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Newton's Law of Gravity and Potential Energy

  1. Dec 27, 2005 #1

    G01

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    2 Jupiter sized planets are released [tex] 1.0 X 10^11 [/tex] m What are their speeds as they crash together?
    I decided to try to do this problem with Potential Energy from newtons law of gravity that is
    [tex] U_{g} = \frac{\-GMM}{r} [/tex]
    I set the 0 of potential energy at the point when the planet's center's crash together. So the change in potential will be the starting point minus the point when the planets just hit (when the distance between them is twice the radius of Jupiter- their outer edges are just touching). Mathmatically this is:
    [tex] \Delta U_{g} = \frac{\-GMM}{1.0014 X 10^11 m} - \frac{\-GMM}{1.398 X 10^8} [/tex]
    Now I should be able to just set the change in kinetic energy equal to the change in potential, but I'm not gettign the right answer. Can someone show me whats wrong with my reasoning. For anyone who has the book this problem is in Knight Chapter 12 #49. Thanks Alot.
     
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  3. Dec 27, 2005 #2

    G01

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    sorry in the second latex line the first term should be negative with the second term positive i think. And in the first line taht should be [tex] 1.0X10^(11) m [/tex] (im not to good at latex yet :rolleyes: )
     
  4. Dec 27, 2005 #3
    did you make sure that both planets have kinetic energy? you may need to show some more work.

    [tex] U_i = \frac {-GMM}{1 \times 10^{11} m} [/tex]

    [tex] U_f = \frac {-GMM}{2(radius of jupiter)} [/tex]

    i think you already have this down so far.

    [tex] U = KE_1 + KE_2 [/tex]
    [tex] KE_1 = KE_2 [/tex]
    [tex] U = 2KE [/tex]
    [tex] \Delta U = 2 KE [/tex]
     
    Last edited: Dec 27, 2005
  5. Dec 27, 2005 #4

    G01

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    the way i was thinking is that both planets move the same distance in relation to eachother and they'd both have the same KE. What your saying though is that each planet would have half of the kinetic energy. I guess that makes sense. Let me try it I'll post when i come up with anything.
     
  6. Dec 27, 2005 #5

    G01

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    OK that was my problem. I don't know why i didn't see that in the first place. Thanks alot andrew you were a big help!
     
  7. Dec 28, 2005 #6
    The system is under the influence of gravity alone, so we one can relate the initial and final state of their velocity, and position using the conservation of mechanical energy.
     
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