Jorrie
Science Advisor
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kev said:We start with the classic Keplerian equation for the orbital period
(Eq 1) T = 2 \pi \sqrt{\frac{R^3}{G M} }
You then applied gravitational and velocity time dilation to get
(Eq 10) T_p = \frac{2 \pi R }{c}\sqrt{(\frac{R c^2}{G M}-3)}
Your result is correct for a circular orbit and I think the steps are correct too. The only non-rigorous part is the 'leap of faith' that one can in fact take the Keplerian period, apply time dilations and get the relativistic period. This only works for perfectly circular orbits in Schwarzschild geometry (other orbits have to deal with spatial curvature too), so it by no means a general truth and should be used with care.
-J
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