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Formulation of orbital kinetic energy

  1. Jul 28, 2012 #1
    For a radial potential proportional to r^k, the virial theorem says [itex]\bar{T}=c(k)\bar{V}[/itex]. My problem says to show that c(k)=k/2 for a circular orbit.

    I actually solved the problem all ready (by setting the gradiant of the potential equal to the centripetal force and solving for 1/2*m*v^2.)

    However, while I was looking around, I came across something.

    I found a formulation of kinetic energy saying that [itex]\bar{T}= 1/2 \vec{∇}(V) .\vec{r} [/itex].

    How could you derive this? Is it true? Why? The original post said this came from the virial theorem.
     
  2. jcsd
  3. Jul 28, 2012 #2

    dextercioby

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    Just integrate with respect to time in the LHS and you'll get it.

    [tex] \frac{m}{2} \int_{t_1}^{t_2} \frac{dr}{dt}\left(\frac{dr}{dt} dt\right) = ... [/tex]
     
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