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Universal Gravitation

  1. Jul 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Given: The universal gravitational constant G = 6.67 E-11, the mass of the earth M = 5.98E24, and its radius R = 6.7 E6. How much work must an external force do on the satellite to move it from a circular orbit of radius 2R to 3R, if its mass is 2000kg? Answer in Joules.


    2. Relevant equations

    I know that E = -GMm/2r, where M is the mass of the earth and m = satellite.


    3. The attempt at a solution
    So I calculated the total energy in each radii,
    for 2R, -GMm/4R
    for 3R, -GMm/6R
    then I subtract one from the other to get the net energy change. I don't know what to do after that. Please help. Thank you!
     
  2. jcsd
  3. Jul 6, 2007 #2

    Kurdt

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    There is a considerably simpler way of doing this. The work done in moving an object from one orbit to another is:

    [tex] W=\int_{r_1}^{r_2} F(r) dr [/tex]

    EDIT: Sorry forgot to say that your method is fine for potential energy not total mechanical energy as you have used.
     
    Last edited: Jul 6, 2007
  4. Jul 6, 2007 #3
    thank you so much! :)
     
  5. Jul 6, 2007 #4

    Dick

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    Actually, I think he is ok. The kinetic energy at radius R is GMm/(2R). The potential is -GMm/R. So the sum is, as he states, -GMm/(2R).
     
  6. Jul 6, 2007 #5

    Kurdt

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    Was just the procedure in the attempt at the solution that made me think it was total mechanical energy.
     
  7. Oct 13, 2007 #6
    I have the exact same question. This is one of the method that I used:
    E intial = Ki + Ui = 0 + (-GMm/2R)
    E final = Kf + Uf = (1/2)(GMm/3R) + (-GMm/3R)
    then Work = E final - E intial

    this make sense right? Then why it doesn't work

    I also try the integrate method, but nothing work! Can somebody please tell me why?
    and how to fix it?
     
  8. Oct 13, 2007 #7

    Dick

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    If it's in orbit at radius 2R, how can Ki=0?
     
  9. Oct 13, 2007 #8
    ok, so Ki is not = O; Then why is it the integration method didn't work either?
     
  10. Oct 13, 2007 #9

    Dick

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    They both work. You just have to do them right.
     
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