The acceleration due to gravity at the poles and the equator is $g_p$ and $g_e$ respectively. If the earth is a sphere of radius $R_E$ and rotating about its axis with angular speed $\omega$, then $g_p-g_e$ is then given by

The correct option is D ${\omega }^2{R}_E$
Acceleration due to gravity at a place of latitude $\lambda$ due to rotation of earth is $g^{\prime }=g-R_E{\omega }^2{\cos }^2\lambda$
At equator, $\lambda =0,\cos 0=1$
$\therefore g^{\prime }=g_e=g-R_E{\omega }^2$
At poles, $\lambda =90,cos90=0$
$\therefore g^{\prime }=g_p=g$
$\therefore g_p-g_e=g-(g-R_E{\omega }^2)=R_E{\omega }^2$