Question

The value of acceleration due to gravity at Earth's surface is \( 9.8 \, \text{m/s}^2 \). The altitude above its surface at which the acceleration due to gravity decreases to \( 4.9 \, \text{m/s}^2 \) is close to: (Radius of Earth \( R = 6.4 \times 10^6 \, \text{m} \))

• A. \( 2.6 \times 10^6 \, \text{m} \)
• B. \( 6.4 \times 10^6 \, \text{m} \)
• C. \( 9.0 \times 10^6 \, \text{m} \)
• D. \( 1.6 \times 10^6 \, \text{m} \)

Hint:

As altitude increases, the acceleration due to gravity decreases. Use the relation \( g_h = \frac{g}{\left( 1 + \frac{h}{R} \right)^2} \) to determine the altitude when \( g_h \) is known.