Question

At what rate should the earth rotate so that the apparent $g$ at the equator becomes zero? What will be the length of the day in this situation?

Answer 1

For apparent g at equator to be 0,

$m{\omega }^{2}R-mg=0$

$\omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400000}}=1.24X{10}^{-3}rad/s$

Also, length of day is equal to the time of one rotation.

Hence, $T=\frac{2\pi }{\omega }=\frac{2\pi }{1.24X{10}^{-3}}=5067.74s=1.41hr$

Answer is $1.41hr$.