Energy needed to create a black hole vs energy needed to run the Large Hadron Collider?

Your question touches on a few points.

First, yes, he was flippant, but the risk was super-low.   The simplest way to explain this is that nothing happens in CERN that doesn't happen all over the universe and in the upper atmosphere of the Earth or on the surface of all the moons, planets and stars every day.  For example, the Oh My God particle was traveling with several million times the energy of the particle collisions inside CERN.   At those speeds, accelerating the particle further doesn't add much velocity, what they're basically doing is adding energy or mass to the particle and only slightly increasing it's velocity as it's speed gradually moves closer to the speed of light.

What makes CERN useful, isn't that the high speed collisions don't happen anywhere else (they happen all the time all over the Universe).  What makes it useful is the million dollar equipment that zooms in on those collisions, recording what happens with the best cameras in the world.   We can't predict when and where a very fast moving cosmic particle will hit the Earth, but they know when and where the CERN particles are going to collide, because that's planned to happen in a specific place and time.

So, if those kinds of collisions happen all the time and no planet or star has ever been observed being eaten by a particle collision created black hole, then such events are, at the very least, extremely rare.

Add to that, Hawking radiation dictates that black holes that small should evaporate before they can eat anything.   So, when Brian Cox laughs at that question, he has good reason to laugh, because it was never a real risk and it's more of a joke question than a genuine risk.

Now, when you say "pressure" to create black holes.   The pressure inside a black hole, or, I should say, pressure inside a Neutron Star about to turn into a black hole, or the pressure inside a collapsing core of a very large star about to create a black hole - those pressures are crazy high, but if we could re-create that pressure on earth (We can't), but if we could, it wouldn't be enough to make a black hole.   Pressure alone doesn't do it, you need mass (at last about 3 solar masses) to fit within a certain size and the less mass you have, the greater the required density (and greater pressure) is needed to achieve the necessary mass to schwarzchild radius ratio.  The pressure required inside the forming black hole actually decreases as the mass of the black hole increases.  (based on simple pressure equations and schwarzchild radii).   Equal pressure in a lab on Earth wouldn't accomplish it.

What makes black holes theoretically possible is the small quantum size and possibility of extra dimensions and quantum rules that only apply to the very small, so it's a very different set of circumstances.

Collisions in CERN are measured in Electron-volts.   It was theorized that if there were extra dimensions on the quantum level, then micro-black holes might form at the energy of CERN's collisions (this would imply that black holes of this type form in Earth's upper atmosphere all the time, implying they're not dangerous).

But so far, no luck.  No micro-black-holes have been observed.  They should dissipate very rapidly but their decay streams should be recognizable so they should get noticed, but so far, none have been spotted and as a result, there's no evidence of extra dimensions.  That doesn't mean they're not there, and that micro-black-holes can't be created, but so far there's no evidence that they've been created up to about 13 trillion electron volts, the current peak of CERN's collision energy.   See here

Hope that makes sense.  Corrections welcome.


When black holes forms out of a supernova do they have a very high velocity?

There is no specific answer to this -- anything from "just sits there" to flys away at high speed is possible.

It all depends on the symmetry of the supernova (SN) explosion. Extensive modelling shows that the explosions can be quite asymmetrical, and if they are the gravitational waves created can give the new black hole (BH) quite a kick.  If the explosion is relatively symmetric, the new BH can more or less stay right where it was formed.

Note that if the exploding star is a member of a close binary, the SN explosion will usually give the companion star a high velocity. Before the explosion, a close pair may be rotating around their joint center of mass at 1000 kps. When one star goes SN, it blows most of its mass right out of the system and high velocity. This doesn't change the velocity of the companion star which suddenly is no longer bound and has escape velocity from the star system -- so it more-or-less goes flying off in a straight line at its orbital velocity.

Some of the runaway stars discovered by Gaia are probably companion stars to SNs.


I forgot changing the unit of the distance from parsec to light year.
This is a simple unit conversion error which should have been avoided.

Ultimate Graph:

Data Used:

  Name,RA(deg),Dec(deg),pm_RA(mas/yr),pm_Dec(mas/yr),Vr(km/s),parallax(mas)
  Wolf 359,164.120271,+07.014658,-3842.0,-2725.0,19.321,418.3
  Proxima Centauri,217.42895219,-62.67948975,-3775.75,765.54,-22.40,768.13
  Alpha Centauri,219.900850,-60.835619,-3608,686,-22.3,742
  Barnard's star,269.45207511,+04.69339088,-798.58,10328.12,-110.51,548.31
  Gliese 445,176.92240640,+78.69116300,743.61,481.40,-111.65,186.86
  Luhman 16A,150.8218675,-53.319405556,-2754.77,358.72,23.1,500.51
  Sirius,101.28715533,-16.71611586,-546.01,-1223.07,-5.50,379.21
  Lalande 21185,165.83414166,+35.96988004,-580.27,-4765.85,-84.69,392.64
  Ross 248,355.479122,+44.177994,115.10,-1592.77,-77.715,316.7
  Gliese 65,24.756054,-17.950569,3321,562,29,373.70

Code:

%matplotlib qt
import numpy as np
from mpl_toolkits.mplot3d import axes3d
from matplotlib import pyplot as plt
from matplotlib import patches

def parseTextFile(file_name, delimiter=",", header=0):
    """ Parse a text file to a list. The file contents are delimited and have a header. """

    with open(file_name) as f:

        # Skip the header
        for i in range(header):
            next(f)

        data = []

        # Parse file contents
        for line in f:

            # Remove the newline char
            line = line.replace('\n', '').replace('\r', '')

            # Split the line by the delimiter
            line = line.split(delimiter)

            # Strip whitespaces from individual entries in the line
            for i, entry in enumerate(line):
                line[i] = entry.strip()

            # Add the contents of the line to the data list
            data.append(line)

        return data

if __name__ == "__main__":

    file_name = 'C:\\Users\\The Wings of Dream\\Desktop\\UWO-PA-Python-Course\\Lecture 5\\Homework 2\\star_data.txt'

#Program Begin:

fig = plt.figure()
ax1 = fig.add_subplot(211, projection='3d')
ax2 = fig.add_subplot(212)

time = np.arange(-60000,100000,10)
count = 1

xdic = {}
ydic = {}
zdic = {}

for star in parseTextFile(file_name, header=1):

    name = str(star[0])
    RA = float(star[1])
    Dec = float(star[2])
    pm_mRA = float(star[3])
    pm_mDec = float(star[4])
    vr = float(star[5])
    mparallax = float(star[6])

    pm_RA = pm_mRA * 0.001
    pm_Dec = pm_mDec * 0.001
    d = 1 / (mparallax * 0.001)

    vta = pm_RA * d * 4.740
    vtd = pm_Dec * d * 4.740

    vx = vr * np.cos(Dec) * np.cos(RA) - vta * np.sin(RA) - vtd * np.sin(Dec) * np.cos(RA)
    vy = vr * np.cos(Dec) * np.sin(RA) + vta * np.cos(RA) - vtd * np.sin(Dec) * np.sin(RA)
    vz = vr * np.sin(Dec) + vtd * np.cos(Dec)

    vx_pcyr = vx / 977780
    vy_pcyr = vy / 977780
    vz_pcyr = vz / 977780

    xi = d * np.cos(Dec) * np.cos(RA)
    yi = d * np.cos(Dec) * np.sin(RA)
    zi = d * np.sin(Dec)

    x = xi + vx_pcyr * time
    y = yi + vy_pcyr * time
    z = zi + vz_pcyr * time

    xdic['x'+str(count)] = x
    ydic['y'+str(count)] = y
    zdic['z'+str(count)] = z

    distance = np.sqrt(x ** 2 + y ** 2 + z ** 2) * 3.26156

    ax1.plot(xdic['x'+str(count)],ydic['y'+str(count)],zdic['z'+str(count)])
    ax2.plot(time,distance,label=name)

    count = count + 1

w_oort, h_oort = 160000, 3.2
ax2.add_patch(patches.Rectangle((-60000, 0.03), w_oort,h_oort,color='slateblue',alpha=0.2))
ax2.annotate('Oort Cloud', xy=(15000,1.6), size=12)

plt.axvline(x=0,color='gray',linestyle='--',linewidth='0.5')

#plotting constraints
ax2.set_ylim(0.0, 10.0)
ax2.set_xlim(-60000, 100000)

ax1.set_xlabel('x axis')
ax1.set_ylabel('y axis')
ax1.set_zlabel('z axis')
ax1.title.set_text('Motion of Stars in Space')
ax2.title.set_text('Distance-Time')
ax2.set_xlabel('Time (Year)')
ax2.set_ylabel('Distance (Light Year)')
ax2.legend()
plt.show()

ax1 subplot will give you a 3d parametricplot showing the motion of the stars.


Dating the age of the asteroid belt is a problem that has to date no general solution.

Rocks are usually dated in the laboratory via isotopical analysis. This is possible for meteorites, but what to do about those rocks that remain in space?

Several methods to date the asteroid belt are known to me, apart from ad-hoc assumptions that it has the same age as the planets:

Creating links between dated meteorites and asteroid families: The meteorite family of L chondrites have possible origin families in the asteroid belt. Those 'families' are small blobs in the eccentricity-inclination diagrams, that probably originated in one common parent-body breakup event. Dating meteorites originating from those families, gives minimum ages for the parent bodies of i.e. 500 Myrs for the Ordovician meteor event.
Dynamical age modeling: After a breakup of a parent body, the cloud of fragments experience different strengths of YORP and tidal forces, which causes their semimajor axis distribution to spread. Taking a given distribution, one can calculate back to an age of when the cloud was one particle, i.e. the moment of breakup. This again is a minimum age and found for the Flora family to be more than $10^8$, but less than $10^9$ years. This does exclude a very young belt, but does not say much about the upper age limit.
Crater ages: For the largest bodies, i.e. Ceres, Vesta, etc. we can assign relative ages to the lunar cataclysm. Those usually give ages in excess than 3-4 Gyrs. For bodies like Vesta, where the surface is saturated with craters, this age is a lower limit.
Dating of crustal ejecta: There is another family of meteorites that can be linked specifically to Vesta via spectroscopy, called the HED meteorites. Those meteorites look like they're coming from a differentiated parent body (melted, low siderophile element content), and they containt CAIs. Thus, we can date Vesta at least back to the origin of the solar system, as CAIs give us the effective $t0$ for the formation of the solar system.


Likelihood of extra planets in systems identified by Kepler?

There are many planets that Kepler couldn't detect because the ecliptic is inclined so no transit occurs from the perspective of Earth. We are much more likely to see a transit if the transiting planet is close to the star, and so in Kepler's sample of stars, a disproportionate number of the planets discovered orbit very close to the star. It is probable that most stars have planets, and many of those planets orbit much further out than the average in the Kepler sample. And, moreover, the stars which have been found to have planets are likely to have others, orbiting further from the sun, and inclined, so they are not detectable by transits. Kepler has not detected Earth size planets orbiting in the habitable zone of sun-like stars.

So far we do not have the technology to detect a planetary system that mirrors the Solar System. Our alien friends, if they are very close to the ecliptic may detect transits of Earth but would likely miss the other planets. And there is no reason to suspect that the Solar system is not typical in the range of inclination of its planets.


Was the whole Universe close to Big Bang very small, or just very dense?

I am sorry this question is probably silly for professional astronomers, of which I am not one.

This question is by no means silly. Your question is a common one about cosmology (the study of where the universe came from, how it is evolving and what its fate will be). The media often butchers these concepts horribly, resulting in a lot of confusion (out of all scientific information, they seem to have the hardest time reporting cosmology accurately). Your inquisition is definitely a good thing.

  When astronomers say early universe was small, do they simply mean "the part of the universe which corresponds to our observable universe, was small"?

Well, they're usually referring to the entire Universe. In my final paragraph, I explain what this implies for the observable universe.

  If it is infinite, then it would seem that it had to be infinite in infancy as well, just very dense. In fact, it seems "in the limit", it had to be infinite even at Big Bang.

You're closer to the truth. When we talk about the expansion of the Universe, we're really saying that space is being created between all matter.

As you mentioned, the Universe may be infinite. It is not like a ball, but rather like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them. Below is a gif I've made to demonstrate this:

A more useful way to describe this is to say the grid is expanding — that space itself, as a coordinate system, is growing. As an analogy, imagine are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

So the same thing is happening with our universe. The grid is in fact growing, and objects are being swept away with it.

OK, now that we've gotten the core concepts down, I'll introduce one more bit of terminology. The "scale factor of the Universe" refers to how much the Universe has expanded, compared to now. For example, if in a billion years the scale factor is 3, that means that every object in the Universe is 3 times farther from each other compared to now. If the scale factor 700 million years ago was 0.8, then everything was closer by a factor of 0.8 at that time. By definition, the scale factor is 1 right now.

So, if the Universe is expanding now, we'd expect it to be smaller as we look further back in time — i.e. the scale factor would be less. General relativity predicts the scale factor to be zero at 13.8 billion years ago. This would mean that every object would be zero times its current distance from us — in other words, there would be no space.

If you think a Universe without space is impossible, you're correct. We apparently have a contradiction. In GR, you can't have a spacetime with zero space.

Our modern physical theories work fine up a few fractions of a second after the moment of contradiction, and our observations do agree with the idea of an extremely dense early universe. However, our theories break down as we try to model the Universe at earlier and earlier times, until they no longer prove accurate, preventing us from explaining the most interesting moment.

This is why the moment of the Big Bang is one of the biggest mysteries in cosmology. Theories like quantum gravity have arisen to try to explain the conditions near the Big Bang, but none are sufficient as of now.

  I often hear at lectures that immediately after the Big Bang, the universe was small, say, the size of grapefruit or something like that.

Indeed, the problem stems from the ambiguity when one says "universe". In this case, they're referring to the observable universe, which is actually spherical. The observable universe was indeed much smaller near the time of the Big Bang, compared to its radius now.

This is because its radius actually depends on our Universe's scale factor*, which means that at the moment GR predicts the scale factor to be zero, it also predicts the size of the observable universe to be zero.

Obviously that can't be the case, since as we've explained above, it shouldn't be possible for the scale factor to be zero. However, we can say with reasonable confidence that the observable universe was likely the size of a grapefruit at one point, if not smaller (although "grapefruit" seems an arbitrary choice for comparison. I can't actually find the paper that first uses this analogy, so what they originally meant is a bit unclear).

*Measuring distances is actually a bit tricky in cosmology; in some cases, we want to talk about distances or motion of objects while neglecting the Universe's expansion. To save you the need to learn a lot of terminology, I'm right now taking into account the expansion of the Universe when talking about the observable universe's size. The observable universe also grows due to factors besides the Universe's expansion, i.e. light from further and further galaxies reaching us.


Will there be public access to the New Horizons data from Ultima Thule?

The raw images1 from LORRI (high resolution greyscale) are available from JHU APL but the rest of the data such MVIC (wide angle in greyscale plus four color bands: near IR, methane, red, and blue), LEISA (infrared spectroscopic imagery), Alice (ultraviolet spectroscopic imagery), as well as other instruments seems not to be available yet.

Some processed imagery is also available elsewhere on APL's site.

Eventually all the data will be released via the NASA PDS2 but this may take some time since it first has to be transmitted to Earth (this will take 20 months) then processed and calibrated (several more months) or (for some images) converted to a public-friendly JPEG.  Once this is done the data will be released since, per this document (p. 15), the New Horizons data is not subject to embargo by the research teams:

  There are no proprietary data rights for the New Horizons mission. Selected uncalibrated (CODMAC Level 2) data, particularly image data, will be publicly released by the New Horizons project over the Internet in close to real time. ... Fully reduced, calibrated and corrected data products .. will be [released as it becomes available].

Quoting further (from Appendix D):

  Selected uncalibrated data, including image data, will be publicly released over the INTERNET in close to real time.
  Fully reduced, calibrated and corrected data products will be published and forwarded to the Planetary Data System as soon as they
  have been generated and validated.

I've asked a question on Space.SE regarding whether there are currently any archives of raw(ish) New Horizons/Ultima Thule imagery besides the one linked above.

1 Converted to JPEG files.

2 Thanks to @astrosnapper for that link.


You don't say how you got your average temperature. If it is a straightforward average, then I think I have your answer.

The temperature of the moon is highly non uniform, ranging from around 100 K on the night side to 370 K on the sunlit side. This gives a crude average of 235 K.

https://space.stackexchange.com/questions/3290/what-is-the-temperature-variation-between-the-lunar-day-and-night-on-the-moon

However, in terms of blackbody radiation, if we treat both hemispheres as separate blackbodies, then together they radiate as much energy as a hotter blackbody because of the $T^4$ factor in Stefan's law. The "effective temperature" will be
$$ T \simeq (0.5\times 100^4 + 0.5\times 370^4)^{0.25}= 311\ K$$
So this is more like your equilibrium temperature and a more appropriate area weighting and/or tweaks to the exact range of temperatures could bring it into close agreement.


I don't think applying Earth data to Mars makes much sense, well, except perhaps at the very beginning.  The reason I say that is that life on Earth took a long time to get what we might consider "interesting", for lack of a better word.

Source - and, OK, it's a children's book, but I like the picture.

4 billion years ago - simple cells.
3.5 billion years ago - photosynthesis.
2 billion years ago - cells with Nuclei
1 billion years ago - bacteria.

So the Earth took 3 billion years to make - bacteria.   Granted it's only one planet as a sample, but if that's standard, then, arguably, most of the life in the solar system might be 1 celled and, you know, kind of boring to you and me, though I'm sure scientists would love to study 1 celled organisms from other planets.

If planets need a few billion years before things like plants, fish, insects, etc, evolve, then many planets won't stay in the Goldilocks zone that long.  . . . or, option B, maybe things happened slower on earth than other planets, but for now, we just don't know.   Finding life on Mars or records of past life, if there was any, could help answer some of those questions.

Mars may have lost it's oceans fairly quickly, in less than a billion years.   That's not a long time for life to evolve, at least based on the Earth Model.

Now, low gravity, probably not a big deal.   Animals and trees on a low-G planet grow grow taller and thinner/weaker - thinner bones, smaller muscles for example.   High-G, just the opposite.  Shorter and stronger with thicker bones.    Prior to lungs, Insect size is determined by Oxygen content in the atmosphere much more than gravity.   (Fish-Fin size point removed as it probably wasn't correct).

Probably none of this happened on Mars cause I doubt life evolved that far but in theory on a low G planet you might get very large animals, perhaps the size of 10 story buildings, because size is a good evolutionary defense mechanism.  (Fun to think about, right?)

Thin Air, that's a bigger problem cause thin air leads to evaporating oceans and once the oceans evaporate, life would have a hard time continuing the evolutionary process   Life needs water, at least here on Earth.   Life can adapt to living with small amounts of water but life needs water.  In ice, primitive Life hibernates, but it doesn't continue to evolve.

Now it's possible that Mars' thinning atmosphere coincided with it cooling off and it could have had oceans frozen at the top but liquid below that for some time, while heat from inside the planet maintained liquid water and life was able to survive in that.   This happened to the Earth, when the Earth was covered by ice during the snowball earth period and when Snowball Earth ended, the Cambrian explosion started.   The problem on Mars was, if Mars had a snowball with liquid water below the ice period, it never ended.   It never went back to liquid oceans and that's a problem for evolution.

It's possible that snowball earth (limiting the sunlight and atmospheric gases available to life inside the ocean) forced bacteria to evolve, hunting for food rather than simply absorbing it via sunlight and air, but that's just speculation on my part but the timing was pretty consistent.  The Cambrian explosion followed snowball earth.

Humanoids on other planets (assuming we ever find any) would probably have characteristics adapted to that planet.  We stand upright because our ancestors found it helpful to run.   That might not be the case on other planets, perhaps that have thicker brush and no open planes.  That's just one example, there's probably others.   Our height likely developed out of the optimal height for hunting.  If we just were gatherers and trappers, not spear throwers we might be quite a bit smaller with shorter arms cause there's be no evolutionary advantage to being bigger or longer arms if we climbed trees to get away from predators.  You can clime trees higher if you're small for example and be safe that way.   Humanoids might have tails on other planets if they had a use for them, like, impressing the opposite sex with them or for balance for tree swinging/jumping.   There's a variety of possibilities that go way beyond gravity (which might just have an effect on body size and shape) or air pressure, which might affect things like wings for gliding, running (as thick air might reduce running speed) and lung size.

On red dwarf planets for example, plants and people might be significantly different colors.  Plants might be blue or black, to absorb more red light, Animals, for example, might have colors into the Infra-red spectrum that we don't see.   One blue sun planets, plants could be red and animals might have colors into the UV spectrum.   (Some birds on earth have UV spectrum colors that we don't see but they do).  Thicker air pressure might affect sound as well, they might talk much quieter perhaps higher pitch and have smaller vocal cords and smaller ears, where as, thinner air, bigger animals might make thunderous yells when they communicate.

There's a variety of possible factors. . . . hope I haven't gone too deep into speculation here, but speculation is fun sometimes.

Personally, I don't believe Mars ever came close to any of the ideas I mentioned above.  I don't see how it could.  it wasn't habitable for long enough.


Calculating Right Ascension and Declination from Latitude / Longitude / Time

The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time.

(Alternatively, the local sidereal time is RA of the observer's meridian.)


They are - somewhat by definition.
One criterion for a planet is that it is large enough that it is in hydrostatic equilibrium so that it takes on spherical shape. Any body large enough for that to occur will to a certain degree be a differentiated body as hydrostatic equilibrium means that masses need to be able to move, thus the heavier elements will sediment towards the center. Due to the speed of geological processes, differentiation (=de-mixing) may initially be faster, but is never really finished, unless a body is completly solid and cold (which no planet is).
The existence or non-existence of hydrostatic equilibrium is one of the differences between dwarf planets and so-called small solar system bodies.


No.

A black hole is characterised by the formation of an event horizon. Matter which travels past an event horizon cannot return to the space outside the event horizon, to do so would require the matter to travel faster than light, and so would require infinite energy.

There is nothing like a regular star "inside" a black hole. The general relativity model of a black hole contains a singularity. Matter which enters a black hole is certain to reach this singularity in a (usually short) amount of time. Adding hydrogen to a black hole would just make the black hole more massive. There is a theorem in mathematical physics that the only properties that a black hole can have are mass, angular momentum and charge. In other words, a black hole made of collapsed hydrogen would be identical to one made of collapsed iron. There is no matter in a black hole, only mass.

The formation of a black hole normally occurs when a core in a massive star has a certain amount of iron, which saps the core of energy causing collapse. If the resulting object has more than about 3 solar masses, then nothing in the universe can stop it from collapsing all the way to a singularity.

As with much about general relativity, your intuition is a very bad guide to how black holes work. However General relativity is a very good model for how black holes should form and behave


It is the other way around.
An isolated, point-like charged particle like an electron cannot absorb a photon. It is forbidden, because energy and momentum cannot be simultaneously conserved. Only scattering is permitted in such circumstances.
However, a charged particle that is within the electric field of another (e.g. an electron in the electric field of a nucleus) can absorb (or emit) a photon.
The way to think about this classically is in terms of electric dipole moments. An atom has no net charge but is can have an electric dipole moment, because the positive and negative charge are not in the same place. This electric dipole moment can interact with incoming electromagnetic waves and can be made to oscillate by extracting energy from the incoming wave.
In the quantum mechanical extension to this picture, the electric dipole will only oscillate in certain modes with discrete energies. Only photons within a certain narrow range of frequencies/energies will be able to excite (or de-excite) these oscillations.
The extension to bulk matter (and by that I assume you mean solids) is that the atoms themselves are arranged into lattices and other structures. These have electric dipole (or higher order multipole) moments that can be excited into oscillation by electromagnetic waves/photons in a similar way. i.e.  They offer further vibrational modes that act as a pathway to the absorption and emission of photons.
PS.
Nuclei themselves can have electric dipole or higher order multipole moments and are able to directly absorb (or emit) gamma rays.


I don't think we have a good estimate of the number of stars not in galaxies -- they're actually pretty hard to find unless they're really bright, since even if you spot it, how do you know that it's extragalactic?

But galaxy mergers (which are very common) tend to throw out large numbers of stars from the merging galaxies, so it seems likely there are many stars in space between the galaxies.

There is no reason why such stars can't have planets, but it's likely that the planets formed along with the star inside a galaxy and got ejected from the galaxy along with the star.

Stars (and planets) form when Giant Molecular Clouds collapse under their own gravity, the collapse frequently being triggered by tidal forces from a passing star or a nearby supernova.  It seems unlikely that there are many GMCs in intergalactic space, and there are certainly fewer encounters which might trigger collapse. So it seems likely that relatively few stars (or planets) actually form in intergalactic space.

Life does not seem to have evolved because of the existence of stars in the sky, so once a planet exists in a habitable zone, life should form regardless of where it is.

The Gaia mission is providing us with superb position and velocity data for billions of stars, and it has identified a handful (maybe more by now) of stars which seem clearly to have come from intergalactic space.  Here's an article which covers the report last year.


How to compute satellite coordinates (lat,long) given antenna's coordinates, angles and satellite height

A way to do this is to first find the xyz in ECEF of the point on the ground (a transform for a spherical Earth is just the spherical to xyz coordinates transform from most physics text books where lat = 90-theta and lon = phi and r = radius of Earth).  Then, find the satellite slant range (which depends only on the elevation angle and the altitude of the satellite).  Then compute the unit vector toward the satellite in local ENU (east, north, up) coordinates (use Gram Schmidt to find local E and local N given U is the local point and the north pole is (0,0,1). Now, reuse your spherical transform to change az and el into local ENU.  That part is a little weird, but if you think about it, the local horizon elevation is just like the theta of a spherical system, and az is phi (U corresponds to elevation = 90deg).  Finally, compute the same unit vector in xyz (multiply each component by your ENU unit vectors and add the terms), and then "walk along that vector" by the slant path range.  You will now have the xyz of the satellite.  Finally, from the xyz of the satellite compute the lat and lon (and as a check, you should get back your same altitude).


www.skymaponline.net might be the one you want.

Main goal when taking images in terms of timing and exposition:
1. Choose time with good sky condition (e.g., clear sky, no wind)
2. Expose long enough for good S/N ratio, but not too long that CCD is saturated.
3. If you want to measure the length of the eclipse, you can follow since it is coming in until going out. Or, you can take a couple of images at different epochs, then assume some trajectory and just doing some math.


How would I calculate the length of the day of this planet?

It's almost completely unconstrained. There's a limit to how fast a planet can spin before it starts to break up, which is $p = 2 \pi / \sqrt{m G / r^{3}}$ (where $m$ is the planet's mass and $r$ is its radius). For something like the Earth (and Venus is pretty similar), this corresponds to about 1.4 hours. So the day has to be longer than that.

But there's no limit to how slowly the planet can rotate. If it's rotating slowly enough so that one rotation period = one year, then as seen from the planet's surface the star would never seem to move in its sky and the solar day (time from one noon to the next) would be infinite. This is what you get if the planet is 1:1 tidally locked to the star.


How loud was the asteroid that killed the dinosaurs?

Some estimates which I found worth sharing:

For fun, I searched for Chixculub TNT equivalent and e.g. ScienceDaily claims

The energy released by the impact that blew out the Chicxulub crater was equivalent to about 100 million megatons, many orders of magnitude greater than the nuclear explosion at Hiroshima, a 15-kiloton blast.

This seems to match up what WolframAlpha uses to rank TNT equivalents.

There is a Noise Prediction Calculator which estimates the safe distance from a blast:

This calculator can be used to predict the distance from an open detonation at which 140 decibels (dB) could be expected to be achieved. The 140dB level is widely used as a "safety cutoff" for exposure to impulsive noises while using hearing protection.

Plugging in $10^{17} {\rm kg}$ (of TNT) in that formula (which is probably a rather crude estimate) gives ${460\,320 \rm km} \approx 1.4 \cdot (2 \pi a_\oplus)$ (with $a_\oplus$ being the Earth's radius), meaning more than equatorial circumference of our planet. In other words, no where on Earth was it safe according to UN safety standards.

So let us approach it the other way round: What is the loudest possible sound?

Strictly speaking, the loudest possible sound in air, is 194 dB. The “loudness” of the sound is dictated by how large the amplitude of the waves is compared to ambient air pressure. A sound of 194 dB has a pressure deviation of 101.325 kPa, which is ambient pressure at sea level, at 0 degrees Celsius (32 Fahrenheit). Essentially, at 194 dB, the waves are creating a complete vacuum between themselves.
You can go louder than 194 dB, but that’s not technically a “sound” anymore. The extra energy starts distorting the entire wave, and you end up with something that’s more a shockwave and less a soundwave. At that level, sounds don’t pass through air — they push the air along, producing pressurized burst (shockwaves).

On the shockwave of Chicxulub, I found the page at the Lunar and Planetary Institute insightful:

The Chicxulub Impact event produced a shock wave and air blast that radiated across the seas, over coastlines, and deep into the continental interior.  Winds far in excess of 1000 kilometers per hour were possible near the impact site, although they decreased with distance from the impact site.  The pressure pulse and winds would have scoured soils and shredded vegetation and any animals living in nearby ecosystems.  An initial estimate of the area damaged by an air blast was a radius 1500 kilometers.  There are several factors that can affect this estimate, so the uncertainty might be better reflected in a range of radii from ~900 to ~1800 km.  The travel times are quite short, so this effect would have occurred in advance of any falling debris ejected from the Chicxulub crater.

Equipped with all this, we can now try to answer "How loud was the Chixculub explosion?" even if you did not specify where you measure loudness: At the radius of $\approx 1\,500 {\rm km}$ from the epicenter, the shockwaves "fades" out to become the loudest possible sound of 194dB.


Dobsonian telescopes: do collapsible scopes have any drawbacks?

I agree with Aaron F about storage space, but transport space is different; a non-collapsible tube wouldn’t fit in my car, while my 10″ collapsible-tube telescope fits nicely in the trunk and could fit on the back seat, whether “lying” on it or “standing” as it wouldn’t hit the car’s ceiling.
The only drawback I have heard of, aside from collimation that you would have to do anyhow (nevermind solid tubes holding it better; after enough road, it’s gone no matter what kind of tube you have), is that different thermal expansion of the truss tubes and their holders sometimes makes it difficult to expand or collapse the tube when it’s cold. Also, be careful when storing or transporting your collapsible tube, as the truss bars can potentially end up in positions that slightly bend them, making the expansion or collapse more difficult even if it’s not cold.


The unit of the equation is "kg" because the unit of $10^6$ is $cm^3/m^3$
The unit of the mass of the proton is in kg. The  "1" is $cm^{-3}$ It represents 1 proton per cubic centimetre


What is the gravitational lensing focal distance of a white dwarf star?

The gravitational focus you are talking about is actually a minimum value, defined by parallel rays of light from a very distant star just skimming past the Sun as they are bent according to General Relativity.

The general formula for such lensing is that light is bent through an angle (in radians) of
$$\alpha = \frac{4 GM}{c^2 r},$$
where $M$ is the mass of the lens (assumed to be a point or spherically symmetric mass)  and $r$ is the closest approach of a light ray to the lensing mass.

To work out where a ring of rays will be focused is just a bit of trigonometry.
$$ d_f \simeq \frac{r}{\alpha} = \frac{c^2 r^2}{4GM}$$

This focal distance is a minimum because it would be larger for a ring of rays that passed the lens with a larger value of $r$.

For the Sun as a lens you use $M=2\times 10^{30}$ kg and $r=6.9\times 10^{8}$ m, and calculate $d_f = 540$ au.

White dwarf stars have a similar mass (actually most are about 60% the mass of the Sun, but Sirius B is almost exactly a solar mass), but have radii about the size of the Earth - i.e. a hundred times less than the Sun.

This means that the value of $d_f$ is going to be about 10,000 times less than 540 au. You can use the formula above to calculate it for any combination of mass and radius.

To use the telescope, you place detectors at your chosen focus and observe the bright "Einstein ring" of a distant source that is exactly behind the lens. The magnification factor (the increase in the amount of light collected from the source) then is $4\alpha/\theta$, where $\theta$ is the angular size of the source without the lens.

For a white dwarf, the magnification at the minimum focus would be 100 times larger, because $\alpha$ is 100 times larger.

Note that the size of the image is modified by the ratio of the focal length to the source distance.
$$ x_i = x_o \frac{d_f}{d_o}$$
Thus the image of a distant object will be 10,000 times smaller than if using the Sun, which is much more convenient!

e.g. Observe an Earth-like planet at 10 ly at a focus of 630 au (= 0.01 ly) from the Sun. The image diameter will be 12.5 km. That's a lot of CCD detectors! Using a white dwarf at a focal length that is 10,000 times smaller gives an image just 1.25 m across.

All this assumes that the telescope is perfectly pointed with the source right behind the lens. Any relative motion has to be corrected or the image will move through the focal plane very quickly (like a planet viewed with high magnification through a normal telescope).


Is gravitational wave frequency always equal to double the orbital frequency?

The gravitational waves (GWs) from a perfectly stable, circular binary system are completely dominated by GWs at twice the orbital frequency.

If the binary orbit is eccentric then the GWs have a frequency spectrum; the waves are a combination of discrete frequencies at integer multiples of the orbital frequency (e.g. Wen 2003).
At low eccentricity, most of the power is still at twice the orbital frequency, but for eccentricities greater than about 0.3, the peak frequency moves to higher and higher multiples.

Despite mmeents objections, I believe what I've written is (approximately) true for the circumstances posed in the question - i.e. a non-evolving binary, where the GW energy losses are small, the binary components are thus widely separated and can be treated as point-like, and the rate of periastron precession is small compared with the orbital frequency (note that if the eccentricity were really high, $e \sim 1$, then the latter may not be true). Closer binaries may have significant higher order multipole emission and modifications to their frequency spectrum caused by the decay of the orbit, precession of their periastron, their spin etc.


You don't say what time you were looking. Here is a screenshot from Stellarium at 10pm Wisconsin time on 25th March 2016. Jupiter is in the ESE, but the altitude is a bit lower than 60 degrees. Seems fairly conclusive. You were seeing Ganymede and a Europa/Io combination.


How did Eratosthenes know that the sun is far away?

The sun and the moon go around the observer once a day, Eratosthenes knew that the apparent size of moon doesn't change.  This must mean that Alexandria is near the centre of the moon's orbit.  But the apparent size also doesn't change when viewed from anywhere.  So everywhere is close to the centre of the moon's orbit.  Thus the moon must be much further than the radius of the Earth. If the moon were 6000 miles from the Earth, then it would seem to grow and shrink in size as it passed by (such an effect can be seen on Mars, where the moon really does orbit close to the planet)
And the Sun is further still. At half moon, the sun seems to be at $90^\circ$ to the moon.  This is only possible if the sun is much further away than the moon.
In conclusion, the distance to the sun must be very very large in comparison to the radius of the Earth, and we can assume that that the rays of light from the sun are parallel.


Thought experiment about our location in the Universe

It all depends on shared knowledge. What you are getting at is known in philosophy as indeterminacy of translation. But note that the alien is already talking to the human: there has to be a language present, and languages do contain a lot of assumptions about the world (tenses imply that there is time, and the question itself implies space, for example).

Now, consider a sequence of larger and larger scale maps of the neighbourhood, running from one showing the solar system to one showing the cosmic web. That seems to settle the "where" question using pattern matching fairly well...

...except that there is the obvious question of what the maps show. In this particular example it is stellar baryonic matter, about 0.002 of the total mass/energy budget. If the alien is made of dark matter this will not make much sense except that it would on the larger scales recognise the topology of the web of halo structures it was familiar with. If it knew about baryonic matter it would infer that this was a map of baryonic stuff. Similarly a gaseous alien might think we have left out the 90% of baryonic matter that is intergalactic plasma. But the point is, the right kind of map or explanation would latch onto something in the knowledge of the alien and allow them to start inferring more meaning.

Since this is not just sending a single message but a conversation there is a lot of room for testing and exploration:

"So you are telling me where you are using a map showing 'pulsars'. Are they [untranslatable1] or [untranslatable2]?"

"What is the difference between [untranslatable1] and [untranslatable2]?"

"[untranslatable1] is spatially localised but extends across time,
 [untranslatable2] is temporally localised but spread out in space."

"Then [untranslatable1] sounds more reasonable. We know them as compact stars that rotate fast and send radio waves."

"Are 'stars' also spatially localised and extend across time?"

...

It won't be easy, and it is quite possible that our map doesn't correspond to a known region (if the aliens are communicating in some weird non-local way; had they communicated in a local way they would have known the answer to the questions already). But pattern matching and interactive bootstrapping from the shared language and knowledge-base would be a way of getting around the indeterminacy of translation.


How does diffraction occur in a reflecting telescope?

Diffraction has nothing to do with whether you are using a lens or a mirror in your telescope. Diffraction occurs for the same reason in reflecting telescopes as it does in refracting telescopes: it is passing through an aperture.

There is no threshold aperture diameter for which you have diffraction. No matter how big the aperture, you will have diffraction. No matter the shape of the aperture, you will also have diffraction. This happens on any aperture.

However, the bigger the aperture, the smaller the diffraction pattern, or Airy disc. In
page 19 of the document you mention in the comments, there is a relation between the size of the aperture and the size of the Airy disc.

To clarify what is said in the first paragraph of page 18

  If the telescope is replaced by one with a narrower objective, the images of the two stars would overlap too much and the observer would not be able to see them as separate stars.

If the two stars are replaced with two stars that are close enough together, the images of the two stars would overlap too much and in this case the observer would not be able to see them as separate stars either.

As a side note, refracting telescopes aren't necessarily smaller than reflecting telescopes. There are refracting telescopes with a lens more than a meter in diameter.  A lot bigger than most amateur reflecting telescopes !


If a rocky world is large enough (not a gas or ice giant) can it have helium from the primordial disk in its atmosphere?

The snow line is the inner radius from the star where the water condenses onto dust and larger particles as ice (ie snowballs).  The hydrogen and helium never freezes out.  Planets start forming by accreting these particles (with or without ice but definitely not with H or He), but when a growing planet reaches a sufficiently high mass, it starts to pull in H and He (and more importantly, can hold onto them as they have low atomic mass and thus tend to escape quickly unless there is enough gravity). Exactly what mass that occurs at depends on a lot of factors (temperature, pressure, etc), but it happens below Neptune's mass and in part of the super-earth mass class.  The upper range in mass of the super-earth are also called mini-Neptunes or gas dwarf planets and from their typically low densities it is clear that they are mostly composed of H and He.

However, beyond the snow line it is more likely for a planet to grow to a large mass since more of the surroundings has condensed out of the gas into small particles, therefore accretion is more efficient.  So, it may be that mini-Neptunes are usually forming beyond the snow line and then some of them migrate closer to its star.  But, I don't see any reason to doubt that some of the many mini-Neptunes with H and He found by Kepler/K2 inside the snow-line were formed in situ.  I am not an expert in this field, and perhaps the experts would disagree.


Are white dwarfs the ones always absorbing mass from a companion?

A white dwarf only draws mass because it is denser that it's companion star. If a nearby object was denser than the white dwarf, and exerts enough gravity on the white dwarf, then it will draw mass from the white dwarf.

If the white dwarf is with a neutron star, and the neutron star is close enough to exert a gravitational pull that can distort a white dwarf, then the neutron star will draw mass from the white dwarf.

Now, you see, stars are way less dense that stellar remnants. No matter how massive, it's the density that counts. So if the white dwarf orbits any star, it will draw mass from the star.


Is it possible to see Saturn from Titan's surface, day and night?

"since I believe the atmosphere's opacity changes in day time and night time" maybe to a very small factor, somewhere in the upper atmosphere, in the UV-wavelengths where plasma opacities play a role.
But not in the optical, where the bulk density of the atmosphere plays a role, wich at 1.6 bar doesn't change much between day and night.

Your assertion "So in visible light, Titan looks like an almost uniform ball of beige, yellowish, golden brown." is incorrect, where do you take that from? The optical depth integral is symmetrical both ways (ignoring scattering), so if we by far cannot see the surface, then one will certainly not be able to see Saturn from the surface.
There is also only one existing photo taken on Titans surface, so not sure where you get this idea from.

Apart from the other statement "Titan is the only moon in our solar system with a thick atmosphere, similar to what our Earth's atmosphere was before life 'appeared'", which is a questionable statement at best.


What keeps the planets of the solar system moving in their fixed orbits?

It is ,of course, gravity which keeps the planets circling the sun. They are actually falling toward the sun, but as the sun is a sphere, its surface keeps curving away, so they never get any closer to it. Because they are in a state of free fall, they are in effect weightless with respect to the sun, just as the space station is weightless with respect to the Earth, and as they use no energy in falling they can do it for ever. If you are wondering why you don't feel weightless, that's because of the gravitational field of the Earth. If the space station was a massive body like the moon.the astronauts wouldn't be weightless,but although the space station has a gravitational field of its own it is so trivial that it has no measurable effect on the astronauts.


Could Planet Nine be a tiny black hole or an exotic compact object such that it escaped detection? What would its bounds be?

I think the answer is included in my answer to Could a black hole pass quiescently through the Oort cloud? The mini black hole would be detected via its accretion from the interplanetary medium, probably at X-ray wavelengths.

A ten Earth mass black hole is a ten thousandth (or so) less massive than the Sun, but if it were in a distant orbit, it could be moving slower with respect to the accreting material than the 10 km/s I assumed in the answer cited above.

The net outcome is that this might still be an X-ray source well above the detection limits of all-sky surveys. What's more, it would move with time, so if followed up would show appreciable proper motion.

The variables here are the distance to the black hole and the speed at which is travels with respect to any medium it is travelling through.


What was that Asteroid/Comet in the 90s?

Was it Comet Hale-Bopp? It was discovered in 1995, but made a very close approach in 1997, earning it the nickname "The Great Comet of 1997".

Wikipedia states

  As it passed perihelion on April 1, 1997 the comet developed into a spectacular sight. It shone brighter than any star in the sky except Sirius, and its dust tail stretched 40–45 degrees across the sky. The comet was visible well before the sky got fully dark each night, and while many great comets are very close to the Sun as they pass perihelion, comet Hale–Bopp was visible all night to northern hemisphere observers.

Image courtesy of Wikipedia user LoopZilla under the Creative Commons Attribution-Share Alike 2.0 Austria license.

My second guess is Comet Hyakutake, which made an extremely close approach to Earth in 1996 - becoming "The Great Comet of 1996".


Simple answer: the material is just too spread out.

Forming a planetary-size object takes LOTS of collisions to build up enough matter to start gravitationally attracting the surrounding stuff.  The Oort cloud is very far from the sun (starting at ~1000 AU, compared to a maximum of ~50 AU for Pluto), meaning that the icy fragments are moving very slow in their orbits.  Furthermore, the material is spread out over a huge volume, especially since the Oort cloud is a spherical shell (the matter is spread in 3 dimensions), as opposed to the 2-d disk that makes up the vast majority of matter in the Solar system (Sun, Planets, Asteroid & Kuiper Belts).  The timescale for these objects to accrete to any graviationally significant size is far beyond the lifetime of the Solar system (or even the universe).


Can we observe what objects exist at the L3 positiion of planets of our solar system?

You asked

  L3 is obscured by the Sun, so is that true for our viewpoint of the other planets?

No. Each planet orbits with it's own period because each planet is at a different distance from the sun. So most of the time we can see the L3 point of each planet's orbit from Earth. Occasionally it goes behind the sun from our point of view.

  For the Hilda's of Jupiter have we observed them in that position or is it that we have calculated their coordinates to be in the L3 position?

The Hilda family do not occupy the Jupiter L3 point. They generally orbit closer to the sun than Jupiter, and orbit three times for every two orbits of Jupiter. Their orbits are interesting; they have eccentric orbits (up to 0.3) that in some cases cross the orbit of some of the Trojans. You may have noticed that some of the Hildas also approach the Jupiter L3 point. At that point, and when other members of the family approach the Tojans at L4 and l5, the Hildas are moving more slowly. So the Sun's gravity gradually accelerates them, pulling them back to the lower, faster part of their orbits.

I have linked to the Wikipedia Hilda family page. The Lagrangian point page is also helpful.


What factors influence a star's temperature and density?

For main sequence stars there is not much choice. If you pick mass, then all the rest is determined; any variation you see is due to different chemistry (thus fractions of different elements).
For the main sequence one generally assumes that $L \propto M^{3.5}$ and $R \propto M^{2/3}$ (and similarily the density is given for a given mass). The Hertzsprung-Russel diagramme is expression of this.
Any stars not on the main sequence, be that the white dwarfs or any giants, are there due to their peculiar chemistry. White dwarfs are degenerate stellar remanents, slowly cooling. The giants have evolved away from main sequence. They have some other fusion process(es) than hydrogen burning in their core so that their luminosity is much bigger than hydrogen-burning stars; at least they have a hydrogen-depeleted and helium-enriched core and started hydrogen-shell burning.


Astrophotography: Is there a rule to the color mapping?

Rule in mapping color is just violet (high frequency) >>> red (low one). For example, it was just yesterday thay we had a speaker who who studied in three IR bands: W1 (high frequency = blue), W2 (green), W3 (red).

So, the image has to provide the description about the mapping to be informative.


What's the reason the Astronomical twilight doesn't have constant time?

Look at the extremes. At any time of the year, it takes only minutes (several minutes) for the Sun to sink well below the horizon at the equator. On the other hand, it can take days (several days) after the September Equinox for the Sun to sink well below the horizon for locations above the Arctic Circle. As an extreme extreme, at the South Pole, it takes multiple days for the Sun to rise from just the top of the Sun showing to the full Sun showing (or to set, six months later). Astronomical twilight lasts for several weeks at the South Pole.


Sunset on [Mountain] from [Viewing Point]

My attempt at an answer of my own question. I have no confidence that it is correct, and I'd like to know whether there's an easier way to do this…

https://www.movable-type.co.uk/scripts/latlong.html tells me that the bearing is 90.3°.

A quick application of middle school trigonometry tells me that the visible elevation of Fuji is about 3413 m or 2.90°.

Next, I went to Stellarium, fixed my position to Enoshima, and exported the Ephemeris (F10 menu) of the Sun in a 5 minute intervals with horizontal coordinates from February till May.

Finally, I used a quick python script to find the line in the Ephimeris export where the sun is closest to my desired coordinates (90.3°, 2.9°)

import re
from pprint import pprint

height = "Height"
azi = "Azimut"

# Quick and dirty parser
pre = re.compile("([+-])(\\d+)°(\\d+)'([0-9.]+)\"")
def ph(v):
    m = pre.match(v)
    if m:
        sign, *nums = m.groups()
        return sum([float(s) * 60 ** (-e) for e,s in enumerate(nums)]) * float(sign + "1")
    try:
        return float(v)
    except:
        pass
    return v

res = []
with open('ephemeris2.csv') as f:
    head = next(f).split(", ")
    for line in f:
        php = [ph(v) for v in line.split(", ")]
        res.append(dict(zip(head, php)))

# This is where the magic happens!
filt = [k for k in res if k is not None and height in k and azi in k]
filt.sort(key = lambda v: (v[height] - 2.9) ** 2 + (v[azi] - 270.3) ** 2)
pprint(filt[:5])

This yielded me

{ 'Azimut': 270.13194444444446,
  'Datum and Time': '2019-03-25 17:39:00',
  'Height': 3.0792222222222225 }

I'm unsure whether I can just convert the Bearing to Azimuth like that, but I suspect that a simple mathematical mistake is even more likely…


How to calculate the moon's illuminated fraction tilt?

Your update is correct.

PA is the position angle of the bright limb (measured eastward from celestial north).
q is the parallactic angle between the zenith and celestial north.

Then the angle of the bright limb relative to a horizontal line ($\alpha$) is PA-q-90.
Keep in mind that PA is independent of the observer's location (as long as they are on the Earth), and it is well defined for any given date and time. The parallactic angle is dependent on the observer's location, date and time. I would say that it is well defined only when the Moon is above the horizon. (Of course it can be calculated when the Moon is well below the horizon, but what does that mean?)
For example, imagine two observers separated by 60 degrees of longitude. The Moon is above the horizon for each observer. The PA of the bright limb is the same for each observer at all times. But the parallactic angle is different for each observer since it depends on where the Moon is located relative to the meridian. As the Moon crosses the sky, the parallactic angle also changes significantly from hour to hour, and therefore the tilt you are calculating is different based on date, time, and location. (The PA changes with time also, but it is a slower change except around the time of New and Full Moon.)


Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2\vartheta$, where $\csc\vartheta = 1+\frac{h}{R}$. Thus, the visible portion has a solid angle of
$$\Omega = 2\pi\left(1+\cos\vartheta\right) = 2\pi\left(1+\frac{\sqrt{h^2+2Rh}}{R+h}\right)$$
steradians. Divide this by $4\pi$ to obtain the fractional area of the the visible sky, compared to what you could have if the Earth wasn't blocking your view, since a full sphere subtends a solid angle of $4\pi$ steradians. That is probably a more natural measure of the visible sky than a literal area.

For an actual area, you need some sort of reference distance $r$ to measure from, with the visible sky a distance $r$ away having area $A = \Omega r^2$.


After the sun sets the sky remains light for a while. The significance is that observations during astronomical twilight may be impaired by the continuing glow of the sky. After astronomical twilight, the sky is as dark as it can be, and you can start your long exposures of faint objects.
There are various definitions of "twilight": When it is "not completely dark".  For civil purposes, it is "dark" when the sun is 6 degrees below the horizon, for ships it is dark enough to see navigation stars when the sun is between 6 and 12 degrees below the horizon.  But for astronomers, who need the sky to be completely dark, twilight doesn't end until the sun is a full 18 degrees below the horizon
I see you've already found TimeandDate's series of articles on the different types of twilight but I'll link it here for convenience.


Spectographically determine surface composition of exoplanet

By the title of your question, I don't know if you're asking about the surface composition or the bulk composition. I will answer both.
Although spectrometry can be used to determine the atmosphere composition of a planet, it might also, in theory, helps us to determine the composition of the surface using, as you said it, the reflected light of the planet. This can be done in Solar System planets or moons, because detecting the reflected light is hard but still feasible. In exoplanets tho, it's still very complicated. Maybe with the James Webb Telescope it may be possible in the future. In fact, there are certain papers from which the process of differentiating oceans from land in exoplanets is described, such as Ford et al. (2001), Cowan et al. (2009) or Kawahara and Fujii (2010). I suppose that the actual composition may also be possible. They use the rotation of the planet combined with the difference of spectrometry to make a mapping of the theoretical surface of an Earth-like planet.
Nonetheless, I doubt it about the bulk composition, basically because as you pointed it,  light can't reach into the Interior of a planet, so it's impossible to study its spectrometry in that regard.
If you want to know about the bulk composition of an exoplanet, for example, you can use other techniques such  as the mass radius relationship, which can be often used to determine its bulk composition. Look at Zeng and Sasselov (2013), for example.


How can I download all of this JPL MRO imaging data at once?

You can do it, I will leave you to decide whether you should... As an experiment, I downloaded the mrox_3374 subdirectory of files and the result came out to 3.7 GB. Assuming the other 99 "volumes"/directories are the same, this multiplies out to 370 GB.

I downloaded these data using wget which is a standard utility on Unix/Linux and Mac computers. It is also available for Windows systems; this Wget for Windows 10 tutorial should be of some help in getting it installed and going.

Once this is installed you can do:

wget -np -nH -c --cut-dirs 4 -r https://pds-imaging.jpl.nasa.gov/data/mro/mars_reconnaissance_orbiter/ctx/mrox_3374/

The key flags are -r which means recurse into directories, -np which means don't ascend to the parent directory (to stop it sucking down the entire website), -c which will continue the transfer if it fails and you restart the wget and the --cut-dirs 4 which cuts 4 levels of the directories off the recursive download (the data/mro/mars_reconnaissance_orbiter/ctx parts)

If you really, really want to suck down the whole thing you could do:

wget -np -nH -c --cut-dirs 3 -r https://pds-imaging.jpl.nasa.gov/data/mro/mars_reconnaissance_orbiter/ctx/

and that should grab everything.


Is there any phenomenon where a supernova doesn't anything leave behind?

You are probably thinking of a pair-instability supernova. Such a supernova releases more energy than the star's gravitational binding energy, completely disrupting the star and dispersing the contents.  Meaning that with nothing else in the universe, over infinite time every particle from the star will get infinitely far away from any/every other particle.  The mass, energy, etc. of the star doesn't vanish from the universe, though; it just gets spread throughout it.  But there will be no dense remnant left behind: no dwarf, no neutron star, no black hole; over long time scales there won't even be a discernible nebula remaining (though note the star may have shed off significant amounts of material prior to this event, which would interact with the ejecta and may result in some sort of permanent structure suggesting a star used to be there).


Why would tidal forces on planets become more intense when a star becomes a white dwarf?

From the paper, p.2:

  The star eventually transitions into a white dwarf, which is comparable in size to the Earth but has a Roche radius which extends outward to about one Solar radius ...How objects can be emplaced at one Solar radius from several au – and then circularised – is still subject to debate.

They are considering planets which have been observed in close orbits (orbital period as short as 2 hours) around white dwarfs.


Any record of the Earth passing through the tail (not trail) of a comet?

Earth passed through the tail of Halley's Comet in 1910. It caused a bit of a panic due to claims that cyanogen detected in the tail would cause harm: https://en.wikipedia.org/wiki/Halley%27s_Comet#1910
Linked there is https://timesmachine.nytimes.com/timesmachine/1910/02/08/104920328.pdf
 (click for larger)


Why do "they" portray colliding black holes like that?

The event horizon is not a physical structure. Rather it is the boundary of a region of spacetime from which no information can escape. If you were (unfortunate enough to be) between two black holes, you may be in a position in which there is no net force, and yet you could not travel from your location to the point distant from the two black holes, no matter how fast you were able to move.

If you were in such a position, you would be inside the event horizon of the two black holes.

Finding the event horizon means solving the Field equations of General relativity. Exact solutions are known in several situations: A single mass concentrated at a point, a mass with a charge and a rotating mass. If there are two masses, then the Field equations have no exact solution, but can be solved numerically (with a fast enough supercomputer) It is the results of these calculations that you are seeing.

So, the shape you see is the event horizon, The event horizon of two nearby black holes can be calculated by a computer. Since you may be trapped between two black holes the event horizon will be deformed towards another black hole's event horizon, giving the "teardrop-> bridge->merger" appearance of the final moments of two back holes.


Why is it impossible for a Quasi Star to exist these days?

This is a highly uncertain prospect, since the physics of the system itself relies on processes that are themselves uncertain to varying degrees, e.g., stellar nuclear evolution of very high-mass stars, accretion of gas onto compact objects, etc...
A relevant paper, cited by the wiki article for quasistars, is the work by Begelman $\it{et~al.}$ (2007). They show that newly formed quasistars should have surface temperatures higher than 10,000 K (9,700 °C). As a quasistar cools over time to a limiting temperature of 4,000 K (3,730 °C), for metal-free (Population III) opacities, at and below which hydrostatic equilibrium is no longer sustainable, the end of the quasistar's life leaves behind an intermediate-mass black hole.
Such quasistars would have a short lifespan, at most ~ 7 Myrs, during which the core black-hole would grow into an intermediate-mass black hole which are suspected to be the progenitors of current-epoch supermassive black holes. This implies that quasistars are transient, and so observing them directly would be very difficult. However, it is possible that we could observe indirect effects of their existence on their environments, for example this is how we infer the existence of common envelope evolution, or on the evolution of intermediate-mass black holes.


How would vehicles travel through the interstellar medium with its such low density?

The answer is in your question.

You said

  there couldn't be any (meaningful) forces interacting between them to provide any sort of way to travel through it.

That's the great thing about outer space. There normally aren't many particles around that could slow down a spacecraft via drag or friction. If you give something a push in outer space - somewhere that unrealistically is free from any gravitational interaction with other objects - it will travel forever. Objects travel through the ISM via Newton's first law - an object in motion tends to stay in motion.


Do some ELF (Extremely low frequency) radio waves pass the atmosphere reach Earth surface and then are reflected passing the atmosphere again to reach outer space?

In space plasmas, ELF is used to describe waves in the frequency band ~3 Hz to ~3 kHz.  Very low frequency (VLF) is used to describe waves in the frequency band ~3 kHz to ~30 kHz.

Generally the cutoff frequency of the Earth's ionosphere prevents waves at such low frequencies from freely passing through it.  In plasmas, this tends to be any wave at or below the local electron plasma frequency, which can be approximated as $f_{pe} \approx 8.9787 \ \sqrt{n_{e} \left[ cm^{-3} \right]}$ kHz (where the $\left[ cm^{-3} \right]$ implies that the electron number density, $n_{e}$, should be in units of $cm^{-3}$).  The densities in the ionosphere are ~$10^{3} - 10^{6} \ cm^{-3}$, which corresponds to $f_{pe}$ ~ 280 kHz to ~9 MHz.

Early studies found a relationship between lightning and a VLF emissions called whistlers.  It is now known that high energy electron beams, associated with the lightning, escape into the magnetosphere.  These beams radiate the whistlers through a plasma instability (i.e., free energy due to a fast beam streaming through an otherwise homogeneous plasma).  So the free energy sources can escape the lower atmosphere, but the electromagnetic emissions at such low frequencies cannot.

  And once in outer space can they travel indefinetely in it or the interplabetary plasma would stop them?

Generally, waves with $f < f_{pe}$ experience damping as they propagate through a plasma.  This is because the wave fields oscillate slow enough to interact with the particles, thus adding an effective inertia term to the system.  ELF and VLF waves generally damp through cyclotron and Landau resonances.  Note, however, that both of these resonances are also the processes responsible for wave growth in the presence of sufficient free energy (e.g., free energy from beams is often related to a Landau-like or Cerenkov-like emission).

  This occurs only when when the plasma is excited?

Not necessarily.  We can artificially induce VLF emissions using ground-based transmitters.  But otherwise yes, there needs to be free energy for the radiation of electromagnetic waves.

  When it is not, these radio waves would be able to make it through?

See my comments above about the cutoff frequency.


Someone else posted a question about this very scope not long ago. Simply put, no, it's little more than a cheap toy and a waste of money.  The optical quality is going to be extremely poor.  In my opinion, it is more likely to dis-interest someone in astronomy.

You didn't mention how old your kids are.  Not knowing, it's hard to make a clear recommendation.  Still, I wouldn't recommend this.

If your kids were interested in music and wanted to learn to play guitar, would you buy them the cheapest, lowest-quality instrument available, one that has poor sound, is difficult to tune and then doesn't stay in tune, and with such poor intonation that even when the strings are open-tuned, you can't produce proper chords?  You wouldn't necessarily want to rush out and get them a professional-quality recording-ready instrument, but you wouldn't get them a toy either (at least, I hope not).

I'd say the same thing about a telescope.  You want a real telescope, with quality optics.  I'm sure you don't want to pay a lot, but you don't want to get something that will be so frustrating as to make them lose interest.

Of course, like playing a guitar, unless you know what to do with a telescope, it will be frustrating.  Too many people buy telescopes with the belief they'll see things like they see in pictures on the internet, only to be badly disappointed.  I've seen so many telescopes put up for sale on Craigslist and eBay or thrown in a closet or garage and ignored because the user had no idea what they were doing and were completely disappointed.

Depending on your kids age, my first recommendation would be to find and join a local astronomy club. The club I belong to, the Houston Astronomical Society, does regular star parties for schools and children.  Many of our members join because of their kids - because they want to be able to teach their kids.  We also have many adolescent and teen members who are active in our club.  Membership is inexpensive (most clubs are $50 USD or less per year) and you can learn a lot.

As for a telescope itself,  I would ignore ANY AND ALL telescopes under \$1,000 USD that are computerized, what we call GoTo technology.  While some people might disagree, my experience, which is fairly extensive, with such telescopes is that they are poorly manufactured and prone to breaking down - at which point they usually are not usable in a "manual" mode.  You also end up paying a lot for the electronics and mechanical parts, and getting less telescope.  On top of that, the majority of those telescopes still require you know your way around the sky to be able to find key landmarks (usually a number of bright, named stars) to set up the telescope.

My general recommendation is an 8" Dobsonian telescope.  A Dobsonian telescope is a Newtonian OTA (Optical Tube Assembly, the telescope-proper portion of the full instrument) with a simple altitude-azimuth rocker-box mount (meaning the motion is up/down and left/right, a natural motion for most people).  Dobsonians or Dobs, offer the best ratio of price to aperture, and are very easy to use.  However, an 8" Dob may be too much for smaller children, may be more than you're willing to spend, and may be larger than you're comfortable with. However, if you can, an 8" aperture is a very good starting place for an amateur telescope as it offers a good balance between cost, physical size, and optical capability.  Most 8" Dobs weigh a total of about 50 lbs, evenly divided between the OTA and base, making the heaviest single piece around 25 lbs, with an overall total height/length of about 55 inches, give or take a couple.  If you live in a light polluted area, they typically fit comfortably on the back seat of a midsize sedan or larger, and often will fit in a compact (though not something like a Smart car).  Brand new, you can sometimes find them under \$400, but usually you can expect between \$400 and \$500 with a decent starting accessory package.  I personally recommend the Apertura AD8, though the Orion 8" Skyline and XT8 are also decent models.

If price and/or Size is an issue, the next step down would be a 6" Dobsonian.  The length/height is about the same, but they're somewhat thinner and a little lighter.  The smaller aperture reduces light gathering ability, which reduces overall capability. Brand new, the price runs under \$300 for a decent instrument, but they can often be found used or refurbished for closer to \$200.

Smaller still, I'd consider the AWB OneSky 130 (known outside of the US as the SkyWatcher Heritage 130p).  This is a 130 mm (5") Newtonian telescope on a simple alt-az mount (often referred to as a Dobsonian, but not a true Dob).  The price is around \$200.  Another option at the same size is the Meade LightBridge Mini 130, which is similar in size and capability.  I've seen this closer to \$175 lately.

Stepping down further, there are a handful of 114 mm (4.5") scopes, including the Meade LightBridge 114 and the Orion StarBlast 4.5"  Again, you are stepping downward in capability, but also price, and these smaller scopes are much lighter and easy to transport, though you Won't see as much with them.

I would not recommend ANY telescopes under this.  There are many available, but most are of poor quality or design.  Under \$150-ish, I'd recommend a decent pair of 7x50 or 10x70 binolculars.

Good luck!


How did "oddball" Valetudo, Jupiter's new prograde moon, end up in a wider orbit?

My questions are: 1.:  why are most of the moons that are further away from Jupiter retrograde orbits, while the ones extremely close to Jupiter are prograde orbits? 2.: What possible situations were required for Valetudo to end up where it is opposed to where the rest of the pro-grade moons are clustered? Pure chance / a series of unlikely events?

Answer 1. Inner moons usually orbit in the same direction as the rotation of their parent planet, this is favored due to one means of their formation, and the effect of tidal forces if they were close to the planet and in a retrograde orbit.
Source 1: Wikipedia: "Moons of Jupiter - Regular Satellites":

"The inner moons are in a 1:2:4 orbital resonance. Models suggest that they formed by slow accretion in the low-density Jovian subnebula—a disc of the gas and dust that existed around Jupiter after its formation—which lasted up to 10 million years in the case of Callisto.$^{[39]}$
[39]: "Formation of the Galilean Satellites: Conditions of Accretion" (Dec 2002), by Robin M. Canup and William R. Ward.

Source 2: Wikipedia - "Hill Sphere - True Region of Stability":

"The region of stability for retrograde orbits at a large distance from the primary, is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.$^{[3]}$
[3]: "Chaos-assisted capture of irregular moons" (May 2003), by Sergey A. Astakhov, Andrew D. Burbanks, Stephen Wiggins, and David Farrelly.

Answer 2. It is possible that Valetudo was a binary companion and was captured in an exchange reaction.
Source 1: Wikipedia: "Moons of Jupiter":

"These moons were probably captured by Jupiter from solar orbits."

Source 2: "Satellite Capture via Binary Exchange Reactions: Application to Triton" (Dec 2005), by C. Agnor and D.P. Hamilton:

"... exchange reactions during planet-binary encounters (i.e. ~ an encounter in which one member of a binary is expelled and its place taken by the planet) represent a new pathway for the capture of planetary satellites. This new capture mechanism may be relevant to origin and evolution of several small body populations in the solar system (e.g.~the icy and irregular satellites of the giant planets).

Additional Info: Robert Walker explains and offers this image showing how contact and overcontact binaries, ternaries, quarternaries, etc. can form stable and unstable systems. Those can be torn apart by the tidal forces of larger (even smaller) objects.


Can two heavy objects circling around their C.M. be separated because of the speed of gravity?

No. Newtonian mechanics does quite well assuming instantaneous gravity. You can get relativistically correct orbits either by doing a complete GR calculation, or using a simple approximation as already eluded alluded to in @antispinward's answer.
This excellent answer to Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch? in Space Exploration SE explains that we get the wrong answer if purely Newtonian mechanics is used except for slowing down the speed of gravity, but that's because one is neither treating gravity correctly nor using Newtonian mechanics correctly.
Many answers to How to calculate the planets and moons beyond Newtons's gravitational force? include a way to treat this problem using a well-accepted approximate treatment of general relativity.
From this answer:

The acceleration of a body in the gravitation field of another body of standard gravitational parameter $GM$ can be written:
$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$
where $r$ is the vector from the body $M$ to the body who's acceleration is being calculated. Remember that in Newtonian mechanics the acceleration of each body depends only on the mass of the other body, even though the force depends on both masses, because the first mass cancels out by $a=F/m$.

and later:

The following approximation should be added to the Newtonian term:
$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$


Why do particles from a comet that result in meteor showers spread out mostly along the comet's orbit?

The cometary particles need not retain the exact same orbit as the comet in order to stay close together near perihelion, thus where their orbit passes Earth's orbit.

Slight changes in the orbital parameters of the released dust particles can include slighly higher or lower aphelion, and thus change the revolution time and the time of perihelion (Earth) passage enough to spread out the particles. Over many revolutions it will eventually spread out over the whole orbit.

Additionally the orbits by the dust particles are subject to changes due to interaction with the solar wind and solar radiation which will over time change their orbits (yorp effect, Yarkovsky effect, radiation pressure etc).

This behaviour also explains that the meteor showers will be stronger in some years when Earth passes through a part where particles have not yet had time to more equally disperse.

Addendum: The other question on tails How can comets have tails if there's no air resistance in space? is related. Yet the (visible) tails are caused by either (ionized) gas or (small) dust particles (nm ... µm) which scatter the light. While these dust particles certainly also can make meteors, the larger the particles are, the less efficiently radiation pressure works on them, so their orbital evolution is much slower. And comets do release also a lot of bigger particles from µm up to cm (excact size distriubtion subject to disussion and analysis especially as follow-up to Rosetta mission, also depending on comet size and age) which are not easily visible in the visible light when observing it from Earth.


Relationship between precession and dating ancient alignments

The answer is fairly simple.   It takes about 22,000 years for the Earth to perform 1 spin or move 360 degrees along it's axial precession.   The 22,000 year cycle means that the temple was lined up at 6,000 years ago and at 28,000 years ago, as well as 50,000 and 72,000.

In other words, if we build a statue that points directly at the North Star, In a couple thousand years it will visibly no longer be pointing at the North star, but in 22,000 years, after a full circle, it will be pointing at the North star again.   More specifically the North Star will apparently move in a circle that takes 22,000 years to complete, moving away from and back towards the celestial north pole.

Any skepticism on the 28,000 year date wouldn't have anything to do with alignment as the alignment works just as well with either date, and it works for any period a full circle in the past, like 50,000 years ago.  The skepticism is likely related to doubt that it's that old.

Similar to the hour hand on a clock that takes 12 hours for the hour hand to return to where it started, the Earth takes about 22,000 years to return to the direction the North Pole is pointing.   That means, no time between 6,000 and 28,000 years ago will line up.  Those are the two most recent alignment periods.   It moves in a circle in 3D, more like the picture below than like a clock.

As an astronomical note, I want to point out that 22,000 is somewhat inaccurate.   The Earth's axial precession in and of itself is about 25,771 years, but that's only part of the movement.  There's also a larger, slower apsidal precession which adds to or subtracts from the 25,771 year wobble.   These together affect the alignment of the Earth to the mostly fixed stars, and that period varies, but it was about 22,000 years between 6,000 and 28,000 years ago.   It can take as little as 20,800 years or as many as 29,000.   Average is about 23,000.

Many Native American structures lined up quite well to astronomical alignments, so dating ancient structures by measuring when it lined up a reasonable approach.    As a related, somewhat similar example, the Mayan 2012 prophecy is also based on an astronomical alignment and they were, as I recall, a year or two late, but it was still a very good estimate given when they made it.   Understanding these alignments requires seeing things in 3 dimensions and that can be a little confusing to try to picture.


Divide a star catalogue to optimize star search

This is really a computing problem, but I suppose the only point from an astronomical perspective is what the RA, Dec distributions of your catalogues look like.

I'm not that familiar with optimal search techniques but I guess that you want roughly similar numbers of stars in each region.

If your catalogue is just of the brightest stars, then these are fairly uniformly distributed around the sky, so splitting them up in equal bands of right ascension would probably do.

However, if you are dealing with much larger catalogues then there is likely to be a significant concentration towards the galactic plane. In these circumstance you might be better off splitting your regions in terms of their galactic latitude. So you would calculate galactic coordinates by transforming the RA, Dec and then plot the distribution of number of stars versus galactic latitude to see where you would split your regions.


How accurate is Stellarium at depicting the Earth, from the Moon, approaching a solar eclipse (from Lunar perspective)?

While a black disk might not be entirely accurate, it is not a too bad approximation. Better would be using orbital night-time images of Earth.
There are images in the NASA archive which show Earth at night, the programme is titled "The Black Marble" which depicts Earth at night. There's a video which covers 22 days. The actual image depends on atmospheric conditions and on which side Earth shows - you might see some lights of civilisation. Similar to new moon conditions where you can also see features on the moon, an observer on the Moon would see some features on the Earth's surface. Additionally, maybe some of the refracted light of the atmosphere or airglow or aurorae is visible - but with the atmosphere being less than 1% of Earth radius, that's not much in comparison to the disk of Earth in the observer's FOV; yet it might be enough to be detectable, depending on sensitivity of the observer.


What direction does 0° in the J2000 epoch point towards on January 1, 2000?

Or is 0 degrees the direction towards the ascending node 'equinox'? Which is NOT towards the Sun on January first...

The answer is yes, more or less. From JPL's HORIZONS, the location of the Sun at Noon Terrestrial Time on 1 Jan 2000 is, ignoring atmospheric effects, a right ascension of 18 hours, 45 minutes, and 9.36 seconds. This is nowhere close to a right ascension of zero hours (or 24 hours, same thing).
The right ascension of the Sun with respect to the center of the Earth is zero at the March equinox.


How much is the zoom of Sun-Pluto eclipse image

The "raw" LORRI image is 1024x1024 pixels. The field of view is about 0.29°, see section 3.6 of this paper, or abstract of this paper.
Pluto is a little less than 250 pixels in the raw image, corresponding to 0.29° x 250/1024 = 4'15'', well consistent with your estimate.


Is there a distinction between NEOs and near-Earth asteroids? Is there a difference?

Yes, Near Earth Objects (NEOs) include asteroids (Near Earth Asteroids (NEAs)) and a few percent of comets (Near Earth Comets (NECs)). As shown by the Update to Determine the Feasibility of Enhancing the Search and Characterization of NEOs (NEO SDT Report) in Section 2.3, the risk from Near Earth Comets is about 1% that of asteroids. So "NEO" tends to be assumed to mean "asteroid" unless otherwise stated.


It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller)

The premise of your question is false. I tried to convince you to separate it out as a separate question, but to no avail. So I'll answer it here instead.

From comment:

  I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise.

To help illustrate that a narrow annular aperture has much worse resolution than the full aperture, I'll do a quick comparison. The first plot is the diffraction pattern (scale is +/- 0.5 arcseconds) for a clear aperture of 2 million wavelengths (e.g. 1 meter radius with  1 micron wavelength), and the second is for the same aperture with a 0.9 meter obstruction.

I use Airy disks and Babinet's principle to subtract the E-field of the obstruction before squaring. For this symmetry there's no need to do an explicit Fourier analysis, the Airy disk is the Fourier transform of the circular aperture step function. (see also)

This is for an annulus with outer radius 1.0 and inner 0.9. If it was 1.0 and 0.99 it would look much much worse!

  Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.

That would give you a giant splat pattern of roughly the same size as a simple 3 cm diameter telescope would.

Python script:

from scipy.special import j1
import matplotlib.pyplot as plt
import numpy as np

def get_E(a, lam, theta):
    ka  = (twopi/lam) * a
    x   = ka * np.sin(theta)
    return a * (2. * j1(x) / x)

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

arcsecs = np.arange(0, 0.5, 0.0001)[1:]
theta   = (arcsecs/3600.) * rads

r1, r2  = 1.0, 0.9
lam     = 1E-06

arcsecs   = np.linspace(-1, 1, 1000) # arcsec (skips zero to avoid singularity)
Xas, Yas, = np.meshgrid(arcsecs, arcsecs)
Ras       = np.sqrt(Xas**2 + Yas**2)
theta     = Ras * pi / 3600. / 180.

E1, E2    = [sign*get_E(r, lam, theta) for r, sign in ((r1, +1), (r2, -1))]
asec      = np.linspace(0, 1.0, 500)[1:]
th        = asec * pi / 3600. / 180.
e1, e2    = [sign*get_E(r, lam, th) for r, sign in ((r1, +1), (r2, -1))]

Etot = E1 + E2
etot = e1 + e2

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(E1**2, cmap='gray')
    plt.title('r = 1.0, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 3)
    plt.imshow(Etot**2, cmap='gray')
    plt.title ('r1, r2 = 1.0, 0.9, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 2)
    plt.plot(1E+06*th, th*e1**2) # weight by th for azimuthal (phi) integration
    plt.xlabel('urad', fontsize=14)
    plt.title('r = 1.0, lam=1E-06, integ theta')
    plt.subplot(2, 2, 4)
    plt.plot(1E+06*th, th*etot**2) # weight by th for azimuthal (phi) integration
    plt.title('r1, r2 = 1.0, 0.9, lam=1E-06, integ theta')
    plt.xlabel('urad', fontsize=14)
    plt.show()


How deep in Mars would a person not need insulation and pressure?

According to this paper, there are various but enough good estimations for the temperature gradient of the Martian soil. Note, direct measurements will we have first with the next Martian lander, the InSight:

Image from here

The for us important part of this estimation is this:

  Depth penetration of the annual temperature wave at $120^\circ E,
> 20^\circ N$, using data from the NASA/MSFC Mars GRAM as the surface
  boundary condition and assuming a planetary heat flow of 20
  $\frac{mW}{m^2}$. The snapshots of the temperature as a function of
  depth are given for the models with

  (a) $k_\infty = 0.02 \frac{W}{m\cdot K}$ and
  (b) $k_\infty = 0.1 \frac{W}{m\cdot K}$.

  Heat flows derived from the soil temperatures given in Figures 3a and
  3b for models with
  * (c) $k_\infty = 0.02 \frac{W}{m\cdot K}$ and
  * (d) $k_\infty = 0.1 \frac{W}{m\cdot K}$.

  The snapshots (gray lines) show the local heat flow. The area inside
  the envelope (black lines) represents possible heat flow values that
  can be obtained by a single measurement. The heat flows derived from
  the annual mean temperatures are indicated by squares. Note that in
  order to calculate heat flows, the thermal conductivity was assumed to
  be known.

What we can see here, is similar to the Earth:

there are significant temperature differences in the upper soil, depending on the geographical position, yearly and daily cycles
however, around from 4m depth, the soil temperature is roughly constant.

Extrapolating the upper two graphs, we can expect a pleasant earth-like temperature not very deeply, roughly around 20-40 m.

However, according to this answer, the Earth-like pressure happens much deeper, around 20-40 km below the surface. This answer uses the Barometric formula to calculate the required depth, which is a very good estimation. It seems probably unreachable with the current technology, although it is not very far away from it: the main problems of the digging of deep holes are

cooling
water incursions

The first is much easier, the second is non-existant on the Mars.

Note, the Martian atmosphere is mainly carbon dioxide, thus spacesuits won't be needed, but some oxygen tanks still will be.

There is no such depth where both the temperature and the pressure would be comfortable for us.

P.s. actually, 16% pure oxygen is already enough for us to breathe, which makes the required depth to around 10-15 km.


What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured?

As defined quite clearly, with equations, here

The total atmospheric water vapor contained in a vertical column of
unit cross-sectional area extending between any two specified levels,
commonly expressed in terms of the height to which that water
substance would stand if completely condensed and collected in a
vessel of the same unit cross section.

edit
There's, for example, an ALMA paper which discusses measurements and comparison with meteorological stations.


Are there expanding-universe cosmological models that do not include inflation?

Variable speed of light (VSL) theories (for instance Petit (1988) and Moffat (1993)) have been introduced in the past to solve the horizon problem you mentioned, i.e. that the observable universe appears to be isotorpic and homogeneous. If certain universal "constants" (including $c$, the speed of light) are allowed to change, then it is possible for interactions between extremely distant sections of the universe to occur, solving the horizon problem. These modifications still fit in with general relativity and the idea of an expanding universe.

VSL theories, though, have not gained much traction, and inflation is still the prevailing theory in the scientific community today. It solves the horizon problem, the flatness problem (dealing with how fine-tuned the initial density of the universe appears to have been), and the magnetic monopole problem (i.e. why we don't see magnetic monopoles). Inflation isn't the only possible solution to these problems, but it works quite well.


What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice?

Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams)

The wikipedia page notes that 'Oumuamua "has circulated the Milky Way several times" so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua  600,000 years to travel here from a local star, the time before it can be expected to have another stellar encounter would be much greater, hundreds of millions of years. And the next star that it meets is unlikely to be a currently local star, it could be almost any star in the galaxy.

Once an interstellar object has left the solar system it will travel in a roughly straight line, (slowly curving with the general gravity of the galaxy). For it to return to the Earth it would have to have a gravitational encounter with another star (or more likely a series of encounters) that send it back towards the solar system. Even given a vast mind-boggling long period of time like 10 billion years the chance of this happening is minute.

Such an object will probably orbit the galaxy several times between stellar encounters, perhaps it has an encounter every 250 million years. And it is equally likely to encounter any of the disc stars (conservatively 50 billion stars). This means that the mean time between encounters of the sun is on the order of ten quintillion years.

It isn't impossible.  It is quite easy to design a free-return flight to the moon.   There is no reason why a free-return flight to another star can't be done. However, space is big "therefore ʻOumuamua will never return".


Why do nebulae turn into accretion disks but planets do not?

The circumstellar disk is perhaps about 2%-3% the mass of the central star.   Large planets may form in similar ways.   The mass ratio of the disk to what it orbits depends a lot on the angular momentum and I would think that only large planets that clear out a considerably larger volume of space are likely to have enough angular momentum to form disks around them.    Jupiter's Galilean moons and most formation moons probably form planetary debris disks during their planet's formation.

This ring system around a brown dwarf that orbits a larger star could be compared to a circumstellar disk.

Disks of this nature, both stellar or theoretical-planetary are temporary, so only very young solar systems might have them and we probably don't have telescopes good enough to get a good look at young planets during their formation period.   A large part of the reason we don't hear about planetary disks is that we might not have good enough telescopes to observe them.

(Note, if someone with better knowledge than me answers this, I can remove my answer).