Spherical Accretion (Bondi Model) I have a question. I studied Bondi's model for spherical accretion. I know the assumptions made in the model are big and that they are rarely representative of what happens in astrophysics. What I wanted to ask is: is there a situation that suits this model even approximately, or it is a total idealization? Is there always an accretion disk, or may there be a situation where the accretion is spherical?

One example is that of the core of a Thorne–Żytkow object, or TŻO. It consists of a high-metallicity M class red giant or supergiant with a neutron star at its core. Generally, TŻOs form when a neutron star collides with a suitable red giant/supergiant. The neutron star settles into the center of the other star.

Eventually, the main source of energy for the TŻO comes from gravitational contraction of the envelope as gas from inside the star falls toward the surface of the neutron star. In this case, the change in mass $\dot{M}$ of the neutron star is (see Thorne & Żytkow (1977)):
$$\dot{M}=4\pi r^2\rho v_{\text{in}}\mathscr{R}\tag{1}$$
The difference between $(1)$ and the underlying equation behind Bondi's formula is the factor of $\mathscr{R}$, a correction factor for relativistic redshift, which is given by
$$\mathscr{R}=\sqrt{1-\frac{2GM_{tc}}{c^2r}}\tag{2}$$
where $M_{tc}$ is the total core mass. This formula may look familiar; it comes from the Schwarzschild metric. You need $\mathscr{R}$ to properly describe accretion onto any compact object. This is something needed not just in the case of TŻOs. However, Bondi's formula can be used nonetheless, if you multiply by $\mathscr{R}$.

That said, Bondi accretion can be used without relativistic compensations in a number of ways:

Mass transfer in certain binary systems, where an envelope moves matter from one star to another.
Protostellar clusters, where many young protostars move about, accreting gas.
Galaxy clusters, where galaxies move through the intergalactic medium (IGM).


What keeps Chthonian planets so dense? The cores of gas giants are kept under incredible pressure by the weight of the gasses of the giant. In a chthonian planet, said gases are stripped away, revealing the core of the gas giant, which should alleviate the pressure compressing the core of the gas giant. In a chthonian planet candidate, Kepler 52-c, its density is over 600 times that of Jupiter. What is keeping the planet at such a high density?

10 Jupiter masses at around 2 Earth radii?
That for sure doesn't exist / would be quite the sensation to discover.

When you look at data of any kind, one should pay attention to the measurment errors at least as much as to the actual value.
A regular physics result (for example for a measurment of the gravitational acceleration $g$ of where you stand) looks like
$$g=(9.81 \pm 0.02) \frac{m}{s^2} $$
or if for any reason you have asymmetric errors
$$g=(9.81^{+0.02}_{-0.01}) \frac{m}{s^2} $$
and errors always give an idea about how uncertain the method is with which the value was derived.
Now if you do take a look at the errors reported for the Mass quoted on the website you see that they are
$$M_{planet} = (10.41^{+0.0}_{-10.41}) $$
or so to say highly asymmetric, which should make one suspicious.
A look then into the original publication makes it clear that this cited mass is in fact only an absolute upper limit.
The authors of the paper were using two methods to estimate the masses of planets.

Looking for transit timing variations of known, and seen transiting systems. That means they had the system Kepler 52, with transiting planets K52b,c . K52b because it transits way more often than c has a well-determined period (Period with small errors!) and because of that any deviation in expected future transit time could be attributed to the \textbf{maximum masses} of K52c.
The more massive and the more compact a system is, the quicker it will destabilize. This fact is often used in reverse, to take the system age and at given distances derive maximum masses below which the system must lie, or else it would have already flown apart.

Both methods can only give maximum masses and I'll just leave here fig. 5 from the original paper with the planet you're interested in:

Now remembering, that $1 M_J \approx 320 M_{\oplus}$, you see where your 10 Jupiter masses for K52c come from: That's the planets possible maximum mass for system stability. The TTV method gives already a constraint that is 100 times lower ($37.4 M_{\oplus} \approx 0.11 M_J$).
Thus $37.4 M_{\oplus}$ is the planets true maximum mass.

This is clearly an error on the side of exoplanet.eu, but then probably there are too many planets and papers to read for whoever puts those data in there.

Summarizing
What we have here is only a maximum mass. Also the wrong one. To say what is now more probable, if $M_{K52c} = 37.4 M_{\oplus}$ or $M_{K52c} = 3.74 M_{\oplus}$ i'm not certain enough if I understand their anticorrelation method for the TTV signals.


Why are radar images of comets shaded only on one side? NASA published a picture of a comet the other day. The image shows the comet being lit from above. See NASA's picture below.

However, since it is a radar image, I would have assumed to get a shading on all sides and grazing angles, like velvet or scanning electron microscopy. Or alternatively to have the sides facing the radar dish being shaded, and the edges being dim. For the velvety look, compare to this picture from Wikipedia:

So why is the comet's image shaded only from above?

There is a very good blog article here that explains this effect but basically the Doppler shifting of the radar return means that certain parts of the asteroid light up more than others in the observing wavelengths from Earth. Its better explained in the article>>

Taken from the article "How radio telescopes get "images" of asteroids"
Posted by Emily Lakdawalla:

  As an asteroid rotates, some parts of it are moving toward us, while other parts are moving away. As the broadcast radio wavefronts hit the part of the asteroid that is moving toward us, the asteroid smacks into each wavefront faster than it would if it were not rotating. The speed of the wavefronts does not change, because the speed of light is constant, so the wavefronts end up being packed closer together. This is a Doppler shift. The asteroid has taken the broadcast wavelength and reflected it at a shorter wavelength from the parts of the asteroid that are rotating toward us. On the other side of the asteroid, which is rotating away, the opposite thing happens; each arriving wavefront smacks into the asteroid a little later than it would if the asteroid were not rotating, so the reflected waves are spread farther apart.

Source: http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html?referrer=https://en.wikipedia.org/


Do astronomers generally agree that the distinction between comets and asteroids is not so clear? edit: I just saw this tweet and find it incredibly relevant :)

begin question: See this answer and then consider if there are known or likely solar system bodies that might be identified as both asteroids and comets.
If so (there seems to be an example there) would this be because there is insufficient information known about the body, and the ambiguity could be cleared up with enough data (say a visit by a spacecraft)?  Or are the categories of asteroids and comets actually overlapping?
If the latter, is there work underway to improve or update these definitions?

If we're going to get technical, Asteroids are not really an official name anymore. In 2006, when the IAU redefined what a planet was (and thus demoted Pluto), they also decided to more formally define other terms to identify objects in our solar system. You can see a diagram of all the official terms and how they relate below. Notice the important factor that asteroids no longer made the cut. There is the collection of "Small Solar System Bodies" (SSSB) and within that falls comets and centaurs, but asteroids are not exclusively called out.

So if you want to get technical, asteroids are really SSSBs now and comets are SSSBs as well, but also happen to fit into the sub-category of comets. What defines an SSSB as a comet is its ability to form a coma based on the sublimation of icy surface volatiles. So you might distinguish an "asteroid" as an object which cannot form a coma (and orbits inside Neptune), but that still leads to an ambiguity because what happens when a comet runs out of surface volatiles? Does it now become an asteroid? Is it still considered a comet? There are no clear answers.


Any known Triple Star Systems with this hierarchy? I'm curious, are there any known STABLE triple systems in which a close inner binary pair is orbited by a more massive star, yet the total mass of the binary pair is greater than that single orbiting large star?
Something relatively compact, like this:

Close Inner Binary: Star A with mass of 1, Star B with mass of 0.8. Combined Mass: 1.8
Outer Star Orbiting Binary: Star C with mass of 1.2.
Inner Binary Orbital Period: months or a couple years.
Outer Star Orbital Period (orbiting binary): Decades to maybe a century or two.

Star A and B are both smaller than Star C, with the combined mass of A and B exceeding Star C. Star C orbits the barycenter of the close binary A+B.
I'm only aware of triple star systems in which the largest star is closest to the combined barycenter of the system. I don't know of any examples in which the largest star is also the star furthest from the system barycenter. Does this arrangement exist?
Thanks!

Short Answer:  Star sytem HD 188753 seems to be an example of the type of star system asked for.
Long Answer:
Part One of Two:  The orbital characteristics.
In any multiple star system, all of the stars will orbit the combined barycenter of the system.  Any pairs will orbit around the barycenter of the pair, and the barycenter of each pair will orbit around the barycenter of the whole system.
If there is a close pair of stars plus a third more distant star in the system, the close pair will act like a single star with their combined mass.  If their combined mass is less than that of the single star, their barycenter will orbit farther from the barycenter of the system as a whole than the single star will.  If their combined mass is greater than that of the single star, their barycenter will orbit closer to the barycenter of the system as a whole than the single star will orbit.
According to the formula found at:  https://en.wikipedia.org/wiki/Barycenter[1] the barycenter of the pair of stars will be at a point closer to the more massive star than to the less massive star.  It will be 0.444444 of the distance between the two stars, while the less massive star will orbit the barycenter at a distance of 0.55555 of the total distance between the two stars.
the more massive star should orbit at a distance at least a few times as large as the distance between the stars in the pair.  Assume that if the distance between the two stars in the pair is 1, the distance to the more massive star is 5 or 5,000 as examples.
According to my calculations, if the third star orbits at a distance of 5, the distance between the barycenter of the pair and the the barycenter of the system as a whole should be 2 times the distance between the two stars in the pair.  The third star will orbit the system barycenter at distance of 3.
According to my calculations, if the third star orbits at a distance of 5,000, the distance between the barycent of the pair and the the barycenter of the system as a whole should be 2,000 times the distance between the two stars in the pair.  The third star will orbit the system barycenter at distance of 3,000 times the distance between the two stars in the pair.
So in star systems with a a pair of stars whith a total mass greater than that of the single star, the more massive pair of stars will orbit closer to the barycenter of the system tha tthe single star will, even if it is more massive than any one star in the pair.
So the question is are there any known trinary star systems with a pair of stars and asingle stars whose mass mass is slightly less than that of the pair, though slightly more than any one star of the pair.
I'm sure there are many trinary star systems where the single star is more massive than the pair of stars let alone one of hte stars in the pair.
I'm sure there are many trinary star systems where one or both stars in the pair, and thus the pair as a whole, are more massive than the third star.
The problem is finding a system with three stars of very similar masses.
Part Two of Two:  An Example.
Using the list of trinary stars at: https://en.wikipedia.org/wiki/Star_system[2]
I found that one of the six trinary stars listed has a pair slightly more massive than the single star.
HD 188753 is a trinary system with a main star HD 188753 A with a mass of  1.06 plus or minus 0.007 that of the Sun, and a pair of stars HD 188753 B & C.  B has amass 0.96 plus or minus 0.05 that of the Sun, and C has a mass of 0.67 plus or minus 0.05 that of the Sun.  So A should be a lot more massive than C, and slightly more massive that B, but the pairr B & C together should have about 1.53 the mass of A.   Although in this case the uncertainty in the masses of A & B makes a lsigh psosibility that B could be slightly more massive than A.
In a 1999 version of a catalog of multiple stars, 551 out of 728 systems listed were trinary stars.
https://en.wikipedia.org/wiki/Star_system#Multiple_star_systems[3]
So presumably seraching multiple star catalogs should find other examples of star systems satisfying the requirements of the question.


How much rock has Jupiter "swallowed"? During its formation approximately 4.5 bya, Jupiter passed through what is now the asteroid belt,gravitationally deflecting some spacewards and some sunwards.
Logically, Jupiter, due to its massive gravity, must have swallowed up a fair portion as well.

Is there any way to calculate how much rock Jupiter contains? Have any estimates been made?

Jupiter is a gas giant, with a mass approximately 317.8 times that of Earth, and its atmosphere has a composition that's typical of gas in space: almost 75% hydrogen and almost 24% helium, (barely different to the primordial gas created by the Big Bang). However, its interior also contains rock and metal components.

Until quite recently it was thought that Jupiter's rocky & metallic core was fairly small and compact. Various models gave estimates of a core mass ranging from 7 to 18 Earth masses. But gravitational data from the Juno mission has changed the picture rather radically. It seems that Jupiter's core is diffuse rather than compact, extending fuzzily to almost half the planet's radius! At present, we aren't quite sure of how such a core was formed, but the leading hypothesis is that Jupiter's primordial core was shattered by an extreme impact event.

From Nature, DOI: 10.1038/s41586-019-1470-2 The formation of Jupiter’s diluted core by a giant impact by Shang-Fei Liu, Yasunori Hori, Andrea Isella:

  The Juno mission has provided an accurate determination of Jupiter’s
  gravitational field, which has been used to obtain information about
  the planet’s composition and internal structure. Several models of
  Jupiter’s structure that fit the probe’s data suggest that the planet
  has a diluted core, with a total heavy-element mass ranging from ten
  to a few tens of Earth masses (about 5 to 15 per cent of the Jovian
  mass), and that  heavy elements (elements other than hydrogen and
  helium) are distributed within a region extending to nearly half of
  Jupiter’s radius.

  Planet-formation models indicate that most heavy elements are accreted
  during the early stages of a planet's formation to create a
  relatively compact core and that almost no solids are accreted
  during subsequent runaway gas accretion. Jupiter’s diluted core,
  combined with its possible high heavy-element enrichment, thus
  challenges standard planet-formation theory.

  A possible explanation is erosion of the initially compact
  heavy-element core, but the efficiency of such erosion is uncertain
  and depends on both the immiscibility of heavy materials in metallic
  hydrogen and on convective mixing as the planet evolves. Another
  mechanism that can explain this structure is planetesimal enrichment
  and vaporization during the formation process, although relevant
  models typically cannot produce an extended diluted core.

  Here we show that a sufficiently energetic head-on collision (giant
  impact) between a large planetary embryo and the proto-Jupiter could
  have shattered its primordial compact core and mixed the heavy
  elements with the inner envelope. Models of such a scenario lead to an
  internal structure that is consistent with a diluted core, persisting
  over billions of years.

  We suggest that collisions were common in the young Solar system and
  that a similar event may have also occurred for Saturn, contributing
  to the structural differences between Jupiter and Saturn.

In contrast, the total current mass of the asteroid belt is roughly 4% that of our Moon, so just under 0.05% of the mass of the Earth. Undoubtedly, Jupiter has swallowed numerous asteroids and comets over its lifetime, but even if it ate all of the asteroids in the asteroid belt that would only amount to less than 1% of its total rock and metal content, compared to the primordial rock and metal that it has had since its formation, including whatever rock & metal it inherited from the core impactor hypothesized by Liu et al.


Why don't black holes quickly eat up the galaxy/universe? It's my understanding that the radius of the event horizon of a black hole is proportional to it's mass.
Which means the surface area of a black hole is proportional to it's mass squared.
Lets just assume that black holes eat stuff that by chance "runs into them".
Using this (perhaps naive) assumption, this would mean black holes gain mass at a rate proportional to their surface area, which is the same as their radius squared, which is their mass squared.
So then we get something like:
$\frac{dm}{dt} \propto m^2$
$m = 1 / (C - t)$
Which means that all black holes should grow to an infinite size in a finite length of time.
Indeed they will eventually get to a point where even small amounts of mass will grow the volume of the black hole so much (because volume increase is proportional to the current mass squared) and even if the density of the universe is very low this will eventually mean any growth will set off runaway growth because each atom the black hole swallows increases it's volume such that on average that new volume contains a new atom.
Is it just that the constant factors here are very high, or are one of my assumption quite incorrect? I say quite incorrect because even if they're slightly incorrect (the growth rate is proportional to mass not mass squared) many of these issues still occur on long enough timescales.

Your rough analysis is basically correct; everything may well end up in black holes eventually. But the timescale for that is very long indeed.
The cross-section for direct interaction as you describe is vanishingly small. The diameter of a ten solar mass black hole is 60 km. The rate at which it scoops up interstellar material from a cylinder of diameter 60 km is totally negligible. e.g. A speed of 10 km/s with respect to interstellar gas and a typical gas density of $10^6$ H atoms per cubic metre, would result in an accretion rate of about 1.5 kg/year.
The chances of direct star to star collisions in a galaxy is exceedingly rare, so the chances of a wandering star crashing directly into a black hole are even smaller.
With supermassive black holes, the issue with your analysis is that they sit in the centres of galaxies and stars do not encounter them by chance. The stars and gas in a galaxy have angular momentum that must be conserved. This takes them on orbits which do not go near the black hole unless perturbed in some way. These supermassive black holes feed while they can, but when all the easily available matter has been consumed, they become dormant. This is one of the reasons that the peak of quasar activity was about 10-12 billion years ago.


Blue, red, and white star? At 6:30pm central time in the USA I saw a rapidly twinkling red blue and white star in the sky. It was about south east and I was really confused. I looked it up it might be Sirius but I'm not sure. I know it's not an airplane cause it hasn't moved in 20 minutes. Does anyone know?

Sirius is the most likely identification.  It is visible in the South East at 6:30 from the USA, it is the brightest star in the sky and it's brightness makes the twinkles caused by the atmosphere more noticable. Similarly its brightness makes the colours caused by twinkles more apparent.
By shaking the tripod a picture of the apparent colours of Sirius can be captured
https://epod.usra.edu/blog/2012/01/sirius-twinkling.html
There are several other bright stars in that region of the sky: Betelgeuse, Rigel and Procyon, any of which could have been your star, but Sirius is the most likely simply as it is the brightest.  The stars will be back in (almost) the same positions tonight.  If you can find Orion and Orion's belt, and follow the direction down you will get to Sirius, which should confirm your identification.


Grammar of stellar classification How to write G2V correctly (I mean space)? I saw in articles $ G\,2\,V$ and  $G2\,V$, on the internet webside it is usually $ G2V$.

One reason to prefer the space between the temperature and luminosity classes is that the hypergiant luminosity class is "0", which could be confused with a subtype. Usually this appears as part of a range with luminosity class "Ia" as opposed to on its own, but this range is often written as "0-Ia". Something like "G5 0-Ia" is more comprehensible than "G50-Ia".

If you're writing a parser, be prepared to handle both. In general, parsing spectral types is a bit tricky (I have written spectral type parsers before, it's "fun") and there are a surprising number of annoying edge cases. It would be nice if astronomical catalogues also included something like the IVOA spectral type encoding, but usually the furthest you get is some bespoke system that only encodes standard OBAFGKM types and provides no way to encode, say, carbon stars or Wolf-Rayet types, even though the catalogue contains examples of these.

Some other issues you may run into:

The ambiguity between "-" as a minus sign and "-" as a range symbol. Uusually the machine-readable files will be ASCII so none of the other dashes in Unicode are available, and given the general state of things I wouldn't trust them to be used reliably anyway!
The previously-mentioned "0-Ia" luminosity class range means that a parser will run into an ambiguous interpretation of "G0-Ia": either an open ended range "G0 or earlier" of luminosity class "Ia", or a "G" type star with no subtype and luminosity class range "0-Ia".
The fact that "+" may be an open-ended range symbol or a multi-spectrum separator "M5+" (star of type M5 or later) vs "M5+..." and "M5+K2" (M5 star with companion).
The peculiar A-type stars, e.g. "kA5hF0mA5 V".
The various two-parameter systems for carbon and S-type stars.
The fact that the entire spectral type or various subcomponents may be bracketed to indicate uncertainties...
...which must be distinguished from the use of square brackets to indicate that the star is the central star of a planetary nebula
If you're really lucky, your source will render the spectral types in ALL UPPERCASE (so that peculiar A-star spectrum from a few bullet points up becomes "KA5HF0MA5") because it's been compiled using some Fortran code that hasn't been updated since the 1960s or something...


Help new Dobsonian owner I need a bit of advice. I purchased a Orion xt8 and a lens kit with a 2x Barlow 6mm 20 mm eyepieces and every time I try pulling things into focus / magnification or focus goes out the window recently purchased a 5x Barlow to try to get the images bigger. But I’m starting to get a bit discouraged as the only thing I can see in the sky is the moon. All others. Clusters. Andromeda. Mars  I can not pull into focus to save my life.

I don’t know what I’m doing wrong. I see all these pictures of Andromeda from the same telescope and on mine I see dots. (Above picture is the most dots I could see near a cluster. But it’s not the cluster. I guess my question would be. How can I see objects in higher mag / detail. Andromeda is a dot like a star. So are all the planets.

And yes I know what to expect. I know things will look very blurry or out of focus. But only seeing star dots (when they are planets) concerns me.

I have experience using this telescope.  It looks like you can focus correctly if you get a good view of the moon.  You probably don't want more magnification (the Andromeda galaxy is comparable in size to the moon).  It is easier to get and keep targets in the field of view with a low power eyepiece, and I suspect that is your problem.

Start with planets.  Use your low-power eyepiece, 20 mm.  Jupiter is the easiest to see as a disk, but it is unfortunately close to the sun right now so it is low in the west in the evenings.  With your 20mm eyepiece, it should be quite obvious as a disk with satellites and you may be able to see bands.

Mars is quite prominent right now in the evening sky but it is smaller than Jupiter.  Still, with your 20mm eyepiece, you should be able to at least tell it is a disk (and a bright orange one) if you have it in the view of the eyepiece.  Once you have it centered in view of your 20 mm eyepiece, you can try switching to your high-power (6mm) and see if that reveals any more detail.  Markings on Mars or quite subtle; often it looks like a featureless orange ball.

If your night sky isn't that dark because you are close to a city with its outdoor lights, it can be a challenge to get a good view of even spectacular "deep sky" objects like the Andromeda galaxy.  Again, use your low power eyepiece and no Barlow lens.  The core of the galaxy is much brighter than the extended glow around it, so it can appear star-like.  Try looking at with "averted vision" (don't look directly at it, to use the more sensitive light-detectors towards the edge of your field of vision) to pick out the faint glow that surrounds the core.  If you go to a dark sky location (one where you can really see the Milky Way) the XT8 can give a pretty impressive view of the galaxy where the elliptical outer glow will fill the FOV of your low-power eyepiece.

If it is hard for you to travel to a dark-sky location, you can try buying a deep-sky filter.  This filter tries to block the specific wavelengths thrown into the sky by outdoor lighting.  It makes everything you see through the scope dimmer and kind of green, but it can improve contrast on galaxies and nebula by making the background sky darker.


Does Saturn raise tides on Titan's lakes? Titan is the only body in the Solar System other than Earth to have large bodies of surface liquid. Since Saturn is much more massive than the Moon, I would expect for tides, if they're present to begin with, to be much larger than the ones here on Earth.

Sort of...  Titan is tidally locked to Saturn, so points the same side toward Saturn all the time.  The tide caused by Saturn on Titan should be nearly unchanging, except that Titan has an elliptical orbit, and is at times closer.  This cases a significant tidal effect, but it doesn't travel like on Earth.  See this article for more information.
This article states that this tide deforms Titan about 33 feet.  I do not know how much the lakes themselves would change though.
The sun should also cause a very minor tide because it is further away.


What are the most valuable meteorites and asteroids in terms of precious metals? There were some suggestions that asteroids could one day be mined for precious metals. I have found an outdated study that says that some siderites contain as much as 0.001% of gold:
http://pubs.usgs.gov/circ/1968/0603/report.pdf

Are there recent studies on precious metals in space rocks, i.e. platinum, rhodium, iridium, rhenium, osmium, ruthenium, palladium, germanium and gold, and is it genuinely a viable mining research subject?

I have found an outdated study that says that some siderites contain as much as .001% of gold. Is it genuinely a viable mining research subject?

Not yet, and not for quite a long time.

Gold is precious because humans only mine about 2.5 million kilograms of the stuff per year. Compare this to the huge amount of infrastructure (and huge cost) needed to retrieve the few kilograms of stuff that has been brought back to Earth from space.

We haven't the foggiest idea how to mine minerals in space, let alone minerals that at best might have a concentration of ten parts per million (another way to look at your 0.001% figure). On the other hand, there are many ideas regarding taking advantage of untapped resources right here at home. For example, there are manganese nodules all over the sea floor. These nodules contain not just manganese , but also iron, nickel, and precious metals. Space mining cannot compete with those untapped local resources until they run out. (And then we'll find yet another untapped local resource.)

Mining stuff in space will be profitable only if there's a need for that stuff in space. If space mining becomes reality, the first stuff to be mined will be rather mundane volatiles such as water and methane rather than metals. Those volatiles are the low hanging fruit in space. How to extract and make use of those mundane volatiles is a good problem for our current generation of young adults. How to go beyond that is a problem for their grandchildren.


Directions to Proxima Centauri I'm writing a sci-fi novel in which humans venture to the stars for the first time. Their "shake down" cruise takes them to Proxima Centauri. I'm trying to understand where Proxima Centauri is located relative to the sun rather than Earth, as that seems a more logical reference for navigation to nearby stars.

The location of Proxima Centauri in Earth's night sky is described with the Earth-relevant coordinates

  Right ascension   14h 39m 36.49400

  Declination –60° 50′ 02.3737

However, the Earth's orbit is inclined 7.155° to the Sun's equator and our axis is tilted at 23.4392811°.

Is there a reference indicating Proxima Centauri's sol-relative position, or can that position be calculated based on the above information?

A system of coordinates requires a fixed plane to define "zero latitude" and a direction chosen in that plane to define "zero longitude".

For directions to the stars, the distances are so great that it matters not if you base your coordinates at the Sun or at the Earth, because the distance between the Sun and Earth is so small in relation to the distances to the stars.

There are a few coordinate systems in use.  The first uses the Earth's Equator as the fixed plane and the direction of the sun on March 21 (vernal equinox) as the direction of zero longitude. For historical reasons the longitude in this system is called Right Ascension and latitude is called Declination. It would be quite suitable for your adventurers to use.

You can also use the plane of the solar system, or the plane of the galaxy as the fixed plane. In these coordinates systems Proxima is located at

-44:45:47.62 Latitude(deg)   +239:06:53.08 Longitude(deg) Ecliptic
-01:55:37.90 Latitude(deg)   +313:56:23.96 Longitude(deg) Galactic

But remember that Proxima is moving, so these figures are not fixed.


Can expansion of space cause the motion of galaxies, does expansion of space do effect motion of our sun around center and earth around Sun? I'm just wondering that expansion is anyway related to in effecting motion of a star under the galaxies ? Planets are currently revolving around sun in helix motion, if galaxies has motion as well reason to expansion of space then it is possible that, that can effect anyway on that planetary motion around stars ?

No. The solar system is a gravitationally bound system so it is unaffected by the large scale expansion of the universe - see this Wikipedia article for further explanation. Similarly, galaxies like the Milky Way, and even groups of galaxies such as the Local Group, which includes the Milky Way and Andromeda, are gravitationally bound systems and are also unaffected.


Is the pulsar map on the Pioneer/Voyager probes only meaningful for earthlings? The Pioneer probes brought plaques (and later the Voyagers two golden discs) with a "pulsar map" imprinted on them. The purpose of the map is to make the position of the Sun clear for an alien intelligence, in the case they encounter the probes in the far future, by creating a frame of reference of known galactic pulsars with specified periods and their distance relations to the Sun.
Pulsars have nice properties for this goal; they can be detected from large distances (which gives a higher chance of some of them been known by both sides), their period is very specific, easily measured with clocks and can be unambiguosly matched to a specific pulsar (if known), and their period changes over time in a predictable manner (which allows for aliens to trace back the date of origin in which the map was made).
But I have a question. If pulsars are detected on Earth when their beams point towards us (the "lighthouse"/"becon" analogy) and there might be many pulsars that are not known by us due to us being ouside of the cone where the beam can point, then wouldn't the 14 pulsar used in the "pulsar map" only be evident for aliens in the immediate solar neighbourhood? Maybe a few pulsars can be observed by them in other parts of the galaxy but the 14 of them would cenrtainly not be the same set of prominent pulsars we know of, right?
What am I missing?

You aren't missing anything. The pulsar map is fairly useless unless the spacecraft is found within the next million years, when it won't have got too far from the Sun (the probes will have travelled about 30 light years in that time).
The problem is not so much that the set of pulsars won't be visible, because they almost certainly would be if you change your vantage point by a few light years, but that the pulsar activity might cease. Pulsars have a limited lifetime and most likely will have stopped being pulsars in a million years or will not be identifiable from their pulse periods. In addition most pulsars have very high velocities, so they themselves will have moved by $\sim 10$ times as much as the probes.
For something of use beyond timescales of a million years they could have used a network of globular clusters, although again the motions of the globular clusters would need to be accounted for.  It's a tricky problem when everything that's close enough that that their positions would change if you move of order hundreds of light years are themselves in motion. And of course the Sun is moving around the Galaxy too.


Hypothetically, could a civilization in a solar system near a supermassive blackhole use the blackhole as a huge gravity lens telescope? I have heard of various proposals that use the massive gravity of stars to sort of focus light at and observe with a suitable telescope placed at such a distance, and I am assuming here that bigger the star in terms of mass, the more focusing and the better images the telescope sees
Assuming a hypothetical civilization, or perhaps humanity developed such a telescope, could they bring it feasibly close enough to a supermassive black hole at the center of say, our galaxy to observe insanely far away celestial bodies and bodies close by with incredible detail?
Typically, uptil what distance could we comfortably resolve details of say, the color and clouds of exoplanets, and view stars with this? A few hundred thousand light years atleast?

It is indeed true that we are contemplating the use of a star: namely, our Sun, as a telescope. The concept is called the Solar Gravitational Lens (SGL). See here for our very latest manuscript, which also contains references to our earlier work: https://arxiv.org/abs/2104.08442
The tremendous light amplification and angular resolution of such a gravitational lens telescope arises from the fact that its light collecting area is effectively the Einstein ring that is seen around the lensing object. The diameter of the Einstein ring determines its angular resolution; its diameter and thickness together determine the light amplification. When using the Sun as a hypothetical telescope, in conjunction with a modest (~1 m aperture) observing telescope, light amplification is of ${\cal O}(10^9)$ and angular resolution is below a nanoarcsecond, though deviations from spherical symmetry in the solar gravitational field tame these numbers somewhat.
These numbers are determined solely by the size of the Einstein ring, i.e., the radius of the Sun (since the Einstein ring has to appear around the Sun; the Sun itself is opaque.) What the mass of the Sun determines is how far away from the Sun images can form. Obviously, the more massive an object is, the more it deflects light, so the sooner light rays can meet. For the Sun, the earliest point where initially parallel light rays passing on opposite sides of the Sun intersect is about 550 astronomical units from the Sun. For a more massive object, this could happen sooner. Or conversely, at a given distance, the Einstein ring seen around a more massive object will have larger radius, corresponding to more light collected and higher angular resolution.
However, the fact that to reach, e.g., Sgr A* at the center of the Milky Way requires travel over several kiloparsecs kind of negates any practical advantage that an SMBH might offer as a practical gravitational lens telescope. Any civilization possessing the technology to travel across the Milky Way (assuming such technology even exists within the realm of physics) probably has other, more practical means at their disposal to make astronomical observations of the deep sky.


Would dropping a white dwarf or a "still-hot" stellar core into a large planet or brown dwarf create a star? In the Wikipedia article about stellar engineering, I found this quote:

In The Saga of the Seven Suns, by Kevin J. Anderson, humans are able to convert gas giant planets into stars through the use of a "Klikiss Torch". This device creates a wormhole between two points in space, allowing a neutron star to be dropped into the planet and ignite stellar nuclear fusion.

Now this is definitely impossible, as neutron stars are degenerate and cannot fuse. But what if we dropped a helium white dwarf or extract the hydrogen-burning core of a low-mass star and put it into a large gas giant or non-fusing brown dwarf? Would this trigger fusion or would this fail as well (presumably because the white dwarf is also degenerate)?

The issue is not whether the dense core can fuse, but what fusion processes can occur on its surface. Remember that novas happen when gas accumulate on the surface of white dwarf stars: the high temperature and density is enough to trigger runaway fusion.
Hence this method, where you to magically make a dense core appear inside a gas giant or somehow drop it onto the core, would likely produce a nova effect. Fusion starts, heats up the hydrogen, this speeds up fusion and increases the temperature, and then the whole thing blows up. Likely not enough for a supernova, but not really creating a star as much as just burning up the planet.
The exception may be a low mass star core, where the gravitational gradient and temperature may be mild enough to allow gradual burning. Here unfortunately a lot depends on the assumed magic making the dumping happen, which presumably makes the answer ill defined.


Why stars are named Gliese? There are exoplanets called Gliese 581c, Gliese 581d etc. I assume their star is Gliese 581. So where does this Gliese come from?

They come from Gliese's catalog of nearby stars (various editions and co-author), the number is the star's catalog number.


What would this moving point of light be? Tonight (10:00 PM EST Middlesex County, Mass.) when looking at a mostly clear sky I observed what I at first believed was a meteor. This point of light was moving east to northeast, and that would be in the vicinity of the Lyrid meteor shower.

In the app Star Walk for iOS it says that the "n-Lyrids" meteor shower peaks on May 9th. However, information I found online about this meteor shower says it already peaked back in April. (I did see a bright meteor a couple weeks ago.)

Can anyone offer insight on possible meteors right now? I highly doubt that this was a meteor though, because it lasted way too long, probably two minutes.

My next guess was the International Space Station but NASA's Spot The Station did not show any viewing times for my area today. The Star Walk app also showed the ISS well below the horizon.

This leads me to conclude the object was a plane, however there were no blinking lights and the only time its light changed was when it passed behind a thin cloud.

It could also have been a satellite glint such as an Iridium flare. I did not see the beginning of the light's appearance, so it is possible I missed the flare. However flares do not last for minutes, but glint from solar panels or other satellites may last this long.

What did I see?

This sounds like a satelite to me. Satelites can also "dissapear" right above you, since the sun no longer reaches them and they fade into darkness.

Also 2 minutes is very well possible for a satelites to be visible that is in earths orbit.

You should also know, that there are hunderds of other satelites that can be as bright when observed from earth as the ISS is.


How do rogue planets orbit around stars in other planetary systems? I got some interesting answers for What would happen if a rogue planet hit one of the planets in our Solar System?

But I have seen some documentaries that state that rogue planets from other planetary systems (which are ejected from the planetary systems in which they formed) have become part of the other planetary systems and orbit around those new stars as planets.

For example, please check this link:

  The latest research suggests that sometimes, these rogue, nomadic worlds can find a new home by being captured into orbit around other stars.

How is this possible? Stars' gravity can pull the rogue planet into them. How do rogue planets escape this and orbit around the new stars. This seems to be a bit mysterious.

Capture of any object is never common. A rogue planet that passes a star will be accelerated by the star's gravity, and provided that it doesn't hit anything will pass the star in a hyperbolic path.

For capture to occur, the rogue planet has to lose momentum, and there are a few ways in which this can happen. The most general way is for the rogue planet to interact with one of the stars existing planets. If the rogue planet passes close one of the star's orbiting planets, it can interact gravitationally, and transfer momentum to the planet, and slow down in the process. This is rather like the reverse of a gravitational slingshot.

It is also possible that a double planet can be captured, if one of the bodies transfers momentum to the other. One of the double planet will escape, the other is captured by the star's gravity.

Capture of rogue planets is very rare. There is no evidence that it has occurred in our solar system (never in 4.7 billion years is a big Never) though some have speculated that Sedna might be a captured object. But we know that captures like this are possible, as Jupiter and Saturn have a collection of moons, some of which seem to have been captured from the asteroid belt.


Does the Sun's light travel fast enough to have a straight path to Earth? When we travel through space, we use a curved path following an orbit to reach our destinations.

As far as I'm aware, we could take a more straight line approach but it would require much more deltaV (both in reaching the speed and in slowing down when reaching the target to be captured.)
However the path would still be slightly curved because of the curvature of space and because of orbital tracjectory when leaving Earth.
Does the light escaping the Sun move fast enough to follow a straight line path until it hits Earth, or is it curved in those 8 minutes?
What about to somewhere further like Pluto?

Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of now' through spacetime, which is of course depends on the reference frame one chooses to use.

For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right)}_{\mathrm{d}S^2}\text{,}$$
where $\Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $\mathrm{d}S^2$. The trajectory of light follows null geodesics ($\mathrm{d}s = 0$), so the light travels as if traveling through a medium of refractive index
$$n = c\frac{\mathrm{d}t}{\mathrm{d}S} = c\sqrt{\frac{1-2\Phi/c^2}{1+2\Phi/c^2}}\approx 1 - 2\frac{\Phi}{c^2}\text{.}$$
A changing refractive index does make light bend; this is exemplified in the simplest case by Snell's law, although the continuously varying case found here needs the more general Fermat's principle.

You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of gravitational redshift not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.

Light deflection is quantified by an impact parameter $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a scattering angle $\theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)

Modeling the Sun as having a spherically symmetrical gravitational potential $\Phi = -GM/r$, the leading-order light deflection is,
$$\begin{eqnarray*}
\theta = 4\frac{GM}{c^2b} &\quad&\left(\text{if}\;\frac{GM}{c^2b}\ll 1\right)\text{,}\end{eqnarray*}$$
which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.


Maximum disturbance to space time of gravitational waves, and will this affect humans? Do we have a theory that can tell us how much change gravitational waves can make to space time. If it is infinite then I think any time we can see biggest change in space time and how much could it affect to humans?

Gravitational waves cause a stretch and squeeze in space-time. It has two polarizations Plus and Cross polarization. This serves as an axis for the stretch and squeeze when a gravitational wave hits a body. However, the order of stretching and squeezing is actually really small. It changes space-time by an order of $10^{-21}$m. That's around the diameter of a proton. There are Numerical relativity simulations where you can see for yourself how much the change in space-time occurs for a certain pair of mass collisions.

Check out the Waveforms for SEOBNR, Taylor etc. You might get a good idea.

FYI : Question belongs in physics.stackexchange


Why do Jupiter impact events look dark brown? When a large asteroid impacts a rocky planet, the impact area becomes molten, looking red. However, when an asteroid or other large object impacts a gas giant, like Jupiter, the impact area turns dark brown. Why does this happen?

The initial impacts are brighter.  The team that observed the shoemaker levy impacts described them as "fireballs" and the initial pictures showed spots glowing bright after impact, though I'm not sure how much those images used the infra-red range, so, grain of salt, because I think you're asking about the visible spectrum.
Jupiter's bands are essentially Hadley cells where the darker bands are the warmer rising atmosphere and the lighter band is the cooling, more icy falling atmosphere.   The whiter bands across Jupiter's latitude lines, what you see is sunlight reflected off ammonia ice.   (the blue color of Uranus and Neptune is methane ice).  When gas giant planets get hotter, they tend to lose their pretty colors and become darker reddish or brown colors.  They hydrogen and helium that dominate the atmosphere are mostly transparent.  The color we see comes from ice-clouds and other molecules in the atmosphere.
Impacts on Jupiter, with the planet's high escape velocity,  creates an enormous amount of heat for any impact big enough for us to see from Earth and perhaps obviously, heat removes ice.
A variety of molecules play a role in Jupiter's reddish/brown colors, I'm not sure which is dominant, but phosphine gas/phosphorus compounds, Sulphur and hydrocarbons all reflect or scatter darker colors.  But the primary answer to your question is that the brown spots you see after impact are due to a lack of atmospheric ice in combination with sufficient mixing so that there's an abundance of colored molecules.  It's not too different from Jupiter's large storms that create it's dark or red spots.
A more recent image of an impact taken by an amateur astronomer.   The initial impact isn't dark.  It's lighter due to the heat of impact reaching above visible temperatures with presumably some atmospheric covering.

If you use an infrared camera, you probably see the spots remain brighter for some time, even as they become darker in the visible spectrum.


Is the dark energy between the moon and Earth measurable in any capacity? What are the experiments, or measurements, that can detect, or account for, dark energy involved in making calculations concerning Earth and its only natural satellite?

Yes and no.

First, the answers here should give you some understanding of dark energy and its effects.

The net effect of dark energy is tiny over small distances. If it is given by a cosmological constant, then its effects on gravitationally bound systems is independent of time (hence 'constant'). Observational evidence puts us at least close to the constant case. Which means that even if we are not actually in the constant case, then it will be many billions of years until there would be a measurable difference.

We can, however, estimate the mass equivalent of dark energy in a given area by measuring the apparent amount in the universe itself.

As estimated here:

  If we take the WMAP value for critical density at $$\rho_{c,0} = 9.47 \times 10^{-27}\, kg/m^3$$
  and presume that dark energy makes up about 73% of that, then the effective density of the dark energy would amount to just over 4 hydrogen atoms ($m = 1.67 \times 10^{-27}$ kg) in a cubic meter of space. If we take $5.9 \times 10^9$ m as a mean radius of Pluto and calculate the volume of a sphere of that radius, then the dark energy in that sphere would be equivalent to just under 6000 kg distributed throughout a space representing the solar system. The density of the asteroid Ida has been measured to be about 2.7 g/cm$^3$. So all the dark energy in the solar system would amount to only about 2.2 m$^3$ of the material of Ida, or a sphere of about 0.8 m radius. This is a hundred times smaller than Ida's tiny moon Dactyl!

Of course, that assumes dark energy is essentially the same on small scales as it is large scales, which is unknown. Nevertheless I think it is pretty clear that it would be extremely difficult to observe dark energy directly within the Earth-Moon system.

See also Stan's comment.

Our understanding of dark energy is minimal at best, to boot, giving us no solid ideas on how to detect it in the first place. We infer its existence from the large scale expansion of the universe, but we're otherwise short on solid ideas on its nature.


Are multiple geostationary satellites required to handle arbitrary locations? Do geostationary satellites need to have the equator as the plane of rotation, and the earth's centre to be the centre of rotation? Can it rotate over, say, the Tropic of Cancer, focusing on a single city?

If not, in which case, we need a chain of geostationary satellites to do a function, why employ geostat satellites at all? Aren't there satellites at lower altitudes which could do a better job?

Earth's center of mass must be at one of the two focus points of a satellite's elliptical orbit, or at the center of a circular orbit such as a geostationary orbit. One cannot orbit a certain latitude, except for the equator.

But there are clever alternatives for different purposes. Geosynchronous (as opposed to geostationary) orbits mean that the satellite will always be within sight from the same location on Earth, but it would move around and probably need antennas which continuously aim towards them. Japan is building its own regional GPS-system in this way.

There are also semi-synchronous orbits which put the satellite at the same place on the sky at the same time every day. This is used for some Earth observation purposes because the shadows on the ground remain the same for the same location on Earth everytime an image is taken.


How Can I Interpolate Ephemerides? I have two ephemerides for orbital elements—that is, each contains $a$ (semi-major axis), $e$ (ecc.), $i$ (incl.), $\Omega$ (long. asc. node), $\omega$ (arg. of peri.), $\tau$ (time of peri.), $T$ (period), and a timestamp $t$.

I want to compute a new ephemeris at a point somewhere between these two ephemerides.  For example, the timestamp ($t$) of the new ephemeris can be calculated by a linear interpolation of the start ($t_0$) and end ($t_1$) ephemerides' times as:$$
    t = \text{lerp}(t_0,t_1, x) = t_0 \cdot (1-x) + t_1 \cdot x
$$

Unfortunately, I can't seem to get an interpolation for the other parameters that doesn't have discontinuities in it.  My question: how should I be interpolating ephemerides?  Maybe I need to do this through some intermediate parameters, like $\varpi$, $\lambda$, or $\nu$?

  For example, if I take $\tau$ from the first ephemeris and lerp everything else, I get a strong discontinuity when mean anomaly is calculated (as $M:=n (t-\tau)=(2 \pi/T) (t - \tau)$).  I suspect this is because the mean anomaly calculation is only valid at the ephemeris, and as e.g. $T$ changes due to interpolation, $M$ changes erroneously.

  Here is an example for Mercury.  The first three ephemerides come from NASA HORIZONS and are correct.  Interpolating the first two ephemerides produces $M=328^\circ$.  However, two seconds later, the interpolation (which is now interpolating from the last two ephemerides) produces $M=294^\circ$.


Observed data: Why so many MPECs about TNOs lately? The Minor Planet Electronic Circulars (MPECs)
for July 15 through July 18 reported an unusually high number of re-observations of known trans-Neptunian objects (TNOs).
These were found in the
Pan-STARRS archive;
most had not otherwise been observed since discovery.
Is this activity part of the search for Planet 9,
another observing program such as OSSOS,
a result of software improvements,
or just someone clearing a backlog?

This is the result of a new way to investigate the images taken by the PanSTARRS survey, see
Weryk et al. 2016.

A second batch of new discoveries was issued on July 26.
This has been planned for several years, and now there's enough data to get useful results.

Note that Gareth Williams from the MPC has currently some technical problems with the most recent MPECs (O360 and higher). If these links don't work for you, it's probably because of that.

For further information, see
the PS1 Science Consortium blog
and Holman et al. 2015.


What is the movement of stars within the nucleus of a barred spiral galaxy I'm looking for visualizations that show the movement of stars in barred galaxies but not having much luck. I've found many that simulate spiral galaxy arms in general, but none that include a bar or a nucleus.

I'm trying to word my question in a way that is not dependent on my own assumptions, but some of my assumptions are:

the area that includes the bar is the nucleus.
the nucleus and the bulge are not necessarily the same thing. (Bars within the nucleus is a result of gravity waves. Bulge is an artifact of how the galaxy was formed.)
the edge boundary of the nucleus is a gravitational "tipping point".
stars are not crossing this boundary (ie: they either belong to the nucleus or to the arms, not both).

What is the movement of stars within the nucleus of a barred spiral galaxy, and how do their galactic orbits differ from stars outside the nucleus?

It's a little hard to know how to answer your question, partly because you're using "nucleus" in a confusing fashion. So let me try answering what might be your general question, which is "how do stars move in the region spanned by the bar?"

("Nucleus" is generally used to refer to the inner few hundred parsecs -- or even, these days, just the inner ten parsecs or so. Since bars extend to anywhere from several hundred parsecs to as much as ten thousand parsecs in radius, they are usually well outside the "nucleus". [I am deliberately ignoring so-called "nuclear bars" -- which are generally only a few hundred parsecs in size -- in order to keep things simple ;-)])

Stars in a disk galaxy generally all orbit in the same direction and in the same plane. Some of the orbits are close to circular, while others are more elongated. These elongated orbits precess -- that is, the axis of the apparent ellipse rotates with time, so the star traces out a rosette. In a barred galaxy, there are a lot of stars which are on elongated orbits which are aligned, and which precess at the same rate. Thus, they maintain a common oval region that is dense with stars (the "bar") which rotates as a whole at a more-or-less constant speed (even as the stars making up the bar are doing their own individual orbits at different speeds). The combined gravity of all these stars in an oval configuration is what keeps the whole thing synchronized and consistent, although there are some stars on more chaotic orbits.

There is a kind of magic radius called "corotation". This is where an imaginary star in a circular orbit would go once round the center of the galaxy in the same time that the bar as a whole make one complete turn. Stars inside corotation go once around faster than the bar turns, while stars outside this radius would go once round slower than the bar. Stars orbiting inside the corotation radius can form part of the bar (and generally have to be part of the bar if they're not on some very chaotic orbit). Stars outside corotation cannot be part of the bar; their orbits will actually tend to be elongated perpendicular to the bar, though as you get further away from the bar, its gravitational influence becomes less and less important, so that stars further out can have orbits that are pretty much circular.

There is good evidence -- both theoretical and observational -- that bars cannot extend further out than the corotation radius, and that bars in fact tend to extend about 3/4 or 4/5 of the way to corotation (with the stars further out but still inside corotation probably having very chaotic orbits).


Why do most meteors vanish just before hitting the ground? Many times we see in the sky that most of the meteor shower trails vanish when they are about to strike the earth several feet above the ground. Why does this happen? They could strike the earth's surface (still emitting light until they strike).

Pic source: http://imgur.com/afiZGCm

Only very few meteors actually make it anywhere near the surface of Earth; most burn up 75–100 km above the surface. From your point of view, however, the curvature of Earth's surface may make it look as if they get much closer, and even fall below the horizon. But depending on where you live, the horizon often has quite a lot more background light (e.g. from cities far away). That means that when they get near the horizon, which you may interpret as "several feet above the ground", the seemingly disappear.


How do we know an event is a gamma-ray burst? How do we know an event is a gamma-ray burst (grb)?

There are two kinds of grbs, long burst and shot burst, which are caused by supernova explosions explosion and the merger of two neutron stars, respectively. However, at least one grb, J1644, is a tidal disruption event. Is there any other grb which was not caused by a supernova or the merger of two neutron stars?

When can we say that a supernova explosion is a grb after observations? How do we know an event is a grb especially when we missed its explosion?

Can we tell that the afterglow is caused by a grb?

Please give me some reference papers since I am a layman of this field.

Yes, on the right track where the energy produced in a supernova can indicate if it was a LGRB or SGRB.

An indicative tools stems from the light curves of the remnants, both the afterglow brightness and it's region of explosion can indicatively be traced to how long the event lasted. Large gamma-ray remnants are typically associated with rapid star formation regions and are a more luminous remnant, where short gamma-rays are typically associated with little or no star formation regions and are a less luminous remnant.

It's not an exact science, as reproducing the effects and duplicating results have proven vastly difficult, however using distribution plots can definitively separate the differences and provide a theoretical model to work on.

Some minor information can be found at Wikipedia, however further reading and a more in depth view can be found at Cornel University Library. It is a bit of a read and attempts to explain the physics behind emission through mathematics, I'm not entirely sure whether is will provide the answer you are seeking.


Can a solar system exist where the second planet rotates fast, and the third planet is tidally locked to their star? Today I read about the Romulan home star system and it looks like it might be impossible for such a star system to exist.

Memory Alpha describes Remus:

  Remus was tidally locked, with one hemisphere always facing its primary while the other always faced away from it. The civilization of the Remans was located on the night side.

Remus is farther from their star than Romulus. Romulus rotates many times during a single orbit of their star while Remus is tidally locked and rotates once per orbit of their star.

If a planet orbits even deeper in the gravity well of their star than Remus it should be even more likely to be tidally locked than plants further out.

Could tidal interactions with neighboring planets and/or its moons prevent a planet from becoming tidally locked while being unable to prevent planets further out from being tidally locked?

The tidal locking timescale depends on several factors: $$\tau_{lock} \approx \frac{0.4 \omega_0 a^5 m Q}{3 G M^2 k_2 r^3}$$ such as the initial spin rate $\omega_0$, the semimajor axis $a$, the mass $m$, the solar mass $M$, the radius $r$ and various dissipation parameters $Q$ and $k_2$.

Two planets that merely differ in $a$ will have the inner one lock much earlier than the outer one. But give them slightly different $\omega_0$, $m$ and especially $r$ and the inner might dissipate more slowly than the outer.

There is also the possibility of spin-orbit resonances. Mercury is in  3:2 resonance, and tidal forces might lock planets into 2:1 or 5:2 resonances. That way the inner planet may be in a resonant lock while the outer is tidally locked 1:1. In fact, Venus is turning very slowly (and retrogradely) so it is not too far off, although the causes of the retrograde spin are unclear.

So, yes, you can have tidally locked planet orbiting outside non-locked planets.


What was the orbital path of earth before collision with a Mars sized planet? There are hypotheses about a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago, about 20 to 100 million years after the solar system coalesced.

I guess this changed the orbital path of earth around the Sun. Do we have any info what was the Earth's orbital path before the collision? (I would prefer some visual explanation.)

The total angular momentum would always be conserved about the Sun (assuming it at rest), as there is no torque about the Sun. Apart from that, I think it is hardly possible to say anything about the orbits. Also, it may be true that the pre-collision orbital plane and the post-collision orbital planes changed due to the collision, as small transverse perturbations must also have resulted from such a collision. However, I would also like to see what other members suggest.


Can spectroscopes identify minerals? Could a spectrometer on a probe or space telescope identify minerals, not only elements and molecules, from the spectral lines? Is there any technology for identifying minerals remotely?

Yes, minerals can be observed using spectroscopy on a telescope, typically from their thermal signature. We can determine what type of mineral based on the elements that make up the mineral as well as the type of crystal structure, which will give a different spectral signature


How do I get the change in angular diameter value? Given that I got the equation relating the angular diameter(θ) and $V$–$K$ band value, which is known as $\logθ_0=(0.262±0.004)(V−K)_0+(0.547±0.006)$ from this paper, how do I get the 'change' in angular diameter value from that equation?

1) Can I just get the max and mix value of the $V$–$K$ value and insert each into the equation and get max θ and min θ and get the difference?

2) Or I need to the whole lists of angular diameter value over each period and insert the lists of value into the equation and plot the graph and then get the difference?

3) Or is there any other method to do this?

You can see from the paper you linked that they followed the procedure you outlined in option (2). Figure 4 of that paper shows the $\theta (t)$ for a few stars. They specifically state towards the end of section 6

  Figure 4 shows in the top panel the relation between the angular diameter predicted by the SB-relation versus radial variation, and in the bottom panel the angular diameter as a function of phase.

The SB-relation they're talking about is the one you cite in your question. So your process should be to take your list of $V-K$ colors over time and convert them to angular size using the surface brightness relation, then fit a function to that (via cubic spline, least squares, etc.) and determine the total angular variation over time.

I will note also that the $\theta$ vs $V-K$ relation you've cited here is equation 10 in the linked paper which is not a result of that paper. Their result is listed in equation 9 with equations 10 and 11 being results from other, previous work for the reader to compare. You should sure of which equation you want to use and cite the appropriate source.


If Pluto is tidally locked with Charon, is its rotation still slowing down? I know that a a planet or moon's rotation will slow down or speed up until it is tidally locked with a body, is Pluto's rotation being affected by the sun or is it being forced by Charon just to to be tidally locked with Charon?

Short answer.  Pluto's rotation is 100% governed by Charon, or, more accurately, the Pluto-Charon system.  The sun is irrelevant to it's rotation.

Small point to add.  Pluto's rotation is pretty much governed by the Pluto-Charon orbit.  If Charon moves closer to Pluto, Pluto's rotation will correspondingly increase.   If Charon moves further away, Pluto's rotation will decrease.  That's how tidal locking works.   I answered the question in regards to the orbit changing, not the rotation changing, but in a tidally locked system the rotation period follows the orbital period.

Long answer:

  I know that a a planet or moon's rotation will slow down or speed up
  until it is tidally locked with a body,

Tidal locking doesn't always happen.  For distant orbits the tidal force that makes tidal locking happen is often very weak and very slow, often requiring more time than the lifespan of the star in that solar system.

Also, for high eccentricity orbits, classic tidal locking is impossible though resonant orbits like Mercury's 3:2 around the sun are probably fairly common.   And many moons will escape their planet or crash into their planet long before mutual tidal locking occurs, though it's not uncommon for moons themselves to be tidally locked.   Also small moons don't have sufficient gravity to pull the object they orbit into mutual tidal locking.

For mutual tidal locking, like Pluto-Charon, there needs to be respectible mass ratio or, perhaps an initial formation where orbital speed and rotation are very close, so it doesn't take much force to lead to mutual tidal locking.  Our moon, at 1/81st the mass of the Earth, would eventually cause the Earth to tidal lock to it, but estimates say it will take longer than it will take the sun to go red-giant for that to happen, and the expected outcome of our moon is that it will crash into the Earth before there is mutual tidal locking.

So, mutual tidal locking like pluto-charon is actually pretty rare because you need the smaller object to have a significant percentage of mass to the larger one.   One-way tidal locking is quite common, but your question addresses mutual tidal locking.

  is Pluto's rotation being affected by the sun or is it being forced by
  Charon just to to be tidally locked with Charon?

For a quick approximation, the tidal force can roughly/bad approximation be estimated by the size of the object in the sky times it's density.   There's other factors, ofcourse, like the size of the object receiving the tidal force, but for a rough approximation, the size in the sky gives an estimate.   For example, the sun and moon each take up about 1/2 a degree of arc when viewed from the surface of the Earth, so they have similar tidal forces due to having similar size.   The major difference being that the Moon is about 2.4 times more dense than the sun so it's tides are correspondingly about 2.4 times as strong.

This visual size approximation doesn't work for very-near-earth orbits, as it's the size that would appear from the center of the Earth, not the surface that should be applied.

Charon from the surface of Pluto is huge.   It spans several degrees of arc, over 10 times the size the Moon appears from the Earth.   The sun, from Pluto is tiny about 1/30th the size it appears from Earth, which is only about 50% larger than Venus appears in the sky at it's largest, or, about 1 minute of arc, but even at that distance the sun is still bright enough to cast visible shadows and would probably damage your retina if you stared at it.   It would be the most impressive light in the sky from Pluto, but still tiny, so the tidal effect on Pluto-Charon would be quite small.  (I'd prefer not to do the math, my math tends to run into errors and I'm not good with mathjax).

But the sun's tidal influence on the Pluto-Charon system is close to negligible, It can be calculated and it's not zero, but it's very small.   Over a very very very long time, the Sun's tidal force will sap energy from the Pluto-Charon system and draw them closer towards each other, but at that distance, billions maybe trillions of years.  Pluto is probably more likely to meet a different fate well before then, such as crashing into Neptune or being thrown out of the solar-system before the Pluto-Charon system becomes unstable.

It would be different if Pluto-Charon orbited at Mercury or Venus distance (ignoring the obvious melting of ices and outgassing in a warmer environment).  That close, the sun becomes a major player in the future of the mutually tidally-locked systems, likely causing them to crash into each other, but at Pluto distance, there's virtually no effect.

When a system is mutually tidally locked with a nearly circular orbit, then it's very stable.  If there's no tidal locking or one way tidal locking like Earth-Moon, then there's a tidal bulge that leads or follows the moon and that creates a force that can slow down or speed up the planet and push the moon away or draw it in closer.   With tidal locking, there's no leading or following with the tidal bulge, so the orbit is less prone to change.   There's always some orbital variation and some orbital energy converted to heat due to no orbits being perfectly circular, but mutual tidal locking is still a very stable orbital system with less change than most, as long as there isn't a 3rd party forcing change, such as a debris cloud creating drag or outgassing or orbiting close to a massive body that saps energy from the tidally locked orbit.

No orbits are stable forever, and considering all the factors, Pluto and Charon are close enough that they exchange some atmosphere and as the sun grows more luminous over time, that exchange will probably increase, and likely both are losing mass slowly over time due to the solar wind.    This very gradual loss of mass, I would think, would cause the orbit to expand, but very slowly.   With no loss of gas by the solar wind, the Pluto-Charon orbit should very slowly contract due to tidal forces from the sun and some orbital energy being converted into heat, but both of those effects are very slow.   Mutual tidal locking is for the most part a very stable system.

Hope that wasn't too long, but that's the gist of it.   My answer could probably be improved with some math and some linked references but that would make it even longer.


Calculating the mass of star I am doing a project which requires that I have data for the mass of several stars. However, the data for mass of stars is obviously not readily available as it's really hard to calculate. In such a scenario, would it be alright to use the mass-luminosity relationship with different values of p(3.5 for a main sequence star) without getting major errors?

Also, what value of p should i use for other stars like white and brown dwarfs?

You don't say what other information you have for the "several stars". Yes, you can use a mass-luminosity relationship if the stars are on the main sequence. In terms of mass uncertainties I would estimate that you might be at the level of 20% unless you can absolutely pinpoint them on a Hertzsprung-Russell diagram, because the luminosity of a (fixed mass) main sequence star changes, even whilst it is on the main sequence.

If the stars are in any other phase of their evolution then your problem is much harder, even if you can place them on the HR diagram. Not least because there is considerable disagreement between different evolutionary models as regards the positions of stars of a given mass at a given age, on both the pre- and post-main sequence. Furthermore, their luminosity and temperature evolution can be influenced by factors such as rotation, metallicity, magnetic fields and mass-loss.

Compact stellar remnants and brown dwarfs do not follow a mass-luminosity relation. In general you can only estimate their masses if they are in binary systems, or for white dwarfs, you can estimate their radii and either their surface gravity or gravitational redshift. Estimating a brown dwarf's mass needs both its luminosity and age.


Moon, center of mass & geometric center, tidal locking Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they? Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)

Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they?

They differ by about two kilometers.

  Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)?

Yes, but that condition is not what I would call "perfectly homogenous". What's needed is a non-spherical mass distribution. For example, a uniform density ellipsoidal body would have its center of mass and center of figure of coinciding, but would still be subject to torques. It's the moment of inertia that is key rather than the offset between center of mass and center of figure. An object whose center of mass and center of figure do not coincide necessarily has a non-spherical moment of inertia tensor.


Is it likely that intergalactic stars would still retain their planets? Given that we've discovered hundreds of intergalactic stars (IGs), and that most stars have planets, what do we know about the likelihood of IGs retaining their planets after being ejected from their galaxy?

Related to that, what are the chances of rogue planets being intergalactic as well? If there's a planet floating out in intergalactic space, is it more likely to be orbiting a star, or on it's own?

While near encounters with supermassive black holes are bad for the stability of solar systems [citation needed?], many intergalactic stars become intergalactic without a close encounter. They are part of streams and tidal tails that emerge when galaxies or globular clusters get tidally disrupted by another galaxy.

Still, even while encounters with black holes are usually disruptive simulations suggest that some planets may remain around the star. This is most likely for planets real close to the star, like hot jupiters.


What can we expect to see with a telescope with a 70mm aperture and a 10mm eyepiece? We purchased my young daughter a beginner telescope for Christmas and she's super excited to look at the large planets. Our telescope has a 70mm aperture and two eye pieces (10mm and 20mm) How much, if any, detail can we expect to see on Jupiter and Saturn? I don't want her going in expecting to see Saturn's rings and come out disappointed! We are very new to this and I want her to have the best experience possible with the equipment we have. Thanks!

I would buy a good 2x barrow lens to go with that. She might see Saturn's rings as a haze. Good for looking at the moon. Maybe Mars. A good tripod is a must.  I would suggest a camera mount. A light weight camera.  Secure the front of the telescope with 3 rubber bands to the tripod. To reduce vibration. 2 seconds pause on the camera. For vibration to end. A telescope that size is a fair place to start on Photos. A 3 second exposure. For stars planets & such. She can get some good photo's. Also other photo's. 70mm is a good daytime telescope size. Make sure she never looks towards the sun with it. For night work you need a 90mm or above. Not into astronomy. But use a 90mm Vexin for photo work. Ships in the bay. Objects up in the mountains that can not be got close to in Asia. A camera makes it so much more fun. You can build a camera mount not hard to do.


Why is right ascension measured in hours? When viewing star maps in Stellarium, I noticed that the meridians of the equatorial grid were denoted in hours and not degrees, and that always puzzled me. Why did astronomers choose this system? I understand it has something to do with being relative to the Earth's rotation which takes 24 hours, but I can't figure out exactly what.

In order to know when a star will be above horizon, you'll need an equation with times, not "celestial latitudes".

So you end up adding RA and sidereal time.

EDIT

In order to find out the position of a star you need to find out how its Hour Angle varies with time.

The Hour Angle is how far the star is from the observer's meridian:

In this image, it is the angle between the site's celestial meridian (North Celestial P ole-Z enith-S outh horizon) and the star's celestial meridian (P-X-Y). It is the same angle as between where X culminates and X.

So it comes than $HA=ST-RA$ and since you may not have a "local sidereal time" clock at hand, you can use Greenwich's Sidereal Time: $HA=GST-Lat-RA$

$HA$ (local) Hour Angle of a star
$ST$ Local Sidereal Time
$GST$ Greenwich Sidereal Time
$RA$ Right Ascension of a star
$Lat$ (local) Geographical Latitude


How does one go about submitting a name for a newly-discovered lunar crater? Suppose I discover an uncatalogued crater on the lunar surface, and decide to name it. How would I submit my proposed name to the IAU?

You use this from https://planetarynames.wr.usgs.gov/FeatureNameRequest. You need to follow the rules as set out in https://planetarynames.wr.usgs.gov/Page/Rules as well as following the themes for naming Moon based features https://planetarynames.wr.usgs.gov/Page/Categories.


Calculation to Convert from Equatorial to Heliocentric coordinate systems Can anyone help me with the formulas to convert Equatorial coordinates (Right Ascension / Declination) into Heliocentric coordinates (Angle and Distance in relation to the Sun) ?
i.e. Ra/Dec of 6.7659/18.4563 ---> 281.98 degrees and 0.4562 AUs
It feels like this is possible with no other inputs, but I can't figure it out.

No, this is not possible. Equatorial coordinates only specify the direction of a certain point in space (as seen from Earth), not the distance. During a solar eclipse, the Moon and the Sun have (approximately) the same equatorial coordinates, but the distance to the Sun is 0 or almost 1 AU.
It is possible to define equatorial or ecliptic coordinates with the Sun as center, but even then there is no general formula. You'll need the full position (i.e. three coordinates) of the Earth (relative to the Sun) or vice versa, and the full position of the object under observation to be able to convert the coordinates.


What would be the practical consequences (on earth) if the Moon was not tidally locked? I was thinking about the fact that all the largest Solar System moons are tidally locked to its primary and this question arose.

If the Moon was not tidally locked, it would mean that the Moon is in the wrong orbital distance from Earth. The Moon is tidally locked because it is close to the Earth. If the moon were closer, it would approach the Earth's roche limit, be torn apart, and its debris would become a ring for ~100-200 million years. If it were too far, it would continue to drift outward until it was free of Earths' gravity.

By being tidally locked, the Moon has been extending the Earth day by slowing down the Earth's spin, to about ~6 hours from an 18 hour day to a 24 hour day. This process will continue into the future until the Earth is also tidally locked with the Moon. The Moon, while in this state, has also stabilized the Earth's seasons by stabilizing the Earths' orbital axis.

Lastly, a possible indirect consequence would be upon human history. Mankind would have probably abandoned the celestial sphere model in favor of the spinning Earth model much sooner had they seen proof that a celestial body could spin and rotate, much like when Galileo saw that moons were orbiting Jupiter.


Can a tectonically inactive planet retain a long-term atmosphere? Can a planet be tectonically inactive and still retain a magnetosphere and protected atmosphere? How does that work?

How else could a planet retain a thick atmosphere like Earth's for extensive periods of time?

Note, I'm not asking on Earth science SE as, well, the question is not about Earth. I was hoping there was some sort of astronomical theory for other potential habitable planets.

Yes, a tectonically inactive planet can retain a long-term atmosphere.
You make the connection that a lack of plate tectonics on a planet indicates a "dead" core and thus said planet has no magnetosphere. As such, I'm going to interpret your question as, can a planet without a magnetosphere retain an atmosphere long-term? As proof, I offer up Venus.
Venus' Magnetic Field
Venus is a planet without a magnetosphere generated by a core. It is thought that the cause for this is Venus' slow rotation rate (of nearly 243 days) and a lack of convection, allowing for bulk motion in the core. As I'm sure you know, you need moving charge to make magnetic fields and Venus' core just isn't moving. As such, we see that Venus is a tectonically dead planet - it's surface is roughly 500 million years old, whereas Earth's surface gets recycled every 100 million years or less due to our plate tectonics.
Now, Venus is not entirely devoid of a magnetic field. Ironically enough, it's lack of a magnetosphere allows for the generation of a magnetic field by it's atmosphere. Because the Sun's radiation is more or less directly hitting the atmosphere, Venus has a strong ionosphere. When you get lots of charged particles moving around in an atmosphere, you get a magnetic field. But on the whole, this field is very, very weak compared to a true magnetosphere such as we have on Earth.
I found this source which talks a lot about this concept and why Venus doesn't have a magnetosphere. Check it out to get a lot more in depth detail.
Venus' Atmosphere
So, Venus has no appreciable magnetosphere (or plate tectonics). Why does it have an atmosphere? And boy does it have an atmosphere. The surface pressure on Venus is estimated to be $\sim93\:\mathrm{atm}$.
In short, the answer is that the inundation of solar wind against an atmosphere is not necessarily the major contributing factor to atmospheric loss. It can be, but not always. For example, Mercury, another planet with a weak (but non-zero) magnetosphere, has no atmosphere (if it ever did) because it is so close to the Sun that the solar wind likely blew away that atmosphere long ago. Venus on the other hand is far enough away that the solar wind just can't strip the atmosphere. Here I'm going to quote wikipedia directly (emphasis mine).

A lack of magnetic field does not determine the fate of a planet's atmosphere. Venus, for instance, has no powerful magnetic field. Its close proximity to the Sun also increases the speed and number of particles, and would presumably cause the atmosphere to be stripped almost entirely, much like that of Mars. Despite this, the atmosphere of Venus is two orders of magnitudes denser than Earth's. Recent models indicate that stripping by solar wind accounts for less than 1/3 of total non-thermal loss processes.

Atmospheric Loss
If the solar wind isn't the contributing factor in atmospheric loss, what is? The answer to that is a process known as Jean's Escape. To put it simply, for gas particles in the atmosphere to escape into space, they need enough energy to climb out of the planet's gravity well. Some particles will have that energy and thus escape into space. Over time, the atmosphere bleeds off, little by little (this is happening for Earth too!).
The factors that contribute to the rate at which a planet loses its atmosphere are such things as the the planet's mass and radius, and the mass of the atmospheric particles. Let's look at Venus. It is comparable in mass and size to Earth and so has a reasonably appreciable gravity well. For something to escape Venus it must be traveling at $10.4\:\mathrm{km/s}$ (compared to the Earth's $11.2\:\mathrm{km/s}$). But, for Venus at least, the important factor there is that the atoms and molecules in its atmosphere are heavy. It is almost entirely ($\sim97\%$) carbon dioxide which has a mass of $\sim44\:\mathrm{amu}$. That means, the chances of such a massive particle getting the energy to escape is pretty small.
Out-gassing
Just one more point to add to this. One may argue that possibly the atmosphere is/can be replenished continuously, but that won't work here because we're assuming the planet is tectonically dead. You can't really have out-gassing on a planet with no active surface.
Conclusion
There are many factors that determine atmospheric escape. Different planets will lose their atmospherics for different reasons. However, it is entirely possible for a planet, under the right conditions, to maintain an atmosphere long-term, despite lacking a global magnetosphere. As we can see from Venus, the conditions are generally that the planet should be sufficiently far from the star, its atmosphere should be sufficiently dense and comprised of heavy particles, and the planet itself should be large enough to have an appreciable gravity well. If these conditions are all met, a planet may retain an atmosphere without having a magnetosphere to protect it.


Is it Possible for Planetary Bodies To Exist In Close Proximity Without Adverse Effects? There are several sci-fi movies where planetary bodies are shown in the sky in order to indicate that the characters are on an alien planet. A good example is in Predators (2010):

Is it actually possible for such planetary bodies to exist so close to each other as to be visible in such detail to the naked eye? Wouldn't gravitational forces cause disruptions on them that inhibit the development of life (whether plant or animal)? In our Solar system there's a good example in Io which has severe volcanic activity resulting from tidal heating due to its proximity to Jupiter.

Quick calculation of Roche limit for two identical earthlike (fluid) bodies.
Roche distance is about equal to 2.44 X radius X(density1/density2)^1/3.
For identical bodies the density conveniently cancels, so we get
Rd = 2.44 X radius.
Earth radius is 6378 km.
So Rd = 2.44 X 6378 = 15500 km (center to center).
The distance surface to surface (15500- 6378) is 9184km.

At that distance, by trigonometry, another earth would subtend about 44° of sky, pretty big!, and just at the bleeding edge of stability.

The nearest planet in the provided picture looks like it might be at about that distance. Evacuation might be in order as the other three planets or moons will degrade the close binary's orbital stability.


Underground sea in Titan(Saturn's moon), water or hydrocarbons? Is the underground sea on Titan formed of water or hydrocarbons. Is anyone certain?

I have read a lot of various documents from NASA and Cassini missions and Huygens missions, and I have become confused.

Titan is believed to have a layer of liquid water under its surface, due to tidal heating of the icy crust.

Titan has a surface formed of ices with some hydrocarbon (methane) lakes. The crust of Titan is mostly water ice.  Below this, there is a liquid water layer and perhaps more ice and silicate rocks below that.

This is described at https://science.nasa.gov/science-news/science-at-nasa/2012/28jun_titanocean


Mountains are higher than the atmosphere? Is it theoretically possible that planets exist with mountains so high that their peaks overtop the planets atmosphere? And which physical laws are relevant for this question? I'm just curious.
The question came while I imagined the atmosphere like a second ocean above our ocean of water. And I thought it would be nice if such a gaseous ocean could have islands as well.

This is a bit of a gray area, as an atmosphere doesn't have a clear boundary. That being said, Olympus Mons on Mars is so tall, the atmospheric pressure on top of it is only 12% the average pressure on the surface of Mars. That's near vacuum by terrestrial standards.

https://en.wikipedia.org/wiki/Olympus_Mons#Description

In general, for this to happen you need:

a pretty thin atmosphere to begin with
some exceptional geology that ends up producing very tall anomalies like Olympus

It's not a very likely combination, but it can happen, as seen on Mars.


How can I crop a fits file and get correct WCS header info without a new plate solve and WITHOUT USING PYTHON? While my question is similar to this one, the suggestions offered there won't work for my specific situation.  For over a month I've been trying to find software capable of this but it's been an unsuccessful search.  I don't think the software exists; however, I haven't asked this group yet. I sure hope I'm wrong.
The Problem

if I crop a fits image, the WCS is lost (which makes sense, the image has changed) but the cropping software won't recompute the WCS for the new area and add it back into the new header

The Data Set

a collection of .fits image files created by a CCD (not tables or any
other kind of .fits file)
each individual image is large in terms of degrees of sky
minimum number of files in set: ~1,150 and max number is: ~15,000
each image has the WCS info in the header
each image is registered on a sub pixel level to the others

Software Requirements

IS NOT PYTHON unless it has a GUI interface hiding all evidence of Python code
runs in local Windows environment, not cloud based
works in a batch or bulk mode
copies the header from the original but updates the WCS entries
updates the other headers as necessary such as NAXIS1

Additional info: why standard methods don't work

the images can't be plate solved by astrometry dot net (they are very out of focus (on purpose with a diffuser) or the plate scale is extreme at over 20 arc-secs per pixel which looks out of focus but isn't)
while the crop with WCS update could be done in Python, this project requires a GUI

I've tried SAOimageDS9, AstroImajeJ, PixInsight, Mira, and MaxIm DL 6 Pro but they don't recompute the WCS entries. My goal is to create a procedure to do aperture photometry on special data sets without using Python but first I need to determine if it's possible.  A crop is required before all else due to the size of the data.
When working with other data sets, cropping issue isn't a problem.  The images are registered, then the crop is performed (as well as any 'rotate 180 deg' or mirror left/right/top/bottom).  It removes the WSC headers but plate solving the new, smaller, postage stamp image puts the corrected ones back in.
If anyone knows of software to crop fits images and return correct WCS headers, it would help tremendously.

Skycat also stores the WCS information when selecting an image region from a fits file. Not sure how exactly this can be done in non-interactive mode but many features can in fact be "batched". Give it a try. Best luck!


Is there a connection between black holes and dark matter/energy? Is there a spatial or gravitational relationship between black holes and dark matter/energy?

No, not really.

As Stan Liou hints at in his comment, massive objects such as black holes (BHs) were previously a candidate to dark matter (DM). But it can be shown that in order to explain the various problems that the existence of DM needs to solve, there would have to be so many, that their gravitational effect ("microlensing") on background sources would be much larger than what is observed. Today, most people think that DM is simply a particles with an extremely small interaction cross section, such that it virtually doesn't interact with other particles, except through gravitational attraction.

You also ask if there's any spatial relationship, and in a way you could say there is. Most, if not all, galaxies sport a supermassive BH, and these tend to lie in the center of the gravitational potential wells created by the galaxies' DM. "Normal" BHs, however, just float around like stars.

BHs also tend to grow by accretion of matter onto them, and of course they can also eat DM, so to some extend BHs also consist of DM. But since the DM doesn't interact, a DM particle would need to fly directly toward the BH, lest it would simply be deflected and escape. Normal matter, on the other hand, is slowed down by friction and spirals toward the BH. Thus, BHs consist mostly of normal matter. It doesn't really matter, though, since after something has fallen into a BH, everything but its mass, charge, and angular momentum will be forgotten.

Dark energy is something completely different, better described as a negative pressure which is a property of space itself. I think. Or we think. Nobody really knows. But it presence doesn't correlate with that of BHs or DM. Ten years ago, there was a hypothesis that a collapsing star should be converted into dark energy rather than a BH, causing a so-called dark-energy star, but I don't think anybody believes this anymore.


Could this be a cosmic ray hit on my cameras sensor (CMOS, DSLR)? While taking pictures of the bubble nebula I noticed a very strange artifact on only one of my pictures. It can't be a satellite since its not a straight line and this is a 90s exposure that should make a line much longer. Also I doubt this is just readout noise from the camera (image is unstretched, unedited) since hot pixels don't form lines. So could this be a cosmic ray hit? If not does anyone have an idea what this could be?

The image was taken with my Canon 7D Mark II @ISO1600, 90s exposure.
Celestron C8 SGT (XLT), Celestron Advanced VX

It's a track about 50 pixels long, at 4.1 $\mu m$ per pixel that's 200 $\mu m$ in the plane of the sensor.

It's about 1 pixel wide, quite lumpy, and seems to change direction a bit. It really fits the description of an ionizing particle. There are multiple possible sources,

I'm pretty sure @JamesK has nailed it as an electron, but what else can we learn? There must be some way to get a very rough idea of how many $e^- h^+$ pairs correspond to a fully exposed pixel at "@ISO1600", then we could convert it to a dE/dx and see if it's minimum ionizing or if it's much higher.

Going diagonally, with a density of 2.3 g/cm${}^3$ that's about 1.3 mg/cm${}^2$ areal density per pixel. A [minimum ionizing particle] at 1.5 MeV / g cm${}^2$ would deposit only about 2000 eV which would make less than 1000 $e^- h^+$ pairs. That might correspond to a bright pixel. It's not out of the question that it's a minimum ionizing particle.