How to find how many degrees from the beginning to the end of the sunset? According to what I what told here there is 0.5 degrees from the beginning the end of the sunset.

Unfortunately I didn't understand two things:

  1) How can I prove in the simplest way that there's 0.5 degrees only
  from the beginning to the end of the sunset.

  2) Why it is not adjust to my past observation on these phases (which
  was much longer (range from 5-7 minutes). which makes it between 1-2
  degrees.

The easiest way to measure the sun would be to use a pin-hole camera.

Use a piece of card with a pin hole. Hold it up so the sun shines through the pin hole and onto a piece of paper.  (Don't look at the sun through the hole - eye damage)

You will see a circle of light on the paper, this is the image of the sun. If the paper is 100cm from the card and the size of the image of the sun is $x$ cm, then you can work out the angular size of the sun.  It is about $\tan^{-1}(x/100)$ which is approximately $\frac{180x}{100\pi}$ You should get an angle of about 0.5 degrees.


Why is solar corona in visible light filamentary? There is a video on YouTube (unfortunately in Czech) where Miloslav Druckmüller talks about structure of the solar corona in visible light. He explains that the filaments, clearly visible on his photographs, delineate magnetic field lines - the light coming from photosphere is scattered by free electrons which move along magnetic field lines.

What I don't understand (and he doesn't explain) how this explains the filamentary structure.
Why is the scattered light visible only along some field lines and not in between them?

The coronal plasma is not spatially homogeneous. It preferentially exists inside heated magnetic structures, which take the form of filamentary loops.
It is better to imagine these structures as spiralling ropes of magnetic flux rather than individual field lines. They then act as tubes that the heated coronal plasma can fill.


The creation of comets How are comet and meteors made, and what causes them to travel through space so fast? I have tried researching, but I decided this was the faster way.

Comet Formation
From Altweg et al., Fred Whipple first came up with the idea of comets being "dirty snowballs" and Oort came up with the idea that comets originated in a very distant cloud around the Sun (now aptly called the Oort cloud).  Matter in the cloud gradually came together through gravity and other processes to form comets.  Now we know some comets also come from the Kuiper belt.
Comets are incredibly important in astronomy because they contain pristine matter from the early formation of the galaxy through the accretion of the protoplanetary disk, as shown in this illustration, also from Altweg et al.

More recent space mission analysis of comets has caused the community to more accurately refer to comets as "icy dirtballs."
Asteroid Formation
Asteroids are not as old as comets in some sense since, according to Raymond and Nesvorny:

The asteroid belt was dynamically shaped during and after planet formation.

The asteroids can differ much more significantly from one another than the comets.  Raymond and Nesvorny go on to say:

A number of processes may have implanted asteroids from different
regions of the Solar System, dynamically removed them, and excited
their orbits. During the gaseous disk phase these include the effects
of giant planet growth and migration and sweeping secular resonances.
After the gaseous disk phase these include scattering from resident
planetary embryos, chaos in the giant planets' orbits, the giant
planet instability, and long-term dynamical evolution.

Here is a cartoon from their paper comparing and contrasting three competing solar system development models for the creation of the asteroid belt:

Why do comets and meteors move so fast?
Comets are very slow moving until they get bumped into eccentric orbits with the Sun.  When we see them from Earth, they are moving at near their fastest speeds as they pass closest to the Sun, as an example of Kepler's 2nd law.
Meteors and meteorites are mostly moving around the sun at similar speeds as Earth before colliding, but in somewhat different directions.  This makes it appear that they are travelling very fast indeed with respect to the Earth as they enter the atmosphere.  In addition, they are sped up by Earth's gravitational pull.


Is this the best non-radio image of whatever's at the center of M87? How was it taken? The BBC News article First ever black hole image released shows the instantly-iconic radio image of the (almost certainly) supermassive black hole at the center of the M87 galaxy, just released by the Event Horizon Telescope collaboration, also shown below.

Though it calls it a "first image", the article also shows a similarly edible-looking† donut-shaped image taken earlier, presumably using a different technology. However the caption doesn't explain what that is.

†https://i.stack.imgur.com/J5TEe.gif

Does anyone recognize the earlier image and/or know what imaging technology and wavelengths were used to construct it? Is it roughly the same scale? Are these comparable donut-hole images?

  Astronomers have suspected that the M87 galaxy has a supermassive black hole at its heart from false colour images such as this one. The dark centre is not a black hole but indicates that stars are densely packed and fast moving. Credit: DR JEAN LORRE/SCIENCE PHOTO LIBRARY

  The first ever picture of a black hole: It's surrounded by a halo of bright gas. Credit: EHT

Overly entheusiastic Wikipedia edit?

Your top image is this from Dr Jean Lorre stock science images.  Just a photo showing stars and the optical jet with a 22 MB CCD camera through a small telescope.  It is unfortunate that he chose a false color brightness scale with black at the top intensity.  But, that black region does not correspond to the true blackness in the interferometric image which has resolution 100,000 times higher.

For comparison, here's a False-col. optical image of spiral galaxy NGC 1097 by the same "artist", also with the dark center where the visible light intensity is actually a maximum.


on/off filter and their photometric calibration I am reading a paper here

What is off band filter for? If we have already known the positions of the sources we are interested in, then off band filter is useless, correct? So an off band filter is to discovery sources?
An off band filter is usually self-designed? There is no unified standard(central wavelength, band width, etc. )?
For the photometric calibration, when is continuum needed? Since there are standard stars, why is continuum needed?

Thank you. My questions are long, so I put them here.
We assume the continuum is flat, right? The off-band filter is for 1) photometric calibration, 2) discovering new sources?
How to do the photometry then? When do we use continuum and when do we use a standard star?

The on- vs. off-band filter is to let you easily measure the strength of an absorption or emission line in the object's spectrum.   "Off-band" doesn't refer to a shift in position, but rather in the wavelength that the filter lets through. It is "off" the wavelength of a particular spectral line.
This illustrates the idea:

(from Michael Richmond's page here, used under CC-BY-NC-SA 2.0 license).
In this case, the light through the filter shown in red would be affected by the H$\alpha$ absorption line in the star's spectrum, while the light through the filter shown in green would not.  So the ratio of the two would give you a very good measurement of how strong the line is, without worrying about how bright the star is overall, i.e. what the overall continuum level is.
In the paper you're reading, they would be looking for emission lines from planetary nebulae, but the idea is the same - the on-band filter measures line+continuum, and the off-band filter only measures continuum, so it lets you correct for the continuum emission and measure only the line strength.
The design of the filters would depend on which lines you're looking for. In this case they are targeting the [O III] line which is normally at a wavelength of 5007 Angstroms, but because they are looking at distant objects, the emission is red-shifted by the expansion of the universe, so they use a custom filter whose wavelength is shifted to match the red-shifted wavelength of that emission line for this particular group of galaxies.


Volcano activity on moon Are the dark grey areas on the moon formed from the last lava flows from volcanoes? And that is why there are hardly any craters in those dark grey areas because the lava has covered the old craters?And If so are there still active volcanoes that are still flowing to this day?

The grey areas are known as "maria" (singular:"mare") and they were formed from flood  basalts. These are a type of volcano and they have formed on earth too, examples include the lake eruption on Iceland, and the Deccan traps, that played a role in the mass extinction 65 million years ago.

On the moon, the flood basalt is old. About 3-4 billion years old, with some regions probably younger, but nothing less than 1 billion. Volcanism is driven by the heat of radioactive decay. On the moon there are no longer enough radioactive elements to melt rock, and so the moon is no longer a volcanic active body.

There are no active volcanoes on the moon, and there is no prospect of any in the future.


Guess inclination angle from radial velocity measurements I came upon this question where the situation is I have radial velocity measurements of a star with two circular orbiting exoplanets and the question is how to find out if the inclination angles of the planets are close to 90°.

From what I understood this question could be resolved using a transit method.

But I don't see how inclination angle can be guessed through radial velocity measurements when they are close to 90°. What difference does it make on radial velocities ?

Update:
I think that the misunderstanding mentioned in the answers was really relevant, indeed in the scope of this problem an inclination of 90° the three points (Star, Exoplanet, observer) are aligned. Which means that a transit can be observed.

Radial velocity is a measurement in the change of speed of the star towards us, due to the planet's rotation around it. Speed variation change is detected thanks to the Doppler spectroscopy.

If the planets rotation plan is orthogonal with our direction of view, then we cannot see the radial velocity change, because the star speed towards us is constant. This would mean the planets you are talking about in your assignment were discovered using another method.

Here is a picture to illustrate.

By User:Zhatt - Own work, Public Domain, Link

In the above, the star speed changes in the plan that is 90° with our own, so there is no radial velocity variation we can detect with Doppler spectroscopy.

edit:
In your case, you are not telling if inclination is 90° to the line-of-sight or to the line perpendicular to the line-of-sight. If it is 90° to the line perpendicular to the line-of-sight (which is the usual in astronomy), then speed changes are maximum.


Does mass create space? Imagine two cubes each with the side 1 light year. One of the cubes is empty of any matter. The other cube contains a massive star.

Does the cube which contains the massive star have a larger volume than the empty cube? (Because the mass makes space curvy)
If the star is replaced by a gas cloud of the same mass but which is evenly distributed inside the cube, does the volume of the cube change?
Is it not enough with geometry to calculate volumes? Is knowledge of the mass (and its distribution?) also necessary?
Does mass (concentrations) create new space volume, or does it just change the properties of the space around it?

This is a tough question to answer because the dimensions of your proverbial cube would be affected by the mass inside (as dimensions can only exist in space, so anything that affects space will also bend your cube), and mass does not bend space, it bends spacetime. You have to keep in mind that when talking about special and general relativity, you are talking about four-dimensional spacetime. This is what is curved by mass. For example, due to the curvature of spacetime due to Earth's gravity, you feel the same effects on its surface that you would if you were accelerating upwards at 9.8 m/s^2 through deep space (where we assume there is no gravity). This doesn't mean the physical space is curved - if you draw a straight line on a piece of paper in space and then travel down to Earth's surface, it will remain straight. Instead, spacetime is curved, which can affect objects' paths of motion, but not the objects themselves. Now to your questions:

The volume inside of the cubes would be exactly the same, but two identical objects flying through each cube would follow two different paths, and if either looked at the other, it would see that the other's clock is moving at a different speed than its own due to time dilation (due to the curvature of spacetime towards mass, time moves more slowly closer to massive objects). We can see evidence of the effect of gravity on paths with gravitational lensing. Light travels in a straight line through spacetime, but when looking at the space around a star, you will actually see objects that should be directly behind the star and blocked from sight. This is because light rays that, in the absence of gravity, would pass next to the star and continue off at an angle are pulled towards the star by it's gravitational field (in reality they're just following the curvature of spacetime around the massive star), so they curve around the star, becoming visible to us, even though, in space alone, they should be blocked by the star.
The distribution of the mass has no effect on the gravitational field around it (same with electromagnetic fields). If you look at any equation that has to do with gravity (gravitational potential energy, gravitational force, etc.), you will find a term for the total mass of the object, but unless you are within the bounds of the control volume containing the mass, the distribution does not matter. This is one of the reasons why black holes are so hard to study - we can tell how much mass is inside of them, but without the ability to see inside, we have no other way of detecting their properties.
Like I said above, the volume is not affected because space is not warped by gravity, only spacetime. If you made a physical box (or we'll say a cube frame) around empty space and then moved it to a space that contained a star, it would remain the same size and shape (assuming that it was strong enough to withstand the force of gravity pulling it towards the massive star).
Neither, it simply curves spacetime.


Any programme that shows actual planet information? I don't know if this question should be here, if it shouldn't just let me know.

My question is, I'm programming in Java and I want to get the actual stats of the planet, like orbit ,actual temperature etc... Is there any method, package in any language that makes this?

This seems like something that would best be provided by a simple data table which you can open as needed rather than as a feature of the language.  There's one at https://nssdc.gsfc.nasa.gov/planetary/factsheet/ that has the basic statistics for the planets (plus Pluto); is that the sort of thing you're after?


Are black holes really black? I don’t think that black holes are really black as they are the only ones who have the capacity to bend light. At the centre of the black hole the laws of physics are totally defied, black holes appear black as space is black in colour.

Black holes really look black.

Things look black because they don't reflect or emit light. A black hole is a region of spacetime which light can enter, but can't leave. Black holes don't reflect or emit light, so they look black.

Black holes are not the only things that bend light. Every object with mass bends light. Black holes bend light more than other objects because there is so much mass in such a small region.

Black holes don't have a surface, so they aren't black for the same reason that soot is black.  They are black because they don't reflect or emit light. Space also appears black because empty space does not refect or emit light.

"At the centre" is a difficult idea for black holes, because space and time are bent so much near the black hole that the centre becomes a "when" not a "where".  The laws of physics aren't defied, but we don't know what the laws are at the time in the centre of a black hole.

(Hawking radiation is predicted to be emitted by black holes. This makes black holes emit light, but for a large black hole Hawking radiation is very low energy and very dim.  It is theoretical, it has never been detected, but the theory is very strong and well supported)

(Around the outside of the black hole matter may be falling in. This matter gets very very hot, and so shines very brightly. This means we can see where a black hole is, because we can see the matter that is falling into the black hole)


Electron Degeneracy Pressure and the Pauli Exclusion Principle I have read that what keeps white dwarfs from gravitational collapse is electron degeneracy pressure. How does this pressure prevent further collapse, and how is it related to the Pauli Exclusion Principle?

Basically, the Pauli exclusion principle says that two fermions (in this case, electrons) can't be in the same quantum state. To expand: No two electrons in an atom can share the same numbers for their four quantum numbers, properties that help describe the state of a particle. What are quantum numbers? The important consequence here is that no two electrons can have the same spin and energy level.

In a white dwarf or a neutron star, the fermions are packed very close together, and there's quite a lot of force due to gravity. However, the exclusion principle triumphs. Fermions near each other must have different energy levels; this leads to energy differences and degeneracy pressure pressure, which counteracts the force of gravity. Above a certain mass limit (the Chandrasekhar limit, roughly $\sim1.40M_{\odot}$), electron degeneracy pressure is no longer sufficient; the white dwarf collapses to a neutron star. There appears to be a similar limit for neutron stars, where neutron degeneracy pressure cannot support the remnant against gravity, and it collapses into a black hole.


Will the TESS telescope be much better than the Kepler telescope? I would like to know please whether or not the TESS telescope will be much better than the Kepler one.

Thanks.

Rather than one being better than the other, I would say that they have different targets.

Whereas Kepler focused on a specific area of the sky, TESS will cover an area 350 times bigger.

Also, most of the exoplanets discovered by Kepler are far away; TESS will focus on stars within 300 light years away.

For more info, feel free to watch this video I recently made: https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&v=7bsxlG-rh1E


What is this planet or satellite in this picture? I took this photo in Shijiazhuang, Hebei Province, China, 38°13'N 115°18'E
(+38.23, +115.31) on the night of February 8, 2020. Here you can see a round object above the moon, and it was moving fast,the first photo was shot on 19:32:24, and the second was shot on 19:31:46. Can you please tell me what it was? Now that I have checked on SkyView Free, it seemed to be in the location of Pollux, but I don't think it was because it was big and it seemed like a planet which did not glow.

As in this question about an image of a partial solar eclipse, the object in question is:

round and about half the apparent size of the Moon
fainter and slightly bluish
directly opposite the Moon relative to the center of the image

I believe it is a reflection of the Moon from coated surfaces in the camera lens.


What does it mean for a black hole to "evaporate"? It says in this Wikipedia's article:

  Entropy, however, implies heat and therefore temperature. The loss of energy also suggests that black holes do not last forever, but rather "evaporate" slowly.

Would a black hole, given enough time, dissapear? Has this happened before?

It's due to Hawking radiation, a black body radiation due to quantum effects near the event horizon of black holes and named after Stephen Hawking that predicted this phenomena. From the same Wikipedia article:

  Hawking radiation reduces the mass and the energy of the black hole
  and is therefore also known as black hole evaporation. Because of
  this, black holes that lose more mass than they gain through other
  means are expected to shrink and ultimately vanish. Micro black holes
  (MBHs) are predicted to be larger net emitters of radiation than
  larger black holes and should shrink and dissipate faster.

I don't think it was ever observed to actually lower mass of black holes that we observe as astronomy phenomena, the process would simply take too long for any measurable effects to manifest, but scientists did try to observe it experimentally. In September 2010, Belgiorno et al. claimed they've:

  ... produced Hawking radiation by firing an intense laser pulse through
  a so-called nonlinear material, that is one in which the light itself
  changes the refractive index of the medium.

But this is yet to be reproduced and confirmed by experiments of others.


What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies? According to the NASA JPL web-report, Gravity Gradients Frame Oceanus Procellarum, a rift-like ring structure surrounding the Moon's Oceanus Procellarum has been detected as a gravity anomaly by the GRAIL scientific mission (as can be as the dark blue features on the image below).

Image credit: NASA/Colorado School of Mines/MIT/GSFC/Scientific Visualization Studio

What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?

Source: JPL Article

"The gravity anomalies along the border structures are interpreted as ancient, solidified, lava-flooded rifts that are now buried beneath the surface of the dark volcanic plains, or maria, on the near side of the moon."

Because of the lava under Oceanus Procellarum, it's heavier/massive so it has a stronger gravitational pull than the area surrounding it.

Hope that helps!


Periodic error correction in automatic telescopes I am starting a study on automatic telescopes and the correction of their periodic error.
I found two ways of correcting it : with a CCD camera and manually by recording the error. I was wondering if it could be possible to model the error by a periodic signal in order to add it to the entry signal then correct it with a low-pass filter or an other filter ?

Quoting the most relevant parts from the Richard McDonald's Wiki on Astrophotography Mounts: Periodic Error Correction:

  Periodic Error can be reduced to an acceptable level using a variety
  of techniques, only some of which are in the range of a beginner or
  mid-level astrophotographer.

  Throw Money at the ProblemVery high-end mounts for astrophotography have very small periodic error because of the time
  and money spent on manufacturing high-precision gears. You can also
  buy higher-precision gear upgrades for some mounts. For the beginner,
  let's call this impractical. Buying a \$1000 replacement worm gear for
  your \$1000 mount is probably not a good balance.
  Don't Use GearsThis is really an extreme case of the "throw money" solution, mentioned just for fun. There are some
  very-high-end experimental drive systems showing up on the market that
  don't use worm gears, and that don't exhibit periodic error. Examples
  include direct drive systems and harmonic drives. 'Way out of our
  price range.
  AutoguidingA second guide camera and computer can be used to make frequent small corrections to the mount's pointing,
  and this can reduce or eliminate Periodic Error if the onset of the
  error is not too sudden. This is the subject of a separate article.
  Periodic Error Correction FeatureTraining PEC with hand-control Most mounts intended for astrophotography include a
  feature called Periodic Error Correction (PEC), which can be used
  alone, or in conjunction with Autoguiding. PEC is used in two phases:

  Training. During this phase you use the control panel or menus to say "Hey! Pay attention to this!" to the mount, then you manually
  keep a star perfectly centred for several worm periods. You do this by
  centering a star with a high-magnification eyepiece that includes a
  cross-hair reticle. You then stare at the star for 10 to 15 minutes
  and use the mount's hand controller to manually make the small
  adjustments necessary to keep it perfectly centred on the cross hair.
  The mount records the error corrections you supply, remembering where
  in the worm position each one was needed. By training through more
  than one worm cycle, the mount can record an average correction, in
  case you reacted slowly or over-reacted.
  Playback. Once you have trained the mount, you can turn on PEC. The mount will "play back" the recorded error correction information
  by slightly changing the speed of the Right Ascension drive to move
  ahead or back each time your manual error corrections did the same.
  This will cancel the periodic error, resulting in a smoother track.

  Training PEC with
  hand-control: Periodic Error Correction (PEC) button on the control
  panel of an equatorial mount.

  The manual training phase is quite tedious and a more modern
  alternative is to use a camera and computer - usually the same one you
  will use for auto-guiding - to track a star while the mount records
  the training information.

  There is even specialized software available to help collect Periodic
  Error data, analyze and smooth it, and upload it to the mount. You do
  not need such software, as the manual techniques mentioned above will
  work just fine. However, it makes a tedious job simple and pleasant,
  and you may find it a worthwhile investment. I use PEMPRO and am
  very impressed with it. It's not free, but it is inexpensive and works
  very well (and includes another feature to help achieve perfect polar
  alignment).

Quoted text and photograph source, copyright and courtesy of Richard McDonald, no copyright infringement intended. I would also highly recommend reading same author's extensive article on Autoguiding.


Wouldn't be "dark matter" better called "massless gravity"? In spite of being a simple words game, that would make me feel less uncomfortable, as there could be a mysterious field still less mysterious than matter that passes through itself etc. Moreover, couldn't be that there exist some high order terms that play a role only when mass reach big values, i. e. those comparable to a galactic mass? How/why this cannot be the case in general relativity?

I don't think the title of your question and your question are really in sync here and I'm not sure I will correctly answer your question. However, I will give it  a try.

To give you a hint of the conclusion. My answer would be, "NO".

There are several clues / reasons why I come to this answer.

Gravity is not necessarily a real thing. When you read the literature concerning general relativity about gravity you will learn that acceleration can be experienced as gravity and in the fabric of space/time gravity is "merely" a curvature in this fabric that. This means that gravity is highly debatable as something that could be seen as an independent force / object in the universe.
When gravity exists as a force, whether it is an exchange of gravitons between two objects with a mass, or whether it is a curvature in space / time due to the mass of an object, the phenomenon "gravity" is always related to the mass of an object and doesn't stand on its own as an independent force of nature.
Dark matter is not a "mysterious" thing that passes through itself. Dark matter is just the container concept of matter out there that doesn't interact a lot with other matter, but just leaves its marks in the universe by showing a gravitational force as interaction with other matter. Since we have only little understanding of gravity and we still have difficulty measuring gravity as a force in the universe, we have difficulty understanding and measuring dark matter directly. This doesn't mean it is not out there.

So, to get back to your second part of the question, "massless gravity", this is not a viable alternative. Gravity exists either by massive objects exchanging gravitons and therefore exercizing gravity upon each other, or a massive object "denting" the fabric of space / time and therefore creating a gravity field. In either case gravity cannot be seen as a farce that exists without the existence of mass.

Secondly, to the first part of your question. Whether we can observe objects directly or indirectly, this doesn't make it more or less realistic. Dark matter is generally accepted to be there. We just haven't found a way to measure it. We can only see that the visible matter in the universe can't explain the gravitational pull on the visible matter and therefore we assume there is some other matter out there we haven't been able to "see". Hence, dark matter.

However, dark matter does interact through gravity and our current successes in measuring gravitational waves might also result in a better understanding of gravity and how to measure the effect of dark matter on the gravitational pull in the universe. So, to answer your question: "No, we call dark matter because we can only observe it through its influence on other matter due to gravity. There is no such thing as gravity without mass."


Could the James Webb Space Telescope detect biosignals on exoplanets? The James Webb Space Telescope (JWST) which is slated to launch in 2018 will bring us a better view of exoplanets, but will it be sufficient to detect signs of life on at least some of those worlds?

A good answer will provide an estimate of the James Webb Space Telescope's capabilities to detect atmospheric data relevant to the presence of life.

The preferred answer would also include an analysis of what it would take to say with a decent level of confidence that life was present on an exoplanet and whether or not JWST will be capable of this.

From what I understand, James Webb, if used in conjunction with a successful starshade (being developed at MIT), should be able to detect close in planets orbiting nearby stars. However, getting good atmospheric spectra of these planets directly (from planet's blackbody IR emission) is unlikely. What we must hope for is that TESS, which should be going up in 2017, will find a few nearby stars with transiting planets. Then, James Webb will be able to look for atmospheric absorption lines from stellar light passing through a planet's atmosphere during transit. This method may still be limited to large (Jupiter size) planets.
In an ideal situation (say looking at absorption lines of a super-earth) there are many "bio-signatures", but one of the easiest to detect would be an ozone line in the infrared. By itself, this would not be proof, even though there would need to be a constant replenishing source of O2 in the atmosphere to maintain the O3. If methane could also be found we could rightly get VERY excited since methane and oxygen don't co-exist very well.


If sun steals comets from other stars, then what is the primary source of comets? I heard some theory that sun may steal materials such as comets from neighbouring stars, is it true? But if is true, what is the primary source of those comets? Especially "who" creates them instead of steals them? And which stars are "willing" to bear our sun steals materials from it?

The primary source of comets for our solar system comes from the Oort Cloud, a smaller amount coming from the Kuiper belt. The Oort cloud is thought to have originated from the remnants of a proto-planetary disk. This paragraph explains this better:

  The Oort cloud is thought to be a remnant of the original
  protoplanetary disc that formed around the Sun approximately 4.6
  billion years ago. The most widely accepted hypothesis is that the
  Oort cloud's objects initially coalesced much closer to the Sun as
  part of the same process that formed the planets and minor planets,
  but that gravitational interaction with young gas giants such as
  Jupiter ejected the objects into extremely long elliptic or parabolic
  orbits. Recent research has been cited by NASA hypothesizing
  that a large number of Oort cloud objects are the product of an
  exchange of materials between the Sun and its sibling stars as they
  formed and drifted apart, and it is suggested that many — possibly the
  majority of — Oort cloud objects were not formed in close proximity to
  the Sun Simulations of the evolution of the Oort cloud from the
  beginnings of the Solar System to the present suggest that the cloud's
  mass peaked around 800 million years after formation, as the pace of
  accretion and collision slowed and depletion began to overtake supply.


How to retrieve from NASA Horizons site the needed data to plot analemmas for other planets? I posted such a question in space.stackexchange ( https://space.stackexchange.com/questions/46077/which-are-the-correct-input-parameters-for-nasa-horizons-query-to-get-the-right?noredirect=1#comment150190_46077 ), but probably it fits better here, being it a "topocentric problem", although for other planets than Earth; we could call it "eso-astronomy" ;-)
I am trying to figure out how to retrieve the right data from Nasa Horizons to plot sun analemmas as seen from other worlds; I found a query which works fine for Earth, but upon changing the observer location from Earth to other planets, things get weird, and I get wrong curves w.r.t "official" curves (although there is not a single analemma curve for each planet, but different curves for different times).
This is the query I am using for Earth analemma centered on Greenwich at 12.00:
Result:

Using same coordinates on mars, a longer period, and 1477 minutes as interval (martian day length), I don't get the right plot (link):

The right analemma on Mars is this:

I noticed that Earth analemmas are plotted against Azimuth, but Mars analemmas are plotted against Equation of Time. Why? And how can I do by myself such a plot in Excel?

You did two things wrong, one minor, the other major. The minor thing that you did wrong was to chose the solar system barycenter as the target rather than the Sun. Use the Sun.
The major thing you did wrong was to use a step size of 1477 minutes. That is (to within a minute) the length of a Mars sidereal day. The Sun will march across the horizon if you use a sidereal day. What you want is a Mars solar day, which is 1479 minutes and 35 seconds long.
This presents a new problem. Ideally your step size should be 88775 seconds. But you cannot choose seconds as the unit of a step size. You're stuck with minutes, and that is suboptimal. There is another approach:

Choose the "equal intervals (unitless)" option for the step size,
Make the time difference between the start and end time be an integral multiple of 88775 seconds, and
Specify the step size as the integral multiple chosen to determine the time span.

For example, choosing 711 as the integral multiple and 2020-Aug-20 12:00:00 as the start time means the end time should be 2022-Aug-21 01:03:45. You might need to tweet the end time to get the analemma to close up properly.

Edit
I noticed that you used the center of Mars. You need a point on the surface to get a proper analemma. Horizons provides several predefined points. If you choose Viking 1 / Chryse @499, you'll have to modify the start and end times so that the Sun is above the horizon. Add 12 hours to the start and end times shown above appears to work very nicely for Viking 1's location.
Below is what I get plotting the Sun's elevation vs azimuth as seen from the Viking 1 / Chryse location, with 711 intervals (712 data points) sampled over the period 2020-Aug-21 00:00:00 to 2022-Aug-21 13:03:35 UT.


Can we apply Kepler's law of planetary motion to the moons motion around the earth? I have been trying to understand Kepler's law of planetary motion for a while. So I was thinking, if it would even apply to the moon orbiting around the earth. Does the moon too orbit in an ellipse too? Does its orbit have the same properties as the earth's orbit around the sun?

Yes. Kepler's laws apply to orbits of the Earth, and the moon's orbit is elliptical, with an eccentricity of 0.055

Or at least nearly elliptical. In Newton's theory of gravity, if there are two particles in orbit around each other then they will have an elliptical orbit. If there are more than two bodies then the gravity of other bodies will perturb the orbits from exactly elliptical. A more accurate description of the moon's orbit is "Elliptical + perturbations"

The most significant perturbations come from the sun, and from the fact that the Earth is not a perfect sphere. Other planets and tides also have effects on the Moon's orbit.

There are two wiki pages you should read: Orbit of the moon and Lunar Theory


Is the curvature of Mars visible from the top of Olympus Mons? Would you be able to see the curvature of Mars from the top of Olympus Mons? And how much curvature would you see if it was visible, would it be only barely detectable, or would you see a lot of curvature?

Olympus Mons is 25km high and the tallest known mountain in the Solar System.

Mars is just over half the size of Earth.

Here on Earth, the curvature of the planet is hardly visible from the height of commerical aeroplane (roughly 10km).  I was wondering, considering the height of Olympus Mons and the size difference of Mars, if the curvature from up there would be very pronounced.

The impression I get from the Wikipedia page on Olympus Mons is that from the top of the volcano, you can't even see parts of Mars that aren't Olympus Mons. Quoted from above:

Because of the size of Olympus Mons and its shallow slopes, an observer standing on the Martian surface would be unable to view the entire profile of the volcano, even from a great distance. The curvature of the planet and the volcano itself would obscure such a synoptic view. Similarly, an observer near the summit would be unaware of standing on a very high mountain, as the slope of the volcano would extend far beyond the horizon, a mere 3 kilometers away.

Emphasis mine.
Wikipedia sources the bold line to this source in the Way Back Machine, which works out the math. Near the bottom of the page, there is a section starting with the phrase:

Years ago someone asked me if a person could see the horizon of Mars from the top of Olympus Mons. That is not as simple as one may think. If Mars was a prefect sphere and the slopes of Olympus Mons were smooth then the distance to the horizon could be calculated fairly easy. However, the surface of Mars has hills and valleys, and Tharsis bulges out from the mean spherical surface of Mars several miles, so one must consider that as well.

That's pretty much your question. The conclusion is:

Second, if this same [6 foot tall] person stood at the top of Olympus Mons (88704 ft) we would find that he or she could see a lot further to the horizon, some 265 miles away, if nothing was in the way:
$\cos A = (11,121,762.53 / (11,121,762.53 + 88704 + 6)$
$A = 7.2^\circ = 0.126\ \textrm{radians}  \leftarrow (0.126 \times 11,121,762.53 ft) / 5280 = 265 \textrm{miles}$
At those angles a person 6 feet high would be looking down at an angle of 7.2 degrees in order to see the horizon of Mars, so they would surely look down into the hump positioned at 50 miles away or at the 2.5-degree grade of the slope instead of over it to the horizon.

So it would seem that the horizon, as viewed from the top of Olympus Mons, would still be part of Olympus Mons. Indeed, you wouldn't even be able to see most of Olympus Mons from the top.


Does the twin paradox work in an almost empty universe? Background
The twin paradox is the popular thought experiment involving twins, one of whom makes a journey to a nearby star in a high-speed rocket (travelling at a velocity close to the speed of light) and when he reaches the star, changes direction and returns home at the same speed.  When he gets home, he  finds that time has slowed down during his journey and that his twin who has remained on Earth has aged more.  (The paradox being that because both twins have travelled at the same speed with respect to each other, why has one aged faster than the other?)
My understanding is that the solution is as follows. When he arrives at the star and changes direction, the travelling twin switches from a reference frame travelling away from Earth at a velocity near the speed of light to another reference frame travelling towards  Earth with a velocity of the same magnitude but in the opposite direction. In contrast the non-travelling twin stays in the same reference  frame. It is often pointed out that the travelling twin experiences an acceleration when he switches from one reference frame to another, whereas the non-travelling twin does not experience an acceleration. (This acceleration could be a Dirac delta function i.e. an instantaneous switch of direction)
Questions
My first question is what is the travelling twin’s acceleration with respect to? Since in special relativity there are no absolute reference frames. I assume it is the Earth??
Secondly, and this is the question which is intriguing me the most, suppose hypothetically there were only two objects in the Universe namely the two twins. If, in this hypothetical universe, one twin travelled on a rocket for a distance of say ten light years and then returned back to meet the original twin. Would the travelling twin’s clock  still run slow with respect to the other twin?  since there would  be no distant background stars/galaxies in this hypothetical universe  to measure the acceleration against? Is it just as valid to say that the “stationary twin” is the one in motion?
Additional comment
In a various replies to this question, it has been mentioned that the moving twin  feels an acceleration and the stationary one does not, but in this hypothetical Universe containing only the two twins I cannot see how this can be the case

Velocity is relative, but acceleration is not: you can feel it (and it feels just like gravity). But as I said in the question comments, acceleration is a red herring.
The real issue is that the travelling twin occupies two (or more) distinct reference frames. That means their worldline (their trajectory through spacetime) is not a straight line, and any inertial observer measuring that worldline will agree on that, although different observers will disagree on the lengths of the segments of that trajectory, depending on their speed relative to the traveller.
So the important factor is that the worldline has a bend in it, and so it cannot be transformed into an unbent line. It doesn't make much difference whether it's a perfectly sharp bend (your Dirac delta function), or a smooth gradual curve.

There's a variant of the twin paradox that requires no acceleration, but it has 3 participants, A, B, and C. A is based on Earth, C is based on Alpha Centauri (which we can assume is at rest relative to Earth).
B's rocket takes off. It's moving at constant speed $v$ (relative to Earth) as B passes A, and at that moment  they use radio to synchronise their clocks to zero. B then travels towards Alpha Centauri.
One year later (according to B's shipboard clock), C sees B approaching. C starts their rocket and heads towards Earth. C's rocket has a speed of $v$ relative to Alpha Centauri. (According to the relativistic law of composition of velocities, B & C have velocity relative to each other of $2v/(1+v^2)$, using natural units, where the speed of light $c=1$). Just as C passes B, C synchronises their clock to B's clock. That is, C "clones" the value of "1 year" from B's clock the instant they pass one another.
One year later, according to that clock, C arrives at Earth, and as C passes A, they compare clocks. C's clock says "2 years". A's clock says "9 years".


Are there galaxies with 2 or more super massive black holes orbiting each other? We now know that most stellar systems have 2 or more stars orbiting each other. Do we know of any galaxies which have 2 or more super massive black holes orbiting each other? Is it possible?

Yes, there are galaxies with two supermassive black holes in the center, see for instance 4C +37.11

Most likely such galaxies are formed by collision and merger of two galaxies, and their cores have not yet merged. Source


Superluminous super nova output I was reading about how powerful supernova were and found out there are even bigger supernovae also called hyper nova.  A large explosion was discovered in 2016 I believe (or was it 2017).   I would like to know on a human scale how much power that is.  Suppose an inventor produced an engine technology that has the output of the largest superluminous super nova to project a space shuttle.  How fast would the space ship be projected forward?  That is the output pushes air to project the space ship forward.  What speed would it reach.  Part of me wants to say faster than light.  I imagine the ship will travel so fast it would disintegrate since matter cannot exceed the speed of light.

This puts us in the realm of silly numbers.

A supernova has enough energy to rip a massive star apart. Putting all that energy into one small rigid body isn't going to end well.

A hypernova gives off about $10^{46}$ Joules. We will need the formula for relatavistic kinetic energy $E=mc^2(\gamma-1)$ where $\gamma$ depends on velocity. The mass of the shuttle is about $2\times 10^6$ kg, and $c^2=9\times 10^{16}$

So $10^{46} = (2\times 10^6) \times (9\times 10^{16}) (\gamma - 1)$. Rearranging, $\gamma = 5.5\times10^{22}$.

In relativity $\gamma = 1/\sqrt{1-v^2/c^2}$ So even this massive amount of energy won't make the shuttle go faster than light. It will make it go at

99.99999999999999999999999999999999999999999998%

of the speed of light.


Pictures of a curious astronomical phenomenon In a recent holiday, a friend made the following picture. It contains a curious green "thing" in the nightsky. The picture was made in Ladakh, India, 6weeks ago.

The "thing" only appeared in one of the pictures (others were made a few minutes earlier and later). He didn't see it with the naked eye (but didnt look at the sky during the exposure).

Camera: Canon EOS 350D DIGITAL
7 sec exposure
ISO: 1600
f-number: 4

What could that be?

In your friends picture are more artifacts than the one you showed in the 2nd picture a little bigger. I marked more of them in the picture below (click to enlarge).

They are all in a perfect line to the bright light. So these artifacts are caused by the bright light and the lens of the camera. A lens is not flat it's, well lenticular (it's where the name came from). The main light source is shining from an angle on the lens. When light hits a surface two things happen (at least in most cases ignoring Brewster's angle). The two things are reflection and transmission (or absorption, but we are talking about a lens so I let this one drop here). Every time the light hits the surface of the lens, a part is reflected and another part is transmitted. But the lens has a thickness so when the light that was first transmitted now hits the back of the lens, a part is transmitted to the CCD sensor of your camera and a (smaller) part is reflected back into the lens. This can happen several times. Normally this effect is small enough it will not be noticed. But this picture is not a normal case. Most of the picture (dark sky) emits no light, the picture is very dark. In contrast to this, there is a bright light source dominating everything (related to luminescence). And we have a 7 sec exposure the picture is taken with. 7 sec are enough time for the CCD sensor to capture some light from the effect described above.

If you now want to know why is it green and not white. White light includes all wave lengths (colors) and every wave length is reflected in another angle on the surface of air-lens. The green wave length just had the best angle to be reflected severals time and still reach the CCD sensor.

So the most accurate answer to your question "what is this green thing“ is: It's the flashlight (or what ever this light source is) that is seen in the bottom of the picture.


What determines the surface material of the ALMA and Spitzer telescope? The transition between metal dish and mirror happens around hundreds of microns. For example, the shortest wavelength the ALMA(metal dish) can detect is about 0.3mm, while the longest wavelength the Spitzer(optical mirror) can detect is about 160 micrometer.

My questions is what determines the surface material of the ALMA and Spitzer telescope? Could a metal dish observe shorter wavelength than a mirror telescope?

They're all mirrors, they just operate in different wavelength ranges. A wire fence could be a mirror if the wavelength is big enough.

The surface is an easy problem to solve. Pick any material and do vacuum deposition with it on the substrate. It's the substrate, the base material that matters a lot - the thick part that keeps the shape.

A metal mirror is easier to make. Standard engineering principles apply. That's why this technology is used for ALMA.

But there's a catch - precision. This is dictated by wavelength. Imagine all the radiation captured by the mirror, focused in one point. If the mirror is perfect, it's all good. If the mirror has deviations from perfection, parts of the wavefront arrive too soon or too late. The wavefront begins to interfere with itself, reducing the amplitude.

Interference is completely destructive (the signal is completely lost) when the difference between the paths is exactly half the wavelength: λ/2. Then the peak wave (maximum) from one side falls into the trough wave (minimum) from the other side, and they cancel each other.

How do you get a λ/2 difference between two parts of the wavefront? If the surface error on the mirror is λ/4. One part of the wavefront begins to be reflected immediately, while the other keeps traveling λ/4 to get to the mirror, and then wastes λ/4 again on its way back up.

Mirrors with λ/4 errors are bad - since the error is never exactly half-and-half, parts of the wavefront are not cancelled so the mirror keeps working somewhat, but poorly. Around λ/8 the mirrors become decent. λ/16 are very good, and beyond λ/20 are essentially indistinguishable from perfection. This is for optical mirrors (visible spectrum).

This is why optical telescope mirrors are made of glass. It takes such a smooth, rigid material to reach the precision necessary for optical observations.

The Spitzer mirror is actually made of beryllium, cooled to 5.5 K. This is fine. Precision requirements are a bit more relaxed, compared to visible light. In fact, solid metal mirrors have been made for visible light telescopes even by amateurs - but in these conditions glass is preferred.

The ALMA would have to be manufactured to a precision better than 75 micron, in reality much better. That's doable with regular metal parts.

Other materials used for visible light telescopes:

Granite. It takes an excellent polish, just like glass, it's very rigid, and the coefficient of expansion is similar to glass. As long as it doesn't have voids, it's a decent material.

Marble - same as above.

Speculum metal - back in the days of Herschel they were casting mirrors out of this bronze-like alloy. Easy to work with but it tarnishes very quickly.

Special ceramics such as Zerodur.

Quartz - like glass, but better (low thermal expansion). Takes more work to grind and polish because it's harder than glass.

TLDR: It's mostly about precision and ease of manufacturing.

Once the mirror is made, it could be coated with anything you like, so it could reflect any wavelength up to hard UV. Like I said, the surface, the reflective layer, is the easy part.

Simply make a list of all materials that provide good reflectivity in the intended range of wavelengths. Most of these materials turn out to be metals, unsurprisingly. Now compare them based on such criteria as: durability, price, ease of coating. Pick one and drop it in the tungsten "boat" in the vacuum chamber. Put the mirror in the vacuum chamber and run some amperage through the tungsten boat until the metal layer on the mirror is thick enough so it's not transparent anymore - but not thicker than that (for precision). Done.

An alternative that is sometimes used in infrared and visible light is dielectric coating. It's a stack of fairly transparent layers, and there are multiple reflections from all the surfaces between layers. The thickness is such that the path difference between reflections is exactly one wavelength, 1 λ. That strengthens all reflections (constructive interference).

Dielectric mirrors have extremely high reflection coefficients, and are used for lasers, etc. Also tend to have preferred wavelengths where they work well (though broadband dielectric mirrors also exist). They're quite expensive and are not commonly used for telescopes.

Finally, after all that, telescope mirrors sometimes get a protective coating to shield the reflective metal layer from corrosion. A commonly used material is silicon oxide (SiOx, not SiO2 which is quartz and it's different). If the mirror operates in a vacuum or it can be recoated often then this is less important.

The vast majority of amateur telescopes nowadays use:

borosilicate glass (e.g. Schott Supremax) for the substrate; it's a low expansion glass, cheap enough. The original Pyrex glass (not made anymore) was almost the exact same thing.
aluminum for the reflective layer; sometimes silver is used, which is slightly more reflective initially, but decays more quickly
SiOx top coat for durability

Some mirrors have "enhanced coating" which is conceptually similar to dielectric mirrors, in addition to the metal layer.


Does the Time of Sunset (Sunrise) Change Over Time Given a Fixed Location and Consistent Atmospheric Conditions? Assumming a fixed location, clear skies, and consistent atmospheric conditions, does the sunset (sunrise) time change over time?
For example, sunset on February 3, 2021 in Tel-Aviv, Israel, is indicated to be 5:17pm. This is a rounded figure I'm sure. Thus, within a minute, would this stated sunset time have been any different historically---or will it be expected to change in the future?
I know that the length of the tropical year is gradually decreasing overall---does this have any effect on the time of sunset (sunrise)?
Thank you.

Shalom! Strictly speaking, yes, the time of sunset does change, but the amount of change is imperceptible to humans over a lifetime. The change is due to the slowly varying obliquity of the Earth on its axis, the slowly varying eccentricity of Earth’s orbit around the Sun, and other such cyclic or secular changes.
Other than that, you’ll notice a slight change from one year to the next, as the year doesn’t last exactly 365 days. This is a four-year cycle, and the difference is around one minute, I would say, on average, depending on the time of the year—for example, when there is less of a difference in sunrise/sunset times from one day to the next, the difference between the same date of two successive years is lesser than for dates when there is a large difference between one day and the next.


Map projections used for star maps I'm working on a star map program and I'm wondering what are the most common map projections used to print star maps...? I want to map of the entire visible sky from a certain point on the surface of Earth. I have all the visible stars and their Altitude-Azimuth, but I need to map this semi-sphere onto a circle in 2D.

I'd like to get something like this:

I've been playing around with equidistant and gnomonic (-1 RE) projections but they feel too distorted. What are the most common projections used for 2D, circular, entire-visible-sky star maps?

The stereographic projection
(Wikipedia,
Wolfram MathWorld)
is commonly used for all-sky maps
(e.g. Skymaps.com,
Your Sky).
As distance from the center of the map increases,
constellations grow larger but keep their shapes.


Is there any theoretical or empirical research about planetary systems with a black hole in their center? I assume such a system would be hard to detect due to a lack of radiation, at least if the vicinity of the black hole does not contain a lot of mass which falls into it over time, thus creating (a lot of) radiation.

Since black hole formation in a supernova explosion is a process which destroys any planetary system around its original star, any planetary system around a black hole must have formed after black hole formation.

A few objections come to mind off-the-cuff: Because black holes have a large mass there is a minimum distance for planetary orbits below which they would disintegrate due to tidal forces. This may interfere with the usual evolution of planets (they clear their orbits and kick out interfering bodies, for example), because their "orbital space volume" becomes too large for them to keep clean, and the orbital period becomes long.

Is anything known or is there any speculation about such systems?

Since planets are known to exist around neutron stars (e.g. see Wolcznan & Frail (1992); and see the list of "pulsar planets" - technically, these were the first exoplanets ever discovered), then it is difficult to see any fundamental reason why there should not be examples orbiting black holes.

In fact, since it is probably feasible to form black holes via direct collapse, without an associated supernova then maybe the hazards for planets are not so severe.

Against this, must be set the fact that black hole progenitors are likely to be more massive than neutron star progenitors and the role of winds and radiation on the formation of planets is still not well established.

Planets orbiting neutron stars are/were found from a timing analysis of pulsar signals. The same phenomenenon is not present for black holes and direct imaging or observing a transit is not remotely feasible at present.


How did Copernicus get the idea of heliocentric model? The heliocentric model marked the beginning of modern astronomy. But what were the reasons which led Copernicus to think that the earth orbited the sun. I've read that his model was based on observations made by some other astronomer. Can anyone describe how he got this idea?

The heliocentric model is historically thought first by Aristarchus. He eventually gave up on the idea due to not being able to give evidence for a concept called stellar parallax: If earth orbits the sun then we should see the closer stars shift in relation to the other stars (strong evidence for this didn't come until Galileo).

Copernicus became to accept the heliocentric model given by Aristarchus due to his study of the Ptolemaic model. His studies showed that planetary motion in a geocentric model was inaccurate.

His model, though, was still based on the fact that "heavenly objects" moved in complete circles. He had to add much complexity to account for this fact which is why the heliocentric model was still not widely accepted. Side note, it was Johannes Kepler, over 100 years later, that claimed that planets move in ellipses rather than circles.


angle to the equator of the Sun on sky at a given time Is there an easy way to tell at what angle the solar equator is wrt the horizon at any given time of the day? I would think that it rotates 180 degrees from sunrise to sunset, but not in a linear fashion. But solar noon should always be 0 degrees wrt the horizon right?

Depends on the precision you need.. The sun's plane of rotation coincides with the plane of rotation of the solar system.
And for the latter, there is already a name in astronomy, the Ecliptic. Depending on your latitude on the planet $\delta$, the ecliptic plane cuts through your local horizon in east and west at an angle of $\alpha = 90°-\delta - 23.5°$.
This is the maximum height of the sun at midday and the sun's height varies thus between 0 and $\alpha$.

I think this would then be the angle you're looking for.

Another recent answer pointed out the angle between solar axis of rotation and the Ecliptic is $7.25°$. Translating that into the angle unter which one sees the solar equator, it is still $\alpha$, up to a small correction factor of $- \frac{R_{\odot}}{1 AU} \sin(\delta + 23.5°)$ which can be maximally $-0.05°$.


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

  describes the collective effects of changes in the Earth's movements upon its climate[.]

By this theory, the Earth is now experiencing an interglacial.

By the theory, when will the interglacial end and the next ice age begin?

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


Is it possible for a star to orbit a planet? Is it possible for a red dwarf to orbit a gas giant?

OR

Has this happened and it is just assumed that the gas giant is orbiting the star?

Pretty much by definition, no.

In order for "object A" to orbit "object B", "object B" needs to be substantially more massive than "object A".  In order for your red dwarf to orbit something, that "something" needs to be more massive than a red dwarf.  With the current composition of the universe, that "something" will be mostly hydrogen, and once you reach a certain level of mass (about 90 times that of Jupiter), you can't prevent nuclear fusion from starting.  The definition of "planet" is a bit fuzzy, but everyone agrees something lit by nuclear fusion is a star, not a planet.

This could theoretically change in the distant future, when there's enough non-hydrogen mass around for stellar-scale rocky bodies.


What's the point with Hawking Radiation? I saw that Black Holes can evaporate because of Hawking radiation.
Because pair of particle, antiparticle can spawn at any point of the space, the antiparticle could be absorbed by the black hole and the particle go outside of it, like if the black hole where "lightning".
But, if one element of the pair can be absorbed by the black hole, isn't there an equal probability of that the particle will be absorbed, too?
I could imaginate that the probability of fluctuation of going into 0 with an equal probability of +1 or -1 mass over time is nonnull, also that's why black hole could evaporate, or something like it?

I cannot show this or prove it but the main idea, in simplest terms, is that a matter-antimatter pair, near the event horizon, will form. A member of this pair will decay back into the event horizon and be absorbed by the black hole. The other particle will decay out into the free space.


Tools for creating a multiwavelength view of the sky Not sure that this is the correct place to ask this question, but here goes.

I am trying to find tools to accumulate image data (preferably in FITS format) on a given (RA, Dec)-coordinate of sky. Right now there are several image services that offer cutouts, for example:

irsa.ipac.caltech.edu holds data from Spitzer, Planck, WMAP, IRAS, WISE
hla.stsci.edu holds Hubble Space Telescope data
skyserver.sdss.org/dr7/en/tools/chart/list.asp holds SDSS data
etc.

I wonder if anyone has written any code for accumulating these images given an argument of RA and Dec? Or if they can point me in the right direction of tools that they know exist? Any help would be appreciated.

Aladin is a good way to do visualization from multiple catalogs. MAST also offers a cross-mission search.


Internet Resource which register all Ultra-long Gamma-Ray Bursts I read in a scientific magazine there was new gamma bursts that last longer than usual.

One was detected during december 2010, also known as Christmas Burst: GRB 101225A
and another one in 9th of December 2011, GRB 111209A

They were both detected by the "Swift" program and this article goes into detail.

I do not know if there is another programs for GRB detection and I would be curious to find a resource on Internet which registers/records all Ultra-long Gamma-Ray Bursts on one single page/site (across all research/programs).

I guess one of them you're referring to was GRB 130427A (not in December of 2010 or 2011 tho, but in April 2013)? But yes, there are several such catalogs, as are different designators and phenomenological classification methods for these gamma-ray bursts (GBR). For example, GRB 130427A is a simple nomenclature that consists of a date of discovery (130427 denotes 27th of April, 2013 and suffix A is just the first letter from the alphabet, increased when there's more than a single such event in a day), but it doesn't indicate e.g. its type or origin. The same GRB event is so cataloged under designator CSS130502:113233+274156 in the Catalina Real-time Transient Survey catalog, SDSS J113232.84+274155.4 in Sloan Digital Sky Survey catalog, and so on.

So, as you can see, it would help if you could clarify which designators you saw, so we can identify the catalog easier. But not wanting to leave you empty-handed, one such catalog that is frequently used and is relatively easily searchable is the Goddard Space Flight Center's GRBCAT (Gamma Ray Burst CATalog) (on top of those already mentioned).

Post update to the question, I've managed to identify SWIFT pages for the two GRB events in question:

101225A
111209A

Catalog numbers, references and data are available on linked pages. And here is a complete SWIFT GRB catalog (note, it's rather long, listing all GRB in a single page).


radial velocity fitting of a binary I have two series of velocities (HJD, $v_{1}$ and $v_{2}$ ), how to fit in detail?

There is any formula on the web? $sine$ curve is easy, but I need to consider an eccentric orbit.

Suppose we can estimate $m_{1}$, $m_{2}$, and inclination angle $\approx80^{\circ}$, if we just know velocities of both of the two companions, which parameters can me derive?

What is the difference between binary velocity fitting and star+planet fitting?

The radial velocity curve of a star in a binary system (with another star or a planet) is defined through 6 free parameters
$$V_r(t) = K\left(\cos(\omega + \nu) +e \cos \omega \right) + \gamma,$$
where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity, $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $\nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $\tau$, the orbital period $p$ and the eccentricity $e$.

To proceed you estimate what all these parameters are - i.e. an initial guess.

Then, for each time $t_i$ of a data point in your RV curve you:

Calculate the mean anomaly
$$M(t) = \frac{2\pi}{p}(t - \tau),$$
Solve "Kepler's equation"
$$M(t) = E(t) - e \sin E(t)$$
numerically (its a transcendental equation, you could use Newton-Raphson or similar) to give $M(t_i)$, the eccentric anomaly.
Use
$$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$
to calculate the true anomaly $\nu(t_i)$.
Calculate $V_r(t_i)$

You then calculate some figure of merit (e.g. chi-squared) for how closely the model and data agree and go through an iterative process to adjust the parameters and optimise the fit of model to data.

A more sophisticated discussion can be found in this paper by Beauge et al.

If you have the RV curves of both stars, then you can fit them both simultaneously. Obviously, they have $p$, $e$, $\gamma$ and $\omega$ in common, but their RV amplitudes $K_1$ and $K_2$ will be different. The ratio of $K_1/K_2$ gives you the ratio of the two stellar masses.

If you only have one RV curve you are limited to estimating the mass function of the binary system.
$$ \frac{M_2^{3} \sin^{3} i}{(M_1 + M_2)^2} = \frac{p K_1^{3}}{2\pi G},$$
where $i$ is the inclination of the orbit with respect to the line of sight.
This can only give you a lower limit to $M_2$ unless $i$ is known.

Taking your specific case study. If you know $M_1$ and $i$ (this could be the case for a transiting exoplanet, or maybe a binary featuring an eclipsed black hole candidate), then the primary radial velocity curve gives you $K_1$ and hence $M_2$. If the masses and $p$ are known then Kepler's laws give the orbital separation.

There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is Systemic Console.

There is no fundamental difference between analysing the RV curves of stars with exoplanets and stars with unseen (stellar) companions.


What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane? What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?
Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?

What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?

From this page, we can see that it is 3.38°.

  Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?

This is a bit harder to estimate precisely, but we can refer to this view from the IRAS satellite to get a rough figure of 63° between the ecliptic and the plane of the Milky Way (for more information on how to calculate that, check out Aitoff projection). Now just add/subtract Mercury's 7.01° of inclination to ecliptic and you have your result.

EDIT: Because the angles of 63 and 7 degrees may not be along the same axis, there are ways to be a bit more precise: you can take Mercury's coordinates in the ecliptic coordinates system, and transform them into galactic coordinates. If you do that with only 3 different points on Mercury's orbit, you can find the angle made by the plane defined by those 3 points and the horizontal galactic plane.

You can use this website to get your 3 sets of ecliptic coordinates for Mercury, and that website to transform them into galactic coordinates. You can transform those into regular Cartesian coordinates since the website that gave you the ecliptic coordinates also gives you the distance from Mercury to the Sun! Once you get there, it's easy to get the angle made by that plane and the horizontal: good luck!


How to find the moons maximum culmination at a location Forgive me if my terminology is incorrect, but is there someway or somewhere when I can calculate based on my location when the moon will be at the highest point in the sky overhead based on my location?  Per year for example?  I've been searching around the internet and haven't found a calculator or something that could do that.
For context.. I'm working on an architectural project and looking into if the moon will ever be visible through a certain skylight, even if it is for only a short period of time per year.  I'm located in Calgary, 51.0447° N, 114.0719° W
I'm currently using some parametric software to show the moons location within the 3D modeling software, but I can only input dates and get the location, not search for the highest point etc, so I can't really pin down when it will be the highest.
Thanks!

Let's suppose, that moon is always located on the ecliptic. Ecliptic ranges from declination $\delta=-23.5°$ to $\delta=23.5°$. That means, that the maximum apparent height of the Moon in Calgary will be $90°-\phi+23.5°=90°-51.04°+23.5°=62.5°$
But moon isn't always located on the ecliptic. Its inclination is around 5.14°. That means, that it can deviate from the ecliptic for such amount. So the maximum height changes to $62.5°+5.14°=67.6°$. But note that this won't happen every month, but rather every year.
This is just a simple geometrical problem. On figure:

Also, the culmination (the highest point) on northern hemisphere always occurs on the south.


Why did the then PhD student Didier Queloz get the Nobel prize? When Didier Queloz did the work that earned him a Nobel prize (along with Michel Mayor), he was still a PhD student.

Jocelyn Bell Burnell stated

  I believe it would demean Nobel Prizes if they were awarded to research students, except in very exceptional cases

What made this case so exceptional that he got the Nobel in spite of having been a PhD student at the time? Why wasn't the Nobel just awarded to his supervisor, Michel Mayor?

He shared the 2019 Nobel Prize "for the discovery of an exoplanet orbiting a solar-type star".  It was shared with Michel Mayor. Maybe Queloz did much of the work and Mayor thought he rightfully deserved some of the credit. Queloz only got a fourth of the money because the prize was split between Mayor-Queloz and James Peebles. Peebles got his half for "for theoretical discoveries in physical cosmology."

This was a very unique case because it involved no new physics really just the discovery of what most astronomers knew existed.


Mass of black holes compared to parent star What is the range of percentage mass of parent star left in a stellar black hole directly after its formation?

What factors determine this number for a specific case?

There is no general consensus on this. Different evolutionary models give different results. The factors (in addition to the initial mass of the star) that effect the final black hole mass would be the rotation rate of the progenitor, its composition (or metallicity) and whether it was in a binary system or not and whether that binary system was able to transfer mass.

Rotation is thought to be important because it affects internal mixing and therefore the rate at which fuel is supplied to the core and the rat at which processed material gets to the surface, affecting the atmospheric composition. It can also enhance mass loss.

The composition is important because mass loss is driven by radiation and radiative opacities are higher for high metallicity compositions.

A set of calculations by Heger et al. (2003) are one of the canonical works on this subject. Below is a plot of initial mass versus remnant mass for stars with big bang primordial abundance (zero initial metallicity) and then the same again for stars of solar metallicity.

The ratio of the red line to dotted "no mass loss" line gives the fraction you are after.
In zero metallicity (primordial) stars it increases from 10-40% for initial masses of 25-100 solar masses and is perhaps even higher for supermassive population III stars. (I stress that these are theoretical results).

For solar metallicity stars the results are a bit different. The ratio of the red line to the dotted line varies from 10-25% for 25-40 solar masses, but then it is not clear whether black holes can even form at even higher masses because of the much higher mass loss rates (see the difference between the dotted line and the blue curve).


What were the odds that ʻOumuamua passed so close to the Sun? ʻOumuamua passed its closest point at about 0.25 UA from the Sun.

Has anyone researched what the odds were that the first ever spotted interstellar object passed so close to the Sun?

Being the interstellar space almost totally empty, I would intuitively suppose that such odds would be extremely low.

How could it be found otherwise? It is currently (November 2018) only at about the orbital distance of Mars and cannot be detected by the world's largest telescopes.

Such objects are small and only seen in reflected light - they have to get close to the Sun (and Earth) to be seen. There may well be a substantial population of interstellar rocks passing through the solar system unseen at greater distances.


Why is the Hubble Telescope in space? Why is the Hubble Telescope in space?

Do we get enhanced clarity and range by placing it in space?

What can it achieve from space that it could not achieve from Earth?

As @astromax observed, one of the primary factors that makes a space based telescope better than an equivalent telescope of the same size is scattering.

Along with scattering, there is also refraction which can be especially problematic when combined with atmospheric turbulence. In the modern era, this problem can be remedied to a certain extent using adaptive optics, but since the Hubble was designed, built, and launched before AO became practical in the 1990's, a space telescope represented the pinnacle of optical clarity for that time.

However, there is another important optical property of the atmosphere, absorption. Although the Hubble is primarily a visible light telescope, it does have instruments that cover both the near UV and near IR, both of which are absorbed by the atmosphere more than visible light.

Furthermore, there are practical advantages of space based telescopes. There is neither weather, nor light pollution in space.


Date of solstices pre-0 AD I’ve written a script based on Meeus’s Astronomical Algorithms to calculate the position of the sun, moon etc. up to about 4000 BC.
Everything looks good, except when I go to 4000 BC the solstice are about a month after expected dates (they fall mid-July and mid-January.
I checked through this forum and came across this: https://astronomy.stackexchange.com/a/13009 . I looked at the results and saw that we have very similar dates.

Does the precession mean the solstices and equinoxes can shift to this extent? Or have I forgotten a step?

Many thanks.

Sounds like Julian Calendar slippage.

The Gregorian Calendar, the one we use now, was created to fix a problem with the Julian Calendar: The fact that the Solar Year wasn't exactly 365.25 days. As a result, compared to the calendar year, the date of the Vernal Equinox (and more importantly at the time, Easter) was slipping forward.

To remedy this, ten calendar days in 1582 (or up to thirteen calendar days in 1918, depending on how long a particular country waited to adopt the Gregorian Calendar) were omitted, and the Gregorian Calendar added a modification to the Julian calendar's rule of One leap year every 4 years: No leap year on century-years that aren't divisible by 400.

If your code automatically switches from Gregorian to Julian Calendar at the appropriate year, the adoption of the Julian Calendar probably winds up at about the expected days when the Julian calendar was adopted mid-first-century BCE.

But every 400 years further back, your calendar is slipping about three extra days.

And at 4000 BCE, that's thirty days.


Calculate Earth's distance travelled in a year This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).

According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:

Calculate by velocity and time: the average velocity is given as $107200\ \mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $\text{distance} = 107\,200\ \mathrm{km/h} \times (365.256363004 \times 24\ \mathrm h) = 939\,731\,570.736691\ \mathrm{km} \approx 939.731\ \mathrm{Gm}$
Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)
Any other way using the orbital data given in the Wikipedia article?

In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?

You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.

An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").

e.g. Define
$$ h = \frac{(a-b)^2}{(a+b)^2},$$
where $a$ and $b$ are the semi-major and semi-minor axes and $b = a\sqrt{1-e^2}$, where $e$ is the eccentricity.

Then the result you want is
$$ p = \pi(a+b)\left[1 + \frac{h}{4} + \frac{h^2}{64} + \frac{h^3}{256} ...\right]$$

All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7\times 10^{-5}$.

Using this and just the first 2 terms in the series, I get a perimeter of $1.999895\pi$ au. So hardly a correction from $2\pi a$.

However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.


How to calculate exact angle to the moon? I am building some device as a gift for my girlfriend.

For this device to work, I need to be able to calculate the angle to which I should look towards (3D angle), in order to see the moon.

This angle is affected by (and maybe not only):

The time of year.
The hour.
The location on earth from which I look.

I'm sure there's some formula that approximates the angle.

I don't need an exact value, I can live with +-1 degree.

For a one-off calculation, the easiest solution would be to use planetarium software, such as stellarium.

For a programming solution, a python package such as pyephem is an efficient way to calculate the position of the moon at multiple times, in a way that can be imported into a spreadsheet.

The apparent motion of the moon is more complex than you might think.  The moon moves about 15 degrees per hour due to the rotation of the Earth. But on top of this fairly simple motion is the Moon's actual orbit around the Earth, which is elliptical (and so this motion is not even, the moon moves faster when it is close to the Earth) and the position and inclination of the orbit is perturbed by the Sun. The perturbations are fairly regular, and can be calculated, but these effects (rotation of the earth, the orbit of the moon, the eccentricity of the orbit, perturbation of orbit) combine to make the actual calculation more complex  than "a formula" that can fit easily in a spreadsheet, which is why a package like pyephem is recommended.


Calculate right ascension in Zenith at this moment in my location I've been searching on Internet about how to calculate which RA is in zenith at a given location in a given time but I haven't find anything (or maybe I don't know how to search).

Using my star finder I have not draw any conclusion.

How can I calculate which RA is on zenith on a given location in a given time?

Right ascension at Zenith should always be your local sidereal time.


What will the last days of Earth be like? The sun is a main-sequence star; as I understand it the sun will expand into a red giant in a few billion years.  I have heard conflicting reports of what this entails - some sources I found say the sun will expand until the current orbit of Earth is contained within it, others say it will stop expanding short of Earth's current orbit.

Every source has agreed that the planet will have long-since stopped being able to support life as we know it.

Assuming humanity takes no action to significantly alter the natural course of events, such as dramatic alterations of our atmosphere, what will the last few decades of Earth's life be like?  Will the planet be heated to the point where its surface is like Mercury's sun-side?  Will the sun's output drop low enough that it instead cools, like Mars?  Will the sun expand and engulf the planet, or will the planet continue to orbit an expanded sun as a lifeless ball?

Will the sun's expansion be gradual or rapid?  How will this expansion affect the rocky planets nearest to it?

A lot of the answer depends on Orbital mechanics and calculating that isn't easy.   If no action is taken, Earth's orbit is still likely to change, but how much and in which direction is hard to say.

Long before the sun goes red giant it will get too hot for the Earth.  If the earth doesn't move, the sun's increased temperature will likely make the earth uncomfortable in a few hundred million years and perhaps scorching in as little as a billion years.   Warmer air and more water vapor in the air will trap heat and at a certain point, you'd get a run-away green house effect, long before the red-giant phase.    Short term, changes in CO2 in the atmosphere might slow that.   We could see ice ages before we see run-away heating, but over time, with no human action, run-away heating is likely inevitable - on a very large timescale, like a billion years or so.   Longer if the Earth drifts farther from the sun.

  what will the last few decades of Earth's life be like? Will the
  planet be heated to the point where its surface is like Mercury's
  sun-side? Will the sun's output drop low enough that it instead cools,
  like Mars? Will the sun expand and engulf the planet, or will the
  planet continue to orbit an expanded sun as a lifeless ball?

  Will the sun's expansion be gradual or rapid? How will this expansion
  affect the rocky planets nearest to it?

I'm going to answer the 2nd part first.   The sun's expansion will first be gradual, but the red giant stage will be fairly rapid.   Over the next 4 billion years or so, the sun will get some 60% more luminous, so significantly much more heat at the same orbit.   A very rough rule of thumb is 10% more luminous every billion years. (that's not for every star, just for 1 solar mass) - and that's a pretty ballpark estimate.

But just 10% more heat from the sun would raise the Earth's temperature uncomfortably, and with the added water vapor in the air and greenhouse effect, the earth might be a permanent steam room, even at the poles, at just 10% more heat from the sun.

asking about the last few decades is the wrong time frame as changes would probably happen much more slowly, though atmospheric/feedback mechanisms in the climate might make changes happen more quickly - ice age speed for example, a few thousand years or so.   Also, as Antarctica drifts away from the south pole, we might see an end to ice ages - there's lots of moving parts to long term climate predictions, but in general, you'd see gradual heating to the point where complex life might have a hard time adapting.

The suns output won't drop until after it goes red giant, it's on a steady but gradual increase, followed by a relatively fast increase at the red giant stage.

Now, if we terraform Titan, that might be a safe distance to watch the sun go red giant from and it would probably be quite a spectacular show, well, over enough time.

As to how fast it will go red-giant, fast for a star means, about 200 million years, so, from the perspective of a human lifetime, watching a star go red giant would be like watching grass grow.

A much more intelligent discussion on that than I can give, here:   https://physics.stackexchange.com/questions/25622/how-fast-will-the-sun-become-a-red-giant


Could the earth become a star if more mass was added? I've heard that the only difference between a star and a planet is mass, meaning that if planets accredited enough mass they would too become stars.

Does this mean that the star started off as a planet?
Stars require hydrogen fusion and earth has little H. Could the earth become a star if more mass was added, but its relatively low abundance of hydrogen remained the same?

Regarding the title: Yes.

  Does this mean that the star started off as a planet?

Yes, a star could technically start out as a planet, if it accreted enough mass. However, this is extremely unlikely, since the planet would need to be 80x the mass of Jupiter for it to undergo nucleosynthesis.

  Stars require hydrogen fusion and earth has little H. Could the earth become a star if more mass was added, but its relatively low abundance of hydrogen remained the same?

The most important components for a star is the composition and mass. As it stands, Earth doesn't have nearly the concentration of hydrogen for a star to form, even if we kept adding mass.

Most stars work by fusing the lightest element, hydrogen. To fuse heavier elements requires a much, much higher temperature, since the more protons the element has, the more it will repel other ones instead of fusing. Yes, massive enough stars can fuse heavier elements.

However, Earth would inevitably grow in hydrogen, so it would eventually have a significant amount. Above a certain size, given ambient conditions (like radiation pressure, light intensity from the star, etc), Earth would become massive enough that it starts capturing hydrogen gas from the interplanetary medium, and becomes a gas giant (as called2voyage pointed out). Where this happens isn't precisely known, but it's roughly 10x Earth's mass. This means that, theoretically, a terrestrial planet could form a star if you increased its mass ever further.

Now, I can hear you thinking, "What if Earth were surrounded by a vacuum instead? If we kept adding mass, would it become a star?" Well theoretically, the planet would eventually reach temperatures where most of its elements could fuse. For example, the "star" could sustain itself on oxygen and silicon fusion, but not on iron fusion.


How is the first detonation in Supernove type Ia triggered? Ok, I read about the Supernova of type Ia and I found out that there are two detonations happening. First one is in helium shell around the white dwarf and second one seems to be triggered by the first one and it is the main part of whole supernova. I did not find how this helium shell explodes at all so that`s what I am looking for.
Thanks in advance!

Nobody really knows how type Ia supernovae detonate (or deflagrate) - there are a number of possibilities. The "vanilla" possibility is not what you state in your question, it is that the white dwarf accretes sufficient mass that it approaches the Chandrasekhar limit and becomes dense enough in its core to commence carbon burning.

However, the emerging diversity that is seen among type Ia supernovae, once thought to be a single population, suggests there may be other possibilities. There is some evidence that white dwarfs may explode at masses well below the Chandrasekhar limit. If a white dwarf in a binary accretes enough He-rich matter, this can become compressed enough to ignite He burning near the surface (this happens at a lower density threshold than Carbon burning). This then drives a shock wave into the white dwarf and the compression caused by this can ignite the carbon.

Why does the He "explode"? Well, the accreted He will form an electron-degenerate layer at the surface. A fundamental property of this degenerate gas is that the pressure is independent of the temperature. Thus, if the He ignites then at least initially, the temperature goes up but the pressure does not. Since the He fusion rate depends on something like $T^{40}$ this allows a runaway reaction to develop that could be characterised as an "explosion" in the surface layers.


Beginning of space-time I have read that the Big Bang created space-time, but how could this be possible? The Big Bang was itself a singularity, and a singularity is a point in space-time where the curvature becomes infinite. So how could the Big Bang create something without which its existence would not have been possible?

It is impossible to measure anything that happened in the Planck Era, so physicists are unsure whether the universe actually began from a singularity or whether the singularity in the Big Bang theory represents an inability to describe the universe at that time. In fact, "Most physicists believe that this is a mathematical artefact and does not describe what actually happened."
It would be more accurate to say that the Big Bang is the beginning of space-time, than that it "created" space-time.

The Big Bang is a description of how the universe evolved in its earliest stages, not exactly a resolution to why the universe came to be. The latter is really a question for philosophy, not something that can be addressed by science per se. Science informs our philosophy, but does not make philosophical claims.


What caused these strange craters on Hyperion? I was looking at pictures of the moons of Saturn and noticed that the craters on Hyperion have a strange shape, somewhat resembling sinkholes. They look a lot deeper than the impact craters on other moons (including ours). Is it an optical illusion (perhaps caused by the presence of dark material at the bottom of the craters?) that somehow makes them look deeper? If not, what process caused these holes to appear?

According to Thomas et al. (2007), these craters are not unusually deep:

It is unlikely that unusual crater depths significantly enhance the sponge-like appearance. Crater depth-to-diameter ratios for the 13 examples that can be measured reliably using shadow lengths average 0.21 ± 0.05 (s.d.). These ratios are similar to values for fresh lunar craters, and are slightly greater than for some small rocky objects. They are slightly larger than that for craters on large icy satellites, 0.14.

What is unusual and gives a "sponge-like" appearance to Hyperion is the high density of craters and lack of intercrater plains:

For craters with diameters between 2 km and 11 km, Hyperion has a cumulative number of craters per unit area twice that on Phoebe; at similar resolution these objects are strikingly different.

To explain this, they proposed that ballistic ejecta from impact events, which usually cover older craters, are lost to space instead.
A more recent study (Howard et al., 2012) suggested that mass wasting processes (landslides) and CO$_2$ sublimation also account for this special morphology, especially for the dark-floored craters. This has been confirmed by Dalton et al. (2012), who called these craters "suncups":

The irregular shapes of the sublimation pits (suncups) support the
hypothesis of extensive mass wasting and sublimation degradation
(Moore et al., 1996, Howard et al., 2011, Howard et al., 2012) largely
associated with the loss of volatiles (including water ice). When the
ices evaporate, entrained low-albedo grains accumulate in lag
deposits, which are subject to solar heating. Warm grains accelerate
both the local evaporation of ices and their own spatial
concentration.


Intuition about Hubble's Law I have recently learnt about Hubble's Law and this is my understanding of it:
Galaxies move away from us at a rate that is proportional to their distance from us. So, considering an arbitrary unit of distance d, if a galaxy at a distance d from us is moving away from us with a velocity v, then a galaxy at a distance 2d is moving away from us with a velocity 2v. This proves that the universe is expanding.
But the part that I can't quite visualize is the fact that it doesn't matter where you are in the universe. Regardless at what point in space we are standing, it will look like the galaxies are moving away from us. How is this possible? Are the galaxies moving outward? Should I visualize this as the galaxies being on the surface of a sphere whose radius is increasing?
How do I gain intuition about this and better visualize this?

Think about it as if you were baking bread (or cake, whatever you prefer). When you bake a bread with raisins, it rises in all directions (due to the yeast). Every raisin in a rising loaf of raisin bread will see every other raisin expanding away from it. Now suppose the raisins as galaxies and the yeast as the Hubble constant.
An illustration from Hyperphysics (Georgia State University) to illustrate this further:

When you now suppose you are the raisin at the edge, from your perspective, you are standing still and the other raisins are receding away. It is all about the perspective, the body which you take as a reference (like with relativity).


What enhances the capture and merge rates of pairs of small black holes orbiting around supermassive black holes? I just tried to read the new New York Times article Two Black Holes Colliding Not Enough? Make It Three which links to the new 25-June-2020 Physical Review Letter Graham et al. Candidate Electromagnetic Counterpart to the Binary Black Hole Merger Gravitational-Wave Event S190521g
The event described is the merger of two black holes that were embedded in the accretion disk of a supermassive black hole in the center of a galaxy; i.e. in a quasar (loosely speaking).
The NYTimes article describes the following and references a previous paper:

Bolstering that hypothesis was the fact that the flare did not become visible until 34 days after the gravitational waves were detected. It would take about that long for any light from a black hole collision to emerge from such a thick disk of gas, according to a model that Dr. Ford and Barry McKernan, her colleague at the American Museum of Natural History, described in a paper last year.
Dr. Ford described the accretion disk as “ a swarm of stars and dead stars, including black holes,” in a Caltech news release.
She added, “These objects swarm like angry bees around the monstrous queen bee at the center. They can briefly find gravitational partners and pair up but usually lose their partners quickly to the mad dance. But in a supermassive black hole’s disk, the flowing gas converts the mosh pit of the swarm to a classical minuet, organizing the black holes so they can pair up.”

That paper is B. McKernan et al 2019 Ram-pressure stripping of a kicked Hill sphere: Prompt electromagnetic emission from the merger of stellar mass black holes in an AGN accretion disk
Question: Is it possible to explain what it is about "the flowing gas (which) converts the mosh pit of the swarm to a classical minuet, organizing the black holes so they can pair up"? What is it about the accretion disk that enhances the capture and merge rates of pairs of small black holes orbiting within it?

I would think that the same reason that a gas disk enhances the growth of planetesimals. Drag from the disk enforces circular, co-planar orbits, which in turn means that objects that get close to each other have small relative velocities.
Edit: So what is thought to be going on is a little bit more complex than the simple answer above. The density of massive objects embedded in the accretion disk can be enhanced at certain orbital radii. These "migration traps" (Bellovary et al. 2016) are caused by torques exerted on the orbiting objects (in this case black holes) caused by gas accretion. For certain disk conditions it is possible to find locations where at smaller radii the torque acts to increase the orbit radius, but at greater orbital radii the opposite is true. This would cause objects to accumulate in these migration traps, giving them more chance to interact as per my original answer.