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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?
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.
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.
The result is a single, larger black hole.
(Yes, this should be a comment. It's too big, though.)
To address Sidney'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.
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.
