Stable Circular Orbits in Planetary Motion: A Homework Solution

AI Thread Summary
Stable circular orbits in planetary motion depend on the effective potential, Ueff, which must have a minimum when plotted against the radial distance, r. The effective potential combines the gravitational potential, U(r) = -A/r^n, with the centrifugal term, L^2/2mr^2. For n=1, while the gravitational potential does not exhibit a minimum, the effective potential does, allowing for circular orbits. The requirement for Ueff to have a minimum is essential for stability, as it indicates that orbits near this point will not deviate significantly. Understanding this relationship is crucial for analyzing circular motion in various potential scenarios.
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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.
 
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Prashant123 said:
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.
 
Orodruin said:
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.
 
Prashant123 said:
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.

Prashant123 said:
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##.
 
Oh ok.. but is there a proof for why should Ueff have a minima?
 
Prashant123 said:
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).
 
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