Suppose you have a cylindrical elevator that resembles a rod of mass $m$, cross-sectional area $A$ and uniform density $\rho$, with a length $L$ spanning the distance between the Earth and the Moon. Where is the centre of gravity of this Earth-Elevator-Moon system?
Here's my attempt.
I used Newton's Law of Gravity to find an expression for the resultant force experienced by a test mass, $m$ between the Earth and the Moon:
$\sum F=\frac{GM_1m}{(R_1+h)^2}-\frac{GM_2m}{(R_2+(L-h))^2}$
where $h$ is the distance from the surface of Earth, $M_{1,2}$ are the masses of the Earth and Moon respectively, $G$ is the gravitational constant and $R_{1,2}$ are the radii of the Earth and Moon respectively.
Then, I used some calculus to get another expression for the resultant force experienced by a test mass, with the eventual intention to equate and solve for $h$:
$\sum F =\frac{GM_1m}{r^2}-\frac{GM_2m}{(D-r)^2}$
where $D=R_1+L+R_2$.
The force experienced by an infinitesimal mass (of the elevator) is:
$\mathrm{d}F=G\left(\frac{M_1}{r^2}-\frac{M_2}{(D-r)^2}\right)\ \mathrm{d}m$
Using $\mathrm{d}m=\rho A\ \mathrm{d}L$ and integrating (I've replaced $\mathrm{d}L$ with $\mathrm{d}r$ here):
$\int\mathrm{d}F=G\rho A\displaystyle\int_{R_1+h}^{R_1+L}\frac{M_1}{r^2}-\frac{M_2}{(D-r)^2}\ \mathrm{d}r$
After integrating, simplifying and equating it with the previous equation, then simplifying some more, I end up with:
$M_1\left(\frac{L-h-\rho A h(R_1+h)}{(R_1+h)^2(R_1+L)}\right)=M_2\left(\frac{2R_2+(L-h)-\rho Ah(R_2+(L-h))}{R_2(R_2+(L-h))^2}\right)$
Then, I'm not sure how to go about making $h$ the subject here. Am I on the right track? Is there a better / faster way of doing this? I'm thinking using energy approach might be better but am not sure how to deal with the elevator that isn't a point mass.
