# Homework Help: Equation of motion sketch

1. Jan 15, 2006

### stunner5000pt

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

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

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

so far so good?

for the second part
$$\dot{r}(t) = \frac{1}{r} \sqrt{\frac{2}{m} (b - \frac{L^2}{2m}}$$
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. Jan 15, 2006

### stunner5000pt

heres the sketch that is missing from the question

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• ###### charge.JPG
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3. Jan 17, 2006

### stunner5000pt

can anyone help!!!