= Solution
Interpret the orbit in the <Schwarzschild metric> as an unforced circular <timelike geodesic> and choose its plane as $\theta=\pi/2$. Put $f=1-2\mu/r$. The radial <geodesic equation> at constant radius uses the <Christoffel symbols>
$$
\Gamma^r{}_{tt}=\frac{c^2\mu f}{r^2},\qquad
\Gamma^r{}_{\phi\phi}=-fr,
$$
and gives the <Angular velocity of a circular Schwarzschild geodesic>
$$
\Omega^2=\left(\frac{d\phi}{dt}\right)^2=\frac{c^2\mu}{r^3}=\frac{GM}{r^3}.
$$
The <proper-time normalization> then yields
$$
\left(\frac{d\tau}{dt}\right)^2=f-\frac{r^2\Omega^2}{c^2}=1-\frac{3\mu}{r}.
$$
Over one revolution, the <Proper period of a circular Schwarzschild geodesic> is therefore
$$
\boxed{\Delta\tau=2\pi\sqrt{\frac{r^3}{GM}}\sqrt{1-\frac{3GM}{c^2r}},\qquad r>3GM/c^2.}
$$
Both the gravitational and orbital kinematic contributions to <time dilation> are included. Circular timelike geodesics exist for $r>3\mu$; they are stable for $r>6\mu$ and unstable for $3\mu<r<6\mu$, with marginal stability at $6\mu$.
If “circular orbit” instead means a forced trajectory with an externally chosen <angular velocity>, the radius alone does not fix the answer. Its elapsed time is $\Delta\tau=\Delta t\sqrt{1-2\mu/r-r^2\Omega^2/c^2}$, or $2\pi\sqrt{1-2\mu/r-r^2\Omega^2/c^2}/|\Omega|$ per revolution. The <geodesic> interpretation supplies the missing <angular velocity>.
Back to article page
