Physics - Gravitation Question with Solution | TestHub

The apparent acceleration due to gravity at the equator when the Earth rotates with angular speed $\omega$ is given by:

$g^{\prime }=g-R{\omega }^2$

The present weight of a person at the equator is approximately $mg$ (neglecting the small effect of the Earth's current rotation). The new weight is given as $\frac{3}{5}$ of the present weight.

$m{g}^{\prime }=\frac{3}{5}mg$

$g^{\prime }=\frac{3}{5}g$

Substituting the expression for $g^{\prime }$:

$g-R{\omega }^2=\frac{3}{5}g$

$R{\omega }^2=g-\frac{3}{5}g$

$R{\omega }^2=\frac{2}{5}g$

${\omega }^2=\frac{2g}{5R}$

$\omega =\sqrt{\frac{2g}{5R}}$

Answer: $\sqrt{\frac{2g}{5R}}$