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Earth-Moon Center Line of Gravity

  1. Oct 12, 2008 #1


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    1. Locate the position of a spaceship on the Earth-Moon center line such that, at that point, the tug of each celestial body exerted on it would cancel and the craft would literally be weightless. Answer in meters from the moon

    2. Moon mass = 7.36 x 10^22 and Earth Mass = 5.97 x 10^24

    3. My professor gave my class the following advice for this problem. You have probably written an equation in which the force exerted by the Earth is set equal to the force exerted by the Moon on the spaceship. That simplies, but leaves two unknown distances. Since there are two unknowns, a second equation is needed. What is the sum of the two distances equal to? Now you have a second equation. Substitute it into the first, eliminating one of the distances. Notice the unknown variable is quadratic. Expand the equation into the form ax^2 +bx + c =0. The apply the quadratic formula to solve for the unknown distance (from spaceship to Moon).

    I am still very confused, please help.
  2. jcsd
  3. Oct 13, 2008 #2
    Hmm interesting, this is how I would have done it:

    [tex] a = \frac{GM}{R^2} [/tex]

    Therefore for the force to be equal (because the spaceship is of constant mass), [tex] \frac{GM}{R^2} = \frac{Gm}{r^2} [/tex]

    However here we have two variables for radius (R and r). One is measured from the moon, and the other is measured from the earth. What we must do is get the radius from the earth, in terms of radius from the moon.
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