Hooke's law, Bertrand's theorem and closed orbits

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  • #1
Kashmir
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Bertrand's Theorem says : the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive inverse square force.

I'm looking at the hookes law ##f=-k r## and try to see explicitly that the orbit is indeed closed.

I use the orbit equation ##\frac{d^{2} u}{d \theta^{2}}+u=\frac{-m}{l^{2} u^{2}} f\left(\frac{1}{u}\right)## with the force given as ##f=-k r## ,therefore I get ##\frac{d^{2} u}{d \theta^{2}}+u=+\frac{mk}{l^{2} u 3}## as the equation defining the trajectory.

However neither can I solve this nor can I see that the equation implies a closed orbit.

Can you please help me.
 

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  • #2
A.T.
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I'm looking at the hookes law ##f=-k r## and try to see explicitly that the orbit is indeed closed.

Hooke's law gives you harmonic oscillation along each axis in the orbital plane, with a period independent of max. amplitude, thus the same for both axes and equal to the orbital period:
https://en.wikipedia.org/wiki/Harmonic_oscillator
 
  • #3
Kashmir
437
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Hooke's law gives you harmonic oscillation along each axis in the orbital plane, with a period independent of max. amplitude, thus the same for both axes and equal to the orbital period:
https://en.wikipedia.org/wiki/Harmonic_oscillator
Thank you. I got it. We've three SHM along three axis which are periodic, hence a closed orbit.
 
  • #4
A.T.
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We've three SHM along three axis which are periodic, hence a closed orbit.

You can see the SHM orbit on the pound note below, where they placed the sun in the middle of an elliptical orbit. Newton doesn't look too happy about this.

One%2BPound%2BNote%2Bwith%2BIsaac%2BNewton.jpg
 

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