Solve Orbital Mechanics Homework: Find Circular Orbit, Stability, Period

[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}=0Using \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 ] =0And 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:

<br /> 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 <br />

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

<br /> 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 <br />

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 ) &gt; 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