Question

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

Answer 1

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

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

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

$g\prime =0$

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

$\Rightarrow w=\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{640\times {10}^3}}=1.25\times {10}^{-3}\frac{rad}{s}$

$T=\frac{2\pi }{w}=\frac{2\pi }{1.25\times {10}^{-3}}=5026.548s\approx 1.39hr$