Circular orbit stability with exponential force perturbation analysis

[kreil]

Homework Statement

I know how to do this problem, I'm just having trouble actually doing it.

A particle moves in a force field described by,

$$F(r)=-k(ar+1)\frac{e^{-ar}}{r^2}$$

1. Obtain the condition for a circular orbit of radius r0
2. Apply a perturbation to the circular orbit and find the condition between a and r0 for the orbit to be stable
3. Obtain the period of the small oscillation about the stable circular orbit

Homework Equations

From previous parts of the problem, I obtained the equation of motion to be,

$$m \ddot r - \frac{l^2}{mr^3} + k(ar+1)\frac{e^{-ar}}{r^2}=0$$

The Attempt at a Solution

1. The condition for a circular orbit is

$$f_{eff}(r_0)=0 \implies k(ar_0+1)\frac{e^{-ar_0}}{r_0^2} = \frac{l^2}{mr_0^3}$$

2. The way to do part 2 is to define $r=r_0+ \rho$, $\ddot r = \ddot \rho$, plug these into the equation of motion, then get the equation into the form

$$\ddot \rho + \omega^2 \rho = 0$$

Then the condition for the circular orbit to be stable is $\omega^2>0$.

In this case, however, I'm having trouble getting the equation of motion into that form:

$$m \ddot \rho - \frac{l^2}{m(r_0+\rho)^3}+k(a(r_0+\rho)+1)\frac{e^{-a(r_0+\rho)}}{(r_0+\rho)^2}=0$$Using $\frac{1}{(r_0+\rho)^n} = \frac{1}{r_0^n} \left ( 1-\frac{n \rho}{r_0} \right )$ and the condition for a circular orbit in 1,

$$m \ddot \rho - \frac{l^2}{mr_0^3} + \frac{3l^2 \rho}{mr_0^4} +k(ar_0+a\rho+1)\frac{e^{-a(r_0+\rho)}}{(r_0+\rho)^2}=0$$

...

$$m \ddot \rho + \frac{3l^2 \rho}{mr_0^4} + \frac{l^2}{mr_0^3} \left [ e^{-a \rho} \left ( 1 - \frac{2 \rho}{r_0} \right ) \left ( 1 + \frac{a \rho}{(ar_0+1)} \right ) -1 \right ] =0$$And if I multiply everything out and get rid of the brackets there are exponentials with rho, as well as terms with rho^2.

Any help or suggestions is appreciated.

 

[fzero]

I think the algebra will be a bit easier if you define $$\rho$$ to be dimensionless via $$r = r_0 (1+\rho),$$ but that's certainly not necessary. You want to linearize the equation of motion for $$\rho$$, therefore you should continue expanding the exponential and drop the quadratic and higher terms everywhere. Requiring stable solutions will lead to an inequality for $$ar_0$$.

 

[kreil]

So I can just approximate the exponential to first order using taylor series?

 

[fzero]

When you find the condition on $$ar_0$$ you can check if $$a\rho \ll 1$$ was a good approximation (having assumed that $$\rho \ll r_0$$ already).

 

[kreil]

The condition I ended up with is

$$2ar_0+1>0$$

which just seems trivial. Is this what you got?

 

[fzero]

I think you might have missed an $$(ar_0)^2$$ term. I found

$$1 +(ar_0) > (ar_0)^2$$

which can be simplified further by finding the roots.

 

[kreil]

I don't see where that squared term comes from. Here is what I did:

$$m \ddot \rho + \frac{3l^2 \rho}{mr_0^4} + \frac{l^2}{mr_0^3} \left [ e^{-a \rho} \left ( 1 - \frac{2 \rho}{r_0} \right ) \left ( 1 + \frac{a \rho}{(ar_0+1)} \right ) -1 \right ] =0$$

Assume ap << 1 so that the exponential is 1. Then

$$m \ddot \rho + \frac{3l^2 \rho}{mr_0^4} + \frac{l^2}{mr_0^3} \left [ 1 + \frac{a \rho}{(ar_0+1)} - \frac{2 \rho}{r_0} - \frac{2a \rho^2}{r_0(ar_0+1)} -1 \right ] =0$$

The 1's cancel and the p^2 term is very small, so

$$m \ddot \rho + \frac{3l^2 \rho}{mr_0^4} + \frac{l^2}{mr_0^3} \left [\frac{a \rho}{(ar_0+1)} - \frac{2 \rho}{r_0} \right ] = m \ddot \rho + \frac{3l^2 \rho}{mr_0^4} + \frac{l^2 a \rho}{ m r_0^3 (ar_0+1)} - \frac{2l^2 \rho}{mr_0^4} = 0$$

Collecting in powers of p,

$$m \ddot \rho + \left ( \frac{l^2 }{mr_0^4} + \frac{l^2 a }{mr_0^3(ar_0+1)} \right )\rho =0$$

$$\implies \ddot \rho + \omega^2 \rho = 0$$

$$\omega^2 = \frac{1}{m} \left ( \frac{l^2 }{mr_0^4} + \frac{l^2 a }{mr_0^3(ar_0+1)} \right ) > 0$$

 

[fzero]

Assume ap << 1 so that the exponential is 1.

You need to expand the exponential to linear order,

$$e^{-a\rho} \sim 1 - a\rho.$$

 

[kreil]

Thanks for the help fzero