The acceleration due to gravity at a height 'h' above the surface of earth is '\(g_h\)'. At the depth 90 km below the earth's surface the acceleration due to gravity is also '\(g_h\)'. The value of 'h' is
Show Hint
Use g_h = g(1 - 2h/R) for small h and g_d = g(1 - d/R) for depth d. Equate them.
Step 1: Fractional drop in g:
Going up by $h$ reduces $g$ by the fraction $\frac{2h}{R}$ (using $g_h \approx g(1-\frac{2h}{R})$).
Going down by $d$ reduces $g$ by the fraction $\frac dR$ (since $g_d = g(1-\frac dR)$ inside a uniform sphere).
Step 2: Same g means same drop:
$\frac{2h}{R} = \frac{90}{R}$, so $h = 45$ km.
Step 3: Sense check:
Gravity falls off faster outside the Earth (twice the rate per km), so a smaller distance is enough. Half of 90 km is 45 km.
Final Answer:
Option (D).
\[ \boxed{45 \text{ km (D)}} \]