Help Solve Integral for Perihelion Advance of Mass m in Potential V(r)

[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...
https://www.physicsforums.com/showthread.php?t=106913
https://www.physicsforums.com/showthread.php?t=106930

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?

 

[Jhero]

I understand your frustration with not receiving any answers to your questions. I can assure you that it is always important to meet deadlines and turn in work on time, so I commend you for wanting to hand in your assignment on time.

To address your first question, yes, the expressions you have for the potential V(r) and W(r) are correct. In this problem, we are looking at a particle of mass m moving in a central potential V(r) and a small perturbation W(r). The perturbation, represented by W(r), is much smaller than the central potential, represented by \frac{\alpha}{r}. This allows us to treat it as a small correction to the central potential.

Now, for the integral, I can see why you might be confused. The key here is to remember that r is a function of phi, as you correctly pointed out. This means that we need to express r in terms of phi in order to properly integrate. The expression you have for \frac{dr}{d\phi} is correct, and it can be derived from the equation of motion for the particle in this potential. However, note that the potential V_{e}^{0} is the effective potential, which takes into account both the central potential and the angular momentum of the particle. So, yes, you can substitute these expressions for r(phi) and solve for beta.

Finally, to verify that your answer is dimensionless, you can simply check the units of each term in your final expression for beta. The terms alpha, m, a, L, and epsilon all have the same units (mass, length, and energy), so they will cancel out and leave you with a dimensionless quantity, as expected.

I hope this helps clarify your doubts. If you have any further questions, please do not hesitate to ask. Good luck with your assignment!