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The observer at infinity (in the plane of the satellite orbit) starts his clock when he sees the satellite cross the line of sight between him and the black hole and stops his clock the next time the satellite crosses the same line of sight.

**He would find the period to be in exact agreement with the prediction of Kepler's original equation for the period of a circular orbit**,

[tex] T_{(Kepler)} = 2\pi\sqrt{\frac{R^3}{GM}} = \frac{2\pi R}{c}\sqrt{\frac{2R}{R_s}}[/tex]

while a non-orbiting non-inertial local observer at R would measure the orbital period to be less than the Kepler prediction by a factor equal to the gravitational time dilation factor.

[tex] T_{R} = 2\pi\sqrt{\frac{R^3}{GM}\left(1-\frac{2GM}{c^2R}\right)} =\frac{2\pi R}{c} \sqrt{\frac{2R}{R_s}\left(1-\frac{R_s}{R}\right)}[/tex]

Agree or disagree?