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According to Wikipedia the acceleration of small mass due to the gravitational attraction of a larger mass is $$a = \frac{G(M + m)}{r^2}.$$ This implies that the force acting on $m$ should be $$F = ma = m \times G(M + m) $$ $$F = \frac{G(Mm +m^2)}{r^2} $$

If for example if $m =$ 0.001$M$ such that $m$ is a small fraction of $M$, the above equation could be written as :

$$F = \frac{G(Mm + (0.001M)^2)}{r^2} = \frac{G(Mm + 0.000001M)}{r^2}$$

it seems that because when $m$ is much smaller than $M$ the $m^2$ term becomes negligible and is discarded to give us the much more well known formula $ F = GMm/r^2$, meaning the well known formula is an approximation even without considering general relativity and is not accurate where $m$ is a significant fraction of $M$. Is that the case?

Qmechanic
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KDP
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1 Answers1

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The link says:

"Dividing by their respective masses and subtracting the second equation from the first yields the equation of motion for the acceleration of the first object with respect to the second"

That is, in an inertial frame:

$$ a \equiv \frac{d^2r}{dt^2} = F/m = G\frac{M}{r^2} $$

but the formula you give is for:

$$\frac{d^2(r-R)}{dt^2} = G\frac{M+m}{(R-r)^2}$$

where $r$ ($R$) is the position of $m$ ($M$).

So we can kind of work backwards. With $r$ and $R$ being affine points in an inertial frame:

$$ \ddot R = \frac 1 M G \frac{Mm}{(R-r)^2} = G \frac{m}{(R-r)^2}$$ $$ \ddot r = \frac 1 m G \frac{Mm}{(R-r)^2} = G \frac{M}{(R-r)^2}$$

so

$$ \ddot r - \ddot R = G \frac{M+m}{(R-r)^2} $$

where I flipped the sign since the forces are equal and opposite .

JEB
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