Physics - Gravitation Question with Solution | TestHub

Let${\omega }_0$be the new angular velocity.

At the equator$g_E=g-R{\omega }^2$

The weight$m{g}_E$should be zero.$g_E=0$

$\therefore R{\omega }_0^2=g$or${\omega }_0=\sqrt{\frac{g}{R}}$

$\therefore {\omega }_0=\sqrt{\frac{9.8}{6.4\times 10^6}}=1.24\times 10^{-3}\frac{rad}{s}$

The present angular velocity of the earth is$\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{24\times 60\times 60}$

$\therefore$$\omega =\frac{3.14\times 10^{-2}}{12\times 36}=7.27\times 10^{-5}$

$\therefore \frac{{\omega }_0}{\omega }=\frac{1.24\times 10^{-3}}{7.27\times 10^{-5}}=17$

$\therefore {\omega }_0=17\omega$

It should rotate faster by 17 times the present value.