Rearranging equation with a square root

[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:
[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(6).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(5).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(4).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(3).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(2).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(1).gif[/URL]

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)

Thanks :)

 

[HallsofIvy]

"

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:
[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(6).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(5).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(4).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(3).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(2).gif[/URL]

[URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(1).gif[/URL]

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)

Thanks :)

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 $r^2/(LmM)$

 

[Automated]

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 $r^2/(LmM)$

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

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

:\

 

[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...