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Central potential problem

  1. Mar 6, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    A particle of mass m is under a central potential of the form [itex]U(r)=-\frac{\alpha }{r^2}[/itex] where alpha is a positive constant.
    At time t=0, the spherical coordinates of the particle are worth [itex]r=r_0[/itex], [itex]\theta = \pi /2[/itex] and [itex]\phi=0[/itex]. The corresponding time derivatives are given by [itex]\dot r <0[/itex], [itex]\dot \theta =0[/itex] and [itex]\dot \phi \neq 0[/itex].
    The total energy is 0 and the modulus of the angular momentum is worth [itex]\sqrt {m \alpha }[/itex].
    1)Write down the Lagrangian of the particule.
    2)Find [itex]r(t)[/itex] and [itex]\dot r (t)[/itex] expressed in terms of m, alpha and [itex]r_0[/itex].
    3)Same as in 2) but with phi(t) and [itex]\dot \phi (t)[/itex] and find the trajectory [itex]r(\phi )[/itex].
    4)Calculate the time in which the particle reach the origin of the coordinate system. How many orbits does it describes before reaching it?
    2. Relevant equations
    L=T-V.
    E=T+V.

    3. The attempt at a solution
    I've made a sketch. Since theta is constant and [itex]\theta =\pi/2[/itex], the motion is constrained into the xy plane. Therefore the angular momentum is with respect to [itex]\phi[/itex], namely it is worth [itex]P_\phi = \frac{\partial L }{\partial \dot \phi}[/itex] where L is the Lagrangian.
    In spherical coordinates, [itex]T=\frac{m}{2}(\dot r^2+r^2 \dot \theta ^2 \sin \phi + r^2\dot \phi ^2)[/itex]. But here [itex]\dot \theta =0[/itex]. So that [itex]p_ \phi =mr^2 \dot \phi[/itex]. I am told that [itex]|r^2 \dot \phi |=\sqrt {\frac{\alpha }{m}}[/itex].
    1)So that the Lagrangian reduces to [itex]L=\frac{m}{2}(\dot r ^2 + \sqrt {\frac{\alpha }{m}} \dot \phi )+\frac{\alpha }{r^2}[/itex].
    I still didn't use the fact that the total energy vanishes...
    2)Euler-Lagrange equation for r gives me [itex]\ddot r +\frac{\alpha }{m r^3}=0[/itex]. I don't know how to solve this DE. Since [itex]\dot r[/itex] does not appear I think the substitution [itex]v=\dot r[/itex] should work, but I don't reach anything with it.
    So I'm basically stuck here and I'm wondering whether I'm over complicating stuff because I'm not using the fact that [itex]\dot r <0[/itex] and [itex]E=0[/itex].
    Any help is greatly appreciated.
     
  2. jcsd
  3. Mar 8, 2012 #2
    Multiply both sides with [itex] \dot{r} [/itex] and integrate wrt t to get a first order equation.
     
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