Question

The height at which the acceleration due to gravity becomes \( \frac{g}{16} \) (where \( g \) is the acceleration due to gravity on the surface of the earth) in terms of \( R \), if \( R \) is the radius of the Earth.

• A. \( 2R \)
• B. \( 3R \)
• C. \( \sqrt{2}R \)
• D. \( \sqrt{3}R \)

Hint:

Use the relation for gravity at height \( h \) to solve for the height when gravity is reduced to a specific fraction of its value on the surface.

Answer 1

The Correct Option is B

The acceleration due to gravity at a height \( h \) above the Earth's surface is given by: \[ g_h = \frac{g}{\left( 1 + \frac{h}{R} \right)^2} \] We are given that \( g_h = \frac{g}{16} \). Thus: \[ \frac{g}{16} = \frac{g}{\left( 1 + \frac{h}{R} \right)^2} \] Simplifying: \[ 16 = \left( 1 + \frac{h}{R} \right)^2 \] Taking square roots: \[ 4 = 1 + \frac{h}{R} \] Thus: \[ \frac{h}{R} = 3 \quad \Rightarrow \quad h = 3R \] Therefore, the correct answer is (B).