66The idea is that the Earth will assume a shape which will be an equipotential surface of the combined potential arising due to gravitational and centrifugal (in the rotating frame of earth) effects. Why the earth does so is simple to explain. You should know that the force arising from the field is normal to an equipotential surface i.e. it has no component along the surface. So if the earth does not assume an equipotential surface, there would be some force on the particles at the surface which would be in the tangential direction; so over a long time, the earth will tend to flow along the surface and this flow would stop only when the tangential force would vanish. But in that case, the resulting surface would be an equipotential.
Note that the centrifugal potential is V_c=-\dfrac{1}{2}\omega^2x^2. Here x is the distance from the center of the circle.
For the present case, if we measure the (variable) distance from the center of the earth as r and the angle θ from the north pole, then at an arbitrary point (r,θ), the total potential is
V=V_g+V_c=-\dfrac{GM}{r}-\dfrac{1}{2}\omega^2r^2\sin^2\theta
Since the deviations from the original sphere would be small enough, I write r = R+h where h is height from the actual sphere of radius R and h << R.
Using this we get
V=-\dfrac{GM}{R+h}-\dfrac{1}{2}\omega^2(R+h)^2\sin^2\theta\approx -\dfrac{GM}{R}+\dfrac{GM}{R^2}\,h -\dfrac{1}{2}\omega^2R^2\sin^2\theta
applying suitable approximations. Using the fact that g = GMR2
the above becomes
V=-\dfrac{GM}{R}+gh -\dfrac{1}{2}\omega^2R^2\sin^2\theta
So the equipotentials are of the form
h=\dfrac{\omega^2R^2}{2g}\sin^2\theta+C
The correct value of C is determined by the fact that average value of h over the (spherical) earth's surface must be zero. That is
\int_0^{\pi}h(\theta)2\pi R^2\sin\theta\ \mathrm d\theta =0
This gives
C=-\dfrac{\omega^2R^2}{3g}
According, we get the height variation as
h=\dfrac{\omega^2R^2}{2g}\sin^2\theta-\dfrac{\omega^2R^2}{3g}=R\left(\dfrac{\omega^2R}{6g}\right)(3\sin^2\theta-2)
