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Gravitation above Black Hole

  1. Apr 13, 2006 #1
    The radius Rhand mass Mh of a black hole are related by Rh = 2GMh/c^2, where c is the speed of light. Assume that the gravitational acceleration agof an object at a distance Ro = 1.001Rh from the center of a black hole is given by Ag = G M / r^2 (it is, for large black holes). (a) What is ag at ro for a very large black hole whose mass is 1.61 × 10^14 times the solar mass of 1.99 × 10^30 kg? (b) If an astronaut with a height of 1.66 m is at ro with her feet toward this black hole, what is the difference in gravitational acceleration between her head and her feet?


    Rh = 2(6.67x10^-11)(3.204x10^44)/ (3.0x10^8)^2
    Rh = 4.75 x 10^17

    So then, Ag = GM/ (1.001Rh)^2
    Ag = (6.67x10-11)(3.204x10^44)/(1.001 x 4.75x10^17)^2
    Ag = 0.0945 m/s^2

    I find myself having trouble with part B. How do i relate them?

    I did Ag = (6.67x10^-11)(3.204x10^44)/(1.66 x 4.75x10^17)^2
    Ag = 0.0344 m/s^2
    So i did the difference btwn answer A and B and got .0601 m/s^2

    But i know its wrong. Can someone help me out? Thanks!
     
  2. jcsd
  3. Apr 13, 2006 #2

    Astronuc

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    Staff: Mentor

    In part B,

    calculate ag at ro and at (ro + 1.66 m).

    This assumes the astronaut's feet are at ro.
     
  4. Apr 14, 2006 #3
    So isnt Ro part A's answer?

    and when you say Ro + 1.66, does that mean...
    Ro =(1.001+1.66)Rh?
     
  5. Apr 15, 2006 #4

    Astronuc

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    Staff: Mentor

    According to the problem, in part A, one is to calculate ag at ro.

    In part B, find the between ag(ro) and ag(ro+1.66m).

    It should be something like A * (1/ro2 - 1/(ro+1.66m)2), where A is some constant.
     
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