Vmag versus VpgMag in GSC 2.3 catalogue I tried to find out the magnitude in V for V* V881 Per.
Here are my results:
SIMBAD:  $11.09$    (Fluxes->V)
GSC 2.3:  $10.9815$  (VpgMag) and  $10.887$   (Vmag)
Which of VpgMag or Vmag in GSC 2.3 correpsonds to SIMBAD V-flux?
For each star GSC 2.3 cites two values for magnitude in V so which of the two should I use?

Your "GSC 2.3" entry is actually from GSC 2.4.1, as shown in the header lines containing the string "GSC241".
According to the record structure for GSC 2.4.1, Vmag is "V-band magnitude" while VpgMag is "Photographic V-band magnitude". The magnitudes are associated with a code that describes what bandpass/filter they are using. In your case, the VpgMagCode field is 1, which corresponds to the emulsion+filter IIaD+W12 (au used by the Palomar Quick-V survey). Confusingly the VMagCode field is set to 3, which corresponds to a Johnson B-magnitude. Perhaps this is an error in the catalogue and should be 4 (Johnson V-magnitude)? The source code of 19 indicates the magnitude is from APASS (the AAVSO Photometric All-Sky Survey). The APASS website provides some further information on the magnitudes used:

Johnson B and V were chosen to extend the Tycho calibration to fainter magnitudes and to match the many archival published datasets. These passbands can be easily used to create sequences for any VSX / GCVS star. Sloan passbands provide a direct link to the SDSS, SkyMapper, PanSTARRs etc. surveys, and provide a homogeneous catalog of brighter stars.  The Johnson B and V magnitudes are on the Vega system; the Sloan u',g',r',i',z_s and Z magnitudes are on the AB system.

A few more details on the Johnson magnitude system can be found on this web page: the system is set up so that the star Vega has a V-magnitude of 0.03, and the colors (e.g. B-V) are 0. Details of the Johnson-Cousins UBVRI filters can be found on the website of the Leibniz-Institut für Astrophysik Potsdam.
The details of the magnitude in SIMBAD can be obtained by going to the reference listed there: Norton et al. (2007). This paper indicates that the magnitude is from the SuperWASP photometric survey. The details of the filter the SuperWASP survey uses can be found on the SVO Filter Profile Service. Norton et al. (2007) describe the value in section 3 of the paper:

the mean SuperWASP magnitude,
defined as $-2.5 \log_{10} (F/10^6)$ where $F$ is the mean SuperWASP
flux in microVegas; it is a pseudo-V magnitude which is comparable to the Tycho V magnitude

A microVega is a millionth of the flux from Vega. Note that the above formula implies a V-magnitude for Vega of 0, in contrast to the Johnson system.
As for which one you should use, it depends on what you are doing. If you're comparing with other SuperWASP results (or perhaps objects in the Tycho survey), use the SuperWASP V-magnitude. If you're comparing with other magnitudes in the Johnson system, go with Vmag from GSC 2.4.1.


Has "GHz-spinning dust" ever been demonstrated in the laboratory? This excellent answer to the question What is the physics of the “spinning dust” contribution to Cosmic Microwave Background measurements? includes the figure below which helps show what it was the triggered the existence-hypothesizing of cosmic spinning dust.

The question Why doesn't the equipartition theorem disallow spinning dust? is my way of starting to wonder if spinning dust is real, or if it really happens.

One thing that would make me more comfortable than an argument gedankenexperiment involving a modified Crookes radiometer with the gas removed would be some kind of experiment showing that simply being exposed to some very low pressure hydrogen, or photons, can make dust spin at anything close to 20 GHz!

These days optical trapping of particles in very high vacuum is readily accessible for example, so I think if someone wanted to try to make GHz-spinning-dust they could certainly have a go.

Question: Has this been tried? Has "GHz-spinning dust" ever been demonstrated in the laboratory?

This poster suggests an answer of yes and no. Essentially the poster reports a demonstration of some of the proposed mechanisms, but the rotation speeds produced are in the $kHz$ not $GHz$ range because the dust partcles are bigger, the vacuum less good and so on.

To quote the summary:

  First laboratory measurements of rotation of the analogs of interstellar
  dust grains induced by radiative torques have been made
  The measurements were made on irregularly shaped SiC particles
  particles of ~ 0.17 to 8.2 $\mu m$ effective radii levitated in an electrodynamic
  balance and illuminated with laser light at a wavelength of 5320 Å, at
  pressures of $10^{-3}$ to $10^{-5}$ torr.
  The rotation rates with radiation intensities of ~ $4-30 Wcm^{-2}$ were
  observed in the range of 1,000-22,000 rot/sec.
  The grain rotation rates were observed to follow the expected functional
  relationships, being directly proportional to the incident radiation
  intensities and inversely proportional to the drag represented by the
  ambient pressure.


Why are main sequence stars more massive than the Sun less dense? e.g. Vega, Spica etc Why do main sequence stars more massive than the Sun have lower densities? e.g. Vega, Spica etc.

Main sequence stars are defined by being hot enough in the core to fuse hydrogen, so their core is at least about 10 million K, and can get up to 20 million K for the more massive ones (because they are more luminous, so their nuclear fusion has to crank itself up a bit more).

The way they get their cores hot enough to fuse hydrogen is via gravitational contraction, but a more massive star does not need to contract as much to reach the necessary temperature.   Since it does not need to contract as much, it is at lower density.

To see the scaling approximately, use the virial theorem, which says that the energy per particle in the core (roughly set by the need for fusion) is about $GMm/R$, where $M$ is the mass of the star, and $m$ is the mass of a proton.  Expressed in terms of density $\rho$ instead of radius $R$, that's proportional to $M^{\frac{2}{3}}\rho^{\frac{1}{3}}$.  Ergo, $\rho$ drops when $M$ rises.


Does the density in the core of a "higher mass" star stay generally constant while it evolves on the MS? We know the CNO cycle produces the majority of energy production in a "higher mass star" of approx. over 2 solar masses and the core is convective due to the large temperature gradient. My professor has noted that radius, luminosity, T_eff, and T_center all increase but rho_center (density) does not. I cannot find anything in my text that explains this, but I assume it is because convection works against gravity. If someone could clear this concept up (as to why rho_center) stays generally constant, I'd appreciate it. Conceptual explanations are helpful!

The most important thing that happens is hydrogen is turned into helium, which essentially means that electrons and protons are getting merged into neutrons (the neutrons inside the helium). That reduces the number of particles.  For example, if you turn a pure hydrogen gas into a pure helium gas, you now have half the electrons you had before.  Also, you have 3 particles (a He nucleus and 2 electrons) where before you would have had 8 (in the case of pure H initially, just as an example).  Losing particles means you lose pressure at a given temperature and volume (it would be proportional to number of particles), so the core must contract.  The temperature does go up a little, but not much because nuclear burning is very temperature sensitive.  But the particle density drops a great deal, so for the density to stay the same, the core pressure must drop quite a bit, perhaps by almost a factor of 1/2.

The reason this might not require the core density to change is that the core pressure scales roughly like the radius to the -4 power (because of the requirements of force balance, which balances the pressure times the surface area, which scales like P*R^2, with the gravity, which scales like 1/R^2).   So that's quite a steep drop in P as R rises a little, and apparently the small rise in R is enough to compensate for the large drop in core P, to keep the density the same (based on your professor's statement that it does stay nearly the same).  It must be essentially a coincidence of how these things work out, it's not obvious to show why the density should be exactly constant but at least we can see that a large drop in P is compensated well by a small rise in R.  The physics of convection is pretty tricky, and there is only an approximate treatment called "mixing length theory" but it's never really been clear how to do this in a formally correct way.


Upgrading telescope from 5" to 8" or 10"? I currently have Celestron Nexstar 130STL.  I am trying to upgrade my telescope to next one.  I examined Meade 8"ACF Goto telescope, but I was not too thrilled about it.  It gave me better view, but the details were not too different from my 130mm.
Maybe my expectation was too big.  I have an option of getting Orion 10" dob motorized alternatively.  I am slightly concerned about the size and the portability.

I would appreciate any recommendation, thanks!

The first thing I would recommend is to join an astronomy club or society.  There's most likely one near you, and membership is generally very affordable (in the US, prices usually are $50 or less per year).  Most clubs conduct freqeunt star parties, which would give you a chance to see a wide variety of telescopes in action and talk to their owners to find out more.

As Alphecca replied, the light gathering power of a telescope is proportional to the square of the radius of the aperture.  For example, your NexStar 130SLT has an aperture of 130 mm (just over 5").  If you use the formula for area (pi * r^2), this gives you an aperture area of about 13,273 mm^2 (square millimeters).  If you were to upgrade to an 8" telescope (203 mm aperture), you'd get an aperture area of about 32,365 mm^2, which is about 240% the light gathering area of your 5" scope.  If you go to a 10" (254 mm), you end up with about 50,671 mm^2, or about 380% as much as the 5" or 157% as much as the 8"

What does this actually mean?  That's a little more complex.

First, there is more light gathering.  Contrary to the common opinion of those not experienced in using telescopes, magnification is not the key concern with a telescope, light gathering is.  But there's more to it than just that, and I'll circle back around shortly.   In the case of light gathering, larger apertures are better.  The telescope, in a sense, acts as a funnel for light, collecting from a larger area and concentrating on a smaller area.  In this sense, a larger telescope can, quite simply, gather more light.

But there's a catch: it's gathering ALL of the light that hits it, not just the light from the target object.  It will gather the light of your target object and any stray light, such as light pollution.  To deal with this, we need to increase contrast to make the dim deep sky object stand out from the sky.  This is heavily influenced, then, by things such as focal ratio and your eyepiece selection.   I spend a fair amount of time on Reddit r/telescopes, and we've had some discussions about this. One of the other regulars there wrote a pretty good article on medium.com about this.

So, to put it mildly, there's more invovled than just getting a bigger aperture.  Though, in fairness, having the biggest aperture you can afford and handle as far as weight and size goes is always the best way to go - you will never have too LITTLE aperture.

The real questions then become what you want to do with it.  If you want to do astrophotography, you're going in entirely the wrong direction.  For visual observation, however, an 8 or 10 inch telescope is a good option.  I personally recommend an 8 inch for any beginner that can afford one.

But, based on what you said about being disappointed with the Meade 8", I think there is a question of expectation.  If you expect to see what you see in photographs, then you're just plain out of luck. Even a very large scope (say a 24" Dob) won't do that for you. Our eyes are simply not capable of long exposure imaging as are cameras.  But a good observer who has learned to train their eye properly can still see some pretty amazing sights.  And this is another good reason to join a club: to spend time observing with other, more experienced observers and learning how to truly see through the eyepiece.  Visual observation is not just a matter of pointing the telescope in the right spot and taking a look.  Observing requires spending time letting your eye and brain work together to really see what's out there.

I jsut said our eyes aren't capable of long exposures.  This is true, but it's also true that extended observing does do something that's not entirely unlike long exposure imaging.  As you spend time at the eyepiece, your eyes collect more and more light and your brain does start to synthesize a more complete image.  It starts to fill in gaps and show you more and more.  This is especially the case when observing from darker skies where there is less light pollution to muddle the view.  A great example in my experience is M51.  From my club's dark site, when it's up in the sky, I can almost alawys make out the two galactic cores, but frequently the spiral structure is hard to see.  But if I have a comfortable seated viewing position and spend a few minutes, my eyes pick up hints of the spiral arms, which my brain starts to fill in.  A good technique for developing this is sketching.  If you sketch the object, your brain process it differently and you start to catch hints and glimpses of more detail which end up in your sketch.

Your current telescope is a GoTo, as is the Meade you mentioned, and you talked about an 10" motorized Dob.  While GoTo technology and tracking can be very helpful, they are also often more of a hindrance than expected.  Most particuarly, the cost of the GoTo mounting inflates the overall cost of the instrument.  An 8" standard Dob can be bought or under \$500 (occasionally under \$400).  An 8" GoTo telescope is likely to run over \$1,000.  The GoTo equipment typically will at least double the price.  For the price of an 8" GoTo dob you could get a 12" manual Trus dob with almost 550% the light gathering area of your 5".  (Not to mention a larger aperture offers greater detail resolution capability for planetary observation).

In the end, my recommendation is to start by joining a club and spending some time learning more.  Then you can make a more informed decision on a purchase (additionally, many clubs have a thriving buy/sell/trade culture with members constantly selling off something to fund their next purchse - I've bought most of my equipment at very good prices this way).

Good luck and clear skies!


Can satellites travel in clusters? Last night we saw a large cluster of what looked like exceptionally bright stars that moved, as a group, across the night sky in Lancaster, PA. It was about 9:15 pm, and they moved roughly from SW to NE. There were 25 or 30 of them. They moved as quickly as a satellite I once saw, but were bigger and brighter. They were soundless and did not blink, and the relationships between the lights changed slightly as they moved.

We don't believe in UFOs, but can't we find any explanation via google.

Can that many satellites move as a group?

Yes, they can.  The ships that visit the ISS often orbit near the station for several hours before docking.  Further, there are several satellites that travel in pairs or trios with other satellites.  The nearer the satellites are to earth, or the farther apart they are, the more difficult the task of maintaining spatial relationships between the spacecraft. This is because the spacecraft must be in the same orbital plane to travel together.  The most common configuration is where they are in the exact same orbit, with one satellites a few seconds ahead of the other along the same path.


Why is the Sun's atmosphere (the corona) so hot? The visible surface of the Sun is 6000 degrees Celsius but the gas in the solar corona is at least 1 million degrees. What heats the solar corona?

It is still an open question, even though it is clear that it is linked to the magnetic field of the Sun. The following hypothesis are the following:

Heating waves mechanism: propagation of magnetohydrodynamic waves can heat significantly the corona. These waves can be produced in the solar photosphere and propagate carrying energy through the solar atmosphere; they turn in the end into shock waves that dissipate energy as heat in the corona.
Magnetic reconnection mechanism: magnetic field lines are tied to the surface of the Sun and even when the plasma moves, they stay tied to their original position (their "foot points"); therefore, plasma movements can drag the magnetic field lines and entangled them in geometries such as loops, that can eventually "reconnect". "Reconnection" corresponds to a modification of the field line topology (see the image that pictures it better than my words). During this process, heat and energy are released quite efficiently.
Nanoflares mechanism: the dissipation of a lot of small scale structures (slow convective motions of magnetic foot points could create current sheets, coutinually dissipated and reformed, providing heating through Ohmic dissipation). You can see it as a kind of a lot of magnetic loops reconnecting together, instead of one big loop reconnecting alone.

Black lines represent the magnetic field lines. They are tied to the surface of the Sun (the foot points). One of the magnetic field lines is looping, and it is "crossing itself". When it reconnects, it forms, on the top, a magnetic loop and, on the bottom, an magnetic arch. As pointed out by this diagram, magnetic reconnection mechanism is also associated to Coronal Mass Ejection (CME), since the energy released by magnetic reconnection is powerful enough to eject a significant amount of matter.

Source:  A Contemporary View of Coronal Heating


Does Triton trigger the active atmosphere of Neptune? As we know the atmosphere of Neptune looks more active than Uranus', is this due to the tidal forces caused by the retrograde orbit of Triton around the planet?

Converting answer in comment to community wiki

Unlikely. Neptune is warmer than Uranus, more internal heat means more turbulent atmosphere. The reason why this is so isn't known, but it's possible that Uranus had a more turbulent past and it emitted heat from its formation more quickly. It's also possible that Neptune had a collision more recently that may have added to its internal heat. Triton is too small to have the effect you describe and the retrograde motion isn't a factor.


Who found the existence of Dark Matter? Dark Matter's observational evidence is widely attributed to Vera Rubin based on her observation of galaxy rotation curves with Kent Ford in 1970s, which was later confirmed by Albert Bosma in 1978.

However, it was in the late 1930s that Fritz Zwicky inferred the presence of dark matter based on his observations of Coma galaxy cluster using the virial theorem. Herein lies my confusion. Why does the scientific community and people, in general, give credits to Vera Rubin instead of Zwicky for inferring/observing the evidence for dark matter? Or, is the recognition given because she independently strengthened Zwicky's observation thereby confirming the existence of dark matter?

You have a clear view of the history, and it wasn't an independent confirmation, it was simply the best evidence.  Whenever one talks about the discovery of dark matter, one should start with Zwicky, and move to Rubin.  One might do something similar with the Big Bang, starting with Hubble, and moving to Penzias and Wilson.  Normally we give credit both to the initial discovery, and to the confirmation, but it is the most conclusive evidence that generally gets associated the largest credit.  After all, it was Edgar Allan Poe who first described the Big Bang model, but he gets no astronomical credit because he provided no evidence.  Zwicky certainly did better than that, so does deserve (and should receive) much credit, but there were still many doubters in his day, much as with Hubble and the Big Bang.  Perhaps Zwicky's role contributed to the reason that Rubin never received the Nobel prize, I hope it wasn't simply the unspeakable reason, but of course it's too late now for both of them.


How many times has the Milky Way rotated? I would think that is a highly naive question, but I still ask it.

Given current experimentally discovered numbers:

the Milky Way galaxy is ~13-14 billion years old
At the galactic radius where our sun is, the rotational period is about 240 million years.

So, with nothing changing at all (the galaxy and stars pop fully formed into existence, the rotation period doesn't change), the galaxy has rotated roughly

  ~14/(1/4) = 50 to 60 times.

Since the assumptions are naive, what is a better approximation for the number of times? How long has the mass that is recognizable now as the Milky Way been rotating (and what has been the speed by radius over time)? Is it a lot more than 50 times and if so, has it been slowing down as a whole or just at the Sun's position, or a less because the coalescence into stars rotating took a few billion years to translate into the faster rotation which it is now mostly set at? Or some other reason which is more articulate?

Note: This question is similar to but distinct from The equation for how many times the Earth has revolved around the galaxy, as the title asks similarly what I'm asking but the content just asks for how to calculate rotational speed.

The question doesn't have a single answer, because the Milky Way's rotation is differential rather than solid-body. This means that stars in the inner part complete their rotation about the center in a shorter period of time than stars at a larger radius. Stars orbiting with a radius of 1 kpc (about 3,000 light years) will take about 30 million years to go once around; stars like the Sun, with orbital radii of about 8 kpc, will take about 220 million years; stars orbiting with a radius of 20 kpc will take about 500 million years.

If you mean "how many times has the narrow region of the Milky Way currently containing the Sun's orbit rotated" -- ELNJ's answer is probably reasonable, given that the last major merger contributing to the Milky Way's growth was probably about 10 billion years ago, so changes in the amount of mass inside the Solar radius (which would change the orbital timescales) were probably fairly small after that.


Time dilation due to the expansion of the universe This is not a homework question. I want some help in being able to clearly perceive the expansion of the universe and the consequences thereof.

We know that the universe is expanding in an accelerated manner. It is my understanding that space alone cannot expand without affecting the local time. Thus as per GTR, depending on the curvature of space-time and the energy density, time too shall suffer a change due to the changing space.

Assuming that there is no curvature of space-time and no mass or energy contained in a chunk of space (assuming the zero point energy of empty space as zero) between two galaxies A and B, let us say that the local time at a point P in this space goes as $t_0$. If we on Earth could somehow observe this point and measure the time at P as $t$,

When we talk about the rate of expansion for example as in Hubble's Law, do we follow the comoving coordinates or the proper coordinates?
How would $t$ and $t_0$ relate to each other if the rate of expansion is uniform and accelerating respectively?
How much is 1s for the comoving observer at the inflation period in terms of the usual 1s now on Earth?

Hubble's law
$$
v=H_0 d,$$
relates the recession velocity $v$ of a distant object to it's physical distance $d$. Today, the physical distance coincides, by definiton, to the comoving distance $\chi \equiv d/(1+z)$, but if you want to know how fast two galaxies at a redshift of, say, $z=1$, you must plug in their physical distance at that time (and use $H(z)$, the Hubble parameter at that time, rather than $H_0$). Depending on how you measure their mutual distance, there are various ways to calculate this.
When we observe a galaxy at redshift $z$, general relativity predicts that we see its time dilated by a factor $1+z$. That is, some physical process which takes a proper time $t_0$, will be observed to take a time
$$
\boxed{t = t_0(1+z),}
$$
which is the formula you're after.

A good example of this is the time it takes for the brightness of a supernova to decline; a supernova at $z=1$ declines half as fast as a local (i.e. $z=0$) supernova (see e.g. Goldhaber et al. 2001).

Calculating the distance to that supernova, or the age of the Universe at the time it's observed, is a bit more complicated, involving a model of the expansion rate of the Universe (see e.g. this question about the Friedmann model).
Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, the corresponding redshift was of the order$^\dagger$ $z\sim4\times10^{25}$, so if we could observe physical processes there, they'd be dilated by a factor $\sim4\times10^{25}$, so 1 second would, at this dilation, take a 3 billion years. However, already 1 second after inflation, the expansion rate of the Universe had already slowed down so much that the corresponding redshift, and hence dilation factor, was "just" $\sim4\times10^9$. So the next second would only, from our point of view, seem to take $\sim140$ years.

Note that these considerations don't really have anything to do with the accelerated expansion, but is just a consequence of the expansion.

Note also that special relativity predicts the same time dilation; that is, if galaxies were flying through space rather than just being carried along with expanding space, we would observe the same relation between redshift and time dilation. However, the special relativistic interpretation of redshifts is incompatible with the relation between redshifts and magnitudes of supernova, and can be ruled out at $23\sigma$ (Davis & Lineweaver 2004).

$^\dagger$This can be calculated as follows: Until the Universe was $t_\mathrm{eq}\sim50\,000\,\mathrm{yr}$ old, corresponding to a redshift of $z_\mathrm{eq}\simeq3400$ (Planck Collaboration et al. 2018)
 its dynamics was dominated by the energy density of radiation, causing the size (scale factor $a=1/(1+z)$) to evolve with time as $a\propto t^{1/2}$. Calculating backward from $t_\mathrm{eq}$ yields $a\sim10^{-26}$ at $t\sim10^{-32}\,\mathrm{s}$, corresponding to $z\sim10^{26}$.


How can the Earth's rotation speed up? It is known as a fact that Earth's rotation is slowing every year, that's why we sometimes have 61 seconds long minutes.

In the Java documentation of Time : http://docs.oracle.com/javase/7/docs/api/java/util/Date.html , it is said :

  Some computer standards are defined in terms of Greenwich mean time
  (GMT), which is equivalent to universal time (UT). GMT is the "civil"
  name for the standard; UT is the "scientific" name for the same
  standard. The distinction between UTC and UT is that UTC is based on
  an atomic clock and UT is based on astronomical observations, which
  for all practical purposes is an invisibly fine hair to split. Because
  the earth's rotation is not uniform (it slows down and speeds up in
  complicated ways), UT does not always flow uniformly. Leap seconds are
  introduced as needed into UTC so as to keep UTC within 0.9 seconds of
  UT1, which is a version of UT with certain corrections applied. There
  are other time and date systems as well; for example, the time scale
  used by the satellite-based global positioning system (GPS) is
  synchronized to UTC but is not adjusted for leap seconds.

Why is that so? How can this occur?

Any body with an irregular mass distribution will experience those wobblings. And more if it is not just irregular but fluid.


What is the diffuse ionized gas? I've been trying to find a clean definition what people mean by when they talk about diffuse ionized gas in the interstellar medium, but I couldn't find anything so far.
Apparently it's supposed to be trivial.

Is it simply ionized gas, mixed with non-ionized gas in the ISM through the process of diffusion?

The "diffuse ionized gas" (DIG) is another term for the phase of the interstellar medium (ISM) usually called the warm ionized medium (WIM). With a temperature of the order $10^4\,\mathrm{K}$, but extenting to lower and higher temperatures, it is hot enough to keep hydrogen ionized, and various metals exist as low-ionization species, such as S II, N II, and O II, and even (weak) O III (e.g. Hill et al. 2012; Zang et al. 2017; Weilbacher et al. 2018).

Rough pressure equilibrium with the other phases of the ISM results in characteristic densities of $0.1\,\mathrm{cm}^{-1}$, but typical ranges are an order of magnitude to both sides.

A review of the DIG/WNM can be found in Haffner et al. (2009).


How hot must a star get before it is considered to be a star? How hot must a star get before it actually becomes a star? Why does it need to get so hot? Please find an official site to quote from, if you can.

From a physics perspective

From a physics perspective an object is a star when it is undergoing nuclear fusion, generally of hydrogen atoms at its core, this is regardless of its temperature!

A star is not determined by its temperature, it is instead determined by it's internal processes.

This does mean that if Jupiter began nuclear fusion it would be considered a star, albeit a minuscule one.

In this case it is a yes/no distinction of if an object is a star.

From an observational point of view once something is classified as a star there are 7 groups it can fall in to determined by its features.

Sourced From: http://en.wikipedia.org/wiki/Star#Classification

  Class Temperature  O:    33,000 K+
  B:    10,500–30,000 K   A:   7,500–10,000 K
  F:    6,000–7,200 K     G:   5,500–6,000 K
  K:    4,000–5,250 K     M:   2,600–3,850 K

Note: Three more classifications L T and Y have been added to the colder end of this list, but I am unsure of the cut off points so omitted them.

But strangely they are not classified by temperature but by their spectrum, it just so happens that their spectrum correlates to their temperature! The temperature spoken of here is of the photosphere of the star (where the photons begin free streaming), not its core (where photons are created from ongoing fusion reactions).

Dwarf stars have their own classification system prefixed by the letter D though.

Quote from Wiki article:

  White dwarf stars have their own class that begins with the letter D. This is further sub-divided into the classes DA, DB, DC, DO, DZ, and DQ, depending on the types of prominent lines found in the spectrum. This is followed by a numerical value that indicates the temperature index.


Are objects in the universe moving away from each other at the same acceleration? This is a follow up question to Does the universe expand at the same rate everywhere in the universe?.

If the universe is expanding, then it would seem necessary for each object (ie galaxy) to be accelerating away from every other object at the same rate of acceleration which is constant (or not?). If that is true, what is the acceleration?

In other words, relative to the Milky Way, does it seem that all the other galaxies are moving directly away from us at the same rate of acceleration? If so, is it possible for us to prove that for any other galaxy that we know, the same observation would be true, were we able to view the universe from that galaxy?

I assume you mean is $\frac{dv}{dt}$ the same for all objects and/or is it constant for all objects, where $v$ is recessional velocity (and $t$ is cosmic time)? The answer is no it is not the same for all objects and it is not constant for all objects.

$$v=\frac{dD}{dt}=H(t)D$$

Where $D$ is proper distance and $H(t)$ is the Hubble parameter. If we take the Hubble parameter as a constant, then:

$$D = D_0e^{Ht}$$

Where $D_0$ is the proper distance at the present time. So,

$$D''(t) = \frac{dv}{dt} = D_0H^2e^{Ht}$$

So the "recessional acceleration" of an object depends on its present proper distance and is exponentially increasing with time.

Now the Hubble parameter isn't a constant and in LCDM cosmology is currently asymptotically decreasing to a constant. In the current epoch though it is fair to say that the "recessional acceleration" of an object depends on its  proper distance and is increasing with time.


Do all planets rotate in the same direction in relation to each other? In other words, does everything spin counterclockwise according to their axes (in relation to each other)? I suppose my question goes for stars, moons and other objects too.

But then again, all you have to do is look at any rotation upside down and it goes clockwise. So depending how you're looking at it, one could say everything rotates in the same direction whether they do or not. It's all a matter of perspective, right? And in space, there is no "right side up". Maybe my question is moot now that I think about it.

If we re-drew all of Earth's maps and globes upside down, New York or Sydney would be on the "west coast" and the Sun would rise in the "west" and set in the "east". Is it just by luck or maybe by the prominent early mapmakers, or that the majority of human population lives in the Northern Hemisphere, that it's given "dominance" on "top"? (I've heard that theory before)

Ugh. This is all so confusing.

This Q can be split into two:

Do all the planets in a stellar system always rotate around the star[s] in the same direction as the star[s] itself/themselves?

Question #1 is a duplicate of Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does?. As you mentioned, calling it counter-clockwise is simply due to the convention of North on top.

How did the convention of North being the top of maps and globes get established?

Question #2 is about earth science history or the history of cartography. I encourage you to ask that as a separate Q either on the Earth Science Stack Exchange or on the Geographic Information Systems Stack Exchange.

In many cultures, East was/is put on top, as a nod to the rising sun. I imagine if the convention had come from the Southern hemisphere, South or East would have ended up on top.


What is this pale blue dot in a daytime photo of a Sun halo? This photo of the sun behind a cloud was taken on the south coast of England in early May 2021 using an iPhone 11. A halo is visible which I think is the 22-degree halo. However, the photo also shows a small, pale blue dot near the middle of the photo, within the photo, left and above the Sun. Please could somebody suggest what the blue dot is?

Yes, I believe this is the internal lens reflection. Lens reflections have hexagonal shape, and if you look closely, it somewhat resembles this shape. Also, see other answers in the comments.
Internal lens reflection happens because the lenses don't have 100% transmission. This often happens when taking bright objects, like Sun in your example.


What is a Seyfert galaxy? I know Seyfert galaxies are types of active galaxies, but I do not understand what separates them from other types of active galaxies.  Are they brighter/not as bright as other types of active galaxies?  Do they emit in different wavelengths relative to other active galaxies?  I've read that 10% of all galaxies are Seyfert galaxies, so it seems to me that understanding what these galaxies are is very important to understanding galaxies overall.

Seyfert galaxies differ from other active galaxies (most notably quasars) in that their galactic nuclei are lower in luminosity compared to the rest of the galaxy. Quasars have nuclei that easily outshine the rest of the galaxy. Seyfert galaxies, on the other hand, host active nuclei that do not outshine the rest of the galaxy by the same amount.

Interestingly enough, in his original analysis, Seyfert focused on emission lines, noting that in this class of galaxies, there are strong high-ionization emission lines present in certain parts of the spectrum. This alternate definition is also used today. LINERS (Low Ionization Nuclear Emission line RegionS) are very similar to Seyfert galaxies if this definition is used; they are differentiated because LINERS also contain low-ionization emission lines.

References:

Seyfert (1943)
Popping (2008)
A Knowledgebase for Extragalactic Astronomy and Cosmology: Seyfert Galaxies


Can some stars not emit any energy in the visible spectrum? Stars convert mass into energy. Even converting the slightest mass into energy is immense because the speed of light is so great and $E = mc^2$. This means that stars have to emit large amounts of energy. So can they emit so much energy in a form of electromagnetic radiation that is not visible light? Resulting in a star that cannot be detected using our senses(+ a telescope). If so what problems could this star pose?

Light that is not light

That's meaningless.  All light is electromagnetic radiation.  A finite part of the infinitely large range of the electromagnetic spectrum is visible light.  So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.

Stars emit lots of energy at frequencies that are outside the visible range.

  Dark Matter

Dark matter does not mean that it's just not radiating in the visible part of the spectrum.  It means it's not radiating in any part of the EM spectrum much.  If dark matter exists (and that's not certain) it won't interact with most matter much at all.

If dark matter were in star-like objects, it sounds very unlikely that there would not be any EM radiation as a result of the interactions that would result in a star-like object.  We'd detect it if it were there.  So I don't think that's likely.


How long for the sun to traverse its diameter at the pole? As the sun rises once per year at the north/south pole, the time taken to rise (and set) must also take significantly more time to complete to rise (and set).

How long does this take at either pole?

Interested to know if it is the same percentage of a day as anywhere else on earth (the day at the pole being several months long).

The Sun's daily motion does not affect its altitude on the poles, because it revolves around the zenith (and the zenith coincides with the celestial pole). This means the Sun's altitude is only altered by its movement on the ecliptic. The ecliptic is tilted at a roughly 23.5 degree angle to the equator (which is the horizon if the observer is on the North pole), and only the vertical component of its yearly motion contributes to its increase in altitude. Therefore, in order to estimate how long the Sun takes to rise/set on the North pole, let us assume the Sun is moving across the ecliptic at a constant velocity and that Euclidean geometry holds for the small area on the celestial sphere around the Sun. If the Sun takes one tropical year to traverse 360 degrees, its "speed" will be $\rho=0.98565 \frac{\circ}{d}$, and the vertical component $\rho\sin(\epsilon)=0.39\frac{\circ}{d}$. The time it takes to traverse a solar diameter is (roughly $0.5^\circ$) is $0.5/0.39 \sim 1.25$ days. The same is for the South pole, everything is rotated around the equator. You can find more details about the topic here.
As for the ratio of sunrise duration to daytime duration, these are affected by different phenomena, and the ratio of sunrise duration to daytime duration is not constant throughout the year and depends on the latitude of the observer so comparing them is quite ambiguous.


Theoretical measure of distance to black hole? Would it be possible to measure the distance to a black hole of known mass by measuring the radius of the most redshifted radiation that is detectable? The idea is that since the point where that radiation vanishes is related to the gravitational potential difference between the observer and the point of emission of that radiation, we should be able to get an estimate of how far we are by observing how far into the black hole we can see (assuming that for a given mass and an observer at infinity that radius is fixed).

Yes. If you could spatially resolve a black hole and measure the redshift of radiation at a given angular radius, then that would yield the distance to the black hole.

This works because the amount of redshift depends on the ratio of radius to Schwarzschild radius, but the latter is known if the mass is known. Thus the redshift yields the radius. The distance is then given by the radius divided by the apparent angular radius.

There will be complications due to gravitational lensing and any orbital velocity around the black hole, but this does not stop it working in principle.


Orbital speed is (vector) sum of tangential and normal speed? Orbital speed definition in wiki does not state clearly - it is just tangential speed component or square root of squares of normal and tangential speed (full speed vector).

When we say that moon orbital speed is 1 km per sec, we don't know its tangential speed (speed along its trajectory), right? We need some complicated math to calculate it (because of normal speed component inside that 1 km/sec orbital speed)?

"For any object moving through space, the velocity vector is tangent to the trajectory."(Citing https://en.m.wikipedia.org/wiki/Orbital_state_vectors).

Therefore, it's both of these components, there is no normal component1 (relative to the central body)... the orbital speed represents the intensity of the orbital velocity which is always tangent to the ellipse (or parabola/hyperbola) the body in orbit describes.

The Moon's orbital speed being 1 km/s means it moves along an ellipse around the Earth at a speed of 1 km/s (this speed is roughly constant, so that ellipse can sometimes be approximated to a circle).

1However you can separate orbital motion into two components normal to each other (the most popular such separation is to the radial component and one normal to it, but this is unnecessary for circular orbits where there is no radial velocity, only radial acceleration which causes the curvature of the trajectory), and in that case the orbital speed will indeed be the square root of sum of the squares of those two.

$$v^2 = v_{||}^2 + v_{\perp}^2$$

But in two-body Kepler orbits there is no velocity component normal to the plane of the orbit, which is the plane containing the $\mathbf{v_{||}}$ and $\mathbf{v_\perp}$, which is also the plane containing $\mathbf{r}$ and $\mathbf{v}$.


How to determine arrival time delay given dispersion measure? In an online textbook, the following formula is given to calculate the arrival time delay between two frequencies, $v_1, v_2$ in a dispersed radio pulse:
$$t_1 - t_2 = 4.15 \cdot DM [(v_1/\text{GHz})^{-2} - (v_2/\text{GHz})^{-2}]$$
Where does the $4.15$ come from, and what exactly is “GHz”? Do I divide by 1 GHz or?

As is the case with light traveling through any medium, radio waves traveling through space experience refraction, which reduces their speed. A wave of infinite frequency will experience no refraction, meaning that we can compute how much a given wave will be delayed compared to such an infinite-frequency wave as it moves through outer space. The group velocity of a radio wave is determined by $v_g=c\mu$, with $\mu$ the refractive index defined by
$$\mu=\sqrt{1-\left(\frac{\nu_p}{\nu}\right)^2}$$
with $\nu$ the frequency of the wave and $\nu_p$ the plasma frequency, a fundamental property of the medium, which depends on its density:
$$\nu_p=\sqrt{\frac{e^2n_e}{\pi m_e}}$$
with $e$, $m_e$ and $n_e$ the charge, mass and number density of electrons. (You're probably more familiar with refractive indices for light that humans can see, but they're important in a much wider portion of the electromagnetic spectrum!) In the ISM, we'd expect to see plasma frequencies in the kilohertz range - much lower than the frequencies of radio waves from whatever source we're observing, which are likely in the hundreds of megahertz to tens of gigahertz range. The delay a wave of frequency $\nu$ experiences over a distance $d$ is
$$t(\nu)=\int_0^d\frac{dl}{v_g}-\frac{d}{c}$$
where the second term requires no integral as the refractive index of a wave of infinite frequency is always $\mu=1$ regardless of the medium. We can use a binomial approximation for the refractive index of our physical finite-frequency wave, since $\nu\gg\nu_p$, and after some algebra, we find that
$$t(\nu)=\frac{1}{c}\int_0^d\left(1+\frac{\nu_p^2}{2\nu^2}\right)dl-\frac{d}{c}=\frac{e^2}{2\pi m_ec}\frac{1}{\nu^2}\int_0^dn_edl$$
Now consider two waves of frequencies $\nu_1$ and $\nu_2$. The difference in arrival times is
$$
\begin{aligned}t(\nu_1)-t(\nu_2)&=\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl-\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl\\
&=\frac{e^2}{2\pi m_ec}\left(\int_0^dn_edl\right)\left[\frac{1}{\nu_1^2}-\frac{1}{\nu_2^2}\right]
\end{aligned}$$
If we define the dispersion measure by
$$\text{DM}\equiv\int_0^dn_edl$$
and define it in units of $\text{cm}^{-3}\;\text{pc}$, write the chunk of constants in front in units of $\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$:
$$\frac{e^2}{2\pi m_ec}=4.15\;\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$$
and write out frequencies in units of $\text{GHz}$, then we indeed recover the expression you give - with the time delay in units of milliseconds. It would be a lot clearer if we rewrote it as
$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{cm}^{-3}\;\text{pc}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$
which has the proper units.
In short: If you write your frequencies in gigahertz and your dispersion measure in parsecs per cubic centimeter, the formula will give you a time delay in milliseconds.


Is it possible that a satellite creates a forever eclipse? So, to add a bit of context, I’m doing the NaNoWriMo challenge, which consists of writing a novel of 50k words in one month. My whole idea is based around the following question: What would happen if the sun were to suddenly disappear?

To make it short, the story is about a colony on a faraway planet, which is settled for few generations already. And one day or let’s say, in less than a month, the sun disappear. So far, so good. Except I need a valid reason as to why the sun disappear… I have found some ideas (dust from volcanic eruption, change of orbit, dying sun, etc…) but none of them looked good enough. (I’m open for suggestion if any.)

I’m now investigating the possibility that the planet would gain a satellite big enough to hide the sun on the whole planet (like an eclipse but lasting forever). This topic lead me to think that if the satellite is in geostationary orbit, it is possible. But there is still some questions I can’t get my head around.

How fast a new satellite can “appear”?
Is it possible that a bigger but “lighter” (with less gravitational force) satellite can orbit around a smaller but “heavier” planet?
This topic reminded me that no sun means no heat. Can the geothermic activity of a planet be enough to prevent the surface from being impossible to live on? (Off topic.)

So, basically, the question is, if it's possible to have a forever eclipse, how ?

I’m sorry in advance if the questions are not clear enough, even though I understand the basics, I don’t necessarily have all the technical terms.

Thank you !!

The only potentially valid way to cause a long-lasting eclipse that I can think of is to have your moon orbit at the planet's L1 Lagrange point between the planet and the star (and the moon has to be the right size). If it had just the right orbit and was the right size, it could then cause a long-lasting eclipse. Note I'm not saying permanent because such a system wouldn't be permanent. I'm not going to run through the math$^1$ to determine if such a system is stable or for how long. I'd guess it may be stable for a few orbits at least.

The main problem is getting your moon there in the first place. One could conceivable say that by some freak chance a third fairly massive body came close to your planet/moon (either from outside the solar system or near the edge) and gravitationally perturbed your Moon's orbit such that it ended up in the L1 point. But the chances of such an event are so slim that it would be incredulous if it actually happened.

And just to add a technicality, once your moon moved to the L1 point, it would now be orbiting the Sun and wouldn't be a moon anymore according to the definition of a moon.

$^1$ Out of curiosity, I ran through some of the math very loosely and found that if you put the Moon in Earth's L1 position, the Moon survives for about 3-5 months before perturbations knock it out of the L1 position. Of course our Moon isn't the right size to cause an eclipse at that distance, but this gives a very rough estimate for how "permanent" the scenario I described above really is.


What causes the Earth to have gravity? I've always wondered how our planet Earth or any other large spheroid-shaped thing in our universe has gravity?

Does it have a small black hole on its core causing it to have gravitational force?

Take our planet as an example: what causes it to have gravity?

Any massive body (i.e., a body having mass) like the stars, planets, moons, and any atom in the universe have, what is called, a gravitational potential which results in its gravity.  This is described by Newton's Law of Gravity. Why masses cause gravity, we do not know.

So, in short, the earth's mass causes it to have gravity.


Is it possible to use a laser to propel a spaceship? Reference: http://adsabs.harvard.edu/full/1968BAN....19..421H
It would be necessary to aim a laser to the target star. How would the density of the laser compare with the interstellar radiation?

Yes, it might be possible to send laser-propelled spaceships to the stars.
There are a whole series of technical challenges in the laser technology and the materials that such a spacecraft might be made of.
The basic idea is to accelerate mini-spaceships with very low mass, but a large, reflective surface area, up to relativistic speeds. The project that I know of is called "Breakthrough Starshot".
Contrary to what you say in your question, the laser would not be aimed at the target star, it would be aimed at the spaceships! These would have to be set off on a carefully calculated trajectory, because during their decades-long journey, the relative positions of the Sun and target star would change significantly. The impulse phase would take place within the Solar System and the craft would cruise most of the way.
The intensity of the lasers required would be immense compared to the light from a star. The baseline plans involve something like $10^9$ W/m$^2$ delivered to the light sail for 10 minutes. This is to be compared with about $10^3$ W/m$^2$ from the Sun at the orbit of the Earth.


how steady over time are right ascension and declination values? I understand that astronomers use right ascension and declination for the position of stars. However both these values strike me as very anthropocentric and, more importantly, unreliable over time as I understand that the Earth wobbles a bit over time so the celestial equator shifts and I guess Sun's rotation around the galaxy and Milky Way's own trajectory should play a role in depreciating the accuracy of those values over time.

So my questions are:

[1] how steady are right ascension and declination values for distant stars over time?

[2] is there a more stable coordinate system or do we lack an absolutely fixed frame of reference?

The values are more steady for objects close to Ecliptic poles, but in genenral, they change by a few arcseconds every year.

Yes, we lack an absolutely fixed frame of reference.


black Holes and Neutron Stars if the universe is constantly expanding then the distance between black holes and neutron stars must be increasing. Then

What makes them to come under the influence of each other's gravitational field and start rotating, or are they rotating around each other from birth?
Why is the rotary motion spiraling one instead of just rotating around each other in a stable fashion like planets around sun?

The expansion is on a scale larger than galaxies, so within a galaxy, two black holes or neutron stars might pass close to each other from time to time as they move under the influence of the galaxy's gravity. If a third object gets involved (eg a star) then they could end up being captured into orbit . Much more likely, however, is both components of a binary star (which are common) evolve inbto black holes or neutron stars so they are orbiting one another "from birth".
This is a difference between general relativity and Newtonian gravity. In GR no orbit is ultimately stable -- energy is slowly lost as gravitational waves and the objects spiral in. If the gravitational fields are not very large, however, this effect is much too small to detect, and overwhelmed by other effects, so it seems like planets around the sun are orbiting stably. Only when you have objects with a very strong gravitational field in a very close orbit is it noticeable. Even then we're only detecting a few such events per year within a billion light years of Earth, so it's not common.


What does "unremarkable transverse peculiar velocity" mean exactly, and how is it calculated here? The abstract of A magnetar parallax (also in MNRAS) contains the following:

Combining our new observations with two archival observations from 2006, we have refined the proper motion and reference position of the magnetar and have measured its annual geometric parallax, the first such measurement for a magnetar. The parallax of 0.40±0.05mas corresponds to a most probable distance 2.5+0.4−0.3kpc for J1810. Our new astrometric results confirm an unremarkable transverse peculiar velocity of ≈200km s−1 for J1810, which is only at the average level among the pulsar population

This reports the first radio astrometric determination of parallax "for a magnetar".
The peculiar phrase "unremarkable transverse peculiar velocity" is vernacularly dissonant, but parenthetical clarifications in the body of the paper help with that:

5 DISCUSSION
As is shown in Table 1, our new proper motion significantly improves on the previous value inferred from the two year-2006 positions; the new distance D = 2.5+0.4−0.3 kpc is consistent with 3.1 ± 0.5 kpc estimated using red clump stars (Durant & van Kerkwijk 2006), while in mild tension with 3.1−4.0 kpc constrained with neutral-hydrogen absorption (Minter et al. 2008), suggesting the  distance to the neutral-hydrogen screen was over-estimated. In models of NS kicks from the electromagnetic rocket effect (Harrison & Tademaru 1975) one might expect magnetars to have higher velocities (Duncan & Thompson 1992). Our new parallax and proper motion corresponds to the transverse velocity vt = 198+29−23 km s−1. Using the Galactic geometric parameters provided by Reid et al. (2019) and assuming a flat rotation curve between J1810 and the Sun, the peculiar velocity (with respect to the neighbourhood of J1810) perpendicular to the line of sight was calculated to be vb = −54 ± 8 km s−1 and vl = −175 ± 26 km s−1. Our refined astrometric results consolidate the conclusion by Helfand et al. (2007) that J1810 has a peculiar velocity typically seen in “normal” pulsars, unless its radial velocity is several times larger than the transverse velocity.

My understanding of that is limited, but I think that "peculiar velocity" is the velocity relative to some larger structure in which it is found and/or believed to be gravitationally associated with, and the "perpendicular peculiar velocity" component comes from some model that includes the assumption of "a flat rotation curve between J1810 and the Sun". But I haven't a clue what that means.
See also Wikipedia's Peculiar_velocity; Galactic_astronomy for a short description.

In galactic astronomy, peculiar motion refers to the motion of an object (usually a star) relative to a Galactic rest frame.

Questions:

How is the transverse component of said peculiar velocity determined here, and what does "a flat rotation curve between J1810 and the Sun" refer to?
Have I got this right, and in general "peculiar velocity" is a velocity relative to some larger structure in which it is found and/or believed to be gravitationally associated with?

"Peculiar velocity" is a fixed term and describes the velocity of an object relative to a defined rest frame.
Astronomy has the problem that you need different methods to measure the 3D motion of an object. Therefor one often only gives the velocity within line-of-sight (from spectrographic data) or the perpendicular velocity as measured from astrometry, thus positional data on a sphere; that is the perpendicular velocity (to the line-of-sight).
As such the "perpendicular peculiar velocity" is the velocity of the magnetar as determined from astrometry - as also stated in your quote.
The reference frame of rest in this context is the expected motion of the stars around the galactic centre - at the distance of the observed magnetar. Galaxies don't have a simple Keplerian velocity where orbital speed reduces with distance to a central heavy central object like in our solar system. Dark matter changes the relationship between orbital distance and orbital speed in a way which is subject to investigation. With a "flat rotation curve between J810 and the Sun" they mean that orbital velocities don't differ, thus they indicate directly their normalization, their reference rest frame.
The peculiar velocity of a star within our Galaxy thus refers to their motion relative to their rest frame which orbits the Galactic center in the way expected for their distance from the center of the Galaxy.
So how was it measured? They say "from annual parallax measurements". That means they measure the positon repeatedly and very exactly. This allows to measure the parallax in reference to the background. But at the same time you get the proper motion, thus the change of the positon when you removed the expected apparent change due to parallax. A typical astrometry series looks like  which I took from http://spiff.rit.edu/classes/phys240/lectures/parallax/parallax.html#nottrivial You nicely see the annual variation (the parallax) and the proper motion which remains when the circular annular motion is subtracted. This velocity is the perpendicular or transverse velocity.
Convert this measured velocity in the image plane to a real velocity by taking into account the distance to the object, reduce it by what you expect it to be for the Galactic distance of the object, and find that it is pretty normal to what you expect and you have an "unremarkable transverse peculiar velocity"


axial tilt of uranus Simple question: I read that the axial tilt of Uranus is 97.77 degrees. I don't understand why it would be this instead of 82.23 degrees the other way. It seems like saying that the Earth's axial tilt is 157 degrees rather than 23, or something. can someone explain why Uranus's axial tilt is bigger than 90 degrees?

EDIT:
this is not a question of how it came to be so tilted, I am sure that is answered elsewhere, but simply why it is accepted as being bigger than 90 degrees rather than less than 90 degrees the other way. it is the same for Venus.

I have found the answer. I just found that venus has an axial tilt of nearly 180 degrees, so I realised that the >90 degree tilts are due to the fact that they are the planets with retrograde rotation, which is equivalent to an 'upside down' (>90 degrees) counter clockwise, normal, rotation.


Hypothetically, if our sun was yellow would we still see "white" colours? Hypothetically,if our sun was yellow would we still be able to see white colour on objects like snow,cotton etc.? Would we have the phenomenon of rainbow?

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

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

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

There are a couple of caveats though:

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


Why do objects appear blue while on a neutron star? I was playing around in Space Engine recently, and I realized I hadn't yet landed on a neutron star. This seemed like a very sensible thing to do, so I did it and chose a (relatively) well-known one called Black Widow on which to conduct my adventure. As I landed, I saw mostly expected things-it bent space-time, was rotating very fast, etc. One thing I didn't expect, however, was that everything was blue. Everything. I don't mean the star itself appeared blue, I mean as I looked off the star into the galaxy all the stars appeared blue. I searched for a while but didn't find anything on the Internet. Anyways, does anyone here know why this is happening?
Edit: I have a theory —probably wrong— but maybe because of the warping of spacetime near the neutron star, photons are getting compressed and blueshifted.

Your suspicion is right, the term is gravitational blueshift.
Going down a gravity well, photons gain energy, pushing their wavelengths in the blue direction. The opposite would be gravitational redshift observed when photons climb out of a gravity well.
In terms of general relativity, when you observe the world from near the neutron star, time appears to go faster outside the gravity well, and therefore arriving wavelengths are blueshifted.


What causes jets from newly born stars? Newly born stars often have bilateral jets of gas ejected from them at high speeds (hundreds of km per seconds), often ending in Herbig-Haro objects. As I understand it those jets would have something to do with the star's magnetic field.

How does this work? How can a magnetic field cause the high-energy ejection of matter?

Did you get this from Wikipedia? If you did, I sympathize; all the entries related to this give no explanation for the phenomenon. If you didn't go there . . . Well, it won't answer your question.

But I think I've found the answer somewhere else. I'll go over some background stuff for anyone out there who doesn't know much about this, to make it a bit clearer (although you should be able to skip all this). Protostars are small pre-pre-main-sequence objects. At first, they aren't more than small clumps of gas and dust in a stellar nursery (a special nebula, perhaps). Soon, though, more matter begins to come towards a clump, and gradually the clump builds up enough matter that it begins a slight gravitational collapse into a definite object. Temperatures build, and eventually the object becomes a pre-main-sequence star.

These pre-main-sequence stars begin to accrete matter (here's where you should stop dozing off) into a circumstellar disk. At the moment, it doesn't matter if the disk will become a protoplanetary disk or something completely different; for now, we can treat it as a circumstellar accretion disk. Matter "falls" into the disk and towards the star; the star absorbs most of it. However, something else comes into play, and that's the star's magnetic field.

A lot of objects in the universe have magnetic fields, and a young pre-main-sequence star is no exception. It's field may be stronger than that of the Earth - so strong, in fact, that some of the matter in the circumstellar accretion disk follows slightly different paths than it normally would - paths along the field lines of the star's magnetic field. While there isn't enough evidence (scarcely any at all, really) to support or, more importantly, disprove this idea, it is though that the magnetic field lines lead the matter towards a point on the star, perhaps at its poles, and then compress it together into an astrophysical jet (or bipolar outflow) that shoots out into space. Such jets are often associated with other cool phenomena, such as Herbig-Haro objects.

So for everyone who considers this post a "Too long; didn't read" answer, here's  a summary:

The truth is that nobody really knows. The current explanations for this phenomenon are that they are caused by matter from accretion disks being concentrated by magnetic field lines onto a single (well, double, one on either end of an axis) point, which then becomes an astrophysical jet.

I hope this helps.

Note: This paper, specifically the second page, was extremely helpful. Other pages that might interest you (or provide more information for those who are curious about this and related topics) are:

Protostar

Herbig Ae/Be star

Herbig-Haro object

Paper #1

Paper #2

Pre-main-sequence star

Circumstellar disk

Astrophysical jet

Bipolar outflow

Softpedia.com

Sci-news.com

IOP


Does General Relativity really predict Black Holes? I'm doing some research on black holes for a science video contest. I want to explain the physics of how they work, but also want to have a little background on how they're formed. As far as I've searched, black holes are predicted by general relativity (GR). But I saw this site that said 'general relativity is inaccurate at very small sizes' and that it kind deviates from GR. So I want to confirm whether black holes really are a prediction of GR or an inconsistency with GR. Can someone please help? Thanks.

Well, yes, but we must be careful with the meaning of "predict".
The Schwarzschild solution, developed by Karl Schwarzschild in 1916 [1], is the first closed-form, explicit solution of Einstein's field equations for gravitation. It describes a spherically symmetric, static, vacuum spacetime. The solution goes singular at a specific radius (the Schwarzschild radius). In the weak field limit, it correctly replicates the Newtonian gravitational field of a compact object.
Though Birkhoff's theorem (general relativity's version of the shell theorem) was not yet known in 1916, the Schwarzschild solution was nonetheless recognized as the general relativistic description of the gravitational field outside a compact gravitating body. The fact that it went singular at the Schwarzschild radius was either ignored or taken to imply that objects that are as small as, or smaller than, this radius cannot exist.
In any case, just because a solution exists in general relativity does not mean that objects described by that solution exist in Nature. (E.g., the field equations admit solutions that blatantly violate causality; yet I don't see any time machines out there.) For all they knew, the Schwarzschild solution was nothing more than a mathematical curiosity, an idealized case that does not describe reality.
For this reason, I'd suggest that a much more groundbreaking paper is that of Oppenheimer and Snyder from 1939 [2]. This paper demonstrates that a spherical cloud of dust initially at rest will undergo "continued gravitational contraction". An observer that is falling with the collapsing matter would see total collapse in a finite amount of time, but to a distant observer, the collapse will continue forever, the object asymptotically approaching, but never quite reaching, its Schwarzschild radius. And it was in 1957 I believe that Regge and Wheeler first demonstrated that a Schwarzschild singularity is stable under small perturbations [3], i.e., perfect symmetry is not required. (Wheeler, of course, was also the first to popularize the name, "black hole".)
Lastly, as I was reminded in a comment, we should not forget Penrose's 1965 singularity theorem [4], which introduces the concept of a trapped surface and shows that gravitational collapse is indeed a very generic feature of general relativity. After all, this is the result that earned Penrose the 2020 Nobel prize in physics.
In light of that, I think we can confidently state that general relativity predicts black holes, but only because we know not only that black hole solutions exist, but also that physically realizable configurations of matter can collapse into black holes (or, at the very least, to objects observationally indistinguishable from black holes, which of course may evaporate in stupendously long but finite timeframes due to Hawking radiation [5]) and that the solutions are stable under small perturbations, i.e., perfect symmetry is not a prerequisite.

[1] https://arxiv.org/abs/physics/9905030
[2] https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455
[3] https://journals.aps.org/pr/abstract/10.1103/PhysRev.108.1063
[4] https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.14.57
[5] https://www.nature.com/articles/248030a0


How many quakes have been detected by InSight? Insight has been on Mars with its seismometer for almost two years now. Has it detected any mars-quakes yet? How often do these quakes seem to occur?
Is there a website where the current number of quakes detected by insight is indicated?

This is the latest catalogue from seismic events recorded by InSight: Mars Seismic Catalogue, InSight Mission; V5 2021-01-04. This report includes seismic events up to October 12, 2020. These documents get periodicaly published, so keep an eye on the page pointed out by @planetmaker on the comments. The robot has measured a total of 1759 Marsquake events as of the report date. Keep in mind that 1034 of those events are of suspicious quality, meaning that the signal may not be attributable to a seism. Setting dubious events aside, in the period of 686 days, the mission  has detected about one quake per day. I could not find any website with live data.


Semi-major axis in a binary star system The semi-major axis a here,

what exactly does it represent? Point it out in the image below.

It is the semi major axis of the relative orbit i.e. the ellipse whose major axis is the sum of the two major axes and whose focus is in the barycentre. Essentially, this is like fixing the reference frame at the barycentre and looking at how the stars' position vectors change. It will turn out (try looking up two body problem) that this is equivalent to an object of reduced mass $\mu = \frac{m_1*m_2}{m_1+m_2}$ orbiting the one with total mass.As uhoh pointed out, this image illustrates semi major axes : https://i.stack.imgur.com/298tF.png .


How does gravity affect a physical clock? In all the explanations for time dilation and how gravity affects time a common example which is used is "Clock near to the ground will tick away faster than clock away from the ground due to gravity".
But what exactly does this mean ? How does a regular clock gets affected by gravity, does the second hand slows down ?
If it has to do something with gravity affecting the inner working of wind up and pendulum clock will this also affect digital clock ?

** I realize I am missing the fundamentals of lotta things in this question, apologies for the stupid question in advance **

If it has to do something with gravity affecting the inner working of wind up and pendulum clock will this also affect digital clock ?

As astrosnapper noted in his comment, gravitational time dilation affects all time-dependent processes. This includes not only clocks (including digital clocks and atomic clocks), but also the rates at which chemical reactions and radioactive decay occur, measured relative to the equivalent process far removed from the Earth's gravitational well.

General relativity is a descriptive theory. It describes what happens, but not why. (Newton's theory of gravitation similarly was a descriptive theory.) What happens is that time flows slowly in the vicinity of a gravitational source compared to how time flows well away from that gravitational source. The time dilation is a function of gravitational potential energy, not gravitational acceleration.

If you want an explanation regarding why this happens, the answer is no one knows. As noted above, general relativity is a descriptive theory. It explains what will happen but it does not explain why.


Does Mars contain more iron than the Earth? Is the reason Mars is red because its surface contains a lot of iron? (When dirt is red on Earth, it is sometimes caused by a high amount of iron).

If so, does Mars contain more iron than the earth?

Mars definitely has far less Iron than Earth. Mars has 10.7% of Earth's mass. On the other hand, Iron comprises 32% of Earth since there is so much Iron in its inner core, outer core, and the mantle. That means if Mars was made entirely of Iron (which it is not), Earth would still have more than 3 times as much Iron.

You are right that Mars has more Iron at its surface than Earth so it looks very red. Earth has many other metals on its surface (not to mention water to make it look blue and plants to make it look green).


How can there be 1,000 stellar ancestors before our Sun? I've heard from a few sources* recently that the Sun is a 1,000th generation star, meaning it had a thousand stars that came before it based on its heavy-element content.

I understand that earlier stars' supernovae created the heavier elements and those were incorporated into the Sun when it formed and we can calculate the generation based on that.

My question is based on the age of the universe and the age of the Sun and the average age of stars in the galaxy.

Basically, the universe is ~13.7 billion years old, the sun is ~4.6 billion years old (and has ~5 billion years of life left), which means it was formed when the universe was ~9.1 billion years old.

For 1,000 stars to live in that span of time, those stars would have to have an average lifespan of ~9 million years (not including time between stars). This equates to each of those thousand stars having an average mass of ~100-150 solar masses so their lives would be sufficiently short to not go over that ~9 million-year lifetime.

So basically, there were 1,000 stars that lived and died before our Sun, all of which were huge monsters. This just all seems so improbable. How can this be?

* TV shows, I don't remember which ones

The Sun is actually a THIRD generation star. What I mean by this is that there are chemical elements in the Sun that were made inside another star, but that star itself can only have made those elements because it had material in it that must also have been made inside previous, second generation stars. Eventually we get back to the first generation stars, born out of primordial gas from the big bang that contained almost no heavy elements (those beyond helium) at all.

That is quite a mouthful, so let me explain using an example - barium.

There is barium in the Sun. We can tell that by looking at the spectrum and seeing absorption lines due to barium. But barium cannot be made in the Sun. The barium is made via the s-process, which involves the slow capture of neutrons onto the nuclei of iron-peak elements. This happens during the asymptotic red giant branch phase of stellar evolution, and the Sun has 6 billion years or so before it gets to that point. [NB: not even half the abundance of chemical elements beyond iron are produced by supernovae explosions$^1$.]

So, before the Sun, there must have existed a star - probably an intermediate mass star - which evolved to become a giant, made barium in its interior, then lost its envelope through a massive wind into the interstellar medium, and that material was incorporated into the protosun. Such stars (between, say, 2 and 10 solar masses) would have much shorter lifetimes than the Sun$^2$, so plenty of time for them to live and die before the Sun was born.

But wait a minute! That previous star must have already had iron-peak elements in its interior to act as a "seed" for the s-process production of barium. These were not and could not be made inside that star. They must have been made in a previous star, probably a massive star, that burned through all the nuclear fusion stages before exploding as a supernova, casting heavy elements, including iron-peak elements, into the interstellar medium. This previous star could also have had its own (metal-rich) ancestors, but ultimately as we go back in time we reach a point where the previous star was a first generation star, made from primordial H/He gas, with almost no heavy elements. These first generation (a.k.a. population III stars, just to be confusing) were probably very massive and short lived - a few million years. They would be born when the universe was a few hundred million years old and we can see no examples of them in our Galaxy today.

To try and define more precisely what I mean by "generation".

First generation - made from primordial big bang material.
Second generation - a star made only from the detritus of dying first    generation stars, enriched in heavy elements but lacking in primary s-process elements.
Third generation - a star made from material already enriched in heavy elements and including elements that are produced in the s-process inside previous second (or third) generation stars.

So that is why I claim the Sun can be classed as a "third generation star" - it contains atoms/nuclei that must have been inside at least two previous stars.

But you should not take this too literally. There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in individual stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside many different stars of different types.

Stellar evolution and nucleosynthesis calculations tell us the same story. For example, whilst most of our oxygen was made in massive stars that underwent a core collapse supernova, such events do not produce that much carbon. The C/O ratio tells us that most of our carbon comes via the winds from intermediate mass AGB stars. Heavy elements like uranium may be dominantly produced in neutron star collisions, but others like barium and strontium are not.

The details of how many ancestors have contributed to the Sun has no simple answer. Much of the solar hydrogen and helium could be pristine; some will have been through more than one star. Heavier elements (bar some lithium) will have been through at least one star. The fact that we have s-process elements like Ba, Sr, La and Ce, which are formed by neutron capture onto iron-peak elements, tells us those have been through at least two stars.

However, these are vast underestimates. Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities, driven by the winds and supernovae from massive stars, should distribute material on galactic-length scales in a billion years or less (Roy & Kunth 1995; de Avillez & Mac Low 2003), though local inhomogeneities associated with nearby recent events can persist over $10^{8}$ years. If this is the case, then the Sun is the product of the $\sim$ billion stars that died before it was born.

The reason you are confused with your lifetime argument is that you have ignored the possibility of the Sun being made from stars that lived at the same time in different parts of the Galaxy. The material that they ejected near the end of their lives has just been thoroughly mixed up.

$^1$ The rest are produced by the s-process in intermediate mass AGB stars; through nova events on white dwarfs; or perhaps, in the case of the heavier elements, through the collision of neutron stars (see this Physics SE question).

$^2$ A rough expression for the lifetime of a star is $10 (M/M_{\odot})^{-5/2}$ billion years.


Spinning black holes acting as a slingshot Could we use spinning black to act as a skyhook giving us enough energy boost to travel the galaxy?

Skyhooks or tethers are most useful for lifting mass from a surface into space, increasing velocity. Kerr blackholes have no appropriate surface for basing mass. Instead a gravity assist maneuver may take advantage of the high spin speed of a Kerr black hole, which induce frame-dragging where the space around the blackhole (within the ergosphere) is dragged at the speed of light in the same direction as the spin.
https://nasa.fandom.com/wiki/Gravity_assist#Limits_to_slingshot_use
The Penrose process also offers a means of extracting energy from Kerr blackholes, if the rotational energy of the black hole is located not inside the event horizon but inside this ergosphere.
https://en.wikipedia.org/wiki/Penrose_process


How did the authors of Surya Siddhanta find the diameters of other planets in the solar system? The Surya Siddhanta, "a Sanskrit treatise in Indian astronomy from the late 4th-century or early 5th-century CE" is truly a great work.

But how was it possible for the writers to find the exact values of the diameters of different planets and the distance between the sun and the earth?

Also, are there any mistakes in this book?

If there's any problem in my question please inform me.

The authors assume a geocentric universe (first thing that is wrong). They then assume that the planet Mars has the same apparent diameter as a globe 30 yojana in diameter (about 150 miles) in the same orbit as the moon, from a perspective at the centre of the Earth. This is just stated, and appears to be supposition. It is an incorrect figure. It is conceivable that some kind of system for sighting through a pinhole could be used for this estimate. This method tends to over estimate the apparent size of planets.
They next claim that the other planets have apparent sizes that would be the same as globes with diameter 37.5 (Saturn), 45 (Mercury) 52.5 (Jupiter) and 60 yojana (Venus). These form an arithmetic sequence with a common difference of 7.5, but apart from that, these values seem to be arbitrary (and are wrong). They might have been based on some kind of pinhole observations, but rounded to a simple sequence for the sake of the poetry or for easier memorisation.
There is uncertainty about the ancient value of the yojana. A different conversion to miles will give different values throughout.
Using this assumption and the distance already calculated to the moon they calculate the apparent angular diameter of each planet. The actual values they get are wrong, In fact, the angular diameter varies as the planet's distance varies.

Planet
calculated value
modern observed value

Mars
2 arcmin
(actually 0.06 - 0.39)

Saturn
2.5 arcmin
(0.25 - 0.34)

Mercury
3 arcmin
(0.08 - 0.17)

Jupiter
3.5 arcmin
(0.51 - 0.83)

Venus
4 arcmin
(0.16 - 1.05)

As you see the calculated values are much too large (ie wrong). They are comparable with values supposed to have been found by Hipparchus, but without a known observational basis. Tycho Brahe also gave similar values, found by sighting through a pinhole. This simple method doesn't give a good estimate of the angular size. The simple pattern in the calculated values is due to the original assumption that the globes were in an arithmetic sequence.
There is then a calculation of the geocentric orbital radii of the planets. They assume that Venus, Mercury and the Sun all orbit at a distance of 3.4 million miles.  This is a significant underestimate, especially for the sun (this is wrong). The distances to Mars, Jupiter and Saturn were also underestimated.
One can then combine the apparent angular diameter (which is too large) with the calculated distance (which was too small) to obtain an estimate of the planetary diameter. When this is done, it gives exceedingly good estimates of the diameter of Mercury and Saturn (within 1%). A good estimate for the diameter of Mars (about 10% out), and a poor estimate for Venus and Jupiter (about 50% of the accepted modern values).
It is clear that there is much that is wrong. A combination of overestimates of some values and underestimates of others cancels out to give some impressive looking figures for two out of five planets. But do note that the estimates for Jupiter and Venus were a long way out, and all the numbers which the calculations were based on were wrong. In total there are 15 values calcuated here 5 apparent diameters, 5 geocentric distances and 5 actual diameters. Of these only 2 values are impressively accurate.
They did better than other pre-telescopic observers, but it isn't clear that they obtained values by anything other than luck.
Source


Open source code for the maths behind a heliostat? Theoretically, using a Raspberry Pi, (at least) one mirror, and two motors, one should be able to build a heliostat, i.e. a device which redirects sunlight to a fixed spot, like a scrub in the shadow of a building.
I am now searching for heliostat (open) source code, ideally in python, hopefully with enough comments. Also: Is my following rough approach correct?

We need to know the exact geographical location of the mirror in terms of longitude, latitude. For simplicity, we assume that the mirror itself always has an obstacle-free view of the sun.
For a given time, we can use celestrial mechanics to calculate the path of the sun on the sky.
Using the reflection law from geometrical optics, we can determine the position of the mirror, since we know the vector from the mirror position to the spot we want to direct light to.

That sound simple enough, at least theoretically. I read that for step 2, many use precalculated tables. Why? Is it numerically so challenging?
References

https://www.sunearthtools.com/dp/tools/pos_sun.php
Jon Henley: Rjukan sun: the Norwegian town that does it with mirrors
Renard Bleu: ICARUS: the Analog Heliostat
Raspberry Forum: 2 axis Heliostat
Open Source Sun Tracking Skylight
Sadly mostly offline, but everybody is linking there: http://www.cerebralmeltdown.com/

I wrote the source you need some years ago:
https://jumpjack.wixsite.com/progetti/sorgenti-ipsun
The Arduino/Processing version was just a demo program to manually control a TENVIS camera by multiple buttons, it lacks the "astronomical algorithm".
The Javascript version contains astronomical calculations and a demo page which connects to a local TENVIS camera.
I also built a prototype...

https://jumpjack.wixsite.com/progetti/ipsun
...then I found, at MakerFairRome 2016, a maker which sold on Kickstarter a ready-made heliostat, initially named "Lucy", then renamed "Caia"... but 5 years passed and it has not yet been born... so I think I'll have to go back to my project and develop my own heliostat.
But in the meantime I found this page about a "static heliostat", and now I am trying to reverse-engineer it to figure out if it could be easier/cheaper:

http://cleardomesolar.com/solareflexpanels.html
Additional resource with dozens of links to heliostats, coelostats, and whatelse...:
http://www.redrok.com/main.htm#gsc.tab=0


How to tell which of the ecliptic cross horizon angles is to the east? Jean Meeus' Astronomical Algorithms, 2nd ed. p. 99 gives a formula for finding the two longitudes where the ecliptic crosses the horizon:

$$\tan\lambda = \frac{-\cos\theta}{\sin\epsilon \tan\phi + \cos\epsilon\sin\theta}$$

where "$\epsilon$ is the obliquity of the ecliptic, $\phi$ is the latitude of the observer, and $\theta$ is the local sidereal time, returns $\lambda$ the longitudes of the two points of the ecliptic which are (180 degrees apart) on the horizon".

The software I am using then applies $\arctan$ to get $\lambda$ and adds $\pi$ if $\lambda < 0$. Then it returns $\lambda$ and $\lambda + \pi$.

I need to know which of the two angles returned points to the easterly direction. Is there any way to tell which is which? I have all of the information needed to perform the calculation (UT, local time, sidereal time, latitude, longitude), but I can't see how they are related.

Thanks much!

These steps can be used to find which of the two horizons is to the east:

The local sidereal time equals the right ascension ($\alpha$) of the meridian. From equation (13.3) that transforms the ecliptical into equatorial coordinates, the ecliptic longitude of the meridian ($\lambda_m$) can be calculated. Since the ecliptic latitude is zero ($\beta=0$) on the ecliptic,

$$\frac{\tan(\alpha)}{\cos(\epsilon)} = \tan(\lambda_m)$$

Note that $\lambda_m$ must be in the same quadrant as $\alpha$. If this is not the case, add 180 degrees. Of the two ecliptic longitudes calculated to intersect the horizon, the one that is approximately 90 degrees greater than the adjusted $\lambda_m$ is the one that is in the east.


What happens if a white dwarf collides with a star like our sun or a red dwarf? Would it explode in a supernova or  will it form a new star?

Fun question.   Without doubt it would be very violent and spectacular, but not much is known about stellar collisions and only a few have ever been observed.

Most stellar collisions happen due to tight orbits where the stars spiral in towards each other or perhaps, 3 or more body chaotic star systems with unstable orbits that lead to an impact.   Space is so large that actual star on star collisions are exceptionally rare.

A spiral in towards each other is almost certainly more common, it also quite different in terms of outcome.  Much of the gravitational kinetic energy of the collision goes into angular momentum, so, while still spectacular, it's quite a bit less explosive than a direct head-on collision.

Black hole/black hole collisions have been observed by detection of gravity waves when two black holes orbit around each other and the orbit decays.

Neutron Star/Neutron star collisions may be responsible for the very heavy elements in the Universe and White dwarf-White dwarf collisions may lead to a supernova similar to a type 1a Supernova (per Wikipedia source).   All these collisions.

In a tight orbit, a white dwarf will steal material from a near-by star and if the orbit is close enough, lead to a decaying orbit and collision, but that would happen over time.  If the white dwarf reaches the Chandrasekhar limit in that scenario, it will undergo type 1a supernova.  If there's not enough mass for that, it should eventually absorb the nearby star, leading to a new star, admittedly a strange one with very high metallicity and perhaps abnormally hot and short lived for it's mass, but it should basically be a star once the collision settles down, but probably a star that burns both hydrogen and helium, but wouldn't collect enough helium for a helium flash.   I don't see any other alternative with that kind of below Chandrasekhar mass spirally in collision than it being a very metallic star, hot for it's mass and with a shorter lifespan.

Now a head on collision - that's very different and, I'm basically guessing, but you have a white dwarf star - about the size of Earth with a mass probably at least 30%-40% of the sun.   White dwarfs don't have a specific minimum size, but for a star to go white dwarf within the 13.6 billion years or so that stars have been around, that star would have needed enough mass to have completed lifecycle and in the process, it loses some mass, so perhaps, 30% of the mass of the sun might be the current lowest possible mass of a white dwarf, but as billions of years pass, that low end will grow smaller.

But if we take a low-mass white dwarf at 30%-40% the mass of the sun, it's still a super-dense object and crashing that into a star at their combined escape velocity, not the more common decaying orbital spiral, that's a messy calculation.   I'd only be guessing.

Escape velocities aren't additive, but still, white dwarf and small star would collide in this theoretical at an impressive velocity.    If it was a glancing blow, the white dwarf would almost certainly punch straight through, blowing a lot of matter off the star and perhaps affecting it's rotation rate.   It would be an impressive site.

If it was a direct hit, I don't know if the white dwarf would punch through the star or not.   It might still punch through.   The White dwarf would pull the surface of the star towards it as it got close and that would be the first contact between the two.   The Sun's escape velocity is about 618 km/s and the mutual gravitation the White dwarf would be traveling to the sun a bit above that, but the White dwarf's surface gravity is much greater, so the mater pulled off the sun into the white dwarf would be hitting the white dwarf at a few thousands of km/s.  (Small white dwarf, so lets say 3,000 km/s).

That's still far too slow for fusion to happen, so this is basically still just collision mathematics.   As the white dwarf plowed deeper into the star, towards the much hotter core, the heat and velocity of impact probably would increase fusion rates, but if it just passed right through the star that would be a pretty temporary and minimal effect.   The larger effect would be the disruption to the star after a very dense planet sized object punched it's way through it.

If the star was viscus enough to stop the white dwarf from making it all the way through, then the combined mass and dense center and added heat from impact should significantly increase the rate of fusion of the original star, even though some material would likely be blown off, the net result should be a heavier star that burns hotter and faster and if the new mass is comfortably above 1.4 solar masses, you likely have a super-nova in the making.   But as to whether or not a white dwarf would be to punch all the way through a star upon direct collision - I have no idea.  I guess that it would, but that's just a guess.

A fun related question is, how long would it take the white dwarf to emerge from the other side, assuming it's a relatively center hit and assuming it passes through (which it might, but I'm not sure if it does).   I'll try to give that a look too.

Anyway, just my 2 cents on this question.   Corrections welcome.


How long can afterglow last after a fireball? I was watching the YouTube video Perseid Meteor Shower - Mojave Desert, California which looks East before sunrise, presumably in mid-August of 2016.  I've made a GIF of a few seconds near the beginning where a fireball appears at about 00:03 in the video, and a faint luminescent trail appears to drift and change shape. I've timed the two stars in Orion Bellatrix and Betelgeuse and the motion is about seven seconds of video to move about 30 minutes in R.A. (near declination 0), so based on that the three or four second visibility in the video works out to about 10-15 minutes.

Is this common? Am I interpreting what I'm seeing correctly? Is such luminosity visible by eye, or only using a camera with a good lens?

This is an persistent ionization trail. As the meteoroid goes through the upper atmosphere at high speed (sometimes more than 45km/s) electrons are stripped from atoms. As the electrons recombine, the gas glows.

Persistent trails are formed by brighter meteors, and can be visible to the naked eye. They may be visible for up to about 45 minutes, but that depends on the meteor. Generally brighter meteors have brighter trails.

The movement and streaking in the trail can reveal the presence of upper atmosphere winds. As radio waves bounce off the ionised trails, they can be used for long distance radio communictions.

Persistent trails were discussed by the "Bad Astronomer"

Image taken during the 1966 Leonid Meteor storm, produced by a Mag -6 meteor