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Potential for particle circular motion

  1. Mar 30, 2010 #1
    Hi everyone!

    I have a small question that bothers me. Consider a test particle in the Earths gravitational field on circullar obrit. Speciffic effective potential is
    [tex]
    \begin{equation}
    \widetilde{V}_{\rm eff(1)} = \frac{1}{2} \frac{l^2}{m^2} \frac{1}{r^2} - \frac{GM}{r},
    \nonumber
    \end{equation}
    [/tex]
    where
    [tex]
    \begin{equation}
    l = m r^2 \Omega = {\rm const.}, \nonumber
    \end{equation}
    [/tex]
    because [tex]$\theta$[/tex] is cyclic coordinate. But we have partice in circular orbit, so [tex]$\Omega = $[/tex]const. too, and we can write
    [tex]
    \begin{equation}
    \widetilde{V}_{\rm eff(2)} = \frac{1}{2} \Omega^2 r^2 - \frac{GM}{r}.\nonumber
    \end{equation}
    [/tex]
    Now we have correct form:
    [tex]
    \begin{equation}
    \frac{ \partial^2 \widetilde{V}_{\rm eff(1)}}{\partial r^2} = \frac{3 l^2}{m^2 r^4} - \frac{GM}{r} = 3 \Omega^2 - \frac{2GM}{r^3}.\nonumber
    \end{equation}
    [/tex]
    and incoorect form
    [tex]
    \begin{equation}
    \frac{ \partial^2 \widetilde{V}_{\rm eff(2)}}{\partial r^2} = \Omega^2 - \frac{2GM}{r^3}.\nonumber
    \end{equation}
    [/tex]

    Why is there a difference between [tex]
    $ \partial^2 \widetilde{V}_{\rm eff(1)} / \partial r^2, $ [/tex] and [tex] $\partial^2 \widetilde{V}_{\rm eff(2)} / \partial r^2, $
    [/tex]?

    Thanks for any replies, have a nice day with physics :-)
    Martin
     
  2. jcsd
  3. Apr 7, 2010 #2
    I found the answer with the help of my friends.
    Second potential is wrong, because $\Omega$ is constant only with time - it depends on $r$.
     
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