Question

Determine the speed with which the earth would have to rotate on its axis, so that a person on the equator would weigh $\frac{3}{5}$th as much as at present. Take $R=6400km$

Answer 1

$g^{\prime }=\frac{3}{5}g=g-R{\omega }^2{\cos }^2{0}^{\circ }$
$\therefore R{\omega }^2=\frac{2}{5}g$
or $\omega =\sqrt{\frac{2g}{5R}}=\sqrt{\frac{2\times 9.8}{5\times 6400\times {10}^3}}$
$=7.8\times {10}^{-4}rad/s$