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Equation of motion sketch

  1. Jan 15, 2006 #1
    For a particle of mass m moving in a potential V(r) = -b/r^2 where the constant b>0 obtain the equation [itex] r = r(\phi} [/itex] of the trajectory for the particular states of motion with total energy E = 0 and angular momenta such that [itex] \frac{L^2}{2m} < b [/itex]
    SKetch the trajectory and discuss the motion for
    [tex] \dot{r} (t=0) >0 [/tex] and
    [tex] \dot{r} (t=0) <0 [/tex]

    Ok so we know that phi and r are related by this equation
    [tex] \phi = L \int \frac{1}{r^2 \sqrt{2m(E - V_{e} (r))}} dr + \mbox{constant} [/tex]
    here [tex] V_{e} (r) = \frac{-b}{r^2} + \frac{L^2}{2mr^2} [/tex]
    also E = 0 so
    [tex] \phi = L \int \frac{1}{r^2 \sqrt{2m(\frac{b}{r^2} + \frac{L^2}{2mr^2}}} [/tex]

    and integrating we get
    [tex] C exp(\phi \frac{\sqrt{2mb - \frac{L^2}{2m}}}{L}}) = r(\phi) = r [/tex]

    so far so good?

    for the second part
    [tex] \dot{r}(t) = \frac{1}{r} \sqrt{\frac{2}{m} (b - \frac{L^2}{2m}} [/tex]
    do i need to find explicit expression for r(t) and phi(t) ?
    for r' > 0 then r > 0 and phi > 0
    for r' < 0 from the relation between r and phi above it does nt look like that could ever be less that zero unless C <0? Do i need to solve for C by the way?

    YOur help is always, greatly appreciated!
    Last edited: Jan 15, 2006
  2. jcsd
  3. Jan 15, 2006 #2
    heres the sketch that is missing from the question

    thank you for your help!

    Attached Files:

  4. Jan 17, 2006 #3
    can anyone help!!!

    this is due tomorrow!! I need to know if what i have is right... please please help! im desperate!
    Note that i posted it about 4 days in advance
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