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I'm trying to show that for the perihelion precession $\Delta\phi$ follows:

$$\Delta\phi=2\int_{r_\textrm{min}}^{r_\textrm{max}}\frac{L}{r^2(E-U_\textrm{eff}(r))^{1/2}}~\mathrm dr$$

where $L$ is the angular momentum (constant), $E$ the total energy and $U_\textrm{eff}$ the effective potential.

What I did so far:

The total energy is given as $$E=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}+U(r)$$ and $$L=mr^2\dot{\phi}$$

$r$ is a function of $\phi$, it follows: $$\dot{r}=\frac{\mathrm dr}{\mathrm d\phi}\frac{\mathrm d\phi}{\mathrm dr}=\dot{\phi}~\frac{\mathrm dr}{\mathrm d\phi}$$

Separation of variables:

$$\begin{align}\phi &=\int\frac{L}{r^2\left(2m(E-U(r))-\frac{L^2}{r^2}\right)^{1/2}}~\mathrm dr\\ \implies \Delta\phi &=2\int_{r_\textrm{min}}^{r_\textrm{max}}\frac{L}{r^2\left(2m(E-U(r))-\frac{L^2}{r^2}\right)^{1/2}}~\mathrm dr\end{align}$$

But I don't know how to write this with the effective Potential $U_\textrm{eff}$ I found that $U_\textrm{eff}(r)=U(r)+\frac{L^2}{2m^2r^2}$ but how could I use that?

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  • $\begingroup$ I am not sure I understand your question. If you plug in your expression for $U_\text{eff}$ in the initial equation, you get your final equation. $\endgroup$
    – Fabian
    Commented Nov 30, 2016 at 4:19
  • $\begingroup$ If I put $U_\textrm{eff}(r)=U(r)+\frac{L^2}{2m^2r^2}$ in $E-U_\textrm{eff}(r)$ I get $E-U_\textrm{eff}(r)=E-U(r)-\frac{L^2}{2m^2r^2}$ but this isnt my final equation... $\endgroup$
    – user137707
    Commented Nov 30, 2016 at 4:36
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    $\begingroup$ Just by looking at the dimensions of the quantities involved in the first equation you get $L$ in units of $kg m^2/s$, $r^2$ in $m^2$, $dr$ in $m$, energies in $kg m^2/s^2$. That means that you are missing some mass quantity to make the angle dimensionless $\endgroup$
    – Andrei
    Commented Nov 30, 2016 at 4:55

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