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Homework Help: Gravitation energy 5

  1. Jul 5, 2014 #1
    1. The problem statement, all variables and given/known data
    At which minimal speed must a stone be thrown from the moon in order to reach earth.
    R is the earth's radius and r the moon's.
    M is the earth's mass and m the moon's.
    I ignore the stone's mass, it cancels

    2. Relevant equations
    R=6.4E6 [m]
    r=1.7E6 [m]
    M=6E24 [kg]

    3. The attempt at a solution
    There is point A at a distance 54R that the forces equal. to reach there:
    $$-\frac{GM}{81r}-\frac{GM}{60R-1.7E6}+\frac{V^2}{2}=-\frac{GM}{81\cdot 6R}-\frac{GM}{54R}$$
    $$\frac{V^2}{2}=GM\left(-\frac{1}{81\cdot 6R}-\frac{1}{54R}+\frac{1}{81r}+\frac{1}{60R-1.7E6}\right)$$
    $$V^2=2\cdot 6.7E-11 \cdot 6E24 \left( \frac{-54-81\cdot 6}{81\cdot 6\cdot 54\cdot 6.4E6}+\frac{1}{81\cdot 1.7E6}+\frac{1}{60\cdot6.4E24-1.7E6}\right)$$
    It should be V=2.26 [km/sec]

    Attached Files:

  2. jcsd
  3. Jul 5, 2014 #2


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    You've already correctly found the (approximate) distance that it needs to travel in order to fall towards earth instead of towards the moon.

    From conservation of energy:
    ΔKE = ΔGPE
    (KE = Kinetic Energy ... GPE = Gravitational Potential Energy)

    ΔGPE will be the change between the moon's surface and the point 6R from the center of the moon

    ΔKE will be (the negative of) the entire KE of the object (because we want it to run out of speed when it gets to that point, because the problem asked for the minimum speed)

    So, essentially the problem comes down to finding the change in Gravitational Potential Energy.

    So what would you say the change in GPE would be?
    Last edited: Jul 5, 2014
  4. Jul 5, 2014 #3


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    I'll try to discern what your math says is the ΔGPE

    I'll rearrange your equation to get:

    On the left side we have ΔGPE and on the right side we have ΔKE
    (except for the common factor of the mass of the object, which we will ignore)

    Seems right to me.

    When I calculate it, though, my answer is 2.02 km/s

    I'm not sure why it's a little different than the given answer of 2.26 km/s

    (I calculated it before my first reply and got 2.31 km/s ... I'm not sure what I did inconsistently)

    Perhaps the discrepensy is from the estimation of the distance (54R and 6R) where the gravity is equal.

    P.S. I think there's a E24 where there should be an E6 in your final equation (the last term)
    Last edited: Jul 5, 2014
  5. Jul 5, 2014 #4
    You are right about the last term:
    $$V^2=2\cdot 6.7E-11 \cdot 6E24 \left( \frac{-54-81\cdot 6}{81\cdot 6\cdot 54\cdot 6.4E6}+\frac{1}{81\cdot 1.7E6}+\frac{1}{60\cdot6.4E6-1.7E6}\right)$$
    This gives V=2315 which is closer.
    The distance to the equilibrium point A is 54R according to the book
  6. Jul 6, 2014 #5


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    Well then that's strange. I'm not sure what else could account for the error.
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