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Period of a satellite

  1. Mar 4, 2008 #1
    [SOLVED] Period of a satellite

    1. Given: G = 6.67 x 10^-11 N m^2/ kg^2
    The acceleration of gravity on the surface of a planet of radius R = 4910 km is 11.8 m/s^2. What is the period T of a satellite in circular h = 8445.2 km above the surface? Answer in units of s.

    2. F = GMm/r^2
    F = ma = mv^2/r
    v = sqrt(GM/r)
    T = 2pi*r/v

    3. i got r = 13355200 m; a = 11.8 m/s^2; that gave me the velocity = 12553.54m/s. i plugged it into the period equation and got T = 6684.425s but its wrong
  2. jcsd
  3. Mar 4, 2008 #2
    nvm i got it
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