Question

Let ‘${{g}_{h}}$’ and ‘${{g}_{d}}$’ be the acceleration due to gravity at height ‘h’ above the earth’s surface and the depth ‘d’ below the earth’s surface, respectively. If ${{g}_{h}}={{g}_{d}}$ then the relation between ‘h’ and ‘d’ is
A. $d=h$
B. $d=\dfrac{h}{2}$
C. $d=\dfrac{h}{4}$
D. $d=2h$

Answer 1

Use the formula for the acceleration due to gravity of earth at a height $h$ from its surface. Also use the formula for the acceleration due to gravity of earth at a depth of d below the surface of earth. Then equate the two equations to get the relation between h and d.

Formula used:
$g=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{2h}{R} \right)$
where g is acceleration due to gravity at a height h above the surface of earth (near the surface of earth), M is the mass of the earth, G is gravitational constant and R is the radius of earth.
$g=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{d}{R} \right)$
where g is acceleration due to gravity at a depth d below the surface of earth.

Complete step by step answer:
The acceleration due to gravity at a point near the surface of the earth and at a height h from the surface is given to be equal to $g=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{2h}{R} \right)$.
In this case, it is given that acceleration due to gravity at height h above earth’s surface is ${{g}_{h}}$.
Therefore,
$\Rightarrow {{g}_{h}}=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{2h}{R} \right)$ …… (i).
The acceleration due to gravity at a point at a depth d below the surface is given to be equal to $g=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{d}{R} \right)$.
In this case, it is given that acceleration due to gravity at depth d below the surface of earth is ${{g}_{d}}$.
Therefore,
$\Rightarrow {{g}_{d}}=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{d}{R} \right)$ …… (ii).
It is given that both the accelerations due to gravity are equal, i.e. ${{g}_{h}}={{g}_{d}}$.
Therefore, equate equations (i) and (ii).
$\Rightarrow \dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{2h}{R} \right)=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{d}{R} \right)$
$\Rightarrow \left( 1-\dfrac{2h}{R} \right)=\left( 1-\dfrac{d}{R} \right)$
$\Rightarrow \dfrac{2h}{R}=\dfrac{d}{R}$
$\therefore d=2h$

Hence, the correct option is D.

Note: The formula for the acceleration due to gravity of earth at a height h above the surface of earth, which is $g=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{2h}{R} \right)$, is only valid when the height h is much smaller than the radius of earth (i.e. h <<< R). If the height h is comparable to the radius of earth, then we cannot apply this formula for the acceleration due to gravity.