Central Forces and Angular Momentum

  • Thread starter azabak
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  • #1
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Imagine a mass in an elliptical orbit around a central force. If such force is inversely proportional to the distance squared then the mass will accelerate when going nearer the focus. At the same time the angular momentum of the orbit is a constant.
Can one show that the work done by the force is equal to the variation of kinetic energy solely due the conservation of angular momentum?
 

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  • #2
tiny-tim
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hi azabak! :wink:
Can one show that the work done by the force is equal to the variation of kinetic energy solely due the conservation of angular momentum?

work done is always equal to the variation of kinetic energy :smile:
 
  • #3
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I'm ok with the basic concepts. My doubt is that since the distance from the focus decreases two things must happen: conservation of angular momentum and work done by the force. If you calculate just the work done due the variation of potential energy angular momentum is not conserved. For it to happen the force must do an "extra" work. In any situation of rotation conservation of angular momentum is the important factor. Therefore accepting that angular momentum is always conserved implies accepting that any central force does more work than the variation of potential energy.
Correct me if I'm wrong.
 
  • #4
Doc Al
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If you calculate just the work done due the variation of potential energy angular momentum is not conserved.
Why is that?
For it to happen the force must do an "extra" work.
What do you mean?
 
  • #5
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The angular momentum of a mass m is L = m*v*R.
The central force is F = -k*m/R² where k is a constant.
The radius varies from R to r.
The work done W = -ΔU and the potential energy U = -k*m*/R.
The final kinetic energy due variation of potential energy is K = (m*v²/2)+(k*m*((1/R)-(1/r))).
To conserve the angular momentum the final kinetic energy should be K = (m*v²*R²)/(2*r²), which will be larger than the calculated just from variation of potential energy.
Therefore there must be an additional work in order to conserve angular momentum.
 
  • #6
Doc Al
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The angular momentum of a mass m is L = m*v*R.
Careful. Angular momentum is a vector quantity:
[tex]\vec{L} = \vec{r}\times m\vec{v}[/tex]
The angle between the vectors v and r is important.

Angular momentum, properly defined as the vector product above, is conserved, but the quantity m*v*R is not.
 
  • #7
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Oh, I see. Thank you very much.
 
  • #8
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The angular momentum of a mass m is L = m*v*R.
The central force is F = -k*m/R² where k is a constant.
The radius varies from R to r.
The work done W = -ΔU and the potential energy U = -k*m*/R.
The final kinetic energy due variation of potential energy is K = (m*v²/2)+(k*m*((1/R)-(1/r))).
To conserve the angular momentum the final kinetic energy should be K = (m*v²*R²)/(2*r²), which will be larger than the calculated just from variation of potential energy.
Therefore there must be an additional work in order to conserve angular momentum.

If you use the apihelion and the perihelion distance for r and R, so as to escape Doc Al's objection, you will find that you can't freely choose v, R and r. If you give the apihelion distance, and the initial speed, the perihelion distance is fixed so that both energy and angular momentum are conserved, and the 2 energies you computed are equal.

(At the perihelion and apihelion the distance vector r is perpendicular to v, so the magnitude of the angular momentum is equal to m*r*v)
 
  • #9
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So this implies that there's a unique world line such that both the variation of potential energy and angular momentum match the "required" kinetic energy.
 
  • #10
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So this implies that there's a unique world line such that both the variation of potential energy and angular momentum match the "required" kinetic energy.

There's always an unique wordline if you specify an initial position and velocity. If you have a central force and a conservative force field, angular momentum and total energy will be conserved.
 
  • #11
tiny-tim
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hi azabak! :smile:
If you calculate just the work done due the variation of potential energy angular momentum is not conserved. For it to happen the force must do an "extra" work. In any situation of rotation conservation of angular momentum is the important factor. Therefore accepting that angular momentum is always conserved implies accepting that any central force does more work than the variation of potential energy.

i don't understand this :confused:

potential energy is defined as minus the work done by a conservative force

angular momentum is always conserved if the force is central

(proof: d/dt(r x mr') = r x mr'' = r x F

which by definition is 0 if F is central)​

and any central force which is a sensible function of r will be the gradient of a scalar, and so will be conservative :smile:

as willem2 :smile: says …​
If you have a central force and a conservative force field, angular momentum and total energy will be conserved.
 
Last edited:
  • #12
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Hi tiny-tim.
I previously considered that the energy and angular momentum were conserved for any value of the radius. Then willem2 cleared me that it can only happen for a certain defined values of the radius. So choosing randomly any value for it would "break" both conservation of energy and angular momentum. The resultant world line is the one that both are conserved simultaneously.
 
  • #13
tiny-tim
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hi azabak! :smile:
… Then willem2 cleared me that it can only happen for a certain defined values of the radius.

no he didn't, he only said that you can't choose all three of v r and R …

once you choose two of them, the third is automatically decided because energy and angular momentum are conserved
 

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