Hello sun18,
It's an interesting problem, for sure! Anyway I worked through it and there are two mistakes in your general approach and perhaps another quibble about your notational semantics. I'll start with notation part.
sun18 said:
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:
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##\rho_C##--------------------##\rho##-----------------------M
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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} ##
Not bad so far. But let me comment on the notation.
Newton's universal law of gravitation for point mass particles is often expressed as, using \vec \rho as the displacement vector,
\vec F = G \frac{Mm}{\rho^2} \ \hat \rho
That said, it's often expressed using \mu = GM as
\vec F = \frac{\mu m}{\rho^2} \hat \rho
where \mu is called the "
standard gravitational parameter" and has the mass of the gravitating body (often Earth when dealing with things orbiting Earth) folded into it along with Newton's gravitational constant.
In the notation you were using, you've simply replaced G with \mu, but didn't fold M into it. The notation that I've seen most often has M folded into the \mu, along with the G.
Of course, this is all just notational semantics, and isn't important on the grand scheme of things. But for what it's worth, I thought I should mention it.
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##
There's one problem right there.
The change in mass does not vary linearly with the change in \rho in this case. It does vary linearly in the case where the rod is parallel to the displacement vector; in that case it would work fine. But here, with the rod perpendicular to the displacement vector, it does not hold true.
Let's define x as the distance from \rho_c to some point along the rod. So you can say easily,
dm = \frac{m}{\ell} dx
You should be able to form a relationship between x and \rho knowing that they form a right triangle. \rho^2 = \rho_c^2 + x^2.
Now for the second problem: you are forgetting about the pesky unit vector \hat \rho.
In the case where the rod was parallel to the position vector, the direction of the force wasn't important; the force on every individual dm all pointed in the same direction, so the unit direction vector didn't really matter. But in this case, where the rod is perpendicular, the each of the individual force directions are different.
You need to break up the \hat \rho into its x- and y-components. It's in the form, \hat \rho = a_x \hat x + a_y \hat y. Due to symmetry you know that the horizontal (x-) components will all ultimately cancel. But the y-components will not. You need to put that into your expression: The way you express the unit vector for a given dm should be should be a function of \rho_c and x. [Edit: or alternatively, express things in polar coordinates involving an angle. Either way will work, as long as you take directionality into account one way or the other.]
Good luck!
[Edit: Looks like
@haruspex and others beat me to the punch. I suppose I should either type faster or write shorter.

]