Question

Use the assumptions of the previous question. An object weighed by a spring balance at the equator gives the same reading as a reading taken at a depth d below the earth's surface at a pole $(d<<R)$. The value of $d$ is

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

Answer 1

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

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 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,
$g_e(1-\frac{d}{R})=g_e-{\omega }^{2}R$
$\Rightarrow \frac{d{g}_e}{R}={\omega }^{2}R$
$\Rightarrow d=\frac{{\omega }^{2}R}{g_e}$