Formulation of orbital kinetic energy

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SUMMARY

The discussion centers on the formulation of orbital kinetic energy in the context of the virial theorem, specifically for a radial potential proportional to r^k. It establishes that c(k) equals k/2 for circular orbits. The user successfully solved the problem by equating the gradient of the potential to the centripetal force, leading to the kinetic energy expression. Additionally, a formulation of kinetic energy as \bar{T}= 1/2 \vec{∇}(V) .\vec{r} is introduced, with a suggestion to derive it through integration over time.

PREREQUISITES
  • Understanding of the virial theorem in classical mechanics
  • Familiarity with radial potentials and their mathematical representations
  • Knowledge of centripetal force and its relation to kinetic energy
  • Basic calculus, particularly integration techniques
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  • Study the derivation of the virial theorem in classical mechanics
  • Explore the implications of different values of k in radial potentials
  • Learn about the relationship between potential energy and kinetic energy in orbital mechanics
  • Investigate advanced integration techniques relevant to physics problems
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Physics students, mechanical engineers, and anyone interested in classical mechanics and orbital dynamics.

Woodles
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For a radial potential proportional to r^k, the virial theorem says \bar{T}=c(k)\bar{V}. My problem says to show that c(k)=k/2 for a circular orbit.

I actually solved the problem all ready (by setting the gradiant of the potential equal to the centripetal force and solving for 1/2*m*v^2.)

However, while I was looking around, I came across something.

I found a formulation of kinetic energy saying that \bar{T}= 1/2 \vec{∇}(V) .\vec{r}.

How could you derive this? Is it true? Why? The original post said this came from the virial theorem.
 
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Just integrate with respect to time in the LHS and you'll get it.

\frac{m}{2} \int_{t_1}^{t_2} \frac{dr}{dt}\left(\frac{dr}{dt} dt\right) = ...
 

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