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I was reading "the science behind interstellar", and found that Kip Thorne purposefully used a SMBH in his simulation which had its maximum attainable spin around its own axis. He argued that this was needed to be done in order to achieve the extreme time-dilation depicted in the movie (an hour on Miller's planet was 7 years of Earth time). I'm really interested to understand the theory behind this. Particularly, why does a rotating black-hole have a stronger gravitational pull than a black-hole with the same mass but no spin.

I think it has to do with its increase in rotational kinetic energy contributing to a further increase in its mass, but I'm not sure about this.

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    $\begingroup$ It is because only a spinning black hole allows circular orbits close enough (and you need to go with the spin too). In a nutshell, following the spin of a Kerr black hole you can get very close to the even horizon and this alone is what makes the potential time dilation factor so large compared to non-spinning (or going against the spin). $\endgroup$ – m4r35n357 Apr 23 at 12:06
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A spinning black hole does not necessarily have a strong gravitational pull than a non-rotating one of the same mass. In fact, at large distances you would not really be able to tell difference. The differences relevant for the increased gravity for an orbitting planet are the following.

  1. The event horizon of a spinning black hole is much smaller than that of non-spinning black hole. For a maximally spinning black hole the radius of the event horizon is $GM/c^2$, whereas for a nonspinning BH it is $2GM/c^2$. This means you are able to go to smaller radii and experience "stronger gravity", while still being able to escape the black hole.

  2. For black holes there is a smallest possible (stable) circular orbit (known as the innermost stable circular orbit or ISCO). The radius of the ISCO depends on the spin of the black hole and "sense" of the orbit. For non-spinning black holes the ISCO radius is at $6GM/c^2$, while for spinning black holes the ISCO for "prograde" orbits (orbits in the same direction as the spin) is smaller (and for retrograde larger) than the non-spinning case. In the prograde case the ISCO actually approaches the event horizon as the spin approaches maximality. This means that for a maximally spinning black hole, you can orbit it arbitrarily close to the event horizon. Since the time dilation diverges at the event horizon, this means that you can arrange for the time dilation to be arbitrarily large on an orbiting planet (which was needed for the plot of Interstellar).

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  • $\begingroup$ Can you cite some resources to learn more about the mathematics of this? I'm really curious to know how those formulas for the orbits are derived. $\endgroup$ – Apekshik Panigrahi Apr 24 at 9:47

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