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Kinetic energy of space probe launched from planet

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Zero, a hypothetical planet, has a mass of 3.0 x 10^23 kg, a radius of 3.0 x 10^6 m, and no atmosphere. A 17 kg space probe is to be launched vertically from its surface.

    If the probe is launched with an initial kinetic energy of 5.0 x 10^7 J, what will be its kinetic energy when it is 4.0 x 10^6 m from the center of Zero?


    2. Relevant equations



    3. The attempt at a solution

    I figure that mechanical energy is conserved, and so

    K1 + U1 = K2 + U2

    With U given by GmM/r.

    So

    [itex]K_{2} = K_{1} + GmM(r^{-1}_{2} - r^{-1}_{1})[/itex]

    With r1 = 3.0x10^6, and r2 = 3.0x10^6 + 4.0x10^6 = 7.0x10^6

    Which gives me the negative value -14794285.71 for K2.

    I have no idea what is wrong with the equation I'm using.

    Any ideas?
     
  2. jcsd
  3. Apr 4, 2012 #2
    gravitational potential energy is negative

    also I think that your r2 is a bit too big
     
  4. Apr 4, 2012 #3
    Yes, so

    K1 - (GmM/r1) = K2 - (GmM/r2)

    And

    K2 = K1 - (GmM/r1) + (GmM/r2)
    K2 = K1 + (GmM/r2) - (GmM/r1)
    K2 = K1 + GMm((1/r2)-(1/r1))

    Which is what I used..
     
  5. Apr 4, 2012 #4
    I did

    K2 = K1 - U1 + U2
    K2 = K1 -(U1 - U2)
    K2 = K1 -GmM((1/r1) - (1/r2))

    and got 21652500 J

    I also got the same number when I used your method

    I think the problem is your value for r2
     
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