Estimating Flattening of the Earth

[Fightfish]

Homework Statement
Estimate the flattening of the earth, given by $$1 - \frac{R_{P}}{R_{E}}$$, where $$R_{P}$$ is half the distance between the north pole and south pole and $$R_{E}$$ the radius of the equator.

The attempt at a solution
Well, I did manage to obtain a solution of sorts, but it involved a lot of approximations along the way, which I am not sure if they are all justifiable.

We start off with the base assumption that the flattening of the Earth is very small, such that there is relatively little deviation from a spherical shape.
Then, Newton's law of gravity is approximately true for this oblate spheroid. Modeling the Earth as a frictionless incompressible fluid, in order for a point at the pole surface to be in hydrostatic equilibrium with one at the equatorial surface, we have:

$$\frac{GM}{R_{P}^{2}} = \frac{GM}{R_{E}^{2}} + R_{E} \omega^{2}$$
$$\frac{R_{E}^{2}}{R_{P}^{2}} = 1 + \frac{R_{E}^{3} \omega^{2}}{GM}$$​

Now, if the Earth were a perfect sphere, $$M = \frac{4}{3}\pi R_{S}^{3} \rho$$. Since the flattening of the Earth is very small, we expect $$\frac{R_{E}^{3}}{R_{S}^{3}} \approx 1$$. Then,

$$\frac{R_{P}}{R_{E}} = (1 + \frac{3 \omega^{2}}{4G\pi\rho})^{-\frac{1}{2}}$$​

As we expect $$\frac{R_{E}}{R_{P}}$$ to be very close to 1, this implies that $$\frac{3 \omega^{2}}{4G\pi\rho} \ll 1$$, and hence

$$\frac{R_{P}}{R_{E}} \approx 1 - \frac{3 \omega^{2}}{8G\pi\rho}$$​

Thus, the flattening of the earth

$$1 - \frac{R_{P}}{R_{E}} \approx \frac{3 \omega^{2}}{8G\pi\rho} = \frac{3 \pi}{2GT^{2}\rho}$$​

Does this look reasonable?

 

[Andrew Mason]

Estimate the flattening of the earth, given by $$1 - \frac{R_{P}}{R_{E}}$$, where $$R_{P}$$ is half the distance between the north pole and south pole and $$R_{E}$$ the radius of the equator.

...
$$\frac{R_{E}^{2}}{R_{P}^{2}} = 1 + \frac{R_{E}^{3} \omega^{2}}{GM}$$

I would stop right there. You can work out $\omega = 2\pi/T$ (T = period of one rotation of the earth) and you can look up M (= mass of the earth). There is no need to make any approximations as you have done.

AM