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Stability of Orbits 
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#1
May712, 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)(1s/(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 )(1s/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. 


#2
May812, 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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