Central force field-condition for closed orbits.

[humanist rho]

Homework Statement

A particle moves in the central force field $\overrightarrow{F}=-kr^{n}\hat{r}$ , where k is a constant, and r is the distance from the origin. For what values of n closed stable orbits are possible?

Homework Equations

The Attempt at a Solution

I thought for stable configuration Kinetic energy = potential energy.

for central force field,

$\frac{mv^{2}}{r}=-kr^{n}$

ie,$KE,\frac{1}{2}mv^{2}=-kr^{(n+1)}$

and $PE = -\int Fdr=\frac{kr^{(n+1)}}{n+1}$

For stabe configuration,

$-kr^{(n+1)}=\frac{kr^{(n+1)}}{n+1}$

n=-2

But the answer says there's two turning points at n=1 and n=-1.
I think my method is absolutely wrong.

Please help.

 

[vela]

I thought for stable configuration Kinetic energy = potential energy.

This can't be correct. For 1/r² forces, for example, the potential energy is negative, so it obviously can't be equal to the kinetic energy. Also, the kinetic energy and potential energy vary over an orbit, so even if they were equal at one time, they wouldn't be equal later.

Any other ideas on what's required for a stable orbit?

 

[humanist rho]

Thank you vela.

Any other ideas on what's required for a stable orbit?

The energy should be a minimum for a stable orbit.is that correct?