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Homework Help: Particle moving in a potential

  1. Jan 15, 2006 #1
    a particle of mass m moves in a potential V(r) = gamma/r^2 where the constant gamma >0

    a) Sketch an effective potential and discuss the nature of the non zero angular momentum (L not zero) orbits without solving for the equation of the trajectory

    i know that [tex] V_{eff} = V(r) + \frac{L^2}{2mr^2} [/tex]
    here [tex] V_{eff} = \frac{1}{r^2} (\gamma + \frac{L^2}{2m}) [/tex]
    so it will look like an inverse square graph as in teh digram


    Calculate teh equation of the trajectories discuss their shapes and sketch a typical trajectory

    one thing thats got me with this question is that i cant solve for r(t) or even r(phi) becuase their formulas involve using E which depends on (dr/dt)^2 and i dont end up getting anything solvable. Have a look
    for r(t)
    [tex] r(t) = \sqrt{\frac{2}{m} (E - V_{eff}(r))} [/tex]
    but [tex] E = \frac{1}{2} m \dot{r}^2 + V(r) + \frac{L^2}{2mr^2} [/tex]
    and that yeilds nothing useful
    am i doing something wrong? Is energy supposed to be zero? But why?

    there are two more parts which i will post later on. they are related to a and b.
     

    Attached Files:

  2. jcsd
  3. Jan 18, 2006 #2

    siddharth

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    stunner,
    Your answer to part A looks right. To find the equation of the trajectory, there is a nice trick you can apply. Make the substitution [tex] u=\frac{1}{r} [/tex] and try finding [tex] u(\phi) [/tex]. You will see that the differential equation involving [itex] u [/itex] and [itex] \phi [/itex] can be solved and you will be able to find [tex] u(\phi) [/tex].

    In more detail, since [tex] r=\frac{1}{u} [/tex]

    [tex] \frac{dr}{dt} = -\frac{1}{u^2} \frac{du}{dt} = -\frac{1}{u^2} \frac{du}{d\theta} \frac{d\theta}{dt} [/tex] where [tex] \frac{d\theta}{dt} = \omega [/tex].

    So, substituting for v ([itex] \frac{dr}{dt}[/itex]) in the energy equation you wrote in terms of [itex] \frac{du}{d\theta} [/itex] and [itex] u [/itex], you get

    [tex] E = V(u) + \frac{L^2 u^2}{2m} + \frac{1}{2}m(\frac{-L}{m}\frac{du}{d\theta})^2 [/tex].

    Now differentiate this equation with respect to theta. Can you see that [itex] \frac{du}{d\theta} [/itex] will cancel?

    The resulting differential equation you get can then be solved which will give you u (and hence r) as a function of theta. Can you take it from here?
     
    Last edited: Jan 18, 2006
  4. Jan 18, 2006 #3
    Thank You !

    alrighty so i had the right idea... the thing is we dont have a text book for this course and there are so many versioins of the same ofrmula! Anyway here are the ther two questions

    If [itex] (r_{m},\phi_{m}) [/itex] are the planbe polar coordiantes of the perigee, evalute phi m (b) and from this calculate the impace parameter b in terms of the angle of scattering theta. Use this result to compute the differential scattering corss setion in terms of gamma, E and theta

    The perigee represents a minimum distance between the mass and thecenter of motion. So to simply differentiate phi(r) w.r.t. r and plug that eqal to szero should yield the answer b.
    if you could have a look at this thread too that would be awesome too
    based on a simlar concept
    https://www.physicsforums.com/showthread.php?t=106913
     
    Last edited: Jan 18, 2006
  5. Jan 18, 2006 #4

    first of all im not quite sure if i am supposd to be using that version of E in this problem
    what does [itex] \frac{dE}{d \theta} [/itex] yield that is useful?

    also i get a differential equation with (u'')^2 which cnat be solved
     
  6. Jan 20, 2006 #5

    siddharth

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    In my post, [tex] \theta [/tex] was the polar angle. So [tex]\frac{dE}{d \theta} [/tex] is 0 because total energy is conserved. I'm pretty sure you don't get a (u'')^2 term. You could post and show where you are stuck.
     
    Last edited: Jan 20, 2006
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