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With a circular orbit, when do I use 1/2mv^2

  1. Jan 17, 2017 #1
    1. The problem statement, all variables and given/known data
    At a circular orbit and at an elliptical orbit when do I use 1/2mv^2 instead of the kinetic equation from which I derived from F=GMm/r^2 which is Ek = GMm/2r

    2. Relevant equations
    F=GMm/r^2, Ek = GMm/2r,
    Ek = 1/2mv^2

    3. The attempt at a solution
    For instance, when I tried to use the other kinetic energy equation (GMm/2r) to solve the kinetic energy of a satellite at perigee, it gave a totally different answer if I use 1/2mv^2
     
  2. jcsd
  3. Jan 17, 2017 #2

    PeroK

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    You need to quote the precise question and show your working.
     
  4. Jan 17, 2017 #3

    gneill

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    Perhaps you're confusing the total mechanical energy which includes both kinetic and potential energy and is a constant over the whole orbit, with the kinetic energy alone at a particular location?
     
  5. Jan 17, 2017 #4
    Your kinetic equation is valid only for circular orbits. It is derived by balancing the forces.

    GMm/r2 = m v2 / r

    The problem is that the r on the left is the distance from the sun (or whatever) and the r on the right is the radius of curvature of the motion. These are only the same if the orbit is circular.
     
  6. Jan 18, 2017 #5
    In
    Is total energy conserved in a circular orbit or just an elliptical orbit?
     
  7. Jan 18, 2017 #6

    PeroK

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    A circle is an ellipse with eccentricity 0.
     
  8. Jan 18, 2017 #7
    In this system you have only gravitational potential energy and kinetic energy, so between those two quantities energy must be conserved for any orbit.
     
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