How old are Janus and Epimetheus? How old are Janus and Epimetheus, and are they believed to have formed out of a collision? (I think would explain their shared orbit).

Not the age of their elements, which should be similar to the age of the solar system, but the age of their surface and formation.

Cassini told us that Saturn's rings formed relatively recently, 100 or 200 million years ago.  Are there similar aging techniques that can be applied to the surface of icy moons and as that been done?   (related, this type of aging technique should apply to Pluto and Charon too, because they are believed to have formed by collision).

Edit:  I came across this which says:

  The extensive cratering indicates that Epimetheus must be quite old.
  Janus and Epimetheus may have formed from a disruption of a single
  parent to form co-orbital satellites, but if this is the case the
  disruption must have happened early in the history of the satellite
  system. From its very low density and relatively high albedo, it seems
  likely that Epimetheus is a very porous icy body.[4] There is a lot of
  uncertainty in these values, however, and so this remains to be
  confirmed.

So, significant uncertainty but probably quite old, early in the history of the satellite system, which would be fairly early in solar-system formation.  Much older than Saturn's rings.

I'll leave this question open in case anyone has better info, the above being from Wikipedia, though they reference a NASA article.

You are correct that the main way to determine planetary surface ages is through crater dating. This comes in two versions, relative and absolute dating. Both require counting the number of craters per area and so you need good resolution imaging since the number of impactors and craters follow a power-law (more small impactors/craters than large). It then basically boils down to older surfaces have had a longer time to be hit by more impactors and so there will be more craters per unit area.

With relative dating, all you can do is put different surfaces in order of amount of craters and say "surface A is older than surface B as it has more craters". If you have another method of dating, then you may be able to say where on the timeline a surface may be but you would still only be able to "surface A is older than surface B (dated to be 250 Million years)" but you don't know much older.

If you have an idea of the rate of cratering bodies and the rate of crater formation in the system you are trying to study, then you can perform absolute dating. This allows the cratering rates to be compared across different bodies. This has been done for the inner Solar System (Mercury, Moon, Mars) but is tricky where the surface can be altered (weather, lava flows etc). There are also problems with crater saturation with very old surfaces where new craters overlap and overwrite older ones and issues with understanding the lifetimes of the craters and what processes can fill-in and erase craters and how long this process takes.

In the case of the Saturn system, determining the number of potential impactors and performing absolute dating is likely to be pretty tricky as the environment is dynamically complex. There are a lot of large Moons and you also have Jupiter and the other gas giants planets affecting things. In addition, the positions of the giant planets are thought to have changed dramatically around 500 Mya after formation (the Grand Tack model and hypothesis).

The images of the Pluto and Charon system returned by New Horizons have also shown some of the difficulties with crater counting (blog article on this by a member of the New Horizons team and crater dating expert). Some parts of the system such as Charon appear to be very old and have large numbers of craters; other areas such as Sputnik Planitia appear to be very young and no-one can find any craters of any size anywhere on that region.


How does the black hole at the center of M87 look like in the visible spectrum? I learned here that the picture of the black hole we saw all over the news was a false color image, based on the radio spectrum emissions. How would the black hole at the center of M87 look like when seen by the naked eye from a close enough distance, or by a large enough optical telescope? (please tell me if the two would be different)

Disclaimer: this is speculative, since no images of a black hole have been taken with enough resolution in visible light.

The black hole represented in the movie Interstellar is moderately realistic with the knowledge we have at present. In fact, the movie makers asked astrophysicists to contribute.

From the article Gravitational lensing by spinning black holes in astrophysics, and in the movie Interstellar, this is a moderately realistic view of a black hole.

The light is not the black hole itself, but the accretion disk around it. The circle of light is the part of the accretion disk that is behind the black hole, the light of which is lensed by the black hole.

Now for a black hole spinning on itself, then it would not look as symmetrical. Doppler shifting would make one side appear more red, and the other more blue.


Does the ecliptic precess? Does the Sun's rotation precess? We know that the Earth, like a top, slowly precesses as it rotates. We know the Sun rotates, but does it precess as well? What about planetary orbits? Is the orientation of the ecliptic stable or will its inclination change over time? If so, would the orbital inclinations of the planets currently in the ecliptic change independently or in unison? We know the ecliptic is inclined with respect to the galactic plane. Would that have been a consequence of "residual" angular momentum when the solar system formed, or did it gradually take up its present orientation?

As you mentioned, the Earth's rotational axis undergoes a slow precession. This results from tidal torques caused largely by the Sun and the Moon. There would be no precession if the axial tilt with respect to the orbital angular momentum vectors was zero or if the Earth was spherical.

The Sun is almost spherical it's oblate to about $10^{-6}$, and is almost perfectly aligned with the ecliptic, it has a relatively slow rate of rotation, and all of the bodies that might cause a precession are far from the Sun, and because the gravitational gradient torque is an inverse cube relationship. All of these factors combined means the precession rate of the Sun is ridiculously small. So the short answer to your question is:

Yes, the Sun precesses.

Planetary orbits are a little different, I'm not exactly sure how to state the process for their precession without waving my hands about the room, so I won't attempt it. However, I will say that, as all things in the universe, I am sure the ecliptic is a dynamic system unto itself.

About the ecliptic (dashed line) and the galactic plane (spotted strip) see the attached image.

If the ecliptic and the galactic plane were aligned, the two would line up on top of one another, as the Sun's and other planet's paths would be roughly the same path.

There is still residual angular momentum within the solar system, which is how we arrived at our current orientation amongst the stars.


A double star consists of two stars of equal brightness at equal distance from the Earth. What is the apparent magnitude of the combined object? The magnitude of each star in the V-band is mV= 8.34. If these stars are so close together that they appear as one object, what is the apparent magnitude of the combined object?
I just want someone to point me in a general direction of how to approach this problem. The magnitude is logarithmic so it won't be double when combined. Or at least that is what I think. How do I apply the formula
$$m=-2.75 \log_{10} F + C$$

The magnitude formula is $$m = -2.5 \log_{10} \left(\frac {F}{F_{0}}\right)$$
Where $F_0$ is the flux of magnitude 0 star.
So if you have two stars of equal brightness, the flux is doubled, and you get
$$m = -2.5 \log_{10} \left(\frac {2F}{F_{0}}\right)\\
= -2.5\log_{10}(2) -2.5 \log_{10} \left(\frac {F}{F_{0}}\right)$$
The first value is about -0.75, so the double star is 0.75 magnitudes brighter.  If each object is 8.34 then the double star appears as a 7.59 object.


How to tell whether an optical spectrum is featureless or just with a low S/N? A featureless spectrum and a spectrum low S/N are very similar.
Is there a way to tell the difference between them?

Convert answer in comment to community wiki

You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course, the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.


Why are all space observatories in Chile? Anytime I hear about astronomy there is observatory in Chile involved. Sounds like there are the best conditions to watch space. Anyone care to explain what is the reason?

[disclaimer: I am also not a true observer, but I have visited observatories before, and I do amateur observations regularly]

Definitely not all observatories are in Chile. Chile, however (the Atacama desert in particular) is an excellent location for an astronomical observatory. The Atacama desert has one of the darkest night skies on Earth, which is excellent for astronomical observations.

As Warrick mentioned, the ideal observatory is located at the highest possible altitude, such that there is as little air as possible between the telescope and the stars. Something else to consider, though, is that if the telescope should be manned by humans permanently it cannot be located too high due both to the difficulty for humans to function at extremely high altitudes and the price tag of getting the telescope parts up there. The two observatories in the Atacame desert of Chile are both located at an altitude of 2.5 km, which is definitely not the highest point on earth (the Mauna Loa and Mauna Kea observatories mentioned by Warrick are both located at higher altitudes).

Another requirement is that the location should be as cold as possible. This is not to torture the poor observers operating the telescope, but because warm air tends to be more turbulent. More turbulence in the air means less precise observations. However, as temperature tends to decrease with increasing altitude, this is usually automatically satisfied. Having a stable temperature also helps. If the telescope heats up during the day, the temperature of the air inside will differ from that of the surrounding air at night. Since the refractive index of air depends (although very slightly) on temperature, this will cause further deviation of light rays from their ideal straight paths, leading to less accurate measurements. For this reason professional telescopes usually contain some kind of temperature regulation.

A combination of the above reasons is why observatories are in the locations they are. Theoretically, the best observatory location has been found to be Ridge A on Antarctica. The reason you always hear about Chile is not its location. There are other locations that are at least as well suited as the Atacama desert for astronomical observations.

The most important reason you always hear about Chile is probably that it is home to the VLT, or Very Large Telescope, which is the most productive ground-based telescope in the optical region (according to Wikipedia). It uses interferometry combining the observations from four individual telescopes. This makes the combined effective aperture of the VLT the biggest in the world. As both the most productive telescope and the telescope with the largest effective aperture it produces a lot of results, and those results are quite likely to be important enough to appear in popular science articles.

However Chile is definitely not the only possible location for observatories. Apart from Mauna Kea and Mauna Loa, the Observatory del Roque de los Muchachos on La Palma is the first example that pops up in my mind. It is home to the GTC, the worlds largest single-mirrored optical telescope.


Why is it strange that outer stars in a galaxy are travelling with the same speed as inner while the total mass is also increasing? Astronomers say that it's strange that the outer stars of a galaxy are travelling with the same speed as the inner stars. Therefore they introduced dark matter to solve this strangeness. The theory that the inner stars 'have to' go faster is that because they are closer to the center so the gravity is more, like the earth is closer to the sun and faster than Neptune.

But what I wonder is, whether the fact that just the outer stars are attracted to more mass, because there is more mass of stars that are within the orbits of the outer stars, explains the equality? So in this view it is not so strange that they have the same speed because they are influenced by more gravity. Or is it that this extra influence is considered to be too little to explain the equality?

This is accounted for in real models of the rotation curves of galaxies. The rotation curve of a galaxy is more complicated than the motion of planets in the solar system, for the reasons that you describe.

A trivial model would consider the galaxy as spherically symmetric and would use the shell theorem to estimate the centripetal acceleration of a star at radius $r$. Thus
$$ m r \omega^2 = G\frac{m M(r)}{r^2},$$
where $m$ is the mass of the star, $\omega$ is the angular velocity and $M(r)$ would be the mass of all gravitating matter at radii $<r$. Note that this only applies to a spherically symmetric distribution of mass.

To make further progress demands that you know the density distribution of the gravitating matter. Let's just assume that density $\rho$ is constant for the moment. Then
$$ r^3 \omega^2 = G \int \rho\ 4\pi r^2\ dr = \frac{4\pi}{3} G \rho r^3$$
$$ \omega = \sqrt{\frac{4\pi G \rho}{3}}$$

Thus the angular velocity would be constant with radius and the rotation speed, $v = \omega r$ would increase with radius. This I think is the situation that your question supposes and so yes, if there was a constant (or perhaps slowly declining) density of material in the Galaxy then this would produce a rotation curve that increased (or was flat).

The trouble is that the density of material in our Galaxy inferred from the matter that we can see is not constant with radius. It declines rapidly and exponentially, such that there is very little visible matter beyond a radius of about 15 kpc. If we take a situation where we go to radii beyond the gravitating matter, then a similar treatment to the above suggests that
 $$ m r \omega^{2} = G\frac{mM}{r^2},$$
where $M$ is now the total mass of the (visible) Galaxy. In this case
$$ \omega = \sqrt{\frac{GM}{r^{3}}}$$
and the rotation velocity $v = \omega r$ should decline as $1/\sqrt{r}$ (like it does in the solar system).

It is the fact that the rotation velocity of stars and gas at large radii ($>15$ kpc) continues to be flat or even increase that leads to the conclusion that the mass that we can see is not all that there is. i.e That we need "dark matter" to explain the high rotation speeds at radii where there is very little visible matter.

Realistic models of the galaxy do not make the assumption that the visible matter is spherically symmetric (it isn't). But the conclusions I qualitatively set out above hold in the same way.


The colour of saturn as it appears to the naked eye I have started to watch the skies in order to learn stars, constellations planets etc. The few references I checked state that Saturn presents itself to the naked eye as a pale yellow tiny little ball, easy to notice if one knows beforehand its position.

So, using Stellarium I have managed to learn that Saturn is near the Scorpio constellation, just below its claws, which is easy to spot because Antares has distinct red-orange colour.

However, when I first spotted the planet I noticed that it was like a blue ball. I should say it looked like a pale, near white blue or a pale blue, near white -- I can not be more precise than this.

I am spotting it correctly? I was taught that planets do not scintillate like stars. The ball I am talking about certainly did not do it. If I did spot it correctly, I would like to ask if and how often Saturn changes its colours.

p.s. I thought it might be a case of city pollution, however I should say I live in a small mountain village with very few if any air or light pollution.

Many thanks in advance.

Planets do scintillate, just much less often - it takes much greater turbulence to make them twinkle. I've seen Venus twinkle quite a few times, especially when it was close to horizon.

Light pollution makes no difference when observing the planets in the Solar System. They are far too bright to be affected by it.

Saturn should be an off-white when observed with the naked eye. That being said however, if the planet is close to horizon and atmosphere is turbulent, it can cause some weird color effects. But then the color would appear to change rapidly.

In any case, as someone else said in comments, no planet looks like a "ball" to the naked eye. They are all points of light just like the stars.

Check this guide, it should tell you the best method to locate Saturn these days (use the image at the top of the article):

http://www.skyandtelescope.com/observing/sky-at-a-glance/this-weeks-sky-at-a-glance-september-4-12/

If the object you're looking at is in the right place with regard to Antares and the other things in Scorpio, then it's probably Saturn, no matter what color it seems to be. Color is very subjective and can be modified by the atmosphere in some cases. Position is not subjective.

Don't worry too much about color. For comparison, Jupiter is a more pronounced shade of butter-yellow than Saturn, so that's less controversial, you don't see people argue about the color of Jupiter. But a lot of people see Saturn as plain white; I see it as off-white, the color of old PC cases.

If you can borrow some good binoculars, you should see the ring. Lean the instrument (or your elbows) against something solid (like a table or wall), try and keep the binoculars very steady, and watch the object. Be patient and hold steady. You should see the ring as a tiny streak across the planet, kind of like the letter phi: Φ. Just regular binoculars should be enough to make this observation, and it's pretty awesome when you see it for the first time.

In a bigger instrument the real shape of the ring becomes evident.


What was the entity/particle/state at the time of the Big bang? At the time of the big bang if there was an original entity, what were it's attributes(if applicable) - mass, electric charge, color charge or spin? Can the current matter and dark energy together account for the original entity or was some of the original stuff lost?

The notions of today's universe don't make much sense for the (hypothetical) pre-Planck epoch of the universe. All four known fundamental forces should have been united. Matter didn't exist. It can form only in cooler conditions. Even cause and effect don't work well below Planck scales.

The structure you eventually get can be treated in terms of a groupoid, as described e.g. in this paper. Groupoids are related to groups (algebra). Groups actually are - in some sense - very simple groupoids. You may construct some groupoids from disjoint groups by connecting them via isomorphisms. So you get kind of a network of abstract entities.

Now try to adjust quantum mechanics and General Relativity, such that it fits with these structures, meaning without the smooth space and time, as usually assumed for standard physics. Let known or extrapolated physics of the Planck- and GUT (grand unified theory) epoch emerge from the pre-Planck physics to start the actual Big Bang.

It's possible to construct physical theories for a pre-Planck era. But it's very hard to find observational or experimental evidence to test these theories with respect to actual physics.

Those pre-Planck groupoid theories can be defined in a way that they don't need a predecessor epoch, e.g. by eliminating the macroscopic cause and effect paradigm, which is based on a notion of an arrow of time.
Causality and a concept of before/after on isomorphic entities doesn't make much sense.

Although, there should be some mechanism which eventually disrupted a strictly non-local pre-Planck configuration.


What is the spectral reflectance of starlight in a close binary? Ignore the Blender Monkey! (from here)

In a close-orbiting binary pair, a small but significant fraction of the light from each star falls upon the other, and the result has to be carefully modeled to simulate the observed spectrum of the system as it rotates.
One effect is that each star's radiation will non-uniformly heat the surface of the other star and alter its thermal radiation, but here I'm asking about light from one star being scattered or diffusely reflected by the other.
How is this scattered/reflected light calculated, and roughly what is the spectral reflectance?
Of course each star will be different, so one example or maybe two would be sufficient.
We often treat stars as blackbodies as an approximation, suggesting that stars have low albedo, but at some depth the electron density will be high enough so that the plasma frequency reaches visible wavelengths, increasing reflectivity, though that may be too deep to matter.

Stars are far from perfect blackbodies due to scattering/reflection.  This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount.  For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, but the latter number is not meant to be taken seriously, they simply don't care if the light is reflected or absorbed and re-emitted because it isn't an important term. But the value of 0.30 for the K star might be meant more seriously, though it is still not regarded as a critical parameter because it only affects the color of the light that is reflected, not the total amount of light (given that stars are in radiative equilibrium, so must ultimately return all the incident light, whether it happens by reflection or heating).

Indeed, stellar emissions are often characterized by "effective temperature," to connect the surface flux of a star to the Stefan-Boltzmann formula for the emission of a blackbody by using a T parameter that is not necessarily the actual temperature.  When using this notion, as is quite common for dealing with stars, there is no essential difference between heating of and reflection from the surface of the star in question. The details of the difference have to do with the shape of the spectrum, but that shape is generally not a Planck function anyway, so as soon as one is using the "effective temperature" concept one has already parted company from a detailed understanding of the shape of the spectrum.  (When you do want the details of the spectrum, you will have to model the situation with some care.)

You are right that the albedo depends on the type of star (hotter stars having high scattering albedo due to all those free electrons), and many stars do show a lot of scattering from their surfaces.  Scattering/reflection elevates the effective temperature above the actual temperature of the layers being looked at, but again, if you stick to effective temperature, it makes no difference if there is scattering or heating of a blackbody, because effective temperature explicitly refers to the outgoing  flux with no claims on the actual temperature.  It is quite common for this distinction to be swept under the rug, and most stellar temperatures are effective temperatures, not actual temperatures.

Another consequence of radiative equilibrium is that the "reflection effect" cannot alter the total luminosity of the binary system seen from all angles, so if the effective temperature of the surface of the stars is elevated by reflection and/or heating, as seen from some directions, this must be exactly compensated by the reduced brightness due to eclipses seen from other directions.  So the reflection effect is part of the study of eclipse light curves, which are quite useful for understanding things like the sizes of the stars.

Added:  a nice paper on the polarization of reflected light was provided in the comments by uhoh, it is  nature.com/articles/s41550-019-0738-7 .  The idea is that if the stars are not too close to each other, there's not a lot of reflected light, but it is still noticeable by virtue of the fact that it is highly linearly polarized in the direction perpendicular to the line between the stars.  So it shows up much better in linearly polarized light, a potentially important new diagnostic of binary systems.


Why don't we have a browsable spacetime augmented virtual solar system? Wouldn't it be nice to be able to render any viewpoint from all the data transformed into a point cloud, an augmented virtual reality of moving objects in our solar system.

I would just like to be able to chose any point in the solar system and view exactly what is there at a specific time.

It's possible to extrapolate 3D data from multiple viewpoints of an object, don't we have enough satellite data to do it?

There is NASA/JPL's Solar System Simulator. It doesn't cover all points, but does include many of the interesting ones.


Intuition about why gravity is inversely proportional to exactly square of distance between objects What is the intuition behind why gravity is inversely proportional to exactly square of distance between two object and not cube or not some multiplier.  Basically how Newton came up that its exactly square of distance? How it was validated it is in fact square of distance ?

Imagine "gravity" spreading out in a sphere, like light from a bulb.
For each doubling of the distance, the sphere has four times the area.  The surface area of the sphere is proportional to the square of the radius.  If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius.
This gives sufficient intuition for most physicists of the time to believe that gravity was probably inverse square.  Newton's real genius was that he was able to prove using mathematics that a planet moving in an inverse square law would obey Kepler's laws, something that his contemporaries had failed to do.
A well known story, related by de Moivre:

In 1684 Dr Halley came to visit [Newton] at Cambridge. After they had been some time together, the Doctor asked him what he thought the curve would be that would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it. Sir Isaac replied immediately that it would be an ellipse. The Doctor, struck with joy and amazement, asked him how he knew it. Why, saith he, I have calculated it.


When the sun explodes, will some of the planets in the Solar System survive and become rogue planets? Say if a large enough star, that has a number of planets, were to explode in a supernova:

Would any of the planets survive?
Would they become rogue planets?

The Sun will not become a supernova, it will never explode. A star must have about 8 times more mass than the Sun in order to cause a supernova explosion. When it starts to fuse helium, the Sun will become a red giant and expand out to about Earth's orbit and throw out gas to form a beautiful so called planetary nebula. Mercury, Venus and maybe Earth will be inside of its photosphere when it is at the largest. I don't know if they will evaporate completely inside the thin hot gas of the red giant Sun.

The remaining planets will keep orbiting the white dwarf remnant, which is not a star (no fusion) but a compact planet sized object. It will shine like a star because of its stored heat. Orbits will change because the Sun will have only 40% of its mass left, the rest having been thrown out to form a nebula, over very long time. Orbits should become larger, and planets maybe will be pushed out a bit by the ejections from the red giant too. Gas planets maybe get their atmospheres a bit stripped off, icy dwarf planets like Ceres and Pluto could melt and evaporate a bit of their mass since the helium burning red giant is hotter than the Sun today. But I'm kind of guessing here, I don't know if those are significant effects.


Would I actually be able to see Ceres without shining a really bright torch onto it? NASA's put out a semi-CGI video of a Ceres flyby using material from the Dawn mission.

Now I realise that visibility is relative, even on Earth: I can walk into my basement and not be able to see anything until I've stood there for a few moments letting my eyes adjust. But I'm wondering, since Ceres is so far from Earth, and despite the images we see beamed back from spacecraft, if I were on a spacecraft in orbit of Ceres, facing the side of Ceres that faces the Sun, would I be able to see its surface with the naked eye? Or is sunlight too dim at that distance to do the job on its own?

Ceres is 2.76 AU away from the sun, so the sunlight is seven times weaker. That is comparable to the levels of illumination 30 minutes before sunset on Earth, so you are not going to have any problem seeing the surface.

The albedo of Ceres is quite low, comparable to worn asphalt. But it is still easy to see it.


Time after sunset until star can be seen Given a star's apparent location and apparent magnitude, how how many degrees below the horizon must the sun be for the star to be visible to the naked eye for an observer on Earth at a specified location and elevation? You can ignore all light pollution, though I understand there will be some error due to air quality/weather/atmospheric aberrations.

Is there such a formula? If not, why not and what methods of approximation are available?

(I figure the brightness of the afterglow of the sun at a certain angle above the western horizon can be calculated for a certain time and place, and if it is less than the apparent brightness of the star in question, the star will be able to be seen.)

It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is 18 degrees below the horizon. Then, you can see all of the ones you're going to be able to see.

You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.


Does the Sun have a numerical designation in any star catalogues? Given the existence of large star catalogues listing millions of astronomical objects with numerical designations (such as Tabby's Star, KIC 8462852) it occurred to me that one of them might contain the Sun. However, I had no luck.

I searched NASA's HEASARC which combines GRB, SIMBAD, VizieR, NED, and BSC.

HEASARC found Sirius in SIMBAD, NED, and BSC (HR 2491), but there were no results for "Sun" or "Sol".

So, is there any catalogue (maybe a stellar evolution catalogue) that puts the Sun on equal footing with other stars, giving it a canonical number designation?

Bonus high fives if that number is 0 or 1. (Sorry, HR1 is a 6.71 magnitude star in Andromeda, also known as HD 3).

It's somewhat difficult to prove it either way given the large number of catalogues out there. The majority (at least of the ones on VizieR) do not include the Sun.
The only case I'm aware of that almost meets your requirements is in the Preliminary Version of the Third Catalogue of Nearby Stars by Gliese and Jahreiß (1991). This catalogue is well-known as the source of stellar designations beginning "Gliese" and "GJ", and is queryable via VizieR.
The first record in the catalogue is the Sun, which includes its spectral type, apparent magnitude, colour indices and space velocity as for other stars. Right ascension, declination and proper motion aren't included for obvious reasons. It doesn't match your requirement for having a numerical designation because the name field is just "Sun", while the other stars in the catalogue have numbers. The "NN" stars originally didn't have numbers either but these were added in 1997: in the literature these objects are still typically referred to as "GJ ####" rather than "NN ####".


If a planet gained too many moons could the tidal forces of those moons rip the planet apart? Or would the planet just be subject to extremely intense tidal forces instead?

No.
There are two simple arguments, each on its own alone enought to refute such thought:
First way: Tidal force is the gradient of the gravitational attraction. Gravitational force goes like $F_g \propto \frac{M}{r^2}$ and the tidal force thus as $F_t \propto \frac{M}{r^3}$. For a moon and its planet the mutual distance is the same. And as the planet has the higher mass, the tidal forces of the planet would be higher on the moon than the one of the moon onto the planet. Materials are not that different that it plays a major role here. So before a moon destroys a planet, the planet destroys the moon (or you mis-named the two in the first place, calling the planet the moon and vice versa). This may happen for moons within the Roche limit of a planet.
Second way to argue:
The extreme case of 'many moons' is actually a ring of moons, thus a complete ring around the planet. However a spherical symetric mass assembly around a center of mass (here the planet) does not excert any tidal force anymore. Thus a circular mass assembly will only create a small gradient in the normal direction to the equatorial plane and increase the equatorial radius a bit.


Is it possible to have a part of space devoid of matter? I have recently read this article and I am finding it a little hard to believe that there would be a place that is without matter. So, if there is a galaxy at the other side of the void, we would be unable to detect it. I am still very much a rookie at this, so please correct me if I'm wrong.

So could this article be telling the truth?
If so, what is the science/reasoning behind this?
What if say, a black hole were to appear in such a void, what would happen?

OK, so I don't think the article is claiming that it is a place "without matter", but that it is a large region of space where the density of matter is lower.

The original research, published in 2007, suggests that a lack of radio sources seen in a particular direction coincides with a "cold-spot" that is seen in the cosmic microwave background by WMAP (which has since been confirmed by Planck). They calculate that if the void were completely empty it would need to be about 280 Mpc across and at redshifts $z\leq 1$.

Galaxies certainly could be seen on the other size of this void. Many of the galaxies in their survey must have considerably higher redshifts and are more distant.

The cold spot could be caused by such a void, because light is redshifted as it travels through gravitational fields. The Sachs-Wolfe effect, as it is known, causes large scale perturbations in the cosmic microwave background as light travels from the distant past, when the CMD was generated, to us in the present day. It is argued that a large void that is comparatively empty of both visible and dark matter, but which has dark energy. The way it works is that photons works against gravity to enter a void, loses energy, but then gain back some of that energy as it leaves the void. However, the expansion of the universe means that in the time taken to cross the void the amount that is regained is not as much as was lost in the first place.

"What if a black hole were to appear"? Not clear what you mean by this. Black holes are black, you wouldn't necessarily see them, however if the "void" or lack of radio galaxies was compensated for by an equivalent amount of mass that was for some reason invisible then it would not explain the CMB cold spot.

The idea of a "supervoid" causing the CMB cold spot has also been supported by other surveys of extragalactic sources (Szapudi et al. 2015), but you are correct that the size of this void is difficult to understand in terms of current cosmological assumptions.


Why does gas form a star instead of a black hole? When a space gas gets pulled together a star is formed. On the other hand, when a massive star dies, it collapses to a black hole.

You would think that the initial mass of the gas would be bigger than the star that had existed for billion years and lost mass in the process.

So what stopped the space gas from forming a black hole in the first place?

Essentially gas does, it just happens to form a star first.

Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from reaching a critical density needed for black hole formation. As these energy producing processes run out of useable fuel the star will eventually collapse onto itself creating a black hole.

So you can't just take a large volume of gas and create a black hole. Other physical processes occur.


Could there be a brown dwarf in our solar system? Is it possible that a brown dwarf object could be orbiting our star and undetected so far by wise surveys?

The nemesis theory proposes that a low mass star or brown dwarf in highly elliptical orbit is a companion to our sun as a solution to the cyclical mass extinction problem (http://www.theatlantic.com/science/archive/2015/11/the-next-mass-extinction/413884/) and (http://www.space.com/22538-nemesis-star.html). Scientists noticed that some mass extinctions follow a seemingly cyclical pattern, so researchers proposed that the orbit of a small companion star (or a brown dwarf) was disturbing comets or asteroids as it approached aphelion and flinging them towards the earth.

As interesting as this would be, WISE's infrared scans of the sky have revealed nothing and there is currently no evidence supporting the theory. Furthermore the orbit of any companion would have to be very large and would therefore be very unstable and would likely be detectable. The Astrobiologist David Morrison stated of the theory: "(T)he Sun is not part of a binary star system. There has never been any evidence to suggest a companion. The idea has been disproved by several infrared sky surveys, most recently the WISE mission. If there were a brown dwarf companion, these sensitive infrared telescopes would have detected it."


How bright would it be on the "face" of Titan If you were planing a maned mission to Titan and wanted to find a landing site that would get the most visible light for the longest amount of time, your first logical choice would be the dark side of the moon since it faces the sun for a longer duration than the face would. ("face" and "dark side" corresponding to the colloquial definitions they have when referring to our moon) Also, unlike the on the face, you wouldn't have to worry about Saturn eclipsing the sun. This would put you in a range of brightness roughly equivalent of twilight on Earth due to the distance of the sun and the thick atmosphere of Titan.

But how bright would Saturn be when reflected on the face of Titan? Would it be bright enough to see anything if you wanted to go there, or would you need flashlights and night vision goggles?

It's not hard to calculated brightness to distance, but that doesn't take into account cloud cover, which is important for Titan.

Just looking at distance first, Titan (based on Saturn's distance) averages 9.6 AU from the sun, or about 1/91st as bright as the sun appears from Earth, assuming equal atmospheres.  1/91st isn't a bright sunny day, but it's probably bright enough to read by.    The sun has an apparent magnitude of about -26.7 from Earth.   91 times less bright works out to -21.8 from Titan.

Saturn is a little more tricky because from Earth, we often see Saturn's rings but from Titan, you mostly don't.   Titan orbits almost entirely in line with Saturn's rings (just 0.34854° inclination), which would make the rings appear like a thin band.   Visible but probably not very reflective.    You could also probably see ring shade and ring shine reflecting off the surface of Saturn from Titan.

How bright would Saturn be?

From Earth, Saturn's brightness varies from (mostly) between 0 and 1 apparent magnitude.  That's a range of 2.5 depending on distance and angle, which is kind of a lot.    Titan would probably receive less variation due to no ring reflection (see Saturn as viewed from Earth).

There would still be variation based on the angle of the sun reflecting off Saturn's surface.   Perhaps an estimate of between 0.2 and 1 is in the ballpark (as a guess) at Earth distance but Titan's viewpoint, almost entirely above the ring.

Using an an average distance between Earth and Titan of about 9.6 AU, and Titan's distance from Saturn, about 1/122 of 1 AU.  Titan is some 1170 times closer to Saturn than the Earth on average, which makes Saturn about 1.38 million times brighter from Titan than from the Earth.   which works out to between 15.3 and 15.4 apparent magnitude units.

Using my rough estimate, Saturn, from Titan would be in the range of -15.2 to -14.3 apparent magnitude, which would be some 400-800 times less bright than the Sun from Titan (it's not possible for a planet to outshine a star from the point of view of the moon), but several times brighter than the full moon from Earth.   You could probably see, well enough on Titan under a "full Saturn" - being some 5-10 times as bright as the full moon (peak -12.9/average -12.6).  It would also be much larger, so it wouldn't appear bright in the sky, but it would provide enough light to see well enough, but maybe not enough light for reading.   Enough to see though.

Permanent Clouds

The numbers above, however, doesn't take into account Titan's cloud cover and it has nearly constant clouds.   Here is an article with a picture of a rare situation of sunlight reflecting off the surface of titan.   Mostly it's under a permanent haze.   I don't know how effective that haze is in blocking sunlight, maybe 90%, maybe even 99%, but it would certainly block a fair percentage, so, while you could probably see under sunlight, I think reading is no longer possible under those clouds.

Here's an interesting article that says Titan is brightest at twilight, not daylight, because it's atmosphere is very effective at scattering light.   It's atmosphere is thicker and much higher than Earths, and also more massive overall.     It's likely for any real work, Astronauts would probably want a permanent artificial light source and not rely much on light from the tiny sun though all those clouds.    But they might be able to see well enough to move around and explore without flashlights a little bit, but flashlights would help.

Night vision goggles would do nothing.   They see heat or longer wavelength IR light outside the visible wavelengths.   They could spot heat sources, like other astronauts, or a rip in a space suit or hut - so they'd have uses seeing where the heat is escaping for example, but they wouldn't help you see the on surface of Titan.

Titan takes nearly 16 days to orbit Saturn anyway, and nobody would want to put up with 8 days of darkness.  Artificial light would be essential which, I would think, makes choosing a location based on sunlight not a key factor.  Perhaps someplace where there's less methane rain would be a priority, or less chance of solid ground becoming liquid under a warm spell like a nice "dry" water ice Island.


What does "located in the Hubble flow" mean? I do not understand that much about the term Hubble flow.
For cosmological studies with type Ia supernova, I often see the sentence like "We used 100 SNe Ia in the Hubble flow". What does the difference between "located in the Hubble flow" and "not in the Hubble flow"? Thank you.

The apparent motions of external galaxies can be broken down into two components and the redshift observed will be due to the sum of both.
The first is an apparent recession velocity that is given by Hubble's law. $v_H = H_0 d$, where $v_H$ is the apparent recession velocity, $d$ is the proper distance and $H_0$ is the current value of the Hubble parameter.
On top of that, a galaxy will have a peculiar velocity $v_P$ with respect to the cosmological rest frame (usually taken to be coincident with the frame in which the cosmic microwave background has no dipole anisotropy). This preculiar velocity is due to the gravitational influence of nearby other galaxies or due to a galaxy's motion within a group, cluster or supercluster of galaxies.
Since the first component grows with distance, whereas the second component is fixed and of order hundreds to perhaps a thousand or so km/s, then if you look far enough away from the Earth then $v_H \gg v_P$. This is what it means to be "part of the Hubble flow" - the velocity that you measure for that object is dominated by the cosmological expansion of the universe and local peculiarities and gravitational perturbations that the galaxy is influenced by play a small enough role that they become a negligible source of uncertainty.
Since $H_0 \simeq 70$ km/s per Mpc, and $v_P \sim 1000$ km/s, then a quick bit of maths tells you that in order for $v_H \gg v_P$, then $d \gg 14$ Mpc for an object to be considered part of the Hubble flow. There is no exact definition.


Is it possible to calculate how much a star's brightness dims when a planet transits in front of it from a viewer's perspective? Last night I got to thinking about what would happen if Jupiter and Venus suddenly switched places.

Since Venus comes the closest to Earth than any other planet and Jupiter is much larger, does that mean that Jupiter (now following the orbit of Venus) would appear large enough to eclipse the Sun from Earth's perspective?

Jupiter's angular diameter in this scenario is given by $$\delta = 2\arctan \left(\frac{d}{2D}\right)$$

where

$\delta$ is its angular diameter,
$d$ is its actual diameter,
$D$ is the distance between both bodies.

The closest Venus and Earth ever get to each other (in the near future) is $d = 39.5 \times 10^6$ km, and Jupiter's diameter is $D = 139,820$ km.

So if Jupiter replaced Venus, its maximum angular diameter would be $$\delta = 2\arctan \left(\frac{139,820}{2 \times 39.5 \times 10^6}\right)$$
$$\delta = 0.202812°, or \ 0° 12  ' 10.12 " $$

The Sun's angular diameter as seen from Earth is about 30 arcminutes, and so Jupiter in this scenario with only 12 arcminutes would not appear to be large enough to completely eclipse the Sun.

However, assuming it is a total transit, it would still cover $≈16.27 \%$ of the Sun's disk as seen from Earth. Surely that would result in a noticeable dim in the brightness of the Sun as seen from Earth, no?

Is there an equation to calculate how much a star's brightness dims when a planet transits in front of it from a viewer's perspective?

You answer your question basically yourself with the calculation - that's the right way to approach it, your math is correct. Stars and planets are big enough, that simple geometric optics work well to calculate the degree of occulatation when one body passes in front of another. This is the method used to estimate the size of exoplanets pretty accurately (we know stelar sizes from the mass-luminosity-radius relations).

Typical drop in luminosity from observed exoplanets is 1% or less - which happens to be exactly the square ratio of solar radius to jupiter radius. You correctly get bigger luminosity drop as you don't put both objects in infinity, thus don't directly compare radii but their apparant angular diameter.

The only not exactly straight-forward part is the influence of the atmosphere of the transiting body as it contributes to absorption depending on density. That is one of two reason that for any brightness curve of exoplanet transits you don't see an immediate drop in brightness (the other is the result of the limb darkening of the star). (See e.g. this light curve) This is used to infer exoplanet atmospheric properties. It can even be taken so far as to make atmospheric spectroscopy by comparing transition spectrum and normal spectrum of the star.

You, however, seem implicitly to ask (also) about the human factor, how we, as humans, perceive the brightness variation. That is an entirely different story as our eyes cover a MUCH larger brightness span as they work logarithmically - thus small changes in linear brightness (eclipsing part of the sun's cross-section is simply that) don't make a big impression. CCDs or CMOS work linearily, thus detect such changes much easier - but are bad when it comes to covering large brightness variation.

Anyone who observed a solar eclipse can tell the story that you don't notice much of a brightness drop until the very last moments when the sun becomes a very tiny sickle just moments before totality. The most notable difference is behaviour of the shadows (they change shape) and the sharpness of shadows. It gets (slightly) colder, as less light arrives - but the remaining light source is as bright per unit area as the already eclipsed part is so that the perceived brightness remains nearly the same.
It's similar when you switch on two identical lamps - the room does not seem twice as bright compared to only one lamp.


Are the Trappist-1 planets in stable orbits? The Trappist-1 planets all orbit very close to each other.  During NASA's press release, they mentioned that these planets are close enough to disturb each others orbits.  Is this system stable over a long time scale?  Or could we perhaps just have imaged this system before one or more planets is ejected or destroyed?

Later into the press conference, one of the scientists says "these planets should have formed further out and migrated inwards".  Could they possibly still be moving inwards, or are their orbits now stable?

We don't know.
Much of what we do know - or at least think we know - about the planets' orbital parameters comes from simulations the team ran via $n$-body methods[1]. Some methods of integration led to short-term disruptions, on the order of less than $\sim10^6$ years. That said, the system is at least $5\times10^8$ years, and it would be odd if the observations came just before the final instability, in the authors' opinions.
Fortunately, using a different (statistical) method, the team got different results. The found that the system has

a 25% chance of suffering an instability over 1 Myr, and an 8.1% chance of surviving for 1 billion years

This does not seem great for long- or short- term instabilities. However, the authors were extremely cautious, emphasizing that there is plenty of uncertainty. There are many other factors that need to be taken into account:

Tidal interactions between the planets and the star and between each other
Possible other planets in the system (although giant planets may be ruled out)
The masses and orbits are not known to great accuracy, and so interactions can not be modeled as ideally as one would like.
Resonances could be important, and the planets in the system fall into near-integer resonances, which is interesting.

Whether or not the planets are still changing their orbits is an interesting question. The authors believe that the planets moved inwards via gas disk migration, which would require interactions with the initial protoplanetary disk. However, as the disk has clearly dissipated, it would seem that the migration has stopped, and has not happened for some time.


What can a lunar-based telescope do? What can a lunar-based telescope do? If the band is ultraviolet and it can do high time sampling exposure, maybe it can be used to get distance for a star with stable pulses. However, the star should be bright and its distance should not be large.

What can a lunar-based telescope do?

First, you would have no atmospheric distortions to interfere with the viewing.
Second, there would be no atmospheric absorption of ultraviolet or infrared light.
Third, there would be no cloud cover, so viewing can be almost continuous (the sun may interfere).

  However, the star should be bright and its distance should not be large.

I don't know what you reason for saying that, but it is unimportant since the Hubble telescope looks at all the stars.

Disadvantages:
The telescope will need to be self cleaning. There is a lot of dust kicked up when as meteoroid strikes.
An extended manned mission to the moon to set up a large array.


Resource Recommendations to systematically develop a foundation via Self study in Astrophysics/Astronomy? I am a high school senior (Grade 12) with a passion for Astronomy/Astrophysics. I already searched quite a bit here on the other questions on resources, but I couldn't find any that shows a concise, systematic order that one can follow. (I might have also missed it if so I do apologize. If it helps I do have quite a decent handle on AP-level Calculus and decent physics background.
So, I would like a systematic order of books, videos, resources of any kind that I could peruse to achieve a significant foundation (and ambitiously say up to a Graduate level?) in Astronomy/Astrophysics.
Thank You!!!

High School Senior

Your first mission is to concentrate on getting a good set of results from high school.  That really means not spending your time chasing astronomy and astrophysics yet.
Yes, this is what your parents and teachers would say, but you know, if you ever want to have the time, resources and maybe the chance to study and/or work in these fields you absolutely are going to need the very best high school results you can manage.  And even if you don't weant to study a B.Sc. or higher then you still need the resources and that means a decent job and career prospects.  Money is important, or as the saying goes "No bucks, no Buck Rogers”.
So focus on that first.

a systematic order

For self study this is not needed.  I'd suggest not being systematic outside of a specific goal, like having to pass exams set by other people.
You need to differentiate between Astronomy and Astrophyics.
Astronomy is a kind of "catch-all" term for every possible thing you could study or observe about anything that goes on  outside of our atsmosphere and planet.  You can include geology, biology, chemistry, physics and just about anything else that can be termed a science, as well as engineering (you have to build stuff to do stuff - rocket science and telescopes of many types, etc.).  Astrophysics is then a subset of Astronomy focusing on the application of physics to astronomy.
Now you want graduate level or graduate level "foundation" as a target.  That means you want a B.Sc. level course.  As you want Astrophysics you basically need a B.Sc. level course in physics.  That's a lot of study.  You'll cover topics like:

electromagnetism
basic quantum theory
basic special and general relativity
basic thermodynamics
mechanics
optics

Keep in mind a systematic B.Sc. course on basic physics for astrophysics typically takes four full years.  You have to learn to think about physics problems and get a feel for how things are done.  That's arguably more important than the actual details of any particular theory.
Don't be distracted by glamor subjects like quantum theory and relativity.  Mechanics, Electromagnetism and Thermodynamics are very, very important.
So let's focus on astrophysics and understand that you have to start with physics and what resources will help.
Libraries
Great things, not loved or used enough.  One set I personally really useful was the Schaum Outline series as a launching point and if you can get anything from your library, borrow them.  You don't need to swallow books whole - read what you need and borrow again when you need more.  There are lots of books out there and for physics and we all have favorites (and ones we'd avoid ever reading again).  Boring old textbooks like e.g. Halliday and Resnick are as good a place to start as any.  My first exposure to quantum theory proper was by a book borrowed from my local library.
Videos
There are multiple courses on physics online.  A couple to consider :

Yale physics courses online
Leonard Susskind physics lectures

There's also excellent YouTube channels like :

PBS Spacetime
FermiLab
Sabine Hossenfelder

By no means the only ones.


Prograde to retrograde orbital path Is there a planet or planetary moon flyby path that could send a solar prograde orbiting asteroid into a retrograde solar orbit?

If an asteroid on an inner orbit catches up with a planet on an outer orbit, and passes close enough to the planet to be strongly influenced by the planet's gravity, it can whip around the planet ahead of the planet and then move in and orbit backwards (retrograde) farther from the sun than the planet.
So the asteroid might be captured by the planet if its retrograde speed is slow enough, and then move in a backwards or retrograde orbit around the planet.  A lot of small asteroid sized moons of the giant planets have retrograde orbits indicating they were captured that way - also Triton, the large moon of Neptune.
I just counted abut 119 natural satelelites with retrograde orbits in our solar system.
https://en.wikipedia.org/wiki/List_of_natural_satellites[1]
And the the asteroid's retrograde speed after passing around the planet is fast enough, it won't be captured by the planet but will continue to orbit the Sun, but in a retrograde orbit.
In fact, a very tiny percentage of all known asteroids have retrograde orbits, only 82 out of more than 726,000.
https://phys.org/news/2017-06-stable-retrograde-orbit-bee-zed-asteroid.html[2]


Can ground based telescopes use a starshade in space? 1) Could a starshade in orbit be used by ground based telescopes? If so, what kind of orbit should it have?

2) Could the same starshade be used by two different space telescopes, or does the positioning make it impractical?

If the starshade was to stay in place over the ground telescope, it would have to be in a geosynchronous orbit, which puts it almost 36,000 km from Earth's surface. To cover an angular distance of 1 arcsecond (1/3600 of a degree), it would have a size of around 10km. Certainly not impossible, but technologically impractical at the moment. Also, the starshade would only shade a small region of the sky, so it wouldn't be useful from other observing sites.

As for the space telescopes using a starshade in geosynchronous orbit; that would be impossible, at least for scopes orbiting Earth, like Hubble is. The starshade is orbiting Earth at a different distance, and therefore speed, than the telescopes. As far as I can see, it would be near impossible to keep them aligned for anything more than a few minutes. I haven't done any number-crunching on this, though.


Does the inverse of gravity exists? I am not talking about an antigravity effect that would cancel gravity, rather if there is a repulsive force, like when you use two magnets and flip one of them, where it becomes obvious that one is pushing away the other.

Hope the question is a good fit for the site, feel free to migrate it otherwise.

No.

No such effect has been discovered. Gravity seems to be always attractive, never repulsive. Nor has a particle with negative mass ever been discovered. In particular, antimatter appears to have positive mass.

It is not theoretically impossible, but a particle with negative mass would be very odd, as the faster it moved, the less momentum and energy it would have.


How can I use my current camera whose zoom lens can't be removed behind an f/6 refractor for as an astrophotography? I am looking at an Astrotech AT60ED (60 mm f/6 with a field flattener) and a Digi-Kit telescope adapter to do astrophotography, mainly deep-sky and I'm wondering if I can use my current Panasonic DMC-FZ50 Type CCD camera body.

The problem is that it has a zoom lens that can't be detached which is why I am worried.

The AT60ED is a nice small refractor. However, its objective lens simply projects an image at the focal plane. Adding a field flattener mimimizes curvature that degrades sharpness at the edges of the image. The flattener requires a specific back-focus, or distance (about 57 mm for the AT60 FF) between its (T2, 42 mm thread) flange and the image plane where the sensor should be located.

If your camera has a non-removable lens, its sensor cannot be placed directly at the focal plane. You cannot just attach your lens to the telescope or field flattener to take an image.

I believe that the only alternative for a camera with lens is to use eyepiece projection, also called digi-scoping. An eyepiece magnifies the image at the refractor's focal plane and projects it, usually so the eye can then focus the image on the retina. Instead, if a camera is attached to the eyepiece using a suitable adapter, its lens can then focus the image.

Eyepiece projection also has disadvantages, including a small field of view, vignetting, and often balance issues due to the camera weight and adapter attached to a small eyepiece, rather than directly to the focuser of the telescope. However, lots of people use eyepiece projection just by holding their cell phone camera at the exit pupil of the eyepiece.

Edit:  Added some comments regarding deep-sky imaging.

Also, deep-sky imaging presents additional challenges. The Moon and planets can be photographed with short exposures. However, images of even the brightest nebulas and galaxies require long exposures. Astrophotographers usually take a large number of individual images, called sub-exposures, each typically lasting tens of seconds to many minutes. These are then aligned and stacked to improve signal to noise by averaging. Total integration time can often be several hours or more, depending on the brightness of the object.

A telescope mount that can accurately track a celestial object, compensating for the Earth's rotation, is necessary to prevent motion blur over these long exposures. The cost of a mount suitable for even a relatively short focal length refractor such as the AT60ED will be several times the cost of the telescope, and much more than a camera capable of prime focus astrophotography.


What exactly is a "moon"? Recently some new moons of Jupiter were discovered. However the moons are pretty damn small indeed - only a few kilometers wide. That makes me wonder if they should even be called moons in the first place. The Wikipedia article on natural satellites says there's no clear lower limit (or upper limit for that matter). But if we accept that, then Jupiter already has countless numbers of moons in its ring system, and even the Earth has multiple moons since the International Space Station is a "moon".

What exactly is a moon? Is it really a vague notion of "something that goes around a planet"? If so, why hasn't the IAU clarified what exactly a moon is? They've already done so for planets; it seems natural to do so for moons as well.

Unlike "planet" the IAU hasn't attempted to precisely define "moon". General usage requires that a "moon" is a natural satellite of a planet (or dwarf planet, asteroid, or perhaps even of another moon?) and it is big enough for us to have seen it as an independent body.

This contrasts with the "moonlets" that have been detected in Saturn's rings by their graviational effects on the even smaller ring particles. We know that a body exists there, but it is too small to be directly imaged.

As such the question "How many moons does Jupiter have" is probably unanswerable, since it depends on the sensitivity of your telescope and how close you are to the planet.

The smallest moons are about 10km across, but moonlets in the rings exist that are only a few hundred metres across. So in current usage, 10k might be roughly where people stop saying "moon" and start saying "moonlet".

Language exists to serve us, the definition of planet was only made to solve a particular naming problem.  No such problem has arisen for moons. So there is no authoritative answer.


How to convert Proper Motion in RA & Dec to Galactic Coordinates I'm trying to parse stellar data in the GAIA 2 release to galactic coordinates and am struggling with the velocity component. I've tried following online documentation and papers, but end up with velocity results with strange biases that indicate I'm probably doing something wrong. Sources I've used are:
http://adsabs.harvard.edu/full/1987AJ.....93..864J
https://gea.esac.esa.int/archive/documentation//GDR1/Data_processing/chap_cu3ast/sec_cu3ast_intro.html#SS
The GAIA 2 dataset has these relevant variables per star:

ra   (from 0-360)
dec  (from -90 to 90)
pmra
pmdec
parallax
radial_velocity

And this is the code I have (C#) to try to obtain the velocity components. Done by first converting pmra and pmdec to cartesian velocities based on the stars position, and then rotating that location to galactic coordinates.
double RA = ra * Math.PI / 180.0;
double DEC = (dec + 90) * Math.PI / 90.0;

Microsoft.Xna.Framework.Matrix positionrotation = new Microsoft.Xna.Framework.Matrix();
positionrotation.M11 = (float)(Math.Cos(RA) * Math.Cos(DEC));
positionrotation.M12 = (float)-Math.Sin(RA);
positionrotation.M13 = (float)-(Math.Cos(RA) * Math.Sin(DEC));

positionrotation.M21 = (float)(Math.Sin(RA) * Math.Cos(DEC));
positionrotation.M22 = (float)Math.Cos(RA);
positionrotation.M23 = (float)-(Math.Sin(RA) * Math.Sin(DEC));

positionrotation.M31 = (float)(Math.Sin(DEC));
positionrotation.M32 = 0;
positionrotation.M33 = (float)(Math.Cos(DEC));

Microsoft.Xna.Framework.Vector3 vector = new Microsoft.Xna.Framework.Vector3((float)(radial_velocity), (float)(pmra / parallax), (float)(pmdec/ parallax));
vector = Microsoft.Xna.Framework.Vector3.Transform(vector, positionrotation);

Microsoft.Xna.Framework.Matrix galacticrotation = new Microsoft.Xna.Framework.Matrix();
galacticrotation.M11 = -0.054876f;
galacticrotation.M12 = -0.87347f;
galacticrotation.M13 = -0.483835f;
galacticrotation.M21 = 0.494109f;
galacticrotation.M22 = -0.444830f;
galacticrotation.M23 = 0.746982f;
galacticrotation.M31 = -0.867666f;
galacticrotation.M32 = -0.198076f;
galacticrotation.M33 = 0.455984f;
vector = Microsoft.Xna.Framework.Vector3.Transform(vector, galacticrotation);

The DEC conversion in the question is incorrect.
Try this:
double DEC = dec * Math.PI / 180.0;

positionrotation looks like a composition of two rotations bringing (1, 0, 0) to the RA and DEC of interest.
It seems correct, but something like this would be easier to verify:
using Microsoft.Xna.Framework.Matrix;

Matrix positionrotation =
    Matrix.CreateRotationZ((float) RA) * Matrix.CreateRotationY((float) -DEC);

The units in vector are mixed up.
According to these Gaia DR2 docs, radial_velocity is in km/s.
pmra and pmdec are in mas/yr; dividing (*) them by parallax in mas (1 au baseline) gives results in au/yr. Try this:
using Microsoft.Xna.Framework.Vector3;

// conversion to km/s
double au_yr = 1.49598e+8 / 3.15576e+7;

Vector3 velocity = new Vector3(
    (float) radial_velocity, // km/s
    (float) pmra / parallax * au_yr,
    (float) pmdec / parallax * au_yr);

Vector3 vel_icrs = Vector3.Transform(velocity, positionrotation);

If galacticrotation is from equation 3.61 in these Gaia DR2 docs,
// Matrix galacticrotation as above except
galacticrotation.M12 = -0.873437f;

Vector3 vel_gal = Vector3.transform(vel_ircs, galacticrotation);

(*) Actual distance in parsecs may differ from the reciprocal of measured parallax in arcseconds, which may be zero or negative.


Where would be the absolute most logical place in USA to observe the total solar eclipse on August 21, 2017? Cloudy weather would defeat the purpose so I am guessing that the western states would be more ideal due to greater chance of clear sky.  Isolation would also be nice.  At a camp site on or near a mountain top.  Also, are there efforts to set up mobile solar observatories in the totality path?

I assume that when considering where to view the Solar Eclipse from, you're looking for the place with the best observance of totality. This elicits two factors, the degree of totality and the time elapsed during totality. The best observed eclipse will be a total eclipse, i.e., 100% totality, and for the longest time possible.

A map of the path of 100% totality is shown below. In it, you can see that grey band that passes across the US. Anywhere within that band, you'll be able to see a 100% total solar eclipse.

If you view this interactive map from NASA, you can see two specific points along the totality band that are called out, one labeled GE and one labeled GD.

The GE marker signifies the point of greatest eclipse which is defined as:

  The instant when the axis of the Moon's shadow cone passes closest to Earth's center.

The GD marker signifies the point of greatest duration, quite sensibly defined as:

  The instant when the length of the total (or annular) phase reaches a maximum along the central path of a solar eclipse.

So, if you're looking for the point of 100% totality that lasts for the longest amount of time, you'll want to be at the point of the greatest duration. Geographically, this is pretty close to Carbondale, IL.


Exactly how long does it take for the exposed core of a star to cool from its starting temperature (several billion K) to ~50,000 K? OK, I didn't know how I should word this question. But the basic point is that most white dwarfs that we have classified fall in temperature ranges from ~50,000 K to 6000 K. However, at the end of a low-mass star's life, the core temperatures reach billions or maybe even trillions of Kelvin. So when the star dies and sheds its outer layers, exposing the core, exactly how long does it take for the core to cool down to the "standard white dwarf temperature?"

The answer is of order 1 million years to cool from a standard end of He burning temperature of just over $10^8$ K to the top end of the white dwarf temperature range you give in your question. The details would depend exactly on the mass and composition of the white dwarf and there are also some theoretical uncertainties in neutrino cooling rates.
The surface of a white dwarf is not its interior temperature. If you "expose" the core to empty space then there will always be a transition (i.e. a temperature gradient) between an interior temperature and a "photospheric temperature", which is the temperature at which the plasma density falls low enough for photons to escape.
For example, white dwarfs with photospheric temperatures of 50,000 K to  6,000 K have interior temperatures of about $5\times 10^7$ K to $3\times 10^6$ K. This interior temperature applies to the bulk ($>99$%) of the white dwarf because electron degeneracy leads to high thermal conductivity and the surface layer where the temperature falls and the electron degeneracy disappears is very thin because of the high gravity.
So the cooling timescale for cores that become white dwarfs may not be as short as you think. The thin, outer, non-degenerate layer acts like an insulating jacket around a hot water tank. The cooling timescale (the time to radiate all internal heat at its current luminosity) changes from tens of millions of years at interior temperatures of $3\times 10^7$ K to billions of years at interior temperatures of $3\times 10^6$ K, thanks to the strong dependence of luminosity on temperature.
However, this story only works once the core has cooled to around $3\times 10^7$ K from its initial temperatures. Above that temperature, the main cooling mechanism is via neutrino emission and the insulating jacket does not stop them. Neutrino emission is very temperature dependent $(\sim \propto T^{10})$. So the initial cooling phase to $\sim 3\times 10^7$ K is very rapid once nuclear reactions have ceased; about a million years. After that the neutrino losses become nmuch smaller quite quickly and certainly photon emission is dominant by the time the interior temperatures are $10^7$ K and the photospheric temperature is <30,000 K.
A similar process occurs in neutron stars. Very rapid cooling by neutrinos for $\sim 10^5$ years, followed by slower photospheric cooling later. The difference is that most of the white dwarfs we see are in the photospheric cooling phase, whereas the neutron stars we can see are cooling with neutrinos. Neutron stars have such small radii that they become effectively invisible once they reach the photospheric cooling phase.


What orbit is the smallest or takes the least time to complete? What orbit in our solar system would take the least time to perform 1 orbit?

What is the fastest possible orbit to complete in the universe?

In general, the closer you are to the primary, the shorter the orbital period, except of course that you can't orbit inside the primary. So, taking the primary to be a sphere of radius $R$ and density $\rho$ we find that the mass is $$\frac{4}{3}\pi R^3\rho$$ so that the acceleration due to gravity at the surface is $$G\frac{4}{3}\pi R\rho$$ For a surface-grazing orbit, this has to equal $R\omega^2$ where $\omega$ is the angular velocity, so we get $$\omega^2 = \frac{4}{3}\pi G\rho$$ Now the orbital period $P$ is $2\pi/\omega$ so we get  $$P = \sqrt{\frac{3\pi}{G\rho}}$$
So the answer, at least for spherical primaries) is going to be a close orbit around the densest object in the solar system. The densest planet is Earth!
There are denser objects, such as the LAGEOS satellites, but I'm not sure if it is possible to orbit LAGEOS under its own gravity, because of the perturbing effects of Earth's gravity.


Are there celestial bodies that are about as far away as the Sun? Just curious. The Sun is 500 light seconds away from the Earth depending on time of the year. The orbit of Mars is about 50% as far as the Sun, the orbit of Vesta about 150%. Is there / Are there celestial bodies inbetween with about the distance of the Sun, their orbit around 93 million miles far from the Earth?

Asteroids with an orbit of between 1.8 and 2 Au (So their closest approach is is about the same distance to the Earth as the Earth is to the sun) are called Hungaria group asteroids, named after 434 Hungaria.

Hungaria itself has an orbit of 1.94 AU, so would be a good example of body. Such asteroids are rather rare. There is a gap in the asteroid belt caused by Jupiter at 2.06 AU, and tighter orbits tend to be disrupted by Mars.

Of course, at any given time, the actual distance to Hungaria will vary between about 1 and 3 AU


Current best-known Drake Equation values? Frank Drake's equation is used to estimate $N$ the number of civilizations we might be able to communicate with in our galaxy:
$$N = R_\star f_p n_e f_l f_i f_c L$$
where:
$R_\star$ is the average rate of star formation in the galaxy
$f_p$ is the fraction of stars with planets
$n_e$ is the average number of goldilocks planets per planetary system
$f_l$ is the fraction of goldilocks planets that actually develop life
$f_i$ is the fraction of life bearing planets that develop intelligent life
$f_c$ is the fraction of civilizations that broadcast their existence into space
and $L$ is the length of time for which a civilization broadcasts their existence
Question
Have the many exoplanets we've discovered in the last several decades improved our estimates of $f_p$, and $n_e$?  Are these values different than what was guessed about prior to the discovery of the first exoplanets?  What is our current best estimate for $R_\star$ for the Milky Way?

Have the many exoplanets we've discovered in the last several decades improved our estimates of , and ?

Absolutely! The quantity $f_p n_e$ is a quantity of much interest in investigations of extrasolar planets, which effectively equates to the fraction of stars that have a habitable planet in their star's habitable zone. Plenty of studies have examined this specifically for Earth-sized extrasolar planets and place the value of $f_p n_e$ between ~$\frac{1}{6}$ (i.e. one in six stars have an earth-sized planet in a relatively close orbit to the host star) and ~$\frac{2}{5}$ [1][2]. A 2013 comprehensive study by Petigura is a great paper to get familiar with the specifics and assumptions of how the Kepler surveys are conducted and the quantities of interest calculated [3].

What is our current best estimate for ⋆ for the Milky Way?

Unfortunately the answer is: it varies. Roughly, we think that the Milky Way puts out about 3 solar masses worth of new star every year [4]. However, these don't have to be in one star. Since we think most stars are probably ~1 M$_\odot$, this works out to ~3 stars/year. However, some estimates go less, and some estimates (based on chemical decay of elements like aluminum) go as high as 7 stars/year [5].


Why does Wikipedia state that the Solar System has one unconfirmed planet? I was looking at exoplanets on Wikipedia, and found this surprising information:

The table states that the Sun has 8 planets (obviously) and one unconfirmed planet. Why is this here? I think it's Planet Nine, but I also have underlying thoughts that this planet is Pluto, a disputed dwarf/real planet. Is this correct?

That was a rather recent (16 January 2021) change to that wiki page made by a user named Mariofeds, who does not have a wikipedia user page and who has just one contribution to wikipedia.
It may well have been vandalism.
The associated talk page explicitly excludes adding the hypothetical Planet Nine as an unconfirmed planet. In line with this discussion, the number of planets listed for the solar system had remained as eight confirmed, none unconfirmed for a long time.


What am I looking at? I am in Northampton England and every night in the SW sky is a very bright, white light that twinkles. It visibly appears to be moving towards the west but appears pretty much the same start place every night. Could anyone possibly tell me what it might be please?

Is it moving straight to the west (i.e. it keeps the same distance from the horizon) or does it 'move with the stars'?

Have you checked an online planetarium like The Sky Live? Based on the current information, it could just be Sirius, the brightest star in the night sky, or even Venus which is currently in her Evenstar period (but planets generally don't twinkle).


.2015: When last did both New/Full moon in a fort-night cause an Eclipse? Going through http://earthsky.org/…/dates-of-next-lunar-and-solar-eclipses I made the (what was for me - momentous) discovery that

March 20, 2015 - New moon - is a Solar Eclipse, and that the succeeding
April 04, 2015 - Full moon - heralds a Lunar Eclipse.

Is it common that both New/Full moon in a fort-night cause an eclipse?
When was the last time an Eclipse occurred back-to-back on a consecutive New Moon/Full Moon?

As written here:

  Another oddity of nature is that solar eclipses and lunar eclipses tend to come in pairs – a solar eclipse always takes place about two weeks before or after a lunar eclipse.

And here:

  Rules of Eclipses (Solar and Lunar) ... Eclipses tend to go in pairs or threes : solar-lunar-solar. Lunar eclipse always preceeded by or followed by a solar eclipse (two weeks between them)

So, seems like a totally common thing.

P.S. The previous pair was first lunar on Oct 8, 2014, and then solar on Oct 23rd.


What's the map on Sheldon & Leonard's refrigerator of? In the Big Bang Theory, there's a map on Sheldon & Leonard's refrigerator throughout the entire show It has false-color shadings of various kinds, with prominent craters, and so is presumably of some non-terrestrial body, perhaps Mars, the moon, or an asteroid. Can anyone identify it?

It's the Beethoven basin region on Mercury.
The MESSENGER image gallery says:

A digital terrain model (DTM) derived from Mercury Dual Imaging System (MDIS) stereo images.
The lateral spacing is 330 m and the map is in stereographic (conformal) projection.
The image is taken from abstract number 1913 submitted to the 2012 Lunar and Planetary Science Conference, by Frank Preusker and colleagues.

NASA / JHU APL / Carnegie Institution of Washington
A collage of MESSENGER photos from 2011 debuted on the Apt. 4A freezer door around the same time.
A statement released 2012-04-05 quotes PI Sean Solomon:

The MESSENGER team is thrilled by the decision of the producers and writers of 'The Big Bang Theory' to weave some of the spacecraft's latest images of Mercury into this week's episode...
We look forward to seeing how the images figure in the story line and camera shots, and we hope that interest in the MESSENGER mission is broadened by this exposure on one of the most popular series now on television.


Velocity of most distant galaxies Taking Hubble's law into account, what is the velocity of the most distant galaxies relative to Earth? Yes, I know that it is actually space that is expanding, (so, it is not their real velocity), but at what speed relative to Earth are they receding from us?

The speed relative to Earth is greater than the speed of light. This is consistent with Relativity as the space between us and the distant galaxies has stretched.

The speed is given by

$$(\mathrm{Hubble\ constant}) \times (\mathrm{radius\ of\ the\ observable\ universe\ in\ Mpc})$$

Which, as noted by Pela in a comment, is roughly 3.3 times the speed of light.


What methods can high school students use to measure the radius of the earth? I suppose they could split into 2 groups and use the shadow method from different latitudes.

Maybe they could also use the stars on the horizon at two different latitudes.

Is measuring a ship below the horizon feasible ?

What other methods could be tried by high school students?

It is hard.  You should consider what you are willing to allow them to research and what you will ask them to calculate.

First consider the classic "Shadow of a stick" method: On a certain day you measure the length of a shadow cast by a stick at two locations. The locations should be at the same longitude. By some trigonometry you can work out the number of degrees of latitude between the locations. By combining this with the distance between the locations you can calculate the radius of the Earth.

The biggest problem is how will your students know the distance between your chosen locations?  Traditionally this would be done by surveying. The process of surveying is a lot harder than measuring the length of a shadow. If your students are allowed to google for the distance, they might (reasonably) argue that they could have just googled for the Earth's radius.

Moreover if students are actually going to use this method, they need to travel thousands of km North or South, this may not be practical, so your students can't actually measure the length of the two shadows.

A second method uses time. The time of sunrise (or sunset, or noon) is measured at two locations on the same latitude. This lets you find the angular distance between the locations and that can be used with the actual distance to find the radius of the Earth.  Again this requires you to know the distance between the two points. And there are the same difficulties in travelling.

Looking at the shadow of the Earth as it crosses the moon can be used to calculate the radius. The is described in a document "Measuring the size of the shadow of the Earth" The difficulty with this method is that it depends on knowledge of the size of the moon (and the distance of the sun, though this can be approximated as infinity) This knowledge is difficult to get directly. However, this can demonstrate that the radius is finite and fixed, ie "the Earth is round".

In other words, to calculate the radius of the Earth, one needs additional information that can't be easily measured: Either the distance of two distant locations, or knowledge of the size of other astronomical bodies.


What is the current accepted theory for Neptune's immense wind speed? According to "Neptune: The Other Blue Planet in Our Solar System" (Choi, 2010), Neptune, despite its great distance from the sun, is the windiest planet in the solar system, with winds reaching 2000km/h (1500 mph) as detected by the Voyager 2 probe in 1989.

Given that Neptune is 30 AU from the Sun, what are the currently accepted theories for where the planet gets energy for these immense wind speeds?

Disclaimer: Most of what is written here is based highly on theory and speculation. Voyager 2 is the one and only trip we have taken to Neptune and most of these theories are built upon the data it has sent back. Leave large room for error

Is Neptune windy?

Before we start answering why Neptune is windy we should first specify if Neptune is even windy in the first place. What proof do we have?

The Great Dark Spot

One observation was of a large abhorrence on the surface of Neptune aptly titled "The Great Dark Spot". Repeated viewings placed the movement of the GDS to be about 700mph.

Sourced from: http://nineplanets.org/neptune.html

Keck Telescope

A lot of observations and images of Neptune have come from the Keck II Telescope, which is able to observe other markings moving at similar to faster speeds on Neptune.

Sourced From: http://www.keckobservatory.org/recent/entry/keck_images_of_neptune_best_ever_captured1

What causes the winds on Neptune?

It's theorised that there are two potential causes to Neptune's winds. Shallow processes in the outer atmosphere or much deeper atmospheric changes that extend into the interior.

http://www.space.com/21157-uranus-neptune-winds-revealed.html (paraphrased for simplicity) Claims that

  The researchers found the windy layers of Neptune occupy the outermost 0.2 percent of It's mass, suggesting that shallow processes drive those winds, such as swirling caused by moisture condensing and evaporating in the atmosphere.

What is the main source of energy for these winds?

So now you are probably wondering where Neptune get's this energy from as it is so much further from the Sun than we are.

http://nineplanets.org/neptune.html Claims that this is due to Neptune's own internal heat source radiating twice as much energy as it receives from the Sun.

Why does Neptune have stronger winds than Earth then?

So theoretically, if it was down to energy, it would make sense that Earth, being so much closer to the Sun, would have much more energy required to reach these speeds.

Yes and No.

The Sun is the main provider for our energy but it also creates turbulence in the atmosphere. It's cooling and heating of the layers keeps air constantly rising and falling, disturbing natural air flows and stemming the size of our storms.

On top of this http://www.space.com/18922-neptune-atmosphere.html claims that 80% of Neptune's atmosphere is molecular Hydrogen.

Hydrogen being the smallest and lightest element requires less energy to reach high speeds than our larger, nitrogen based atmosphere. This may not be a main factor to consider but it is certainly contributing in no small amount.


In what manner is the spectrum of a black body really a continuum? Usually they say the spectrum of a black body is a continuum and emits radiation at every frequency. Is it really a continuum like a line or are there (small) intervals?

For example, can a body emit at 700nm and also at 400nm and everything in between, or at very small intervals like at 400.000001nm and 400.000002nm etc.? So is it really like a wave or line which implies a kind of infinity of wavelengths or is it more or less like discreet lengths although very close to each other? Or is it not really known because measuring is in a certain way always with discreet numbers (like 400.0001)?

As I understand it, it should always be at certain discreet numbers, firstly because the electrons from which the radiation comes only have a discreet values, and secondly because collisions between atoms/protons/electrons create energy at certain set wavelengths. But that is just my interpretation.

So what is the reality of black body radiation?

A blackbody spectrum is continuous.

As Dean comments, there are no true blackbodies. But this is not really the issue here:

It's true that most emitted photons from, say, a star, which is quite close to a blackbody, is the result of some discrete process. E.g. an electron is brought to an excited state by a collision, the atom de-excites, and a photon with the wavelength corresponding to the given transition is emitted.

But two mechanisms make the total spectrum of the star truly continuous:

Firstly, ionized atoms recombine with free electrons. The electron will go to any state, ground or excited, and with various probabilities. The energy of the photon emitted is equal to the energy it would take to remove the electron from the atom (i.e. ionize it again), plus the kinetic energy it had before recombination. And this kinetic energy can have any value (although if it's too large, the probability of recombination decrease).
Secondly, even without recombinations, you will not observe only the discrete lines. The reason is that the atoms of the blackbody are not stationary, but move with a ("thermal") velocity $v=\sqrt{k_\mathrm{B}T/m}$, where $k_\mathrm{B}$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the atom. This motion causes a Doppler broadening of the emitted lines, resulting in a continuous, and Gaussian$^\dagger$, probability distribution.

$^\dagger$Actually the probability distribution is a convolution of the Gaussian random motion, and the Lorentzian natural line profile, resulting in a so-called Voigt profile. This profile is dominated by the Gaussian in the line center, and only becomes Lorentzian in the wings.


Measuring time and distance in space When reading through articles about discovering gravitational waves, I got basic understanding of events that happened. But one thing wasn't clear for me. They say that two black holes collided 1.3 billion years ago.

My question is, how can we know, that this collision happened 1.3 billion years ago?

A model is fitted to the signal that depends on various input parameters. The input of the model that fits the signal best is then considered the "most favored" model, while the input of models that fit better than a given threshold are used to infer confidence intervals on the various parameters.

The model has to simultaneously reproduce as well as possible the frequency and the amplitude of the signal. Among the input parameters are the black hole masses ($36_{-4}^{+5}\,M_\odot$ and $29_{-4}^{+4}\,M_\odot$ ) and (luminosity) distance ($410_{-180}^{+160}\,\mathrm{Mpc}$), corresponding to a light travel time of roughly 1.3 Gyr, depending on the assumed cosmological density parameters (they used those of Planck 2015).


How can I tell if my solar glasses are up to ISO standards? Like many, we bought solar glasses to view the eclipse, and are now in a strange place where we can't tell if they're safe or not.  The big fuss seems to be about ISO 12312-2 compliance.  Apparently there are some glasses which are claiming to be certified and putting the logo in the glasses when they are, in fact, not up to spec.

Now the only "solution" that has been given is to buy from one of the "reputable" vendors documented by the AAS.  Many smaller vendors are crying foul, saying the big guys are just trying to push out competition.

So my question is this: if I had a pair of glasses, what would be the proper way of determining if they met ISO 12312-2 standards?  Is there supposed to be a way for a customer to trace their purchase back and verify that the company that made them at least has their paperwork in order? (A company that has their paperwork and is still counterfeiting anyway is a different challenge).

Obviously the astronomy community is interested in ensuring their products are safe.  How were they supposed to do that in the ideal circumstances?  If I can't confirm their certification as a small-time buyer, how would a large-scale buyer of, say, 100,000 glasses or viewers determine that the product they are buying is genuine before someone burns their eyeballs?  Surely when there are hundreds of thousands of dollars on the line, there would be a process to demonstrate compliance.

EDIT: I am specifically looking for how one is supposed to verify these glasses without relying on a whitelist provided by a single group such as the AAS (other than perhaps the ISO community itself).  Surely there is a way to verify claims of ISO compliance without relying on the good intentions of a professional society.  Otherwise, what is the point of having certifications in the first place?

if I had a pair of glasses, what would be the proper way of determining if they met ISO 12312-2 standards?

The process involves a lot of lab equipment, calibrated sources of infrared, visible light, and ultraviolet, and calibrated detectors. Extremely unlikely you'll be able to whip something up in your garage.

Buy from vendors with good reputation; these are online stores that sell nothing but astronomy equipment, and they know what they're dealing with. If you're an astronomer, you know who those are. If you aren't, go on various astronomy forums and see which vendors people are recommending. Or see the AAS page for a list of vendors.

Amazon is now cracking down on bad eclipse glasses, but realistically speaking they are not an online store specializing in astronomy, so their product vetting may or may not be perfect.

It's getting a little late now, so better hurry up and get your eclipse glasses already.

You can always dash to the local welding supplies store and get a #14 welding glass. It's completely safe and passes the ISO standard. Maybe get the whole mask to go with it, so your face won't get sunburned.


Why are galaxies distributed as walls and voids, or spider webs? I have wondered why the "walls and voids" of the cosmological galaxy distribution seem to resemble a "foam", like rising bread that's ready to fall, with "voids" corresponding to empty bubbles in the foam. James K and Rob Jeffers https://astronomy.stackexchange.com/a/18610/14485 seem to be saying local matter tends to relatively "overrule" the cosmological expansion, and that the bubbles ("voids") get relatively more cosmic expansion. This may explain the "foam" appearance: local large areas of slightly lower relative galaxy density will tend to cosmic expansion more, and will grow (less dense) more than denser regions, like bubbles in the foam, effectively driving all the galaxies into the walls between the bubbles.

Sorry I have not enough reputation to comment. I am not going to give a full answer as I am on my phone.

I guess there is a (at least seemingly) neat mathematical explanation for the geometry. But it is simply a result of gravitational collapse. The degree of 'formy-ness' depends on cosmology (for example, need more / less time to form structures depending on cosmology). If you have random distribution of matter in infinite 3d with slight density fluctuation, some patch of the space become denser faster than other patches and become even denser and so on. Every matter pulling each other while clustering themselves results in 3d to 2d, 2d to 1d, 1d to 0d structure of overdensity. So most dense regions are point-like structures (super clusters), and next most dense regions are filamentary, then sheets. The rest are voids. Cosmic expansion changes the strength of gravitational pulling but does not really alter the 'shape' or direction of force, hence little effect on the 'formy-ness' of voids. As long as it is a cosmology with collisionless darkmatter, the same formy structure will form given enough time.
 Cosmological N body simulations use fixed(comoving) volume where only the magnitude of gravitation changes due to the cosmic expansion. And in those simulations you can see (and that's where astronomers learned about) the cosmic structures. There are many movies of cosmological simulations you can easily find.


Spaghettification of hard-things/Elements Is it possible for a black-hole to easily stretch or turn any hard object like (Diamond , Gold bar , Platinum bar , Brass) into spaghettification state in just a seconds or minute once it gets near nor reached the black-hole's event horizon??

It depends on the size of the black hole and the size of the bit of material you are trying to pull apart, as well as what it is made of. What is important is the difference in gravitational pull between one end of your sample and the other. This difference force grows as the inverse of distance cubed as one moves in towards the black hole.

However, nothing special marks crossing event horizon and if we were talking about a supermassive black hole (a million solar masses or more), you might not even notice it had happened, since the force difference, sometimes called the tidal force, would be quite small. At the event horizon of a smaller black hole, the tidal force would be much larger. To see this, we note that the difference in force between two ends of a bar of length $\Delta r$ is approximately
$$ F = \frac{8GM}{r^3}\Delta r,$$
where $M$ is the black hole mass. At the event horizon $r = 2GM/c^2$, so
$$ F = \frac{c^6}{G^2M^2}\Delta r.$$

The spaghettification force continues to increase as the object heads towards $r=0$, which will occur in a finite proper time for the falling object. So it will always increase sufficiently to rip something apart, no matter what it is made of; the question is merely at what radius does this happen.

It is of interest however to calculate whether it would happen to an object outside the event horizon, simply because then an external observer (at a safe distance) might witness it. This in turn depends on the material making up the object, it's size and the mass of the black hole.

Consider a material of tensile strength $T$, size $\Delta r$ and cross-sectional area $\sim (\Delta r)^2$. For this to break up before reaching the event horizon then the tidal force at the event horizon must exceed $T (\Delta r)^2$ and thus
$$ \frac{c^6}{G^2M^2}\Delta r > T (\Delta r)^2 $$
$$ M < 2\times10^5 \left({T \Delta r}\right)^{-1/2}\ M_{\odot}$$

Diamond has $T\sim 10^{11}$ N/m$^2$. If it was of size 1 cm, then a black hole would need to be larger than 6 solar masses for this chunk of diamond to survive being ripped apart outside the event horizon. Other materials or larger bits of material are easier to break and would not survive outside the horizon of even more massive black holes. e.g. The tensile strength of bone is about $10^{8}$ N/m$^2$, so a 0.1m piece of bone would be pulled apart prior to falling through the event horizon if $M< 60M_{\odot}$.