Are there stars in the first photo from space?

I really doubt it. (Note: I'm answering this more as a photographer than an astronomer)

From this vantage point, Earth is far brighter than any stars you would see beyond it (save the Sun). When you set a camera to expose Earth, if you set it such that you can expose Earth without blowing out the details you need to set the camera accordingly. This, however, would mean that it's unlikely the camera settings would be sensitive enough to also get stars. As we can see rough details of the atmosphere in your photo (Despite the grain), it would suggest that the exposure settings would not be sensitive enough to depict stars.

For example, here is a photo of Jupiter:

There are a few things to note here:

Jupiter is entirely blown out. It's over exposed and you can't see any details of the planet.
You can see several moons. If you properly expose Jupiter, you generally won't also get the moons, because your camera settings won't be sensitive enough to pick them up. (at one time I had a more properly exposed photo that lacks the moons. Apologies, I'm not sure where it's gone off to)
You aren't seeing any stars in the background. Again, the exposure settings to pick up the moon are too sensitive to properly expose Jupiter, but not sensitive enough to pick up the stars behind it. (A brighter star, had it been in the frame would be visible in the photo, but there are certainly some stars that were in a shot this wide. Regardless, there are stars far brighter than the Galilean moons, so this isn't a particularly solid point)

There is certainly something in the photo. Judging by the amount of grain in the photo and by the fact that JPG images will always contain some artifacts it would seem more likely that what you're seeing is the affect of noise or artifacts. Dirt on the lens is unlikely - this usually doesn't show up as specks in the final photo, as this is before the light has not been distributed yet. You see dust if it's on the lens exit or between the lens and the sensor, but in that situation you're going to see it generally reducing light from the final image.

This answer may change if, by chance, the photo was taken when it was night on Earth. However, this seems unlikely for an early rocket launch and certainly for a first attempt at a photo of Earth from space.

As this is really more about analyzing a photograph than astronomy, if you want to probe further you may want to inquire in Photography.SE.


Can the next possible ice age be projected based upon the Milankovitch cycle?

This paper seems to be discussing what you've asked: http://www.sciencemag.org/content/297/5585/1287.full. The next ice age is expected to begin about 50,000 years from now. However, I think it's still debatable how accurate the predictions of Milankovitch cycles are, and how much of an effect humans are likely to have on this earth system.


Will we start seeing galaxies disappear due to Universe expansion?

It is a common misconception that galaxies receding faster than light cannot be observed. There are two versions of this misconception:

Galaxies that are now receding faster than light cannot be seen.
If we observe a galaxy today, it may recede faster than light now, but when it emitted the light we see now, it did not.

Both are incorrect.
Intuitive explanation
This is most easily seen in a spacetime diagram. But before showing that, a somewhat intuitive way of understanding this apparent paradox is to consider the journey of the photon from the reference frame of the emitting galaxy: In this frame, the photon leaves the emitter with a velocity $v=c$ (as it should), but since its velocity is always $v$ locally, then in the frame of the emitter, its velocity increases without bounds. This is not a proof, but may help to show that "light may travel faster the the speed of light".
Spacetime diagram
Now to the spacetime diagram. My favorite version is from the classic paper by Davis & Lineweaver (2004), but I actually like better this version from Pulsar's excellent (!) post:

Real example of a galaxy that has always been, and always will be, receding with superluminal velocity
Such a diagram shows age (or, equivalently, size) of the Universe, as a function of the corresponding comoving distance$^1$ of events. A lot of things are going on in this diagram, and I'll refer to Pulsar's answer for an interesting discussion of all this. For our purposes, consider the only thing the I have added to the figure: the world line — i.e. the "path" taken through spacetime of some object — of the galaxy GN-z11, seen as a vertical dashed line at a comoving distance $d_C \simeq 32.2\,\mathrm{Gly}$. It is vertical, because it doesn't move in comoving coordinates. We observe it at a redshift of $z=11.09$, which means that the light we see (moving at 45º toward our worldline which is at $d_C=0$) was emitted when the Universe was roughly 400 Myr old. At that time, GN-z11 was receding at a velocity $v\simeq4c$!
Hubble sphere and event horizon
The innermost green bubble (labeled "$v_\mathrm{rec}=c$") is the Hubble sphere, defined such that at any point in time, everything outside recedes faster than $c$. Meanwhile, the red line shows the event horizon, which defines the boundary between everything that can eventually be observed (light red region), and everything that cannot. A subset of this is out past lightcone (orange region) which is stuff that we have already observed (or at least have had the chance to).
Everything lying in between the Hubble sphere and the event horizon consists of events that we have been, or will eventually be, able to see, but which recede faster than the speed of light when it emits the light we see.
As you can see the region becomes narrower with time$^2$, meaning that the accelerated expansion of the Universe makes it harder and harder for photons emitted far away to reach us. We will never be able to see galaxies that today are farther away than roughly 17 Gly. But galaxies that are nearer will be seen (if we look), and a subset of these — namely the ones with $14\,\mathrm{Gly} \lesssim d_C \lesssim 17\,\mathrm{Gly}$ — are currently receding faster than the speed of light.
The size of the observable Universe always increases
The observable Universe is defined as the region within which light has had the time to reach us. The "edge" of this is called the particle horizon, and as seen from the blue line, it always increases its distance from us (in comoving coordinates, and thus even more so in physical coordinates). In comoving coodinates, it reach a maximum size of roughly 63 Gly. That means that more and more galaxies will enter our observable Universe, but as time goes they will do so at an ever-decreasing pace, asymptotically reaching a finite value (of 5 trillion galaxies, more or less). Once inside the particle horizon, a galaxy will never leave. It will, however, keep increasing its distance from us in physical coordinates, as well as its speed, and thus redshift without bounds and fade to black.

$^1$Comoving coordinates are defined so that they "expand with the expansion of the Universe". That means that in these coordinates, galaxies lie approximately still wrt. each other. To get the physical coordinates — i.e. what you would measure if you froze the Universe and laid out measuring rods — multiply the comoving coordinates by the scale factor $a$ seen on the right $y$ axis. Since today $a\equiv1$, comoving and physical coordinates coincide today.
$^2$The fact that the diagram is in comoving coordinates makes it a bit hard to see, but this is still the case in physical coordinates.


A few questions regarding the transit of planets

When we talk about a transit, we are normally talking about one that is visible from Earth, or at least from some planet in the solar system.

But if you are willing to go to a point is space, it is always the case that every planet is transiting the sun from somewhere. So the Earth is right now transiting the  sun from any position that is directly opposite the sun and more than about 1600000km from Earth (otherwise the sun won't be transited, it will be eclipsed.

The transits of all the planets can be seen from exterior planets, though the transits of (say) Saturn from Neptune are exceedingly rare

If you are in the orbital plane of the planet, you will eventually see a transit.
But you may be waiting a long time. The semi major axis doesn't have any effect.

If the planet in a planetary system are in the same plane, then you may see more than one transit. But you are less likely to see the outer planets as they are further from the star.


Reflector vs refractor in terms of magnification, which one wins?

Magnification is (mostly) done by the eyepiece. You can change the magnification by using a different eyepiece. Whether it is a reflector or refractor doesn't make much difference.

There is a limit to the useful magnification that can be done. At a certain point, you are just magnifying the blur. Larger apertures can resolve better, and so can allow greater magnification. Using more magnification also makes the view dimmer. A bigger aperture collects more light, so also allows for more magnification.

The choice between reflector or refractor is not about magnification. Lenses work by refraction, and this can create coloured fringes (which can be compensated for, but this increases the cost). Thicker lenses also absorb light, but so do mirrors to some degree.  Generally, bigger is better, but other factors (stability of mount, ease of use, ease of transport, computer control etc.) may be significant. "Magnification" is often not a good measure of quality.


I am not sure what you are referring to; possibly it is intensity interferometry using the Hanbury Brown & Twiss method, but even this requires two telescopes. This kind of interferometry was used to estimate angular diameters for some tens of stars but is not generally used these days for this purpose. Instead, one constructs interferometers from separated telescopes using delay lines to compensate for path-length differences and maintain coherence between the different paths that the light takes. This then uses amplitude interferometry to reconstruct the angular diameter of a star from the interference pattern seen in different telescope baseline configurations.

The radii (or to be more exact, the angular size) of many nearby stars and also of many giant stars at rather larger distances, have been measured using interferometric techniques. An accurate distance is then required to estimate an absolute radius.

Relevant papers are by Boyajian 2010 and Boyajian et al. (2013). There are a number of such interferometric experiments, mostly working in the infrared.

There are also direct radius measurements for stars that are occulted by the Moon and other solar system bodies - the time-dependent diffraction pattern can yield an angular size for the object.

Absolute radii are also provided by eclipsing binary systems (independent of whether they have a known distance).

See also How do astronomer measures the size of any celestial objects?


If the density of nebular distribution increases symmetrically in all directions, how does space curvature explain it?

Let me preface my answer by saying that you're reading a book from 1937 which references the scientific knowledge from the early 1900's. In case you're unfamiliar with astronomical history, I'll say that we've come a looong way since the early 1900's. For example, this book discuses "Nebula as Island Universes". At this point, astronomers had only just discovered that these mysterious "nebulae" were actually entire galaxies that were millions of light years away. Hubble is still referring to them as nebula though, rather than galaxies. That being said, it may actually be kind of difficult to "translate" some of this science from almost 100 years ago into modern theories and terminology. You may even find things which are flat out wrong. Hubble did make a lot of important discoveries, but there were many details he was mistaken on.

Do more recent observations still show a symmetrical nebulae density increase?

Again, nebulae here refers to galaxies, not what we've come to consider as nebulae. What you're asking here, and what Hubble was referring to, before it had an official name, is about the Cosmological Principle.

I think the appropriate answer to this question though, is to point out that I believe you're misinterpreting the text. In the Comparison of Observations with Theory section, Hubble is showing that observations of nebulae (sic, galaxies) indicate they are uniform through space and that his uniformity is easily explained by a static, non-expanding universe.

  If both of the extra terms (for recession and for curvature) were absent, the surveys would clearly fit the formula because the situation would be precisely that in a stationary universe.

The "formula" he's talking about here is the one which indicates

  nebulae are uniformly distributed through a non-expanding universe in which red-shifts are not primarily velocity-shifts, [and] then the numbers should be proportional to the volumes

However, he also knows that the Universe is expanding (he was the one to discover this after all) so he took the standard formula and threw in universal expansion. What he finds though, is that the equations now predict that

  the numbers of nebulae increase faster than the volume of
  space through which they are scattered

The important point here is that an expanding universe (with no curvature) predicts this is the case, not that it actually is the case. His previous sections and accompanying plot indicate that observations show the universe is homogeneous. Hubble makes the point (and you reference it) that this is discrepancy can be solved by including the curvature of the universe into the mix as well in which case the predictions now match up with the basic "formula" which didn't include expansion or curvature.

Just as an addendum, I'll expand on this and add that current observations do in fact show that galaxies are isotropic and homogeneous on large scales. One just survey showing this is the 2dF Galaxy Redshift Survey. Shown below is a plot from their results. While full statistical analysis is needed to says something conclusive, you can see that to the naked eye, the galaxy distribution does appear to be homogeneous and isotropic.

Note that this only covers part of the sky because the rest of it is blocked by the disk of our galaxy.

I understand how an expanding Universe explains the symmetrical redshift, but I fail to understand how it explains the symmetrical increase of nebulae density.

I'm interpreting this to mean, you're wondering how the "formula" with expansion included can produce the result that "the numbers of nebulae increase faster than the volume of space through which they are scattered". This I think is more or less a math question. If you're really interested in seeing how it was derived, feel free to look at Hubble's original paper: Hubble, E., Tolman, R. C., Astrophysical Journal, 82, 302, 1935. I think you'll find the summary equations in Section 12 (page 328 in the paper, page 27 in the pdf) with lengthy derivations before that.


Precision in the measurement of the distance to the Sun

This is a partial answer too long for a comment, but it may help to get the ball rolling...
Starting from this answer we can see that the standard gravitational parameter of the Sun used by JLP in their development ephemerides is 1.32712440040944E+20. While that doesn't mean we know the Earth's orbit to 1 part in $10^{14}$ it hints that it's believed to be known surprisingly well!
It's possible that something like this might shed some light on the subject, but it's not an easy question to answer. The Planetary and Lunar Ephemeris DE 430 and 431, IPN Progress Report 42-196 (February 2014)  Part of the problem is that we don't know exactly where the solar system barycenter is with respect to the Sun because there may be bodies far away we haven't detected yet. That's only a small effect on the Sun-Earth orbit. If I had to guess, I'd say it's somewhere between 1 and 100 meters uncertainty between the Earth and the center of the Sun. Where the edge of the Sun falls is a horse of a different color!


How can cosmological bubble universes "collide" with each other?

If time started in our Universe at the Big Bang, then in what time dimension (outside of ours) do collisions between Universes occur?

In inflationary cosmologies, time did not start at the Big Bang. There are many models, but in at least some of them, there is an "inflating" universe which extends infinitely in past (and possibly future) time, constantly creating new space within itself. Within that inflating universe, at any point with some low probability per unit of time, a phase transition (like water freezing to ice) can start to happen, fundamentally changing the nature of space and time at one point. That effect propagates out (again like supercooled water freezing) creating a growing bubble of this changed space-time, which in an abuse of language, we call "a bubble universe".

In what dimension are the other "Universes" moving relative to ours?

Of course there is no reason why this needs to happen just once in the whole of the infinite inflating universe. Indeed, it will happen infinitely often. If it happens at two points in space and time which are close enough together, the two bubbles may grow into one another (before they are pulled apart by the inflationary expansion of the space between them. This is (again perhaps by an abuse of language) called a "collision". Nothing is really moving, except the edges of the bubbles.
If I can suggest, the real answer to your question is that, in this branch of cosmology, "universe" does not mean quite what you think it does. Instead of referring to the totality of existence, it refers to one of these non-inflaying "bubbles" in inflating space-time.


What is the closest candidate for Dark Matter currently?

This arxiv post says if there is dark matter in the solar systems, its density should be below $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$.

They tried to find the dark matter in the solar systems by measuring its possible gravitational effects.

Note, it is a negative result, i.e. they didn't find anything, but they are suspecting, if the dark matter density would be at least $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$, they had found it.

There are various candidates what could it be, the currently considered most probable are the WIMPs (particles interacting only gravitationally and maybe weakly).


You don't need to do it that complicated. The sunlight's spectrum is the same everywhere in the solar system. And in order to get the light intensity, you can simply apply the inverse square law that intensity drops to 1/4 if you double the distance to the sun - like you did for getting the irradiance. The spectral sensitivity of the human eye is also the same everywhere in the solar system.

What is different, is the local conditions: On the surface of bodies it is, of course, influenced by the atmosphere - and that is likely the most important aspect when you want to model different colour; this does include possible aerosol content (like you see for instance on Mars during dust storms). The 2nd most important impact probably is the colour of the light reflected from the body you are "standing" on, thus a feature of the surface material.

So in order to get the colour impression you have to account for these two spectral filters, applied to the (nearly) black-body distribution of the sunlight. Thus in your equation you need to insert these factors as well - or just look at these separately as your equation states the default. Unfortunately this opens a VERY wide field as you can have a variety of atmospheric compositions - just think of the different light conditions you can have here on Earth - and now add the factor of completely different chemestry in the atmosphere. Modelling that is definitely an ongoing scientific endeavour.

If your quest is "just" to get some artists impression, play around with different materials and adjust the spectral transmission and scatter properties of the atmospheres (especially that) - and the ground material.


Does the "photo of a black hole" taken by EHT actually depict a visible light-based scenario?

You are right that the image is not a visible light image. It was taken using very, very short wavelength microwaves -- the wavelength is 1.3 mm which is only a bit shorter than the Far IR!

I think you may be reading a bit too much into the word "photo" which, as you say, has the everyday meaning of "taken with a camera using lenses in visible light."  I think most astronomers would use the term "image" rather than "photo" -- if you look at the NSF's press release, for example, you'll see that it uses the term "image".  (Links to all six of the papers the EHT team has published are in the NSF press release if you want further details.)

(I'd bet that "image" was changed to "photo" somewhere in the journalistic process as being "more meaningful" to the average reader.)

The appearance to the naked eye would probably not be terribly different, since the photon orbits don't depends on frequency.

As far as whether it got "boosted" or not, you're right to be very skeptical -- university publicity offices are prone to exaggerate and sometimes pump relatively ordinary science into amazingly important-sounding "breakthroughs."  In this case it all seems pretty straightforward, though. And there's nothing special about a visible-light image over a microwave image -- both are useful and neither tell us everything. (It is not uncommon for images in the visible to be among the least useful -- with X-ray or radio actually telling us much more.)


The spots could be mold or fungus.
If a gentle cleaning with distilled water, a little detergent, and sterile cotton (no abrasives!) fails to remove them, there are several vendors who can strip the old aluminum coating and apply a new one.
If you don't want to go that far, you can probably find a buyer who would.
Criterion scopes have a considerable fan base.


Are there events in universe which we receive first from their neutrino's instead of their photons?

Yes.  When a core collapse supernova happens, it takes a few hours for the shock wave to reach the surface and "break out" (when we first "see" it). Meanwhile, the neutrinos can escape nearly directly once they reach the top of the core.

Of course, as you wrote, photons travel slightly faster, so if the SN is far enough away, the photons might overtake these neutrinos.  However, I have not done the calculation to see if the universe is big enough.


False-color wavelength assignments in this "drop-dead gorgeous" image of NGC 2903?

Spacetelescope.org
indicates under "Colours & filters" that the blue channel of the image is from a 658 nm filter (red)
and the red channel is from an 814 nm filter (near infrared).
The cyan and orange rows of the table also list these filters; presumably the green channel is a blend of the two.

The ACS instrument handbook says filter F658N is narrow, passing only NII or slightly redshifted Hα emission (Table 5.1, Figure 5.4).
Filter F814W is wide, passing wavelengths between 710 nm and 960 nm (Table 5.2, Figure 5.1).

NGC 2903 is one of the brighter galaxies not in the Messier catalog.
Here is a wider angle, more natural RGB image by Bob Franke, rotated for easier comparison.
The reddish emission nebulae here appear blue in the HST image.


The actual density depends on the mineralogy, we don't have a crystalline iron core and silicon crust. You do have a lot of oxygen available, too when you look at the overall elementary abundance. So the abundant materials are Fayalite (${\rm Fe_2SiO_4}$), Olivine (${\rm (Fe,Mg)_2SiO_4}$), Fosterite (${\rm Mg_2SiO_4}$) etc. which make up most of the crust. These have a bulk density between ${\rm 3g/cm^3}$ and ${\rm 4.5g/cm^3}$ at normal pressure and of course a somewhat higher density under pressure.
In the core we are talking about some iron nickel sulfide alloys which have a density of less than pure iron, too (typically ${\rm FeS}$ is around ${\rm 4.8g/cm^3}$ at norm pressure). So conversely knowing the total mass from celestial mechanics and satellite date, the volume from imaging and the surface composition from spectroscopy we can actually make density estimates on the core's composition (with some further guesstimates on the equations of state on the alloys in question).
Add to these figures that we will have in the interior some high-pressure phases which have somewhat higher density, the estimated mean density of ${\rm 5.4g/cm^3}$ seems to fit quite well.
Also mind, whether a source talks about weight% (thus referencing the weight contribution of single elements. One iron atom weighs nearly 4x as much as one oxygen and twice that of magnesium) or whether the source talks about number fractions (thus counting actual atoms, disregarding weight). For one single molecule of Olivine we have ${\rm Fe:Mg:Si:O}$ in the weight ratio 56:24:28:64 while we have the number ratio of 1:1:1:4.


How did astronomers determine the path of 'Oumuamua so quickly?

Three accurate observations are sufficient to fix a Keplerian orbit (ie an elliptical or hyperbolic orbit with the sun at the focus)

In practice, observations are not perfectly accurate due to limitations of the equipment and observations over a short time are particularly prone to observational error being magnified. Moreover the orbit will be perturbed by the gravity of the planets, so won't be perfectly Keplerian. For this reason pre-discovery images will help fix the exact orbit more accurately.
As the time difference between first and last observation is critical in the quality of the orbital determination, we talk about the length (in days) of the observation arc as a measure of how well defined the orbit is.

Multiple observations can reduce error by an averaging effect (the Gaussian curve is so named from Gauss's use of it in orbital determination)

However once you have three or more observations of a body you can determine its orbit almost immediately, using a computer to do the calculations for you. It would have been immediately obvious that the object was on a very unusual trajectory.

There is no need to use space probes for the observations. The relative motion of the Earth and the Object provide sufficient parallax to determine distance, position and velocity.


What is the direction of a comet's dust tail before and after perihelion?

Comet tail always aims away from the sun, as it's torn from the comet's gas cloud by solar wind. It trails slightly behind, as the comet moves along its orbit while the gas travels directly away, but that's relatively minor - the speed at which the gas and particles are pulled away forming the tail is much higher than the comet's orbital velocity.

Your visualisation shows it trailing behind the comet as if it was moving in some gaseous/liquid medium that stops the lighter particles while letting the heavy comet head travel ahead. This is not the case - as the comet is far away from the sun, the gas cloud just forms its atmosphere and travels with the comet. As it approaches though, the intensity of solar wind increases and it pulls away the atmosphere which can't be protected against it by magnetic field or strong gravity, as the comet has neither.


Could it have been possible for the Moon to be split without any observable effect on Earth?

No the energy required to do this would cause the moon to melt.
However the story of the spitting of the moon supposes a supernatural cause, and one which is not bound by the physical laws.
God could split the moon and put it back together by his word. And God could arrange so that only some people saw it and others looked away.  When you bring God into your explanations the answer is always "yes he can".  That's what being omnipotent means.


How do we know that the discovered M87 black hole isn't just a star surrounded by a dust disk?

Black holes are often studied (and discovered!) by observing their effects on objects around them. Stellar-mass black holes, for example, can be found by determining the orbit of any luminous companion. Supermassive black holes, by comparison, affect the motion of numerous stars and clouds of gas in their immediate vicinity. By fitting the motions of those stars, astronomers can determine that there must be an extremely massive object in that location, and typically a supermassive black hole is the only possibility.
In the case of M87, these measurements were first conducted in the late 1970s (Sargent et al., Young et al.). Both groups noted that the velocity dispersions near the nucleus required a central mass on the order of $\sim5\times10^9M_{\odot}$. Mass/luminosity ratio profiles were also calculated based on photometry, and both sets of observations noted a steep rise in $M/L$ near the center. Neither group was able to rule out other possible explanations, like a compact star cluster, but a supermassive black hole was - to quote Young et al. - "the most attractive of the models considered". Further observations over the last four decades have ruled out those other options.
The image produced by the Event Horizon Telescope is consistent with a supermassive black hole, as the EHT Collaboration wrote in the first of their papers on the observations:

It is also straightforward to reject some alternative astrophysical interpretations. For instance, the image is unlikely to be produced by a jet-feature as multi-epoch VLBI observations of the plasma jet in M87 (Walker et al. 2018) on scales outside the horizon do not show circular rings. The same is typically true for AGN jets in large VLBI surveys (Lister et al. 2018). Similarly, were the apparent ring a random alignment of emission blobs, they should also have moved away at relativistic speeds, i.e., at ~5 μas day−1 (Kim et al. 2018b), leading to measurable structural changes and sizes. GRMHD models of hollow jet cones could show under extreme conditions stable ring features (Pu et al. 2017), but this effect is included to a certain extent in our Simulation Library for models with Rhigh > 10. Finally, an Einstein ring formed by gravitational lensing of a bright region in the counter-jet would require a fine-tuned alignment and a size larger than that measured in 2012 and 2009.

There are other arguments you can test yourself. For example, the photon ring matches calculations from general relativity, assuming the now-accepted mass of the black hole.
In short: Stellar and gas dynamics require the presence of a large mass in the center of M87, and the image rules out many non-compact objects.


@Calc-You-Later's answer is correct in places where the path of the Sun is perpendicular to the horizon (at the equator during the equinoxes, somewhere between the tropics the rest of the time). There, the sunset lasts 2 minutes. However, at other latitudes, the sunset may last a bit longer.

Let $\alpha$ be the angle the path of the Sun makes with the horizon. This angle depends on your latitude and the date.

The speed of the Sun along its path is $\vec{v_s} = $ 0.25°/min. This speed can be projected along the vertical and horizontal axes. The vertical speed is now $\vec{v_s} . \sin (\alpha)$. The sunset length is still the apparent angle of the Sun divided by the vertical speed of the Sun.

So at higher latitudes, if the Sun sets, for instance, at a 45° angle, then the sunset will last 2 min / sin(45°) = 2.85 minutes = 2 minutes 51 seconds. If the Sun sets with a 60° angle, the sunset will last 4 minutes.

To get the value of $\alpha$, you can add your latitude and the solar declination:

Determine how many days have passed since December 21st (i.e since the winter solstice). Let's call this number $x$
Multiply this number by the amount of turning the Earth does in one day
$ y = x * 360/365.25$
Get the cosine of this number and multiply it by the tilt of the Earth $z = -23.5 * cos(y)$. This is the solar declination
Add the latitude $\theta$ of your location $\alpha = z + \theta $

This does not account for atmospheric effects. The atmosphere will bend the rays of light, and can make it possible to see the Sun even though it is under the horizon.

The atmospheric effects, however, depend on the atmospheric conditions (temperature gradient,...), and are more difficult to quantify.


How many sun-like stars are there in the universe?

This is a question that concerns the initial mass function (IMF) - an empirical (that is, defined by observations rather than theory) function that describes the statistical distribution of stellar masses.

Edwin Salpeter (1955) was the first to describe the IMF, though if you read Chabrier (2003) there are some reasonably comprehensive explanations of the theory and history. However, these lecture notes are a fair bit more accessible.

From the approximations in the UCSC lecture notes I linked above, I get that around 4% of stars are between 0.7 and 1.3 solar masses (92% are between 0.1 and 0.7 solar masses!).

There are perhaps 100 billion stars in a galaxy and 100 billion galaxies in the observable universe, giving something on the order of $4\times 10^{20}$ (400 billion billion) stars that are about ($\pm 30\%$) one solar mass.


Watching the Mercury transit with improvised devices

Mercury's angular diameter on transit day will be 12 arcseconds.
A camera obscura using a 12 mm aperture could resolve it; one lens from +0.75 diopter reading glasses, if you can get them, will project a bright 12 mm image of the Sun at a distance of 1.33 m.
Note that a larger aperture or a shorter focal length will make the Sun image hotter than direct sunlight unless you add a filter.

Test with sunspots before relying on it for Mercury.
If it counts as improvised, projection with 7x35 binoculars easily showed me the 2012 transit of Venus.

Update: I tried both of the above with the Nov 2019 Mercury transit.
With handheld wobbling, neither showed it.
Someone else's 50mm binoculars did, probably because they were mounted on a tripod.


Have any planets been discovered in Orion constellation?

Yes, there are  exo-planets discovered in the Orion constellation, for instance, the stars HD 37605 and HD 38529 both have two known planets.

HD 38858 is also interesting, as it has a known planet, and also a disk of comets.


Does the sky darken proportionally during a partial solar eclipse?

Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, $I(d)$ as:
$$I(d) = 1 - \frac{\mathrm{Area}_{\mathrm{overlap}}}{\mathrm{Area}_{\mathrm{Sun}}}$$
Where $\mathrm{Area}_{\mathrm{overlap}}$ is the overlap between the Moon and the Sun (the green area).

(source: uga.edu)
Some calculation finds the overlap to be:
$$\mathrm{Area}_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$
Where $d$ is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:
$$\mathrm{Area}_{\mathrm{Sun}} = \pi r^2$$
Putting this together, you get the graph:

Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is $d$ (ranging from $2r$ to $0$).
I hope this answers your question: the brightness does not change proportionally to coverage/time.
Note:
This assumes that $d$ increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).


Apart from the field-theoretical standpoint presented by Stan, one can repel objects in a sense, when taking orbital mechanics into account.

The slingshot maneuver extracts angular momentum and energy of an orbiting mass by the use of gravity. The trick here is that the probe's velocity just gets redirected in the planet's reference frame. But this results in an acceleration in the solar rest-frame that the probe essentially tries to leave.

As you were asking about curvature, I don't know how and if centrifugal potentials that play a role here are described by general relativity.


How can black holes be sometimes so gaseous?

A black hole is not a solid or gaseous body. It is a shape that space-time can take. Now the event horizon is sometimes treated as being the "edge" of the black hole, but there is nothing at this edge. The black hole can't shrink its event horizon, since the event horizon is not a physical object

There is no matter inside the black hole (for this reason the shape of space-time around a black hole is sometimes called a "vacuum solution" of Einstein's equations). The mass of the black hole is concentrated at the singularity (which, due to the curving of space time you can imagine as more like a point in time that you can't avoid if you cross the event horizon.

Since the radius of the black hole's event horizon is proportional to the mass, the volume of the black hole is proportional to the cube of the mass. If you double the mass, you increase the volume by a factor of 8.

Density is mass/volume so if you double the mass you change the density by 2/8=0.25, so the more massive a black hole is, the lower is the average density of the region inside the event horizon.

This does mean that if you did manage to find a gas cloud with 17 billion times the mass of the sun (much bigger than any known nebula - Orion Nebula M43 is only a few thousand solar masses) and you were to somehow compress it to a sphere with 0.1 atm pressure, it would undergo gravitational collapse and form a black hole.


What would it feel like to be orbiting near a merging black hole?

It could do serious damage if things were just right.

Three solar masses of energy is about 5x1047 Joules.  1 AU is about 1.5x1013 cm. The area of the sphere is thus about 4x1027 cm2.  That's a flux of 1020 joules cm-2.  That's plenty of energy to do damage...if it can couple to you.

The strain of the GW measured at Earth is around 10-22.  Strain drops off as the inverse distance, and the ratio of a gigaparsec to an AU is about 1014, so the strain at 1AU from the merger would be about 10-8. By itself, this would probably not harm you.

But it's not a simple one-time strain, it is oscillating with the orbital frequency of the merging black holes, and anything which had a physical resonance at that frequency could pick up energy, possibly significant energy, from the GW fields.

(Note that this depends on orientation. The mechanical resonance needs to align with the GW, and GWs are not emitted in all directions, so you'd want (or, perhaps, not want!) to be in the plane of the merger.)

The argument that GWs can't deposit energy is plainly wrong -- if it were true, LIGO couldn't detect them! -- but if you believed that, you were in superb company: It took Einstein decades before he became convinced that GWs actually carried energy!


Distance from the sun can equal a hospitable planet?

It's actually the opposite.  Venus, Mars and I'll include the Earth as well, are likely getting more solar energy per square meter, not less, over time, even as they slowly move away from the sun.

1.5 cm per year, from the Rob Jeffries' answer to the 2nd question you linked is too slow to make much difference.  That's 1 km every 66,000 years and about 70,000 km over the age of the solar-system at that rate.   Now, planets may have moved away at different rates, especially when the solar-system was young and orbits were more chaotic and when Jupiter was thought to be doing it's dance in and back out, so it's not precisely known how close Mars was to the sun, say, 3.7 to 4.1 billion years ago, but we do know that the Sun was a little smaller and not as hot at that time, perhaps about 25-30% less luminous overall, though it was also more active with solar storms.   But Mars probably got less solar energy 4.1 billion years ago than it gets now, and it will get more solar energy a billion or two billion years in the future, even though it will probably be a little further from the sun.

The primary factor is how much of an atmosphere it had.   Planets can't have liquid water oceans without a reasonably thick atmosphere and Mars is believed to have had an atmosphere long ago.   How close it was to the sun matters too, but the thickness and composition of the atmosphere matters more.  Heat from the planets interior also matters and all the planets had more internal heat after formation, in the 3.7 to 4.1 billion years ago range.

Its also not 100% certain that Mars ever had relatively permanent liquid oceans and what we'd consider moderate temperatures.   Mars' geology suggests it had floods, but flooding could have been caused by local or temporary heating from volcanic activity or large meteor impacts.   It might also have been driven by a a more extreme axis leading to greater seasonal variation with glacial melt every summer (687 days in a year so summer lasted longer), and perhaps re-freezing every winter.

The Mars Ocean Hypothesis is a good hypothesis, don't get me wrong, but it's worth pointing out that its not a certainty.  It's possible that the oceans were ice covered on top but liquid underneath from heat coming from inside the planet.   Nobody knows for certain how temperate Mars was 3.7 to 4.1 billion years ago.


How powerful a telescope would allow me to view the astronauts aboard ISS do a space-walk?

This has been done before, so I don't have to go through all the heavy calculations using Rayleigh criterion accounting for atmospheric diffraction and visible light wavelength. Ralf Vandebergh, a Dutch astronomer, professional photographer and veteran satellite spotter has been busy trying to do exactly this since the 2007 and has indeed succeeded on several occasions by now using a 10 inch (25.4 cm) Newtonian reflecting telescope that has a resolving power (angular resolution on the CCD sensor) of roughly one pixel per meter at the distance to the International Space Station (ISS) that is currently in a 230 miles (370 km) orbit above the Earth:

   Ralf Vandebergh’s detail of an image he took on Mar. 21, 2009 showing astronauts working outside the ISS. Credit: R. Vandebergh

Vandebergh's personal page also hosts all kinds of other successful observations of the ISS through his telescope and recorded in both photographs as well as some short videos. Why short? Because targeting the ISS as it moves at the speed of 4.8 mi/s (7.7 km/s) is rather tricky, and the atmospheric conditions and times at which the ISS passes over some area on the surface of the Earth don't make it any easier either. But perseverance and hard work have paid well for this individual astronomer.

          Raw video of the ISS as seen through the air turbulence. Note the good visibility of the Lira antenna at the Russian          Zvezda Module in the lower part of the image. Credit: R. Vandebergh

So again, skipping the math to calculate the required angular resolution of a telescope and apply that to some arbitrarily selected image sensor size and resolution, we can see that using a well collimated 10 inch Newtonian or a Dobsonian telescope on a clear night can, with some near perfect targeting, produce a direct proof of a spacewalker doing his or her job 230 miles (370 km) high during an EVA. More powerful telescopes would of course produce better resolution images, but the atmospheric effects limit their use and are of course much harder to target an object moving fast over the skies with.


The luminosity mass relation is inexact. The luminosity also depends on the composition of the star, particularly in and around the nuclear burning regions.
The composition changes during the main sequence lifetime as hydrogen gets turned into helium. The average mass per particle goes up and the number of electrons per mass unit goes down.
The former means the temperature increases to maintain a similar pressure, thus increasing the fusion rate. The latter decreases the opacity of the gas leading to a shallower temperature gradient and so a bigger star. A star like the Sun gets about twice as luminous over the course of its main sequence life.
The loss of mass due to radiation from the star's surface is negligible for the Sun, but more important in higher mass stars.
For the Sun, the mass loss rate due to this process is about $6\times 10^{-14}$ of its mass per year, so only about 0.6% of its mass is lost in this way over its main sequence life. The mass lost via the solar wind is probably a bit more important but the total is insufficient to counteract the compositional change effect. The radiation mass loss rate goes up as $M^{3.5}$ for higher mass stars, but of course their lives are also shorter by $M^{-2.5}$. The net effect is that the fraction of mass lost is proportional to the mass, so about 6% for a ten solar mass star. On its own this would only reduce the luminosity by about 20%; the compositional change effect is much stronger.


Does every object in the Universe have gravity? Space has no gravity, why?

A quick search up on google you would find this:

  Gravity causes every object to pull every other object toward it. Some
  people think that there is no gravity in space. In fact, a small
  amount of gravity can be found everywhere in space. Gravity is what
  holds the moon in orbit around Earth.

Why is this? That's because:

  Outer space is the closest known approximation to a perfect vacuum. It
  has effectively no friction, allowing stars, planets and moons to move
  freely along their ideal orbits. However, even the deep vacuum of
  intergalactic space is not devoid of matter, as it contains a few
  hydrogen atoms per cubic meter.

If there's matter in space, thus this makes this statement valid.

  Yes, everything that has mass will cause a gravitational pull on other
  objects.


Formation of a magnetosphere for gas giants / small stars

We do not know whether Jupiter has a solid core. That is one of the primary goals of the Juno mission.

In general it is clear that a solid core is not required to generate magnetic fields. The Sun does so and magnetic fields have been measured (with average surface field strengths of around 3 kG) on very low mass stars. Incidentally, the Sun's magnetic field does have a large scale dipole component.

More recently, there have been observations of cyclotron and auroral emission suggesting that even old, cold brown dwarfs have strong, organised magnetic fields. See for example Kuznetsov et al. (2012); Hallinan et al.(2015) and references therein.

http://arxiv.org/abs/1111.7019
http://arxiv.org/abs/1507.08739

At the moment, nobody can say for sure how exactly these fields are generated, sustained and organised.

In fact you could do worse than look at a paper that popped up at the top of today's astro-ph listings by Williams et al.  The abstract begins:

"The well-studied M9 dwarf TVLM 513-46546 is a rapid rotator (P_rot ~ 2 hr) hosting a stable, dipolar magnetic field of ~3 kG surface strength."

http://arxiv.org/abs/1511.05559


Can the explanation of Tabby's star KIC 8462852 be an intervening planet in orbit around another star?

The answer is no, because the light curves simply do not match up.

Astronomers have detected planets before using the transit method. Large planets cause dips in a star's light curve when they move between their parents star and Earth; if these dips are regular and strong enough, we can conclude that there is a planet-like body orbiting the star. It's not perfect (best used on conjunction with other methods), but if there is a planet orbiting a star that is detectable like this, we should see a couple key features:

Periodicity, corresponding to the planet's orbital period.
Relative regularity in the depths of the transits.

Transits aren't always easy to see, but sometimes you can rule them out as the cause of light curves. The light curves of KIC 8462852 are one of these cases. Boyaijan et al. (2015) provided some of their light curves from the observations of the star. Here's one from Figure 1(b), showing a number of "dips":

There is neither regularity nor periodicity in the data, meaning that your proposed explanation is not possible. Looking at the dips in more detail (see the rest of the paper) confirms the conclusion.


How is it known that Callisto has no core?

It may have a (silicate) core, but it's very small. There also is the possibility of an internal liquid water "ocean". Currently the most probable composition is that of rock and ice, with the density of rock steadily increasing as one digs deeper (This is in stark contrast with planets like Earth, where there are discontinuities in the density between layers). Planetary differentiation has not occurred appreciably because Callisto is not tidally heated.

The composition can never be completely determined — the composition of a spherical body cannot be fully discerned from just rotation and translation1. However, we can make reasonable guesses. Basically, we can calculate its total mass and moment of inertia by observing its rotation and revolution. Then, assuming that the density ρ varies radially, we get the following two equations:

$$M = \int \rho \pi r^2dr$$

$$I = \int (\rho \pi r^2dr)r^2$$

These do not completely determine ρ, however with other reasonable assumptions on the composition and the amount of planetary differentiation it can have, we can get a good estimate for ρ.

1. It can be determined if we take a larger gravitational field and pivot it, but this ventures into the Aristotlean "Give me a place to stand on, and I will move the Earth." zone of thought experiments.


Terminology: Is there a name for the points on the surfaces of tidally locked parent/satellite bodies that always face each other?

For tidally locked binary stars, the two points in question are known as the substellar points.

For a tidally locked exoplanet, the point closest to the star would also be known as the substellar point. If the star was also tidally locked to the planet, then there would be a subplanetary point.

For a moon locked to a planet, the point on the moon would also be the subplanetary point. If the planet was mutually locked, it would have a sublunar point (not sure whether this latter only applies to the Earth's moon).


What will a lunar eclipse look like from moon?

It would look like this (actual picture of the Earth, seen from the Moon, during a lunar eclipse):

Link to full page: http://apod.nasa.gov/apod/ap140407.html

The Earth would appear surrounded by a bright ring, even though the Sun is completely hidden behind it. The ring is sunlight refracted through the atmosphere. It's basically all the sunrises and sunsets, seen at once.

The Earth would never appear fully dark. The bright ring around it will always be there. This is due to the atmosphere.


The answer, for the universe we live in, is almost certainly No. The best guess is that the universe, on the largest scale, is very nearly flat and probably extends infinitely in all directions.

The physics of general relativity does permit a hypothetical universe that curves back on itself in the way you describe, and if it were small enough, we could indeed see ourselves. But I don't know whether stars and planets could evolve in such a universe.

Incidentally, it doesn't have to even have curvature to wrap around as you describe. The universe could be like the game Asteroids, where you wrap off one edge and come back on the other. That has the topology of a torus, but is geometrically flat -- for instance, the angles inside a triangle still add up to 180 degrees.

But it doesn't look like our universe is like that.


How many members of the IAU typically attend votes?

One of the talking points of people how think that the definition of a planet should include Pluto is that during the 2006 vote on the definition of a planet, only a small percentage of the members of the IAU actually voted.

This argument is very disingenuous or rather naive, or both. One the most disingenuous of all is Dr. Alan Stern, the principal investigator of the New Horizons mission to Pluto who said

  "I'm embarassed for astronomy. Less than 5 percent of the world's astronomers voted." Stern called it "absurd" that only 424 astronomers were allowed to vote, out of some 10,000 professional astronomers around the globe. "It's patently clear that Earth's zone is not cleared," Stern told SPACE.com. "Jupiter has 50,000 trojan asteroids," which orbit in lockstep with the planet.

These are blatant lies.

Regarding the small number of voting members who bothered to vote, Stern was high enough up in the technical food chain to know exactly how professional societies work. Across all societies, voting on resolutions almost uniformly occurs on the last day of the society's general assembly. Most of the members of a technical society do not attend that society's general assembly, and of the few who do attend, most do not attend the final day of the proceedings. A few societies have started to transition to electronic voting, but this is very recent, much more recent than 2006.

Regarding "clearing the neighborhood," Stern either knew or should have known exactly what that phrase meant. If he did know, his statements were a blatant lie. If he didn't know, his statement makes him guilty of academic fraud, which is even worse than lying.

Stern was the lead author of a paper presented to the IAU six years before the vote occurred titled "Regarding the criteria for planethood and proposed planetary classification schemes." This paper developed a metric that shows a 5.5 order of magnitude gap between the eight planets and objects such as Pluto and Ceres. Either he knew what was in that paper / presentation (which makes his later statements a blatant lie) or he didn't, in which case he was guilty of a very serious form of academic fraud, that of forcing a subordinate to add the supervisor's name as an author. That fraud is even worse when the supervisor forces the subordinate to make the supervisor the lead author.

Here's a typical schedule for almost every professional organization's large meeting:

A meet & greet session on the evening before the meeting officially starts. This is a great time to have informal chats, to grab some free food, and to pocket the per diem. One  should carpe per diem whenever the opportunity arises.
An opening plenary session with lots of pomp and circumstance, and lots of blah-blah-blah. This a great time to either meet people in the hallways or to play hooky from the conference and tour the boondoggle of a locale where the conference is being held.
A number of small paper sessions, poster sessions, society committee meetings, and plenary sessions for invited speakers and invited panels. These meetings, along with the vendor exhibits, are why people typically go to these large conferences.
A closing session with lots more pomp and circumstance and lots more blah-blah-blah. This is almost universally where proposals / rule changes / resolutions are voted on. This is also almost universally a great time to take one last look at the vendor exhibits, to play hooky and tour the boondoggle city, or to be already flying home.

Most of the members of a professional organization do not attend the organization's general assembly; they are members for other reasons. Of the fraction of members who do attend, most do not attend the opening session, and even fewer attend the closing session. Voting has traditionally been very low. This tradition is changing as some societies now use electronic voting, but this is very recent.

Since pictures are worth a thousand words, here are a couple of pictures from the 2006 IAU General Assembly voting session, one from the rear and one from the front:

Below are a couple of pictures from the 2015 IAU General Assembly, the first from the opening session and the second from the voting during the closing session. Note that the closing session didn't even use the full auditorium.


Take for instance radio, when listening to radio you gather static on the frequency. A passband filters the static which in turn provides a clearer signal. Objects radiate in multiple frequencies/wavelengths, for instance heat/thermal (infrared) and visible. Not all object radiate in a range of frequencies, humans for example can be seen well in infrared and visible, but our skin and circulatory system are opaque in x-ray and radio. Spitzer MIPS 24µm is the wavelength corresponding to mid-infrared (between 3µm-50µm).

What the statement is saying, is multiple imaging sources have been used for the same frequency. When choosing the Spitzer MIPS 24µm result, it additionally includes results from other imaging sources such as the IRAS 25µm result. It does this because the filters of the passband overlap and return the same result, as a result the process is changing where it only displays the result of the filter you choose.

All radiation is apart of the electromagnetic spectrum, spectral energy distribution is a way to present the data results as a plot, in turn that can be used to identify the source of the radiation.

The observed passband, is just stating which frequencies are being filtered, such as perhaps within a small range. For example, a Quasar is a highly active radio source typically found in the center of young galaxies. In correlation, you can determine what kind of stellar objects you would expect given a certain range of frequencies or the observed passband.

Frequency is the simple measure of how frequent something happens, for example if you flash a torch once a second, the time between each flash is 1 second long, the frequency 1Hz. If you flash a second torch twice a second, the time between each flash is a 1/2 second long, the frequency is 2Hz. More relevant in physics, is the correlation between wavelength, frequency and photon energy. Waves oscillate and light acts as waves, and the frequency between each wave correlates to how fast the wave is moving and what wavelength the radiation is.

Homogenized results are just a collection of results which are made to fit a pattern, sometimes to better represent the data (this may be filtering results in a database to exclude results outside a range of study). Published results on the other hand are raw results which include a full spectrum of the data. I am unsure about the mode aperture, this may refer to the sensitivity and correlate to how much interference/static is captured with the wave. Best if someone else can explain this.


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

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


Yes, absolutely! The performance of any telescope depends on its diameter. A larger instrument not only collects more light, but also provides better resolution than a smaller instrument. The resolution limit is calculated from:
$R_{arcsec}\ = {1.22\ \lambda \over D_{mm}}$
whereas limiting magnitude is calculated from
$L_{mag}\ =\ 2.5\ +\ 5\ log\ D_{mm}$
I suppose there are eventual limits to $R_{arcsec}$, but I have never heard of any. As for limiting magnitude, as it’s a logarithmic formula, it grows very slowly as $D_{mm}$ increases, but a larger telescope is still a gain.
The whole thing is to balance $R_{arcsec}$, $L_{mag}$, and $$$, of course!


Amazon has solar film sheets if that's what you're looking for. It is sold by Amazon for Thousand Oaks Optical. Amazon sometimes has next day or 2 days delivery.


Can somebody explain heisenberg uncertainty principle to 14 year old me?

I've watched a TED video about the Heisenberg uncertainty principle, but I couldn't fully understand it and I'm looking for a fairly easy explanation if possible.

A comment points to Can the Heisenberg Uncertainty Principle be explained intuitively? and notably there are 17 answers there! It is possible that a few will be helpful to any given person.
Quantum mechanics is a set of theories and equations that predicts remarkably well what happens. That doesn't mean that anybody actually understands it. Many people know how to use it, and many many more people kind-of know roughly how to use it. And at least 17 (if not 17,000) people think they know how to explain it.
But don't let them fool you.
Take it from the master explainer. Per Richard Feynman:

Nobody understands Quantum Mechanics!

They behave in a way, that is like nothing you've seen before! Your experience with things you've seen before is inadequate, incomplete...

The behavior of things that are small is so fantastic, is so wonderfully… different. So marvelously different than anything that behaves on a large scale.


How accurately could late 50s - early 60s humans have mapped the solar system?

I recall hearing Professor Lovelock about 10 years ago, describing a discussion with his students in the 1960s - what observation would indicate life on another planet? They concluded that spectroscopic analysis would show that the chemistry of the planet was inconsistent with the planet's age. A slower than expected rate of entropy would indicate the presence of life, because the chemistry of Earth life uses photosynthesis to build giant molecules which store energy by absorbing solar radiation
Spectroscopic analysis was used to examine of the atmosphere of Mars in the 19th century, many decades earlier than the context of your question. Before the space race began, scientists had already seen enough to rule out the possibility of life on other planets in the solar system
Lovelock's discussions about extra-terrestrial life led to his Gaia hypothesis, a set of models which demonstrate that Earth's current atmosphere and ocean chemistry emerged from billions of years of life on Earth


Hypothetically, if our sun was yellow would we still see "white" colours?

Every "normal" star, produces a more or less "black-body" spectrum, at some temperature (more-or-less because there are brighter and darker lines caused by specific elements in the upper layers of the star), so there will always be a range of colours of light (or electromagnetic radiation at least) in its light, which can be spread to give a rainbow.

A white object (like ice crystals, or bleached cotton) reflects all colours of light more or less equally (over some range of wavelengths of interest, anyway).

So, if the Sun suddenly cooled, so that it was emitting a 5000K black body spectrum instead of its current 6000K, with more red light and less blue light, that change would be reflected in the light reflected from a white object, which would look more yellow or orange to our eyes.

There are a couple of caveats though:

As @badjohn pointed out in a simultaneous answer, our brain processes visual information in odd ways. If the colour of sunlight changed, the brain might try and compensate for the overall shift and you might see everything more as you expect it, and less as it now is.
We evolved to perceive sunlight as white. If we had evolved around a cooler star, we would probably perceive a different colour as white.


What are the benefits of a large refracting telescope?

This is a really interesting question!
The Dragonfly telescope described below takes advantage of the fact that on Earth, the optical surfaces of some lenses can generate less diffuse scattering than the optical surfaces of reflecting telescopes. If you were worrying about scattering from secondary mirrors you could make an off-axis primary, but at a nanometer scale a metalized mirror will have more roughness than a glass lens with antireflection coatings (or so the links say). I suppose there are broadband dielectric reflective coatings for telescope mirrors, but that's probably a topic for a new question.
note that in many cases surface brightness limitations are often the result of  skyglow (artificial) and airglow (natural) rather than telescope optics, so applications for large lenses might be found above the atmosphere.
If we ignore the fact that in the solar system space is filled with micrometeorites, then a lens in space might maintain it's lower diffuse scattering for quite a long time! But we can't, so it may not. It may get beaten up fairly quickly in fact. A long tube in front of the lens will decrease the solid angle of primary impacts, but impacts inside the tube may still cause trouble, and if you don't shade the tube from sunlight, diffuse scattering from the close-to-but-not-completely-black coating might also cause trouble.
So far there are no answers to How fast do optical surfaces get dirty or damaged in space? that I can point you to.

Below's discussion of and links to the Dragonfly Telescope is taken from the question What (actually) is the “deprojected half-light radius” of this almost-all-dark-matter Galaxy?

The recent news of the Ultra Diffuse Galaxy (UDG) Dragonfly 44 is an excellent example of what could be termed 'observe different' thinking. The dragonfly telescope is noted not for the size of its collective aperture, but for the absence of the diffracting effects of secondary mirrors and surface roughness that limit the contrast of dim objects in conventional telescopes when brighter sources are nearby. See here and here and here.

above: image of a Dragonfly refractive array telescope from here. Image: P. Van Dokkum; R. Abraham; J. Brodie

above: The Dragonfly 44 ultradiffuse galaxy from here. "Dragonfly 44 is very faint for its mass and consists almost entirely of dark matter. (Pieter van Dokkum, Roberto Abraham, Gemini Observatory/AURA)"


What could be past the particle horizon?

It sounds like your question is about the single most important principle of cosmology, called the cosmological principle.  This asserts that as long as all our observations are consistent with the idea that the universe is the same everywhere at a given age, we will continue to use a model with that attribute.  There are lots of good reasons for adopting that principle, the least of which is not that it allows us to have a model in the first place.  So what this means is, our answer to what is beyond what we can see is that we don't know, but we are going to model it in a way that makes sense to us, so we will use the cosmological principle.   You have to be able to distinguish what we know in the sense of having direct information about, versus what we know about our own best models.


What exactly was the Moon's "Evection Resonance"?

The orbit of the moon is elliptical.  But the direction of the major axis of this ellipse isn't fixed. Perturbation by the sun causes the axis to rotate towards the East.  The time for the moon to go from perigee to perigee is slightly longer than the time it takes for the moon to orbit the Earth.
The direction of the major axis of the moon slowly rotates around the Earth.  Now it takes about 8.85 years to make a complete 360-degree rotation.
But early in the history of the Earth-moon system, it might have made a complete rotation in less time. There could have been a time when it took one year to make a 360-degree rotation.  This could lead to a resonance that would affect the orbit of the Earth around the sun.  In particular, it could transfer angular momentum from the Earth-Moon system to the Earth-Sun system.  This can, perhaps, explain why the Earth-Moon system has less angular momentum than would be predicted by a system that formed from a planetary collision.
The paper investigates how such a resonance could form, how it could transfer angular momentum, and how the resonance could eventually be broken.


It's a "ripple" in spacetime. Imagine the traditional picture of planets as marbles sitting on a sheet--heavier bodies push the sheet down farther and deeper. If you place a bowling-ball on one end of the sheet, the entire sheet is affected, but not instantly. It takes an amount of time for the sheet to be moved--this movement can be seen as a wave, with a wavefront, a speed, etc.

In reality, massive gravitational bodies don't magically appear in space, (that we know of... maybe God is planning to drop a giant marble in our Solar System) but gravitational waves are still generated by any change in placement of mass or any acceleration (Einstein's theories that show that gravity and acceleration are indistinguishable).

So you get gravitational waves caused by movements of planets around the sun, etc. That's going to be relatively smooth though, so hard to detect. What we really need is something that creates sharp waves--two giant black holes which are orbiting each other at an incredibly fast RPM is perfect since they're massive and will create a lot of high/low waves as they rotate that we can then detect.

As we improve the detection technology, we may be able to use gravitational waves to detect all sorts of things besides black holes, it just depends on how much we can measure. The nice thing is that you don't need "line of sight" to measure with a gravity wave. So if God drops a giant marble on the other side of the Sun, we could detect it even though we couldn't see it.


What is the underwater temperature of Europa?

No, nothing on Europa could possibly be photosynthesizing as we know it. Jupiter doesn't emit light, and what it reflects from the sun is not enough, plus there's no significant amount of carbon dioxide there.

However, the first life on Earth was not plants, it was chemotrophs. These bacteria gained their energy from the heat of hydrothermal vents, and if there's life on Europa, it would almost certainly take this form.

About the temperature: Europa's heat comes from friction due to tidal forces, so it comes from within. The water temperature increases as you go down, but the pressure is crushing down there and it's unknown whether life could form under those conditions.

The tenuous oxygen atmosphere comes from charged particles hitting it's cold surface and creating vapor. It does not indicate biological processes.


Could the largest SMBH swallow an entire galaxy?

This question might be related to Is there a maximum size for a black hole?
Typical SMBHs do not contribute much to the overall mass of a galaxy, much less than the sun is of the total mass of our solar system. And as we can see, even such mass centered quite nearby does not swallow all objects in the solar system very easily.
While there is no theoretical maximum, in practice - based on simulations - maximum mass for a black hole seems to be around 50 billion solar masses†, which will give a radius of 0.01 light years. This is very small compared to the size of the galaxy, so it will not affect the situation much. Thus the main effect of huge black hole vs galaxy would be via plain old gravitation.
The main impediment for a black hole to swallow anything is angular momentum. Any object a black hole is going to consume must shed its angular momentum in order to get closer to event horizon. This is not very easy thing to do, and the main reason black holes are surrounded by discs of extremely hot gas - friction will turn angular momentum into heat, allowing matter to descend towards black hole.
†practical maximum of 50 billion suns is due to fact that accretion disc becomes unstable due to new stars forming from it. See for example Andrew King's 2015 paper How Big Can a Black Hole Grow?

Abstract:
I show that there is a physical limit to the mass of a black hole, above which it cannot grow through luminous accretion of gas,  and so cannot appear as a quasar or active galactic nucleus. The limit is $M_{max} ≃ 5 \times 10^{10}$ M⊙ for typical parameters, but can reach $M_{max} ≃ 2.7 \times 10^{11}$ M⊙ in extreme cases (e.g. maximal prograde spin). The largest black hole masses so far found are close to but below the limit. The Eddington luminosity $≃ 6.5 \times 10^{48} \text{erg s}^{-1}$ corresponding to $M_{max}$ is remarkably close to the largest AGN bolometric luminosity so far observed. The mass and luminosity limits both rely on a reasonable but currently untestable hypothesis about AGN disc formation, so future observations of extreme SMBH masses can therefore probe fundamental disc physics. Black holes can in principle grow their masses above $M_{max}$  by non–luminous means such as mergers with other holes, but cannot become luminous accretors again. They might nevertheless be detectable in other ways, for example through gravitational lensing. I show further that black holes with masses ∼ $M_{max}$  can probably grow above the values specified by the black–hole – host–galaxy scaling relations, in agreement with observation.


Options for first time astrophotography w/ 80/900 refractor?

There is no "best" camera as it really depends on the subject you intend to image and the optical instrument you intend to use with it.  This could turn into a very long answer, so I'll try to offer a short response first and then offer a bit more to consider.

For planetary imaging, use a camera capable of doing video frames and this does not need to be a camera with a large chip or high resolution.  Jupiter, for example, is just fractionally under 44 arc-seconds diameter when viewed at opposition. (compare to the moon at roughly 30 arc-minutes or more than 40x the angular width).

Typically you'd capture perhap 30 seconds to a couple of minutes worth of frames and then feed those frames into stacking software.  Ideally the video frames use a "lossless" format.  Several image processing applications can stack the frames (e.g. AutoStakkert & Registax are popular and free).

A camera such as the ZWO ASI120MC-S is very popular and in your budget.  There's the ASI290MC for those with deeper pockets (and versions for those with much deeper pockets).    From QHYCCD there's a QHY5R-II ... more to the entry-range with lower resolution and a smaller chip.

Ideally your scope would be on a motorized equatorial mount so that it keeps the object in your field of view (otherwise the object will quickly drift out of the camera frame.  For this scope & field of view combination, that would happen in about 45 seconds for a planet in the "center" of the frame to drift out of frame completely).  If using a 2x barlow, it would happen in about half the time.

A barlow is usually used for planetary imaging (the simulated field of view did not use a barlow).  To understand why, you'd want to read about Shannon-Nyquist sampling theory and how it applies to astrophotography.


Does the Milky Way belong to a Galaxy cluster

The Local Group contains 54 plus galaxies.
Don't know that that counts as much of a cluster.
Next up in scale, the Milky way is part of the Laniakea Supercluster
That contains about 100,000 galaxies, so it's a bit on the large size to call a simple cluster.
The well known Virgo Cluster contains about 1300 galaxies.
So it's clearly a thing of intermediate scale.


Why does radiation not need a medium to travel?

Electric and magnetic fields can and do exist in a vacuum. Electromagnetic waves are just fluctuations in electric and magnetic fields. In a vacuum (meaning no charges or currents, which do require the presence of matter), then the solutions to Maxwell's equations are electromagnetic waves.


Is it possible to observe if there is traces of life on an exoplanet?

Direct observation of exoplanets (and then of traces of life on them) is still a big challenge.

Up to now, there are several  indirect methods (effects on the parent star or on other stars), and direct detection is performed by using devices such as coronagraphs on faint stars, where very bright planets are imaged.

A way to image things such as as "forests" is very far in the future, and could be achieved by using extreme coronagraphy in space interferometers .

Another way is to use spectroscopy to find traces of biological activity in the athmosphere. This solution is very tricky due to the faint light reflected by extrasolar planets.

It is anyway a field where progresses are very quick, so stay tuned!


Difference between 3D real space and 3D redshift space?

  Why is 3D real space ever even considered? Is it some kind of highly idealistic cosmology-dependent only scenario, ignoring peculiar velocities?

Simple explanation:

3D real space is the actual distance, position, velocity, (even spin), etc. of an object; this can be used to measure Earth's distance to our Sun or another planet's distance to its Sun.

3D redshift space is the apparent distance, a distorted view by which one can compare one object to another. The distortion is caused by varying density of the interstellar medium, gravity, motion of Earth and the objects being viewed, even dark matter, etc.

The real distance between two objects can, for example, affect each other by gravity; this alters the trajectory of the objects. The apparent distance has no effect on gravity, only the real distance.

By comparing the real distance, obtained by making calculations, with the apparent distance, obtained by making observations, we can compare what we see with what we calculate; this permits checking the theory, finding new objects (or phenomena, such as suggesting that there is dark matter). This gives us the path lengths for each direction and distance (because distance varies depending on direction and distance due to gravity).

A more complicated explanation is offered in:

"A Spherical Harmonic Analysis of Redshift Space" (Sept 13 1994), by A.F. Heavens and A.N. Taylor

  "1 INTRODUCTION
  An accurate three-dimensional map of the galaxy distribution would be enormously valuable for cosmology, but the lack of an accurate estimator of distance precludes this. Despite best efforts, the errors in the Tully-Fisher and Faber-Jackson distance indicators mean that the best one can do is to use the recession velocity $v$, or redshift, of a galaxy, and to assign its distance $s$ by using the Hubble expansion law $v = H_0s$, where $H_0$ is the Hubble parameter. The resulting `redshift space' map is of course not perfect, because galaxies are not necessarily moving precisely with the idealised expansion of a homogeneous universe. Density inhomogeneities have associated with them peculiar velocities, which introduce a radial distortion between the true (real-space) map and the redshift space map. This distortion is, of course, inconvenient: for example, extracting the power spectrum of density fluctuations from a galaxy redshift survey becomes a non-trivial task. However, the distortion itself can be exploited, as the magnitude of the distortion is dependent on the density parameter of the Universe $\Omega_0$, if fluctuations grow by gravitational instability. Constraining $\Omega_0$ is one of the main aims of this paper.".

"Design and analysis of redshift surveys" (May 27 1997), by A.F. Heavens and A.N. Taylor

  "1 INTRODUCTION
  ...  [This omitted portion: Similar to the above quote, but not identical]
  It is clear that the longest wavelength which can be measured is limited by the size of the survey, so it is attractive to consider surveys which are essentially one- or two-dimensional, to maximise at least one dimension without incurring prohibitive cost in observation time (e.g. Broadhurst et al 1990). The difficulty with such an approach as a method for measuring the power spectrum is that a low-dimensional power measurement at a given wavenumber will have a contribution (which may be dominant) from much smaller scales in three dimensions (e.g. Kaiser & Peacock 1991). The interpretation of the observed power spectrum can therefore be difficult. For surveys which do not correspond to the ‘distant-observer’ approximation (cf. Kaiser 1987), the power spectrum measurement and the redshift distortion become linked, and this further complicates the analysis.

  The ease with which the parameters of interest may be extracted depends on the choice of coordinate system and basis functions in which the density field is expanded. For two reasons the choice of spherical polar coordinates is compelling. Firstly the survey is almost certain to be defined in terms of a fixed areal coverage (independent of depth), and secondly, a flux limited survey will have a selection function φ (or, equivalently, a mean observed density
  $ρ_0(r)$) which is dependent on distance, but not on direction. The mean density of the survey is then separable in spherical coordinates $\overline{ρ}(r) = ρ_0(r) M(θ, ϕ)$, where $M$ is either 1 or 0 depending on whether the direction $(θ, ϕ)$ is in the survey or not.
  The second reason is that, unless $β ≪ 1$, it is impossible to ignore redshift distortion effects, and since the distortion between the real-space map and the redshift-space map is purely radial, it is straightforward to include the distortion in a power-spectrum analysis in spherical coordinates (cf Zaroubi & Hoffman 1996).".

More simply, this webpage describing Halton Arp's image:

Our position (the Earth) is at the bottom point in all cases. Distance (away from the Earth) is measured along the straight edges. In the top left image, we show what a galaxy cluster in Arp's universe would look like without the big bang perspective.  It is a family of galaxies and quasars and gaseous clouds of mixed redshifts (in the top diagrams, the large dots are low- redshift, the medium-sized dots are medium-redshift, and the small dots are high redshift). At the center, there is a dominant galaxy -- it's usually the largest galaxy, and the galaxy with the lowest redshift of the cluster. This galaxy is surrounded by low-to-medium redshift galaxies, and toward the edges of the cluster we find the highest redshift galaxies, HII regions, BL Lac objects and quasars.

The image to the right shows what happens if we try to force the same galaxy cluster into a redshift-equals-distance relationship. The cluster becomes distorted. What was once a sphere becomes an elongated bubble. The central dominant galaxy drops to the front of this bubble, followed by a spike of low-to-medium redshift galaxies stretching away from the earth and "bubble and void" of high redshift objects.

Every cluster in the sky does this, like fingers of god pointed at the earth from every direction. The third image is a 90 degree slice of the sky showing all galaxies arranged according to their redshift-  determined distances. The Fingers of God distortions show clearly, each representing a single galaxy cluster. (The bubbles and voids are not as clear, because this chart cuts off before it gets to high redshift.) Everything points at the Earth.

Without the redshift-equals-distance distortion, a new picture of galaxy clusters and the universe itself is revealed. The age of the universe is no longer known, because we no longer have a constant expansion to backtrack to a bang. The size is also unknown. Most quasars and some galaxies that we see are closer than we thought they were, because they have been distorted by the Fingers of God. But we have no idea how far the universe stretches beyond our telescopes' limits.

Compare the real space in the upper left with the observed image in the upper right.


Can Mercury's core "boil over" if it gets to close to the Sun?

I was reading about how mercurys core makes up most of the volume of the planet. Im guessing this is both because of the small size and the distance to the sun.

There are at least two hypotheses about the causes of Mercury's composition:

Early in the Solar System's history, Mercury may have been struck by a planetesimal of approximately 1/6 that mass and several thousand kilometers across. The impact would have stripped away much of the original crust and mantle, leaving the core behind as a relatively major component.
Alternatively, Mercury may have formed from the solar nebula before the Sun's energy output had stabilized. It would initially have had twice its present mass, but as the protosun contracted, temperatures near Mercury could have been between 2,500 and 3,500 K and possibly even as high as 10,000 K. Much of Mercury's surface rock could have been vaporized at such temperatures, forming an atmosphere of "rock vapor" that could have been carried away by the solar wind.

But if the second hypotheses is right Mercury should have been relocated to wider orbit later. Planetary migration is very interesting topic and I would say it's not solved completely yet.
At current orbit Mercury is not hot enough to evaporate silicates at any significant rate.

If mercury got too hot could it completely "liquidify"?

A lot of "hot jupiter" exoplanets have benn observed in recent years and results are very interesting. For example HD 209458b "Oziris" has a comet-like tail of evaporating atnosphere. Clouds of iron and titanium dioxide were observed too. Iron rain (literally) could be one of exciting effects.
Compared with "hot jupiters" Mercury is much smaller with less gravity. If closer to Sun it undoubtedly would have comet-like tail of evaporating elements, mostly silicates and metals. If intensive enough the tail could be seen by eye with black glass during Mercury transits (also the transits would be much more often at closer orbit).
At closer orbit Mercury should be tidally locked. The Sun-faced surface would be completely melted lava, but dark side would probably be solid because of lack of stable atmosphere. Evaporated rocks would cool down by thermal radiation and dowfall behind the terminator line.

Would it join the sun and completely merge with it, gradually crack
apart and spread, or something else entirely?

The planet should be affected by some force to migrate even close and close to Sun. I can't tell surely will evaporation slow down the planet and move it closer to Sun but I suppose it will not. I suppose Mercury will not fall to Sun but will be evaporating. I can't do math now to estimate how close to Sun it should be to evaporate completely in less than 5 billion years (after that Sun will become a red giant and will envelope Mercury by its atmosphere in "coup de grace" act :) ).
I don't see reasons for Mercury to crack apart if it is not close to Roche limit (again with assumption that evaporation only is not enough to lower its orbit).

Would it cause other problems in terms of life on earth?

I don't think so. Maybe Solar system space would be more dusty because of Mercury's evaporating tail. Even cracking of Mercury would not affect Earth because the debris would stay close to Mercury's orbit (much closer than current Mercury's orbit, don't forget!). Theoretically some of debris pieces could develop more eccentric orbits in millions of years  to reach Earth (because of collisions, uneven evaporation, orbital resonanses) but I bet all them would evaporate before that.
P.S. Most of this answer consists of educated (or not so educated) assuptions. So objections, amendments, corrections are welcome.


The simplest and shortest answer to this question is that since our universe is flat (to within our error bars), its density must be that of the critical density - that's what it means to be flat! In that case the overall density of the universe, accounting for matter, dark matter, dark energy, etc. is (assuming $H_0=100\ \mathrm{km/s/Mpc}$)

$$\rho_c = 1.8788\times 10^{-26}\ \mathrm{kg/m^3} = 1.8788\times 10^{-29}\ \mathrm{g/cm^3}$$

If you want to be more precise about it and actually use measurements of our universe, you can go about calculating it by using the density parameters. By definition, the density parameter is defined as

$$\Omega \equiv \frac{\rho}{\rho_c}$$

where $\rho_c = 1.8788\times 10^{-26}h\ \mathrm{kg/m^3}$ (notice that little $h$ now included). Numerous missions such as WMAP and Planck have measured these density parameters for the variety of mass-energies in our universe and you can look them up and put them all together to get the total mean density of our universe. Overall, you should find that

$$\rho = (\Omega_m + \Omega_c + \Omega_\Lambda + ...)\rho_c$$

where the various $\Omega$s are defined for the various mass-energies in our universe. You can look up values for the density parameters from recent results such as from WMAP or Planck. Just decide on your chosen model and pull out the $\Omega$ values $-$ or more properly the $\Omega h^2$ values if you want to use your own value of $H_0$. It's not too difficult to plug everything in. You'll pretty much always find though, that the sum of the density parameters very nearly equals one and to an order of magnitude you can say $\rho=\rho_c$.


From what distance would our sun have an angular diameter of 7 arc seconds?

From us, the Sun is visible around a half grad, which is $\approx$ 30 arc min = 1800 arc seconds.

To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.


Are accretion disks the only way stellar planetary systems form?

The current paradigm for understanding the formation of the solar system (including the planets) is nebular theory, which is thought to be applicable to the rest of the Universe.

Question: Now that we've discovered many extra-solar planets, are there theories differing from the accretion disk theory for the formation of planetary stellar systems?

Not in terms of modern scientific theories - at least not that I am aware of, but this is outside of my expertise. I am willing to be wrong here. My point here is that if you consider the long history of proposed hypotheses of the formation of the solar system (which includes the planets) as "alternatives," then you could consider some of them in some special cases, perhaps, but this is pure speculation. Most of these historical hypotheses have been ruled out. There are issues for the contemporary nebular theory, for examples, explaining the origin of planetesimals, and explaining the discrepancy between the observed and theoretical angular momentum of the solar system.
Regarding exoplanets, I am also not aware of any system that currently definitively challenges the nebular/accretion formation paradigm (again I could be wrong here). However, mechanisms such as disk migration are used to explain the properties of many exoplanets.


Is a Procyon - brown dwarf - Jupiter - superearth - Mars - Pluto system theoretically possible?

Thanks to this answer linked by antispinwards I learned that such system most probably would not be stable, as I calculated it with proximity to upper limit of Hill sphere for each body.

But, if instead, I would have taken distances closer to 1/3-1/2 Hill sphere, it could be theoretically feasible.

  The orbit of a satellite's satellite will be stable if it's deep
  enough inside the Hill sphere, within the so-called true region of
  stability. The limits are a bit fuzzy, but the true region of
  stability is typically the lower 1/3 to 1/2 of the Hill sphere.

On a personal note, if it's so, it surprises me this type of systems are not more pondered on and discussed, with maybe even some probabilities calculated. As it could have a great impact on thinking about abundance of potentially habitable worlds in the universe (imagine how great a number of earth-like satellites could live in such a system in the habitable zone of its host star (it being short-lived on cosmological time-scale notwithstanding)).


Why can't this be the simple (and obvious) explanation for the dimming of KIC 8462852?

The simplest reason why your proposed explanation is not really that obvious, and thus was not considered is that the dimming of KIC 8462852 was not an isolated event. The dimming is a pattern that has been observed over years. There is no known phenomenon in interstellar space that would cause a repeated obstruction of that star, so it doesn't make any sense to consider that possibility when there are other phenomena (such as those captured in Boyajian's paper) that are more likely to be the source of the dimming.


Are lunar occultations visible to the naked eye?

Yes - especially as the occultation does not have to occur on the bright side. Planets may also be occulted (see http://www.lunar-occultations.com/iota/planets/planets.htm)

Incidentally the symbol is not originally Islamic. It was the flag of Constantinople when conquered by the Ottomans and then adopted by them. The Ottomans later became the Caliphs - the nominal leaders of the world's Sunni Muslims and so the flag was widely adopted across Islam.


Is it possible to mount a telescope on a plane? Is it beneficial?

This has been done.

SOFIA is an infrared observatory built into a Boeing 747 SP:

SOFIA takes advantage of the fact that some infrared bands are visible at atltitude, these are attenuated by water in the atmosphere so they're less visible on the ground.
There have been infrared observatories before SOFIA:

The first use of an aircraft for performing infrared observations was in 1965 when Gerard P. Kuiper used the NASA Convair 990 to study Venus. Three years later, Frank Low used the Ames Learjet for observations of Jupiter and nebulae.[20] In 1969, planning began for mounting a 910 mm (36 in) telescope on an airborne platform. The goal was to perform astronomy from the stratosphere, where there was a much lower optical depth from water-vapor-absorbed infrared radiation. This project, named the Kuiper Airborne Observatory, was dedicated on May 21, 1975. The telescope was instrumental in numerous scientific studies, including the discovery of the ring system around the planet Uranus.[21]
The proposal for a larger aircraft-mounted telescope was officially presented in 1984 and called for a Boeing 747 to carry a three-meter telescope. The preliminary system concept was published in 1987 in a Red Book. It was agreed that Germany would contribute 20% of the total cost and provide the telescope.

Other airborne astronomy is more incidental. Around solar eclipses, you'll often see some astronomy flights. Some of these are tourism, others perform science. These are usually passenger aircraft temporarily modified (scopes installed that look through the existing windows). These take advantage of the fact you can lengthen the eclipse by flying along its path, and you can reach eclipses in places otherwise inaccessible.


Is there any cosmological evidence that neutrinos travel at a speed other than c?

A paper by Wie et al. (2016) uses the coincidence between gamma rays from Gamma ray bursts and individual 3-30 TeV neutrinos to estimate the difference in propagation speed between neutrinos and photons.

They find
$$ \frac{|v-c|}{c} < 2.5 \times 10^{-18} $$

They say this is about 7 orders of magnitude more precise than any previously published limit. However, it does rely on the genuine coincidencce between the GRBs and these individual neutrino events, which does seem to be in question or even doubtful in some cases.

The mass limit this places on the neutrinos appears to be a rather uninteresting $<6.8$ keV, but it is a direct measurement of speed rather than an extrapolation of the speed from a mass limit and special relativity.


Why do some celestial bodies have atmospheres, and not others?

The amount and kind of gases a body can trap depends on the object's surface temperature, and its density & radius (which refers to it's gravity).

An object with high gravity and low surface temperature will be able to hold more gases in it's atmosphere. In the case of the Moon, due to its low gravity it could barely trap an atmosphere of Xenon.

You can check this thanks to this amazing plot: http://astro.unl.edu/naap/atmosphere/animations/gasRetentionPlot.html

But this is not all you need to hold an stable atmosphere like the Earth does. The object will also need some protection against the solar wind. In the Earth we have a magnetic field that prevents the solar wind to reach our atmosphere, but in the case of Mars it has a weak and poor magnetic field to defend itself against the solar wind that rips it's atmosphere.


Can we calculate the orbit of exoplanets?

The characteristics of the orbit of an exoplanet can be determined very accurately using measurements of the reflex motion (the "Doppler wobble") of its parent star. These measurements yield the orbital period and eccentricity of the orbit. If we can estimate the mass of the parent star, then the orbital semi-major axis is known through Kepler's 3rd law. Orbital periods, and hence semi-major axes, can also be measured very precisely from transit data, and eccentricities can also be estimated provided one has a good idea of the mass and size of the parent star (e.g. van Eylen & Albrecht 2015).

Additional details can be gleaned from transiting exoplanets, such as the inclination of the orbit to the line of sight and any difference between the orbital axis and the spin axis of the star it orbits. These measurements exploit something called the "Rossiter-McLaughlin effect", as explained in an answer to the question you linked to.

The answer to your broader question is yes, of course people are looking at the distribution of orbital parameters: the behaviour of eccentricities versus planet mass and semi-major axis; the prevalence of planets of particular mass at certain distances around different types of stars; the frequency of occurence of multiple planets and orbital resonances, the misalignment between the spin of a star and the orbits of its planets, and so-on, in order to try and understand how planetary systems form and evolve.

Some resources:

The HEASARC Exoplanet orbital parameter database

exoplanets.org


What was the composition of the primordial asteroid belt?

We can be pretty sure that the asteroid belt once held one or more sizable planetesimals,as there is evidence that it or they were large enough for gravity & internal heating to drag the heavy metals (mainly iron & nickel) to the centre,forming a metallic core. Bodes Law decrees that there should be a planet where the asteroid belt is now,but it seems that as this planet was forming it was shattered by an impact & the remnants scattered. Some reached Earth,which is how we know that their materials had differentiated into metallic core & rocky mantle. It is likely that some of the huge asteroids which have struck the Earth in the distant past were fragments thrown out of the asteroid belt & that some hit Mars & Venus as well. There may be more material in the asteroid belt than we think,because some fragments are so small & widely dispersed that they don't register in our telescopes. I think we can assume that the primordial planetesimals had a similar though not identical composition to the asteroid belt of today.