Does General Relativity imply greater accelerations than Newtonian gravity in strong gravitational fields, such as at 2 m/s^2?
Do the general relativistic corrections add up to more "gravity" in strong gravitational fields?
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1Does this answer your question? What is the weight equation through general relativity? – John Rennie Jan 16 '24 at 07:39
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Yes they do; there is an "extra" inverse cubic term in the effective potential which overcomes the inverse square "centrifugal repulsion" at very close distances, see https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity#Effective_radial_potential_energy – m4r35n357 Jan 16 '24 at 13:04
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Yes, instead of the Newtonian $\rm F=m \ a=G \ M \ m/r^2$, under Schwarzschild we have $\rm F=m \ a=G \ M \ m/r^2/\sqrt{1-r_s/r} \ $ for a stationary observer, and at $\rm r=1.5 \ r_s \ $ you'd need $\rm v=1 \ c \ $ in order to orbit, while under Newton you'd need only $\rm 0.577 \ c$, see here.
Yukterez
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@m4r35n357 - Sure, it's the circumference divided by 2π as per the usual convention. – Yukterez Feb 07 '24 at 19:22
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It is an identification, but it is clearly not the same thing. Anyway, I have done this myself in order to "compare" orbits in Newton with GR, so I am pragmatic about it, I just don't take it seriously as the "same thing". – m4r35n357 Feb 08 '24 at 09:09