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Energy in orbital motion

  1. Feb 25, 2015 #1
    Hey, so I have a question about motions of planets and their energy basically.

    When we have a circular orbit, why is it that the kinetic energy is just the opposite of potential energy? (Assuming it's a closed orbit)

    Like if we have U = something, than the kinetic energy T = -1/2U? This would be saying the kinetic energy doesn't change for a circular orbit but the potential energy does, and than I would think this would be a parabolic orbit as energy would then equal to 0 and epsilon (eccentricity) is equal to 1.

    I hope that made sense, I'm having trouble understanding such motion.
  2. jcsd
  3. Feb 25, 2015 #2

    Simon Bridge

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    Have you followed the derivation?
    http://www.pha.jhu.edu/~broholm/l24/node1.html [Broken]

    No - if U changes, the T will also change. If U does not change, then neither does T.
    Note: that should be T=-(1/2)U
    Last edited by a moderator: May 7, 2017
  4. Feb 25, 2015 #3
    Interesting, because recently I did a problem, for which the kinetic energy remained the same and the potential energy had changed, so that is where most of the confusion comes from.
  5. Feb 25, 2015 #4

    Simon Bridge

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    It is possible T to remain the same and for U to change - this happens, for eg, when you lift an object at a constant speed - I'm not saying that cannot happen. I am saying that the relation T=-(1/2)U does not indicate that either U or T will change or remain the same. Instead it tells you the relationship between U and T for a circular orbit.
    See the link in post #2.
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