Question

What is that imaginary angular velocity of the earth at which a body placed at the equator has zero weight? The radius of the earth is 6400 km and $g=10m/s^2$.

Answer 1

The apparent weight of a person on the equator (Latitude $\alpha =0$) is given by

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

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

As $g^{\prime }=0$

$g-R_e{\omega }^2=0$

$\omega =\sqrt{\frac{g}{R_e}}$

$\omega =\sqrt{\frac{10}{6400\times {10}^3}}$

$\omega =1.25\times {10}^{-3}rad/s^2$