Black hole and acceleration

  • #1
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Homework Statement



A man is hovering above a Schwarzschild black hole. He is 1 meter tall. What would be the radius of the black hole if the man's head is accelerating away from his feet at a rate of 10 m/s^2.

Homework Equations





The Attempt at a Solution



The force on the man would be equal to GMm/r^2, where M is the black hole mass and m is the man's mass and r is the Schwarzschild radius. This means the acceleration is GM/r^2 ... but I've hit a snag.
 
  • #2
1) the radius of the feet and head aren't the same. You are doing tidal forces here. 2) GMm/r^2 is not very relativistic. Is this really a relativity problem?
 
  • #3
I wasn't certain where to post this question and this appeared to be the most appropriate. I can post it elsewhere, if you suggest. I did realize that it's a tidal forces problem, but I came into difficulties. The force at the man's feet is different from the force at his head, but I don't know how to take the next step.
 
  • #4
Maybe I should ask you what snag or difficulty you hit? You have a (Newtonian) equation for the acceleration as a function of radius. So the inner radius is r and the outer radius is r+(1 m). Taking the difference and setting it equal to 10m/sec gives you an equation in M and r. If r is the Schwarzschild radius of a hole of mass M, you know another relation between M and r. Solve the two simultaneous equations. I would wonder if using GM/r^2 would be good enough, but if the differential acceleration is as small as 10m/sec, it probably is.
 
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  • #5
I did think about the simultaneous equations, and I also thought about differentiating with respect to r, but I'm not clear how to do it. I feel like there is not enough information, but there must be.
 
  • #6
You can approximate the acceleration difference with a derivative if you wish. But you don't have to. You can still solve the system directly. So just do it. What did you get? There is no missing information.
 
  • #7
Do you mean the following:

equation 1: GM/r - GM/(r+1) = 10
equation 2: 2GM/c^2 = r

Do you mean all I have to do is to solve these 2 equations? Then it's no problem.

How would I go about doing it with a derivative instead?
 
  • #8
Hi again,

Thanks for the help. I found that doing it with a derivative makes it easier to do the simultaneous equations. I've solved it, and it was easier than I thought. Thanks again
 

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