Physics - Gravitation Question with Solution | TestHub

The effective acceleration due to gravity at the equator due to the rotation of the Earth is given by:

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

The weight of a body at the equator is:

$W^{\prime }=m{g}^{\prime }=m(g-{\omega }^{2}R)$

According to the given condition, the new weight becomes $\frac{3}{5}$ of the initial weight ($mg$).

$m(g-{\omega }^{2}R)=\frac{3}{5}mg$

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

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

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

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

$\omega ={\left(\frac{2g}{5R}\right)}^{\frac{1}{2}}$

Answer: ${\left(\frac{2g}{5R}\right)}^{\frac{1}{2}}$