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When you calculate the gravitational force $F_G = G \cdot\frac{m\cdot M}{r^2}$.

Is this equation precise for relativistic big masses, given that $m$ and $M$ do not and can not move? If not is there an "easy" accurate equation?

What does not moving mean: I am interested in the force purely caused by the existing masses, if a mass moves its motion alone will have an effect on spacetime that also changes the force on other masses. So if we could fix both masses in places so both are in rest from perspective of observer and both will have magical force countering every gravitational force acting on them, preventing them from moving toward each other. Is there still an error in newtonians equation for i.e. gigantic masses maybe?

aludebe
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2 Answers2

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So I assume $m$ and $M$ don't move means you are interested in the force at fixed radius. The post-Newtonian expansion of the potential (a concept that works for weak fields with no motion) is:

$$ V(r) = -\frac{GM}r - \frac 1{c^2}\big(\frac{GM}r\big)^2 - \frac 4 3 \frac 1{c^4}\big(\frac{GM}r\big)^3+\ldots $$

The 1st term is the regular Newtonian potential.

In terms of the Schwarzschild radius $R=\frac{2GM}{c^2}$, that's

$$ V(r) = V_{\rm Newton}\Big( 1 + \frac R {2r} + \frac 4 3 \big(\frac R {2r}\big)^2 +\cdots\Big)$$

The series expansion is for $\ln(1+x) = x + \frac 1 2 x^2+\cdots$ for $|x|\lt 1$, with $x = -R/r$, so it diverges at the event horizon.

JEB
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Your question is a little muddled; you say "given that $m$ and $M$ do not and can not move", but then ask about "relativistic big masses" -- what do you mean by that? Any mass at all will seem to be "moving" in some reference frame

In general Newton's equation for gravity is only an approximation. It's a good approximation if the masses involved are not moving very quickly relative to one another, and are not too dense. The exact solutions (so far as we know) are given by Einstein's field equations of general relativity. These are quite complicated. Some non-Newtonian approximations to the EFE might work in some situations (i.e. give reasonable, and more accurate, answers than Newtonian gravity) but it will depend on exactly what the configuration is that you're trying to model.

Eric Smith
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