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 the given value into the equation yields:\[\frac{g}{2} = \frac{g}{\left( 1 + \frac{h}{R} \right)^2}.\] Simplifying the equation:\[\left( 1 + \frac{h}{R} \right)^2 = 2.\] Taking the square root of both sides:\[1 + \frac{h}{R} = \sqrt{2}.\] Rearranging to solve for \( h \):\[\frac{h}{R} = \sqrt{2} - 1.\] With \( R = 6.4 \times 10^6 \, \text{m} \):\[h = (\sqrt{2} - 1) \cdot 6.4 \times 10^6.\] Further simplification:\[h = (1.414 - 1) \cdot 6.4 \times 10^6 = 0.414 \cdot 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m}.\] Final Answer:\[\boxed{2.6 \times 10^6 \, \text{m}}\]