How close would a Sunlike star have to be in order to be dangerous to look at it through a spyglass? Alpha Centauri A which has an about 50% higher luminosity than the Sun is obviously undangerous to look at through a handheld telescope, at about 4.4 ly distance. How close would a yellow dwarf have to be in order to make it dangerous to look at it through an average spyglass? Or to feel some heat on the eye through the spyglass? This question can also be seen reversed as "how far from the Sun would you have to be in order to look at the Sun through a spyglass without danger".

This is more a question of "what is the brightest light I can look at", and subsequently this answer is more physiology, geometry than astronomy which only gives us the distances.
According to the wikipedia page on laser safety it is around 1mW/cm². A binocular - or any optics - again change the equation as so far that they accumulate all light falling onto its entry pupil into your eye.
Stars are sufficiently far away so that they are point sources (that's why they wobble around in the viewfinder due to atmospheric disturbance, unlike solar-system objects).
Energy emitted by a star is obtained by a simple black-body assumption $E=4\pi R^2_\star \sigma T^4$ and energy $W$ per on an area in distance $r$ thus is $W_\star = \frac{4\pi R^2_\star}{4\pi r^2}\sigma T^4 = \frac{L_\star}{4\pi r^2}$ (using the stellar luminosity $L_\star$ further on). Let's assume that your pupil has 0.5cm opening, thus the area there is $A_e = \pi 0.25^2 = 0.2cm^2 = 2\cdot 10^{-5}m^2$. Let's assume a binocular with 5cm aperture, thus the area there is $A_b=\pi\cdot 0.05^2m^2 = 8\cdot 10^{-3} m^2$, thus the ratio $A_e / A_b = 2.5\cdot 10^{-3}$. Thus the light at the entry of the binocular needs to be $W_{max} = 1mW/cm^2 \cdot 2.5\cdot 10^{-3} = 2.5\mu W/cm^2 = 2.5\cdot 10^{-2} W/m^2$ at most to not become dangerous. So, what distance does that imply? Let's equate:
$W_{max} = W_\star $ and solve for $r$: $r = \sqrt{\frac{L_\star}{4\pi W_{max}}}$. So for a sun-like star ($L_\star = 4\cdot 10^{26} W$) we get a minimum distance to safely look at its disk of roughly r=240 AU or 0.004 light years.
You can modify that equation for other stars with higher or lower energy output. Alpha centauri is not much brighter than our sun. But for a star like Vega ($L=40L_{sun}$), the minimum safe distance would already be more than 6 times as distant - yet still very near for any distances when considering distances to other stars. Thus any star becoming dangerous to look at with the naked eye or a binocular would pose a threat to our solar system.
And even if you consider not binoculars, but telescopes, say 1m diameter, thus 400x the light gathering power compared to the considered binocular, we would start to get distances of the order of a light year... even then all stars are further away.
(this still sounds crazy... is there an error in the estimates? But likely this errs on the side of caution. While you should not look at the sun even with the unaided eye, you're not immediately blind - and it's intensity is $1400W/m^2$ compared to the assumed safe power of $0.025W/m^2$, thus there's a difference of $10^5$)


How do we know pulsars have two beams? Pulsars are neutron stars the emit a beam of electromagnetic radiation that is not aligned with their rotation axis. So we can only see a neutron star as a pulsar if the beam crosses Earth's path.

The common representation is that there are two beams, one on each side, but unless the magnetic axis and the rotation axis are at 90°, we can only ever see one of the two beams. How do we know that there are two beams and not one, or four, or any other number?

Several pulsars have double-peaked profiles: you get two peaks per rotation, with one peak stronger than the other. An example of this is the Crab Pulsar. This suggests that at least some pulsars have two beams. Remember that the gravitational bending of light in the vicinity of the neutron star makes it possible to see a larger fraction of the surface, which increases the chance of both poles being visible.

The single-peaked light curves likely represent the case where the second pole is not visible. Of course, if you wanted to play Devil's advocate, you could argue that there might be single-beam pulsars lurking in that sample... you'd probably need to come up with some kind of physical mechanism by which such an object could exist, given the physical theories that have so far been used to explain pulsars predict two beams.


How much photon is still in flight? Since space is mostly empty, one might first think that the majority of the photons, e.g. emissions from the stars since the beginning of the universe, are still in flight and not absorbed by anything.

Is this true?

Is there any estimates that the proportion is relative to the total amount of photons ever created?

(Let's consider the photons that have made it into space first, e.g. exclude the photons that gets absorbed and re-emitted inside stars for now).

Peebles and Fukugita's excellent "The cosmic energy inventory" estimates that about $10^{-4.3}$ of the total mass-energy is in the form of cosmic background radiation, and $10^{-5.7}$ in the form of radiation from stars.

The estimate that the perturbation due to plasma and other interactions of the background radiation is around $10^{-8.5}$, so clearly most photons have never been captured by anything and are still in flight.


Can a black hole reach a limit where it can no longer attract more matter? Can a black hole reach a limit where it can no longer attract more matter? If so, what happens with that black hole? Does the black hole die? Does the black hole decrease its size?

There is no known upper limit for the mass of a black hole.
A black hole stops attracting matter, when there is no matter within reach.
In this case the (macroscopic) black hole is assumed to evaporate by Hawking radiation over very long time scales.

The estimates for the currently (as of early 2014) known most massive black holes (ultramassive black holes) are below about 40 billion ($4\cdot 10^{10}$) solar masses. A black hole of that mass would have a Schwarzschild radius of about 780 times the distance Earth-Sun, or about 4.5 light-days.

(Hypothetical) micro black holes are assumed to be able to evaporate faster than they can accrete matter.


What's the critical mass of a rocky planet before its gravity traps enough gas in its atmosphere to start becoming a gassy planet Looking at the latest theories, I've noticed that most rocky planets are below a certain mass range. I speculate that maybe the way gas giants and gas dwarfs form is a rocky planet during the planetary formation phase of its history accumulates enough mass to begin gaining a hydrogen atmosphere that keeps building until it's no longer a rocky planet.
What would be the critical mass of a rocky planet before it would start down the path of becoming a gassy planet?

In most planet formation theories, the boundary is around 10 Earth masses - the build up of the core mass before that is relatively slow, but once it crosses that threshold, the planet gains mass quickly via attracting gas from the surrounding nebula via the core’s gravitational pull, a process called “runaway accretion.”  As this summary shows, you can still end up with some planets in between.
One interesting result that has emerged from recent studies of exoplanets is that small planets seem to be divided into two groups, those with radii around 1-1.5 Earth radii, and those with radii of ~ 2-3 Earth radii, but with a gap in between where there are few planets:

The current theory explaining this is that the larger group of planets managed to accrete some atmosphere, while the smaller group are bare, rocky worlds with little or no atmosphere.


Would things be exactly the same if the Sun was orbiting the Earth and the Earth was in the center of the Solar System? Would things be the same if the Sun was orbiting Earth and the Earth was in the center of the Solar System, or would there be any big difference?

It seems like why people before thought the Earth was the center of the Solar System since it didn't conflict with their experience.

It depends on what you mean by "in the center" or "switch places."  If you physically moved the Earth to where the Sun is, and the Sun to where the Earth is, and gave each the speed that the other had while keeping the same reference frame, this would do tremendous violence to all the other planets in the solar system (though the Earth would continue to orbit more or less the same).

However, if what you are talking about is simply the language of what you call the center, then you are talking about changing to a different set of coordinates where the Earth is at the stationary center of your coordinate system.  Doing that presents no problems, and nothing is different except the language you are using to talk about what is going on.  One of the great things Einstein did was to reframe the laws of physics in a form that was independent of coordinates, allowing us to forever separate what the laws say is happening from the language we use to talk about what is happening (i.e., our chosen coordinates).  So it all comes down to whether you think "being in the center" is a real thing, or just the language of some coordinate system.  If the former, then no, things would not be at all the same, if the latter, then yes, they'd be exactly the same, if you are careful to use coordinate-independent laws like general relativity.


Meteor shower shooting stars Shooting stars are actually meteors falling down on the surface of earth. How come some of them appear shooting up and not down?

Shooting stars are indeed meteors falling down, but consider this: meteors which are part of a meteor shower actually appear to come from a point in the sky called the radiant point:
google image

Since you're looking upwards, you may be looking straight at the radiant point, and therefore the meteors are 'radiating' away from that point in all directions; not only 'downwards' towards the horizon, but 'upwards' from your point of view.

You can see that, even for a meteor which is not part of a shower, it's just as likely for a meteor to be going 'upwards' across the sky, as it is to be going 'downwards'.


Is a star powered by fission possible? Stars can easily fuse atoms to give of heat and radiation. But at Wikipedia it said that only sub-iron atoms give of energy when fused and take energy when split, and post-iron atoms is the exact opposite. So, if enough heavy elements got together could a "reverse star" be made in which it gives of light and heat through means of fission? How long would such a star last, and why aren't they common, if they even exist?

I think it's an interesting question.    The trick would be a sustained fission reaction, faster than half life, but slower than a chain reaction.   A chain reaction could hardly be considered "star like" - it would just explode.

Lets say you had a planet sized object, maybe 100 parts Iron, basically inert, to 1 part Uranium  , which would generate heat through half life decay and some neutrons would spread from uranium to uranium - many wouldn't.  And as the object got hotter, more and more uranium would drift towards the center.  You'd have gradual heating, and the object, in time, would probobly begin to glow.     I'm guessing it's not possible using anything close to pure uranium, as the neutrons released in a uranium rich ball of matter would do 1 of 2 things - either blow the thing apart, or increase the rate of reaction and cause a chain reaction.   But i think, with the proper mix of uranium and inert elements, and the uranium spread out, not all in the center, you probobly could get something sort of star-like.   It would have much less available energy so it wouldn't burn nearly as long as a star.    At least, that's my thought experiment answer to this question.

The problem with nothing but heavy, neutron rich elements is that, I don't see any way that wouldn't cascade and burn or blow up quickly.   The nuclear reaction is driven by neutrons hitting the nucleus with potential energy and as the reaction happens, more free Neutrons are produced which speeds up the reaction significantly.   A ball of of Uranium many miles across would undergo fission and explode.   A ball heavy enough to not explode - say, jupiter sized, it would heat up and burn quickly and then settle down and cool off slowly. - perhaps similarish to a star going Nova then settling into a white dwarf.

If you have a material inside that slows the reaction, absorbs and re-admits Neutrons you could probobly generate a sustained burn - similar to what happens inside a nuclear plant while maintaining enough gravity to keep the object in one piece and over time, That would heat up and glow like a star in time and last for a while.   Planet sized, even moon sized would probobly be more than sufficient, but you'd need the material to slow the reaction and keep the Uranium spread out.   That's my take at least.


What causes the antipodal bulge? The Wikipedia article on tides shows a diagram with a tidal bulge both towards the Moon and on the antipodal point (the Sun isn't included in the diagram). Surely that's wrong - what would cause the antipodal bulge?

Image courtesy of Wikipedia user Jhbdel under the Creative Commons Attribution-Share Alike 3.0 Unported license.

That is because the moon attracts both the water and the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is

Greatest at the point nearest to the moon (called sub-moon point)
Is lesser on the earth's center, and'
Is the least in the opposite side of the earth.

Assume that earth is falling to wards the moon. Now, as the force on the point near the moon is the maximum and it has 'fallen' the maximum distance, while the earth 'falls' a lesser distance and the farthest of all, the antipodal point 'falls the least'.

"Tidal-forces". Licensed under CC BY-SA 3.0 via Wikimedia Commons.

The above figure shows the forces acting on earth due to moon, The top figure shows the actual gravitational forces acting, while the bottom one shows the forces after subtracting the force on the earth itself. That should explain the antipodal bulge. See the wikipedia ariticle in Tidal forces


What happens if gravity shuts down for only one second? What would happen if gravity completely stopped for 1 second?

I expect stars would belch mightily, neutron stars and black holes would explode.

Stars are only held at their steady state size by the opposed forces of gravity and radiation, so eliminating gravity for 1 second will cause expansion, followed by collapse as gravity kicks in again.
A Neutron star's compactness gives it a surface gravity of up to 7×10^12 m/s², and expanding from 20km to ~70 billion km (or 300000km when c limit is considered) over the course of 1 second constitutes an explosion.
Black holes tend to have higher gravity values at the event horizon (same ref) so the result would be even more cataclysmic.


What is equation of straight line for a ideal blackbody on the color color diagram? The color-color diagram is a plot between $U-B$ vs $B-V$ values.

For a ideal blackbody, what is the equation of this line?

In other words, what does the linear relation between $U-B$ and $B-V$ look like?

Use Planck's law: $$F(\lambda,T) = c  \lambda^{-5}  (e^{d/(\lambda  T)} - 1)^{-1}$$.

Since magnitude is defined as, $M = -2.5 \log_{10}(F) + \textrm{constant}$, the color index $A-B$ is
$$(A - B) = -2.5 \log_{10}(F(\lambda_A,T)/F(\lambda_B,T))$$.

To answer your question, use an appropriate effective wavelength of $U,B,V$. Link $U-B$ and $B-V$ via temperature, you should get $U-B = f(B-V)$.


Looking at the Expanding Universe through the lens of relativity The basis and most observable metric for how the universe is understood to be expanding at an accelerating rate is the red shift in observable light.  I am assuming it has been thought about or maybe it is just obviously wrong to people with more knowledge, but why can we not say the universe is technically always collapsing on itself under gravity's pull when considering relativity?

We know that space and time are affected by gravity so if we think in relative terms if everything is accelerating at an accelerating rate towards everything else this could explain the red shift (appearance of moving away). Things behind me would be accelerating slower towards the singularity things in front of us would be accelerating faster. Therefore relative to us it would actually look like it is all expanding at an accelerating rate, explaining the red shift.

In this very simplistic thinking, this would work for a universe that expands and contracts or is in a never ending state of contraction, which relative to us could go on infinitely since time and space is all intertwined with gravity.

If you work out the equations of motion for the gravitational free-fall collapse of a clouds of particles (or galaxies), you will find that, from the point of view of any of these particles, all other particles appear to be approaching. This is true in both Newtonian theory and general relativity.

The calculation is found in numerous text books and lecture notes, e.g. Susanne Höfner's, but a qualitative argument goes like this:

From your point of view, sitting on one of the particles, a particle farther from the center of the cloud does "feel" a smaller gravitational pull from the central particles. This is, I think, the reason that you think it should accelerate more slowly. However, any particle feels the gravity from all particles that are closer to the center, so the farther you are from the center, the more particles will attract you. On the other hand, the gravity from all particles that are farther away from the center cancels out, so you don't feel those.

This is, in essence, Newton's shell theorem, and it turns out to imply that all particles reach the center at the same time. The distance between each particle is getting smaller and smaller, until the cloud collapses into an point after a a time inversely proportional to the square root of the density.

This is exactly opposite to what is observed.


Do moons experience periods of no sunlight when orbiting "behind" their planet? I'm writing a story that takes place on the Saturn moon Iapetusand I was thinking that especially since Iapetus only has an inclination of about 17 degrees from the ecliptic, wouldn't there be a point during its orbital period where it orbits "behind" Saturn — that is to say, Saturn is directly between Iapetus and the Sun — and doesn't get any direct sunlight?
Is that correct or am I missing a detail here?
If it is true that moons like Iapetus experience these periods of no sunlight, how difficult would it be to calculate roughly how long this period would be using the inclination and the orbital period?

This is basically a repeat of Aristarchus’ experiment of trying to figure out the distance between the Earth and the Moon from a lunar eclipse. See for example https://pwg.gsfc.nasa.gov/stargaze/Shipprc2.htm
So, we know that Iapetus orbits Saturn in 79.3215 days, at an average distance of 3,560,820 km. Saturn’s diameter is 116,464 km, and it orbits the Sun at an average distance of 1,433,530,000 km. The Sun’s diameter is 1,392,684 km.
Using this data, you can now use the formulas given on Wikipedia https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_(Aristarchus)#Lunar_eclipse to calculate the size of Saturn’s shadow at Iapetus’s distance. Knowing how long Iapetus spends in it is then only a matter of dividing the size of Saturn’s shadow by the circumference of its orbit, then multiplying by Iapetus’s orbital period.
I’ll let you do the math. We’re here to help find the answers, not give all the answers. ;-)


Divide a star catalogue to optimize star search I'm continue working on my Planetarium software and I have a doubt.

At this moment my star catalogue is very small, but I'm planning to use a bigger one. To optimize the search inside this catalogue I think I have to divide it in regions.

These regions will be delimited by right ascension and declination.

Using stars' right ascension and declination, how can I divide the sky to optimize my search?

I want to retrieve all stars visible at this moment. Maybe not all of them, only the visible by the user. For example, if user is looking to the North, he/she won't see the stars at the South. The region will be the area visible by the user at this moment, and I want to get all stars for that given region using also a limit magnitude.

This is really a computing problem, but I suppose the only point from an astronomical perspective is what the RA, Dec distributions of your catalogues look like.

I'm not that familiar with optimal search techniques but I guess that you want roughly similar numbers of stars in each region.

If your catalogue is just of the brightest stars, then these are fairly uniformly distributed around the sky, so splitting them up in equal bands of right ascension would probably do.

However, if you are dealing with much larger catalogues then there is likely to be a significant concentration towards the galactic plane. In these circumstance you might be better off splitting your regions in terms of their galactic latitude. So you would calculate galactic coordinates by transforming the RA, Dec and then plot the distribution of number of stars versus galactic latitude to see where you would split your regions.


What is the term for astronomical objects outside the solar system that are smaller than dwarf planets? Asteroids and comets only seem to refer to masses within the solar system, and it seems unclear whether planetesimal does.

Just add the prefix "exo-". At least exocomet and exomoon seem to be established.


Calculating right ascension for two planets and their separation using python I am experimenting with the Python library PyEphem for astronomy, and I am trying to recreate/understand some of the basic calculations there.

There is a function called separation that, given two planets and a date/time, it calculates the separation (angle) between those 2 planets with respect to their "x" projection in a plane.

So for 2018/1/1 for planets Mercury and Mars, we have:

import ephem
import math

mercury = ephem.Mercury('2018/1/1')
mars = ephem.Mars('2018/1/1')

s1 = ephem.separation(mercury, mars)
print(math.degrees(s1))

Which returns

33.792384499568264

But if I wanted to calculate this without the separation function, then I think that the calculation would be as simple as the "right ascension" of 1 minus the "right ascension" of the other:

math.degrees(mercury.ra) - math.degrees(mars.ra)

Which returns

35.114532008671574

Why are the angles different? Since I am not including a latitude and longitude of the observer, all the calculations are supposed to be geo-centric, according to PyEphem.

Is anybody familiar with the calculations going on behind PyEphem, or another library with built-in ephemeris that can produce consistent results for separation?

The issue appears to be in your understanding of the separation function. You state that

  it calculates the separation (angle) between those 2 planets with respect to their "x" projection in a plane.

however, the documentation states

  The separation() function computes the angle in degrees between two bodies, as measured by their right ascension and declination. In general separation() can measure the angle between any pair of spherical coordinates, so long as the elements of each coordinate are spherical longitude (angle around the sphere) followed by spherical latitude (angle above or below its equator). Each pair should be provided as a two-item sequence like a tuple or list. Appropriate coordinate pairs include right ascension and declination; heliocentric longitude and latitude; azimuth and altitude; and even the geographic longitude and latitude of two locations on earth.

This means that this function is calculating the angular separation between the two objects, not just the separation in a particular plane. If you want to perform the same calculation (as the documentation says they're performing), it looks like you actually want the general angular distance formula.

$$\theta = \cos^{-1}\left[\sin(\delta_1)sin(\delta_2)+\cos(\delta_1)\cos(\delta_2)\cos(\alpha_1-\alpha_2)\right]$$

where $\delta\in[-\pi/2,\pi/2]$ is the declination, and $\alpha\in[0,2\pi]$ is the Right Ascension.


How well supported are the theories of black-holes containing their own universes? The thought crossed kidnapped my mind one night almost a year ago. I was laying down, but I was very interested in thinking about the secrets of a black-hole.

Nat. Geo. - News article

What does the current state of research tell us about the idea that black holes might form universes?

Don't forget to backup your answer!

More and more of modern theoretical physics deals with unphysical models and ideas. The only basis we have of predicting what is happening beyond an event horizon, ANY event horizon, is what we understand about what happens around us, on this side of that horizon. Feel perfectly safe in making up anything you wish about what the Physics is like beyond the event horizon, we'll never be able to contradict you. We'll never be able to falsify any claim made for events on the other side.

Purple people eaters? Why not? Ghosts and Gods? Sure. 1+1 = 3? Yup, that too. The event horizons in nature are the natural limits to our Universe. The edge, if you will. Unless and until we are able to observe consequences of the claims you make (called verifiable predictions) you're not dealing with Science, at least not what I regard as Science.

Right now, we know of 4 types of edges to our Universe:

The edge at ~1E-44 seconds after the hypothetical Creation Event
The event horizon of any black hole
the event horizon at about 15-20 billion light-years distance (at which space is expanding at greater than the speed of light and
the edge of our Observable Universe (from which no signals will ever reach us).

Physics is about what goes on on this side of these "edges". Metaphysics, philosophy, religion, opinion are what you may use to deal with those things beyond, but don't call it physics or science.

P.S. I am sympathetic to the idea that there may be an (as yet unknown) physical law which prevents singularities from actually forming. In fact, we have only weak indirect evidence that event horizons actually exist, and absolutely none that singularities form. Generally, when a physical theory and its mathematical expression leads to infinities, we change the theory - see renormalization.

You want to know about multiverses or what is beyond the event horizons of black holes? OK. I want to know about the color of the nail polish that the tortoise carrying the Universe on its back uses. Both are equally valid areas of inquiry. See for example https://en.wikipedia.org/wiki/Multiverse#Proponents_and_skeptics ... and links found there.

I should also mention that you know an article which talks about events prior to time zero (which I call the Creation Event) are metaphysics and/or religious. None of them, I repeat: none of them are based on known Physics. (Why? because our known physics breaks down at about 10E-44 seconds, extrapolating back to t=0 isn't possible.) Sure they make an interesting story, but they should be regarded in the same way as any fiction. You get to choose which you like and which you don't without any regard for ever having to justify your choices.

Much of human knowledge is about narratives which allow us to understand our place in the Universe. It is part of our human nature to be curious and seek out narratives that make sense of our world. The set of scientific narratives differ from this set because the goal is different. They doesn't require our understanding (for instance, Feynman said: "I think I can safely say that nobody understands quantum mechanics." (1965) Little has changed since then.), they only require our ability to apply them to make/verify observations.

Creating a satisfying narrative does not require the scientific method(s) be enforced. But ignore them at your peril.


Near-future astronomic events that could threaten Earth What kinds of astronomical events can pose a serious threat to life on Earth in the near future (1000-10,000 years). If there are none in the near future, when is the soonest event that is predicted to threaten life on Earth?

I am ignorant about astrononomy and so far I have only thought of an asteroid collision, solar flares, and effects from the sun dying in several billion years (and a visit by some hostile aliens).

Essentially, anything could happen! We can't really predict events such as you've described.
The only event we can really be certain of is that the Sun will inevitably die in about 5 billion years. When it's almost run out of hydrogen to fuse it will expand into a red giant, becoming so large it will take out the Earth. There's a good video demonstrating this here: http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm

You could add geomagnetic reversal and gamma ray bursts to your list. Also see: http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm


Time according to the gravity of Sagittarius A*? This may be a really dumb question (I'm more of a Biologist than an Astronomer) so I apologize in advance for my little knowledge pertaining to Astronomy, but, if I'm not mistaken, time is effected by gravity, right? So what is Sagittarius A*'s time compared to ours since it has so much stronger gravity? Do we know specifically the difference?

Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, time passes "normally".

But to answer your question, we must specify what is meant by "the black holes's time" (let's call the black hole $\mathrm{BH}_\mathrm{Sgr\,A^*}$; see note below on the nomenclature), since it depends on how far from Sgr A* we are talking. The time pace at a distance $r$ from the center of a BH is given by
$$t = t_\infty \sqrt{1 - \frac{r_\mathrm{S}}{r}},$$
where $t_\infty$ is the time "at infinity", i.e. far from the BH, and
$$r_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 3\,\mathrm{km}\,\times \left( \frac{M}{M_\odot}\right)$$
is the so-called Schwarzschild radius (the "surface" of the BH), which is where not even light can escape. Here, $G$ is the gravitational constant, $M$ is the mass of the BH, $c$ is the speed of light, and $M_\odot$ is the mass of the Sun.

The last equality shows that a BH with the mass of the Sun would have a radius of 3 km. The mass of $\mathrm{BH}_\mathrm{Sgr\,A^*}$ is some 4.1 million Solar masses, so its radius is $r_\mathrm{S} = 12.1$ million km.

Plugging in the other numbers, we can see that at a distance from $\mathrm{BH}_\mathrm{Sgr\,A^*}$ of

1 lightyear, time runs slower by a factor of 1.00000064, i.e. unnoticeably.
1 astronomical unit (the distance from Earth to the Sun), time runs 4% slower.
1 million km from the surface, time runs slower by a factor of 3.6.
1000 km from the surface, time runs slower by a factor of 110.
1 km from the surface, time runs slower by a factor of ~3500.
1 m from the surface, time runs more than a 100,000 times slower.
At the surface, time stops.

Note that this time dilation is what a distant observer (i.e. the guy with the $t_\infty$ time) would measure for an observer at the distance $r$. The person at $r$ would just measure his/her own time as usual. For instance, according to point 5 above, if you were hovering 1 km from the surface, waving your hand every second, then I, choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see you wave approximately once every hour. And when you run out of fuel and plummet into the BH, then when you cross the surface you wouldn't notice anything particular, but I would see you frozen in time. This is the concept of relativity.

Finally, let me use this chance to clarify something that people, including myself, often have gotten wrong: Sagittarius A (without an asterisk) is a radio source in the center of the Milky Way. It consists of three parts: Sagittarius A East (a supernova remnant), Sagittarius A West (dust and gas clouds), and Sagittarius A*, or Sgr A*, which is a very bright and compact radio source believed to be formed by a supermassive BH. Sgr A* isn't actually the BH itself. I think the BH doesn't really have a name, so I'll call it $\mathrm{BH}_\mathrm{Sgr\,A^*}$. Maybe that's a bad name…


How to understand the meaning of this International Occultation Timing Association (IOTA) lunar occultation of Mars map? Earthsky.org's Watch Mars disappear behind the moon on February 18 as well as Moon occults Mars morning of February 18 include the image below and credit it to The International Occultation Timing Association or IOTA.

But I can't understand the meanings of the various colored lines on the map. There's the big blue droopy infinity sign, and attaching to both ends an upward-facing hyperbola with three colored segments.

I went to https://occultations.org/observing/occultation-predictions/ and clicked the Lunar Occultations link, but it only returns error 404 file not found.

Is it possible for someone to explain the meanings of these lines?

Unfortunately, there was a server problem in the days leading up to the exciting Mars occultation. (It didn't matter to me because it was, unfortunately, cloudy where I live.)

This description of the colored lines is from a temporary page provided by IOTA:

  The turquoise curves show where the disappearance or reappearance occurs at moonrise (left side) or moonset (right side); there is no northern limit as that misses the Earth to the north. The southern limit of the occultation, from which a partial occultation will be visible in a strip about 5 miles wide, crosses the n.e. Pacific Ocean and southern Mexico (white line, event at night), Central America (dark blue line that’s hardly visible, event during morning twilight), and northern S. America and the Atlantic Ocean (red dotted line, event during daylight).

If I am interpreting the image correctly, the limits are as follows (tl;dr 5: Entire occultation is visible)

No occultation is visible because the Moon has already set.
Mars disappears at moonset.
Mars disappears before moonset (so is visible), but reappearance occurs after moonset (so is not visible).
Mars disappears with the Moon above the horizon, but Mars reappears at moonset. (Entire occultation barely visible.)
Entire occultation is visible.
Mars disappears at moonrise.
Mars disappears before moonrise (so is not visible), but reappearance occurs after moonrise (so is visible).
Mars reappears at moonrise.
No occultation visible because the Moon has not risen.
Mars grazes southern limb of the Moon (occurs during daylight).
Mars grazes southern limb of the Moon (occurs during twilight).
Mars grazes southern limb of the Moon (occurs during night).

Sometimes the colors along the graze line (the southern limit in this case) represents whether the graze occurs on the dark limb of the Moon or the bright limb, but that does not appear to be the case here.


What is the apparent magnitude of the weakest stars we can see in one telescope on the ground with a diameter of 2.4m? All I know is its main mirror diameter; 2.40 m

I also know that the eye, which its pupil is 5mm in diameter can see up to an apparent magnitude of 6.

How can I calculate the apparent magnitude that the telescope can see?

I have used the equation of

$5⋅log(\frac{D_1}{D_0})$
However, it gave the answer of 13.40, while the answer should be 19.40

You just need to add the 13.4  that you got to the 6 that you can see without a telescope and that's it.  The formula is telling you the difference in magnitude that arises from the different apertures of the eye and the telescope.


Why don't all perseid meteors follow the same trajectory? Since these bits of debris crossing Earth's orbit have the same relative speed, why aren't they drawing parallel paths in the upper atmosphere? (consider one hour of observation to minimise Earth's rotation influence)

They are all following parallel paths. They are not on exactly the same trajectory but they are all travelling parallel to each to other in their orbit around the sun, in a stream of particles.

What you see is then multiple meteors on parallel paths. When you look in the direction that the meteors are coming from you see them appearing to radiate from a point

These meteors are all travelling parallel to each other. It is only perspective that makes them appear to be radiating from a point.

This is similar to the "flying through space" screensaver that used to be popular on windows.


What would the night sky look like if the interstellar medium didn't exist to absorb or block light? I was curious about how much brighter it would be and how much further we could see.

The main effect would be that the Milky Way would become much more prominent and asymmetric.

At the moment, our view into the Galactic plane is limited to around 1000-3000 parsecs by dust. If you look at the Galactic latitude distribution of naked eye (Aren't there more naked-eye-visible stars in the Milky Way plane? ) you see that most naked eye stars are much closer than this and there is only a modest concentration towards the Galactic plane.

What I think this means is that there would not be a big increase in the numbers of resolved naked eye stars. It is the sensitivity of our eyes, rather than dust that limits how far we can see them. However, the numbers of unresolved stars in the Milky Way would be greatly increased. The number of stars in the resolution element in your eye would increase as distance squared, but the light received decreases as distance squared. This means each "shell" you add contributes equally to the observed brightness.

The Sun is in a 30000pc diameter Galactic disc, about 8000pc from the centre. Thus I estimate that going from being able to see stars to 2000pc, seeing them to anywhere from around 7000pc (away from the Galactic centre) to 23000pc (through the Galactic centre) will increase the brightness of the Milky Way by factors of 3-10, strongly concentrated towards the Galactic plane (more so than now), depending on which way you look. This asymmetry will be increased by the increasing stellar density towards the Galactic centre.

In addition, the "bulge" would become more prominent. This is a pseudo-spherical region of diameter 4000pc, with a greater density of stars, centred on the Galactic centre. The bulge is mainly hidden by dust now, but without dust we would see a brighter, roughly circular blob towards the Galactic centre (Saggitarius), with an angular diameter of 30 degrees.

Some idea of the change in the visual appearance to the Milky Way can be gained by comparing an all-sky image at visible wavelengths (affected by dust) with a 2MASS infrared version (much less affected by dust). See how the "width" of the Milky Way becomes more concentrated and the distribution of stars becomes smoother, with a prominent bulge at the Galactic centre.


Why does Jupiter atmosphere does not get thinner on edges Looking at Jupiter image   the atmosphere edge of planet is very sharp as if it has solid surface which is not as it’s gases giant and there should be some thinning of atmosphere like we see around earth . Or there and could be seen on higher resolution images?

Why does Jupiter's atmosphere not get thinner at the edges?
Looking at an image of the atmosphere's edge it is very sharp as if it has a solid surface, which it has not since it’s a gas giant. There should be some thinning of the atmosphere like we see around Earth. Would this be seen on higher resolution images?

When you look at a spherical object from a distance it appears round, sometimes when you view it from close up it doesn't appear quite so smooth (or round). The wavelength that you look at an object with also affects what you are able to see.

$$\scriptsize \text{Figure 6. High resolution SEM image of CN ball bearing inner ring fabricated from 52100 steel showing} \\ \scriptsize \text{fine globular cementite. Presence of fine precipitates of carbides and nitrides uniformly aggregated in the} \\ \scriptsize \text{fine martensitic microstructure, which act as obstacles to the motion of dislocations.}$$
The above scanning electron microscope image of a smoothly ground ball bearing illustrates this, from a relatively far distance it appears smooth; from much closer the particles constituting its makeup become visible. Image Source: "Effect of Carbonitriding on Endurance Life of Ball Bearing Produced from SAE 52100 Bearing Steels", by Karthikeyan Rajan, Vinod Joshi, and Arindam Ghosh.
A similar principle applies to viewing Jupiter from so far away. While its clouds are far greater than 2000 km thick most of the gases are opaque thus not allowing you to see very far into the atmosphere using visible light.

The upper left image shows Jupiter from a distance using visible light. Source: "PJ14 Image 26 - Hot Spot (Enhanced)".

The upper right image  shows the different layers of Jupiter's atmosphere. Source: "How deep is the atmosphere?".

The lower left image from NASA's Juno spacecraft approaching Jupiter on Aug. 27, 2016, the Jovian Infrared Auroral Mapper (JIRAM) instrument captured the planet's glow in infrared light. Source: "Juno Captures Jupiter's Glow in Infrared Light".

The last two images are from: "NASA's Juno Mission Provides Infrared Tour of Jupiter's North Pole", there is also a close-up video flythrough of the atmosphere created from infrared imaging. A second video on that webpage, titled: "Jupiter's Dynamo", shows magnetometer imaging overlaid on visible spectrum imaging. The YouTube links for those videos are: "Low 3-D Flyover of Jupiter’s North Pole in Infrared" and "Jupiter's Dynamo".


Why aren't all planets in the same plane? Obviously all planets are not in same perfect plane. Because if at all they were in the exact same plane, it would mean that Mercury transits and Venus transits would not be that rare.

Any explanation would be much appreciated.

Your reasoning is correct: if Mercury orbited in the same plane as Earth, we'd see it transit the Sun every 4 months or so. In fact these orbital planes are inclined 7 degrees to each other, and the other major planets' orbits are inclined 1 to 3 degrees relative to Earth's.

The planets perturb each other's orbits slightly, so no planet's orbit is perfectly planar.
However, the Solar system average plane of all orbits and rotations is
invariable,
and most individual planets' orbits will remain near it for millions of years.


Limiting magnitudes for different telescopes I live in an extremely light polluted place. On a new moon night,  magnitude 3.5 stars are at the limit of  naked eye visiblity on clear nights, and the Orion nebula is visible as a smudge through my binoculars and I can see 6-7 stars of the Pleaidies (apparent magnitude about 5.5 I guess)
If I buy a 130 mm reflector telescope, roughly what will be the limiting magnitude of the objects I can see?
I used $5log(130/6.5)=6.5$, which means I would be able to see upto about magnitude 10 even from a light polluted city.
This sounds too good to be true...
Is my estimate roughly accurate and do you think spending a few hundred dollars on the telescope is worth it if I never intend to leave the city for a while?

Your calculated estimate may be about correct for the limiting magnitude of stars, but lots of what you might want to see through a telescope consists of extended objects-- galaxies, nebulae, and unresolved clusters.  The magnitude estimates of these objects tend to consider them as point sources, but as extended objects they have much less of a contrast difference with the sky.  In my experience in an equally light-polluted location with a similar-size telescope, the sky brightness makes the limiting magnitude of extended objects between 7 and 8, even with a light-pollution reduction filter in use.
That isn't to say a telescope is worthless in the city; you can still observe the Moon and planets, resolved clusters are pretty, and you can observer the brighter extended objects.  But bringing the telescope to a dark-sky location makes an incredible difference with many more objects visible.


Are there collected data about the direction of rotation of black holes and the direction of the magnetic field? Is there a relation between the direction of rotation of black holes or neutron stars and the magnetic dipole moment of BH or neutron stars? BTW, are there different directions of this two parameters for suns?

Edit

I want to be sure there are or there are not the direction of rotation (which one can expressed by an arrow on the rotation axis) and the direction of the magnetic field (expressed by an arrow from the south to the north pole (both arrows are simple conventions)) parallel only. I'm not sure that the direction of the magnetic field is observable at all. And the comment from Zibadawa Timmy about pulsars perhaps a good extension to my question about BH, neutron stars and Suns.

For pulsars the relation is complex. The magnetic field tends to "freeze in place" after the star collapses, but I'm sure there are all kinds of perturbation mechanisms that may change its initial orientation.

For blackholes (the Kerr-Newman metric, rotating and charged black hole) the relation is very simple: the rotation axis of the BH and the symmetry axis of its magnetic field always coincide.

https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric#Some_aspects_of_the_solution

There is a simple and intuitive justification, although perhaps not rigorous. If the magnetic field was not aligned with rotation, the BH would appear to be "blinking". But a BH cannot "blink", nor can it otherwise send any signals to the external universe. Therefore, its magnetic field axis must coincide with its rotation axis.


Why does the outer layer of the sun have differential rotation? I don't feel that way myself but some might find it really strange and wierd that it has differential rotation and feel that it's a contradiction. They may say "How is it possible?" I am asking on behalf of others who might have that question what assumptions they are making and how to break them.

I think they're not taking into account the Coriolis force in the frame of reference of the part of the outer layer of the sun at each latitude. I can't quite think of a proof that the outer layer of the sun would be predicted to have differential rotation. However, when you realize about the Coriolis force acting on the convection currents, you see that it does not give you a proof that there shouldn't be differential rotation. Maybe the part at one latitude then exerts a force on the part at an adjacent latitude until the rates of the two layers satisfy a certain equation.


How many nested stable (1 Mio years) orbits are theoretically possible? If we look at our solar system we have the sun which has planet orbits and planets have moon orbits.

If we look at the possible size difference from hypergiants to small asteroids why shouldn't it be possible to nest orbits to an astounding degree: supergiant orbit a hypergiant, star orbits supergiant and so on ?

My first guess is that the size and distance must be so small in contrast to the primary that the gravitational gradient of the primary does not cause too much disturbance, but it should be possible.

Has anyone tried to find it out with celestial mechanics ? While it may be completely theoretical, it is IMHO a solid question and it would be nice to find out how deep celestial bodies can be nested and how it limits stability.

Some conditions:

The orbits should be (very probably) stable for one million years.
Hypergiants are burning fast, so I limit the stability for a relative short timeframe.
The barycenter of both satellite and its primary are inside the primary.
No double system.
The first primary should be a star, not a black hole or a neutron star.

Yes, it's possible to have many nested levels, all stable. Each nested orbit needs to be inside the Hill sphere of its central body in order to be stable.

You can't nest any further when the Hill sphere gets too close to the Roche limit. Below the Roche limit the satellite disintegrates.

The exact number of nested levels depends very greatly on the characteristics of all bodies involved.


What exactly is meant by "expanding universe"? What exactly is meant when we say the universe is expanding? I understand stars and galaxies are getting farther and farther away from us, but are individual particles (atoms, molecules, anything found on Earth, etc.) drifting away from each other as well? If no, then why not? If yes, then shouldn't it, at some point, get to a point when particles can no longer bond with each other?

When we say the Universe is expanding, we really mean that space is expanding. The Universe could be infinite in size, but there continues to be more and more space between objects. Essentially, objects are moving away from each other because more and more space is being created between them.

In the early Universe, matter was much closer together than it is now; this density caused extreme temperatures and no hadronic matter could form. The first expansion is thought to have occurred because of how hot and energetic everything was, and spacetime itself was expanded in an event we call the Big Bang.

As objects spread out from each other during the inflationary epoch, the Universe began to cool down. Now, dark energy is the main culprit for its accelerating expansion.

To answer your question, yes, dark energy is making all particles move away from each other, although farther objects move away from each other at a faster rate. Why is that?

Well, the only thing holding all atoms, stars, planetary systems and galaxies together are the four fundamental interactions — the strong force, weak force, electromagnetic force, and gravity. In space, the main attracting interaction is gravity. As such, dark energy will cause farther objects (which are less affected by our gravity) to move away at a faster and faster rate.

Think of it like you are holding a dog with a leash, and the ground begins to spread out from below you. You and your dog are being moved away in opposite directions, but you both stay close together because of the strong leash.

Now, your mention of atoms being torn apart is actually the basis of the Big Rip hypothesis. It states that dark energy will become more and more abundant until even closeby objects separate from one another. First, galaxies will be disbanded as the stars move away from each other in all directions.

Much later, the gravity of planetary systems won't be enough to hold them together. Later still, planets and stars will be disbanded into their molecules. This will eventually continue as the amount of dark energy increases, until not even atoms can be held together by the electromagnetic force, and the subatomic particles will be moved apart.

By the time the expansion of the Universe exceeds lightspeed at the subatomic scale, no particle will ever be able to interact with one another. The Universe would dissolve into countless lonely particles that won't be able to do anything. Now, the Big Rip hypothesis is only one of the three most known fates of the Universe. Currently, the most accepted theory is the heat death of the Universe.


Why are most discovered exoplanets heavier than Earth? Looking at all discovered exoplanets (4393 exoplanets), I found than only 17 of them (less than one percent!) have masses less or equal to Earth's mass. Why so?

Is it because it is very difficult to discover an exoplanet of a low mass?
Is it because of the mass distribution, so that Earth-mass planets are very rare?
Is it because of the some other physical limitations?

According to Wikipedia:

The minimum mass/size required for an extrasolar object to be considered a planet should be the same as that used in our Solar System.

From another article:

A dwarf planet, by definition, is not massive enough to have gravitationally cleared its neighbouring region of planetesimals: it is not known quite how large a planet must be before it can effectively clear its neighbourhood, but one tenth of the Earth's mass is certainly sufficient.

So, where are all these planets that are lighter than Earth? Personally, I suspect that it's very difficult to detect these (relatively) low-mass planets. If so, are there any theoretical limitations that prevent formation of low-mass planets?
Note 1: most of the planets (around 70%) from the mentioned catalog do not have masses (i.e. there's no estimate for the mass of a planet). Most of the rest have $\sin i$ mass estimates. That might be one of the reasons.

There are a number of methods of detecting exoplanets, but all of them favour detection of larger planets over smaller ones, albeit for slightly different definitions of large:

Radial velocity measurement — this detects the small movement of the star towards and away from us as the planet and the star orbit their mutual barycenter. This movement is fastest when the planet is massive (so the barycenter is further from the center of the star) and close to the star (so the orbital velocity is highest). I also needs the planet's orbit not to be "face-on" to the Earth. This method produces measurements for the $mass\times \sin(i)$ since a more massive planet in a less inclined orbit produces the same motion as a less massive planet in a more inclined orbit

Transverse displacement — this detects the small movement of the star from side to side (against the background of distant stars) as the planet and the star orbit their mutual barycenter. The displacement is largest when the planet is massive and far from the star (although distant planet require observation over a long period of time).  It works best on stars close to us.

Transit — this detects the tiny reduction in the brightness of the star when the planet moves between us. It is more likely to detect large planets, and more likely to notice if the orbital period of the planet is fairly small.


How can the 13.8 billion years old universe have a radius of 46 billion light years? The universe is about 13.8 billion years old. Assume at the big-bang it starts from a small region and the maximum possible speed according Einstein is the speed of light how can the universe got a radius of 46 billion light years? According my understanding it could have a max. radius of 13.8 billion light years only. Can someone explain to a non-astrophysicists.

You assumption is wrong. Universe can( and is ) expand faster than the speed of light.

The photon emitted towards our planet in the early universe, had to traverse a universe that is expanding. That photon experienced redshift, that means his wavelenght increased and frequency decreased. We measure that photon's wavelength, compare it to a photon that was not redshifted and calculate the scale, which is then applied to the radius of the observable universe.

Reading this Wikipedia page should give you more complete understanding of the problem.

http://en.wikipedia.org/wiki/Observable_universe


Why does the sensitivity to GWs drops off inversely proportional to the distance? This answer makes me wonder why the sensitivity to gravitational waves decreases proportionally to the distance.

Since gravitational waves extend in all directions, my (uneducated) guess would be that the same argument can be made as for the decrease in sensitivity for electromagnetic waves. Why is this not the case? Why does the sensitivity to GWs decrease in a linear fashion?

EDIT I'm leaving the original, highly upvoted answer below, but I've had a fundamental rethink about this, prompted by questions from Keith McClary and a helpful clarification from a Physics SE question.

The original answer I gave is the reason that we can detect gravitational waves (GWs) at all. Their coherent nature as single oscillators, means that despite their comparatively low powers, they can be detected right across the universe. In comparison, electromagnetic sources are usually the superposed light from countless incoherent emission sources. On average this has a destructive interference effect that reduces the power (intensity) received; and due to the rapidly changing signal, it is usually intensity that is measured.

However, the answer to the present question is actually just how "sensitivity" is defined.

In order to detect a source, we must identify it in a background of noise. This is done by defining a signal-to-noise ratio. The signal is the product of your source strength (more on that in a minute) and how long you observed it for. The noise is a property of your instrument. The sensitivity of the instrument then is something like the minimum (signal $\times$ observation time) that will produce a significant detection.

In astronomy it is conventional to express the signal in terms of power received since, due to the arguments given above and in the original answer, power (intensity) is generally what is measured. The sources of noise are therefore also defined in terms of a power and the sensivity has units of something like Watts $\times$ seconds, or more conventionally, W/Hz.

In gravitational wave astronomy, because it is the amplitude that is directly detected, GW astronomers express their source signal in terms of amplitude (which is proportional to the square root of detected power) and their sensitivities are expressed in terms of gravitational wave amplitude (which is dimensionless) divided by $\sqrt{Hz}$ as a result.

i.e. we are not comparing like with like. Doubling the sensitivity of a gravitational wave detector is actually like quadrupling the sensivitity of an electromagnetic wave detector. Thus there is no fundamental difference here, the apparent difference in behaviour is merely a result of how sensitivity is defined. The reason for the different definition is as per my original answer below.

Original answer

The difference is that usually when we detect sources of electromagnetic waves, we are detecting intensity, which obeys the inverse square law.

In contrast, we are detecting the amplitude of gravitational waves, and amplitude only scales as the inverse of distance

Why the difference? Sources of gravitational waves are coherent oscillators. A  merging binary produces a single coherent wave train with an amplitude that can be defined and measured. By contrast, when we look at a distant star or galaxy in electromagnetic waves we are seeing the incoherent contribution from countless accelerating particles and atoms and all we can detect is the resultant summed intensity. There is no coherent electromagnetic wave with an amplitude that can be measured.

This difference in behaviour is fundamentally because whilst there are positive and negative electric charges, which require contrived circumstances in which to behave coherently (e.g. in a laser), gravitational waves are produced by accelerating masses, and since there is only one sign of "gravitational charge", the individual parts of a gravitational wave source are able to act in concert quite naturally to produce a coherent waveform that has a wavelength larger than the body itself.

An excellent discussion of these points can be found on the first page of the review article by Hendry & Woan (2007).

In principle, if we were looking at a single coherent source of electromagnetic waves then we can detect the amplitude (for example by the force it exerts on charged particles), and then the sensitivity would just reduce as the inverse of distance. At optical frequencies the electric field varies so rapidly that this cannot be done, but it is possible at radio frequencies. Unfortunately the coherence length and coherence time (the time over which the phase of the wave is predictable) are so short that this is rarely practical in laboratory, let alone astronomical, sources.


Does one need to take into account finite gravity speed in N-body simulations? Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use  classical Newtonian gravity.

Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.

My question: if I were to make a large-scale N-body simulation using Newtonian gravity, would it be enough to "correct" that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?

If you're asking whether it's sufficient to use a retarded (time-delayed) positions to calculate gravitational forces, then no, that would be much worse than Newtonian gravity. For example, that would predict that the Earth should spiral into the Sun on the order of about 400 years. See also answers to

Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch?
Is the influence of gravity instantaneous?

  Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use classical Newtonian gravity.

This is true, but some simulations use relativistic corrections.

The post-Newtonian expansion works in order $\epsilon\sim v^2\sim U$, where $U$ is (the negative of) Newtonian potential, and including the first-order post-Newtonian correction to the N-body equations of motion forms the Einstein–Infeld–Hoffmann equations. Because they contain $3$-body terms, the simulation would now be $\mathcal{O}(N^3)$ instead of the Newtonian simulation $\mathcal{O}(N^2)$.

The post-Newtonian three-body equations are known to 2PN, while the two-body problem to 3.5PN. Effects beyond the first order can be important in e.g. binary simulations because gravitational radiation losses are seen at 2.5PN order. However, they are not translatable to an N-body problem.

A technique of making the EIH equations more tractable by making a second expansion that depends on the problem at hand can be found in Will (2014), which is applicable when most the relativistic dynamics are dominated by a few contribution (e.g., a supermassive central black hole in galaxy simulations).

  Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.

As Walter says, most of them use Newtonian gravity on a expanding background. Not necessarily directly computer particle-particle Newtonian force, but perhaps modeling Poisson's equation through mesh methods, or perhaps some combination thereof (e.g., direct force calculations for close particles, mesh methods for larger scales).

  My question: if I were to make a large-scale N-body simulation using Newtonian gravity, would it be enough to "correct" that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?

If you discount cosmological expansion, it will simply be wrong. However, first-order corrections to the Newtonian gravity part of simulations do exist in the post-Friedmann formalism. I'm not really familiar with PFF, but according to Bruni et al. (2014), the leading-order correction to Newtonian gravity introduces a vector potential in the $g_{0i}$ term in the metric that can produce weak lensing effects but doesn't affect matter dynamics. Which is entirely sensible because that kind of gravitomagnetic effects should be suppressed by a further $v/c$ term for matter, just like magnetism.

For references on how low order GR corrections to Newtonian gravity simulations are often made, see answers to How to calculate the planets and moons beyond Newtons's gravitational force?

References:

Will, C. M. "Incorporating post-Newtonian effects in N-body dynamics". Phys. Rev. D 89, 044043 (2014) [arXiv:1312.1289]
Bruni, M., Thomas, D. B., Wands, D. "Computing general-relativistic effects from Newtonian N-body simulations: Frame dragging in the post-Friedmann approach." Phys. Rev. D 89, 044010 (2014) [arXiv:1306.1562]


Galaxy discovered lacking Dark Matter Recently a Galaxy was discovered that contained no Dark Matter. I was wondering what explanation could be offered for this ? This Galaxy also contains only 1% as many Stars as our Milky Way does . Is there be a connection between these two things, lack of many Stars and lack of Dark Matter ?

The paper describing this finding is van Dokkum et al. (2018). In a galaxy, the ratio of the stellar mass $M_\star$ to dark matter mass $M_\mathrm{DM}$ is normally very small, increasing with mass until Milky Way-sized galaxies where it reaches $\sim1/30$, then decreasing again.
But the dwarf galaxy NGC1052–DF2, seems to contain little or no dark matter. Galaxies of that mass ($M_\star\sim2\times10^8\,M_\odot$) typically has several hundred times more DM than stars.

How the galaxy was formed is not known, but the authors speculate on a few different mechanisms that all have to do with the fact that gas, in contrast to DM, may cool and condense and thus form clouds of very high $M_\mathrm{gas}/M_\mathrm{DM}$ ratios:

NGC1052–DF2 is located neat an large elliptical galaxy (NGC 1052) which could have gone through a merging event, tidally stripping a chunk of gas from one of the mergers. This is consistent with NGC1052–DF2 having a large velocity wrt. the elliptical.
NGC1052–DF2 could have formed from low-metallicity gas that was swept up in quasar winds (as described in Natarajan et al. 1998).
Lastly, NGC1052–DF2 could have formed from the fragmentation of gas accreting onto the elliptical, possibly aided by shocks.

EDIT (thanks to @WayfaringStranger): A fourth possibility is that the authors have misinterpreted the data. Shortly after the paper was put out, several other papers criticized the statistical methods used by van Dokkum et al. to infer their result (Martin et al. 2018;
Famaey et al. 2018;
Laporte et al. 2018).
van Dokkum wrote in a very long reply on his blog how at least the first of these papers actually confirm, rather than refute, his results. I am not enough of a statistician to comment on who's right and who's wrong, but note that there is currently an ongoing debate on the Facebook group astrostatistics.

You also ask whether the lack of DM is related to its "lack of stars". I wouldn't say that NGC1052–DF2 "lacks stars", anymore than the Milky Way "has too many stars" — it's just a small galaxy. But in general, the smaller a galaxy is, the larger the scatter in the $M_\star/M_\mathrm{DM}$ ratio is. Small galaxies, or small clumps of gas, have shallow gravitational potentials, so a small galaxy can more easily get stripped of gas, and a small chunk of gas can more easily escape a galaxy, or accretion stream, without attracting dark matter. In contrast, if would be very hard to conceive a very large galaxy having a considerably different $M_\star/M_\mathrm{DM}$ ratio, and indeed the scatter for massive galaxies is less than a factor of two (e.g. More et al. 2010).


Does the Perseid peak cause enough light pollution to be a problem? The annual Perseid peak occurs between August 9-14. During this time, about one meteor burns up in the atmosphere per minute.

When these meteorites burn up, they generate light. When you have one of these per minute, does it produce enough light pollution to be a problem for those making astronomical observations?

Tought to answer, because the amount of airborne dust due to meteor showers and its stability in the upper atmosphere will vary significantly enough. We had many meteor showers, Perseids included, in the past that were a bit of a letdown and didn't produce as many shooting stars as initially predicted, and also the other way around of course.

But here's the thing tho, that discounting for any additional atmospheric glow during a few crucial hours of dusk and dawn, this airborne meteor dust might even improve the average atmospheric refraction with mostly extremely fine powder silicate dust settling onto the lower atmosphere and dispersing water molecules and binding to other solids (read: decrease humidity and pollutants), by being highly electromagnetically charged from the atmospheric reentry and electrostatically binding to larger atmospheric or pollution molecules, or simply for being hygroscopic and binding really good to water molecules, making them heavier enough to fall in a light spray back on the ground and remove the pollutants from the lower atmosphere in the process, too.

There might be other physical phenomena going on that I'm not aware of, but what I wanted to illustrate is, that there might even be positive side-effects in terms of light pollution vs increase in meteors, which is how we call meteoroids entering Earth's atmosphere.


What exactly is a stellar association? What exactly is a "stellar association" in the strict kinematic sense?

A stellar association is a loose cluster of stars, that formed at the same time from the same molecular cloud, and so have the same proper motion. Unlike open clusters, they are not gravitationally bound together, so the stars in a stellar association will gradually separate, forming a moving group of stars.

An example is the Scorpius–Centaurus Association


Before the 1761 transit, what was our best estimate of the distance to the Sun? In 1761, many expeditions were launched to determine the distance to the Sun using parallax during the transit of Venus. Prior to the 1761 transit, what was the best estimate for the Earth-Sun distance?
In antiquity, Aristarchus found that the Earth-Sun distance was 20 times the Earth-Moon distance, which is a very gross underestimate. Were there any better measurements than Aristarchus' measurement in the early 18th century?

Aristaurus gives a calculation which would result in a value for the Earth-Sun distance  of about 0.016-0.065AU, clearly that's a long way out. The true value is 1.00 AU
Archimedes gives a value that would correspond to 0.43AU, other Ancient authors gave a range of values, all less than 0.5AU and many were close to Aristaurus's value.  So among ancient writers, all we can really say is that "They had no idea how far the sun was".
Kepler and his laws of planetary motion could show that values of 0.02au were far too small (since then a very significant parallax would be visible in venus) but couldn't say with precision how far out.
In the 17th century, telescopic measurements of the Earth-Sun-Moon angle allowed for Aristaurus's method to be used with greater precision and yielded values of 0.5 or more
Huygens made a number of assumptions on the size of the planets and came up with a very good value of 1.068AU,  but his reasoning was faulty.  This value is more luck than good maths. However, Cassini used observations of Martian parallax (and a good estimate of the size of the Earth) to get a well-calcuated value of 0.925AU
This was the best estimate prior to the measurement using transit of Venus.


On (minuscule) dark matter production in supernovae It is believed that dark matter is made of particles, which interact with matter only weakly and gravitationally. One common candidate for dark matter are so called WIMPs. WIMPs, specifically, are heavy and may be their own antiparticles.

And as any other particles dark matter particles can be produced at sufficiently high energies. The mass of dark matter particles is unknown, but is estimated to be of order $1$-$100 \textrm{GeV}$, which corresponds to temperatures of $T_{DM}\approx 10^{13}$-$10^{15}\textrm{K}$, at which these particles may be expected to be produced.

Such enormous temperatures are barely attainable in any reasonable astrophysical processes, but say in core-collapse supernovae newly formed core has temperatures of $T_{SN,after}\approx 10^{11}\textrm{K}$, and probably more during the collapse phase. Then a crude estimate would suggest that the amount of dark matter produced is $M_{DM}\approx e^{-T_{DM}/T_{SN,max}}M_\odot$. Or, in number form $\log_{10}(M_{DM}/\textrm{kg})=30.3-0.43(T_{DM}/T_{SN})$. This means that at $T_{SN}=1.4\cdot 10^{-2}T_{DM}$ the amount of dark matter produced during a supernova will be around one kilogram. Such temperatures are fairly reachable for $1 \textrm{GeV}$ DM particles. So one can optimistically expect few kilograms of dark matter produced per supernova.

Now the question. What is a typical dark matter production in core-collapse supernovae? A good answer, I imagine, would be a more robust expansion on the existing estimate. Any constructive comments are welcome.

The most favoured WIMPS at the moment are probably neutralinos, see
http://en.wikipedia.org/wiki/Neutralino

These particles are purely hypothetical at the moment.
The mass estimates in the above Wikipedia article for the lightest neutralino range between
10 and 10,000 GeV, meaning that the production rates in SNs will be much lower than with an assumed 1 GeV.
Higher production rates should have already been detected at LHC.

Hence from the non-detection (in the form of energy loss) of WIMPS at LHC an estimate of an upper bound of the production rates in SNs should be possible.


Do we still use the term "astronomical unit" nowadays? An astronomical unit is defined as the measurement of distance between Earth and our Sun, my question is since distances between celestial objects beyond our solar system are vast and unimaginable astronomer adopted another familiar term "light-year" instead. My question is do people still use astronomical unit in the present days?

Certainly. Astronomical Unit is probably one of the most used distance units used in astronomy. It is of course only used when discussing the distances within a stellar system, such as the distances between the Sun and its planets or other bodies in the solar system. It is also used to discuss distances in other stellar systems, e.g. the distances between stars and their (exo)planets.

It is easier to get a sense of the distances between bodies in the solar system if you use astronomical units. It doesn't say much when we say that the distance between Neptune and the Sun is $5\times10^{-4}$ light year. You would have to know the distance between Earth and the Sun ($1.5\times10^{-5}$ light year), and then calculate the difference (i.e. 30 times = 30AU) to get a sense of scale.

As a side note, in professional astronomy, the light year is hardly ever used. Professional astronomers use the parsec (=3.26 light year), kilo parsec (kpc), megaparsec (Mpc), and Gigaparsec (Gpc) to specify distances between stars, galaxies, and galaxy clusters.


Why don't small stars end up as a black holes? I have recently done some research into black holes, and realized that big stars form black holes, whilst smaller ones don't. Is this because the gravity isn't strong enough for it to fall in on itself, or something else?

The actual answer has nothing at all to do with the temperature.

Even low-mass stars would form black holes if they ran out of nuclear fuel to burn, and simply cooled whilst being supported by "standard" gas pressure in their centres.

That is because that gas pressure would be proportional to the temperature, but the star is able to cool so it would need to shrink and use gravitational energy to stay hot, but of course eventually it would disappear inside its event horizon and become a black hole.

The real reason this does not occur is electron degeneracy pressure. This is a quantum mechanical effect, related to the Pauli exclusion principle, that does not allow two electrons to occupy the same quantum state. As the gas gets squashed in the core, the electrons are forced to fill higher and higher momentum states in order to "avoid" each other. Because these electrons have large momentum they also exert a large pressure - degeneracy pressure. This degeneracy pressure does not depend on temperature and so the star (it is a white dwarf when a low-mass star arrives at this stage) can cool and remain in hydrostatic equilibrium without shrinking any further. Thus it avoids becoming a black hole (or even a neutron star) and simply cools at nearly constant radius.


Why is the moon's motion more complicated than that of those planets? I am of course not an expert on astronomy. But from the literature, I have the impression that the lunar motion is very complicated. It seems that Kepler can make sense of the motion of the planets but cannot do that for the moon.

So, why is the moon's motion more complicated than that of planets like Mars?

Keplerian motion would be a correct way to explain the Moon's orbit about the Earth if the Sun, Venus, Jupiter, and all the other planets weren't present, if the Moon wasn't slowly receding from the Earth due to tidal acceleration, and if Newtonian gravity properly described the underlying physics. None of these is the case.

In the case of modeling the behaviors of the major objects that comprise the solar system (i.e., the Sun and the eight planets), perturbations involving planets interacting with one another, particularly with Jupiter, make the orbits of the planets slightly non-Keplerian. That the universe is relativistic rather than Newtonian famously makes Mercury's orbit precess in a way that even accounting for planetary perturbations cannot explain. That said, Keplerian motion is a fairly good approximation with regard to the Sun and the eight planets.

The same is not true for the Moon. It's orbit is notably non-Keplerian, even over the course of just a month. Ptolemy (after the fact) and Brahe (also after the fact) both noticed irregularities in the Moon's motion that cannot be explained in terms of Keplerian elliptical orbits in which area swept out per unit time is constant.

Newton explained some of these effects by noticing that the Sun is a huge perturber of the Moon's orbit about the Earth. Newton was not quite successful in this endeavor; the mathematical tools needed to model the Moon's motion hadn't been developed. The full development of classical lunar theory took two hundred years after Newton's time.

Another complicating factor is that the Moon is gradually receding from the Earth due to tidal acceleration. That the Moon is slowly receding from the Earth was first hypothesized by Halley, one of Newton's contemporaries. A century after the first publication of Newton's Principia, Laplace erroneously attributed this lunar recession solely to perturbations from the Sun. In the next hundred years, others corrected Laplace's calculations and found that solar perturbations are unable to explain this recession. George Darwin (Charles' son) explained the recession in terms of the tides. The full explanation? People are still writing papers on this subject.

Relativity makes for yet another complication. Relativistic effects on the Moon are very hard to see given the many orders of magnitude larger complications due to perturbations by the Sun, Venus, Jupiter, ..., and tidal acceleration.


Rings around a smaller, close-in planet? Is it possible for a small, close-to-star planet to have a stable ring system--icy or otherwise?

A smaller planet means the rings would have to be closer to the planet, which means the radial shear (difference in orbital speed vs. radius) in the rings would be higher. A higher radial sheer would be more likely to cause turbulence...or so I would think.

Obviously one would not expect icy rings around a planet that's inside the frost line....

I think the related answer that James Kilginger provided is very good and worth a read.  I'll touch on a few points related to your question.

There's 2 parts to this, how do ring systems form and where are they more likely to be stable and long lasting.   I doubt that any ring system would be permanent, but some should be much longer lasting than others.

So, lets talk stability first.

As you pointed out, as ices melt inside the frost line (somewhere around Ceres, though different ices have different melting points, but in general, around Jupiter distance or greater ice particles are largely stable and can be part of a ring system (I'm not sure about Jupiter's gonzo magnetosphere though).   Closer to the sun, certainly by the time you get to Mars, probably a bit further out than that, solar radiation melts ices and it turns into gas which the solar wind blows away from ring systems pretty quickly.   What's more, any smaller rocky debris mixed with ice could experience spin or acceleration as ice is melted and escapes from one side as gas.   Those local accelerations could blow some material from a ring system too.

Smaller dust particles in a ring system near a star would get blown away too by the Poynting-Robertson effect which is much more significant closer to a star.   Some of these problems would be avoided if the star is in it's white dwarf stage and quite cold, but a main sequence star is much more friendly to distant ring systems than closer ones.

Ring systems are only stable inside the Roche Limit.  That's covered in the other question and outside the Roche Limit, the material would tend to reform into a moon.   Solid objects however can orbit inside a liquid or fine particle Roche limit and maintain their shape.

Planets close to the sun still have Roche limits usually well inside their Hill Spheres, but long term stability is still a problem for planets close to their stars.   Here's an article that touches on this regarding Mercury and one on Venus.  The same principal would apply to ring systems where the significantly larger solar tide would have a destabilizing effect on all objects in orbit around a planet close to it's star.   It wouldn't destabilize immediately but over many orbits, objects could either spiral into the planet or outside the Roche limit, so Ring systems would be significantly shorter lived as planets get closer to their star.  The tidal force grows with the square of the distance.  Earth, for example experiences 90 times the tidal force from the Sun that Saturn experiences.

Saturn is also about 95 times the mass of the Earth, so it's not hard to see why Saturn is a much better candidate for a long term ring system than one of the inner planets.   Much more mass and much further form the sun.

Formation of ring systems can happen by 3 methods, (that I can think of).  Large collision, break-up of a moon that travels inside the Roche limit or accretion.   The collision can be on a planet's surface where debris is blown into orbit or similar high speed impact on a moon in orbit around the planet.    Collisions of this size are rare, but likely more common in young solar systems.

A moon that moves too close to a planet can get broken up and form into a ring system if it travels inside the Roche limit see pictures top right.   There's no reason why that couldn't happen from time to time on a number of planets.   It might one day happen to Phobos as it spirals in closer to Mars.   Moons tend to move towards or away from planets quite slowly so this doesn't happen often, but it can happen from time to time, even when solar-systems are billions of years old and mostly settled into stable orbits.

Finally, accretion, Enceladus see here is feeding Saturn's e-ring and may have fed other rings in the past.   Enceladus is unique among all the moons in our solar-system in that it's able to steadily feed a ring system.   This is possible because it's quite small, so volcanoes on it's surface are able to eject material at greater than it's escape velocity and because it probably has a liquid water-like layer under it's surface, maintained by tidal heating because it's close enough to Saturn to experience enormous tidal forces.   Enceladus is likely responsible for only a tiny part of Saturn's rings, but this is still a viable method for ring formation.

Europa on Jupiter experiences similar consistent volcanism due to large tidal forces that also keep some of it's inner layers liquid, but Europa's a much larger moon, so far less material gets ejected at or above escape velocity than with Enceladus.   Only small moons close to large planets can supply material to a ring system in this way.

But, because moons may play a role in ring formation and Large planets are more likely to have more moons due to greater gravity and Planets far away from their star are also more likely to capture more moons because they have larger Hill Spheres and objects don't orbit the sun quite as fast, that's probably the gist of it.

Large planets are more likely to have moons and rings than small planets and further away from the sun, a planet is more likely to maintain a ring system then close to the sun.    It's not impossible that planets close to a star or smaller planets could have a ring system, but quite a bit less likely.

  A smaller planet means the rings would have to be closer to the
  planet, which means the radial shear (difference in orbital speed vs.
  radius) in the rings would be higher. A higher radial sheer would be
  more likely to cause turbulence...or so I would think.

High orbital speed, I suspect, makes little difference, in fact, a larger planet would have higher orbital speed in it's rings than a smaller one.  A heavy Jupiter, 8 or 10 Jupiter masses and 2 or 3 times denser than the Earth could have a large orbital region inside it's Roche limit and potentially, very large, very fast orbiting rings.

Hope that wasn't too long.


Can we draw a min(distance) and max(mass) relation based on current observation data of nearby astronomical objects within several parcsecs? The data precision now can exclude a >>$1M_{\odot}$ black hole companion, which has a, say 10000 year orbital period with our sun?

Can we draw a two dimensional diagram(Mass VS. distance) to show the region that is still a grey area? Within 1 pc, there is no stellar mass black hole? If there is one, the orbit change of objects in solar system may be detectable.

How to calculate its proper motion or its planets' proper motion?

I am serious to ask. Please give some reference/technical method to calculate PM. For example, is there some reference paper about the distance limitation of the black hole based on the motion of asteroids/comets?

First lets make sure we know the scale of the solar system. Neptune, the most distant major planet, has an orbit of 165 years. Sedna, the object with the longest known period, has an orbit of 11400 years.

Black holes come in different sizes. A black hole that has formed from the collapse of a giant star would have a mass more than three times the mass of the sun. If a black hole of this size were inside the orbit of Neptune, it would not be orbiting the sun, the sun would be orbiting it.

How about further out? Blackholes can be spotted in X-rays when matter falls into them forming a disc of matter orbiting at very high speeds, and heating up to enormous temperatures.  A black hole without an accretion disc would be very dark and hard to detect from a distance. But we can be reasonably confident that there are no in our immediate neighbourhood.

The motion of a blackhole in orbit with the sun would be described by Newtons laws of gravity. The sun and the black hole would orbit on elliptical orbits around their common centre of gravity. Planets of the sun or the black hole would be stable within their Hall radii. Close to the blackhole one would need General relativity to describe the gravitational field, but from further out, its gravity would be almost exactly as predicted by Newton. You could model it in any gravity simulator.

It is possible that micro black-holes exist in the universe. These could pass through the earth without doing much damage, as the are sub-atomic size and since they interact only with matter by gravity, and don't weigh much, they don't have much effect. They might be "hot" with Hawking radiation.

None of these are much of a risk to the survival of the solar system. Black holes are rare.


Why does the sun have a convection layer and a radiation layer? Why is the outer section of the sun convective but the inner section stable and only radiative? They are both made up of the same kind of matter, both heated by the core.

The onset of convection is controlled by temperature gradients (see the famous experiment from Rayleigh & Bénard. When those become too strong, convection sets in.

How strong a temperature gradient gets, depends on the local efficiency of energy transport inside the star. But for the sun (and other lower mass stars), the surface radiates energy away efficiently, dropping the temperature there compared to the rest of the star, making the outer layers convective. Lower temperatures (not gradients) compared to high-mass stars also play a role in controlling the energy transport for lower mass stars. Cool temperatures will allow the formation of $H^{-}$ (negatively charged hydrogen ions) that give high opacities and thus prohibit energy transport.

For high-mass stars it is the enormously fast fusion processes in the core that 'overheat' this region, again letting convection take over.


Could Eris have a permanent atmosphere? In Wikipedia they write that Eris has an atmosphere when around perihelion (close to the Sun) which collapses when Eris moves away from the Sun. So it behaves similar to a comet's tail. However, they thought the same of Pluto's atmosphere, and the information on Wikipedia is from prior to the New Horizons Pluto flyby which proved that Pluto has a permanent atmosphere.

Could they be wrong on Eris too or does Eris really go far enough from the Sun so that its atmosphere collapses? It is possible that they are right with Eris for its aphelion is about twice as far as Pluto's, but are we sure about its atmosphere?

As far as I'm aware, there's no direct evidence of an atmosphere at the current time. The abstract of Sicardy et al. (2011) "A Pluto-like radius and a high albedo for the dwarf planet Eris from an occultation" gives a limit of ~1 nanobar on the presence of methane, argon or nitrogen atmospheres during the stellar occultation observed in November 2010.

Hofgartner et al. (2019) "Ongoing resurfacing of KBO Eris by volatile transport in local, collisional, sublimation atmosphere regime" argue that at aphelion there would be a local atmosphere over the warmest region, which they state is a similar situation to the atmosphere of Io. They predict that this local atmosphere could still have significant effects on the volatile distribution on Eris:

  The model results indicate that volatile transport (VT) on Eris, even at
  its aphelion distance of nearly 100 AU, can be significant. The nitrogen-ice
  temperatures are < 30 K and vapor pressures are < 10 nbar but the significant pressure gradients in the local atmosphere regime result in transport of
  nitrogen mass, that integrated over the long timescales associated with such
  a distant orbit, can be significant as compared to the column mass of the
  atmospheres of Triton and Pluto.

So depending on where you are on Eris, the atmosphere (if there is one) may or may not last for the entire orbit. It will take a long time to confirm this, though.


Why does M101 search string have RA variable from 0 to 360 and Dec variable from -90 to +90 in the Hubble Legacy Archive? I was scraping some images off the Hubble Legacy Archive and I found that searching M101 with RA in range 0-360 and Dec in range -90 to +90 yields results. Now I understand that Messier 101 galaxy has the co-ordinates RA 14h 3m 13s | Dec +54° 20′ 57″.
So what images am I getting if HLA has images for M101  RA = 52.000000  Dec = -13.000000  r = 1  [03:28:00.000 -13:00:00.00]
Could someone explain what's going on?

I think that somehow that string “M101” got inserted into the search box but has nothing to do with your query results.   For example, if you click the “M101” link right below the box it will insert that into the search box without changing your search results until you click “Search”.
The images shown in your search output are labeled with their object names, RCS0327-1326.
If you’re still getting odd results, the editing your question to show your full URL from your query would be helpful.


If the axial tilt of Uranus (97°) was caused by a giant impact, then why do its moons orbit around its equator at the same tilt (≈97°)? So the most accepted explanation for Uranus' axial tilt is that it once had a normal tilt like the other planets bar Venus (0° to 30°), but then a certain massive object collided with it and flipped it on its side.

But by that logic then shouldn't Uranus' moons be orbiting 0° to 30° relative to the orbital plane? (or whatever mild axial tilt Uranus had before the impact)

If they were formed before when Uranus still had a mild axial tilt then it makes no sense for them to be orbiting directly over present-day Uranus' equator.

The obliquity of Uranus is a tricky one to explain, the issue with keeping the satellites in the equatorial plane is one of the problems. Another issue is that giant impacts would be expected to randomise the rotation, yet Uranus has ended up with a rotation period similar to Neptune despite its tilt.

As noted by Morbidelli et al. (2012), tilting Uranus from zero obliquity with a single giant impact runs into the problem that while the circumplanetary disc can realign with the planet's equatorial bulge after the impact, the sense of its rotation would be retrograde. This does not match the present-day system, where the satellites are prograde.

Rogoszinski & Hamilton (2020) investigate the possibility that Uranus was tilted via resonances between its spin, orbit and the other giant planets. This is the mechanism that appears to be responsible for Saturn's obliquity, via a resonance with Neptune. The process would likely be gradual enough to bring the satellites along with it as they tend to align with the equatorial bulge, while avoiding the rotation period randomisation of a giant impact. In the case of Uranus they find that with the right initial conditions, Uranus can be tilted to about 70°, requiring a much smaller impact to push it into the current configuration, so you could avoid having to reverse the sense of rotation of the circumplanetary disc. This has the further advantage that there were likely to be more Mars-sized objects than Earth/super-Earth sized impactors, so such collisions are less improbable.

In their follow-up study, they investigate whether the right initial conditions are compatible with the Nice model for the formation of the Solar System. Unfortunately getting to the current situation through the spin-orbit resonance mechanism appears to take too long.

Currently there does not seem to be a particularly satisfactory scenario to explain the rotation of Uranus, but it does appear that the process of tilting the planet was more complex than a single giant impact, and the satellite system is one of the indications that the simple scenario is unlikely to be correct.


Are exoplanets which are above escape velocity of particular interest? I was perusing the exoplanet database and looking for some of the more extreme orbits, when I happened upon HD 20782 B:

http://exoplanets.org/detail/HD_20782_b

Which is also listed (in two sentences) on wikipedia, so there's not a whole lot of information. However, the first thing I noticed was that it was almost at escape velocity!

I guess not much is known, but my overall question is:

Would an exoplanet on an escape trajectory offer us any opportunities for research that have yet to be performed? What would those opportunities be? The answer can be a justified no (E.G. once the planet leaves the host star, we can't observe the doppler shift anymore).

Imaginary Bonus Points: have there ever been any observed as escaping/becoming rogue planets?

The problem is that planet ejection is something that takes place over a very short timescale compared to the ages of the systems in question, so we're unlikely to see it actually take place. The main prospect of detection would be imaging of the ejected planets, for which you'd want a young giant planet as these retain enough heat from their formation to be sufficiently luminous in the infrared for detection even when not illuminated by a nearby star.

There has actually been a claim for the detection of a (proto)planet in the process of being ejected from a binary star system: the case of TMR-1C as reported by Terebey et al. (1998) "A Candidate Protoplanet in the Taurus Star-forming Region". Unfortunately the nature of the object is not confirmed, see Terebey et al. (2000) "The spectrum of TMR-1C is consistent with a background star". There has been some back-and-forth on this: Riaz & Martín (2011) "Large-amplitude photometric variability of the candidate protoplanet TMR-1C
" reject the background star hypothesis, while Riaz et al. (2013) "A near-infrared variability campaign of TMR-1: New light on the nature of the candidate protoplanet TMR-1C" regard it as a strong candidate for being a young stellar object (YSO).

Another relevant case might be the circumbinary planet HD 106906 (AB) b, which may have been saved from ejection by a stellar flyby, see De Rosa & Kalas (2019) "A Near-coplanar Stellar Flyby of the Planet Host Star HD 106906".

Ejected planets are of interest due to their contribution to the population of low-mass objects in star clusters and the field (this population may well contain both ejected planets and low-mass brown dwarfs), and the impacts on the orbital architecture of the surviving planets after episodes of planet-planet scattering. One example of the latter I found is Carrera et al. (2019) "Planet-planet scattering as the source of the highest eccentricity exoplanets", there are many other papers on this topic.