If Gargantua is a black hole in Interstellar, why didn't it swallow the planets?
closed as off-topic by Valorum Feb 17 '15 at 12:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
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From the Black Hole FAQ on a physicist's website:
If a black hole existed, would it suck up all the matter in the Universe?
Heck, no. A black hole has a "horizon," which means a region from which you can't escape. If you cross the horizon, you're doomed to eventually hit the singularity. But as long as you stay outside of the horizon, you can avoid getting sucked in. In fact, to someone well outside of the horizon, the gravitational field surrounding a black hole is no different from the field surrounding any other object of the same mass. In other words, a one-solar-mass black hole is no better than any other one-solar-mass object (such as, for example, the Sun) at "sucking in" distant objects.
There is a complication in that this only applies to objects at a greater radius from the solar-mass black hole than the radius of the Sun's surface--once you get in close enough to a black hole you may find effects that are specific to very dense objects (black holes and some others like neutron stars), for example black holes have a photon sphere inside of which it's impossible even for light to remain in orbit. But physicist Kip Thorne, who wrote the original script treatment for Interstellar and was the science consultant on the movie, explained in The Science of Interstellar that he assumed Gargantua was rotating very rapidly, and the faster a black hole is rotating, the shorter the distance between the event horizon and the radius of the "innermost stable circular orbit" or ISCO, the formula for which can be found on this page. On this thread from physicsforums.com I asked about some of the formulas for a rotating black hole, and with the ones I was given I was able to write up with equations for the innermost stable orbit, the time dilation factor, and the orbital velocity, in a form that can be plugged into an online calculator like this one--see post #8 on that thread for the actual equations.
Using these equations, I found that to get the time dilation factor mentioned in the film and book (where 7 years pass for observers away from the black hole for every 1 hour spent on Miller's planet), given the the mass of Gargantua was 100 million times that of the Sun (the figure given in the book), it worked out that the rotation rate would differ from the maximum possible rotation rate for a rotating black hole (beyond which it would become a naked singularity) by only a factor of 1.33266 * 10^(-14). This is approximately equal to the figure Thorne gives in the technical notes at the end of the book about ch. 6, where he says "we obtain alpha = 1.3 * 10^-14; that is, Gargantua's actual spin is less than its maximum possible spin by about one part in a hundred trillion." So based on this mass and rotation rate, the equation I found for the innermost stable circular orbit would be only 1.000037636343 times the "gravitational radius" GM/c^2 (the gravitational radius is equal to 492.7 light-seconds, or 147707744 km, so 1.000037636343 times that would be a radius of 147713303.18 km). Meanwhile as I mentioned in the first post on that thread, the event horizon would be located at r = m + sqrt(m^2 - a^2), where m is the gravitational radius and a is the gravitational radius times 1 - 1.33266 * 10^(-14) (the rotation rate as a fraction of the maximum), which works out to about 492.70008 light-seconds or 147707720.02 km. Subtracting 147713303.18 - 147707720.02, I find that the distance between the event horizon and the innermost stable circular orbit would be only 5583.16 km--for any distance larger than or equal to that, stable circular orbits should be possible.