• Support PF! Buy your school textbooks, materials and every day products Here!

Algebraic Manipulation

  • #1

Homework Statement


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

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:
[tex]
\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}
[/tex]

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

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
72
What've you done in your first line? i.e. how does the original equation become x2(r-x)2=GM(Gm) ?
 
  • #3
118
0
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.
 
  • #4
danago
Gold Member
1,122
4
Could use the quadratic formula
 
  • #5
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:

[tex]
\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}
[/tex]

thanks once again for the help from those who replied
 
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
72
Why not cancel the G on both sides in the first line? There's no need to carry it through the calculation then.
 
  • #7
oh true, thanks
other than that are my steps right?
thanks
 
  • #8
cristo
Staff Emeritus
Science Advisor
8,107
72
oh true, thanks
other than that are my steps right?
thanks
Not quite:
[tex]
\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}
[/tex]
You missed a minus sign in the last line: it should read (without the G's)[tex]x=\frac{-2Mr\pm\sqrt{4r^2M^2+4(m-M)Mr^2}}{2(m-M)}[/tex]
 
  • #9
184
0
kill the G!
 
  • #10
okay thanks cristo, that was great help thanks
unique_pavadrin
 

Related Threads for: Algebraic Manipulation

  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
3
Views
764
Replies
5
Views
1K
Replies
3
Views
756
Top