= Solution
{c}
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 $g_{tt}=-fc^2$, $g_{rr}=1/f$, $g_{\phi\phi}=r^2$, with $f=1-2GM/(rc^2)$. The radial <geodesic equation> at constant $r$ is
$$
\Gamma^r_{tt}(u^t)^2+\Gamma^r_{\phi\phi}(u^\phi)^2=0,
\qquad \Gamma^r_{tt}=\frac{GMf}{r^2},\qquad \Gamma^r_{\phi\phi}=-rf.
$$
Consequently $\Omega^2=(u^\phi/u^t)^2=GM/r^3$. The <four-velocity> normalization gives
$$
-c^2=-fc^2(u^t)^2+r^2\Omega^2(u^t)^2
=-c^2\left(1-\frac{3GM}{rc^2}\right)(u^t)^2,
$$
and hence
$$
\boxed{u^0=u^t=\frac{dt}{d\tau}=\left(1-\frac{3GM}{rc^2}\right)^{-1/2}.}
$$
Here $t$ 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 $c\,dt/d\tau$.
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 $\Omega=\tfrac12\sqrt{GM/r^3}$; this valid accelerated circular trajectory has $dt/d\tau=\sqrt{40/31}$, instead of the displayed $\sqrt{10/7}$.
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