Stable Circular Orbits in Planetary Motion: A Homework Solution

[Prashant123]

Homework Statement

Consider a particle moving in the potential U (r)= -A/r^n, where A>0. What are the values of n which admit stable circular orbits?

Homework Equations

The Attempt at a Solution

I tried to solve by putting dr/dt=0 in the total energy equation E= T + Ueff. But it didn't work. Then I came across a solution which said that for the orbit to be circular, Ueff(r) needs to have a minima when plotted against r, where Ueff is the effective potential (L^2/2mr^2+ U (r)). But I don't understand why it has to, because when n=1, where circular orbits are possible, Ueff does not have a minima since it varies with 1/r.

 

[Orodruin]

Ueff does not have a minima since it varies with 1/r.

Yes it does. Ueff is the effective potential, not the actual potential. The effective potential includes the L^2/2mr^2 term in addition to the potential itself.

 

[Prashant123]

Yes it does. Ueff is the effective potential, not the actual potential. The effective potential includes the L^2/2mr^2 term in addition to the potential itself.

Yes. But I want to know if the statement "circular motion is possible in this case when Ueff has a minima when plotted against r" is true and how, as for n=1, Ueff= -Gm1m2/2r and this does not have a minima.

 

[Orodruin]

I want to know if the statement "circular motion is possible in this case when Ueff has a minima when plotted against r" is true

It is true.

and how, as for n=1, Ueff= -Gm1m2/2r and this does not have a minima.

You are wrong. For n=1 the potential is $U(r) = -Gm_1m_2/r$, but the effective potential is $U_{\rm eff}(r) = L^2/(2mr^2) + U(r) = L^2/(2mr^2) - G m_1 m_2/r$.

 

[Prashant123]

Oh ok.. but is there a proof for why should Ueff have a minima?

 

[Orodruin]

Oh ok.. but is there a proof for why should Ueff have a minima?

This is the definition of the orbit being stationary. It means that an orbit close to circular will not go too far away from being circular (the effective total energy must be larger than the effective potential).