How can I find the shape of an orbit under an inverse cube law of force?

[bushdayroses]

Homework Statement

I've found the a general form of equation for the explicit integration of the shape, $r(\theta)$, of a one-body-problem (a particle rotating about a set point, O) for arbitrary $\mathbf{f} = -f(r)\hat{\mathbf{r}}$, and used it to find $r(\theta)$ for an inverse square law $f(r) = \mu / r^2$. Where $u = (1/r)$.

I was able to get $$r = r(\theta(t)) => \dot{r} = {{dr} \over {d\theta}} {{d\theta} \over {dt}} = {{dr} \over {d\theta}} \left({{h_0} \over {r^2}}\right)$$

Where $h_0 \neq 0$ is the constant angular momentum.

Can anyone help me to figure out the shape under an inverse cube law of force, $f(r) = {{\nu} / {r^3}}$, and show that if $\nu \le h^2$, the orbit is unbounded?

I know that this solution will depend on the relative sizes of $\nu$ and $h$, but I can't seem to get this to work.EDIT:: Got LaTex to work

 

[chrisk]

One way to approach this problem is to use the substitution

$$r=\frac{1}{u}\mbox{ and u =}\frac{1}{r}$$

then differentiate r with respect to t giving

$$\dot{r}=-\frac{1}{u^2}\frac{du}{d\theta}\frac{d\theta}{dt}$$

and

$$\dot{\theta}=\frac{L}{mr^2}$$

This will give

$$\dot{r}=-\frac{L}{m}\frac{\mbox{d}u}{\mbox{d}\theta}$$

Differentiate again with respect to t to find the second derivative of r in terms of u and theta. Write the general equation of motion for r in terms of F(r) and "centrigugal" force (L²/(mr³), and substitue the second derivative of r with respect to t. You will arrive at an equation of motion in terms of u and theta. This differential equation is easier to solve than straightforward methods. Once u is found substitue 1/r and solve for r.

 

[bushdayroses]

Got it figured out. Thanks