At what angular rate should the earth rotate so that the apparent acceleration due to gravity at the

Question

At what angular rate should the earth rotate so that the apparent acceleration due to gravity at the equator becomes zero? Also, find the length of the day in this situation. ($R = 6400~km$, $g = 9.8~m~s^{-2}$)

Professor PC Thomas Classes · Answer

Apparent gravity $g' = g - R\omega'^2$. For $g' = 0$, we need $\omega' = \sqrt{\frac{g}{R}} = \sqrt{\frac{9.8}{6.4 	imes 10^6}} \approx 1.24 	imes 10^{-3}~rad~s^{-1}$. The new length of the day is $T = \frac{2\pi}{\omega'} = 2\pi\sqrt{\frac{R}{g}} \approx 5075~s \approx 1.41~h$.