# Algebraic Manipulation

## Homework Statement

Hey I need to rearrange the following, and find x in terms of G, M, m, r
$$\frac{{GM}}{{x^2 }} = \frac{{Gm}}{{\left( {r - x} \right)^2 }}$$

2. The attempt at a solution

I haven't manged to get far with this problem as I am confused about the powers of x and how to manage them. This is where I have manged to get to:
$$\begin{array}{c} \frac{{GM}}{{x^2 }} = \frac{{Gm}}{{\left( {r - x} \right)^2 }} \\ x^2 \left( {r - x} \right)^2 = GM\left( {Gm} \right) \\ x^2 \left( {r^2 - 2rx + x^2 } \right) = G^2 Mm \\ r^2 x^2 - 2rx^3 + x^4 = G^2 Mm \\ \end{array}$$

Any help is greatly appreciated, many thanks in advance,

cristo
Staff Emeritus
What've you done in your first line? i.e. how does the original equation become x2(r-x)2=GM(Gm) ?

I have an idea.. multiply both sides by (1/G) or divide by G..
Then you'd get mx^2=M(r-x)^2 and work from there.

danago
Gold Member

Thanks cristo for having pointed out that stupid mistake.Pugfug90, your method doesn't seem to work, but thanks anyhow. Danago, thanks for your suggestion, as i have used it. Here is what i have managed to come up with:

$$\begin{array}{l} \frac{{GM}}{{x^2 }} = \frac{{Gm}}{{\left( {r - x} \right)^2 }} \\ Gmx^2 = GM\left( {r - x} \right)^2 \\ Gmx^2 = GM\left( {r - x} \right)\left( {r - x} \right) \\ Gmx^2 = GM\left( {r^2 - 2rx + x^2 } \right) \\ Gmx^2 = GMr^2 - 2GMrx + GMx^2 \\ - GMr^2 = - 2GMrx + GMx^2 - Gmx^2 \\ GMr^2 = 2GMrx - GMx^2 + Gmx^2 \\ 0 = \left( {Gm - GM} \right)x^2 + 2GMrx - GMr^2 \\ x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ x = \frac{{ - 2GMr \pm \sqrt {\left( {2GMr} \right)^2 - 4\left( {Gm - GM} \right)\left( {GMr^2 } \right)} }}{{2\left( {Gm - GM} \right)}} \\ \end{array}$$

thanks once again for the help from those who replied

cristo
Staff Emeritus
Why not cancel the G on both sides in the first line? There's no need to carry it through the calculation then.

oh true, thanks
other than that are my steps right?
thanks

cristo
Staff Emeritus
oh true, thanks
other than that are my steps right?
thanks
Not quite:
$$\begin{array}{l} \frac{{GM}}{{x^2 }} = \frac{{Gm}}{{\left( {r - x} \right)^2 }} \\ Gmx^2 = GM\left( {r - x} \right)^2 \\ Gmx^2 = GM\left( {r - x} \right)\left( {r - x} \right) \\ Gmx^2 = GM\left( {r^2 - 2rx + x^2 } \right) \\ Gmx^2 = GMr^2 - 2GMrx + GMx^2 \\ - GMr^2 = - 2GMrx + GMx^2 - Gmx^2 \\ GMr^2 = 2GMrx - GMx^2 + Gmx^2 \\ 0 = \left( {Gm - GM} \right)x^2 + 2GMrx - GMr^2 \\ x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ x = \frac{{ - 2GMr \pm \sqrt {\left( {2GMr} \right)^2 - 4\left( {Gm - GM} \right)\left( {GMr^2 } \right)} }}{{2\left( {Gm - GM} \right)}} \\ \end{array}$$

You missed a minus sign in the last line: it should read (without the G's)$$x=\frac{-2Mr\pm\sqrt{4r^2M^2+4(m-M)Mr^2}}{2(m-M)}$$

kill the G!

okay thanks cristo, that was great help thanks