What would it take to capture a passing asteroid or other celestial body into Earth's orbit? Every once in a while, we have a celestial body that passes through our solar system. What would be needed to reel one of these in? Not crashing down to Earth, that would be bad, but say into a high earth orbit (above the satellite belt)?

There have been proposals to capture a small (boulder-sized) asteroid into Earth orbit (more or less) as several other answers have discussed. This needs a lot of planning, a big rocket and a great deal of patience, but all of the asteroids being considered already orbit the Sun, and generally are among the relatively small number of asteroids that periodically come close to Earth.

Capturing an extra-solar object like Oumuamua would be much harder. Firstly we'd need to detect it a long time  (decades, ideally) before it gets anywhere near the Sun, which means spotting something that is far away, very dimly illuminated, and coming from a completely unpredictable direction.  We'd want to get to it as quickly as possible, break a lump off if it was too big to redirect as a whole, and then use something, probably nuclear explosives to change the course of the lump enough for it to come close to one of the giant planets, ideally Jupiter. That encounter would have to be arranged so that the planet's gravity deflects our (now slightly radioactive) lump of rock into a closed orbit around the Sun.

Once that is done, the time pressure is off, and we can use less drastic measures (rockets, light-sails,...), over possibly many decades, to steer our target lump, via whatever gravitational slingshots seem useful until it encounters the Earth and Moon at a fairly low relative velocity. Then a combination of more rockets and the Moon's gravity and maybe aero-braking in the upper atmosphere can be used (over several passes if necessary) to actually capture it.


Why pulsar pulses are so extremely regular? It's often said that the pulses of the pulsar have a very precise period, which helps to detect easily exoplanets that cause fluctuations in the time of arrival (TOA) of the pulses. Anybody knows why?

Pulsars are precise because... in essence, there isn't much to cause them to be imprecise. It's as if you spun a top in intergalactic space and left it there for a hundred years - it would likely still be spinning (at pretty much the same rate).

In essence, pulsars are super-dense neutron stars that are rotating really fast, and their emission axis happens to line up correctly with Earth so that we see it as a pulsating source of radiation. Because there isn't really much to slow their rotation down, it stays relatively constant (though there can be what we call pulsar glitches). However, pulsars do gradually "spin-down", and their period decreases over time (very slowly, mind you), and we can calculate this - we call it the p-dot, or period-derivative - the rate of change of the period, if you will. (As James K pointed out in his comment, this emission of radiation is also part of what causes this spin-down.)


Is starquake unique to neutron star? I read from somewhere that astronomer use X-ray to observe the "shaking" not to be confused with wobbling of a distant neutron star, I think it is similar to what geologist using microphone to detecting sound wave (vibration) inside the earth (no pun) to find out its composition. How does a tightly packed super dense object like neutron star generate starquake? is this neutron star signature move... i mean trait?

I'm not sure what the current status of neutron starquake theory is, but last time I looked there were a couple of ideas.

The outer parts of a neutron star consist of a solid crust of neutron-rich nuclei (plus electrons, and a little deeper, free neutrons). Neutron stars are born rotating extremely fast, but they spin down due to the loss of rotational kinetic energy through magnetic dipole radiation. As a result, they are born oblate, but as the spin rate decreases this is not the lowest energy configuration. It used to be thought that starquakes (pulsar glitches) were caused by cracking of the crust and rearrangement towards a more spherical configuration.

The second idea is that it could be due to changes in the very intense magnetic field of the neutron star. The magnetic field is tied very tightly to the crustal material, so if the magnetic field shifts, possibly associated with the neutron star spin down, then it will put the crust under stress, which can be relieved by a starquake.

White dwarf stars also contain solid, crystalline material, however they do not spin so fast and do not have similar ultra-strong magnetic fields. I have not heard that they experience starquakes.

"Normal stars" do not have a crust or any other solid layer, so there can be no similar starquake phenomenon.


What's the likelihood of the existence of unknown elements in the Solar System? What's the chance that there might be undiscovered chemical elements in the Solar System - either on planets or around the Sun or on asteroids of the Oort-cloud?

As far as elements (e.g. on the periodic table) go, I would say the odds are very slim.  We already discovered or produced all the elements of the Periodic Table up to atomic number 112 at least.  As the number increases, the half lives of the elements generally decreases, and is very short for elements above 102. If this trend holds true as the number increases, practically all the "undiscovered" elements should have turned into the lower known atomic number elements.

However, there is hope.  There is a theorized "island of stability" where a narrow range of yet to be discovered high atomic number elements may be stable: http://en.wikipedia.org/wiki/Island_of_stability
I would say there is a slight chance this element could be discovered in the solar system.


Which lens and magnification is best to see planets? I just bought a Celestron Travel Scope 70 and it comes with a 20mm and a 10mm eyepiece. I get a good view of the Moon but I am not able to see Jupiter, Saturn or Mars as expected. I read about the Barlow lens but I am not sure how much magnification is required to view these planets. So based on my telescope, which lens should I go for? And should I also get Red/Blue filters to view them?

Eyepieces

The highest useful magnification of the Celestron Travel Scope 70 is about 165x. However, you are unlikely to find that much magnification to be usable except under very ideal conditions ("good seeing" i.e. clear atmospheric conditions, dark skies with no light pollution, viewing on a night with a new moon, etc). Even then the image at the eyepiece is going to be dim at that magnification.

Any telescope can in theory achieve an unlimited magnification. But in practice, its magnifying power is limited by the amount of light the telescope is taking in, which is to say, its aperture. The telescope is only collecting so much light. Every increase in magnification is zooming in on a smaller and smaller area, which means it is looking at a smaller and smaller fraction of the original amount of light.

On your telescope, the aperture is 70mm. This is the main thing that dictates how bright of an object you can hope to see at the eyepiece. There is another stat that is important to figuring out what eyepieces are likely to show you, though, and that is the telescope's focal length. Your scope has a 400mm focal length. Total magnification is the focal length of the telescope divided by the focal length of the eyepiece.

Your telescope came with two eyepieces:

400mm (telescope focal length) / 20mm (eyepiece focal length) = 20x magnification
400mm (telescope focal length) / 10mm (eyepiece focal length) = 40x magnification

As you can see, the smaller the eyepiece focal length, the greater the final magnification. If you tried an 8mm eyepiece, you'd get a 50x magnification. To get a 65x magnification, you'd need somewhere around a 6.2mm eyepiece.

(Interesting aside: Since the magnification also depends on the telescope's focal length, note that each eyepiece will have a different magnification depending on the telescope it is in.)

Before you go buying any extra eyepieces, though, I'd strongly recommend you go see what you can see with the equipment you already have! You should certainly be able to see Jupiter, Saturn, and Mars.

Jupiter will appear as a disc. You can see its four Galilean moons, Io, Europa, Ganymede, and Callisto, so-named after the first person to see them, Galileo. They appear as points of light around the disc of Jupiter.
For Saturn, you should be able to make out the rings poking out either side of Saturn's disc. They look like tiny little teacup handles. You can also see Saturn's largest moon, Titan.
Mars is a little bit less exciting, but it should be visible as a reddish disc.

If you do opt to buy another eyepiece, one thing to pay attention to is eye relief. Eye relief is how far away from the eyepiece you can be to still see a decent image. Eyepieces with shorter eye relief are cheaper, but they are also harder to use. An eye relief around 6mm to 8mm is much more comfortable for me.

As for me, I can never help myself: for every telescope I've owned, I always buy an eyepiece that pushes the limits of my telescope, just to see what it looks like. But from experience, I can tell you it is not a great view. Still, it can be fun to see what your equipment can do. But before you jump all the way to an eyepiece that gives you 130x magnification (or whatever), I'd say to shoot solidly for the middle range of your scope (in your case, maybe 60x - 80x). If you can afford it, buy a mid range eyepiece with decent eye relief and field of view (FOV). You'll be much happier with the view through this eyepiece than you will an eyepiece that maxes out the magnification.

Filters

Do yourself a favor and at least pick up a lunar filter. They filter a certain percentage of the light and allow the rest through without modifying the color. The moon is so bright, it will make your eyes water trying to look at it through even a modest telescope. Lunar filters make looking at the moon a much more enjoyable experience.


How would latitude affect day/night lengths on a planet with no axial tilt? How would the length of day and night change with changes in latitude on a planet with no axial tilt?

Would the poles of the planet be in constant sunlight?

The length of the day would be the same on (almost) every latitude, and just a bit longer than the night. The reason for that is that the planet's star (let's call it 'Sun') is a disk, not a point. So even if the middle of the Sun is below the horizon, part of it might still be above the horizon and it's technically 'day'. For the same reason, the length of an Earth day at Equinox is slightly more than 12 hours.

At the poles, and a small region around them (depending on the apparent size of the Sun), it will always be day. At the North Pole of the planet, you will always see exactly the northern half of the Sun; the southern half will always be under the horizon. It will still take the Sun exactly the same amount of time to go 'around' the sky 360 degrees as it will on the equator or any other place on the planet.

The diagrams in the Wikipedia article are a good visualization how every day on such a planet would look like (click to enlarge):


Would a viewer see Earth from the rim of Shackleton Crater? I've been researching this but so far have not found an answer.  If you were standing on the top of the rim of Shackleton Crater at the lunar south pole, would you be able to see any of Earth? I can't find anything about the height of the rim, although nearby Malapert Mountain is apparently visible from Earth so you could see Earth from Malapert.  Shackleton, though, I'm not so sure.  Any insights appreciated. KPRedmond

The moon has a monthly motion called libration Of interest to you is libration in latitude. The orbit isn't exactly in the plane of the ecliptic, and the rotation axis is slightly different from each. This means that about a week after passing its ascending node, you can see up to about 6.5 degrees past the moon's South pole. Two weeks later, the south pole is invisible, but then north pole can be seen.

From the point of view of an observer on Shackelton it would mean that over the course of a month the Earth would slowly rise and then set again.


Lucky imaging with Celestron 14 - is this result reasonable? I attempted to do some lucky imaging of Jupiter using a permanent installation (Celestron $14$" SCT + Paramount ME II) and Nikon D5600 attached directly to a Baader Hyperion 8-24mm eyepiece (without a filter). The atmospheric conditions weren't ideal ($65$% humidity, $2$ m/s winds), the telescope was properly acclimated prior to use, and Jupiter was around $17^{\circ}$ elevation during imaging.
After processing $3600$ frames (from 1 minute of video1 at $60$ fps) with Autostakkert 3 and then aligning the RGB channels and recomposing I got these:

Left (or top if your screen is small) is using $20$% of the stack, right (or bottom) is using $50$%. Since this is my first foray into planetary imaging my question is: are my results reasonable given the setup used and atmospheric conditions described?2
I'm asking because I've seen much better results from people using smaller telescopes. I think the discrepancy is the result of one (or several) of the following:

Something is wrong with the equipment I used.
I'm using the wrong equipment.
I'm not processing the images correctly.
This is the best the equipment could do under those conditions.
I did something wrong in my setup or imaging.

and I'm trying to rule out number 1.

Here is a single, unprocessed frame for reference:

 Update
I have acquired a Bahtinov mask to rule out the focus as the issue, here is an exposure of Manubrij (Omnicron Sagittarii) with the mask:

At the time of imaging, the angular separation between Jupiter and this star was less than $6^{\circ}$. Conditions were much better this time, humidity was $38$%, winds were less than $1$ m/s. The telescope was acclimated for about 2 hours prior to use.
Again using $3600$ frames af $60$ fps I got these:

Left (or top) is with $10$% of the stack, right (or bottom) is with $20$%. This looks slightly better to me, but I'm still not sure if this is what I should expect from the equipment.

1 Note that the video in the link has been compressed through ffmpeg using the H.265 codec with CRF of $25$.
2 This result is just one example - I did $30$ minutes of imaging in $3$ minute segments over the course of several hours, applying the same process to other sets of exposures yielded similar results.

I don’t think it’s any of the points you suggest.  The scope and camera are fine for what you are doing.
I’d say that the focus is slightly off. It takes a lot of effort to get it spot on


How small could an orbital system be in our solar system? Answers to How does the Sun's gravity have so much force and pull on the solar system? How does it scale? Newton's law of gravity scales on orbits so that we can theorize very small orbits with very small masses.
Could two hydrogen molecules orbit one another or would external gravity prevent them from having a stable orbit? We know there are quite small orbital systems in our own asteroid belt, but is there a practical limit to how small an orbital system (with the little orbital system also orbiting the Sun) can be in terms of total mass of the orbital system?

If we only consider gravity, one answer may be found using the Hill sphere. This is the distance the gravity of a body dominates over the sun: $$r_H \approx a \left(\frac{m}{3M_\odot}\right)^{1/3}$$ where $a$ is the semi-major axis, $m$ the mass and $M_\odot$ the sun's mass.
Now, an actual body has some nonzero density $\rho$ and $m=(4\pi/3)\rho r^3$. If the Hill sphere is inside the body there will not be any orbits around it (they would be dominated by the sun's gravity). So, we get the equation $$r = a\left(\frac{(4\pi/3)\rho r^3}{3M_\odot}\right)^{1/3}$$ which simplifies to
$$\rho = \frac{9M_\odot}{4\pi a^3}.$$ Objects less dense than this have Hill spheres inside themselves: at 1 AU this density is  $4.3\cdot 10^{-4}$ kg/m$^3$ (a thin gas), while at 0.1 AU it is 0.4255 kg/m$^3$ - about a third of sea level air density.
For hydrogen atoms, if we calculate the density for a 25 picometer atomic radius I get a density of 25,570 kg/m$^3$ (in actual hydrogen gas the atoms are spread out way more). Hence there the Hill sphere argument actually allows them to orbit each other!
In practice this does not happen. The orbital period at (say) 3 atomic radii is $\sqrt{4\pi^2r^3/Gm}\approx 3.4$ hours and the binding energy is $1.5\odot 10^{-27}$ J. This is $4\cdot10^{-5}$ of the thermal energy of the cosmic background radiation: even if there wasn't any sunlight or other radiation from inside the solar system it would jostle the atoms enough that they would split.
This suggests an apparent way of answering the question: if the binding energy $Gm/r$ is less than typical disrupting energy the orbit will not be possible. Actually calculating the forces is non-trivial (there are many kinds, from Jupiter's gravity to solar heating) and weaker forces can sum up over time. Knowing the disruptive background also just gives an upper bound for $m/r$, one could have smaller orbits.
So the true answer will be given by how small dense objects we are willing to consider, and (as the other answer points out) local forces. In the solar system the most relevant may be electromagnetic charging due to the solar wind: if the objects are metallic and close they can even attract each other if they have the same charge (!). Things like magnetic fields, infrared radiation and solar wind will play a role, making the true answer somewhat undefined.


How are black holes doors to other universes? I'm watching MIT OpenCourseWare lectures on general relativity and not too long into the first lecture the professor stated that the Kerr Black Hole solution allowed for travel between universes. How is that possible to know? How does one derive this/come to this conclusion?

It is correct that the Kerr black hole solution of GTR allows travel between universes. However, that does not mean that if you actually jump into any kind of black hole that you could go to another universe.

To motivate the resolution to this conundrum, let's start off very easy: suppose you stand on the ground with a ball in your hand, and you throw it with some initial velocity. For simplicity, let's ignore everything except a uniform gravity. Mathematics will then tell you that the ball follows a parabolic arc, and when and where the ball will hit the ground. And if you take the resulting equations too literally, then it will also tell you that the ball hits the ground twice: once in the future, once in the past. But you know the past solution isn't right: you held the ball; it didn't actually continue its parabolic arc into the past.

A morally similar kind of thing occurs for, say, a Schwarzschild black hole. If you look at it in the usual Schwarzschild coordinates, there's a problem at the horizon. Mathematics will then tell you that the problem is just with the coordinate chart, and that there's actually an interior region to the black hole that becomes apparent in different coordinates. And if you do this generally enough, it will tell you that there's more to it than even that: there's also white hole with a reverse horizon and its exterior region--another universe. This full "maximally extended" Schwarzschild spacetime has this other universe connect to ours via an "Einstein-Rosen bridge" and then "pinch off", producing separate black and white holes.

Of course, that too is an artifact of mathematical idealization: and actual black hole is not infinitely extended in the past and future; it was actually produced by something, a stellar collapse. (And the "bridge" isn't traversable anyway; one will be destroyed in the singularity if one tries.)

Finally, on to the Kerr solution, it's a bit better because formally the singularity is avoidable, unlike the Schwarzschild case. However, it's still physically unreasonable: in addition to the fact that actual black holes aren't eternal, the interior of the Kerr solution is unstable in regard to any infalling matter, which will perturb the solution into something else entirely. Therefore, it cannot be taken as a physically meaningful. Still, it is true that full Kerr spacetime contains a way into another universe--in fact, infinitely many of them, chained one after another.

If you're interested in the details of its structure, you could look at some Penrose diagrams of those black hole solutions.


Calculate mass of exoplanet from transit method Is there any way to calculate (or at least estimate) mass of an exoplanet from transit method? I know that mass can be calculated with radial velocity method, however I would like to create program for processing Kepler light curves (without radial velocity data).
The data I have are:

radius, mass, surface temperature and luminosity of star,
orbital period, semi-major axis and radius of exoplanet,
duration and depth of transit.

For the case of a single planet, you can't. The transit method is sensitive to the radius of the planet, not the mass.
For a multi-planet system, you can use variations in the timing of the transits caused by the gravitational interaction between the planets to infer the masses, for example as done in this paper by Nesvorný & Morbidelli (2008). These timing variations can be very significant if the planets are located close to mean-motion resonances (i.e. the ratio between the orbital periods being close to a fraction with small denominator, e.g. 1:2 or 2:3).


Why are relativistic jets perpendicular to the accretion disk? From looking at many images (artist's rendering and computer-generated) at black holes, I've noticed that the relativistic jets are always perpendicular to the accretion disk. Is this always true, and if so, why does it happen?

We can think of the magnetic field lines as part of its host gas in the jet, in the sense that when the gas moves, the magnetic lines of force must move with it - and vice versa. Magnetic field lines passing through an accretion disc are thus forced to rotate with the disc. The particles of ionised gas above and below the disc are then urged to rotate with the field lines and consequently feel a centrifugal force that flings them outwards. Beyond a certain distance, however, the magnetic field is not strong enough to force the particles to rotate with it. The gas particles lag behind and bend the magnetic field lines backwards. As the field lines become more and more twisted around the axis of rotation, they begin to exert a strong squeezing force that may be responsible for the initial focusing of the beam. This tendency of a helical magnetic field to contract towards its axis may explain why jets are so tightly focused. In some circumstances, the screwed magnetic field can also act like a compressed spring and help to accelerate the gas even further.

The energy source for a jet is probably the accretion disk itself. As the infalling material gains speed, it begins to grow into a disc of matter that revolves around the black hole. In the accretion disk, the density becomes greater and greater the closer the matter gets to the black hole. This results in increasingly violent collisions between the particles. The enormous increase in particle collisions suddenly heats up the gas so much that it explodes and drives the matter back out. However, this material cannot shoot through the surrounding material in the disc. It can only escape through the central hole of the donut-shaped accretion disc: Two symmetrical jets form that shoot outwards, which are focused by the aforementioned "magnetic field tunnels" This mechanism would be suitable for radio galaxies; however, their jets only have a speed of about a quarter of the speed of light.
The jets of black holes reach speed of close to the speed of light (99% of c).
Jets in AGN, however, reach speeds far above the maximum orbital velocity of an accretion disc: they are accelerated to a fraction of a percent of the speed of light. Such a flow can therefore not only be generated with gravitational energy released by the infalling matter. But there is another source of energy that can be tapped: the rotational energy of a rotating black hole.
The surface area of a black hole can never decrease. Since a rotating hole has a smaller surface area than a non-rotating hole of the same mass, there is no principled reason why rotational energy cannot be extracted. The mechanisms that explain this process are based on the high similarity between a black hole and an ordinary electrical conductor. This analogy can be illustrated by imagining a charge being brought closer and closer to a non-rotating black hole. When the charge is far from the hole, the electric field lines protrude radially from the charge. But as the charge moves closer to the horizon, the field lines become more distorted: They "wrap" around the hole, so that a distant observer would see the electric field as arising at the surface of the horizon. It is as if the charge has spread across the surface of the hole, as it would if the "surface" were an electrical conductor.

A rotating black hole behaves like a rotating conductor. Let us assume that the gas approaching a rotating black hole has a magnetic field. When a charged particle, such as an electron, moves in a magnetic field, it is acted upon by a force that is perpendicular to its direction of motion and perpendicular to the lines of the magnetic field.  The force causes the electron to deviate from the direction of motion. The electrons in an accretion disc are free because the gas in it is in a plasma state in which the electrons are no longer bound to their atomic nuclei. The accretion disc rotates with its axis parallel to the lines of the magnetic field. The free electrons in the disk will tend to move either towards the centre, or away from the centre depending on the orientation of the magnetic field and the direction of rotation of the disk. In either case, the accumulation of electrons in one part of the disc and a corresponding minority of electrons in another part of the disc will create an electric potential. Conducting gas bound to the magnetic field lines transfers this electric potential to a circuit between the poles of the black hole and its equator. The jets use this circuit, which resembles a battery, as an energy source. The energy that is extracted from the hole accelerates charged particles and ultimately drives the jets. This creates a drag on the rotation of the black hole, which is thus slowed down.

The huge voltage drop in this circuit could actually be responsible for the creation of the matter that flows in a jet. At such a large voltage, electrons are accelerated so strongly that particles of matter and antimatter - mostly electrons and positrons - are created from the vacuum of space.
Mainly took this from here. Went on a little bit of a tangent but I hope I answered your question. I myself am actually not sure about why these magnetic field lines diverge from a straight course and "spread" themself across the surface of a black hole, but maybe someone in the comments can explain that.


What is the effect of the axial precession on the orientation of the planet's orbits? I am working on an online application to visualize the positions of the planets and to calculate their ephemerides (you find it here, but it is not complete yet). I do not have much knowledge in astronomy and I am learning new stuff every day while developing the code. A few questions came to my mind that I hope to get answers for here.

The first thing I do not quite understand is the effect of the earth's axial precession. I understand the way it causes the vernal point to move along the ecliptic within the period of the precession of about 25770 years. But since the longitude of the ascending nodes of each planet's orbit depend on the vernal point and remain almost constant over time it seems like all the orbits change their position relative to the stars at the same rate as the vernal point does. But I cannot believe that the orientation of the orbits is related to the precession of the earth's axis.

You can see the effect in the application: When you move the topmost time slider for years you can see the zodiac rotate and the orbits staying in place because the vernal point is fixed with the scale. Shouldn't the orbits move together with the stars? But then the longitude of the ascending nodes wouldn't be constant anymore... I am confused. Thank you.

When talking about the longitude of ascending node you must be very careful to define the reference plane that you are using.  As you state, due to the Earth's axial precession, the First Point of Aries moves along the ecliptic over ~26,000 years.  This is because the celestial equator is slowly precessing about the ecliptic.

Now, in the case of the orbital elements of other planets, the longitude of ascending node is again with respect to the First Point of Aries.  As your intuition suggests, the orbits of the outer planets won't be influenced by the precession of the Earth's equinox.  This means that the longitudes of ascending node will not be constant over long time periods.  If we suppose that the physical orbits are fixed (i.e., the planets are not perturbing each other), the change in the longitudes of ascending node will be entirely due to the Earth's precession.  When you take both these changes into account, you will find that they cancel each other out and the position of these orbits remains fixed with respect to the distant stars.

You can see in the orbital elements provided by NASA that they are given with respect to the J2000.0 epoch.  In other words, the orbital elements are provided for a single point in time, and to get the orbital elements today you must take into account the Earth's precession.

http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt


Is there an upper limit on the mass of black hole mergers we can detect? From the LIGO website, black hole mergers have been observed between black holes with a mass up to roughly 50 $M_\odot$.

Are there no black holes with a mass above 100 $M_\odot$ or is this an observational bias? Why haven't we observed any mergers between black holes with a mass in the 100 - 1000 $M_\odot$ range?

It is quite likely there is an astrophysical upper limit to the mass of a black hole that can be produced during the core collapse of a massive star, caused by the pair instability supernova phenomenon. There isn't an observational bias against detecting more massive black holes in the range of 100 to a few hundred $M_{\odot}$.

Details:

The frequency of the gravitational waves is twice the orbital frequency of the binary system. The general scenario is that early in their evolution, a mrging binary system will be orbiting at relatively long periods (seconds !), but as gravitational waves take energy out of the orbit, the system becomes more compact, the orbital period gets smaller and the frequency of the emitted gravitational waves goes up. This continues until the black holes get so close together that their event horizons merge.

Very roughly, we can derive (from Kepler's third law, not going into detail), using Keplerian orbits
$$ f_{\rm max} \sim \left( \frac{GM}{\pi^2 a_{\rm merge}^3} \right)^{1/2}\ ,$$
where $f_{\rm max}$ is the peak frequency at merger (when the gravitational wave signal is also maximised), $a_{\rm merge}$ is the separation of the masses at merger and $M$ is the total mass of both black holes.

If we let $a_{\rm merge} \sim 2GM/c^2$, the sum of the two Schwarzschild radii of the black holes, then
$$f_{\rm max} \sim \frac{c^3}{GM} \left( \frac{1}{8\pi^2}\right)^{1/2} \sim 2\times 10^4 \left(\frac{M}{M_{\odot}}\right)^{-1}\ {\rm Hz}$$

Now, LIGO is limited to observing frequencies above about 20 Hz. The sensitivity drops off rapidly below that because of seismic noise and other factors. If the mass of the merging black holes exceeds some critical value then the frequencies of the gravitational waves they produce never get into the sensitivity wndow of LIGO. Using the expression above, we can estimate that this happens only if the total mass exceeds $1000 M_{\odot}$. Observing the mergers of more massive black holes would require a detector that is sensitive to lower frequencies, probably beyond the surface of the Earth (e.g. LISA).

This calculation is only good to a factor of 2 or so, but we can check it. GW150914 had a total mass of around $65 M_{\odot}$ and merged at frequencies of about 120 Hz. Since $f_{\rm max}$ scales as $M^{-1}$ this suggests 360 solar mass mergers should be just about detectable, but clearly demonstrates that LIGO could detect black holes of 100-200 solar masses. What's more, at a given distance and frequency, the signals from such mergers would be more powerful than for less massive black holes -- something like $h \propto M^{5/3}$, which means the volume in which the mergers would be visible goes as $M^5$. Thus more massive black hole binaries would have to be extremely rare in order to have evaded detection.

The astrophyscal reason for an upper limit is the pehenomenon of  pair instability supernovae (e.g. Farmer et al. 2019), which blows the star apart rather than leaving a black hole (or any other kind) of remnant. This likely happens for stars with initial masses of $130+\ M_{\odot}$, and means that leaving behind black holes with $M > 50M_{\odot}$ is very difficult, with even lower mass limits for stars with a metallicity more similar to the Sun, since they lose more mass in stellar winds during their lives.

For initial masses of $250+ M_{\odot}$ it is possible that the pair instability supernova mechanism ceases and direct collapse to a black hole becomes possible. In which case their might be a population of $300+ M_{\odot}$ mergers that are just below LIGO's sensitivity window. New Earth-based gravitational wave detectors like the Einstein Telescope and Cosmic Explorer aim to push their low frequency response down to a few Hz and might be capable of detecting mergers in the 300-1000$M_{\odot}$ range.

This means you could not get a merger pair between about $100 M_{\odot}$ and $300 M_{\odot}$ (unless they themselves were the products of a merger).


Elemental abundances in the crust of exoplanets and accessibility to them Suppose I am orbiting a rocky planet (density = 4.4 g/cm3) two-thirds the mass of Mars in the Habitable Zone of Gliese 251. This planet has a thin atmosphere, presumably from out-gassing, still has internal heating, has a weak magnetic field, but has no sign of tectonic activity. I need to mine precious and radioactive elements to replenish my ship. Gliese 251 only has about 8% of the metallicity of the sun. Can I assume that the planet (theoretically) would have about 8% of the elements (per volume of crustal material) as the Earth? Would lack of tectonic activity make them much harder to get to?

No, it doesn't work like that. A rocky planet, to first order, will probably contain a similar mix of elements to the Earth.

The most likely reason that the star has a low metallicity is because it is old, and formed when the interstellar medium was less metal rich. But because rocky planets do not incorporate (much) hydrogen and helium, what you are really asking is how do the abundances of other elements scale to the iron abundance (which is often what is meant by metallicity)? The r-process events, such as type II supernovae that produce the heaviest radioactive elements (beyond lead), enrich the interstellar medium "promptly", because the progenitor lifetimes are short compared to the age of the Galaxy.

However, because a lot of the iron and nickel in the interstellar medium is not produced/disseminated by type II supernovae, but by type Ia supernovae with longer-lived, lower mass progenitors, then if anything I would expect a larger ratio of elements produced/disseminated by type II supernovae to iron/nickel. This will also be true of the "alpha" elements like Silicon, Oxygen and Magnesium.

The planet might be a bit short in those elements predominantly produced by the s-process in lower mass giant stars - so elements like carbon (!), nitrogen, barium, rubidium, strontium, zinc and many others may be scarce.

For a star of this age/metallicity I would expect these variations to be plus or minus a factor of two or so compared to the Earth.

Note, I am discussing bulk abundances. The accessibility of the material, where it is etc., is more of a speculative geology question, and I don't know the answer. A further complication is where exactly the planet forms compared with the condensation temperature of various elements and materials. The above discussion implicitly assumes that the planet forms in a protoplanetary disc material at a similar temperature and density to the conditions in which the Earth formed. However, a change in these conditions will alter the detailed compositional mix.


Estimating the angle covered by the star trails and deducing how long the exposure lasted How would you estimate the angle covered by the star trails and deduce how long the exposure lasted in the following image?

Here is the photo's caption:

  Star trails beyond the Gemini Observatory on Mauna Kea, from which a laser beam forms a guide star on the Earth's ionosphere that enables the use of adaptive optics. In this long exposure, the laser beam tracks across the sky, making the visible arc. (The object in the sky at which the laser was pointed changed several times during the night.) Note that stars rise and set at an angle (not perpendicular) relative to the horizon, because Hawaii is about 20 degrees north of Earth's equator. Also, the farther a star is from the pole, the more its path looks straight, over a short distance.

I read the answer to this question about how to measure star trail angles

  The only star trail angles I'm familiar with deal with astrophotography.
  You set up a camera, put the film in, open the shutter for a time exposure of the stars. As the stars appear to revolve around the earth, they make "arcs" on the film.

  To calculate the angle of the arc, use the formula: 1 minute of time = 15 minutes of arc. If the answer is over 60, divide by 60 to get degrees of arc. So: 12 minute exposure = 180 minutes of arc or 3˚ of arc length for your star trail.

I'm not sure how to approach this problem since the answer uses time to calculate the angle of the arc, but no time is given here.

The answer is given here. One minute of time corresponds to 15 arcminutes (written as 15'). This is because in 24 h the Earth revolves 360º, so
$$\textrm{angle per time} = \frac{360º}{24 \textrm{ h}} =\frac{21,600'}{1440 \textrm{ min}} = 15'/\textrm{min}.$$
If you turn this fraction upside down, you see that 1' corresponds to 1/15 min, or 4 seconds.

That is, you measure the angle (let's call it $\theta$) of any of the star traces, as seen from the center (notice that the Northern Star is not exactly at the center, so that it itself traces a tiny arc instead of a dot). From the picture below, I get roughly $\theta = 135º$. The exposure time is thus:
$$t_\mathrm{exp} = \frac{\theta}{\textrm{angle per time}} = \frac{135º}{360º/24 \textrm{ h}} \sim 9\textrm{ h}.$$

By the way, if you mark the position of the ends of the trails, you can recover the stellar sky. I found Ursa Major, marked by the yellow dots.


Was the Universe expanding before the beginning of inflation? My point of view is that first there was the Big Bang singularity, and then the period of inflation which resulted in the observable Universe becoming many times bigger. But was the Universe expanding between the Big Bang singularity and the beginning of inflation?

This is likely unanswerable.  Inflation was in part a resolution to a fine tuning problem: without it, it seemed we needed the early universe in a very specific and precisely balanced state to get to where it is today, and there was no solid scientific way to explain why things were so perfectly arranged (other than to simply assert that they were).

With inflation, the state of the universe before the inflationary epoch is fairly irrelevant.  Mostly all it needs is that the region that inflates into what we will know as the observable universe had enough time to achieve thermal equilibrium (which we need to explain why the universe looks pretty much the same in all directions, exactly as if things well outside of light speed communication had nonetheless achieved thermal equilibrium at some point).  Curvature and such get "smeared out" by the inflation to give us the mostly flat and homogeneous universe we see today.

On a pedantic note, some researchers consider inflation and the big bang to be the same thing. At least in the sense that talking about "before inflation" is scientifically meaningless, so if the Big Bang is the (scientific) beginning then we might as well take it to be inflation.


Orbiting supermassive black hole or galactic center of mass? One of the ways they measure the (supposed?) supermassive black hole at the galactic center of the milky way is to measure those tens of stars right at the galactic center that are orbiting what appears to be nothing, but what is said to be a supermassive black hole.

My question is why there is a distinction between all other stars in the galaxy orbiting (for instance, our own sun) and those few right near the galactic center orbiting.  My point is, if you ask what our solar system orbits, an astronomer wouldn't say "the supermassive black hole at the center," he would say "the galactic center of mass."  My assumption is that you couldn't even calculate the mass of the supermassive black hole by knowing our sun's mass and orbit - people would probably say we are too far out, or we are orbiting the galactic center of mass rather than the black hole, yadda yadda.

Questions:

1) Why the distinction?  How do we know that the stars right at the galactic center aren't also orbiting the galactic center of mass vs. the supermassive black hole?

2) What would happen if you magically removed the supermassive black hole in an instant?  Would the stars right at the center continue to orbit the center?  If so, how is that different than what they were originally doing?

Note that I know that center of mass of the galaxy and the supermassive black hole are two different things.  I also know that the center of mass of the galaxy is either superimposed on the supermassive black hole, or quite close to it.  I just want to know how or why some objects feel the center of mass of the galaxy and respond to it, while others feel the pull of the supermassive black hole and respond to it.  How far away do you have to be to be considered "orbiting the center of mass" rather than "orbiting the supermassive black hole?"

To a first approximation when we calculate how fast an object is orbiting around some mass distribution we can assume that the gravitational attraction it experiences is only that due to the mass interior to its orbit.

This approximation, known as the shell theorem should only be applied when the mass distribution is spherically symmetric, or most of the mass is concentrated well inside the orbit of the object.

The bulge of our Galaxy can be considered roughly spherically symmetric, and it appears that the mass distribution is strongly concentrated towards the centre.
Hence we can say that the centripetal force on an orbiting object is provided by the gravitational force due to the mass interior to its (again, I am assuming a circular orbit for the sake of argument, but this doesn't greatly matter) orbit is
$$ m\frac{v^{2}}{r} = Gm \frac{M(r)}{r^2},$$
where $m$ is the mass of the object (which cancels out) and $M(r)$ is the mass inside radius $r$. Thus the orbital speed $v$ is given by $\sqrt{GM(r)/r}$.

Thus by observing orbital speeds we can calculate roughly what mass exists interior to the orbit.

In the case of stars near the centre of our Galaxy, that turns out to be 4 million solar masses, packed into a space that is extremely small. The reason it is thought to be a black hole is not because of the mass, but because it is mass that is (i) densely packed and (ii) cannot be seen.

Even if it were a bunch of massive stars shrouded in dust, so that you could not see it in visible light, you can easily calculate that the heated dust should be kicking off huge amounts of infrared radiation; and that is not seen.

In the case of the Sun, well, the mass interior to the Earth's orbit is much larger than 4 million solar masses, it is many billions of solar masses. The contribution of the black hole at the centre is completely swamped by all the other mass in our Galaxy, so we are not directly influenced by the central black hole.

If you removed the central black hole, the stars that were near it would be orbiting way to fast for the new $M(r)$ that they experience. They would fly off and assume highly elliptical orbits with much larger semi-major axes, though I think they would still be part of the Galactic bulge.

Stars further out, including our Sun would have their orbits changed imperceptibly, because $M(r)$ will hardly have changed for them.


Understanding The Fluctuations In The CMB Maps If I'm understanding this correctly the fluctuations in CMB are a result of the "last scatter" of photons when electrons joined together with nuclei that before this formed plasma. CMB is among other things used to see the early matter density fluctuations of the early universe, right?

So, how is this seen? Are the warmer parts in the maps where the matter is most dense? And are these spots warmer because more photons are coming from that spot, or because of some other reason?

The anisotropies in the CMB are caused by four effects; three at the surface of last scattering (SoLS), and one after:

Temperature differences

Denser regions will be more compressed and thus hotter, on average. Hence, an overdensity will result in a hotter spot, with a fractional fluctuation $\Delta T/T_0$.

Gravitational redshift

Photons climbing up (or falling down) the gravitational potential $\phi$ of their SoLS will lose (or gain) energy. Hence, an overdensity will result in a colder spot, and an underdensity in a hotter spot. This effect is also called the non-integrated Sachs-Wolfe effect and changes the temperature by $\Delta T = \phi/3c^2$, where $c$ is the speed of light.

Doppler shift

The SoLS has a bulk motion $\mathbf{v}$. Generally, overdensities have gas infalling, and hence result in a colder spot. In the direction $\hat{\mathbf{n}}$, the temperature difference will be $\Delta T = T_0 \mathbf{v}/c$.

The integrated Sachs-Wolfe effect

If a potential well is getting shallower in time (which happens during dark-energy domination), photons receive a net blueshift in crossing the well and the CMB appears hotter. This gives a temperature difference of $\Delta T = 2\Delta\phi/c^2$, where $\Delta\phi$ is the change in the potential while a photon traversed the potential well.

The total temperature fluctuation is the sum of all these terms. In general, the gravitational redshift will dominate over the temperature differences, which in turn dominates over the Doppler shift.

A review of the CMB fluctuations is given by Challinor (2013).


Why is the Sun so bright, but you can feel it far away? First of all, the Sun's surface temperature is only about 6000 K. Years ago I was working at at a melt work with furnace temperatures 0f about 2-3000 K; of course you could not be near a furnace without protection, when it was open, but you could still stay some 20-30 metres away from it. Likewise, a light bulb has no long reach, till it fades away. But how about the Sun? Why can experience it at 150 000 000 km away?

Besides, I wonder why space is not heated up, because the heat radiating from all the stars and galaxies is so enormous, and actually the matter density in space is so low. How, actually, can heat be absorbed in space? What temperature is measured in space - is it just the wavelengths of radiations, from which temperature is calculated?

The answer to your first question has to do with luminosity. It's a measure of power, the energy given off by an object in a certain amount of time, which you can think of as brightness. The more luminous the object, the brighter it appears.

We can treat the Sun as an idealized object called a black body, which emits thermal radiation according to something called the Stefan-Boltzmann law. Assuming the Sun is spherical, with a radius $R$ and temperature $T$, then its luminosity $L$ is
$$L=4\pi\sigma R^2T^4$$
where $\sigma$ is the Stefan-Boltzmann constant. The important thing here is that $L\propto R^2$. If the Sun was the size of one of your furnaces, it would not be very bright; if the furnace was spherical and could be treated as a black body, the two would have the same luminosity.

The Sun, however, has a radius on the order of about $7\times10^8$ meters. I don't know how big the furnace is, but I'm assuming it can't be larger than ten meters, to an order of magnitude. First, that's a size difference of about $10^7$. Second, we square the radius when determining luminosity, meaning that there's a difference of a factor of $10^{14}$. To make this a bit more meaningful, the Sun is then essentially one hundred million million times more luminous than the furnace, even though the temperatures are the same.

I feel like your second question may be a duplicate of What is the temperature of outerspace? (see also the questions linked therein), and Run like hell has already tried to address it. barrycarter's comment, though, is spot-on. The flux from the source, $F$, drops off via the inverse-square law - that is, at a distance $r$ from the source,
$$F\propto\frac{1}{r^2}$$
Remember how we just established the power (pun intended) of squaring a large quantity? Interstellar distances are enormous; the nearest star system, Alpha Centauri, is roughly $10^{16}$ meters away, or ten thousand million million meters. Now square that distance. That's a tiny factor - even when multiplied by the luminosity of all three stars, which combined is still on the same order of magnitude as the Sun.


Why did Mars lose its magnetic field? I had heard that Mars once had a planetary magnetic field, but that it is now gone? What happened to it?

Our own magnetic field is generated by convection currents in Earth's liquid outer core.
A useful summary from Physics.org:

Differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal as cool, dense matter sinks whilst warm, less dense matter rises.
This flow of liquid iron generates electric currents, which in turn produce magnetic fields.
The spiralling caused by the Coriolis force means that separate magnetic fields created are roughly aligned in the same direction.

Mars used to have a liquid iron core, but it was never as extensive as Earth's and has long since solidified.
Once our own core cools enough to solidify, we too will lose our magnetic field.


What is the most populated/numerous stellar system in which the orbits of all objects are known? I posed a question or two on here a couple months or so ago about the orbital patterns of each of the seven stars in the two septenary star systems Nu Scorpii and AR Cassiopeiae. They remain unanswered so I have to assume the answer is unknown. So, now I want to what the most numerous star system is for which we actually do know the orbital pattern, and obviously I want to know the orbital pattern.

I am NOT talking about a star with multiple planetary or other smaller bodies. I am talking about multiple STAR systems. Do we know the orbit for systems of multiple stars. Again, I'm not talking about every rock larger than a school bus. I mean STARS. Stars that have nuclear fusion occurring.

What is the most populated/numerous stellar system in which the orbits of all objects are known?

The answer is none. Other than our own solar system, astronomers don't know if they know all of the large bodies (aka planets) orbiting any given star system. Presumably other star systems have asteroids, comets, and other stuff. The orbits of those small bodies in star systems is unknown, and will remain unknowable for a long time.

We don't even know that for our own solar system. If we did, we wouldn't have to worry about killer asteroids and comets.


How is the distance of a star from Earth calculated in the HYG star database The HYG star database can be found here

From the hygdata_v3.csv file, how are the cartesian (x,y,z) coordinates calculated? (Is there a mathematical formula for this?)

And in the dso.csv file, there aren't any x,y,z coordinated. How can I calculate that?

The transformation to $x,y,z$ coordinates, starting from the RA, Dec and a distance (as given in the HYG catalogue) is a simple trigonometric exercise, since the $x,y,z$ (in this case) is referred to a coordinate system aligned with the equatorial coordinate system.

$$x = d \cos({\rm RA}) \cos({\rm DEC}),$$
$$y = d \sin({\rm RA}) \cos({\rm DEC}),$$
$$z = d \sin({\rm DEC}),$$
where $d$ is the distance. How that distance has been obtained is not clear from the README file accompanying the database. It may have been by (incorrectly) inverting Hipparcos parallaxes.

The dso.csv appears to be a catalogue of deep sky objects. There is no information in the catalogue itself which allows you to calculate or estimate a distance.

You could try cross-referencing the catalogue against the  NASA/IPAC Extragalactic Database (NED). https://ned.ipac.caltech.edu/
If the recession velocity there is bigger than about 1000 km/s, then you will get a rough distance by using Hubble's law
$$ d = v/H_0$$
 with $H_0 \sim 70$ km/s per Mpc (or whatever your preferred value is). If the recesson velocity is smaller than this then the object is too close for Hubble's law to be valid.

e.g. The first line inthe catalogue appears to be a galaxy known as IC 5370, that has a recession velocity of 10372.00 km/s (no error bar given and I doubt any more than 4 significant figures is warranted and in any case $H_0$ is not known accurately to more than 2 significant figures), which gives an approximate distance of 150 Mpc.


What is the next lens I should buy? I have a question about purchasing another lens that's right for my situation. I own the Orion Skyquest XT10 Dobsonian, a classic case of buying a scope that is WAY to big to be carrying around everywhere, and yet I manage to get it out periodically throughout the year. I am very amateur obviously.

I enjoy using my stock 25mm but find the stock 10mm just too blurry. I do admit, I rarely get out of the neighborhood, the seeing is not great in my valley, there's city light, and I must admit I long for those hubble images popping out of my own lens.

With that said, I just feel like I should be able to get some more clarity with a new lens. I'd really like to be able to see jupiter with some more detail, right now the best I get is a small orange blur with some minor bands visible if you use your imagination.

What wouuld you recommend as the next lens in the $50-$200 dollar range that will provide a bit more detail than my 25mm but with good clarity?

It's a pretty common tendency to fix problems by buying more eyepieces, but that's rarely how it works in practice. There are a few steps you need to take before buying more glass.

Collimation

Check collimation first. This should be your first item on the list, full stop. The performance of the scope decreases rapidly when it's out of collimation (for solar system objects). Check the documentation that came with your scope. Also check these links:

http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/

http://garyseronik.com/a-beginners-guide-to-collimation/

If the scope is not in good collimation, a new eyepiece is a waste of money. All the eyepiece does is magnify the real image generated by the scope in the primary focal plane - so if that image is out of whack, then no eyepiece will give you the performance you're looking for.

Check collimation every time you use the scope. With practice, it will end up taking only a couple minutes.

Seeing

It varies with time and location. Sometimes it's better, other times it's bad. Go on the CDS site, search for your location and use the Seeing forecast; good seeing is represented by dark blue, bad seeing by light blue or white.

http://www.cleardarksky.com/csk/

Plan your observations according to the forecast.

Seeing only affects fine-detail objects like the Moon or the planets. It does not affect faint DSOs (galaxies, nebulae, etc) because your eye does not resolve fine detail on DSOs anyway.

Thermal equilibrium

If the primary mirror is not at the same temperature as the air, "local seeing" conditions (vortices of air on the primary) will blur the image. Take the scope outside 1 hour before you start observing and let it acclimate.

For a scope 8" or bigger it's often a good idea to install a fan on the back of the primary to accelerate the process - but get everything else done first.

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

Light pollution

Only affects DSOs (galaxies, nebulae, clusters). It does not affect solar system objects such as planets. The only remedy is to load the scope up in a car and drive away from the city.

However, all Messier objects are visible in most amateur telescopes even from very light-polluted places like major cities. They just don't look as spectacular as from a dark sky place.

Finding focus

Make sure the scope is actually focused properly. This might be more tricky than it seems. Use a star nearby and adjust until the star is as sharp as possible.

You could also make or buy a Bahtinov mask, which will give you proof positive of perfect focus (adjust focus with the mask on, until all 3 lines intersect each other in the center).

https://en.wikipedia.org/wiki/Bahtinov_mask

https://www.astrozap.com/scripts/prodList.asp?idCategory=74

With all of the above taken care of you should definitely see Jupiter's equatorial belts. They become visible even in 2" of aperture, and a slightly larger instrument should show them pretty clearly, even in a modest eyepiece. Under good seeing, with a perfectly collimated instrument, you can even start noticing the curls and the vortices in the belts even with an aperture as low as 4" ... 6".

Jupiter getting more and more blurry as you increase magnification is normal. It's not a sign that the eyepiece is bad. What's not normal is not seeing the belts in a 10" scope. That indicates that either collimation is bad, or seeing is exceptionally bad, or both; typically it's collimation.

Your eyepieces should actually give decent performance in the center of the field of view.

You also need to let your brain-eye system adapt to the image. You don't have to observe from a very dark spot, Jupiter is bright enough to not require perfect dark adaptation. But your brain and your eye need a bit of time until they can extract all details from the image. Let it drift through the field, stay relaxed, follow the image and just take it in slowly. You will start seeing details after 3 or 4 minutes that were not visible at first (assuming good collimation, seeing, etc).

Now on to the eyepiece. Your instrument is a f/4.7 newtonian, 1200 mm focal length. A 10 mm eyepiece will give you 120x magnification. This should not be too much even under bad seeing - provided the scope is collimated.

I've observed Jupiter a lot at 1200 mm f.l. in a smaller instrument. I've found that a 8.8 mm eyepiece (136x) was a reliable workhorse, never needed anything with less magnification. When seeing got better, I could push it up to 6.7 mm (180x) or even 4.7mm (255x). But speaking in general, Jupiter is low contrast and does not benefit from pushing the magnification too hard.

Mars OTOH does very well under very high magnification. But that assumes your collimation is top-notch and seeing is very good.

Provided that you take care of collimation, etc first, you could try and get better eyepieces. The ones that came with your scope are not that bad actually, they do fine in the center, it's the edge performance that's weaker. But if you do get new eyepieces, you might as well get something better. I've found that the Explore Scientific 82 degree series provides a good bang for the buck - not too far from the performance of top glass, but much cheaper than top prices.

https://explorescientificusa.com/collections/82-series-eyepiece

But the options here are very numerous, so do your own research.

You might also benefit from having a barlow, maybe a 2x or a 2.5x. However, the much higher magnification generated with the barlow will not work unless collimation, seeing, etc are all taken care of!

It's a little difficult when you juggle collimation and seeing at the same time, one you can control (collimation), the other you can't (seeing), but realize that this will take some time to figure out, so go one step at a time.

Also understand that this is not the kind of problem that is solved by simply throwing money at it.


Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this? Trying to wrap my head around some concepts involving the very large gravitational fields of black holes, and what the gravitational gradients look like on a cosmic scale.

I'm familiar with the Great Attractor, and wondered if it were possible if the Milky Way and it's neighbors are already "doomed" to fall into whatever the anomaly is.

Per recommendation, I've moved the second half of my question to a new Question here: Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?

Original Text:

  Extrapolating further, I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low, on the order of thin atmosphere - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration within the event horizon matter?

This isn't anything near a proof that we don't live inside a black hole, but it's a bundle of evidence that certainly goes against it, and that the Great Attractor is not, in fact, the singularity.

First off: Expansion of the universe.

As you no doubt know, the universe is expanding. In fact, the expansion is accelerating. Do black holes expand? Yes. As they suck in more matter, they can get bigger. But if that was the case, we should notice more matter coming into the universe (well, I suppose it could be from outside the visible universe, but we should still see lots of matter coming towards us). Also, the universe's expansion is driven by dark energy, and it pushes (on a large scale) everything away from each other. In an expanding black hole, there would be no reason for the matter inside the event horizon to move away from each other; only the event horizon expands.

You also made a good point in a comment below about Hawking radiation. Eventually, in the far future, when there is nothing left in the universe but black holes, black holes will evaporate via Hawking radiation (okay, they do that now, but they can still take in more matter). If our universe is a black hole, it should then contract. But we see no reason why it should. In fact, the theory that predicts the universe's eventual collapse into a singularity (i.e. the opposite of the Big Bang), the Big Crunch theory, predicts that the universe's contraction will match its expansion. Contraction due to Hawking radiation wouldn't necessarily mirror the black hole's growth. Also, the Big Crunch theory is supported only by a minority of scientists due to the evidence against it.

Second off: The motion of galaxies by the Great Attractor.

First off, see When will the Milky Way "arrive" at the Great Attractor, and what all happen then? (and not just my answer! I dearly wish @LCD3 would expand his/her comment into an answer!). The general gist of things there, as pertains to your remark about the Great Attractor, is that the galaxies aren't all moving towards it. There are doubts (see the papers I mentioned) that the galaxies previously thought to be moving towards it are, in fact, moving towards more distant objects - other superclusters. If the Great Attractor was indeed the singularity, a) all the galaxies in the universe should be accelerating towards it, which is not the case, and b) we should be moving directly towards it, and due to its gravity, not that of the superclusters beyond.

Like I said in my comments, I don't quite think your last part relates to your first part, but I'll try to address it. First, I'm not sure where you got your sources, but I can say that we don't really know what goes on inside a black hole, and I don't now how someone got that density figure (I am, of course, not the authority on black holes - see @JohnRennie for that, on Physics SE, and I could be wrong about this). The density inside the event horizon would not be very thin, however, in regions where there is lots of matter. For example, in a black hole with an accretion disk, the material that is being absorbed might not have that low a density. Also, a large volume of that density would not necessarily collapse to become a black hole, because it would not be compact enough.

Hope it helps.


Is Anti-Dark Matter a thing? Could there be such thing as anti-dark matter? If there is anti regular matter, why not anti-dark matter? If there is no such thing as anti-dark matter, why?

Yes, it is a thing.

Weakly Interacting Massive Particles (aka WIMPS) are thought to come in matter and anti-matter forms and have a self-annihilation cross-section in order to produce the "correct" amount of dark matter (in relation to photons) that we measure today.

As the opening paragraph of the wikipedia article on WIMPS states - there are a number of experiments searching for the photons associated with Annihilation of WIMP anti-WIMP pairs, from which it appears that there is expected to be a neutrino and a gamma ray signature or even the presence of anti-matter (positrons) that might be detected in cosmic ray experiments.

Of the other dark matter candidates - whether sterile neutrinos have an anti-particle depends on whether they are Majorana fermions or not. The hypothetical axion I think is its own anti-particle (a chargeless boson) and there are no annihilation-type experiments for their detection that I can find.


Could Sgr A* be a core of dark matter instead of a supermassive black hole? A recent paper (related news) shows that Sgr A* at the center of the Milky Way galaxy could be a dark matter core, instead of a supermassive black hole, according to their calculation using the astrometry data of the 17 best-resolved S-stars and the mysterious G2. The dark matter particles could be 56 keV fermions, which they call "darkinos".
The team also proposes that, above a critical mass, a dark matter clump could gravitationally collapse into a supermassive black hole, like the one in M87.
Is this dark matter model possible? Do the flares from Sgr A* observed in recent years support or contradict this model? Can LHC or other smaller particle colliders or detectors on Earth help to confirm or rule out the particle?

I'm going to say, "probably not", because (as tfb points out) we have very good evidence for SMBHs in other galaxies -- in particular galaxies with active nuclei where the phenomenon associated with accretion disks around BHs are seen. The most extreme case is probably NGC 4395, where a Seyfert 1 nucleus and associated radio jets exist. The best estimates for accretion-disk+SMBH models suggest a SMBH mass of only a few hundred thousand solar masses; a recent direct dynamical measurement (den Brok et al. 2015) indicate a central "dark object" gravitational mass of $M_{\rm BH} \sim 400,000$ solar masses. There are numerous other galaxies with active nuclei where the estimated or measured central-object masses are $M_{\rm BH} \sim 10^{6}$--$10^{7} M_{\odot}$.
This is rather difficult to square with the claim that "above a critical mass, a dark matter clump could gravitationally collapse into a supermassive black hole" because if you follow up the associated reference to Argüelles et al. (2021), it ends up predicting a critical mass of $\sim 10^{8} M_{\odot}$. This is consistent with something like M87 ($M_{\rm BH} \sim 6 \times 10^{9} M_{\odot}$), but not with all the evidence for accreting SMBHs in galaxy centers with $M_{\rm BH} < 10^{7} M_{\odot}$.
The "flares observed from Sgr A* in recent years" are, I believe, generally supposed to be evidence for a very weak, transient accretion disk around the SMBH (e.g., the discussion on this web page), and so would be difficult to explain with this model.


Is it possible to see a moonrise or moonset twice in a day? I've been wondering this for quite some time.  Due to the variations in tilt from the earth we know that moonrise/moonset times vary.  Is it possible that in any location on the earth one might see more than one moonrise or moonset in a given 24 hour period?

This can only happen if the moonrise at a certain date is earlier than the moonrise at the previous day. There are two reasons why this could happen:

The body (Moon, Mars,...) moves in the opposite direction of the Sun viewed along the right ascension (or ecliptical longitude) axis. The sidereal day is 4 minutes shorter than 24 hours which means that a static celestial body (a star) will rise 4 minutes earlier than the day before. However planetary bodies move (more or less) along the ecliptic, which is the Sun's path during the year along the celestial sphere. The Sun moves on average 4 minutes in ecliptical longitude (and hence in right ascension with a bit more variation) every day. So for a body to rise earlier, the body would have to move slower along the celestial sphere than the Sun (or move in a retrograde direction). This is true for the outer planets (including mars) but not for the Moon, which moves in the same direction as the Sun and much faster (one orbit per month).
If the observer is at a sufficient high latitude on Earth (either north or south of the equator), the inclination of the celestial equator will be small enough so that a body that increases its declination rapidly enough will be able to rise earlier on the second day.

See for instance the figure below where each drawing is at the same time of day. The moon will increase its right ascension during the day (along the celestial equator downwards, the red arrow in the figure). If the observer is on the Earth's equator, moonrise will be later the next day. It does not matter whether the declination (the position above or below the celestial equator (to the left or right in the figure)) will increase or decrease. All objects with the same right ascension will rise at the same time on the equator. If the observer is not on the equator but near one of the poles (for instance at 65° latitude north), the increase in declination may be sufficient to counteract the increase in right ascension (see bottom panel). Note that even though  the right ascension increases (the moon moves left along the celestial equator axis), the Moon will rise earlier during the day.
This only works (in the Northern hemisphere) when the Moon increases its declination (the height above the celestial equator) very rapidly. This happen only when the Moon is near the Vernal Equinox (i.e. in Aries or Pisces). And the observer needs to be at a high latitude (e.g. near the North pole such as Thule as mentioned by @barrycarter)


Why are distant galaxies not visible in the observable Universe? Assuming that the Universe is and has been expanding ever since its creation (Big Bang theory), the galaxies that are now at the 'edge' (not visible theoretically) must have been (at some point in time) at a place around where the Earth is at now.

If it is assumed to be true, then the photons emitted by these galaxies from this point onwards should be around for us to see these galaxies (and even the first light emerged from the Big Bang must be visible).

Then why is the observable so small? It should depend upon the direction in which we look. That is, if we look in the direction of the center of the expanding sphere, it should be a small observable universe. If we train our telesopes in 'exactly' opposite direction of the center, it should be 'fully' observable.

Does it work this way? Or are my basics wrong?

Quick answer: Because they didn't entered our event horizon. Some never will. And some will move out of our event horizon - their last photons that'll be received here being sent right now.

Let's do some fact checks first:

  [...]the galaxies that are now at the 'edge' (not visible
  theoretically) must have been (at some point in time) at place around
  where the Earth is at now[...]

According to current calculations, the first galaxies may have formed around 200 million years after the Big Bang - older estimates went with the 400-500 MY range. For a long, long while, there were no stars to be seen. So if you go back in time you won't be seeing the same structures we see today.

Second, and that may be an awkward mental experiment, nothing else was occupying our place other than ourselves. I'll ask you to excuse the cliché, but the old balloon example is very apt to explain this:

As the universe expands, the distance between celestial bodies increases. Now, here's a way to put it: space is being generated in between the objects.

And not only that - while there's a limit on how fast you can move on the balloon's surface, it doesn't apply to the amount of space being generated.

As a direct consequence there's a bubble around us that basically works in the very same way as a black hole's event horizon does:

In a black hole, gravity is so strong that its pull on photons exceeds C (the speed of light, or 299,792,458 m/s);
On a fringe object, the amount of space generated by the expansion, per second, may exceed C; its photons will never reach Earth.

And it seems that we live in a accelerated universe - that is, the spatial acceleration ratio is actually going up. If that's correct, some fringe objects will  slip away form our event horizon, disappearing (from our point of view) into a sample of the heat death of the universe.

Scary, huh?


Could we estimate the age of the universe based on the planar property of the Solar System? The Big Bang scattered planets and stars everywhere in three dimensions.

But after billion years of moving and interacting with each others through gravity, planets moved on the same plane.

Given the very small but non zero friction in space, could we give a good estimate of the age of the universe?

First, the Big Bang did not happen at a point. I cannot emphasize this enough. It happened everywhere in space at the same time. You can't think of it as an explosion from one location pushing everything else away.

Furthermore, planets and stars didn't exist at the time of the Big Bang. The Solar System, for example, formed about 4.5 billion years ago, while the oldest stars - Population III stars - formed within a few million years after the Big Bang. At the beginning of the universe, everything was in an exotic soup-like state: a quark-gluon plasma.

Therefore, your hypothesis is flawed in two places.


The shortest angular distance between two points on the celestial sphere is measured along a great circle that passes through both of them.

The equation your program is using is
$$ \cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2), $$
where $(\alpha_1, \delta_1)$ and $(\alpha_2, \delta_2)$ are the right ascension and declination of the two points on the sky (expressed in degrees, then converted to radians).

In terms of output from your program its harder to help, especially as you have not given the full output. The first three columns should be self explanatory - columns 2 and 3 are just the coordinates expressed in degrees. I am guessing that the third column from last, that you haven't shown, gives the separation between your object and the catalogue objects and that the penultimate column is the angle that the great circle between the points makes on the sky. That looks right because the fourth object in the list has almost the same RA as your input object but  is 67 arcseconds due south (i.e. at 180 degrees).


trying to understand stellar nucleosynthesis Trying to create a periodic table that shows "where" elements are made. For example big bang, cosmic rays, stars, supernovae etc...but reading a lot about gold,and other heavy neutron rich elements, being formed in colliding neutron stars. Is there a definitive list of elements that may have been formed in neutron stars?

The elements you are talking about are created by the r-process - the rapid neutron capture onto heavy elements that can occur in neutron-rich environments. The r-process is probably responsible to some extent for all the most neutron rich (i.e. more neutrons than protons) nuclei with atomic masses above that of iron and is probably responsible for all elements heavier than lead and bismuth. However, there are three broad peaks in the solar-system abundances, centred at Germanium, Xenon and Platinum (the first, second and third r-process peaks), where the r-process is thought to dominate the production over competing processes such as the s- and p-processes (plot taken from here).

If one subtracts the expected contributions from other sources one can estimate the fraction of a given element that comes from the r-process (plot taken from here).

It is a contemporary astrophysical problem (i.e. there is no definitive answer) for the source of the r-process. In other words, it is an unsolved mystery as to what fraction of each r-process element comes from which astrophysical location and it may well vary as a function of cosmic time or location. The two main candidates for the sites of creation for r-process elements are core-collapse supernovae, the neutrino driven winds from supernovae and neutron star mergers.

The main problems to be solved are the detailed physics of supernova explosions and neutron star mergers, the relative populations and numbers of these events as a function of cosmic time and location and even the nuclear physics ingredients that go into calculating the yields of such elements.

A recent paper by van de Voort (2015) performs simulations that suggests most of the r-process elements in the universe could be created by neutron star mergers, and they point out that it is especially difficult for core-collapse supernovae to produce what are known as the "third-peak" r-process nuclei. These claims are all contentious, although the latter point about the heavier r-process elements is echoed in a number of previous works. (e.g. Wanajo 2013). Nevertheless, even this is disputed; for example Nishimura et al. (2015) perform simulations of rapidly rotating supernovae progenitors with strong magnetic fields and find that they can reproduce the abundance pattern of all the r-process elements seen in the Sun.

So if I wanted a vague stab in the dark at directly answering your question, I would choose the group of elements in the third r-process peak, with mass numbers $191 < A< 197$ as being most likely to have been (mostly) produced by neutron star mergers. This includes Osmium (192), Iridium (191, 193), Platinum (194, 195, 196) and possibly gold (197).


How do we get the angular size of a crater without lens? Suppose I want to calculate the diameter of the crater. I apply the small angle formula $$diameter=\frac{\theta_{diameter}}{206265''}\times D$$ where $\theta_{diameter}$ is the angular size of the crater, in arcseconds, and $D$ is the average surface-to-surface distance from the Earth to the Moon, which is around $375,900$ km. We know that the moon is about $\frac12$ degree, which is $1860$ arcseconds. But how do we estimate the angular size of the crater without using lens, with only  eyes? I was thinking using ruler and applying scale but that way, I think, is inaccurate because the Moon has a round surface.

Most lunar craters are too small to resolve with the naked eye; I would measure a published photograph. A crater near the limb appears as an ellipse whose minor axis is foreshortened but whose major axis measures about the same as if the crater were centrally located. Surface curvature has little effect on this measurement because most craters are small compared to the lunar radius.


How to determine the age of an old galaxy? Recently a galaxy A1689B11 was discovered, having an age of 11 billion years. How that age was determined? News sources reported that some cutting edge technology with gravitational lensing was used. Can anyone explain the exact procedure?

When the article speaks of the "age", what it means is that we are seeing that galaxy as it was a long time ago. We are seeing it as a very young galaxy. The galaxy is very far away, so light has taken a long time to reach us from the galaxy. The light from the galaxy is 11 billion years old. We measure the age of the galaxy by finding its distance from us.

To find the distance we use the Hubble Law, which states that the further away a galaxy is, the faster it moves away from us, due to the expansion of the universe. And we can measure the speed that the galaxy is moving away from us because when an object moves very fast, the light from that object becomes shifted towards longer wavelengths (red-shifted). By measuring light that is known to have a fixed wavelength when not red shifted, we can find the speed of the galaxy accurately, and then use the Hubble Law to find the distance, and hence the age of the galaxy. (source)

A1689B11 has a redshift z=2.54 which corresponds to a light travel time of 11.1 billion years

Gravitational lensing makes the galaxy brighter, so it can be observed at these immense distances, it isn't directly used for finding the distance, but without gravitational lensing, the galaxy wouldn't be visible.


From what distance would our sun have an angular diameter of 7 arc seconds? If an observer travels away from our sun, at what distance would our sun occupy 7 arc seconds of space?

From us, the Sun is visible around a half grad, which is $\approx$ 30 arc min = 1800 arc seconds.

To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.


What point does Earth actually orbit? If we consider the two largest masses in our solar system - the Sun and Jupiter, by themselves, they will orbit a common barycenter which is somewhere offset from the Sun's center in the direction of Jupiter (just above the Sun's surface). If we add Saturn, things get more complicated, but there is still a barycenter, even if it follows a complex path with respect to the Sun's center (or the Sun has a complex wobble around the barycenter). Add Neptune and all the other masses in the solar system, and things get even more complicated. Neverthess, there is still a barycenter (often outside the limb of the Sun) about which the Sun does a complex dance.

Now, what is Earth actually orbiting? The Earth's orbit is described to be an ellipse; the "center" of an elliptical orbit occurs at one of the focal points of the ellipse. What is at the focal point of the elliptical path the Earth follows? Is it the solar system barycenter or is it the Sun's (wobbling) center of mass? In other words, is the Earth's orbit shifting as the Sun is tugged around, our distance from the Sun never varying by more than the eccentricity of the ellipse, or do we orbit the solar system barycenter, and our distance from the Sun varies by the sum of our orbital eccentricity plus the amount the Sun wobbles in its own orbit around the barycenter?

Short answer is "The Sun".

As Conrad notes, since when you include the effects of Jupiter, the Earth orbit is non-keplerian, notions of centre are not really defined. But you can ask, in Newtonian gravity, where the Earth's acceleration vector points. Now the acceleration of the Earth is due mostly to the Sun, partly to the moon and slightly to the other planets.

Let's ignore the moon (ie consider the motion of the Earth Moon barycentre)

The acceleration due to the sun is 4 orders of magnitude greater than that due to Jupiter. So if Jupiter is at right angles to the Earth, the acceleration vector of the Earth is slightly pulled away from the centre of the sun. But not by much, in fact it points to a point about 4000km from the centre. The sun has a radius of 700000km, so the point the Earth is orbiting is well inside the sun, and isn't the Sun Jupiter barycentre.

To see why the Earth doesn't orbit the barycentre, consider the motion of the Sun in a three body system (Sun-Jupiter-Earth) The Earth doesn't orbit the Sun-Jupiter barycentre just as the Sun doesn't orbit the Earth-Jupiter barycenter.


Is it at all possible for the sun to revolve around as many barycenters as we have planets in our solar system? Though it is understandable that the sun and the earth may be revolving around a barycenter, but, if so, not only the sun and Jupiter should also be revolving around some barycenter, the same should be true about the other planets as well? So it has to be true that the sun revolves around as many barycentres as the number of the planets we have in our solar system. I am quite confounded — how any object may at all revolve around multiple barycenters? And if an object can't revolve around more than one point, does it not nullify the theory that the sun and the earth revolve around a barycenter?

The short answer is no; there is only one barycenter. Yes, you can count the Sun/Jupiter barycenter or the Sun/Saturn barycenter, or whichever barycenter you want, but the net effect of all Solar System bodies is to be considered when you calculate the actual barycenter of the Solar System. (And yes, that would include counting all the small asteroids and moons, even those yet unknown to humans, even though their combined effect is negligible.)
One could see it in such a way that yes, there are many barycenters, but the movement of the bodies is around the “average” barycenter. Somehow. But that’s not a good way to describe the system.


Is there an established distinction between a geyser and a cryovolcano in solar system bodies? Is there an established distinction between a geyser and a cryovolcano in the context of cold solar-system bodies such as the moons of Jupiter and Saturn, or Kuiper belt objects (e.g. Pluto)?

Both can lead to spreading of new water across the surface, and both can result in plumes of water vapor above the surface. Do they have distinctly different mechanisms?

Or is it unclear or in a state of flux, a bit like the diagram in @zephyr's answer to Do astronomers generally agree that the distinction between comets and astroids is not so clear?

An educated guess: on Earth, the working fluid in a "geyser" is water, which remains a fluid more or less permanently after each eruption.  This means that geyser eruptions are not associated with buildup of new geological material around the geyser (apart from the much slower process of dissolved minerals accumulating as the water evaporates).

Contrariwise, on Earth, the working fluid in a "volcano" is molten rock, which solidifies after reaching the surface.  Each new eruption permanently deposits new solid material in the vicinity of the volcano.

If I were writing definitions, I would define a "cryovolcano" as a geological vent system where the working fluid is water, but where the environment is cold enough that the ejecta from eruptions accumulates in a permanent mountain of ice around the vent.


Light levels on other celestial bodies What I'm trying to achieve

For a series of realistic renders I am currently working on I am trying to calculate the light level on different celestial bodies, in terms of the luminous flux (in lumens), as this is the parameter used in the rendering software. (as it's made with earth in mind). Specifically, I am now looking to find the luminous flux on Saturn's moon Enceladus.

Thanks in advance for any help provided :)

What I've tried/found so far

Now the solar irradiance is easy to find as lying somewhere inbetween 16.7 and 13.4 W/m^2. However, in order to derive the luminous flux the frequency contents must be known as the luminous flux weights the light by a sensitivity function for human sight.

Wikipedia gives a formula for the luminous flux as

$$\Phi_v = 683.002 \int_0^{\inf}V(\lambda)\Phi_{e,\lambda}(\lambda)d\lambda $$

Where $\Phi_v$ is the luminous flux in lumen, $\Phi_{e,\lambda}$ is the spectral radiant flux in W/nm and $V(\lambda)$ is the luminosity function describing the sensitivity of the human eye to different wavelengths of light (dimensionless).

$V(\lambda)$ was relatively easy to find, but I have difficulty finding the spectral radiant flux $\Phi_{e,\lambda}$ of sunlight at this distance. Can this be calculated from the irradiance?

I did manage to find the solar irradiance spectrum in space at the earth in (W/m^2)/nm, and I assume this could simply be scaled by $R^{-2}$ ($R$ being the distance from enceladus to the sun) to get the same spectrum at Enceladus, but I am not sure how this relates to the spectral radiant flux.

You don't need to do it that complicated. The sunlight's spectrum is the same everywhere in the solar system. And in order to get the light intensity, you can simply apply the inverse square law that intensity drops to 1/4 if you double the distance to the sun - like you did for getting the irradiance. The spectral sensitivity of the human eye is also the same everywhere in the solar system.

What is different, is the local conditions: On the surface of bodies it is, of course, influenced by the atmosphere - and that is likely the most important aspect when you want to model different colour; this does include possible aerosol content (like you see for instance on Mars during dust storms). The 2nd most important impact probably is the colour of the light reflected from the body you are "standing" on, thus a feature of the surface material.

So in order to get the colour impression you have to account for these two spectral filters, applied to the (nearly) black-body distribution of the sunlight. Thus in your equation you need to insert these factors as well - or just look at these separately as your equation states the default. Unfortunately this opens a VERY wide field as you can have a variety of atmospheric compositions - just think of the different light conditions you can have here on Earth - and now add the factor of completely different chemestry in the atmosphere. Modelling that is definitely an ongoing scientific endeavour.

If your quest is "just" to get some artists impression, play around with different materials and adjust the spectral transmission and scatter properties of the atmospheres (especially that) - and the ground material.


Brightest most colorful star in eastern morning sky not near any constellations; rises about 4 or 5 am? I have been believing this is Saturn, but perhaps Antares instead? It is so colorful it shines blue, red, green and white.

It is largest in sky in the East before dawn rises about 4 or 5 AM, seen from the San Joaquin valley, central California.

From 38° north latitude, Sirius rises in the east-southeast just before dawn in late August.
Most of the naked-eye stars in its constellation Canis Major rise later.
Orion's belt points toward Sirius and appears above it when they are rising.
As the Skymaps.com evening sky map shows, Antares sets in the southwest several hours earlier.
Saturn is currently in Sagittarius, which is in the south on August evenings.

Refraction by the changing atmosphere makes stars twinkle and even appear to change color.
This effect is more pronounced when a star is bright or low in the sky.
Planets generally do not twinkle because they span wider angles than stars do.


Propagate Earth trajectory for 1 Myr What is the most accurate way of propagation of Earth trajectory relatively to the Sun? I need to calculate the Earth-Sun distance. Is there any library on Python?

I've used before JPL Horizons, however, it's not useful for propagation for 1 Myr.

The La2010 long-term ephemeris Laskar et al. (2011), which is based on the INPOP numerical ephemeris integrated for 1 Myr, is valid for 250 million years before the present day and a unknown distance into the future. The authors note that due to chaotic motion in the Solar System, the accuracy degrades significantly beyond -50 Myr and likely a similar time into the future. The ephemeris is intended for paleoclimate and similar studies and provides the classical orbital elements which will need to be transformed into an Earth position or an Earth-Sun distance.


Are rising sea levels caused by the moon? If the moon is moving away at a distance of 1.48" per year (BBC), than wouldn't that mean the gravitational pull by the moon would weaken over time causing the tidal bulge to shrink?  Thus causing sea levels to rise?

You are correct in saying that the moon is receding from the Earth. However, as you can see, this rate is very very small.

1.48 inches = 0.037592 meters

The Moon is 384400 kilometers away (on average), or 384400000 meters. That means every year, the Moon's distance increases by 0.00000000009779%. That's such a small difference, it is negligible on human timeframes.

Also, although the tides would shrink, this would not mean the sea level would rise. When we refer to rising sea levels, we refer to an average rise in sea levels that have been seen across the world, mostly due to climate change. During high tide, a beach may experience 4 meter higher sea level, but at low tide, it will experience 4 meter lower sea level. If the moon doubled its distance, then gravity would be a quarter its strength (gravity decreases with the square of the distance), so you would have 1 meter above during high tide and 1 meter below during low tide. As you can see, there is still no net change in sea level, it's just that the changes between high and low tide would be smaller. Again, this is negligible on human timeframes.


Why does exoplanets' naming start from "b" and not from "a"? An exoplanet named after its host star receives the suffix letter "b" if it was the first planet discovered to orbit that star, or "c" if it was the second, or "d" if it was the third, and so on.

Why this naming system starts from "b" and not from "a"? Is the letter "a" reserved for another object in the planetary system? If yes, for what is "a" reserved?

As an example, here is a list of the major objects in the Gliese 667 triple-star system:

Gliese 667A (a star)
Gliese 667B (a star)
Gliese 667C (another star)
Gliese 667Cb (a planet orbiting star C)
Gliese 667Cc (another planet orbiting star C)

A  quick lmwtfy reveals the answer:

  In the WMC naming system, the brightest member of a star system
  receives the letter "A". Distinct components not contained within "A"
  are labeled "B", "C", etc. Subcomponents are designated by one or more
  suffixes with the primary label, starting with lowercase letters for
  the second hierarchical level and then numbers for the third.[2] For
  example, if there is a triple star system in which two stars orbit
  each other closely with a third star in a more distant orbit, the two
  closely orbiting stars would be named Aa and Ab, whereas the distant
  star would be named B. For historical reasons, this standard is not
  always followed: for example Alpha Centauri A, B and C are not
  labelled Alpha Centauri Aa, Ab and B.


Does the radius of your cosmic horizon decrease as velocity increases? If I was moving at c-1 meters per second, and was observing an object moving 2 m/s away from my starting point as a result of the expansion of the universe rather than classical velocity, would I ever see the light coming from this object? Does that mean that as you move faster, the radius of your cosmic horizon decreases?

Your past and future lightcone surfaces are invariant: no matter what speed you have, at a certain point in space and time you can see the same things (although strongly Lorenz contracted and with aberration). So if you could see the object when at rest, you can see it near lightspeed.
The cosmic horizon is tricky since it is non-local: the distance where galaxies appear redshifted beyond detection changes with time because of the changing expansion rate of the universe. There are also visible galaxies that recede faster than light we can still see (!).
In some sense the horizon may seem to shift when you speed up because the redshift makes nearby parts of the universe undetectable, but this would be true even in a cosmology without any horizon. If you accelerate at a constant rate you can even get a true event horizon behind you where light cannot catch up to you (a Rindler wedge), but it does not have anything to do with the cosmology. Still, if the object is in that region it would be invisible. But that is because of acceleration, not speed.


Can the Gaia telescope detect small temporarily captured asteroids near its Lagrange orbit? The Gaia space telescope is in a Lissajous orbit around the Sun-Earth Lagrange point 2. The orbit period is about 180 days and the size of the orbit is 263,000 x 707,000 x 370,000 km. It has been theoretically concluded that there should be a population of temporarily captured orbiters (TCO's) around that Lagrange point. All but a handful of them should be less than a meter in diameter and too small to be observed from Earth. Only one TCO, 2006 RH120, has ever been observed, a 5 meter diameter one.

Since Gaia will be in a somewhat similar orbit as the TCO's and will scan the entire sky several times over the coming years, and an IR telescope is sensitive to asteroids, I wonder if it will be able to detect such TCO's. And if so, what could we find out about them? Gaia will not track any asteroids nearby, if it catches the same TCO more than once it would be by chance, I suppose. Could Gaia learn about their sizes, orbits, surface compositions?

Yes.

Detection of inner Solar System Trojan Asteroids by Gaia (pdf):

  The Gaia satellite, planned for launch by the European Space Agency (ESA) in 2013, is the next generation astrometry mission following Hipparcos. While mapping the whole sky, the Gaia space mission is expected to discover thousands of Solar System Objects. The se will include Near-Earth Asteroids and objects at Solar elongations as low as 45 degrees, which are difficult to observe with ground-based telescopes. We present the results of simulations for the detection of Trojan asteroids in the orbits of Earth and Mars by Gaia.