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Black hole and acceleration

  1. May 22, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. 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.
     
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  3. May 22, 2008 #2

    Dick

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    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?
     
  4. May 22, 2008 #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.
     
  5. May 22, 2008 #4

    Dick

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    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.
     
    Last edited: May 23, 2008
  6. May 22, 2008 #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.
     
  7. May 23, 2008 #6

    Dick

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    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.
     
  8. May 23, 2008 #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?
     
  9. May 23, 2008 #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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