28
$\begingroup$

Is there any stable orbit around a black hole so that the spacecraft will remain in orbit without any disturbance over a long period of time ?

$\endgroup$
  • 3
    $\begingroup$ space.stackexchange.com/questions/1909/… writes to say > ... It's perfectly safe to orbit a black hole - as long as you don't cross the event horizon, you're fine ... $\endgroup$ – Everyone Sep 12 '13 at 11:04
  • 2
    $\begingroup$ A quick reminder that our Sun, and thus our entire solar system, orbits around a black-hole right now and we have a stable orbit. $\endgroup$ – EtherDragon Sep 12 '13 at 21:50
  • $\begingroup$ @EtherDragon - I believe I've already mentioned that in my answer several hours before your comment. And indeed, the question is not specifically asking for close proximity orbits, so they can be considered stable. $\endgroup$ – TildalWave Sep 12 '13 at 22:27
  • 1
    $\begingroup$ Can I add to this question? Or should I ask this on another question? All answers seem to say that yes, there is a stable orbit. BUT: considering that a black hole is that huge ever-hungry monster, won't it (given time) accumulate more matter and mass and increase in size, eventually turning what was supposed to be a stable orbit into an orbit of "decay"? $\endgroup$ – msb Sep 13 '13 at 21:51
  • 1
    $\begingroup$ Black Hole Orbits: Zoom-Whirls and Four-Leaf Clovers: "Orbits around a black hole can be fascinatingly intricate." $\endgroup$ – Simon Woodside May 28 at 20:15
26
$\begingroup$

The answer is "yes", and there are a surprising number of ways to argue this. You will probably want to look at the question about small orbits around black holes in Physics Stack Exchange.

At sufficient distances, black holes are not special

Black holes behave the same as any other spherical collection of matter. This is a conclusion of the (intimidating) equations of general relativity (GR), but it's not a surprising result. Newtonian gravity ("classical" gravity) states that any spherically symmetric collection of matter will behave the same. This means that the radial distribution of the sun's mass doesn't affect its tug on the Earth. The vast majority of the sun's mass is contained below 1/4th of its visible radius. But if it were uniform, it would behave exactly the same! This is a very strong statement, and general relativity follows suit.

In short, we only need general relativity when a combination of parameters puts the system in the regime of highly relativistic effects. For gravity regarding a point mass orbiting a large mass, this is often dictated by the gravitational potential, which you should be familiar with as $\frac{G M }{r}$. When this value starts to get close to $\frac{c^2}{2}$, then you have to worry about these strange relativistic distortions.

The story of Mercury's precession is a common anecdote about GR, but this only deals with the accumulation of small change over a long period of time. Specifically 43 arc-seconds over a century, which is something like 1/100th of a degree... in 100 years. However, since the solar system has been around for billions of years, this can still have an impact on orbital stability.

That's a very minute correction to our conclusions. Otherwise, it generally stands to say that whatever orbits are stable for any large body are also stable for a black hole. Actually, a black hole will be even more stable. All the planets are stars have a complex gravitational tapestry due to density changes due to composition changes. If you collapse any of these bodies to a black hole, they'll have to expel their angular momentum. At long distances from a black hole, the gravitational field will be exceptionally constant, and this leads to greater orbital stability.

When we're talking about Sagittarius A* and things like that, we're generally still in this regime.

Truly relativistic orbits are weird

Once you get close enough such that the potential is close to the relativistic limits, the dynamics change. You can't make a blanket statement that "all orbits are stable/unstable". Things you can say:

  1. obviously anything that passes within the event horizon is gone
  2. all orbits that pass within the "IBCO" are dead unless "rockets" are used
  3. unless you're beyond the "ISCO", your orbit will be highly elliptical and precess
  4. If your orbit precession hits an angle that divides $2 \pi$ evenly, then it's "stable"

My apologies for the jargon here. It's fairly hard to say any of this concisely without the specific terminology.

Let me just establish that IBCO (Innermost Bound Circular Orbit, which is 1.5 times the event horizon radius) is basically the line of death. If you cross this point, you can only escape if you have "powerful rockets". I use this terminology because the physicists use it, but it's really a lie. No rocket would be powerful enough to bring you back out unless you were right next to the IBCO. Nonetheless, there are other ways to get an exception - rotating black holes, for instance. Without these, venturing beyond the IBCO will plunge you to the event horizon. But now, the crazy thing about the IBCO is that you can dance right on the edge of it for basically an unlimited amount of time, and hop right back out.

Now, my points #3 & 4 are basically that orbits that come close to the IBCO have "zoom-whirl" behavior. This can be avoided fully, but only beyond the "ISCO" (Innermost Stable Circular Orbit). Beyond that point, you can orbit in a circle and it will be stability until infinity. If you don't meet this criteria, the orbit moves over time like Mercury. But the shift over every revolution can be any angle. That means that you can spin around the IBCO a few times, and then come back and retrace your previous path to the ISCO. You can repeat this exact path for infinity.

This demonstrates the types of stable orbits you can obtain, but real life will have a more complicated collection of parameters. Most black holes are rotating, and they apparently rotate at a large fraction of their theoretical maximum. Plus, there's other material around it. But both of these open up avenues to get energy from it. These complications will likely prevent "stable" orbits, but since they're non-conservative, you can sort-of "surf" the environment around the black hole until you deplete all its usable energy (hint: that's a lot).

$\endgroup$
  • 2
    $\begingroup$ """The vast majority of the sun's mass is contained below 1/4th of its visible radius. But if it were uniform, it would behave exactly the same!""" - I'd emphasize the original point here by mentioning that if it were all within its Swartzchild radius of 3 km (making it a black hole) it would also be the same. $\endgroup$ – Random832 Sep 12 '13 at 18:47
  • $\begingroup$ I remember I found a black hole orbit simulator somewhere on the Web, alas! I forgot the linky. $\endgroup$ – Deer Hunter Sep 12 '13 at 20:02
  • $\begingroup$ One simple way to think of orbital instability around black hole is: Relativistic mass is a gravitational mass. The faster you move, the heavier you are. The lower the orbit the faster you move. The heavier you are the more the black hole pulls you, speeding you up even more. $\endgroup$ – SF. Sep 16 '13 at 11:25
  • 1
    $\begingroup$ @SF.: Your argument is incorrect. Inertia also increases. $\endgroup$ – Ben Crowell Jun 10 '14 at 18:46
  • $\begingroup$ Is the Innermost Bound Circular Orbit (IBCO) the same as the photon sphere? $\endgroup$ – Simon Woodside May 28 at 20:20
14
$\begingroup$

Well yes, and the best proof I can think of are the stars orbiting our galaxy's galactic center black hole, which is a supermassive black hole in the center of the Milky Way in the Sagittarius A* region:

              enter image description here

 Inferred orbits of 6 stars around supermassive black hole candidate Sagittarius A* at the Milky Way galactic centre. (Source: Wikipedia)

It could also be argued, that we all orbit around this black hole anyway, since the whole galaxy orbits around its barycenter, where this supermassive black hole in the galaxy nucleus is located.

$\endgroup$
  • $\begingroup$ Note that the elliptical orbits must have some precession -- they shouldn't overlap themselves $\endgroup$ – Manishearth Sep 12 '13 at 13:27
14
$\begingroup$

Yes, there are, to a reasonable approximation1. For a nonspinning black hole, there are exactly 4 things that can happen (assuming that you haven't thrown the object directly at the black hole):

  • The object will be going too fast, and it will just continue to infinity with a slight deflection in path.
  • The object will be going too slow, and it will spiral into the center
  • The object goes at a perfect speed at an angle perpendicular to the position vector, giving you a circular orbit
  • The object goes at a (different) perfect speed at an angle not perpendicular to the position vector, giving a rosetta orbit:

enter image description here

This is similar to an elliptic orbit, except that the major/minor axes themselves rotate. Mercury also prominently displays this kind of orbit (nowhere near as much as one orbiting a black hole, though), as it exhibits a prominent precession of the perihelion — the ellipse itself seems to orbit around the sun.

In the case of a spinning black hole, things become more complicated as a spiraling object may actually reverse the direction of rotation due to frame dragging.

1. See Mark Adler's answer. Similar to orbits arising from the electromagnetic force, two orbiting bodies emit gravitational waves, which leads to an eventual loss of energy and inspiral. However, this process is very slow except near the end.

$\endgroup$
13
$\begingroup$

Technically, in General Relativity, there are no stable orbits in any two-body system, period. Regardless of mass or black holiness. Bodies rotating about their mutual center of mass will emit gravitational radiation. By conservation of energy, the orbits will get smaller, and given enough time, they will crash into each other.

Practically, this takes so long to happen in most circumstances that we find in the universe, that it is not a consideration on the time scales of age of the universe. However in unusual circumstances with large masses orbiting each other tightly, in this case two neutron stars, this has been observed:

The orbit has decayed since the binary system was initially discovered, in precise agreement with the loss of energy due to gravitational waves predicted by Einstein's general theory of relativity.

$\endgroup$
0
$\begingroup$

In general relativity the energy of a "test-body" in motion around a Schwarzschild (spherically symmetric, non-rotating) black hole can be written as:

$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2\right)}}}\right)$$.

This can be written as:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\right)$$.

In the special case of pure circular motion you have:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}\right)$$.

Just as classically (it can be shown somehow) for a pure cicular orbit you have: $v=\sqrt{GM/r}$ and we can thus write:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{3GM}{rc^2}}}\right)$$.

I believe (I have not checked it out) that by differentiating it can be seen that this expression has a miminum at $r=6GM/c^2$, which is known as the radius of the "innermost stable circular orbit". This means that any circular orbit with $r>6GM/c^2$ is stable in the sense that circular orbits infinitesimally closer to the black hole requires less energy. However, at $r=6GM/c^2$ it actually requires more energy to sustain a circular orbit closer to the black hole and are because of that unstable. If you try to sustain a circular orbit closer than this distance you will inevitably crash into the black hole.

From the expressions above we also see that it requires "infinite energy" (an object must travel at the speed of light) to sustain a spherical orbit at the "photon sphere" ($r=3GM/c^2$).

Answer: Circular orbits are stable for $r>6GM/c^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.