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Central force field-condition for closed orbits.

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data


    A particle moves in the central force field [itex]\overrightarrow{F}=-kr^{n}\hat{r}[/itex] , where k is a constant, and r is the distance from the origin. For what values of n closed stable orbits are possible?


    2. Relevant equations



    3. The attempt at a solution

    I thought for stable configuration Kinetic energy = potential energy.

    for central force field,

    [itex]\frac{mv^{2}}{r}=-kr^{n}[/itex]

    ie,[itex]KE,\frac{1}{2}mv^{2}=-kr^{(n+1)}[/itex]

    and [itex]PE = -\int Fdr=\frac{kr^{(n+1)}}{n+1}[/itex]

    For stabe configuration,

    [itex]-kr^{(n+1)}=\frac{kr^{(n+1)}}{n+1}[/itex]

    n=-2

    But the answer says there's two turning points at n=1 and n=-1.
    I think my method is absolutely wrong. :cry:

    Please help.
     
  2. jcsd
  3. Jan 29, 2012 #2

    vela

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    This can't be correct. For 1/r2 forces, for example, the potential energy is negative, so it obviously can't be equal to the kinetic energy. Also, the kinetic energy and potential energy vary over an orbit, so even if they were equal at one time, they wouldn't be equal later.

    Any other ideas on what's required for a stable orbit?
     
  4. Jan 30, 2012 #3
    Thank you vela.

    The energy should be a minimum for a stable orbit.is that correct?
     
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