Rearranging for x in Algebraic Manipulation Homework

[unique_pavadrin]

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:
$$\begin{array}{c}<br /> \frac{{GM}}{{x^2 }} = \frac{{Gm}}{{\left( {r - x} \right)^2 }} \\ <br /> x^2 \left( {r - x} \right)^2 = GM\left( {Gm} \right) \\ <br /> x^2 \left( {r^2 - 2rx + x^2 } \right) = G^2 Mm \\ <br /> r^2 x^2 - 2rx^3 + x^4 = G^2 Mm \\ <br /> \end{array}$$

Any help is greatly appreciated, many thanks in advance,
unique_pavadrin

 

[cristo]

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

 

[pugfug90]

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]

Could use the quadratic formula

 

[unique_pavadrin]

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:

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

thanks once again for the help from those who replied

 

[cristo]

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

 

[unique_pavadrin]

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

 

[cristo]

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

Not quite:

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

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

 

[theperthvan]

kill the G!

 

[unique_pavadrin]

okay thanks cristo, that was great help thanks
unique_pavadrin