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Urgent and desperate for help!

  1. Jan 17, 2006 #1
    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 [itex] V(r) =\frac{-\alpha}{r} + W(r) [/itex] (alpha>0) where [itex] W = \frac{-a}{r^4} [/itex] is a small perturbation ( in a sense that [itex] W(r) << |\frac{\alpha}{r}| [/itex] for r not too small), calculate the advnace of the perihelion
    [tex] 2 \Delta \beta = 2 \frac{\partial}{\partial L} \int_{0}^{\pi} \frac{m}{L} r^2 W d \phi [/tex]
    Express your answer in terms of alpha, m, a, L and the eccentricity [tex] \epsilon = \sqrt{1 + \frac{2EL^2}{m \alpha^2} [/tex] 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
    [tex] \frac{dr}{d \phi} = \frac{r^2 \sqrt{2m(E - V_{e}^{0})}}{L} [/tex]
    in this is [tex] E = \frac{1}{2} m \dot{r}^2 + V(r) [/tex]?
    also what about [tex] V_{e}^{0} = V_{c} (r) + \frac{L^2}{2mr^2} = \frac{-GMm}{r} + \frac{L^2}{2mr^2} [/tex]
    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?
     
  2. jcsd
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