[QUE/CM-05016]

When a small correction  $\delta U(r)$  is added to the potential energy  $U=-{\alpha\over r}$, the paths of finite motion are no longer closed, and  at each revolution the perihelion is  dipslaced through a small angle  $\delta \phi$. Show that $\delta \phi$  is approximately given by  $$ \delta \phi = {\partial \over \partial M} \int_{r_1}^{r_2}  { 2M\delta U dr\over \sqrt{ 2m(E +\alpha/r)-M^2/r^2  }} =  {\partial \over \partial M}\left(\frac{2m}{M} \int_{0}^{\pi}   r^2\delta U d\phi \right)  $$  where $r_1, r_2$ are the minimum and the maximum values of $r$. Find $\delta \phi$ when $\delta U = {\beta \over r^2  }$.