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Stability of Orbits

by DuckAmuck
Tags: orbits, stability
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DuckAmuck
#1
May7-12, 08:15 PM
P: 20
So I saw a video on youtube: http://youtu.be/uhS8K4gFu4s

And so I thought I'd try to understand the whole stable orbit thing.

So first you have a simple energy equation of E = K + U.

K = (1/2)mv^2 (kinetic energy)
U = -GMm/r^2 (gravitational energy)

r = radius of orbit
v = velocity of orbiting object
M = mass of center object
m = mass of orbiting object

And we want it to be a stable orbit in the first place, so we have:
v=sqrt(GM/r)

Now, let's say we slightly bump the object in orbit. So r -> r + s, where s is much less than r.
(also w is much less than v)

So we can adjust the stable orbit equation:

v=sqrt(GM/r)

v+w = sqrt(GM/(r+s))

approximate for small distances:

v+w = sqrt(GM/r)(1-s/(2r))

So then, you can subtract the original equation out and have:
w = -sqrt(GM/r)s/(2r)
This makes sense directionally, if you bump the orbiting object inwards, the velocity with increase.

So now, back to the energy equation: E = K + U
Since it's a stable orbit being bumped only slightly, you expect it to be able to eventually return to the same state, so E can't change.

So initially you have:

E = (1/2)mv^2 - GMm/r^2

Then apply the bump:

E' = ((1/2)mv^2 )(1+2w/v) - (GMm/r^2 )(1-s/r)

This then reduces to the form:

E' = E + mvw + (GMm/r^2 )s

Since E' must equal E, the two extra terms should add to zero:

E' = E + (mv)sqrt(GM/r)(-s/(2r)) + (GMm/r^2 )s

E' = E + (-1/2)(GMm/r^2 )s + (GMm/r^2 )s

So I am doing something wrong. I have a missing factor of two somewhere, but I don't know why.

If you guys can help, it will be appreciated. Thanks.
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K^2
#2
May8-12, 10:24 AM
Sci Advisor
P: 2,470
Stable orbit need not be circular. So if you bump it a little, it can become slightly elliptic and v=sqrt(GM/r) no longer holds. I think that's where things go wrong with your derivation.

Best check of orbit stability is looking at effective potential.


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