# Rearranging equation with a square root

1. Sep 15, 2011

### Automated

"
Solving this for κ, substituting into (1), and rearranging for G, the result is:
"

I am trying to rearrange the first equation to make κ the subject and I get:

But somehow I think I've done something wrong :S Please could someone point out where I've gone wrong (or if I am indeed actually right, then how I substitute it into
equation (1) which is [PLAIN]http://upload.wikimedia.org/math/2/c/a/2ca1bb59fb981ea80bd5c3d642d26949.png) [Broken]

Thanks :)

Last edited by a moderator: May 5, 2017
2. Sep 15, 2011

### HallsofIvy

Yes, what you have done is perfectly correct. Now, substituting that into your "eq. 1", you have
$$\frac{mL^2(2\pi \theta^2)}{T^2}= \frac{LGmM}{r^2}$$

Solve for G by multiplying both sides by $r^2/(LmM)$

Last edited by a moderator: May 5, 2017
3. Sep 15, 2011

### Automated

Sorry, but I'm not totally sure how you substituted that in because when I tried it, somehow I managed to get this:

$$\frac{mL^2(2\pi^2)\theta}{2T^2}=\frac{LGmM}{r^2}$$

:\

4. Sep 15, 2011

### Automated

The result it says I should get once "Solving this for κ, substituting into (1), and rearranging for G" is:

But I'm not sure how to get there...