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

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SUMMARY

Bertrand's Theorem establishes that the only forces leading to closed orbits are those described by Hooke's Law and the attractive inverse square force. The discussion focuses on demonstrating that the orbit defined by Hooke's Law, represented by the equation f = -k r, results in closed orbits through the analysis of the trajectory equation d²u/dθ² + u = (mk/l²u³). The conclusion reached is that the harmonic oscillation along each axis in the orbital plane leads to periodic motion, confirming the existence of closed orbits.

PREREQUISITES
  • Understanding of Hooke's Law and its mathematical representation
  • Familiarity with Bertrand's Theorem in classical mechanics
  • Knowledge of differential equations, particularly in orbital mechanics
  • Concept of simple harmonic motion (SHM) and its implications in physics
NEXT STEPS
  • Study the implications of Bertrand's Theorem in various force fields
  • Explore the mathematical derivation of closed orbits under Hooke's Law
  • Investigate the properties of simple harmonic motion (SHM) in multiple dimensions
  • Examine the relationship between orbital mechanics and harmonic oscillators
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Students of physics, researchers in classical mechanics, and anyone interested in the mathematical foundations of orbital dynamics and harmonic motion.

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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Kashmir said:
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
 
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A.T. said:
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.
 
Kashmir said:
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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