Question

If the rotation of the earth about its axis speeds up such that a man on the equator becomes weightless, in such a case, what would be the duration of day?

• A. $2\pi \sqrt{R/g}$
• B. $\frac{1}{2\pi }\sqrt{R/g}$
• C. $\pi \sqrt{R/g}$
• D. $\frac{1}{4\pi }\sqrt{R/g}$

Answer 1

The correct option is A $2\pi \sqrt{R/g}$

$g_{equator}=g-{\omega }^{2}R$
For weightlessness, $g_{equator}=0$
$g-{\omega }^{2}R=0\Rightarrow \omega =\sqrt{g/R}$
$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{R/g}$