Question

Find the angular speed of rotation of the earth for which a body on the equator will have zero apparent weight. What will be the length of a day at that time?

Answer 1

Apparent weight for a body on the equator $(\alpha =0)$ is given by,

$W^{\prime }=W-m{R}_e{\omega }^2$

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

$g^{\prime }=0$

$\therefore g-R{\omega }^2=0$

$\Rightarrow \omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6400\times {10}^3}}=1.25\times {10}^{-3}$ $rad/s$

$T=\frac{2\pi }{\omega }\approx 1.39$ $hr$