0
$\begingroup$

Scenario:

Consider a huge black hole, with the following properties:

$M = 10^{12}M_{sun}$

$r=19741AU$

$g=15.2m/s^2$ (measured at event horizon)

Consider then an orbiting platform just outside the event horizon, with some sort of external long range force carrying mechanism used to provide it the energy required to keep its orbit from decaying into the event horizon.

Finally, consider that this orbiting platform dangles a small weight on a strong thread, such that this weight passes below the event horizon of the black hole.

What mechanism would prevent the platform from then reeling in the dangling weight? A $1kg$ weight (assume the weight of the strong thread is negligible) would only experience a force of about $15N$ due to the gravity of the black hole and because of the very large radius, this would not increase noticeably a kilometer or two upwards or downwards of the orbiting platform's just-barely-outside orbit.

Alternate scenario:

What if an object fell into a black hole, passing just barely below the event horizon, and a huge gravitational wave from a nearby merger event travelled through it? I understand vaguely that black holes are inescapable because of the curvature of space, but gravitational waves are themselves curvatures of space. Could a sufficiently strong gravitational wave "catapult" an object safely out of a black hole's event horizon by briefly curving space against the pull of the singularity?

$\endgroup$
  • $\begingroup$ I got $g=1.52\times 10^{-11}$, much less gravity than you have. I think you could use a more realistic black hole size to formulate the question. $\endgroup$ – JMLCarter Feb 10 '17 at 3:42
  • $\begingroup$ I redid the math a few times (for which I confessed, I used wolfram alpha) but I am still getting about 15. I get 1.5*10^-11 for one solar mass. Can you verify? $\endgroup$ – Wug Feb 10 '17 at 8:42
  • 1
    $\begingroup$ Oh, yes, 10^12 solar masses, ok. I would like an answer too, tbh. (10^10 seems to be more likely and doesn't diminish the question.) $\endgroup$ – JMLCarter Feb 10 '17 at 21:25
  • $\begingroup$ Multiply the g by the time dilation factor, I mean multiply by a large number. Because surface gravity of a black hole is a high altitude person's view of the acceleration at event horizon. $\endgroup$ – stuffu Feb 14 '17 at 15:28
1
$\begingroup$

Short answer:

Because the force required to keep an object at the constant distance from the black hole approaches infinity as the object approaches event horizon.

Long-ish answer:

This is a consequence of geometry of spacetime around the horizon. Locally, event horizon is like a surface going outward at the speed of light (also called light-like surface). In order to avoid falling in it, you have to run away from it. You have to accelerate. And closer it is, faster you have to accelerate.

Locally, this is like hyperbolic motion.

As you approach the light-like surface, the required acceleration approaches infinity. Which means that required force also approaches infinity, hence the statement in the short answer. Once you loose the battle and the surface intercepts you, there is no return - for the surface is moving away from you at the speed of light.

$\endgroup$
0
$\begingroup$

As the upper end of the thread is close to an event horizon, it feels a large force pulling it.

As the lower end of the thread is even closer to an event horizon, it feels an enormous force pulling it.

Now let me explain why the different parts of the thread feel what they feel:

The job of a thread is to transfer momentum. If we ask the upper end of the thread, it says that the rate at which the black hole is pouring momentum on the 1 kg mass is large. The lower end says that the rate is enormous.

Gravitational time dilation is the reason for those different opinions about that rate.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.