Question

"The height from Earth's surface at which acceleration due to gravity becomes \( g/4 \), where \( g \) is acceleration due to gravity on the surface of Earth and \( R \) is the radius of Earth?"

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

Hint:

The acceleration due to gravity decreases with height and follows the relation \( g_h = g (R / (R + h))^2 \).

Answer 1

The Correct Option is B

Step 1: Gravitational variation with altitude. The acceleration due to gravity at an altitude \( h \) from Earth's surface is defined by: \[ g_h = g \left( \frac{R}{R + h} \right)^2, \] where \( g_h = g/4 \), \( R \) represents Earth's radius, and \( g \) is the surface acceleration due to gravity.

Step 2: Substitution of \( g_h = g/4 \). 
\[ \frac{g}{4} = g \left( \frac{R}{R + h} \right)^2. \] Divide both sides by \( g \): \[ \frac{1}{4} = \left( \frac{R}{R + h} \right)^2. \] Apply the square root: \[ \frac{1}{2} = \frac{R}{R + h}. \] Rearrange the equation: \[ R + h = 2R \quad \Rightarrow \quad h = R. \] \[ \therefore \text{The altitude is: } R. \]