Potential for particle circular motion

  • Thread starter MartinK
  • Start date
  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
2
0
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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