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Cow jumps over the moon

  1. Apr 1, 2007 #1
    A. How much energy must a 450.Kg cow expend in jumping over the moon?(excluding the moon's gravity and the earths atmosphere),

    B. Assuming all the work is done during the jump itself, what would her initial velocity have to be?

    C. What would her final velocity be upon her return to Earth?

    D. Again assuming all the work would be done during the initial jump, i.e. for a distance approximately the length of her legs ( 1 meter ), what would her acceleration need to be.

    2. Relevant equations
    EL = G Me m(1/r0 - 1/r1)

    3. The attempt at a solution
    Substituting values in:
    G = 6.67 x 10^-11 Nm^2 / kg^2
    Me = 5.98 x 10^24 kg
    m = 450. kg
    r0 = 6.38 x 10^6 m
    r1 = 3.85 x 10^8 m

    Using all those in the big equation I get this:
    2.77 x 10^10 J This would be her kinetic energy for A hopefully.

    For part B I just used KE = 1/2 mv^2
    Substituting everything in and solving for V I get 5550 m/s

    I don't quite get question C, how would she even return to earth with that velocity?

    For part D I haven't quite worked it out yet but I'am going to use
    v^2 = V0^2 + 2a Delta t and just solve for the acceleration. But this assumes my velocity is right, which i'am not sure of.

    Have I missed anything important in my thinking here? I can't really check my answers for reasonability :rolleyes:
    Last edited: Apr 1, 2007
  2. jcsd
  3. Apr 1, 2007 #2
    are ro and r1 the raddii from earths center to surface and the additional distance to the far side of the moon. If so those numbers look off.

    I'm guessing the distance to the farside is something like 250,000mi plus another 2160 for the moons diam. thats about 250,000/.6 Km. you may have right numbers, not sure, but the moon energy term should be expressed as r0+r1 where r1=distance from surface of earth to far side.

    Elsie would get back the same way she got there, and will have the same velocity profile on the way down as up, just as a rock hurled upwards. (ignoring the obvious orbital mechanics issues)

    another way to calulate the acceleration which avoids messing with time, uses V^2=2a*x where x = 1 meter.
    Last edited: Apr 1, 2007

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