Question

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

• A. \(180\) km
• B. \(120\) km
• C. \(90\) km
• D. \(45\) km

Hint:

Use g_h = g(1 - 2h/R) for small h and g_d = g(1 - d/R) for depth d. Equate them.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The acceleration due to gravity decreases both above the surface and below the surface, but at different rates. For small heights compared with the Earth's radius \(R\), the first order formulas hold.

Step 2: Key Formula or Approach:
1. Height \(h\): \(g_h = g\left(1 - \dfrac{2h}{R}\right)\).
2. Depth \(d\): \(g_d = g\left(1 - \dfrac dR\right)\).

Step 3: Detailed Explanation:
Set the two values equal, with \(d = 90\) km:
\[ \frac{2h}{R} = \frac dR \Rightarrow 2h = d \Rightarrow h = \frac{d}{2} = \frac{90}{2} = 45 \text{ km} \]
The height does not depend on the value of \(R\) here, because \(R\) cancels. Since \(g\) falls twice as fast with height as with depth, we need only half the distance above the surface. Options (A) 180 km would need a depth of 360 km, and option (C) 90 km would give a larger drop than the depth does.

Final Answer:
The height is 45 km, option (D). \[ \boxed{45 \text{ km (D)}} \]