<p>As described in <a href="https://physics.stackexchange.com/a/673697/101892">this answer</a> on the physics SE, mathematically it is possible to have bound orbits/geodesics inside the inner horizon of the Kerr solution for an eternal rotating black hole. However, the Kerr solution inside the inner horizon is known to be unstable to (almost) any perturbations. Consequently, one should not necessarily expect this to be a feature of astrophysical black holes. Instead these solutions are expected to develop singularities at/near the inner horizon.</p>
<p>There are no bound orbits/geodesics between the inner and outer horizon of Kerr. All trajectories cross from the outer to the inner horizon in a finite amount of proper time.</p>

CC BY-SA 4.0
2024-09-25T08:59:42.890Z
2024-09-25T08:59:42.890Z
TimRias
<p>No. There are no closed orbits inside the event horizon.</p>
<p>In fact there are no closed orbits just outside the event horizon either, but an object that is moving fast enough can escape from this region outside the event horizon: there are some paths in space-time that lead away from the black hole.</p>
<p>But inside the event horizon it is a different matter.  All directions lead towards the centre of the black hole.  Its not just that an object can't go fast enough to escape, it is that spacetime is so curved that all directions just lead towards the centre.  (If you are already at the centre, it's not clear what happens,  it seems that there are no directions at all at this point)</p>
<p>As all directions lead towards the centre, it is impossible for something to orbit, it is not even possible for it to have a &quot;powered&quot; orbit, using rockets to maintain altitude. Every object inside the event horizon must inevitably fall to the centre.</p>

CC BY-SA 4.0
2024-09-25T05:38:41.400Z
Any idea if this (all directions lead towards the center) is also the case for rotating black holes?
Allure
CC BY-SA 4.0
2024-09-25T08:51:46.817Z
Not all black holes are Schwarzschild.
TimRias
CC BY-SA 4.0
2024-09-25T19:59:32.133Z
@allure, No, but the headline "you can't orbit inside a black hole" is still the case.  Kerr black holes are weird inside, and I don't want to say more for fear of inaccuracy. See TimRias answer
James K
CC BY-SA 4.0
2024-09-27T20:51:39.827Z
There are events that cause particles to leave a black hole. Such as the recent discovery of a beam leaving a black hole. I was thinking that something like this might be a cause of such an event
Michael Mcgarry
CC BY-SA 4.0
2024-09-27T21:29:09.367Z
No, the beam is generated outside the event horizon.  Hawking radiation is also produced outside the event horizon.   Beams do not leave from inside the event horizon. Nothing leaves a black hole after it has passed the event horizon.
James K
CC BY-SA 4.0
2024-09-24T22:35:14.293Z
2024-09-24T22:35:14.293Z
James K
<p>I am imagining something like a small, fast-spinning, highly magnetic, neutron star.
And if it was, what would be the observable consequences, if any?</p>

CC BY-SA 4.0
2024-09-24T18:12:42.407Z
No--within an event horizon there is no path a particle can follow through spacetime that is not towards the singularity (or through the wormhole), analogously to how outside the event horizon there is no path through spacetime that is not forward in time.
antlersoft
CC BY-SA 4.0
2024-09-24T19:05:50.110Z
There can be no observable consequences of something inside the event horizon. That is almost a definition of the event horizon.
ProfRob
CC BY-SA 4.0
2024-09-24T22:05:07.347Z
If something is inside an event horizon, then it **is** a part of a black hole.
Mithoron
<black-hole><event-horizon>
Would it be theoretically possible for something to orbit a black hole inside its event horizon?
Michael Mcgarry
# Would it be theoretically possible for something to orbit a black hole inside its event horizon?

(Asked by: username_0 on 2024-09-24)

I am imagining something like a small, fast-spinning, highly magnetic, neutron star.
And if it was, what would be the observable consequences, if any?

**Tags:** black-hole, event-horizon
> **Comments:**
> **username_1**: No--within an event horizon there is no path a particle can follow through spacetime that is not towards the singularity (or through the wormhole), analogously to how outside the event horizon there is no path through spacetime that is not forward in time.  
> **username_2**: There can be no observable consequences of something inside the event horizon. That is almost a definition of the event horizon.  
> **username_3**: If something is inside an event horizon, then it **is** a part of a black hole.  

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## 2 answers

### Answer by username_4 on 2024-09-25. Score: 6

As described in [this answer](https://physics.stackexchange.com/a/673697/101892) on the physics SE, mathematically it is possible to have bound orbits/geodesics inside the inner horizon of the Kerr solution for an eternal rotating black hole. However, the Kerr solution inside the inner horizon is known to be unstable to (almost) any perturbations. Consequently, one should not necessarily expect this to be a feature of astrophysical black holes. Instead these solutions are expected to develop singularities at/near the inner horizon.

There are no bound orbits/geodesics between the inner and outer horizon of Kerr. All trajectories cross from the outer to the inner horizon in a finite amount of proper time.


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### Answer by username_5 on 2024-09-24. Score: 4

No. There are no closed orbits inside the event horizon.

In fact there are no closed orbits just outside the event horizon either, but an object that is moving fast enough can escape from this region outside the event horizon: there are some paths in space-time that lead away from the black hole.

But inside the event horizon it is a different matter. All directions lead towards the centre of the black hole. Its not just that an object can't go fast enough to escape, it is that spacetime is so curved that all directions just lead towards the centre. (If you are already at the centre, it's not clear what happens, it seems that there are no directions at all at this point)

As all directions lead towards the centre, it is impossible for something to orbit, it is not even possible for it to have a "powered" orbit, using rockets to maintain altitude. Every object inside the event horizon must inevitably fall to the centre.

> **Comments:**
> **username_6**: Any idea if this (all directions lead towards the center) is also the case for rotating black holes?  
> **username_4**: Not all black holes are Schwarzschild.  
> **username_5**: @allure, No, but the headline "you can't orbit inside a black hole" is still the case.  Kerr black holes are weird inside, and I don't want to say more for fear of inaccuracy. See TimRias answer  
> **username_0**: There are events that cause particles to leave a black hole. Such as the recent discovery of a beam leaving a black hole. I was thinking that something like this might be a cause of such an event  
> **username_5**: No, the beam is generated outside the event horizon.  Hawking radiation is also produced outside the event horizon.   Beams do not leave from inside the event horizon. Nothing leaves a black hole after it has passed the event horizon.  

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