New earth, How does it look like?

The condition is when all tiny masses on Earth surface are equal in potential, So first I will calculate the gravitational potential then I will calculate the angular rotating potential.

first

$U_m=∫_R ^\infty \frac{GM_E m}{R^2} dr =-\frac{GM_E m}{R_E}$

second the Extra potential caused by rotation:

to calculate this we should use "Angular momentum conservation" so the masses when they come from infinity (the angular velocity is zero) to distance R from the center of earth they should gain angular velocity :

$L=I\omega =constant$

$L=mr^2 \omega \rightarrow \omega_m = \omega_f \frac{(Rsin\theta)^2}{r^2}$

|| spherical coordinates

$dU_{extra} = (\omega_m)^2 xdx ={(\omega_f)^2} ( \frac{Rsin\theta}{x})^4 xdx$

$U_{extra}=m∫_{x_0} ^\infty (\omega_f)^2 ( \frac{Rsin\theta}{x})^4 xdx$

$U_{extra}=-m\frac{1}{2} (\omega_f)^2 (Rsin\theta)^2$

$U=U_m +U_{extra} =const$ for all the earth surface so

the potential when $\theta =\pi/2$ is $-\frac{GM_E m}{R_E} -m\frac{1}{2} (\omega_f)^2 (Rsin\theta)^2$

this is equal to that one in the pole when $\theta =0$ so

$-\frac{GM_E m}{R} -m\frac{1}{2} (\omega_f)^2 (R)^2= -\frac{GM_E m}{R_p}$

we calculate $R_p =7521.515km$