The acceleration due to gravity at the poles and the equator is ${\mathrm{g}}_{\mathrm{p}} \mathrm{and} {\mathrm{g}}_{\mathrm{e}}$ respectively. If the earth is a sphere of radius ${\mathrm{R}}_{\mathrm{E}}$ and rotating about its axis with angular speed
$\mathrm{\omega }$, then ${\mathrm{g}}_{\mathrm{p}}-{\mathrm{g}}_{\mathrm{e}}$ is given by

Acceleration due to gravity at a place of latitude $\mathrm{\lambda }$ due to the rotation of earth is

${\mathrm{g}}^{\text{'}} = \mathrm{g}-{\mathrm{R}}_{\mathrm{E}}{\mathrm{\omega }}^2{\cos }^2\mathrm{\lambda }$

At equator, $\mathrm{\lambda } = 0^0, \cos 0^0 = 1$

$\therefore {\mathrm{g}}^{\text{'}} ={\mathrm{g}}_{\mathrm{e}} =\mathrm{g}-{\mathrm{R}}_{\mathrm{g}}{\mathrm{\omega }}^2$

At poles, $\mathrm{\lambda } = 90°, \cos 90° = 0$

$\therefore {\mathrm{g}}^{\text{'}} = {\mathrm{g}}_{\mathrm{p}} =\mathrm{g}$

$\therefore {\mathrm{g}}_{\mathrm{p}}-{\mathrm{g}}_{\mathrm{e}} = \mathrm{g}-(\mathrm{g}-{\mathrm{R}}_{\mathrm{E}}{\mathrm{\omega }}^2) = {\mathrm{R}}_{\mathrm{E}}{\mathrm{\omega }}^2$