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Potential on the Earth surface

  1. Feb 3, 2007 #1
    1. The problem statement, all variables and given/known data

    G = 6.67*10^ -11 Nm^2kg^-2
    Radius of Moon = Rm = 1.73*10^6 m
    Radius of Earth = Re = 6.37*10^6 m
    acceleration due to gravity = g = 9.8m/s
    Mass of moon = Mm
    Mass of earth = Me
    Distance between the centre of the earth and the centre of the moon = D = 3.8*10^8 m

    Find the gravitational potentials on the moon's surface.

    3. The attempt at a solution

    V = - GMm/Rm
    V = - g * Re^2 / Rm
    V = - 9.8 * (6.37*10^6)^2 / 1.73*10^6
    V = - 2.30*10^8 J/kg

    The solution said that V should be
    -G* [ Mm/Rm + Me/(D - Rm) ]
    I don't understand why it should not be "Mm/Rm - Me/(D - Rm)" :confused:

    Thank you very much!
  2. jcsd
  3. Feb 3, 2007 #2
    Because the earth contributes a potential too. I guess you are supposed to assume that the problem wants the potential on the light side of the moon.
  4. Feb 4, 2007 #3
    Thank you! In the past I thought that some potential of Earth will be cancelled out by potential of the moon :tongue2:
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