Question

At a height $h$ above the surface of earth, the change in value of $g$ is same as that at a depth $x$ below the surface on the earth. Both $x$ and $h$ are very small in comparison to the radius of the earth. Then,

• A. $x=h$
• B. $x=2h$
• C. $2x=h$
• D. $x=h^2$

Answer 1

The correct option is B $x=2h$

Let, $g_h$ be the acceleration due to gravity at height $h.$
$g_x$ be the acceleration due to gravity at depth $x.$
The acceleration due to gravity on the surface of Earth is
$g=\frac{GM}{R^2}$
where$,R$ is the radius o Earth$,M$ is mass of Earth and $G$ is gravitational constant.
$\therefore g\propto \frac{1}{R^2}$
$\Rightarrow g_h\propto \frac{1}{(R+h{)}^2}$
Therefore
$\frac{g_h}{g}=\frac{R^2}{(R+h{)}^2}$
$\frac{g_h}{g}=\frac{1}{(1+\frac{h}{R}{)}^2}$
$\frac{g_h}{g}={(1+\frac{h}{R})}^{-2}$
$\frac{g_h}{g}=(1-\frac{2h}{R})$
$g_h=g-\frac{2gh}{R}$
$g-g_h=\frac{2gh}{R}$
Also, the acceleration due to gravity at depth $x,$
$g_x=\frac{4}{3}G\rho (R-x)$
$\Rightarrow g_x\propto (R-x)$
Therefore
$\frac{g_x}{g}=\frac{R-x}{R}$
$\frac{g_x}{g}=1-\frac{x}{R}$
$g_x=g-\frac{gx}{R}$
$g-g_x=\frac{gx}{R}$
If the change in the value of $g$ at height $h$ above earth surface is the same as that at depth $x$ i.e.
$g-g_h=g-g_x$
$\frac{2gh}{R}=\frac{gx}{R}$
$\therefore x=2h$