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Solve for $m$ in $F=G\Big(\dfrac{mM}{r^2}\Big)$

I ended up with $\frac{F}{G}\frac{r^2}{M}=m.$ Can I simplify it further or did I completely mess up the answer? I got to where I was by dividing both sides by $G$ then multiplying it by $r^2$ and the n dividing by $M$.

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Your answer is correct. I guess that's it. There are no further simplifications. – user2468 Aug 27 '12 at 0:21
One thing though, if $G$ was a function, then things would be slightly different. Let $G^{-1}$ denote the inverse of $G$. Then $m = \dfrac{r^2}{M} G^{-1}(F).$ – user2468 Aug 27 '12 at 0:23

$$ \begin{align*} &F=G\left(\frac{mM}{r^2}\right) \\ &\implies \frac{F}{G}=\frac{mM}{r^2} \space \text{Dividing both sides by }G \\ &\implies \frac{F}{G M}=\frac{m}{r^2} \space \text{Dividing both sides by }M \\ &\implies \frac{r^2 F}{G M}=m \space \text{Multiplying both sides by }r^2 \end{align*} $$

Note also that, for your answer

$$\frac{F}{G}\frac{r^2}{M}=\frac{F r^2}{GM}=m$$

which is correct.

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Ok thanks for clarifying that for me – Dragoon2375 Aug 27 '12 at 0:20
@user38799 No problem! Glad I could help. – Argon Aug 27 '12 at 0:21

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