# Stability of Orbits

1. Apr 1, 2013

### CAF123

1. The problem statement, all variables and given/known data
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.

Last edited: Apr 1, 2013
2. Apr 1, 2013

### voko

Seeing how you got that would be helpful. I am not saying this is incorrect, though.

Not so fast. What does $$u'' = -\frac{\ddot{p}}{a\mu e^{-ka}/m u^2}$$ really mean? Is that $$u'' = -\frac{\ddot{p}}{\frac {a\mu e^{-ka}} {m u^2}}$$? If so, then $$-\frac{\ddot{p}}{a\mu e^{-ka}/m u^2} = -\frac{m u^2\ddot{p}}{a\mu e^{-ka}} = -\frac{\ddot{p} m}{a(a + p)^2\mu e^{-ka}}$$

3. Apr 1, 2013

### CAF123

First I computed u'. $$u' = -(a+p)^{-2} \dot{p} \frac{dt}{d\theta} = -\frac{\dot{p}}{(a+p)^2} \frac{1}{\sqrt{\frac{a\mu e^{-ka}}{m}}} u^2,$$ where $$\dot{\theta} = h/r^2 = \sqrt{\frac{a\mu e^{-ka}}{m}} \cdot \frac{1}{r^2}$$.

Then u'' = d/dθ (u') = $$-\frac{1}{\sqrt{\frac{a\mu e^{-ka}}{m}}} \frac{d}{d\theta} \frac{\dot{p}}{(a+p)^2} (a+p)^2 = -\frac{1}{\sqrt{\frac{a\mu e^{-ka}}{m}}} \frac{d}{dt} \dot{p} \frac{dt}{d\theta} = -\frac{\ddot{p}m}{a\mu e^{-ka} u^2}$$

I made a typo in the OP, it should be m/u^2 not /mu^2

Edit: is my method correct?

Last edited: Apr 1, 2013
4. Apr 1, 2013

### haruspex

Something wrong in the exponent. Should it just be -k(1/u - a) = -kp?

5. Apr 2, 2013

### voko

How did you go from

$$u' = -(a+p)^{-2} \dot{p} \frac{dt}{d\theta} = -\frac{\dot{p}}{(a+p)^2} \frac{1}{\sqrt{\frac{a\mu e^{-ka}}{m}}} u^2,$$

to

$$u'' = -\frac{1}{\sqrt{\frac{a\mu e^{-ka}}{m}}} \frac{d}{d\theta} \frac{\dot{p}}{(a+p)^2} (a+p)^2$$

given that

$$u^2 = \frac 1 {(a + p)^2}$$

?

6. Apr 2, 2013

### CAF123

For some reason, I had a few initial problems writing the above out in latex, so a typo has crept in. u^2 should be on the denominator.

7. Apr 2, 2013

### CAF123

Indeed it should.

8. Apr 2, 2013

### voko

I think this $\frac{\mu e^{-ka}}{ma^4} p$ should be $\frac{\mu e^{-ka}}{ma^3} p$. Then the units seem OK. What equation do you get when you fix the exponent and linearize?

9. Apr 2, 2013

### CAF123

That is exactly the case. The final equation I attain after Taylor expanding is $$\ddot{p} + \frac{\mu e^{-ka}}{ma^3} p = \frac{k\mu e^{-ka}}{ma^2}p,$$ which I can subsequently write as $$\ddot{p} + p(\frac{\mu e^{-ka}}{ma^2} \left(\frac{1}{a} - k\right) = 0$$ Here I identify the harmonic solution, with $$w^2 = (\frac{\mu e^{-ka}}{ma^2} \left(\frac{1}{a} - k\right)$$

This can further be simplified using $h^2$ given in the problem statement. Hence $w^2 = \frac{h^2}{a^4} (1-ka)$ which is incorrect by a factor of $h/a^2$.

Last edited: Apr 2, 2013
10. Apr 2, 2013

### voko

$\exp -kp$, linearized, should have a constant term. Which is actually quite bizarre, because that means p = 0 is an equilibrium point for the perturbed system.

11. Apr 2, 2013

### CAF123

When I linearised $e^{-kp}$ I got $1 - kp$. Is this not correct?

12. Apr 2, 2013

### voko

Yes, this is correct. But that means that the equation is $\ddot{p} + \lambda p = c$, which means that $p = c/\lambda$ is the equilibrium point. Not $p = 0$ as expected.

13. Apr 2, 2013

### CAF123

I brought everything together on the LHS to get the familiar harmonic equation and then identified $w^2$. Is my final eqn correct that I posted in my last two posts? When I rearrange I get 0 on the RHS and so p = 0 would correspond to the equibilrium soln, no?

14. Apr 2, 2013

### voko

You got the equation in the form $\ddot{p} + \lambda p = 0$. How is that possible if the original RHS had a constant term?

15. Apr 2, 2013

### CAF123

The eqn I got before rearranging was:
The RHS depends on p, so when I bring it over I have the form $\ddot{p} + \lambda p = 0$. (Unless ofcourse I made an error). I am just off to a tutorial now.

16. Apr 2, 2013

### voko

In #11 you said that the RHS should have a constant term. In #15, however, there is no such term.

17. Apr 2, 2013

### CAF123

I had the form $$-\frac{\ddot{p} (a+p)^2m}{a\mu e^{-ka}} + \frac{1}{a+p} = \frac{1}{a} e^{-kp}$$

Now linearise: $$-\frac{m}{a\mu e^{-ka}} \ddot{p} \left[a^2(1 + 2\frac{p}{a}) \right] + \frac{1}{a} (1-\frac{p}{a}) = \frac{1}{a}e^{-kp}$$

When I bring that single 1/a term on the LHS over to the RHS, I get $\frac{1}{a}(e^{-kp} - 1)$. Now expand the exp and the 1's cancel, leaving only a term in p.

Edit: I neglected the term containing $\ddot{p}p$

Last edited: Apr 2, 2013
18. Apr 2, 2013

### voko

I can't see any mistake in your derivation (assuming the original equations for u and h are correct).

Unless I am mistaken, $\frac {\mu e^{-ka}} {m a^3} = \Omega^2$, where $\Omega$ is the angular velocity of the circular motion. So $\omega = \Omega \sqrt {1 - ka}$. Can't this be what you are really asked about?

19. Apr 2, 2013

### CAF123

Yes, the question as stated in the OP is the one I need to answer. I see that $h/a^2 = \dot{\theta}$ so $\omega = \dot{\theta} \sqrt{1 - ka}$

I don't see how the dimensions of $\sqrt{1-ka}$ are correct anyway. I believe k has dimensions of 1/length. So $\sqrt{1-ka}$ would be dimensionless. $\dot{\theta} \sqrt{1-ka}$ has units of 1/s, as required. The question is a mistake maybe?

20. Apr 2, 2013

### voko

They may have asked a bogus question to have you defend your "wrong" answer, they do that sometimes.

They might also have meant the "relative" period of the perturbed motion (in units such that the orbital period = 1), but then the answer is still off by a factor of $2 \pi$.