Let ${g_h}$ and ${g_b}$ be the acceleration due to gravity at height h above the earth’s surface and at 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

- Hint: Acceleration due to gravity at height h above earth surface = $g(1 - \dfrac{{2h}}{R})$ and acceleration due to gravity at depth d below earth surface = $g(1 - \dfrac{d}{R})$ .Use these two formulas to get the relation between d and h.

Complete step-by-step solution -

Formula for acceleration due to gravity at height h above earth surface is,
 ${g_h}$ = $g(1 - \dfrac{{2h}}{R})$
Here $g$ is acceleration due to gravity at the surface of earth and R is radius of earth,and h is height above earth surface at which acceleration due to gravity is being calculated.
Formula for acceleration due to gravity at depth d below earth's surface is,
${g_d}$ = $g(1 - \dfrac{d}{R})$
Here $g$ and R are the same as above, d is depth below earth's surface at which acceleration due to gravity is being calculated.
Since it is given that ${g_h}$ = ${g_d}$ so,
$g(1 - \dfrac{{2h}}{R})$ = $g(1 - \dfrac{d}{R})$
Canceling out $g$ from both sides,
$1 - \dfrac{{2h}}{R} = 1 - \dfrac{d}{R}$
Canceling out 1 and negative sign from both sides,
$\dfrac{{2h}}{R} = \dfrac{d}{R}$
Cancelling out R from both the sides,
2h=d
Hence option D is the correct option

Additional Information:
As the earth is in the shape of an ellipsoid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Note: Acceleration due to gravity decreases two times faster when we go above the earth surface than when we go below the earth surface and it becomes zero both at infinite distance from earth as well as at the centre of the earth.