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Jupiter's Equatorial Rotation Period

  1. Nov 19, 2009 #1
    How would I find Jupiter's equatorial rotation period when it begins to tear the planet apart?

    Is it true that if the centrifugal force is greater than the gravitational force the material can be torn apart?

    If this is right, can I just set up an inequality for the centrifugal force > gravitational force

    Fg = GM/R^2 Fc = (4(pi^2)R) / P^2

    and then just solve for P throughout the inequality....

    P > sqrt[ (4(pi^2)R^3)/ GM) where R(Jupiter) = 7.14*10^7 and M (Jupiter) = 1.9*10^7

    If I do this, I will get a rotation period of 7.39 minutes. I am wondering if the way I went about the problem is right. As far as Jupiter's current rotational period of 590.5 mins, the period at which it would be torn apart is nearly 80x shorter. Any other suggestions for doing this problem?
  2. jcsd
  3. Nov 19, 2009 #2


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    Sounds reasonable, what's the problem?
  4. Nov 19, 2009 #3
    I just am never very confident with my answers. Thanks for responding though!
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