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To a pretty good accuracy, the elapsed time [itex]\delta \tau[/itex] on a clock in orbit is given by:I am happy with them being brought together though, else there might be something to do with the elliptical orbit that has not been made clear. As you seemed to be able to conclude that B's clock would have gone faster, but in the bit you had quoted of what I had written, I had not mentioned which was in lower orbit, or their respective velocities. Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?

[itex]\delta \tau = (1 - \frac{GM}{c^2 r} - \frac{1}{2} \frac{v^2}{c^2}) dt[/itex]

where [itex]t[/itex] is coordinate time (using the most natural coordinate system). For a perfectly circular orbit, we have, approximately:

[itex]\frac{v^2}{c^2} = \frac{GM}{c^2 r}[/itex]

So [itex]\delta \tau = (1 - \frac{3}{2} \frac{GM}{c^2 r}) dt[/itex]

So higher orbits correspond to smaller [itex]\frac{1}{r}[/itex], which correspond to more elapsed time. You can think of the factor of [itex]\frac{3}{2}[/itex] as being a sum of a factor of [itex]1[/itex] due to gravitational time dilation and a factor of [itex]\frac{1}{2}[/itex] due to velocity-dependent time dilation. I don't know if that answers the question, or not.