Question

Let $\omega$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth a below earth's surface at a pole $(d<<R)$ The value of $d$ is

• A. $\frac{{\omega }^2{R}^2}{g}$
• B. $\frac{{\omega }^2{R}^2}{2g}$
• C. $\frac{2{\omega }^2{R}^2}{g}$
• D. $\frac{\sqrt{Rg}}{g}$

Answer 1

The correct option is A $\frac{{\omega }^2{R}^2}{g}$

Let $g_e$ be the acceleration due to gravity on surface of earth, $g^{\prime }$ be the acceleration due to gravity at depth $d$ below the surface of earth and is given by
$g^{{}^{\prime }}=g_e(1-\frac{d}{R})$ ............................(1)
The acceleration due to gravity of an object on latitude(below the surface of earth) is
$g^{{}^{\prime }}=g_e-{\omega }^{2}R\cos \varphi$
where, $\varphi$ is the solid angle at the center of earth sustained by the object.
At the equator, $\varphi =0^{\circ }$
$\therefore \cos \varphi =1$
Hence,
$g^{\prime }=g_e-{\omega }^{2}R$ ...............(2)
From (1) and (2), we can write
$g_e(1-\frac{d}{R})=g_e-{\omega }^{2}R$
$\Rightarrow \frac{d{g}_e}{R}={\omega }^{2}R$
$\Rightarrow d=\frac{{\omega }^{2}R}{g_e}$