Okay, here is my derivation by request
This is Part 1 (out of 2). Here I'll derive an exact relation between radius and latitude. In part 2 I'll make approximations to get a solution for r.
In a reference frame rotating with the planet, the forces on a test mass
m are the gravitational force and a centrifugal force:
<br />
F_{grav} = -\frac{GMm}{r^2}, directed inward toward the planet's
center
<br />
F_{cent} = m\omega^2 r \cos(\theta), directed outward from the planet's
axis
where
M is the planet's mass
r is the distance from the planet's center to a surface point
\omega is the planet's rotation rate (rad/sec)
\theta is the latitude at a surface point (0 at the equator, 90 degrees at the poles)
The potential energy of the test mass is then
<br />
U = -\int\vec{F}\cdot\vec{dr}<br />
<br />
= -\frac{GMm}{r} - \frac{1}{2} m \omega^2 r^2 \cos^2(\theta)<br />
Note we picked up an extra cosine factor in the centrifugal term, because force and radial vector make an angle to one another:
Work = force x displacement x \cos(\theta)
The planet's surface conforms to an equipotential surface. In particular, the potential energy at a pole equals the potential energy anywhere else on the surface:
<br />
-\frac{GMm}{r_{pole}} = -\frac{GMm}{r} - \frac{1}{2} m \omega^2 r^2\cos^2(\theta)<br />
Multiply both sides of this equation by the factor
-\: \frac{r}{GMm} :
<br />
\frac{r}{r_{pole}}= 1 + \frac{r^3 \omega^2}{GM}\cos^2(\theta)<br />
or
<br />
\frac{r}{r_{pole}}= 1 + (\frac{r}{r_{pole}})^3 \cdot\frac{r_{pole}^3\omega^2}{GM}\cos^2(\theta)<br />
This is an exact equation, using dimensionless variables and parameters:
<br />
\frac{r}{r_{pole}} ,
<br />
\frac{r_{pole}^3\omega^2}{GM} ,
and
<br />
\theta<br />
While cubic equations are analytically solvable, the formula involved is quite cumbersome. In part 2 I'll use the approximation r \approx r_{pole} to get a solution for r.
