1. Acceleration due to gravity at height $h$

At a height $h$ above the earth's surface,

$g_h = g\left(\frac{R}{R+h}\right)^2$

where $R$ is the radius of the earth.

For $h \ll R$, use binomial approximation:

$\left(1+\frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R}$

So,

$g_h \approx g\left(1 - \frac{2h}{R}\right)$

Hence, the decrease in $g$ at height $h$ is

$\Delta g_h = g - g_h = g - g\left(1 - \frac{2h}{R}\right) = \frac{2gh}{R}$

2. Acceleration due to gravity at depth $d$

At a depth $d$ below the earth's surface,

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

for $d \ll R$.

Hence, the decrease in $g$ at depth $d$ is

$\Delta g_d = g - g_d = g - g\left(1 - \frac{d}{R}\right) = \frac{gd}{R}$

3. Given condition: changes are equal

According to the question,

$\Delta g_h = \Delta g_d$

So,

$\frac{2gh}{R} = \frac{gd}{R}$

Cancel $g$ and $R$:

$2h = d$

Thus,

$d = 2h$

4. Checking options

• A: $d = \frac{3h}{2}$ ❌
• B: $d = \frac{h}{2}$ ❌
• C: $d = h$ ❌
• D: $d = 2h$ ✅

So the correct option is D.