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yuiop
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Imagine we have an ideal Schwarzschild black hole. A small satellite of insignificant mass (so that the Shwarzchild geometry remains ideal) is orbiting at coordinate radius R just outside the photon orbit in a perfect circular orbit. For the purpose of this experiment ignore the orbital decay due to loss of energy by gravitational waves and "the observer at infinity" is taken to mean that the observer is at sufficient distance from the black hole that gravitational time dilation on his local clock is insignificant.
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?
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?