Question

The depth at which the value of acceleration due to gravity becomes $\frac{1}{8}$ times the value at the surface is

• A. $\frac{R}{8}$
• B. $\frac{R}{64}$
• C. $\frac{R}{7}$
• D. $\frac{7R}{8}$

Answer 1

Answer: (D)

$\frac{7R}{8}$

Answer 2

The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:

$g' = g \left(1 - \frac{d}{R}\right)$

where $g$ is the acceleration due to gravity at the surface of the Earth and $R$ is the radius of the Earth.

Given $g' = \frac{1}{8} g$, we set $\frac{1}{8} g = g \left(1 - \frac{d}{R}\right)$. This simplifies to $\frac{1}{8} = 1 - \frac{d}{R}$.

Solving for $\frac{d}{R}$, we get $\frac{d}{R} = 1 - \frac{1}{8} = \frac{7}{8}$.

Therefore, $d = \frac{7R}{8}$.