2
$\begingroup$

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 $M$.

  • 0
    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).$2012-08-27

1 Answers 1

2

$ \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.

  • 0
    @user38799 No problem! Glad I could help.2012-08-27