A body is situated on the surface of the earth becomes weightless at equator when the rotational kinetic…
A body is situated on the surface of the earth becomes weightless at equator when the rotational kinetic energy of the earth reaches a critical value ' $K$ '. The value of $K$ is given by [ $\mathrm{g}$ = gravitational acceleration on earth's surface, $M=$ mass of the earth and $\mathrm{R}=$ radius of the earth]
$\frac{1}{2} \mathrm{MgR}$
$\frac{1}{3} \mathrm{MgR}$
$\frac{1}{4} \mathrm{MgR}$
$\frac{1}{5} \mathrm{MgR}$
Solution
A body will become weightless at equator if $R \omega^2=g$ or $\omega^2=\frac{g}{R}$ Kinetic energy of the earth, $\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^2$
For a solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$
(M is mass of the earth)
$\therefore \mathrm{K}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \times \frac{\mathrm{g}}{\mathrm{R}}=\frac{1}{5} \mathrm{MgR}$