It's common to derive the lapse of proper time in circular orbits by splitting up the problem into separate components, one to account for the velocity effect in accord with special relativity, and another to account for the gravitational effect in accord with general relativity. However, the general theory subsumes the special theory, and it's actually easier to treat the problem holistically from a purely general relativistic standpoint. (This is more often the case than one might think, and the persistent tendency to artificially bifurcate problems into "special" and "general" components is partly just a consequence of the two-stage process by which Einstein arrived at the final theory.)
In the vicinity of an isolated non-rotating spherical body whose Schwarzschild radius is 2m the metric has the form
where f = longitude and q = latitude (e.g., q = 0 at the North Pole and q = p/2 at the equator). Let's say our radial position r and our latitude q are constant for each path in question (treating r as the "radius" in the weak field approximation). Then the coefficients of (dt)2 and (df)2 are both constants, and the metric reduces to
If we're sitting on the Earth's surface at the North Pole, we have sin(q) = 0, so it follows that ds = 
dt where r is the radius of the Earth.
On the other hand, in an equatorial orbit with radius r = R then we have q = p /2, sin2(q) = 1, and so the coefficient of (df)2 is simply R2. Now, recall Kepler's law w2 R3 = m, which, as we saw in section 5.5, also holds exactly in general relativity (provided that R is interpreted as the radial Schwarzschild coordinate and w is defined with respect to Schwarzschild coordinate time). Since w = df/dt we have R2 = m/(w2 R) = (dt/df)2 (m/R). Thus the path of the orbiting particle satisfies
(dt)2 = (1 - 2m/R) (dt)2 - (m/R) (dt)2 = (1 - 3m/R) (dt)2
Now for each test particle, one sitting at the North Pole and one in a circular orbit of radius R, the path parameter s is the local proper time, so the ratio of the orbital proper time to the North Pole's proper time is
To isolate the difference in the two proper times, we can expand the above function into a power series in m/r to give
The mass of the earth, represented in geometrical units by half the Schwarzschild radius, is about 0.00443 meters, and the radius of the earth is about 6.38(10)6 meters, so this gives
dtorbit = dtearth + (6.9)10-10 (1 - 3k/2) dtearth
which shows that the discrepancy in the orbit's lapse of proper time during a given lapse DT of proper time measured on Earth is
Consequently, for an orbit at the radius R = 3r/2 (about 2000 miles up) there is no difference in the lapses of proper time. Thus, if someone wants to get a null result, that would be their best choice. For orbits lower than 3r/2 the satellite will show slightly less lapse of proper time (i.e., the above discrepancy will be negative), whereas for higher orbits it will show slightly more elapsed time than the corresponding interval at the North Pole.
For example, in a low Earth orbit of, say, 360 miles, we have r/R = 0.917, so the proper time runs about 22.5 microseconds per day slower than a clock at the North Pole. On the other hand, for a 22,000 mile orbit we have r/R = 0.18, and so the orbit's lapse of proper time actually exceeds the corresponding lapse of proper time at the North Pole by about 43.7 microseconds per day. Of course, as R continues to increase the orbital velocity drops to zero and we are left with just coordinate time for the orbit, relative to which the North Pole on Earth is "running slow" by about 60 micro-seconds per day, due entirely to the gravitational potential of the earth. (This means that during a typical human life span the Earth's gravity stretches out our lives to cover an extra 1.57 seconds of coordinate time.)
By the way, it's interesting to note that equation (1) goes to zero when the orbit radius R equals 3m. That's because 3m is the radius of the orbit of light. This suggests that even if something prevented a massive object from collapsing within its Schwarzschild radius 2m, it would still be a very remarkable object if it was just within 3m, because then it could (theoretically) support circular light orbits, although I don't believe such orbits would be stable (even neglecting interference from infalling matter). If neutrinos are massless there could also be neutrinos in 3m (unstable) orbits near such an object.
The results of this and the previous section can be used to clarify the so-called twins paradox. In some treatments of special relativity the difference between the elapsed proper times along different paths between two fixed events is attributed to a difference in the locally "felt" accelerations along those paths. In other words, the asymmetry in the proper times is "explained" by the asymmetry in local accelerations. However, this explanation fails in the context of general relativity and gravity, because there are generally multiple free-fall (i.e., locally unaccelerated) paths of different proper lengths connecting two fixed events.
For a simple example, consider two test particles in the gravitational field of a spherically symmetrical mass m. One of these particles follows a perfectly circular orbit of radius r, while the other particle follows a purely radial (up and down) trajectory, beginning at a height r, climbing to R, and falling back to r, as shown below.
We can arrange for the two objects to initially coincide, and for the first object to complete one circular orbit in the same (coordinate) time it takes for the second object to rise and fall. Thus the objects coincide at two fixed events, and they are each in free-fall continuously in between those two events. Nevertheless, we will see that the elapsed proper times for these two objects are not the same.
Throughout this example, we will use dimensionless times and distances by dividing each quantity by the mass m in geometric units. For a circular orbit of radius r in Schwarzschild spacetime, Kepler's 3rd law gives the proper time to complete one revolution as
Applying the constant ratio of proper time to coordinate time for a circular orbit, we also have the coordinate time to complete one revolution
For the radially moving object, the usual parametric cycloid relation gives the total proper time for the rise and fall
where the parameter a satisfies the relation
The total elapsed coordinate time for the radial object is
where
In order for the objects to coincide at the two events, the coordinate times must be equal, i.e., we must have Dtcirc = Dtradial. Therefore, for any given value of r, the parameter a must satisfy
bearing in mind that q is a function of both r and a. Once we have determined the value of a for a given r, we can then determine the ratio of the elapsed proper times for the two paths. This gives
For fairly small value of r the ratio of proper times behaves as shown below.
Not surprisingly, the ratio goes to infinity as r drops to 3, because the proper time for a circular orbit of radius 3m is zero. (Recall that the "r" in our equations signifies r/m in normal goemetrical units.)
The a parameters and proper time ratios for some larger values of r are tabulated below.
For large values of r the ratio of proper times clearly approaches ek/r where k is a constant equal to approximately 0.825... As an example, consider a clock in a circular orbit at 360 miles above the Earth's surface. In this case the radius of the orbit is about (6.957)106 meters. Since the mass of the Earth in geometrical units is 0.00443 meters, we have the normalized radius r = (1.57053)109, and the total time of one orbit is approximately 5775 seconds (i.e., about 1.604 hours). In order for a radial trajectory to begin and end at this altitude and have the same elapsed coordinate time as one circular orbit at this altitude, the radial trajectory must extend up to R = (1.55)107 meters, which is about 5698 miles above the Earth's surface. The value of k/r in this example is extremely small, so we can approximate ek/r by 1 + k/r. Thus we have
and so the difference in elapsed proper times is given by
This is the amount by which the elapsed time on the radial (up-down) path would exceed the elapsed time on the circular path.