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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1$\begingroup$ Does this answer your question? What is the weight equation through general relativity? $\endgroup$– John RennieCommented Jan 16 at 7:39
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$\begingroup$ 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 en.wikipedia.org/wiki/… $\endgroup$– m4r35n357Commented Jan 16 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.
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$\begingroup$ Only if you identify r in Newton with r in Schwarzschild! $\endgroup$ Commented Jan 16 at 17:46
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$\begingroup$ @m4r35n357 - Sure, it's the circumference divided by 2π as per the usual convention. $\endgroup$– YukterezCommented Feb 7 at 19:22
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$\begingroup$ 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". $\endgroup$ Commented Feb 8 at 9:09