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Circular Orbit Stability

We now show that a circular orbit is possible whenever the effective potential energy given is either a maximum or minimum. That is, a circular orbit is possible if,

or, F ( r ) = – mr ω²

The above equation is the familiar relation for a circular orbit, viz. the radial force must be equal to necessary centripetal force. (Negative sign shows that force must be attractive.)

For a given value of L, the radius of a circular orbit is given, for gravitational force, this gives (C = GMm),

the corresponding minimum (or maximum) value of U’ ( r ) is,

where we substituted the value of r₀. Since for a circular orbit, dr/dt = 0, shows that the total energy E is same as U’ ( r ). That is,

Recall that the central force motion is analytically equivalent to a one-dimensional motion along radial direction under effective potential U’ ( r ). In case of a circular orbit, the radial velocity is zero and hence the particle stays at equilibrium position r = r₀. Now this equilibrium would be stable if U’ ( r₀ ) is minimum; the equilibrium will be unstable, if U’ ( r₀ ) is maximum. To check the above, we determine the second derivative of U’ ( r ) at r = r₀, for an inverse-square force [U ( r ) = – C/r]:

Putting r = r₀ and using r₀ = L²/mC’, we get

Hence, circular orbit is always stable for an inverse-square (attractive) force.

If the equilibrium is stable about r = r₀, a small displacement along r will result in simple harmonic motion in radial direction, about equilibrium position. The resultant motion still looks almost circular.