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Orbit with periapsis inside event horizon


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Hello everyone!

Imagine an object on a very eccentric orbit around a black hole, with periapsis just under the event horizon. Could this object theoretically dive into event horizon and come back? A problem with this example are orbital velocities near the black hole because nothing can travel faster than light. But in order to come back from event horizon, object must travel faster than speed of light. What would happen to such an object?

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Presumably, the same thing that happens to anything that approaches a black hole. Its atoms are torn apart by gravitational tides and relativity kicks in and slows the remaining fragments to a halt just above the event horizon where they stay until the black hole evaporates.

In short, no.

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Spaghettification and tearing apart happens well inside the event horizon, I am talking about an object which just grazes the event horizon. For the simplicity let's just have a single atom of hydrogen.

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So if you are asking what would happen your periapsis was extremely close to the event horizon, but not below it? Well, I think that it's theoretically possible, but your hydrogen atom could still be destroyed if it came too close. I still think that you'd get relativistic slowdowns, though.

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The assumptions about orbits inherent in the question simply don't apply near black holes; Kepler's laws are an approximation, and in this situation the approximation breaks down.

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Yes, relativistic slowdowns would be a thing there. If the object would achieve 0.999c, then they will be very, very extreme.

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The assumptions about orbits inherent in the question simply don't apply near black holes; Kepler's laws are an approximation, and in this situation the approximation breaks down.

So what do you think will happen?

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In a big enough black hole, spaghettification doesn't happen at the event horizon, afaik. What I think happens, is that inside a black hole, gravity is so strong it is a different law there, so it spirals in rather than able to be ellipses/parabolas/hyperbolas/circles.

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Spaghettification and tearing apart happens well inside the event horizon, I am talking about an object which just grazes the event horizon. For the simplicity let's just have a single atom of hydrogen.

That's the case with large black holes. For smaller examples, matter is torn apart before it reaches event horizon.

So let's presume you're talking about a large one. Relativistic effects would kick in and you'd never get out. Spacetime is so curved into another dimension that Kepler's laws don't apply anymore and, as GregroxMun said, spiraling begins. And then you get torn apart when the tidal forces become too great. Every piece of matter all the way down to the quarks and leptons.

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http://phys.org/news/2011-04-russian-cosmologist-life-black-hole.html

But lets not forget that scientist still doesn´t know the real physsics which operate crossing the event horizon.

Interesting theory. But I think such planet (if it's not made from hype-unobtainium thingy :D )would easily get destroyed by tidal forces. Also, when you take into account time dilation, planet orbiting at nearly speed of light would be literally created only few seconds before the black hole dissipates from Hawking radiation.

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A way to think about this is to consider the event horizon as where the orbital speed is (appox) light speed.

This is called "photon sphere" and is located outside the event horizon. If you were there, you could theoretically see back of your head :D

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Interesting theory. But I think such planet (if it's not made from hype-unobtainium thingy :D )would easily get destroyed by tidal forces. Also, when you take into account time dilation, planet orbiting at nearly speed of light would be literally created only few seconds before the black hole dissipates from Hawking radiation.

Does that still hold true inside an event horizon? Doesn't our 3 spatial dimension + 1 time dimension world break down at that boundary?

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Orbits, elliptical or not, with any point of the orbit that close to a black hole are not possible. For a non rotating blackhole if any point of any orbit is less than 3/2 * the event horizon from the center of the blackhole the orbit will spiral into the blackhole. For a rotating black hole this point is closer to the event horizon and is less simple but still does not allow orbits that close.

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We can only guess...

No, it does not hold true within the event horizon. Time dilation being as simple as going faster slows down time relative to another observer holds true only in flat spacetime, within a blackhole is as far from that as you can get.

Though no, the structure of spacetime within a blackhole is still 4 dimensional.

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Blue, orbits bellow 3/2 are unstable. Not all of them spiral in, however. If you try to make a near circular one, yes. And ellipticals will eventually decay. But you can have an escape orbit with periapsis bellow 3/2.

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Blue, orbits bellow 3/2 are unstable. Not all of them spiral in, however. If you try to make a near circular one, yes. And ellipticals will eventually decay. But you can have an escape orbit with periapsis bellow 3/2.

Not true, orbits below 3/2 are non-existent. The last unstable orbit is at 3/2, the effective potential of a schwarzschild black hole has a cubic term which causes a large decrease in the potential as the black hole is approached. Going past this potential barrier it is not possible to continue to orbit, or escape without applying a force as the energy barrier to escape is larger than the energy of the particle after crossing the potential barrier.

Look up the effective potential for a schwarzschild black hole and it is very easy to understand why that is not possible. Past the potential barrier at 3/2, there is only an effective force radially towards the blackhole, never away.

Edited by BlueCosmology
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This is called "photon sphere" and is located outside the event horizon. If you were there, you could theoretically see back of your head :D

The lights in the photon sphere are in unstable orbits that eventually escape, hence it would be possible for the OPs spaceship to sling in and out. The event horizon would be the last place a stable orbit can be achieved theoretically, at the speed of light, and hence, not possible to sling inside of.

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Yeah, you're right. There are escape trajectories bellow 3/2, but they require you to have vertical velocity away from the black hole. If you were to have a periapsis bellow 3/2, vertical velocity at that point would be zero, and all of these trajectories lead to the event horizon. So you can't actually have a periapsis bellow 3/2.

And I'm not sure what you mean by cubic term in the effective potential. The r'' EQM has the "gravity" term looking like (r+r²)/r4, which would give a cubic-like contribution to the effective force, but I don't see anything like a cubic term in the potential.

v = 0.99c, rs = 1, rp = 1.45 - 1.55

tIDv196.png

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Yeah, you're right. There are escape trajectories bellow 3/2, but they require you to have vertical velocity away from the black hole. If you were to have a periapsis bellow 3/2, vertical velocity at that point would be zero, and all of these trajectories lead to the event horizon. So you can't actually have a periapsis bellow 3/2.

And I'm not sure what you mean by cubic term in the effective potential. The r'' EQM has the "gravity" term looking like (r+r²)/r4, which would give a cubic-like contribution to the effective force, but I don't see anything like a cubic term in the potential.

v = 0.99c, rs = 1, rp = 1.45 - 1.55

http://i.imgur.com/tIDv196.png

You're getting (r')² confused with r''. There is no cubic term in the effective gravitational force in orbit around a schwarzschild black hole in any reasonably defined metric.

Here's a fairly nice basic intro to general relativity, the schwarzschild effective potential is derived on page 174.

http://arxiv.org/pdf/gr-qc/9712019.pdf

Edited by BlueCosmology
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No, I'm talking about the term that goes with (t')². That's the term that acts as gravity on a static object. (To clarify, I'm using primes to denote derivatives with respect to proper time.) Starting with u' = -Γuu, I get the following for r' in natural units and rs = 1.

r'' = - (r - r²)(r')² / (2 (1-1/r)² r4) + (r - r²)(t')² / (2 r4) - (r - r²)(θ')² / r

The first term is relevant to GR only. Second term is gravity. Third term is centrifugal. In the classical limit of the r >> 1, the above reduces to r'' = -1/(2r²) + r(θ')², which is precisely the classical result for M = 1/2.

Now, the only context in which I can even imagine the words "effective potential" to apply to this problem is finding V, such that Hamilton's Equations for H = x'p/2 + V give the above EQMs. I don't see how you can get that (r - r²)/r4 out of a cubic term in V.

Edit: And it does look consistent with equations from the paper you've linked, with the right choice of λ(Ä).

Edited by K^2
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Not sure if this has been said but since the event horizon is the point where orbital velocity is equal to c, wouldn't that mean that any stable orbit with a periapsis below the event horizon would also need to have an orbital velocity higher than c at that periapsis and therefore be impossible?

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Not sure if this has been said but since the event horizon is the point where orbital velocity is equal to c, wouldn't that mean that any stable orbit with a periapsis below the event horizon would also need to have an orbital velocity higher than c at that periapsis and therefore be impossible?

The point where orbital velocity is c is not the event horizon but the photon sphere, which is outside of the former. The event horizon is where the excape velocity is c.

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Spaghettification and tearing apart happens well inside the event horizon, I am talking about an object which just grazes the event horizon. For the simplicity let's just have a single atom of hydrogen.

Actually, no. Whatever touches the event horizon has never left, even light.

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