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Question: Kepler's third law of planetary motion

  1. Mar 13, 2007 #1
    The mean distance between the earth and the moon is 3.84x10^8 m, and the moon has an orbital period of 27.3 days. Find the distance from Earth of an artificial satellite that has an orbital period of 8.5 days.
    a. 1.76x10^8 m
    b. 1.76x10^4 m
    c. 1.76x10^10 m
    d. 5.24x10^8 m

    So far I got:
    I used the formula: (Ta/Tb)^2 = (rA/rB)^3
    Plugged in these values:
    Ta= 8.5 days
    Tb= 27.3 days
    rA= ?
    rB= 3.84x10^8 m

    This obviously is incorrect. Please help thanx!
  2. jcsd
  3. Mar 13, 2007 #2


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    Is (10^24)^(1/3)=10^24? Something is going seriously wrong with your cube root.
    Last edited: Mar 13, 2007
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