Question

If $g$ on the surface of the Earth is $9.8$ $m{s}^{-2}$, then it's value at a depth of $3200$ $km$ (Radius of the earth $=6400$ $km$) is

• A. $9.8$ $m{s}^{-2}$
• B. $zero$
• C. $4.9$ $m{s}^{-2}$
• D. $2.45$ $m{s}^{-2}$

Answer 1

Answer: (C)

$4.9$ $m{s}^{-2}$

The value of gravity changes as we move away from or towards the centre of the Earth.
This is given by: $g_R=\frac{GM}{R^2}$, where $M$ is the mass of a planet of radius $R$
So, $M=\frac{4}{3}\pi R^3\rho$ ; substituting in above equation
$\Rightarrow g_R=G\frac{4}{3}\pi \rho \times R$
Since we want the value of $g$ at depth($d$) from the Earth's surface, we replace $R$ by $(R-d)$
$\Rightarrow g_d=G\frac{4}{3}\pi \rho \times (R-d)$
$\Rightarrow \frac{g_R}{g_d}=\frac{R}{R-d}$
$\Rightarrow \frac{g_d}{g_R}=(1-\frac{d}{R})$

The given depth in the problem is $d=3200$ km, substituting we get,
$\frac{g_d}{g_R}=(1-\frac{3200}{6400})$
$g_d=9.8\times (1-\frac{3200}{6400})$
$g_d=9.8/2$
$g_d=4.9m{s}^{-2}$