# Central Forces and Angular Momentum

1. Mar 2, 2012

### azabak

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?

2. Mar 2, 2012

### tiny-tim

hi azabak!
work done is always equal to the variation of kinetic energy

3. Mar 2, 2012

### azabak

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. Mar 2, 2012

### Staff: Mentor

Why is that?
What do you mean?

5. Mar 2, 2012

### azabak

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. Mar 2, 2012

### Staff: Mentor

Careful. Angular momentum is a vector quantity:
$$\vec{L} = \vec{r}\times m\vec{v}$$
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. Mar 2, 2012

### azabak

Oh, I see. Thank you very much.

8. Mar 2, 2012

### willem2

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. Mar 2, 2012

### azabak

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. Mar 2, 2012

### willem2

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. Mar 2, 2012

### tiny-tim

hi azabak!
i don't understand this

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

as willem2 says …​

Last edited: Mar 2, 2012
12. Mar 2, 2012

### azabak

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. Mar 2, 2012

### tiny-tim

hi azabak!
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