Question

The rotation of the earth about its axis speed up such that a man on the equator becomes weightless. In such a situation, what would be the duration of one day?

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

Answer 1

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

Here balancing forces,

Normal force+centrifugal force due tot earth's rotation=gravitational force.

Here, normal force is zero.

So, $mg=m{\omega }^{2}R\Rightarrow \omega =\sqrt{g/R}$

where $m$ is the mass of the object, and $R$ is the radius of planet.

The time period or duration of one day is then, $T=2\pi /\omega =2\pi \sqrt{R/g}$