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Graviatational force field

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data

    By considering the centripetal force acting on a man, calculate the minimum period of rotation that the Earth would need to have about its axis if a man at the equator were to experience zero normal contact force.
    Take radius of Earth=6400km

    2. The attempt at a solution
    this means the g=Fc(centripetral force)
    to find v= square root of ( radius of the Earth and g)
    g=9.81 N/kg and then sub it to T= (2pie x radius of Earth )/v
    Where is the error? Does a zero normal force means the person is experiencing weightlessness due to a=g?
     
  2. jcsd
  3. Aug 28, 2009 #2

    Doc Al

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    Staff: Mentor

    OK, but you mean mg = Fc.
    I don't see any error. What your final expression for T?
    Yes, that's what "weightlessness" means.
     
  4. Aug 29, 2009 #3
    . where radius of the Earth is 6400km
    But I can't get the ans=1.41hrs
    I can only get 84.3hrs
    square root of{Rg}=square root of (6400km X 35.3km/h)= 475
    (R multiply by 2pie)divide by T= 40212/T
    cross multiply them and I get 84.3hrs. Do u know where is the error?
     
  5. Aug 29, 2009 #4

    Doc Al

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    Staff: Mentor

    When you converted g from m/s^2 to km/h^2, you made an error.

    Instead, use standard units: meters and seconds, not km and hours. g = 9.8 m/s^2

    When you find the answer in seconds, then convert to hours.
     
    Last edited: Aug 29, 2009
  6. Aug 29, 2009 #5
    Thx Doc Al!:smile::biggrin:
     
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