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I need Help. Rotational Kinetic Energy of Earth and the Moon.

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the ratio of the rotational kinetic energy of the Earth to the rotational kinetic energy of the Moon as they spin about their axes?

    3. The attempt at a solution


    Me = Mass of Earth (5.97 × 1024 kg)
    Re = Radius of Earth (6.37 × 106 m)
    Mm = Mass of Moon (7.35 × 1022 kg)
    Rm= Radius of Moon (1.74 × 106 m)

    this is all i know I don't know where to start. I know KE=(1/2)(moment of inertia)(angular velocity)
     
  2. jcsd
  3. Mar 22, 2010 #2
    The moment of inertia for a sphere is
    [tex]I = \frac{2}{5}MR^2[/tex]

    And you can find the angular velocity for the Earth because you know it has a period of one day. I think you'll have to find the length of a day on the moon.

    After that just divide the two energies, you've got the right equation and all the rest of the right info though.
     
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