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Centripetal Acceleration Part II

  1. Mar 3, 2008 #1
    [SOLVED] Centripetal Acceleration Part II

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

    (From previous question) In order for a satellite to move in a stable circular orbit of radius 6689 km at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius of the orbit.

    What is the speed of the satellite? The G is 6.67259e-11 and the mass of the earth is 5.98e24kg.

    The answer to this was 7723.55 m/s.

    Part II:

    Find the time required to complete one orbit. Answer in units of h.

    2. Relevant equations

    The book only gives one equation for T referencing orbits, which is v=2pir/T

    3. The attempt at a solution

    7723.55 = 2pi/T
    7723.55T = 2pi
    T = 2pi/7723.55
    T = .000814

    I think the homework system even laughed at me when I entered that answer :)
  2. jcsd
  3. Mar 3, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor



    Compare the two equations. (You left off the r.)

    And don't forget to convert your answer to the desired units.
  4. Mar 3, 2008 #3
    *sigh* I keep making really silly mistakes.

    So I input the correct equation but it came out correct.

    7723.55 m/s = 2pi * 6689000m/T
    7723.55T = 4.20282e7
    T = 5221.57 seconds

    Convert to hours

    5221.57/60 = 90.6928 minutes
    90.6928/60 = 1.51155 hours

    Thank you so much.
    Last edited: Mar 3, 2008
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