 #1
sun18
 16
 1
Homework Statement
Consider a slender rod satellite in a circular orbit. Show that the position of the center of gravity, ##\rho_G##, can be written in terms of the position of the center of mass, ##\rho_C##, as: ##\rho_G = \rho_C (1+\frac{l^2}{4\rho_C^2})^\frac{1}{4} ##, where ## l## is the length of the rod.
I'm not sure how to best describe this, but the rod is oriented like this, perpendicular to the position vector:


##\rho_C####\rho##M


Homework Equations
I know the total force on the rod acts on the CoG:
## F = \frac{\mu mM}{\rho_G^2} ##
m is the total mass of the bar, M is the mass of the attracting body, ##\rho## is a position vector.
I need to set this force equal to an integral, which I believe goes something like this:
## F = \int_{0}^{m} \frac{\mu M dm}{\rho^2} ##
The Attempt at a Solution
I changed to a length integral (half of the bar multiplied by 2), and have this:
##F = 2 \int_{\rho_C}^{\sqrt{\rho_C^2+l^2/4}} \frac{\mu Mmd \rho }{l \rho^2} ##
using the fact that ##dm = \frac{m}{l} d\rho##
I'm not confident that this is correct, as I can't seem to get the right answer. Something seems off about this integral, but I'm not sure what.