This is going to be a long calculation, be warned in advance
OK, let's make life easy and assume that the satellite is either orbiting at grouind level, or that there is a tower raised up above ground level so that we compare the satellite's clock to a clock that's not only at the same altititude, but that passes very close by the satellite on every orbit.
The "planet" is not-quite a black hole, because the photosphere where light orbits is at r = (3/2) r_s.
I don't think the orbit would be stable, but that's not a fatal objection, the satellite can have thrusters and a program designed to stabilize the unstable orbit.
The equations govering the orbit around a black hole apply directly to this case (even though the planet isn't of itself quite a black hole), because we are assuming the planet is a spherically symmetrical body, and the gravity of any such body is given by the Schwarzschild metric (the same metric as that of a black hole) above the surface of the body.
We will work out the problem in Schwarzschild coordinates, r and t. We have
r = constant
which means dr/dtau must be zero. From eq 25.16a in MTW, this means
V^2 = E^2. By 25.16b this means
eq #1
(1-2m/r) (1+L^2/r^2) = E^2
here E and L are conserved quantiites of orbits in the Schwarzschild geometry that are the relativistic equivalents of the energy per unit mass and the angular momentum per unit mass of the body. Note that the energy and angular momentum of a body in orbit are also conserved under Newtonian gravity.
Now dt/dtau_sat = E/(1-2m/r) by 25.18
here dtau_sat is the time measured on the satellites clock, and dt is an abstract quantity that has no direct relevance to the problem, the schwarzschild coordinate time interval, which can be thought of the time interval "at infinity".
Now we have to find dt/dtau_ground, the ratio of the schwarzschild coordiante time interval to the proper time of a clock on a tower. This will be
dt/dtau_grnd = 1/sqrt(1/(1-2m/r))
by the standard gravitational time dilation formula.
We can take the ratio and find
(dt/dtau_sat) / (dt/dtau_grd) = (dtau_gnd / dtau_sat) = E/sqrt(1-2*m/r)
Using the results of eq #1, we can simplify this to
dtau_gnd / dtau_sat = sqrt(1+L^2/r^2)
This means that the "ground" clock on the tower will show more time elapsed than the orbiting clock.
Also of interest is eq 25.20, which gives the local tangential velocity (ordinary velocity, not 4-velocity) of the orbiting satellite as measured by the ground-based observer. This is
v = L/(r Elocal), where Elocal = E/sqrt(1-2m/r)
I think it ought to be possible to express all the results in terms of v rather than E and L, but I haven't done the manipulations to do so. I also badly need to recheck all the steps to see if I've made any errors, the calculations are rather long.