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Homework Help: Minimum black hole mass to survive fall to event horizon

  1. May 25, 2010 #1

    cepheid

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    I'm having no end of trouble with this seemingly simple problem:

    1. The problem statement, all variables and given/known data

    What's the minimum mass of a black hole for which you could survive a fall through the event horizon without being ripped to shreds? Why would you be ripped to shreds for smaller black holes?

    2. Relevant equations

    [tex] \textbf{F} = -G\frac{mM}{r^2}\hat{\textbf{r}} [/tex]

    [tex] R_s = \frac{2GM}{c^2} [/tex]​
    (Schwarzschild radius)

    3. The attempt at a solution

    Assume person of height h (= 2 m) can survive a tidal stretching force of 5mg over the length of his body (g being the acceleration due to gravity on Earth). Assume further that he is to be just barely surviving when his feet touch the event horizon. Then, the difference between the gravitational acceleration on his head and his feet should be:

    [tex] \Delta a = -GM\left[\frac{1}{R_s^2} - \frac{1}{(R_s + h)^2} \right] = 5g [/tex]​

    When I try and solve this, I get some cubic equation:

    [tex] \frac{c^2 h^2}{10g} = R_s^3 + 2R_s^2h + R_s\left[h^2 -c^2 \frac{h}{5g}\right] [/tex]​

    I spoke to other people who said they solved this easily and got something on the order of 104 solar masses as the lower limit. Have I done something wrong with my algebra (or worse yet, with the physics)?
     
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  3. May 25, 2010 #2

    Redbelly98

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    Without checking your math over, what if you assume h«Rs and drop all the small terms in your equation? (Then double check the h«Rs assumption after you calculate the final answer.)
     
  4. May 25, 2010 #3

    cepheid

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    Yes, if you drop h2 terms everywhere, as well as dropping one rh term when it appears added to an r2 term, you get the same expression as you do if you approximate the change in force by:

    [tex] \Delta F = \frac{dF}{dr} \Delta r [/tex]​

    I got a result of M >= 20 496 solar masses, which corresponds to a Schwarzschild radius of 60 536 km (much greater than two metres!).

    Thanks for the tip. Also, I noticed that dropping h2 terms is equivalent to Taylor expanding (r + h)2 to first order.
     
  5. May 26, 2010 #4

    Redbelly98

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    Cool, and I get the same Rs as you. (Didn't check the mass calculation.)
     
  6. May 26, 2010 #5
    I used the following equation for tidal forces-

    [tex]dg=\frac{2Gm}{r^3}dr[/tex]

    set dg=5g where g is Earth's gravity, r=rs where rs is the Schwarzschild radius [itex](r_s=2Gm/c^2)[/itex], dr=2 and solved for m, the answer I got was 20,495 sol.
     
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