 Solution

The intended orbit is a freely falling timelike circular Schwarzschild geodesic. Use the Schwarzschild metric with $t$ rather than $ct$ as the zeroth coordinate, and put the orbit at $\theta =\pi /2$. Thus $g_{tt}=-f{c}^2$, $g_{rr}=1/f$, $g_{\varphi \varphi }=r^2$, with $f=1-2GM/(r{c}^2)$. The radial geodesic equation at constant $r$ isConsequently ${\Omega }^2=(u^{\varphi }/u^t{)}^2=GM/r^3$. The four-velocity normalization givesand henceHere $t$ is Schwarzschild time, $\tau$ the orbiting clock's proper time, $G$ the gravitational constant, $M$ the central mass, $r$ the constant areal radius, and $c$ the speed of light. The dimensionless $u^0$ follows the question's time-valued zeroth-coordinate convention; with $x^0=ct$ the zeroth four-velocity component would be $cdt/d\tau$.

The condition is $r>3GM/c^2$ for a massive circular geodesic; radial stability further requires $r>6GM/c^2$, with equality marginal. The phrase “any circular orbit” is false without free fall: a powered circular trajectory has $dt/d\tau =[f-r^2{\Omega }^2/c^2{]}^{-1/2}$ with independently chosen $\Omega$. For example, at $r=10GM/c^2$ choose $\Omega =\frac{1}{2}\sqrt{GM/r^3}$; this valid accelerated circular trajectory has $dt/d\tau =\sqrt{40/31}$, instead of the displayed $\sqrt{10/7}$.