Question

What should be the angular speed with which the earth must rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$th of weight measured without considering earth's rotation ? (Take $R$ as the equatorial radius and $g$ as acceleration due to gravity)

• A. $\sqrt{\frac{2g}{5R}}$
• B. $\sqrt{\frac{2g}{R}}$
• C. $2\sqrt{\frac{g}{R}}$
• D. $\sqrt{\frac{g}{R}}$

Answer 1

The correct option is A $\sqrt{\frac{2g}{5R}}$

The apparent weight of an object at any place on the Earth due to the effect of rotation of the Earth, is given by$$m{g}^{\prime }=m(g-{\omega }^{2}R{\cos }^2\lambda )$$ Where, $\lambda$ is the latitude,
$R$ is the radius and
$\omega$ is the rotational velocity.


At the equator, $\lambda =0^{\circ }$. So, apparent weight of the body at equator

$m{g}_{equator}=(mg-m{\omega }^{2}R{\cos }^2{0}^{\circ })$

$\therefore m{g}_{equator}=(mg-m{\omega }^{2}R)$

Given,
$m{g}_{equator}=\frac{3}{5}mg$

$\Rightarrow \frac{3}{5}mg=(mg-m{\omega }^{2}R)$

$\Rightarrow m{\omega }^{2}R=\frac{2}{5}mg$

$\Rightarrow \omega =\sqrt{\frac{2g}{5R}}$

Hence option (a) is the correct answer.

Why this question? To familiarize students with the change in acceleration due to gravity due to the rotation of the earth.