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It only recently was specifically pointed out to me that Einstein was right, and Newton was wrong.

Newton wasn't wholly wrong. He theorized that gravity was a force that acted between bodies, and came up with an equation that fit observed experimental data:

$$ F = G*m1*m2/r^2$$

Where $G$ is an emperically determined constant.

Einstein did away with forces, and instead theorized that gravity is the consequence of bodies moving in a straight line through spacetime, where mass bends space.

And he provided his own formulas based on things like geometry and the speed of light.

Does that mean that Einstein's equations can simplify down to Newton's equation when you're far away from massive objects? Kind of like how terms the Lorentz transformation go to zero when you're not traveling near the speed of light?

And if so, does that mean we could come up with an algebraic expression for $G$?

And if not, why not?

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    $\begingroup$ Related: physics.stackexchange.com/q/118750/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 11, 2021 at 12:47
  • $\begingroup$ @PaulT It does not; that's why i was hoping to ask physics.stackexchange. $\endgroup$
    – Ian Boyd
    Nov 11, 2021 at 17:14
  • $\begingroup$ Can you edit your question elaborate on how that other question differs from the one you're asking? The answer does seem to be in there, to me: the value of $G$ in general relativity is chosen so that it agrees with Newtonian gravity. That implies that there's no new information contained in GR where $G$ can be derived independently. $\endgroup$ Nov 11, 2021 at 19:51

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Newton isn't wrong at all. It is as you write: Einstein's GR simplifies to Newton's law of gravity in the respective non-relativistic limit. But no, we still can't come up with an algebraic expression for $G$ because that constant also appears in Einstein's equations and simply carries through to the Newtonian limit.

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    $\begingroup$ Is there an an relativistic-corrected equation that gives Force = ...? The linked answers in comments give equations that look nothing like "Force, in Newtons, equals ...". In other words, if F = Gm1m2/r2 is incomplete, what is the updated formula to calculate force? $\endgroup$
    – Ian Boyd
    Nov 11, 2021 at 15:59
  • $\begingroup$ @IanBoyd The geodesic equation. $\endgroup$
    – Andrew
    Nov 11, 2021 at 17:10
  • $\begingroup$ one answer is here, although not as nice as you would like it to be spelled out: physics.gmu.edu/~rubinp/ires/Vartak_Summer_Report.pdf $\endgroup$
    – rfl
    Nov 11, 2021 at 17:10
  • $\begingroup$ Based on those two comments i take it the answer is "no"? I don't know what "second derivitive of X superscript mu with respect to T" is; but it doesn't sound like "force in Newtons". $\endgroup$
    – Ian Boyd
    Nov 11, 2021 at 17:13
  • $\begingroup$ @IanBoyd The answer is yes. The "second derivative of $X$ with respect to $T$" is acceleration. Multiply both sides by mass and you have force. (You need to multiply both sides by mass to compare with Newton's laws because mass always cancels out of $F=ma$ in gravity because of the equivalence principle). As stated in the wikipedia article I sent, the geodesic equation can be used "to compare General Relativity with Newtonian Gravity." $\endgroup$
    – Andrew
    Nov 11, 2021 at 19:19

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