= Solution
Write $A(r)=1-2\mu/r$, with $\mu=GM/c^2$. On the equatorial plane, the <geodesic Lagrangian> is
$$
L=\frac12\left(c^2A\dot t^2-A^{-1}\dot r^2-r^2\dot\phi^2\right).
$$
The coordinates $t$ and $\phi$ are cyclic, giving $\boxed{A\dot t=k,\quad r^2\dot\phi=h}$. The conserved physical <energy> per unit <rest mass> is $c^2k$, while $h$ is the signed <angular momentum> per unit <rest mass>. The radial <Euler-Lagrange equation>, with $A'=2\mu/r^2$, is
$$
\boxed{A^{-1}\ddot r+\frac{\mu c^2}{r^2}\dot t^2
-\frac{\mu}{r^2A^2}\dot r^2-r\dot\phi^2=0.}
$$
These are the requested <geodesic equations>. The equatorial condition is consistent with the polar equation by the spherical symmetry of the <Schwarzschild metric>.
The normalization of the <four-velocity> is $c^2=c^2A\dot t^2-A^{-1}\dot r^2-r^2\dot\phi^2$. Insert the two conserved quantities and multiply by $A$ to obtain
$$
\boxed{\dot r^2+\frac{h^2}{r^2}\left(1-\frac{2\mu}r\right)-\frac{2\mu c^2}r
=c^2(k^2-1).}
$$
This is a one-dimensional radial energy equation. Half of its left side is radial kinetic energy per unit mass plus an <effective potential>: the Newtonian centrifugal and gravitational terms acquire the extra attractive contribution $-\mu h^2/r^3$.
A <timelike circular Schwarzschild geodesic> requires $\dot r=\ddot r=0$. Equivalently, set the derivative of $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$ to zero. At $r=R$ this gives $h^2(R-3\mu)=\mu c^2R^2$, and substitution in the radial energy equation yields
$$
\boxed{h^2=\frac{\mu c^2R^2}{R-3\mu},\qquad
k^2=\frac{(R-2\mu)^2}{R(R-3\mu)}.}
$$
Such timelike circular orbits require $R>3\mu$. Their coordinate <angular velocity> satisfies $\Omega^2=\mu c^2/R^3$, and their proper clock rate is $d\tau_s/dt=A/k=\sqrt{1-3\mu/R_s}$.
For the static North-Pole clock, $d\tau_E/dt=\sqrt{1-2\mu/R_E}$. Consecutive overhead passages occur one coordinate orbital period apart, and the repeated Earth-satellite geometry gives equal signal travel delays. The <comparison of circular-orbit and static Schwarzschild clocks> therefore gives
$$
\boxed{\frac{\tau_s}{\tau_E}=\sqrt{\frac{1-3\mu/R_s}{1-2\mu/R_E}}.}
$$
This includes both gravitational and orbital-motion effects; a separate Newtonian time-dilation correction must not be added again.
For an estimate, <Kepler's third law> gives $R_s=42100\,\mathrm{km}\,(3/24)^{2/3}=10525\,\mathrm{km}$. The same approximate geostationary radius and period give $\mu=4\pi^2R_{\rm geo}^3/(c^2T_{\rm geo}^2)\simeq4.4\,\mathrm{mm}$. Expanding the ratio to first order,
$$
\frac{\tau_s}{\tau_E}\simeq1+\frac\mu{R_E}-\frac{3\mu}{2R_s}
\simeq\boxed{1+6.4\times10^{-11}}.
$$
The satellite clock consequently gains about $0.7\,\mu\mathrm s$ over a three-hour orbit in this approximation. The surface observer is taken to be static, and the spacetime model is the spherical <Schwarzschild metric> specified in the problem.
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