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.

Answer 1

The Correct Option is A

The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is: \[ g_h = \frac{g}{\left( 1 + \frac{h}{R} \right)^2}. \] Given: \[ g_h = \frac{g}{2}. \] Substituting into the equation: \[ \frac{g}{2} = \frac{g}{\left( 1 + \frac{h}{R} \right)^2}. \] Simplifying: \[ \left( 1 + \frac{h}{R} \right)^2 = 2. \] Taking the square root: \[ 1 + \frac{h}{R} = \sqrt{2}. \] Solving for \( h \): \[ \frac{h}{R} = \sqrt{2} - 1. \] Substituting \( R = 6.4 \times 10^6 \, \text{m} \): \[ h = (\sqrt{2} - 1) \cdot 6.4 \times 10^6. \] Simplifying further: \[ h = (1.414 - 1) \cdot 6.4 \times 10^6 = 0.414 \cdot 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m}. \] Final Answer: \[ \boxed{2.6 \times 10^6 \, \text{m}} \]