Question

The value of the acceleration due to gravity is g, at a height $h = \frac{R}{2}$ (R = radius of the earth) from the surface of the earth. It is again equal to g, at a depth d below the surface of the earth. The ratio $(\frac{d}{R})$ equals

• A. $\frac{5}{9}$
• B. $\frac{1}{3}$
• C. $\frac{7}{9}$
• D. $\frac{4}{9}$

Answer 1

Answer: (A)

$\frac{5}{9}$

Answer 2

Let $g_0$ be the acceleration due to gravity at the surface. At height $h = \frac{R}{2}$, the acceleration due to gravity is $g_h = g_0 \left(\frac{R}{R+h}\right)^2 = g_0 \left(\frac{R}{R+\frac{R}{2}}\right)^2 = g_0 \left(\frac{R}{\frac{3R}{2}}\right)^2 = g_0 \left(\frac{2}{3}\right)^2 = \frac{4}{9} g_0$. At depth $d$, the acceleration due to gravity is $g_d = g_0 \left(1 - \frac{d}{R}\right)$. Given $g_h = g_d$, we have $\frac{4}{9} g_0 = g_0 \left(1 - \frac{d}{R}\right)$. Thus, $\frac{4}{9} = 1 - \frac{d}{R}$, which gives $\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}$.