= Solution
The <geostationary gravitational clock gain> uses only the potential contribution, as requested. In the weak field, $d\tau/dt\simeq1+\Phi/c^2$ and $\Phi=-GM/r$. The geostationary <radius> follows from $GM/r_g^3=\Omega_\oplus^2$. Taking $GM\simeq gR^2$ with $g=9.81\,\mathrm{m\,s^{-2}}$, $R=6.4\times10^6\,\mathrm m$ and <sidereal day> $86164\,\mathrm s$ gives $r_g\simeq4.23\times10^7\,\mathrm m$.
$$
\frac{\Delta\dot\tau}{\dot\tau}\simeq\frac{GM}{c^2}\left(\frac1R-\frac1{r_g}\right)
=5.93\times10^{-10},\qquad
\boxed{\Delta\tau_{\rm day}\simeq+51\,\mathrm{\mu s}.}
$$
The satellite clock gains time because it is at higher <gravitational potential>. This is not the combined gravitational-plus-kinematic correction; the orbital and surface rotational <velocities> are deliberately outside the requested potential contribution.
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