Can the period of the planet transit across the sun be calculated in the same way the eclipse year is?

I apologize for the length of this in advance! Before getting to the answer it should be said from an orbital mechanics point of view all transits are eclipses and all eclipses are transits.  There are a lot of different types of both and the words are used differently but they mean the same thing - something is passing in front of something else from the observers point of view.  Not many would say "Venus eclipsed the Sun" during a Venus transit nor would they say, "the Moon transited the Sun" during a solar eclipse.  But the way cycles are determined is the same for eclipses or transits.
In short, the answer to this is almost yes:

The period of a planet's transit across the sun as seen from an outer
planet as it crosses the outer planet's orbit plane at its certain
node = The beat period of the period of the transiting planet's being
in inferior conjunction with the sun from the outer planet's
perspective and the period of the transiting planet’s hitting that
node

For us to see Venus or Mercury transit/eclipse the Sun, three things have to happen.  The inner planet's nodes must be at the same ecliptic longitude as the Sun, the planet must be crossing the node when this occurs, and it needs to be on the same side of the Sun as us.  (Oh, and the observer needs to be on the Sun side of the Earth with clear skies which is NOT what happened to this observer during the last Venus transit and I'm still bitter about it.)
Various cycles of different lengths have to come together to make this happen which is what I think you mean by "beat period" though I've never heard it phrased this way in astronomy.  It's as if I ate dinner on Friday every two weeks with one friend and every three weeks with another - then the "beat period" of six weeks would mean a scheduling conflict or a dinner party.
The reason I said your conclusion was almost correct is explained by looking at the transits of Mercury on this Wiki page.  Sometimes a transit doesn't happen even though it is on the "beat period" (which astronomers just call cycles).  Lots of smaller cycles line up, or come together, every so often on a larger cycle.  You got it right when you said the two things have to happen - the nodes lining up with the Sun during an inferior conjunction but as Mercury shows the transit can be missed.  If a larger "beat period" was used to leave out those missed transits, there would be extra transits happening mot on the beat if that makes any sense.
Determining these cycles is fairly easy for us on Earth due to the frame of reference of the ecliptic.  Neither the Earth nor the Sun have "nodes" to the plane of the ecliptic because it's the Earth and the Sun which define the ecliptic.  Nodes are where tilted (inclined) planets cross this plane.  It's a handy reference since we live on Earth, but for the observer freezing on Pluto wondering about all of the transits they will get to see with their telescope, they would want the ecliptic be the plane defined by Pluto and the Sun.  Then the Earth and all of the planets would have nodes as they crossed above and below the plane of the Pluto - Sun ecliptic.  From their perspective, there would be inner planet cycles which would repeat with a "beat" when certain cycles lined up again so that Jupiter, for example, transited the Sun.
On the Wiki page you linked talking about eclipses, I found it awkward the way they kept referring to the "Sun being near/at the node" which technically isn't wrong because we know they mean "from the observes viewpoint, the planet's node lines up with the Sun (same ecliptic longitude)" but I think it could give people the wrong impression as if the Sun was at or near that location in 3D space. That's just my opinion, though.
Here is a good Sky & Telescope article on Venus transits and how they will become more rare in the future. And here is one of the most enjoyable websites on the internet - Orbit Simulator. Look through the list until you see something that interests you and then click it to see fantastic browser based orbit details.  They also have a Windows program you can download to have a lot more control over the simulations.
I apologize again for the lengthy reply - orbital mechanics is a subject I love.  And good on you for putting so much thought into the orbits and cycles, not many want to dive that deep into the subject.


Why not us interferometry to take a picture of Pluto?

Radio interferometry can combine observations over very large baselines.  But optical interferometry cannot.  According to a list of interferometry instruments on wikipedia, the largest baseline for optical measurements is less than a kilometer.  We can't take optical measurements with continent-sized instruments.

Then if you drop down to radio where the instruments do have that capability, I think you'll find Pluto is quite dim (it's not a radio source, and there's no strong radio emissions that it can reflect to us). There's no radio signal from Pluto that can be imaged.

From a page on optical interferometry:

  Interferometers are seen by most astronomers as very specialized
  instruments, as they are capable of a very limited range of
  observations. It is often said that an interferometer achieves the
  effect of a telescope the size of the distance between the apertures;
  this is only true in the limited sense of angular resolution. The
  combined effects of limited aperture area and atmospheric turbulence
  generally limit interferometers to observations of comparatively
  bright stars and active galactic nuclei.


To answer quickly, the answer is that no, an object cannot be made to collapse as such.

There are multiple reasons as to why. Pela brings up a great argument here, reminding us that if a black hole could be formed this way, then I could be made to collapse into a black hole just by something moving by me at extreme velocities. Clearly, this cannot occur.

To understand why this is, we must first completely throw away the idea you have of relativistic mass. The entire concept of relativistic mass was created to allow for Newtonian mechanics to still hold in relativistic situations. By modifying the mass of an object at extreme velocities, equations such as $F=ma$ can be made to still hold. To paraphrase Philip Gibbs, gravity couples not only to mass but also to momentum and itself. Therefore, a description of spacetime in the vicinity of an object will be incomplete without momentum considerations. From this relativistic momentum, we can compute the relativistic energy.

While these quantities are useful for giving detailed descriptions of the body, the answer is actually somewhat simpler. We must remember that an object must behave the same in all inertial reference frames. If the object does not become a black hole in its own frame (as it clearly will not), then it cannot appear to be one in any other.

If you are wondering how to reconcile this with a relativistic mass that approaches infinity, the answer is simply that relativistic mass is not real mass. The metrics which describe spacetime in the vicinity of these objects refer to the invariant mass, which is the rest mass of the body.


Could a "burping" supermassive black hole be responsible for a spiral galaxy's look?

Does the supermassive black holes inside the Milky Way still "burp" or
  it is already cured

Technically, black holes don't burp it's the material that falls into a black hole, as it gets tightly squeezed and under enormous spiral and magnetic forces it shoots out gamma rays at the poles perpendicular to the accretion disk.   More details here.

The super-massive black hole (not holes as there's just one) in the center of the milky way isn't currently "burping" but the next time it eats a star it's expected to "burp" again.   I remember reading that this may happen sometime in the next 10,000 years or so, but I couldn't find the article just now.

"Burping" or Jets of material are more active when galaxies are young.  The quasars observed by large telescope operate in that way.

  I like to think that a supermassive black hole that "burps" lies in
  the core of every spiral galaxies. Am I mistaken?

This is essentially correct.   Quoting this article

  Astronomers believe that supermassive black holes lie at the center of
  virtually all large galaxies

Black holes are surprisingly good at shooting material back into space, strange as that may sound.  Source.


Partial answer only.

120 degrees of sky in 5 seconds or so.

120 degrees in 5 seconds is about 0.42 radians per second. At a distance of 70 km (roughly the distance where an overhead meteor would start to glow visibly) that would be roughly 29 km/sec.
That's a nice number for a meteor.
But it's quite a challenging number for an originally earth-launched object to achieve during reentry unless it's done some planetary flyby's to change it's orbital energy relative to the Sun.
I know it's been almost 6 years now, but if there's a chance that the apparent rate was a lot less than 0.4 radians per second, it would open up the possibility of a reentering artificial object.
If not, I think it is more likely to be a meteor, though reentry of a deep-space bit of space junk is not impossible.


The main reason appears to be to prevent any icing up and contamination of the mirrors and other optical coponents with water and other "non-volatile residues", which would be extremely detrimental to the UV performance.

Emissivity at 270K is not too much of an issue since the instrument only works out to 2 microns.

The issue is briefly discussed in section 8.1.10 of the LUVOIR final report.


Yes, cosmic-ray astronomy is a thing, and it has been for a while.
More generally, there is multi-messenger astronomy since there are now multiple windows through which we are able to observe the universe via direct observation:

electromagnetic (i.e., light rays, photons)
neutrino
cosmic rays (i.e. high-energy protons and atomic nuclei moving at relativistic speeds)
gravitational radiation

Naturally, we built telescopes for observing light first. Galileo used an optical telescope and it turned out to be very fruitful for him, and it took off from there. Lens crafting became an art that was intimately tied with the progress of astronomy until the 20th century, where the discovery of quantum mechanics opened up the world of particle physics. Neutrinos sourced from nuclear reactions in the Sun, processes elsewhere in the Milky Way, and in other galaxies arrive at Earth and have been detected since the 1960s (although a large fraction of extra-galactic neutrinos are generated in supernovae, it is interesting to note that geoneutrinos are sourced from the decay of radionuclides in the Earth's interior, which helped to confirm the radiogenic theory of Earth's heating).

Over the past decade we have been able to begin doing neutrino astronomy, identifying distant neutrino sources.
Is there any reason that this cannot progress to other particles, I.E. general cosmic-ray astronomy?

There are numerous detectors all over the world designed to detect cosmic rays. There are direct (involving balloons or satellites in the atmosphere making measurements) and indirect (involving ground-based detectors searching for the shower of particles that cascade from the decay of cosmic rays in the atmosphere or searching for photons that are created from that cascade moving through the atmosphere) detection methods. I am generalizing, there are other methods and it is an active area of research (as Benny listed many examples in their comment).
For example, muons were originally discovered in cosmic rays shortly after antimatter was discovered.   Muon detectors are of many types (i.e. Cherenkov-radiation water-tank detector) and can be used for all kinds of interesting purposes. For example, exploring the insides of pyramids and volcanoes without people getting hurt.
The recent direct detections of gravitational radiation by the LIGO/Virgo collaboration reveals that parts of the Universe we used to think we closed-off from observation are now see-able (or may soon be), for examples, at the early Universe prior to the cosmic-microwave-background radiation, and inside of compact objects such as white dwarves and neutron stars.
There could be other undiscovered windows into the Universe, for instance, if there is a "dark force" and thus a "dark photon" associated with dark matter, but that is speculative. We generally need reasons to suspect that something exists before we start making detectors to see them, but the early bird can get the worm, or sometimes it doesn't (like Joseph Weber!!).
Before the advent of large particle accelerators, i.e. 1950s, particle physicists relied on cosmic ray observations. These large accelerators, like at Fermilab or the LHC, are built to discover particles, and they have had varied success which also helped in the development of modern cosmic-ray detections that you could consider as being "general cosmic ray telescopes."


How strong is the gravity from half of the distant universe?

Not very strong at all. I get a rough figure of $3.725\times 10^{-9} m/s^2$.

To perform that calculation I made a few simplifying assumptions.

Assume that we can ignore everything outside the observable
universe.
Assume that the observable universe is a static homogeneous sphere
with a radius of 46.6 billion light-years, and a mean density of
$\rho = 0.85\times 10^{-26} kg/m^3$, that figure is the critical
density
needed for the global curvature of the universe to be zero (i.e., on
the very large scale spacetime is flat).
Assume that we can use Newtonian physics to calculate the
acceleration due to gravity.

I didn't bother removing all the stuff closer than 10 billion light years.

We're standing at the centre of the flat surface of a hemisphere of radius $r$ and we want to know how strong the gravity is. We can pretend that all the mass of the hemisphere is concentrated at its centre of mass, which is $\frac{3r}{8}$ below us.

The Newtonian formula for gravitational acceleration is

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

Mass is volume times density, i.e., $M=V \rho$ and the volume of a hemisphere is

$$V = \frac{2}{3}\pi r^3$$

Combining everything, not forgetting to use $\frac{3r}{8}$ as the distance term in the denominator of the acceleration formula, we get

$$a = G \rho \frac{128}{27} \pi r$$

Plugging in the values I mentioned earlier for $\rho$ and $r$, the Google Calculator says $3.72484086\times 10^{-9} m/s^2$. That's about $3.8\times 10^{-10}$ standard gravities.


Can two planets in an empty universe meet/be pulled together?

Yes, they would experience gravitational attraction. It would take a long time for them to collide... the formula (derived here and shown here) is:

$$t = \frac{\pi}{2} \sqrt{\frac{d^3}{2G(m_1+m_2)}}$$

where $d$ is the initial distance between the two planets of mass $m_1$ and $m_2$ and $G$ is the Gravitational constant. This gives a time of about $10^{23}$ years, much much longer than the universe has existed. This assumes Newtonian mechanics. Relativity would not change the conclusion much.

There is no known upper limit to gravity, and plenty of indirect evidence that it has no upper limit.


Which kernel does HORIZONS get the Geometric Albedo from?

As barrycarter commented, codes_300ast_20100725
has ephemerides for 300 large asteroids, generated in 2010 and supposedly valid from 1800 to 2199.
For other asteroids, a Celestia developer forum shows how to how to request an SPK from HORIZONS's telnet interface.
However, they appear to use the SPK for ephemeris only, adding albedo in Celestia's own SSC file format.
In SPICE documentation, neither the tutorials nor the "required reading" seems to mention albedo.

A more direct approach to your goal might be the JPL Small Body Database Search Engine.
For example, if you query for numbered NEO asteroids with absolute magnitude H < 16, you get a table of 169 rows.
Diameter and albedo are known for some asteroids but not others, even though their orbits and rotational periods may be well established.


What is the difference between LMC and SMC?

The LMC has an apparent size of about 645x550 arc mins, the SMC 320x205.

Both contain several hundred million stars each. The LMC is about 14000ly in size, and is about 10 billion solar masses; the SMC is 7000ly in size, and is about 7 billion solar masses.

The visual magnitude of the LMC is +0.28, the SMC is +2.23.

Both feature a number of interesting clusters, nebulae, and supernova remnants. Notably, LMC is home to NGC 2070, the Tarantula Nebula, and Supernova 1987A.
L
The LMC is usually considered an irregular galaxy, though it has a prominent bar, somewhat warped, and a spiral arm. The SMC is a dwarf irregular galaxy, or maybe a barred disc.

LMC is the fourth largest galaxy in our local group, and the third closest to us. SMC is the fifth? largest galaxy in the local group, and the fourth closest to us.

{In a previous version of this response I said: A number of websites proclaim that the LMC and SMC are gravitationally bound to the Milky Way. This has been known to be incorrect since 2007 when Hubble observations showed they are travelling too fast to be orbiting the Milky Way. See also.}

A number of websites proclaim that the LMC and SMC ARE gravitationally bound to the Milky Way. This was based on old information. More recent measurements have challenged this assumption, though further analysis has left open the door for the possibility that the LMC and SMC are gravitationally bound to the Milky Way. We can, however, safely say that the LMC and SMC may be gravitationally bound to the MW.


You're correct, except that you should multiply, not divide, by 3600 (you're converting from degrees to arcsec, not the other way round). You get $5.7\times10^{-5}{''}$, but the correct result is 36002 times higher, i.e. $738.6''\!\!\!$.
But why not just let astropy do the job:
z     = 0.3
x     = 3.29 * u.Mpc
theta = x * cosmo.arcsec_per_kpc_proper(z)

print(theta.to(u.arcsec))
---> <Quantity 738.61092463 arcsec>


the length of second, minute or hour, what defines the time of exo planetary bodies

You have 2 separate concepts here. The length of a day and a year are unconnected with how you measure seconds, minutes and hours.

The length of a day is defined as how long it takes for the planet to rotate on its axis, and a year is how long it takes to orbit its primary.
Seconds, minutes and hours are a measurement chosen by us back in history, and chances are we will still use the same wherever we go in the universe. The way we measure it is designed to let us always measure seconds consistently.

Even on Earth, we say that a day is 24 hours, but it isn't really, as the two sets of concepts don't match up. We haven't defined a second as a fraction of the time it takes the planet to rotate - it is defined based on physical constants. So if you were to say that Gliesa has a day that is 30 units long - it would be astonishing if those units were exactly hours!


Universe as 2d sphere surface - a round trip?

This is another case of taking an analogy too far (the other one I can think of is assuming that the rubber sheet analogy indicates that there is some extra spatial dimension$^1$). In this case, I'm assuming you're talking about the "balloon" analogy - the universe is expanding, like a balloon, but since there's no center, all bits of matter are on the surface of the balloon. This naturally leads to the idea that if you travel in one direction, you'll eventually come back to where you started. This is, unfortunately, not true.

Scientists often talk about the shape of the universe - how it is curved on a local or global scale. There are three possible categories, depending on one parameter, $\Omega$:

Positive curvature $\to$ $\Omega>1$
Zero curvature $\to$ $\Omega=1$
Negative curvature $\to$ $\Omega<1$

This diagram gives a good visual of each curvature:

I should add that these are merely examples of universes (more accurately, manifolds) with each $\Omega$. There are other cases with these curvatures, and so other universe where you can start at one point and end at the same point, as in a 3-torus.

All of these shapes could exist. However, in our universe, the global curvature is one where $\Omega \approx 1$ - a flat universe. The measurements that essentially clinch this are from the WMAP probe, which has analyzed the CMB and determined the age of the universe and its curvature. Measurements show, to a decent degree of accuracy, that our universe is flat - and so if we started in one direction, we would never return to where we started. Check out the NASA page for a full rundown of WMAP's results.

$^1$ Good questions dealing with this can be found here, here and here; John Rennie's answer on Physics is especially good.


How can I determine line luminosities from equivalent width measurements?

If you also have photometry, then there is a reasonably accurate conversion between the apparent magnitude and the continuum flux (per unit wavelength interval) at the wavelength of the photometry.

Once you have this conversion factor, you multiply your equivalent width by it to get a flux.


How much of the sun needs to be blocked in order to see a noticeable drop in brightness

Based on the CNN's reporting it appeared to get noticeably darker at near 95% coverage. That's just from watching the onset of darkness for locations experiencing totality.


Could this three moons system be stable?

I see no reason why this wouldn't work.  The innermost Galilean moons are in a 1:2:4 resonance so it's clearly a stable orbital configuration.  They could all be tidally locked if you want, but the planet itself can't be tidally locked with all of them.  If the planet were tidally locked it would likely be locked with the innermost moon.

Also of note, the Galilean moons all have eccentricity less than 0.01 and inclinations less than 0.5 degrees.


In classical General Relativity, there is no limit to the compression in a black hole, hence you get a singularity. However, many astrophysicists feel that's unphysical, and that a theory which unites General Relativity and Quantum Mechanics will impose some kind of limit, perhaps something connected with the quantization of spacetime itself.

We don't have a working theory of quantum gravity, so at this stage we don't exactly know what happens in the core of a black hole. OTOH, we're fairly confident that the core has to be very small, since quantum gravity effects probably don't kick in until a scale much smaller than the size of an atom, and probably smaller than a proton, somewhere around the scale of the Planck length.


Is it possible for a planet to have multiple moons in a nearly stationary orbit?

You may want to try the Worldbuilding site https://worldbuilding.stackexchange.com/

Anyway, what you're describing is more or less possible, with some caveats. A major question is that of stability - how long those configurations remain in place.

There are several notions you need to think about:

Geostationary orbit: it's the orbit where satellites have an orbital period equal to the planet's rotational period, so they appear stationary in the sky.

Roche limit: the lowest orbit where natural satellites remain in one piece. Any lower than that and they break up into pieces due to tidal forces.

Hill sphere: is the region where the planet's attraction dominates and satellite orbits remain stable (as opposed to being stolen by the star, etc).

Usually Roche limit < Geostationary orbit < Hill sphere.

Satellites at the geostationary orbit appear fixed. Those close to the geostationary orbit would indeed appear to move extremely slowly across the sky. Orbits higher than geostationary would appear to move backwards.

With regards to the long-term stability of multiple moons in various orbits, they would be stable if the orbits are far enough from each other, or if they're sitting in each other's Lagrange points. Also see here https://worldbuilding.stackexchange.com/questions/36929/stability-of-multiple-moons


I happen to have represented my nation at IOAA

Start with "Roy and Clarke's" book : "Astronomy: Principles and Practice"

Move on to the references mentioned in the book.


It doesn't look like an Einstein ring.  An Einstein ring shows highly distorted shapes: stretched and multiple images of the distant galaxy, and a massive foreground object (with different redshift) which would have to be a galaxy itself as nothing else is big enough.
Hoag's object shows none of these distortions, and the central bulge is at exactly the same redshift (so the same distance by Hubble's law).  The areas of star formation in the ring look rather like those in more normal galaxies. It is just an unusual shape.


How early could we detect an asteroid the size of the one that caused the extinction of the dinosaurs?

It could be as little notice as a few days.
There is a large variation in the amount of warning we would have before an asteroid strike similar in energy to Chicxulub.  Let’s examine the case in which we would have the least warning.
The lower bound to the estimated energy released by the Chicxulub collision is 1.3e24 joules https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics.
We know the object ‘Oumoamoa was moving at about 50km/s at 1 au from the sun https://en.wikipedia.org/wiki/%CA%BBOumuamua , which is the distance of the earth from the sun.  The earth itself is moving at about 30 km/s with respect to the sun, so if they struck in opposition,  the difference in velocity could be 80 km/s.
For kinetic energy impact, we know $e=1/2*mv^2$, or energy = 0.5*mass times the square of the velocity.  So $m=2e/v^2$, so m is approximately 4e14kg.  If we assume an asteroid a bit denser than iron at 1e4 kg/m^3, then the volume of the asteroid is 4e10m^3.  If we assume the asteroid is a sphere, then since the volume of a sphere is $V=4/3\pi r^3$, then the radius is $r^3 = (3/4*V/\pi)$ or $r$= 2.121km.
Let’s assume the asteroid is matte black so it hardy reflects any light, give it an albedo of 0.02.  Then the absolute magnitude of the asteroid of this size and albedo at 1 AU from an observer is 17, using the tables here https://cneos.jpl.nasa.gov/tools/ast_size_est.html.  Let’s suppose that we spot the asteroid when it is much much brighter than at one AU (or 1.5e8 km), with the same apparent magnitude as Pluto (magnitude 14).  Since magnitude is a log scale, that is $2.5^3 = 16$ times as bright.
Brightness varies as square of distance, so the asteroid will be at $1.5e8/4$, or 3.75e7 km away from earth when it’s spotted.  At 80 km/s relative velocity, the time from seeing it to impact is 4.6875e5 seconds, or about 5.4 days.
Notes:

The above outlined scenario is almost impossibly unlikely. A much more likely scenario is that an asteroid from our known asteroid belt strikes earth, in which case we would probably know it was going to collide years before the strike.

The object Oumoamoa was classified as an asteroid rather than a comet due to its lack of corona, but now it has an ‘I’ designation for the first Interstellar object which is neither a comet nor an asteroid.  One could argue that I shouldn’t use it as an example given that the question asks about asteroids, but I think it is perfectly reasonable.

We don’t ever know a collision with 100% accuracy.  There is always error in our orbit estimations.  We can’t even tell if satellites are going to collide in earth orbit, it’s just probabilities.

I don't have a good idea of how long a new object of a certain apparent magnitude will be in the night sky before it's discovered by someone.  If someone thinks we would notice this object sooner due to transits or some other reason, I may amend this answer.


Why all the photos from 67p are black and white?

True fact is that most of the images seen so far are taken from the navigation camera (NAVCAM) of the orbiter, Rosetta, and that takes black&white pictures.

Here is a very interesting article about how these pictures are taken (ironically it is on a NASA blog). They state that they show pretty much what you would see with a human eye from there, even thought the comet is actually all very dark. The white parts are due to the contrast with the very dark surroundings.

I'm not sure about the images received by the lander. They are captured by the CIVA cameras, where CIVA stands for Comet Infrared and Visible Analyser. So I guess they might show some infrared light not visible to the human eye, then processed to be in b&w. Unfortunately, I could not find a precise explanation on ESA's website so far. (here their page for Philae instrumentation).

Anyway, as a general rule you should not expect that all images from spacecraft are in true colors.


Has star KIC 8462852's 3 year long Kepler-fading continued after the primary Kepler mission ended?

Well "if no known astrophysical model can explain it" then nobody told Wright & Sigurdsson (2016) who, cognisant of Montet & Simon's results, explore a number of astrophysical models. They conclude by saying that the most "plausible model" is that of small scale intervening material between us and the star that may be responsible for the short-term dips and the secular dimming. Lacki (2016) appears to agree.

Wright & Sigurdsson also consider and largely rule out or class as unlikely a number of other possibilities, including artificial structures.

The Gaia satellite has been observing this star (and a billion others) since December 2013, but so far the photometry time series has not been released. This should become available towards the end of next year and will probably exceed the accuracy of the Kepler data. In the meantime, it has been claimed by Hippke et al. (2016) that the star was relatively constant between 1934 and 1995 in contradiction to previous claims of a progressive dimming over the last century.


How to calculate the redshift of a line in a bunch of line spectra?

While @planetmaker's comment is true if the lines come from the same source, you can have lines emerging from different physical processes which still appear to come from the same location.
An example is absorption (or more rarely emission) lines from galactic winds, which are typically blueshifted with respect to the "systemic" redshift, i.e. the "average" redshift of the galaxy. You can also have strong emission from an object — e.g. a supernova — that has a large peculiar velocity inside a galaxy.
In general, to distinguish lines with different redshifts, you need to have an idea of where you expect the lines to be (well, the same is true even if you just want to identify spectra where all lines are redshifted uniformly). In this case, the authors know which lines may be expected from the neutron star, namely highly ionized iron and oxygen lines, but the absorption features are then redshifted with respect to the rest of the spectrum.
For instance, the O VII Ly$\alpha$ line has a rest wavelength of 19 Å, but is seen at ~25.6 Å, i.e. a factor 1.35 higher, so its redshift is $z = 0.35$.


Is the material that could have formed a "Super-earth" divided among the four terrestrial planets

I think the answer to your question lies with Jupiter, be it directly or indirectly, the gas giant is now believed to have had a large influence on the way the inner planets formed.

Many planetary scientists believe that the reason we only have 4 small inner planets today and not any "super-Earth" sized ones is because of Jupiter, and how it interacted with the circumstellar disk and protoplanets at the time of the solar system formation. There may well have been one or two super-Earths that formed in the early solar system, but models suggest that they would have been sent crashing into the Sun by the inward advances of Jupiter.

I found this blog article which explains it in more detail. And there is also another MNRAS article that goes into more detail about the physics and orbital mechanics behind what might have happened in the early solar system.


Our Solar System’s interstellar movement in relation to the Alpha Centauri System

If you were looking down on the galaxy from above with 12 o'clock pointing from the Sun to the Galactic center, then using SIMBAD we can get the galactic co-ordinates as approx (315,-0.6) in longitude, latitude respectively. Longitude is measured anti-clockwise from the line from the Sun to the galactic center, so Alpha Cen would be approximately between 1 and 2 o'clock. ESO have a map here showing the nearby stars within 12.5 light years.


How did scientists come to the conclusion that it is space that is expanding?

The simplest explanation relies on the statistical unlikelihood of us being at a special place and time in the cosmos.  Let's say, for a moment, that we see galaxies moving away from us because of some cosmic explosion that set them in motion.  In that case, we would expect to see the galaxies in one direction moving faster than the galaxies in another.  Looking towards the center of the explosion, we would see galaxies on the other side moving much faster than if we looked in the other direction.

This, however, is not what we see.  What we observe is all galaxies moving away from us, with a speed proportional to their distance from us.  That means that the further away they are, the faster they are moving.  It doesn't matter what direction we look in; a galaxy 500 Mpc away in one direction will have a recessional velocity approximately the same as one 500 Mpc away in the other direction.

This forces us into one of two conclusions:  either we are at the exact center of the universe, which is statistically impossible, or space itself is expanding, so everything is moving away from everything else.  The animation below shows how this would work.  From any given point in space, it would look like everything is getting further away from you.

Explaining the recession of galaxies due to an explosion also raises many questions:  What caused the explosion?  Why would it start at one particular point?  What were things like before the explosion?  All of these are questions which do not have accepted answers.  Instead, the expansion of space ties directly into the idea of the Big Bang.  The expansion of space itself is what drove the Big Bang.  It is not that the universe is expanding into something.  Instead, space itself is expanding.  However, the topic of the Big Bang is a different subject, so you can look up information on that yourself.


Precision of geocentric gravitational constant

Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $\mu_\mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.

Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.

The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.


Can the moon appear to be in any position in the sky?

Like the sun, the moon, from our perspective on the surface of the Earth, rises in the East and sets in the West. However, it does not rise exactly due East and set exactly due West. If you were to track the position of the moonrise along the horizon over the course of a year, you would notice that is varies considerably. It cannot, however, be located everywhere in the night sky, even over the course of a year. To understand why, we need to understand some celestial geometry.

The ecliptic plane is the plane which a hypothetical line between the Earth and the sun sweeps out over the course of a year. In the image above, you can see
how the Earth's orbit exists within this plane.
Also notice that the line from the Earth's geographic south pole through its north pole (its rotational axis) is not perpendicular to this ecliptic plane. In fact it is 23.5 degrees off this perpendicular direction.
Notice that the Earth's rotational axis points in the same direction regardless of which side of the Sun the Earth is currently on. This rotational axis always points towards the star Polaris. This means that on different parts of the year, the sun will rise at varying points along the horizon.
The Earth's equatorial plane is a plane defined to extend out from the Earth's equator in all directions. This plane is inclined relative to the ecliptic plane by 23.5 degrees (the same amount as the rotational axis is tilted relative to the perpendicular direction of the ecliptic plane).

Now consider the Earth moon system. In order to determine how far North or South the moon can appear to rise from, you need to consider the angle between the Earth's equatorial plane and the moon's orbital plane. Since the Earth's rotational axis is inclined 23.5 degrees relative to the ecliptic, and the Moon's orbit is inclined 5.14 degrees relative to the ecliptic, the highest the Moon's orbit can be relative to the equatorial plane of the Earth is 28.64 degrees.

Thus, the Moon cannot be directly overhead of any point on the surface of the Earth if that point is 28.64 degrees above or (by an symmetric argument) below the equator.

Thus, the Moon cannot appear in every location in the sky for a given location on Earth. There are parts of the sky that the geometry of the Earth moon system simply will not permit the Moon to exist.


In 31.5kyr, Epsilon Eridani and Luyten 726-8 will be < 1ly close; but how distant from the Solar system?

These stars are currently moving away from us, in the case of Epsilon Eridani, at 15.5 km/s. In 31500 years they will be further away than they are now, the distance can be calculated by applying Pythagoras' theorem to the distances and velocities you mention.

It turns out that Eps Eri will be about 12.2 light years from Earth, and Lutyen 726-8 will be 12.6 light years.

They won't be particularly close together in the sky, I estimate about 3 degrees, and they will be even dimmer than they are now, though Eps Eri will still be a naked eye object.


Can I see the star of Bethlehem at Christmas?

The most recent suggestions are that it was a nova or supernova, however no one really knows what the object was. The verbatim records describe it as a star, though it is not certain whether or not it was a "true" star (in those times, a star being any object that had a fixed right ascension and declination). Perhaps it was just a planet.

In any case I suspect it is unlikely you will be able to see it with your naked eye today. Even if you could, its true identity has certainly been lost in history.

Some references for further reading (click on the "Full Printable Article (PDF/Postscript)" link for a PDF of the articles):

What was the star of Bethlehem?, S. Hagar (1918)
Interpretations of the Star of Bethlehem, E. Little (1968)
Can we find the Star of Bethlehem in Far Eastern Records?, C. Cullen (1979)
The Star of Bethlehem: a type Ia/Ic supernova in the Andromeda galaxy?, F. Tipler (2005)


Planetary phase angle of Hohmann transfer

Let's assume that Earth and Venus have circular, coplanar orbits.
The spacecraft advances 180° in heliocentric ecliptic longitude in 143 days; meanwhile, Venus advances 241° and Earth advances 141°.
When the spacecraft departs from Earth (left), Venus should be 61° behind Earth as seen from the Sun.
When the spacecraft arrives at Venus (right), Venus should be 39° ahead of Earth.

The Sun-Earth-Venus angle marked in green in your second figure is Venus's elongation, 43° east of the Sun as seen from Earth at departure time.

In another context, "phase angle" could refer to the Sun-Venus-Earth angle.
At departure time, an Earth-based observer would see Venus in a 76° phase angle or 62% illuminated.


Hypothetically, would we be able to see the moon from Saturn's North Pole?

First of all, at that distance seeing the Moon and seeing the Earth amounts to the same thing. At its closest, Saturn is around 3000 times as far from Earth as the Moon is, so viewed from Saturn, the Moon is never more than about a minute of arc away from the Earth.

If you can see Earth from the North Pole of Saturn, you can see the Moon, also.  (Though it would be a lot dimmer.)

But if you can See the Earth from Saturn's pole, then you can also see Saturn's pole from Earth.  So your question can equally well be phrased as "Can Saturn's north pole be seen from Earth and if so, when?"

If you look at pictures of Saturn from Earth, the rings are sometimes tilted so far that they cover one polar region, though it's also true at those times that polar region tilted away from Earth, anyway!  Most of the time the rings are either not tilted enough to cover the poles or tilted to cover the south pole.

So the answer is that during most of Saturn's orbit around the Sun, the rings would not obscure the sightline from Saturn's north pole to the Moon.


There are several theories about the future of the universe, but I don't think we can be completely certain about any of them. The favourite at the moment, and probably the one with the most evidence to support it, is the dark energy universe which will go on expanding for ever, with the galaxies getting further and further apart and more isolated until they burn themselves out into total darkness. This raises problems about a unique, one-off beginning, with no cause, and vast quantities of matter and energy created magically out of nothing, thus violating the First Law of Thermodynamics.

My preferred option is the Big Bang-Big Crunch-Big Bounce universe in which the expanding universe is brought to a halt by gravity and collapses on itself to form a blue-shifted, shrinking universe which ends in a fireball and singularity similar to the one it was created from. Except it doesn't end. It erupts again (the Big Bounce) in a gigantic white hole which we call the Big Bang, and this cycle is repeated endlessly. It had no beginning, it will never end, and it doesn't violate the Fist Law of Thermodynamics.


Trying to understand the way Saturn's ring look in this famous Cassini image

What's going on with the distortion of the rings on the upper half when they (presumably) cross in front of Saturn?

The brownish areas you see on Saturn are ring light, analogous to seeing the Earth by moonlight. Saturn's rings light up Saturn's night sky, particularly just after sunrise and just before sunset. The two dark bands across the face of Saturn are the A ring (upper dark band) and B ring (lower dark band). Of all of Saturn's rings, these two are the most opaque. The parts of Saturn underneath those two dark bands are brilliantly lit by ring light, but Cassini can't see those parts of Saturn because the opaque A and B rings block that light.

For a much better and more thorough explanation of all of the features in this incredible image, I suggestion you spend eleven minutes watching this youTube video by Emily Lakdawalla.

This is one of the very few times I will post a link to a youTube video. It is worth every eleven minutes.


Does cosmological horizon grow or decrease as it radiates?

I think there's a misinterpretation here that's causing your confusion. There's a difference between the event horizon of a black hole and the cosmological horizon.

The event horizon of a black hole is the place inside which nothing can escape from the black hole. The cosmological horizon, on the other hand, is the place beyond which an observer cannot "retrieve information." The difference between the two is basically that the event horizon of a black hole is in the same place for observers in different locations, while the cosmological horizon differs for observers in different locations.

Also, if the cosmological horizon was the edge of a black hole, the black hole would have to be centered right where we are - which is highly unlikely!


There are a couple of misunderstandings here. Let's take one at a time:

  The Hubble telescope can see the light from the big bang.

The Hubble Space Telescope (HST) doesn't actually see light from the Big Bang. HST has several instruments on board, both for imaging and spectroscopy, but they all operate in the infrared, optical (i.e. "visible to humans"), and ultraviolet wavelength range. When you mention "light from the Bing Bang", I suppose you're thinking of the cosmic microwave background radiation (CMB), which is light emitted 380,000 after Big Bang. This is "the closest" we can get to Big Bang (yet). HST is unable to detect microwaves; instead we have other telescopes for that, the most recently-launched being Planck.

  My question is if our universe is expanding that would make everything
  that's in front of our solar system be in the future?

The expansion of the Universe is one of the most difficult concepts to get one's head around. But firstly, Big Bang wasn't an explosion, hurling matter outwards from a central point in an otherwise empty Universe. Galaxies, on average, don't travel much through space. Rather, they lie relatively still in space, but space itself expands. The distances between galaxies increase all the time, and previously, before galaxies were formed, the distance between atoms and other particles increased. An often-used analogy — which you shouldn't take too far — is a balloon with dots painted on. The dots are fixed on its surface, and when you blow it up, the distances between the dots increase despite the dots still being fixed.

Secondly, your question seems to confuse space and time. Even if the expansion were an explosion, we would travel through space, and yes, you could say that we travel through time as well, but saying that the future is "everything in front of our Solar system" doesn't really make sense (to me, at least). What's in front of you when you travel through space is just more space; you can look at it before you get there, but that doesn't mean you look into the future (in fact you look at the past, since the light you detect with your eyes has spent some time traveling toward you).

  And also can Hubble see ahead of our solar system

It depends what you mean by "ahead". It can definitely see things outside the Solar system. The Solar system consists of one star and it satellites (planets, comets, asteroids, etc.) out of the few hundred billions that make up the galaxy we call the Milky Way. Our Galaxy is itself only one out of at least some hundred billion — and quite possibly infinitely many — galaxies that float around in our Universe. And yes, HST can see many of these. But if by "ahead" you mean "into the future", then the answer is no.

  and does the bang big light ends behind our solar system?

As mentioned above, the Big Bang light — or the CMB — was emitted shortly after Big Bang. That means that it has been traveling for 13.8 billion years through space. Light travels one lightyear per year, but since the Universe is expanding, the distance that the CMB has traveled is more than 13.8 billion lightyears; in fact it's some 46 billion lighyears. This is the most distant "thing" we can see, way, way beyond the Solar system which for comparion is of the order of one lightyear large.


Does Big Bang red shift affect the frequency of both light waves and individual photons?

Yes it affects both. If you treat light as waves then the wavelength from a source at a redshift $z$  is "stretched" by a factor of $1+z$.

Photons emitted from a source at redshift $z$ have their energies reduced by a factor of $1+z$.


It looks like CFITSIO has an interface for F77.

I'm always thinking about programmer time vs. computation time in research settings. If your data is reasonably small and simple in structure, I'd be very tempted to have Fortran write to a text file, and then use something like astropy to turn the text file into a FITS file. It's slow and icky, but it takes fifteen minutes of thinking if you've used astropy before.


What is the frequency distribution for luminosity classes in the Milky Way Galaxy?

Here is how you do it "properly" for the Hipparcos data. As Warrick correctly points out, what you have done in your question is massively biased towards giant and supergiant stars, which actually form the very small minority of stars.

You must form a volume limited sample. To do this, sort the stars by distance (1/parallax) and choose a cut off point. Your sample will always be incomplete, but the greater your distance cut off, the more incomplete it will be and it will become incomplete for more intrinsically luminous stars.

You have a philosophical problem to solve here in terms of what you are trying to achieve. The vast majority of stars in the Galaxy are faint M dwarfs with an absolute magnitude of $>10$. As Hipparcos is only complete to around magnitude 10-11, you will only get these M dwarfs in your subsample if you limit yourself to 10pc. But you will find that this sample contains no evolved stars (too rare) and no white dwarfs (too faint).

EDIT:
This has piqued my interest again, so I have a practical (approximate) solution, based on a two-part process. The first part involves a paper I wrote (actully an undergraduate experiment) based on the nearest 1000 stars to the Sun (from the Gliese & Jahreiss catalogue CNS3). This sample is roughly complete down to mid M-dwarfs, so everything else I say, and the results I give, only applies to a sample of stars more massive than that.

If you look at this volume-limited sample of 1000 nearby stars you can straight away say something about the relative numbers of different types of stars in the Galactic disk (saying something about stars anywhere else in the Galaxy is filled with much more uncertainty). A colour-magnitude diagram is shown below, and from this we see that:

The Sun is among the brightest of stars - brighter than 95% of other stars.

About 6% of the population are white dwarfs (though a few faint, old white dwarfs may still be missing from the sample). This makes sense. If you integrate a typical mass function assuming only stars more massive than about 1$M_{\odot}$ have even had time to become white dwarfs, then this is what you get.

Only 0.9% of the population are giants. The reason for this is that only a small proportion of stars are massive enough to have evolved into giants on the lifetime of the Galaxy. But once they are there, their lifetimes are short compared to the main sequence phase and most have become white dwarfs (see above).

There are a handful of objects, maybe 0.5%, that could be classified as subdwarfs, between the main sequence and the white dwarfs.

So, in broad terms: 92.5% of stars (above $\sim 0.2M_{\odot}$) are main sequence (class V), 6% are white dwarfs, 1% are giants (class III) and 0.5% are subdwarfs (class VI).

There are no very massive stars or supergiants in the immediate solar neighbourhood. That is because they are very rare. To get a better estimate we need to look at a larger volume limited sample. To do this I took all the stars (about 7000) from the Hipparcos catalogue that are closer than 50pc and assumed this was complete to just below a solar luminosity and assumed that those stars with an absolute magnitude brighter than the Sun (1949 stars with $M_V<4.5$ represent 5% of the total population in this volume. The absolute magnitude vs colour diagram for this sample is shown below.

Of these 1949 luminous stars, I find roughly 190 are giants - giving a giant fraction of $5\times 190/1940=0.5$%, in reasonable agreement with the nearby star sample based on smaller numbers. There are still NO supergiants, even in this larger sample. Thus supergiants have a frequency of $\leq 5 \times 1/1949 = 0.0025$%. i.e. less than 1 star per 40,000 is a supergiant.


Deep space could be simply defined, as a part of the universe, which is lying beyond the Earth-Moon system. However, when speaking about deep space objects we generally mean stars, star clusters and nebulae, which are outside the Solar System.


What do you call it when a satellite goes once around its planet?

Definitions:

Rotation can be defined as the circular motion of an object around its own axis, or upon itself.
Revolution can be defined as the circular movement of an object around an external axis, or around another body.

So a satellite moving around a planet once is called a revolution. Or an orbit. "The satellite stayed active for 3 orbits, and then the batteries ran out".
To disambiguate the case 'the satellite having gone once around the Sun': this depends. Is the satellite in orbit around a planet, or does it orbit the Sun directly?
If the satellite is in orbit around a planet, you have to describe what you mean. There's no separate word for this case, AFAIK.
If it's in orbit around the Sun, it's again one orbit or one revolution.


What does it mean when there are two plus-or-minus signs for a measurement?

The first error is due to the precision of the measurement, the second is an uncertainty associated with methodology used.


Can I do M.Sc in physics to become an astronomer after doing B.Tech in aeronautical engineering?

It is good to see that you are interested in astrophysics at an young age.
No problem with you doing the clichéd BE that everyone in India is forced with, you can still get your PhD in Astrophysics.

Your 4 years of B.Tech aeronautical is going to be tough, so be prepared for it. In the mean time, keeping in mind your aim of becoming an astrophysicist, prepare yourself equipped with programming languages, especially Python/C++. The reason I say this is because you don't need too have much of an astronomy background to do M.Sc Astronomy, but you definitely need to have a good knowledge in a programming language.

There are websites like Course Era or EdX and plenty more on the internet that can help you with learning python/C++ in an interactive way. Go for it, and trust me it will help you a lot.

As a final year student doing my PhD in Astrophysics (hailing from India) let me tell you the information I have acquired over the years.

With regard to the Masters/PhD in Astronomy/Astrophysics, the procedure is quite different in different countries.

Procedure within India

There are plenty of Universities in India that offer you a Masters course in Astrophysics.

Almost all the details required are summarised neatly in THIS PAGE,
so I am not going to go through the list of things required.

Procedure for Europe

You would need:

All your original mark sheets attested by the college/university (Bachelors, Masters, etc)
Two/three recommendation letters, and the number of letters depends on the university you are applying to.
A completely filled application form.
IELTS scores (depends on country)
Statement of purpose, as to why you want to do the course.

This is pretty much all you need for Europe. For certain universities, you might be having a Skype interview, wherein they might ask you about the stuff you know.

Procedure for USA:

When it comes to the USA, it gets quite complicated!

You will need:

All your original mark sheets attested by the college/university (Bachelors, Masters, etc)
Two/three recommendation letters, and the number of letters depends on the university you are applying to. For most universities it will be online forms that your professor needs to fill after he/she receives the mail.
A completely filled application form. This is going to cost you somewhere from around 50-80 dollars depending on the university.
GRE scores (both general GRE and subject (Physics,Chemistry etc.) GRE) that you can send to the university directly online by paying money (25 Dollars for each university). Only a few universities accept you without a subject GRE.
IELTS/TOEFL scores. TOEFL scores can be sent online, but if you are taking IELTS you have to go to British council and have them send it.
Statement of purpose, as to why you want to do the course.

So it depends on where you want to apply. I would recommend you to apply to both, to increase your chances. For Europe, the chances of you getting through depends on the country you apply. For USA, the chances of you getting through solely depend on your GRE scores. It doesn't matter whether you are a gold medallist in your Bachelors, it all boils down to GRE. If you fail there, your entry into USA will be difficult.

One advantage of Europe is that, mostly, the PhD is only for 3/4 years, depending on the country. Whereas in the USA, it is mandatory for the student to write exams in the first year, teach/be a research assistant in the second year, and only in the third year start the PhD project.

To search for PhDs in Astrophysics depends on the country you are applying for. But as a start, you can have a look at http://jobregister.aas.org/. PhD calls, internship calls, post-doc positions, etc. related to astrophysics will be posted here during the start of every month.

With regard to your question

  can I do aeronautical engineering so that I can even get employed in ISRO?

Yes you can definitely do that. ISRO not only employs aeronautical engineers, but also people who have completed their BE in electrical and electronics and also computer science.

Have a look at THE CAREERS WEBSITE OF ISRO to get more information.

Good luck with your future!


How could universe inflate itself out of the very dense and curved early spacetime? Could it happen in a black hole too?

Great question! Sorry for this huge response, and it might not be a satisfying answer, but it'll address your questions.

Sadly, as with most of astronomy, the Big Bang is surrounded in mystery. It is one of the biggest uncertainties out there, and you'll come to this realization if you've researched it enough.

Most of what we can get out of the Big Bang is through extrapolation. If we observe the expansion of the Universe, we can presume that at one point, there was far less space between all the matter in the Universe. This is where things begin to break down, though.

Through General Relativity — which explains our modern understanding of gravity — and our extrapolation, we would reach the phenomena you're describing; the entire Universe would have originated as an infinitely small point (called a singularity), from which matter was incredibly hot.

However, General Relativity begins to break down at those levels, since its equations cannot explain the conditions of such a universe accurately. For one, as you hint at, an infinite density and potentially infinite mass would likely have an infinitely strong gravity.

There is no place for such infinities in a proper mathematical description of the universe, and at that point, Einstein's equations — the basis for predicting the evolution of the cosmos — collapse. These equations provide us with unreliable results and we run into some of the problems you are describing.

Quantum mechanics is more fit for such small scales, but it fails to explain how gravity acts and what its role was at the time. Thus, this tells us that QM and GR are necessarily incomplete. Their equations cannot describe the Big Bang singularity in its entirety, especially since GR (which deals with large, high-mass phenomena) and QM (which deals with small, low-mass phenomena) are incompatible and cannot describe small, high-mass phenomena.

Thus, we're left with a lot of unanswered questions: did the Big Bang create the Universe, or was it simply an event in the Universe's history? Why would the Universe be compressed to such a small scale? Would the idea of an infinitely dense universe go against the presumption of an infinite universe?

There has long been hope that these questions can be explained within a theory of quantum gravity. Quantum gravity attempts to unify general relativity and the concepts of quantum theory (QM and QFT). These hopes have recently taken on a more concrete shape, with the help of what is called loop quantum gravity, and its applications to cosmology.

Quantum gravity is still unproven and hypothetical. String theory also tries to unify GM and QFT, but is unfalsifiable. Thus, the Big Bang is a point in time that astronomers really know little about. It's unfortunate and surprising, but it's something physicist are trying to overcome.


Gift for 16 year old Aspiring astrophysicist,,,,

there is a series of educational Japanese manga comics books called "The Manga Guides" they explain a particular subject of science and maths (like Physics, Relativity, Regression Analysis, Statistics, etc) and they are fun to read, I gifted the one on Physics to my 14-year old nephew and he liked it. Also the book has actual exercises between chapters and explained examples, and each book are independent, so you are not obligated to read all of them to finish the story.
https://en.wikipedia.org/wiki/The_Manga_Guides#The_Manga_Guide_to_Physics
There is one on the Universe and General Relativity too. I read the one on Relativity out of curiosity and it was very entertaining


Discrepancy between my results of Great Conjunction 2020 and Wikipedia

There are no bugs in your codes. But you can make it more accurate with the following corrections (additions):

Using new ephemeris series (de430 or de440) most recent one is de440.
Use True equinox of True epoch of the date. Adding epoch='date' as below:

.radec(epoch='date')
.ecliptic_latlon(epoch='date')

use the following codes for loading timescale instead of ts = load.timescale(builtin=True).

url = load.build_url('finals2000A.all')
with load.open(url) as f:
    finals_data = iers.parse_x_y_dut1_from_finals_all(f)

ts = load.timescale()
iers.install_polar_motion_table(ts, finals_data)

These codes load the updated values of ΔT and also apply polar motion. Of course, because your calculations are geocentric, the polar motion does not affect the result.
You can reach the following result by the corrections and some extra calculations for other attributes.
JUPITER -- SATURN Conjunction 2020:

A.D. 2020-Dec-21 18:20:37.3129 UT
Angular Separation: +  0° 06ʹ 06.38154ʺ
JUPITER is -  0° 06ʹ 06.98608ʺ of ('Below', 'Right') of SATURN in direction of (+  0° 00ʹ 02.56845ʺ, N)


Why was the (small) Hubble better able to find KBO targets for New Horizons than large adaptive optics ground telescopes?

I suspect it's a combination of two things:

Stable, guaranteed high-resolution imaging across the entire field of view, something not possible with ground-based adaptive optics;
Very low background in the optical for HST (Hubble), versus a very high background for ground-based AO in the near-infrared.

Most adaptive optics systems are only able to correct in a small region (the "isoplanatic patch") around a bright "guide star" (say, half an arc minute in radius at most); even with artificial laser guide stars, you still need a moderately bright guide star for so-called "tip-tilt" correction. This means you can only do adaptive optics searches in limited parts of the sky.

HST, on the other hand, delivers high-resolution imaging throughout its entire field of view (several arc minutes across), all the time, regardless of where it's pointed.

To make matters worse, the trajectory of New Horizons is close to the Galactic plane, so there are lots of faint background stars. This makes it harder to pick out possible Kuiper Belt objects, making a very accurate and stable point spread function (such as that of HST) even more important.

These searches are best done in the optical, to minimize the sky background. The absence of atmospheric sky glow (mostly scattered light from the Sun and Moon) for HST makes it easier to quickly detect faint sources like KBOs. The fact that adaptive optics systems, as you and Rob Jeffries note, work almost entirely in the near-infrared, where the atmospheric background is much higher, makes it even worse for them.


Why doesn't Jupiter increase the chances of an asteroid to strike the Earth?

With regard to collisions with earth, or inner planet collisions in general, you're right that Jupiter has both a positive and negative effect on asteroid and comet trajectories. On balance though, the effect is mainly positive.
From earthsky.org:

Some astronomers believe that one reason Earth is habitable is that the gravity of Jupiter does help protect us from some comets. Long-period comets, in particular, enter the solar system from its outer reaches. Jupiter’s gravity is thought to sling most of these fast-moving ice balls out of the solar system before they can get close to Earth. So long-period comets are thought to strike Earth only on very long timescales of millions or tens of millions of years. Without Jupiter nearby, long-period comets would collide with our planet much more frequently.
...
But Jupiter creates both good and bad conditions for earthly life. Consider that its powerful gravity prevented space rocks orbiting near it from coalescing into a planet, and that’s why our solar system today has an asteroid belt, consisting of hundreds of thousands of small flying chunks of debris... today, Jupiter’s gravity continues to affect the asteroids – only now it nudges some asteroids toward the sun, where they have the possibility of colliding with Earth.


Pluto- Current theories regarding its geological nature?

Geothermal activity can originate from a combination of:

Residual heat from planet formation: During the formation of a planet, half of the potential energy of
the parent region of dust and gas can be theoretically converted to
kinetic energy---this is known as the Virial Theorem.
The kinetic energy of the particles translate to a temperature.
Internal radioactivity: Around half of the $4.5 \times 10^{13}$W of heat that come from the Earth's interior is driven by radioactive decay.  Should Pluto have the same proportion of radioactive materials in
its composition? Some would assume so, but others would say that the
heavier radioactive materials should have fallen towards the centre
of the solar system during its formation and would thus exist mostly in the inner planets. Note that geothermal activity on
Mars has now ceased despite it being much larger (Mars has a radius of $3400$km while that of Pluto is $1200$), so a combination of residual heat and radioactive decay is
perhaps unlikely in the case of Pluto.
Tidal effects: Tidal forces basically apply whenever a part of a planetary body experiences a changing gravitational field with time (see tidal heating)---a number of processes can cause this:

the moon having an elliptical orbit;
extra moons in the vicinity (relevant to Io);
moon's orbital period and planet's rotational period not being synchronised;
non-uniform density in planet.

With time, the New Horizons mission should shed more light on these $3$ possibilities.

Note that the geothermal activity on Pluto likely refers explicitly to substances such as Nitrogen, Methane, etc., where far less energy is required to drive activity; that is to say, the ability to melt silicates has long since passed.


The moon has just the right speed not to crash on the Earth or escape into space. What are the odds?

There isn't a "Goldilocks speed" for orbit.  If you put two objects in space, and give them a velocity relative to each other, then provided that velocity is less than the escape velocity (at their relative distance) the two objects will orbit each other.
Those orbits will be elliptical, and it is possible that the ellipse is skinny and "eccentric" enough for the two bodies to collide when they are closest to each other. But for an object that is several hundred thousand km from Earth, there is a quite a wide range of possible elliptical orbits.
So when (and if) the grand collision happened, there was a huge amount of matter that was ejected up into space.  Some probably was moving so fast that it escaped, Some certainly went into orbits that didn't have enough energy and so were small skinny ellipses and the matter fell back to Earth.  But there was a lot that ended up in some kind of elliptical orbit.  This matter was not all in the same orbit, but it started to coalesce, and form into a single ball, under its own gravity.
Other moons weren't formed like this, they either formed at the same time as their planets as a "mini solar system" (such as the four major moons of Jupiter) or they were captured from the asteroid or Kuiper belts).  Initially, the captured moons may have had rather elliptical orbits.
But most moons are in rather circular orbits.  Even if the moon was originally in an elliptical orbit, tidal effects will tend to make the orbit more circular.  A planet and moon system has a certain amount of angular momentum and a certain amount of energy.  The angular momentum can't change, but energy can be converted into heat and since tides dissipate some energy as heat, the orbit will tend to change to a shape that minimizes energy, for a given amount of angular momentum.  That shape is a circle.  (See Is the moon's orbit circularizing? Why does tidal heating circularize orbits?)
So the effect of tides is to give moons the "Goldilocks speed" that keeps them in a circular orbit.


Supermassive black holes at the center of galaxies

This is a very well posed question, thanks!
The problem is, we still don't know.
What we suppose, is that dark matter came first.

Primordial fluctuations of dark matter made the right conditions to accrete enough mass to build SMBHs, and then the surrounding galaxy.

But still we don't know if, the SMBH comes from merging, of other SMBHs or galaxies, or from high massive BH that are called seed (of the order of $100\,M_{Sun}$).
Perhaps, there are very recent developments, which could be known by the experts, but until few years ago, this was the scenario.

About the mass issue, we could have a better reasoning if we think to the influence spheres. We discovered a relation between the SMBH mass and many bulge properties (i.e., the luminosity, the velocity dispersion, and others). This means that an object of the order of $\sim1$ pc, influences an object of the order of $\sim1$ kpc (maybe $\sim3$). Plus, the bulge structure is essential to the formation and characterization and classification of the host galaxy.


Sum of the Masses of All Moons in the Solar System

Here you can find a list of all the natural satellites in our Solar System. You can check one by one (good luck!) OR you can check this webpage, and just add the terms.

Please, keep in mind that the latter website is kind of unknown, so double-check at least some of the masses, before to trust it.

Perhaps, you can cross check with this list as well, and see if other parameters (size, distance, etc.) agree.


What is the present day radius of the observable universe?

The answer is basically #1. The observable universe contains objects whose light has taken up to the age of the universe to travel to us.

Thanks to the expansion of the universe, the things that emitted that light are about 46 billion light years away now.
The exact figure depends on the details of the cosmological parameters and also whether you take the first time that the light can have started out as being the "epoch of recombination" (when the cosmic microwave background was emitted and the universe became transparent to radiation) or the big-bang itself (which would add about 0.5 billion light years to the figure).

Light emitted just after the epoch of recombination has a redshift of about 1100. That means the scale factor has increased by a similar amount. Thus something that is now 46 billion light years away was about 42 million light years away when the light was emitted (of course neither the Earth or our Galaxy existed then).

The answer cannot be #2 since light from something 45 billion light years away cannot have reached us in the age of the universe.


Stars don't "burn", they undergo nuclear reactions that do not involve atoms and chemistry at all. The temperatures in the interiors of stars, and certainly in the interiors of stellar remnants like C/O white dwarfs, are far too hot (millions of degrees) for electrons to bind to nuclei and far too hot for molecules to survive. The energy required to break up carbon dioxide is 5.5 eV, which is easily provided unless a gas is cooled well below 10,000K.

The only parts of a star where chemistry can occur is in the outer atmospheres where temperatures can drop to thousands of degrees and where atoms and partially ionised atoms can exist. Here, yes, then it is possible for carbon atoms and oxygen atoms to interact, but that would mostly produce carbon monoxide. This molecule is produced and can survive between temperatures of about 1100K and 3500K in the atmospheres of cool M-dwarfs and brown dwarfs and the atmospheres of red giant stars.

These chemical reactions are utterly negligible in terms of their energetics compared with the fusion reactions that power a star.


At what size do objects burn up in the atmosphere when falling from orbit?

The Earth atmosphere protects us from small impacts from both asteroids and man made objects. This is well known from meteoroids, where meteoroids as large as a few tens of meters in diameter usually fail to penetrate into the lower atmosphere because they get fragmented and dispersed at high altitude. Fragmentation height depends mainly on the meteoroid strength, only strongest irons reach the surface in one piece.

We could extrapolate from meteoroids to man made objects that even large do not have the same physical strength than a massive meteoroid body. This means than nearly all man machine objects will disintegrate before reaching earth. Heavy metallic ones (iron) will disintegrate at lower altitudes than lighter ones such glass and plastic.

Two comments:

This is not true for objects not having reached orbit, e.g. rocket 1st or 2nd stage engines will fall without disintegrating.
Even if they disintegrate, nuclear powered satellites will cause some level of radioactive pollution, however widely scattered in the atmosphere.

Although the risk to human beings walking on earth is next to nil, there is a real risk of impact from man made objects littering Earth orbits to other spatial vehicles.

Reference: "The Impact Hazard", David Morrison, Clark R. Chapman and Paul Slovic, in "Hazards due to Comets and Asteroids", T. Gehrels, Ed., 1994, The University of Arizona Press


I agree with David Hammen.   Hyperphysics is mostly a very good site but they dropped the ball on that page IMHO.   Hope you don't mind a partially speculative answer, but here goes:

  Why does it matter if there are some areas of a planet with extreme
  temperatures, as long as there are other spots on the planet that are
  not extreme?

It shouldn't matter if part of the planet is uninhabitable.   There are deserts on Earth which are all but uninhabitable but that doesn't effect life elsewhere.   Prior to 5.3 million years ago the Mediterranean sea evaporated and that entire region could have had a salty basin and been hugely hot, but I've not read of life on Earth having any problem with that.   Source

  Why are "moderate" seasons required for life to exist?

There's a boatload we don't know about the evolution of life on other planets but seems perhaps universally true that life adapts, so I find it difficult to believe that moderate seasons are necessary.    Very extreme changes could be difficult, but change can force adaptation.

  If humans can live at the equator on Earth where there is the least
  amount of tilt, why would an exoplanet with less tilt or no tilt be
  necessarily non-inhabitable?

The tilt is planet wide but the lowest variation happens near the equator, but animals that thrive near the poles adapt by hibernation or migration and smaller stuff can be frozen and then come back to life, so, I don't agree with the article on this point.

  Even if humans could not live on a planet without axial tilt, are
  there no other forms of known "advanced" life that can? We know that
  extremophiles exist, such as tardigrades' ability to survive in the
  vacuum of space. What is the most "advanced life" that could live on a
  planet without axial tilt?

One of the interesting historical facts of life on Earth, at least to me, is how long it took what we might consider advanced life to develop.   One celled life in various forms was around for over 3 billion years but the first fossils are about 650 million years old.   It took life a very long time on earth to get from too small to see to large enough to leave a footprint . . . but, I digress.

I agree 100%, one celled life or Tardegrades could live on a planet with no tilt or 90 degree tilt.  Easy.   Ocean life in general should be fine cause oceans are more adaptive.  Evaporation keeps ocean surfaces colder than land gets during peak heat and while a completely frozen over ocean isn't great for life, cold oceans hold more oxygen and CO2 which can be good for life.   Oceans also circulate as an effective means of temperature moderation and fish don't really care how windy it is or how much or little it rains.   The tilt question, I think, is really just about life on land.

Land life could be more vulnerable to high wind, extreme temperature shifts, droughts or floods, which could be driven by greater axial tilt, but I find it hard to believe that Axial Tilt is the be-all and end all.   Day length and year length are key factors too.

One point I agree with the article on, is that a close to 90 degree tilt might not be ideal with one part of the planet always facing the sun and the other part never facing it but outside of extreme tilts, I don't see why it would be a big deal.

A thick cloud cover, for example, reduces seasonal changes.   There's a number of factors.


How far apart is the dust in the Sombrero Galaxy's dust lane?

Based on 1995A&A...303..673E,

How far apart they are?

No idea. Somebody else might help to provide a density distribution.

Are they grouped in clumps or relatively even spaced?

This is what I really wanted to answer. They are in a concentric ring-like structure around the galaxy. The maximum optical depth is 0.8 with a mean value of 0.3. By the way, optical depth gives you a nice idea of how far apart they might be, not a good density distribution but you can get the physical picture in mind.


It's called a black hole merger, or coalescence. Here a simulation video.

Even the formation of the event horizons of the two initial black holes takes "super long" in Earth's time. That's similar with the merger. On the other hand we are very close to the completed merger within a short "Earth's" time seen from a distance, as soon as the merger starts. General relativity as well as quantum theory are incomplete with what will happen very close to a presumed singularity or at the presumed event horizon; this will remain disputed until a satisfying theory of quantum gravity is found.

Mergers of black holes are likely to occur, e.g. when two galaxies collide, the momentum of the central supermassive black holes (SMBH) is slowed down by consumption of gas, dust and stars, until the SMBHs merge to the central SMBH of the merged galaxy.
Here a galaxy merger simulation.

Here a simulation of the coalescence of two black holes within a collapsing star.

More on black hole binaries on Wikipedia.


Satellites just add moving lights to the sky, they do not obscure stars. However, some may find that disrupting their view of what a sky should look like.
The visual magnitude of starlink satellites is about 5.92 at zenith, and usually darker (but also sometimes brighter). The new darker satellites are about 0.77 magnitudes fainter. This is in the lower range for naked eye vision under good conditions.
Light pollution is likely to make the satellites invisible to most people, since they already cannot see much of the sky. You need to be outside suburban skies on the Bortle scale to be able to notice them. So if you are concerned with the aesthetics of the sky, city light pollution is a far, far greater problem.


Lunar eclipses occur when the moon is full.  When the moon is full, the far side of the moon is unlit. With neither the Sun nor the Earth in the sky, the only ambient light would come from stars, and this is not bright enough to see anything by.  The moon would look completely black. The fact that a lunar eclipse is happening is largely irrelevant. The far side would be unlit during any full moon.

The near side of the moon would be lit by light refracted around the Earth. This would be bright enough to see the moon surface. It may be bright enough to read by. The exact brightness would depend on lots of unpredictable factors such as the amount of volcanic dust in the stratosphere. It would certainly be red just like we can see the moon during a lunar eclipse, so would the astronauts orbiting it.

In previous passes the Earth would at most partially cover the moon and the nearside surface would be well lit. The far side would be unlit and dark.

In fact the landing happened when the moon was just less than half full and so much of the far side would have been well lit.


Closest path of solar eclipse central line

Your methodology seems to be based on an algorithm that uses successive approximations to obtain the desired values. I was not able to arrive quite at the same values as you mentioned, but I noticed an error in the posted script, zeta = sqrt(ro^2 + x^2 + y^2) clearly should be zeta = sqrt(ro^2 - x^2 - y^2), so perhaps slightly different values or operations were used somewhere else too, but I couldn't find out where. I also noticed that you did not take into account the difference between Ephemeris Longitude and True Longitude in your calculation, True Longitude shifts towards the East by 1.002738 * At (Delta t) / 240 degrees. Moreover, executing more iterations could possibly converge to a better result.

However, to compute the intersection of the Moon's shadow axis from Besselian Elements, I suggest using the following closed-form solution, which is derived from solving the intersection of a parametric line (Moon's shadow axis) and the Earth ellipsoid.

f = 1 / 298.257223563
b = 1 - f
sin_d = Sin(d * PI / 180)
cos_d = Cos(d * PI / 180)
y0_r = y * cos_d
z0_r = -y * sin_d
y1_r = y0_r + sin_d
z1_r = z0_r + cos_d
v = y1_r - y0_r
w = z1_r - z0_r
a_quadratic = v ^ 2 + b ^ 2 * w ^ 2
b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v
c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)
p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)
y_i = y0_r + v * p
z_i = z0_r + w * p
dist_axis = Sqrt(x ^ 2 + z_i ^ 2)
lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI
lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240

Here is the explanation of the formulas:

We use the flattening value to define the WGS84 ellipsoid, and calculate the semiminor-axis:

f = 1 / 298.257223563
b = 1 - f

Convert declination angle d to radians and execute sine and cosine functions:

sin_d = Sin(d * PI / 180)
cos_d = Cos(d * PI / 180)

Select 2 points (points 0 and 1 in the code) on the shadow axis, at z = 0 and z = 1, and apply the declination rotation (_r) on them:

y0_r = y * cos_d
z0_r = -y * sin_d
y1_r = y0_r + sin_d
z1_r = z0_r + cos_d

The x value always remains the same, so we don't need to recalculate it.
Next, evaluate the parametric line's vectors v,w (delta y, z) from point 0 as a base:

v = y1_r - y0_r
w = z1_r - z0_r

By substituting the parametric line's equations in the ellipsoid equation, and rearranging to quadratic form and solve:

a_quadratic = v ^ 2 + b ^ 2 * w ^ 2
b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v
c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)
p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)

'p' here represents the distance from point 0 along the parametric line, expressed in Earth's equatorial radius.
The contents of the square root must ideally be tested beforehand for negative values to avoid exceptions if the line does not intersect the ellipsoid.
With this, we find the coordinates of our points of intersection:

y_i = y0_r + v * p
z_i = z0_r + w * p

Then we convert this to geographic latitude and longitude:

dist_axis = Sqrt(x ^ 2 + z_i ^ 2)
lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI
lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240

Direct conversion for the Latitude is possible here because our point is on the surface of the ellipsoid.

At the end, you can use the following logic to capture out-of-bounds longitude values:

If lon <= -180 Then
    lon += 360
ElseIf lon > 180 Then
    lon -= 360
End If

In terms of accuracy, you can expect pretty good results with this method, but it does not take into account some factors, like local topography, geoid height, refraction, etc. Topography can lead to differences of a few kilometers, the geoid, +-100 meters, and refraction can bend the shadow by a few meters to a few hundred meters depending on the incident angle. Also, there is no correction for the moon's center of figure, that can shift the umbral center by several hundred meters as well.

With the above formulas for UT1 = 18h and delta t = 72.3s, the computed intersection point is at 58.26828°N, 21.57331°W, which matches F. Espenak's values.


What escape velocity would quark stars have?

The escape velocity (in terms of Schwarzschild coordinates) from the surface of a quark star will be given by
$$ v = c\left(\frac{r_s}{r_q}\right)^{1/2}, $$
where $r_s$ is the Schwarzschild radius, $2GM/c^2$, and $r_q$ is the "radius" of the quark star. Strictly speaking, $2\pi r_q$ is the circumference of the quark star.
Since $r_s < r_q$, this does not exceed the speed of light.
However, this applies to objects launched radially outwards. The escape velocity is higher at other angles. Not even light can escape if launched tangentially at $r = 1.5 r_s$ and quark stars could be smaller than this.
If the quark star is spinning (and it would be) the escape velocity becomes even more complex.


Is there any man-made satellite orbiting our moon?

The Lunar Reconnaissance Orbiter is still up there. At a compact 2000kg, it's likely too small to see via earth based telescope. Mission Page

I think the Chinese orbiter is back on earth now, but there may be others.


Isn't it likely that most extra solar planets don't transit their star in line with the Earth?

Yes this is certainly true. The chances that a planet's orbital plane is directly along our line of sight is pretty small. And you are also correct to assert that it is more likely that most planets will orbit in the same plane. So if we don't see a planet around a star, it doesn't mean the planets don't exist. Absence of evidence is not evidence of absence.

In the early days of exoplanet detection, we had only a handful of known planets to study. This hardly precludes itself to statistical analysis. However, as the years have progressed, the number of known and confirmed planets has piled up, allowing astronomers to start inferring information based on statistical arguments.

One can argue that if we've spotted X number of planets after observing Y number of stars under Z conditions for T time then with a little bit of statistical analysis and math, there should be A number of planets in our galaxy, many of which we can't observe with current techniques. In fact, that is precisely what many astronomers have done.

Cassan et al. (2012) looked at microlensing detections and were able to infer, based on known detections and statistical analysis, that:

Planets are at least as numerous as the stars in the Milky Way.
17% of stars host a Jupiter-mass planets ($0.3$-$10\:M_J$).
52% of stars host "Cool Neptunes" ($10$-$30\:M_\oplus$).
62% of stars host an Earth-mass planet ($5$-$10\:M_\oplus$).

Swift et al. (2013) performed a similar type of analysis but focused mostly on M Dwarfs. They inferred that M Dwarf stars are expected to have, on average, $\sim2$ planets per star.

There are certainly many more such analyses and almost all of them come to the conclusion that there are hundreds of billions of planets in our galaxy. This can be confidently stated based on only the ~3500 exoplanets found so far.


How was all the matter curled up inside a singularity during big bang ?

The very short answer is "no one knows". The available evidence very strongly suggests that the Universe 13.8 billion years ago was hot, dense and expanding and the further back we look, the more indirect the clues we get are, but the hotter and denser it seems to have been.

There are multiple theories for what happens as we follow this process back and back, but theories is all they are. Scientists work to make testable predictions that would follow from some of them but not others, and allow us to whittle down the pool, but it's hard.

All that said, at the broadest level there seem to be two common current views as to how to address the question "what happened before the big bang".

One is to argue that it is simply a bad question: time, like space, is "curved" by the presence of a lot of mass or energy and the question may be analagous to asking "what is North of the North pole". There simnply is no "before" direction from "the Big Bang", or to put it another way, it is the nature of this universe to have a time dimension which only goes a finite distance in the "past" direction. This, very loosely, is the "no boundary" approach of Hawking and others.

The other view, which has names like "eternal inflation" or "cosmic inflation" is that the original state of the universe was a smooth and fairly featureless, but extremely rapidly expanding (exponentially rapidly) one, which was only meta-stable. Part or all of this universe, spontaneously "changed phase" to produce normal space and time, and as a result of the energy released by the phase change, was filled with very hot, very dense, matter and energy, still expanding, but no longer exponentially. Indeed, gravity then started to slow the expansion.  In some theories just a small region would undergo this phase change, in others it would be the whole universe. Either way, all of the universe we see comes from a small region of the inflating universe.


Which techniques are used to convert radio signals received by antenna to images?

I can think of at least four very different ways that "astronomers convert radio signals received by their antenna to images."

1. Scanning

This is the oldest and easiest to understand way to image. The answer to the question How did single dish (or single receiver) radio telescopes originally generate images? sums it up concisely:

  They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.

Here is an early 1D projection of scanning the plane of our Milky Way galaxy from The galactic system as a spiral nebula Oort, J. H.; Kerr, F. J.; Westerhout, G.  MNRAS 118, (1958) p. 379. This image is reversed tone for better clarity.

For more information see answer(s) to Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)

2. Interferometry (most common technique today)

If you receive an incoming wavefront from two or more different locations, which in this case means two or more antennas, you can use math to reconstruct an image. This is called interferometric imaging and these days it's done by feeding the signals from each antenna to a large, fast computer known as a correlator. However in the early days of interferometric imaging the interference was actually done electronically, with amplified signals from each antenna combined in a mixer and other electronics.

In fact, one of the earliest techniques of radio interferometry used a single antenna which collected both a direct signal from the source and a signal taking a longer path by reflecting off of water! This is called Sea interferometry and in this case the water acts as a Lloyd's Mirror. This is one dimensional imaging only.

Read more about this in Radio Astronomy at Dover Heights as well as in the paper Galactic Radiation at Radio Requencies V. The Sea Interferometer J. G. Bolton and O. B. Slee, Australian Journal of Physics, vol. 6, p.420, December 1953.

above: Very Large Array (Source)

You may have seen images of (material around) a black hole recently. In this case the signals from each antenna were recorded on to hard drives, which were then shipped to a central location before being put into a computer correlator, but mathematically it is the same process. See answer(s) to How does the Event Horizon Telescope implement the interferometry?

3. Direct imaging using focal plane arrays

The question What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE? shows the first image below, of the Parkes 21cm Multibeam Receiver and seems to be dated 1997. In this answer links to First image with Apertif: a new life for the Westerbork radio telescope which is shown in the second image.

4. Delay-Doppler radar

This is the coolest method! It is based on radar, which means that from one dish astronomers beam a carefully modulated radio signal towards a planet or asteroid, and receive the reflected beam in a separate dish. This kind of radar is called delay-doppler because it extracts two kinds of information from the returned signal, the total delay time giving distances, and the doppler shift caused by the velocity of the surface doing the reflecting.

If the body is rotating, different areas will be moving towards or away from us depending on their location, and different parts will be closer or farther away from us, also depending on location. If they receive enough signals over a long enough period of time, they can reconstruct a hypothetical image of the shape of the object.

See The Planetary Society's Senior Editor and Planetary Evangelist Emily Lakdawalla's How radio telescopes get "images" of asteroids and this answer for more on this technique.

Left: A very slow GIF Source.

Right: A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. Here is an animation that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From here. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?

Note: These two images do not necessarily have the same orientation.


What would the night sky look like if the interstellar medium didn't exist to absorb or block light?

The main effect would be that the Milky Way would become much more prominent and asymmetric.

At the moment, our view into the Galactic plane is limited to around 1000-3000 parsecs by dust. If you look at the Galactic latitude distribution of naked eye (Aren't there more naked-eye-visible stars in the Milky Way plane? ) you see that most naked eye stars are much closer than this and there is only a modest concentration towards the Galactic plane.

What I think this means is that there would not be a big increase in the numbers of resolved naked eye stars. It is the sensitivity of our eyes, rather than dust that limits how far we can see them. However, the numbers of unresolved stars in the Milky Way would be greatly increased. The number of stars in the resolution element in your eye would increase as distance squared, but the light received decreases as distance squared. This means each "shell" you add contributes equally to the observed brightness.

The Sun is in a 30000pc diameter Galactic disc, about 8000pc from the centre. Thus I estimate that going from being able to see stars to 2000pc, seeing them to anywhere from around 7000pc (away from the Galactic centre) to 23000pc (through the Galactic centre) will increase the brightness of the Milky Way by factors of 3-10, strongly concentrated towards the Galactic plane (more so than now), depending on which way you look. This asymmetry will be increased by the increasing stellar density towards the Galactic centre.

In addition, the "bulge" would become more prominent. This is a pseudo-spherical region of diameter 4000pc, with a greater density of stars, centred on the Galactic centre. The bulge is mainly hidden by dust now, but without dust we would see a brighter, roughly circular blob towards the Galactic centre (Saggitarius), with an angular diameter of 30 degrees.

Some idea of the change in the visual appearance to the Milky Way can be gained by comparing an all-sky image at visible wavelengths (affected by dust) with a 2MASS infrared version (much less affected by dust). See how the "width" of the Milky Way becomes more concentrated and the distribution of stars becomes smoother, with a prominent bulge at the Galactic centre.


Conversion from Equatorial Coordinate to Horizon Coordinates

It depends from which celestial coordinates you are converting. I suppose you are working in equatorial coordinates. Then, you can convert as following:

$$\tan A = {\sin h \over \cos h \sin\phi_o - \tan\delta \cos\phi_o} $$ $$ \begin{cases}
 \cos a \sin A = \cos\delta \sin h \\
 \cos a \cos A =  \cos\delta \cos h \sin\phi_o - \sin\delta \cos\phi_o
\end{cases}$$ $$\sin a = \sin\phi_o \sin\delta + \cos\phi_o \cos\delta \cos h$$

Where $A$ is azimuth, $a$ altitude, $\alpha$ right ascension, $\delta$ declination, $h$ hour angle and $\phi_0$ the observer's latitude.

Source: Wikipedia


One criterion that works well is the presence or absence of smaller craters inside. If the crater is clean, it's probably new. If it's full of smaller craters, it's old. This applies to planets and satellites with no atmosphere, or with a rarefied atmosphere.