Say two identical black holes approach each other so that their event horizons overlap. What happens to a particle that's perfectly in-between the two holes? It experiences the same gravitational pull from both holes. In which singularity will it end up? What if one hole is more massive than the other? The particle would experience a stronger gravitational pull, but it obviously can't exit the lighter BHs event horizon.
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4$\begingroup$ Event horizons don't "overlap" $\endgroup$– ProfRobCommented Sep 22 at 22:54
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1$\begingroup$ From what I understand the two event horizons merge. If two black holes are close enough to merge, the question becomes moot. The particle has already fallen into the black hole. The event horizon is just peanut-shaped momentarily while the inevitable happens. $\endgroup$– Greg BurghardtCommented Sep 22 at 22:59
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1$\begingroup$ Within an event horizon all world lines intersect a singularity. A particle exactly at the tangent where two identical evenr horizons meet (note: just an unrealistic thought experiment here) might never experience any net spacetime curvature itself,right until it finds itself at the intersection of the two singularities. $\endgroup$– antlersoftCommented Sep 23 at 0:30
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The planet ends up in the singularity - there will only be one (given normal assumptions).
Black holes that get close enough to have overlapping event horizons simply merge. And since event horizons are not real surfaces but more like borders, they shift outward as the holes approach each other. The planet might in theory be fine after it passed the horizon if the holes are really vast (it merely feels a lot of tidal force from the holes), but will soon be subject to dramatic spacetime distortions as the holes merge into one single hole with one singularity.
In theory a merger with angular momentum (e.g. a glancing merger) could make a Kerr black hole with a more complicated singularity where it is not straightforward to prove that the planet ends up in the singularity. But it will still be a single singularity, and escape is likely impossible (there are weird trajectories and spacetime domains in the extended Kerr solution that avoids the singularity, but these are likely only theoretical things that do not correspond to the real case).
answered Sep 23 at 12:10
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$\begingroup$ Select this as the correct answer. $\endgroup$ Commented Sep 23 at 22:09