_size=5, default=0),
  beta= slider(-180, 180, step_size=5, default=20),
  gamma=slider(-180, 180, step_size=5, default=20),
  frame=False, perspective=False, auto_update=False):
    a = 1
    f = a * ecc
    b = sqrt(a^2 - f^2)
    print("b =", b)

    # Build rotation matrices
    alpha = radians(alpha)
    beta = radians(beta)
    gamma = radians(gamma)
    Rx = matrix(RR, [
     [1, 0, 0],
     [0, cos(gamma), -sin(gamma)],
     [0, sin(gamma), cos(gamma)]])
    Ry = matrix(RR, [
     [cos(beta), 0, sin(beta)],
     [0, 1, 0],
     [-sin(beta), 0, cos(beta)]])
    Rz = matrix(RR, [
     [cos(alpha), -sin(alpha), 0],
     [sin(alpha), cos(alpha), 0],
     [0, 0, 1]])
    R = Rz * Ry * Rx

    # The z=0 plane, origin, & axes
    P = plot3d(lambda x,y: 0, (-a, a), (-a, a), color="#888", opacity=0.2)
    P += point3d([(0,0,0)], size=ps, color="#888")
    P += pline([(-a, 0, 0), (a, 0, 0)], "#aaa")
    P += pline([(0, -a, 0), (0, a, 0)], "#aaa")

    # The 3D ellipse
    fxyz = R * vector((a*cos(t), b*sin(t), 0))
    fxy0 = proj(fxyz)
    P += parametric_plot3d(fxyz, (t, 0, 2 * pi), color = "green")
    # The projected ellipse
    P += parametric_plot3d(fxy0, (t, 0, 2 * pi), color = "#7f7")

    # The vertices & co-vertices of the 3D ellipse and their projections
    # Major
    vx0, vx1 = fxyz(t=0), fxyz(t=pi)
    P += pline([vx0, vx1], "red")
    P += pline([proj(vx0), proj(vx1)], "red")

    # Minor
    vy0, vy1 = fxyz(t=pi/2), fxyz(t=3*pi/2)
    P += pline([vy0, vy1], "blue")
    P += pline([proj(vy0), proj(vy1)], "blue")

    # The foci of the 3D ellipse and their projections
    f0 = R * vector((-f, 0, 0))
    f1 = R * vector((f, 0, 0))
    P += pline([f0, f1], "green")
    P += pline([proj(f0), proj(f1)], "green")

    # Find the true vertices & co-vertices of projected ellipse
    # rf is the radius function of the projected ellipse
    rf = abs(fxy0)
    # Major
    ap, t0 = find_local_maximum(rf, 0, pi)
    _, t2 = find_local_maximum(rf, pi, 2*pi)
    vpx0, vpx1 = fxy0(t=t0), fxy0(t=t2)
    P += pline([vpx0, vpx1], "magenta")

    # Minor
    bp, t1 = find_local_minimum(rf, 0, pi)
    _, t3 = find_local_minimum(rf, pi, 2*pi)
    P += pline([fxy0(t=t1), fxy0(t=t3)], "cyan")

    # Find true foci of the projected ellipse
    eccp = sqrt(ap^2 - bp^2) / ap
    P += pline([vpx0*eccp, vpx1*eccp], "#7f7")

    P.show(frame=frame, projection="perspective"
      if perspective else "orthographic")

Here's a screenshot:

It's a bit hard to see what's going on in that screenshot. It's a lot easier in the interactive 3D view. This view is created using three.js, and responds to the usual OrbitControls: you can pan & rotate the camera using the mouse, use the mouse wheel to zoom. On a touch screen, one finger rotates, use two fingers to pan & zoom.

The script now allows you to easily set the frame off or on, and to choose between a perspective or orthographic camera.


Astronomy detective question: what part of the sky are these photos of? What are a few of the stars? The 9to5Google article Exclusive: Official Pixel 4 camera samples show astrophotography, selfies, more [Gallery] shows some surprisingly nice (at least to me) photos of the Milky Way.

I've asked Milky Way with a cell phone, how is this possible? in Photography SE, but here I'd like to double check the scale and FOV of these images.

I'd like to establish the FOV and scale in pixels per degree for these images, so I need to know what part of the sky are these photos of and what are a few of the stars? I understand that a wide field view with a cell phone camera may have some distortion so I'll do the corners separately from the center.

I wanted to ask for an estimate of the latitude. I guess that's only possible if we assume the bottom edge is parallel to the horizon.

click for full size:

Astrometry.net successfully plate-solved both images.

The first image is of the Cygnus region with north to the left.
It is 73° wide and 55° high, with 65 arcseconds per pixel or 55 pixels per degree.
Bright stars include Deneb (α Cyg) left of center and Altair (α Aql) at upper right.

The second image is of the Sagittarius region with north up.
It is 37° wide and 66° high, with 80 arcseconds per pixel or 45 pixels per degree.
The bright object left of center is Saturn, which was at that position around 2019-07-29.


Refinement to sun/wristwatch estimate of N/S direction As discussed in a 2015 question here
and shown in a 2013 illustration in Primer Magazine,
it's possible to estimate the north/south direction on the ground using a correlation between the hour hand of an analog watch and the azimuth of the Sun.

As Wikipedia:Analog watch suggests, the relation between hour angle and azimuth depends on latitude:

The method functions less well as one gets closer to the equator.

Also the Sun's declination affects its azimuth around 6 AM or PM.
Assuming that the observer has already determined the local apparent solar time,
what further correction could they apply to estimate the north/south direction more accurately?
I expect that this would be a function of the observer's latitude φ, the Sun's declination δ, and the Sun's hour angle h.

The method is generally reliable between 9 am and 3 pm.
The other consideration is when one is using the method in either of the tropics: Cancer or Capricorn.
It is generally accepted that in the northern hemisphere, the sun is located in the southern part of the sky and the opposite in the southern hemisphere. This applies to latitudes greater than the latitudes of the Tropics. Between the equator and the Tropics, depending on the time of year, the position of the sun in the sky alternates between north and south.


Have any more "white dwarf pulsars" been discovered or searched for? Back in 2016, Marsh et al. reported that the binary system AR Scorpii exhibits complex radio signals similar to those observed from traditional pulsars. In particular, pulsed synchrotron emission appears to be produced by interactions between the magnetospheres of a white dwarf and a red dwarf, powered in part by the spin-down of the white dwarf. The system has been dubbed the first "white dwarf pulsar" based on the signals.

The system is thought to be similar to, if not an extreme case of, intermediate polars; we know of a number of these systems. Therefore, I'm wondering: Have there been any concerted searches for other intermediate polars exhibiting this pulsed emission, and if so, have any candidates been discovered (possibly serendipitously)?

This is partly belatedly in honor of February's focus tag.

Edson et al. (2017) list two other candidates that may be white dwarf pulsars.

AE Aquarii has been described as pulsar-like (e.g. Ikhsanov 1998) and was the best case for a white dwarf pulsar before AR Scorpii came along. From what I understand, the case for AE Aqr containing a white dwarf pulsar is not watertight. Blinova et al. (2018) explain the high spin-down luminosity as being the result of interactions with an accretion disc and the white dwarf having a magnetic field typical of that in intermediate polars, rather than a pulsar-like mechanism.

Zhang & Gil (2005) proposed that the radio source GCRT J1745-3009 is caused by a white dwarf pulsar. As far as I can tell, the case for this one isn't particularly secure either because there are other possible explanations. Follow-up observations have not yet confirmed the nature of the source.

I'm not aware of any specific surveys for such objects but so far it seems that AR Scorpii is the best example of a white dwarf pulsar.


Why are asteroids with zero orbital inclination rare? This is a plot of orbital inclination ($i_p$) vs. semi-major axis ($a_p$) of 96944 asteroids in the Main Belt, done by Piotr Deuar.

Some structure can be seen in this diagram; clumps are asteroidal families and vertical empty spaces are Kirkwood gaps due to mean-motion resonances with the planets (mosly Jupiter).

Recently I've noticed that there are fewer points towards the ecliptic in the first $\small{\frac 1 2}$ degree of orbital inclination. Is this due to some selection bias? or Is it because of some kind of resonance that eliminates asteroids with orbital inclinations too close to zero?

Short answer: The scarcity of asteroids with an inclination near zero is an expected result of normally distributed inclinations in 3-dimensions about the normal vector to the reference plane, rather than selection bias or an orbit "clearing" around zero degree inclination.
Long answer: Orbital inclination is typically defined as the angle between a reference plane and the orbital plane:

An equivalent definition is the angle between a reference plane normal and the axis of revolution of the orbiting body overlay-ed here:

Notice that the axis of revolution can be mapped to a point on the Celestial Sphere. We can generate a Gaussian distribution of points on the unit sphere centered with mean at the reference normal using the von-Mises Fisher distribution.  The specific algorithm for generating sampling points is given on Stack Overflow.
Here is a group of 20 revolution axis intercept points with their corresponding circular orbits:

We can generate and plot a thousand points on a unit sphere (that I rotate to show the structure):

Note that only the very top points of this distribution correspond to inclination angles of nearly zero. As the inclination increases, the points get less dense, but the areas on the unit sphere are larger. The below plot is looking down from above the North Pole, with the orbits with inclination less than 2 degrees colored red, and the orbits with inclinations between 2 and 4 degrees colored green.  Note that even though the points are the most dense at the pole, the area covered is the smallest, so 13 points are red, while 50 points are green.  This is the explanation for why we see very few asteroid orbits with inclinations of less than 1/2 a degree from the original question.

Below is a plot with the y-axis as inclination and x-axis as longitude compared with the OP's post. I chose a standard deviation that was much too large, but the 1000 point Monte Carlo demonstrates the same statistical effect asked about by the OP.

Notes:

The above example is for circular ellipses all with the same SMA for convenience, but I don't think we lose any generality.

I assume the reference plane in the picture used by the OP is the Earth's ecliptic, but the above answer works for the Sun's equatorial plane, or Jupiter's orbital plane, or any other close reference plane.  We can assume that the asteroid inclination distribution 2D mean is actually going to be a bit offset from any choice of reference plane normal.

Orbital Mechanics can be very non-intuitive.  Inclination is a 1D value, but its representative underlying probability distribution is best modeled on a 2D spherical surface in 3D.

It's somewhat ironic that I claim this is strictly a math problem, but there are no equations in the above post.  The power of the Monte Carlo approach is in its simplicity and resultant appeal to mathematical intuition.

Statisticians/Probabilists should immediately recognize the distribution of orbit inclinations as a variant of the Rayleigh distribution which should lend a hint that a compression from 2D variables to 1D has occurred.

My Matlab code is available on request.


What if a Black Hole passed through an Irregular type galaxy? So, most of you all are probably (well, why wouldn't you be) familiar with The Magellanic Clouds, and hour they are Irregular type galaxies. You should also be aware that they both orbit the Milky Way, so they act sort of like moons to a planet. Since they (most likely) do not have a central Supermassive Black Hole, I wonder, out of all ridiculousness, what are the answer to these two questions?:

1) What would happen to the Small or Large Magellanic Clouds if a Black Hole (maybe 100 time bigger than the sun) passed though them?

2) What would happen if Sagittarius A* passed through either the Small or Large Magellanic clouds. If you are not aware of Sagittarius A*, it is the Supermassive Black Hole that lies at the center of the Milky Way.

Black holes (even SMBs) are actually very small objects on an astronomical scale and in practical terms probably very little would happen.

There would be some perturbations of stellar orbits within the formation they were in, but unless there was a fairly close approach  very little would happen and on the scale of the entire formation the changes would affect only a small number of objects and not the overall "shape".

To put this in perspective, Sag. A* has a mass of about 4,000,000 solar masses.  Sounds big (is big :-) ).

But the Large Magellanic Cloud has a mass of about 10,000,000,000 solar masses (or about 2500 time larger than Sag. A*).  The dominant gravitational field is still that associated with the LMC itself, not that of Sag A*.

I think a good analogy would be a bullet fired into a lake.  Sure, it can cut a path through one isolated region and cause disturbance close to that path, but the bulk of the lake will be undisturbed apart from a slight ripple.

The Small Magellanic Cloud is still a pretty large mass at about 6,500,000,000 solar masses, so still roughly 1,400 times more massive than Sag. A*.  Again, we're talking bullet in lake territory.