# Urgent and desperate for help!

1. Jan 17, 2006

### stunner5000pt

this is due on wednesday i would really liketo hand it in on time
couple of my questions are in these threads which have not gotten any answers...

For a particle of mass m moving in a potential $V(r) =\frac{-\alpha}{r} + W(r)$ (alpha>0) where $W = \frac{-a}{r^4}$ is a small perturbation ( in a sense that $W(r) << |\frac{\alpha}{r}|$ for r not too small), calculate the advnace of the perihelion
$$2 \Delta \beta = 2 \frac{\partial}{\partial L} \int_{0}^{\pi} \frac{m}{L} r^2 W d \phi$$
Express your answer in terms of alpha, m, a, L and the eccentricity $$\epsilon = \sqrt{1 + \frac{2EL^2}{m \alpha^2}$$ and verify that your answer is dimensionless. (rrepresnets an angle in radians)

problem with the intergral is that r depends on phi.
i know that this is true
$$\frac{dr}{d \phi} = \frac{r^2 \sqrt{2m(E - V_{e}^{0})}}{L}$$
in this is $$E = \frac{1}{2} m \dot{r}^2 + V(r)$$?
also what about $$V_{e}^{0} = V_{c} (r) + \frac{L^2}{2mr^2} = \frac{-GMm}{r} + \frac{L^2}{2mr^2}$$
are those two correct? Do i simply substitute those two expressions in the integral for r(phi) and solve for r(phi) thereby i can solve for beta?
Is this correct?