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Centripetal force and kepler's laws

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    I'm not sure about these two

    2. Relevant equations
    f=mv^2/r
    F=Gm1m2/r^2


    3. The attempt at a solution

    for the first one since the centripetal acceleration always points towards the center of circle do you just do. mg-mv^2/r=F

    second one the equation is G=m1m2/r^2 So dont you just divide the original force of the astronaut on the moon by 1.3 squared? It was wrong so I'm not sure what to do
     
  2. jcsd
  3. Oct 2, 2008 #2

    cepheid

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    For the first one, both forces are upward. Normally, if the man were stationary, the ground would only have to counter his weight. But now, it must also provide the centripetal force necessary to keep him moving in a curved path. Hence:

    F = mg + mv^2/r

    Intuitively, as he "bottoms out", he'll feel heavier than he normally does. His larger apparent weight is due to this extra force.
     
  4. Oct 2, 2008 #3

    cepheid

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    For the second one, I think you're correct. You may not be given m1, m2, or R_moon, but you don't really need that information because you HAVE the force at one distance, and the force at the larger distance can be inferred from the fact that this is an inverse-square law force. All other parameters mentioned remain the same.
     
  5. Oct 2, 2008 #4
    i got .48244 for the first one and it was wrong and the second one i did what i said but the answer was also wrong

    I just called a friend and for the first one it's just mv^2/r divided by a thousand...lol
    and the second one you have to add 1 to 1.3 cause the orbit is 1.3x not the whole radius. tricky tricky
     
    Last edited: Oct 2, 2008
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