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Homework Help: Finding distance of a planet whose orbital period is 2x Earth

  1. Jul 28, 2008 #1
    A small planet has an orbital period that is twice that of Earth. What is the planet's orbital distance?

    From Kepler's 3rd law of planetary motion, I can assume that:

    rp3 / rE3 = Tp2 / TE2

    where rp is the orbital radius of the planet and Tp is the orbital period of the planet


    rp3 = (Tp2 / TE2) rE3

    = [ (2 x TE)2 / TE2 ] rE3


    rp = 3√ ([ (2 x TE)2 / TE2 ] rE3)

    = 1 060 071 250 m

    which differs from the given answer. What error have I made?
  2. jcsd
  3. Jul 28, 2008 #2


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    Gold Member

    I imagine this is a shortcut but, if the period is 2x, isn't the radius simply √2x?

    (No, it's other way round. If the radius is 2x, then the orbital velocity is √2x.)
  4. Jul 28, 2008 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    The equation looks okay, did you use the right values?

    Also, reducing the equation further might help as it reduces chances for math errors.

    (hint: by some slight re-arranging, you can eliminate a variable from the equation.)
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