#### CAF123

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**1. Homework Statement**

A particle of mass m moves under a central force ##\mathbf{F}(\mathbf{r}) = -\frac{\mu}{r^2} e^{-kr} \hat{r}##.

The particle undergoes motion in a circle if ##h^2 = (a\mu/m)e^{-ka}##. I have shown that if ##u(\theta) = 1/r,## then the orbit eqn for ##u(\theta)## becomes $$u'' + u = \frac{1}{a}e^{-k(1/u - a)}.$$ This is a show that so I know it is correct.

I can't make further progress with the following:

Assume that the motion is slightly disturbed. Use the orbit eqn for ##u(\theta)## to show that in the perturbed orbit, the furthest distance from the centre occurs with a period of ##2\pi/w## where ##w = \sqrt{1-ka}##

**3. The Attempt at a Solution**

Assume the perturbation, p(t), is small. That is p(t=0) << a. Let r = a + p(t) => u = 1/(a+p).

Differentiate twice wrt theta using the chain rule and I get $$u'' = -\frac{\ddot{p}}{a\mu e^{-ka}/m u^2}. $$ Sub into the orbit eqn: $$-\frac{\ddot{p} (a + p)^2 m}{a\mu e^{-ka}} + \frac{1}{a+p} = \frac{1}{a} e^{-k(1/(a+p) - a)}$$

Now Taylor expand the ##(a+p)^2##, ##(a+p)^{-1}## and exp terms. Doing this, putting everything together, I get that $$\ddot{p} + \frac{\mu e^{-ka}}{ma^4} p = \frac{e^{1/a}}{ma^2}$$

I was thinking my next step would be to solve this for ##p##, find the maximum of this (i.e the furthest distance it is perturbed) and then add this maximum to a. Does it sound good? (Can anyone confirm what I got above before I continue - it looks somewhat strange since when I find the solution, the arguments of sin/cos will contain an exponential). Actually something has had to have gone wrong since there will be a dimension associated with the arg of the trig terms.

MAny thanks.

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