<p>It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be "ripped apart", since there is nothing to be ripped apart there - just a giant warped tangle of spacetime.</p>

<p>Instead, two black holes could merge, if they get close enough and lose enough mutual orbital motion.</p>

<p><a href="https://www.youtube.com/watch?v=4m-ZVsLf070" rel="nofollow">merging black holes</a></p>

<p>The result is a single, larger black hole.</p>

CC BY-SA 3.0
2015-01-08T19:59:40.983Z
2015-01-09T19:27:34.590Z
2015-01-09T19:27:34.590Z
Florin Andrei
<p>Because the density of the matter at the center of a black hole is infinite (or nearly so) the Roche Radius is 0 (or nearly so).  Two black holes in orbit spiral in towards each other because they radiate gravity waves and form a mutual event horizon without any disturbance to the matter at their centers.</p>

CC BY-SA 3.0
2015-01-08T19:59:47.180Z
Is this simply a property of black holes, or does this imply that a Roche limit for a celestial body is variable depending on the gravitational pull of the body it's interacting with? Ex: The Roche limit of the Sun for Jupiter would be different than the Roche limit of the Sun for Earth.
Sidney
CC BY-SA 3.0
2015-01-09T00:32:16.860Z
"The density of a black hole is infinite"--this simply isn't true, or at least depends on taking peculiar interpretations of the terms that don't have any clear relevance here.
Stan Liou
CC BY-SA 3.0
2015-01-09T18:58:36.727Z
Once an object has been compressed to its Schwarzschild radius it will continue to collapse until it becomes a singularity. The density of the matter in a black hole in GR is infinite, but with QM it may have a radius of its de Broglie wavelength or so.  So the black hole, the region within the event horizon, is just empty space except for a nearly pointlike object at the center.
eshaya
CC BY-SA 3.0
2015-01-10T05:55:00.737Z
@EdShaya what you just said is true (under reasonable assumptions about matter, and except for necessarily being pointlike), but not relevant to the question. A black hole is defined by the presence of an event horizon, which in turn depends on the behavior of test particles in the region. Thus, a black hole would be destroyed by the gravity of another (without merging) whenever test particles in it could escape to infinity (i.e., there would no longer be a horizon). Of course, that doesn't happen, but that's what destroying a black hole would mean. ...
Stan Liou
CC BY-SA 3.0
2015-01-10T05:55:36.667Z
Another way to illustrate the point: if you violate the relevant energy conditions, you could have black holes without any singularities at all. But the answer in that completely-hypothetical case would be completely the same: the other black hole wouldn't be destroyed because its horizon wouldn't be destroyed. What happens deep inside doesn't matter to the question.
Stan Liou
CC BY-SA 3.0
2015-01-08T19:50:58.190Z
2015-01-09T19:00:03.850Z
2015-01-09T19:00:03.850Z
eshaya
<p>(Yes, this should be a comment. It's too big, though.)</p>

<p>To address Sidney's comment to Ed Shaya's answer:</p>

<p>The Roche limit equation can be expressed as
$$1.26 \times R_{\text{secondary}} \times \sqrt[3]{\frac{\text{Primary}}{\text{Secondary}}}$$</p>

<p>Since the radius of the secondary is zero and zero times anything is zero, the Roche limit is likewise zero.</p>

<p>When you think about what the Roche limit really means this is self-evident. The Roche limit is the point where the primary is raising a tide on the secondary that's stronger than it's own gravity. Tides are based on the fact that while the object is at a certain distance it really occupies space, parts are closer (which experience a greater pull) and parts are farther (which experience a lesser pull.) There is no point on the black hole that's closer or farther, thus there is no tide, thus it won't be torn apart even if it had no self-gravity.</p>

CC BY-SA 4.0
2020-11-28T12:20:00.737Z
This Roche limit is calculated using Newtonian gravity. It is inapplicable to a pair of black holes.
ProfRob
CC BY-SA 3.0
2015-01-09T03:29:22.707Z
[Here's](http://physics.stackexchange.com/help/search?q=notation) the notation used. For some reason, the information isn't in the Astronomy help center.
HDE 226868
CC BY-SA 3.0
2015-01-09T03:53:06.643Z
A black hole is, by definition, a region enclosed by an event horizon. Thus, there are definitely points on the black hole that are closer or farther.
Stan Liou
CC BY-SA 3.0
2015-01-09T19:09:08.900Z
But there is nothing in particular at the event horizon.  It is just the point at which light can't escape.  The matter is at the center and the Roche limit is about tearing apart the matter.
eshaya
CC BY-SA 3.0
2015-01-09T23:58:19.553Z
@EdShaya the argument in this answer is not in general sound even if you replace 'black hole' with 'singularity'; e.g., rotating black holes have a spatially extended singularity that's not (spatially) pointlike.
Stan Liou
CC BY-SA 3.0
2015-01-11T17:04:21.900Z
True.  Rotating black holes are thought to form a ring rather than a point.  But, a) the ring is inside the event horizon, so external observers never see what happens to these rings and b) the density along the ring is infinite, so it is still hard to disrupt. Now, an interesting question is what happens exactly when two matter rings of infinite density merge? Just how does that proceed?  It is interesting even if we can never actually see it (and report back).  Well, unless Penrose (1969) is wrong about the Cosmic Censorship of naked singularities.
eshaya
CC BY-SA 3.0
2015-01-12T06:24:44.963Z
@EdShaya a) that's exactly why it's not even relevant to the question, so it's strange that you keep bringing it up. b) the Kerr interior is extremely unstable to perturbations, so no. It doesn't even *matter* how much the internal structure is or isn't affected, because if it's all behind an event horizon, there will still be a black hole, and if it leaves the horizon, then there's no horizon. Hence the existence of the horizon is a necessary and sufficient condition for the black hole to be there, and the structure of the singularity is a complete red herring to this question.
Stan Liou
CC BY-SA 3.0
2015-01-12T06:54:28.540Z
@StanLiou I disagree--I was saying that since it's a point there would be no tide and thus it wouldn't come apart even without it's own gravity.  Thus the non-point nature of a Kerr black hole is relevant.
Loren Pechtel
CC BY-SA 3.0
2015-01-12T07:14:47.497Z
@LorenPechtel sorry, but you're confused about what a black hole is, i.e. what the term means. But at this point I'll be repeating myself again, so I'll stop here.
Stan Liou
CC BY-SA 3.0
2015-01-09T02:41:23.040Z
2015-01-09T03:27:43.867Z
2015-01-09T03:27:43.867Z
Loren Pechtel
<p>The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body.</p>

<p>However, the gravitational tidal forces of a black hole are always finite, except at the internal singularity. This is a problem because the self-gravity of a black hole, in the sense of the acceleration a mass would need to remain stationary on its surface, is infinite<sup>1</sup>. Thus, we shouldn't expect for a large black hole to destroy another through gravitational tidal forces.</p>

<p>Put another way, the Roche limit occurs when particles from the smaller body can escape them... but they can't escape the event horizon of a black hole. Thus, the black holes will either orbit or merge, which is what happens in numerical simulations.</p>

<p><sup><sup>1</sup>There is a separate concept of <em>surface gravity</em> of a black hole that's essentially this re-scaled by the gravitational time dilation factor, and thus kept finite.</sup></p>

CC BY-SA 3.0
2015-01-09T00:44:09.410Z
2015-01-09T00:44:09.410Z
Stan Liou
<p>Consider black hole A, a super massive black hole at the center of the galaxy. Orbiting it is black hole B, a much less massive black hole.</p>

<p>If some passing body were to modify black hole B's orbit such that it fell within the Roche limit of black hole A, what would happen to black hole B?</p>

<p>If it were to turn into a ring, would the black hole matter re-inflate since it wouldn't be under such high gravity? Do black holes even respond to Roche limits like regular matter does?</p>

CC BY-SA 3.0
2015-01-10T00:19:15.780Z
The idea of the Roche limit does apply to black holes if the secondary body *isn't* a black hole. For example, if an asteroid approaches a black hole too closely, it will be torn apart. The distance at which this happens is related to the radius of a body with the mass of the black hole and the density of the asteroid.
Keith Thompson
<black-hole><roche-limit>
Do Roche limits apply to black holes?
Sidney
# Do Roche limits apply to black holes?

(Asked by: username_0 on 2015-01-08)

Consider black hole A, a super massive black hole at the center of the galaxy. Orbiting it is black hole B, a much less massive black hole.

If some passing body were to modify black hole B's orbit such that it fell within the Roche limit of black hole A, what would happen to black hole B?

If it were to turn into a ring, would the black hole matter re-inflate since it wouldn't be under such high gravity? Do black holes even respond to Roche limits like regular matter does?

**Tags:** black-hole, roche-limit

> **Comments:**
> **username_1**: The idea of the Roche limit does apply to black holes if the secondary body *isn't* a black hole. For example, if an asteroid approaches a black hole too closely, it will be torn apart. The distance at which this happens is related to the radius of a body with the mass of the black hole and the density of the asteroid.  

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## 4 answers

### Answer by username_4 on 2015-01-09. Score: 8 | Accepted answer

The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body.

However, the gravitational tidal forces of a black hole are always finite, except at the internal singularity. This is a problem because the self-gravity of a black hole, in the sense of the acceleration a mass would need to remain stationary on its surface, is infinite1. Thus, we shouldn't expect for a large black hole to destroy another through gravitational tidal forces.

Put another way, the Roche limit occurs when particles from the smaller body can escape them... but they can't escape the event horizon of a black hole. Thus, the black holes will either orbit or merge, which is what happens in numerical simulations.

1There is a separate concept of *surface gravity* of a black hole that's essentially this re-scaled by the gravitational time dilation factor, and thus kept finite.


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### Answer by username_2 on 2015-01-08. Score: 7

It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be "ripped apart", since there is nothing to be ripped apart there - just a giant warped tangle of spacetime.

Instead, two black holes could merge, if they get close enough and lose enough mutual orbital motion.

[merging black holes](https://www.youtube.com/watch?v=4m-ZVsLf070)

The result is a single, larger black hole.


---

### Answer by username_3 on 2015-01-08. Score: 3

Because the density of the matter at the center of a black hole is infinite (or nearly so) the Roche Radius is 0 (or nearly so). Two black holes in orbit spiral in towards each other because they radiate gravity waves and form a mutual event horizon without any disturbance to the matter at their centers.

> **Comments:**
> **username_0**: Is this simply a property of black holes, or does this imply that a Roche limit for a celestial body is variable depending on the gravitational pull of the body it's interacting with? Ex: The Roche limit of the Sun for Jupiter would be different than the Roche limit of the Sun for Earth.  
> **username_4**: "The density of a black hole is infinite"--this simply isn't true, or at least depends on taking peculiar interpretations of the terms that don't have any clear relevance here.  
> **username_3**: Once an object has been compressed to its Schwarzschild radius it will continue to collapse until it becomes a singularity. The density of the matter in a black hole in GR is infinite, but with QM it may have a radius of its de Broglie wavelength or so.  So the black hole, the region within the event horizon, is just empty space except for a nearly pointlike object at the center.  
> **username_4**: @EdShaya what you just said is true (under reasonable assumptions about matter, and except for necessarily being pointlike), but not relevant to the question. A black hole is defined by the presence of an event horizon, which in turn depends on the behavior of test particles in the region. Thus, a black hole would be destroyed by the gravity of another (without merging) whenever test particles in it could escape to infinity (i.e., there would no longer be a horizon). Of course, that doesn't happen, but that's what destroying a black hole would mean. ...  
> **username_4**: Another way to illustrate the point: if you violate the relevant energy conditions, you could have black holes without any singularities at all. But the answer in that completely-hypothetical case would be completely the same: the other black hole wouldn't be destroyed because its horizon wouldn't be destroyed. What happens deep inside doesn't matter to the question.  

---

### Answer by username_5 on 2015-01-09. Score: 3

(Yes, this should be a comment. It's too big, though.)

To address username_0's comment to Ed Shaya's answer:

The Roche limit equation can be expressed as
$$1.26 \times R\_{\text{secondary}} \times \sqrt[3]{\frac{\text{Primary}}{\text{Secondary}}}$$

Since the radius of the secondary is zero and zero times anything is zero, the Roche limit is likewise zero.

When you think about what the Roche limit really means this is self-evident. The Roche limit is the point where the primary is raising a tide on the secondary that's stronger than it's own gravity. Tides are based on the fact that while the object is at a certain distance it really occupies space, parts are closer (which experience a greater pull) and parts are farther (which experience a lesser pull.) There is no point on the black hole that's closer or farther, thus there is no tide, thus it won't be torn apart even if it had no self-gravity.

> **Comments:**
> **username_7**: [Here's](http://physics.stackexchange.com/help/search?q=notation) the notation used. For some reason, the information isn't in the Astronomy help center.  
> **username_4**: A black hole is, by definition, a region enclosed by an event horizon. Thus, there are definitely points on the black hole that are closer or farther.  
> **username_3**: But there is nothing in particular at the event horizon.  It is just the point at which light can't escape.  The matter is at the center and the Roche limit is about tearing apart the matter.  
> **username_4**: @EdShaya the argument in this answer is not in general sound even if you replace 'black hole' with 'singularity'; e.g., rotating black holes have a spatially extended singularity that's not (spatially) pointlike.  
> **username_3**: True.  Rotating black holes are thought to form a ring rather than a point.  But, a) the ring is inside the event horizon, so external observers never see what happens to these rings and b) the density along the ring is infinite, so it is still hard to disrupt. Now, an interesting question is what happens exactly when two matter rings of infinite density merge? Just how does that proceed?  It is interesting even if we can never actually see it (and report back).  Well, unless Penrose (1969) is wrong about the Cosmic Censorship of naked singularities.  
> **username_4**: @EdShaya a) that's exactly why it's not even relevant to the question, so it's strange that you keep bringing it up. b) the Kerr interior is extremely unstable to perturbations, so no. It doesn't even *matter* how much the internal structure is or isn't affected, because if it's all behind an event horizon, there will still be a black hole, and if it leaves the horizon, then there's no horizon. Hence the existence of the horizon is a necessary and sufficient condition for the black hole to be there, and the structure of the singularity is a complete red herring to this question.  
> **username_5**: @StanLiou I disagree--I was saying that since it's a point there would be no tide and thus it wouldn't come apart even without it's own gravity.  Thus the non-point nature of a Kerr black hole is relevant.  
> **username_4**: @LorenPechtel sorry, but you're confused about what a black hole is, i.e. what the term means. But at this point I'll be repeating myself again, so I'll stop here.  
> **username_6**: This Roche limit is calculated using Newtonian gravity. It is inapplicable to a pair of black holes.  

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