For a circular orbit, the derivative of

V_{eff} (r) = - \frac{m}{r} + \frac{h ^{2}}{2 r ^{2}} - \frac{m h ^{2}}{r ^{3}}

vanishes. Thus

\frac{m}{r ^{2}} - \frac{h ^{2}}{r ^{3}} + \frac{3 m h ^{2}}{r ^{4}} = 0, h^{2} = \frac{m r ^{2}}{r - 3 m} .

A real finite angular momentum therefore requires

r > 3 m .

ii

Solution

For a circular orbit, normalization and the value of

h

give

E^{2} = \frac{( r - 2 m ) ^{2}}{r ( r - 3 m )}, \dot{t}^{2} = \frac{r}{r - 3 m}, \dot{ϕ}^{2} = \frac{m}{r ^{2} ( r - 3 m )} .

Consequently

(\frac{d ϕ}{d t})^{2} = \frac{ϕ ˙ ^{2}}{t ˙ ^{2}} = \frac{m}{r ^{3}} .

This is the relativistic form of the circular-orbit Kepler third law.

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