(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Homework Help: Equation of motion sketch

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