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}
:\
"
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:
http://www.adamrapley.com/eqn1/CodeCogsEqn%20(6).gif
http://www.adamrapley.com/eqn1/CodeCogsEqn%20(5).gif...