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: Variation of gravity with height. 
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by: \[ g_h = g \left( \frac{R}{R + h} \right)^2, \] where \( g_h = g/4 \), \( R \) is the radius of Earth, and \( g \) is the acceleration due to gravity on the surface. 

Step 2: Substituting \( g_h = g/4 \). 
\[ \frac{g}{4} = g \left( \frac{R}{R + h} \right)^2. \] Cancel \( g \) from both sides: \[ \frac{1}{4} = \left( \frac{R}{R + h} \right)^2. \] Taking the square root: \[ \frac{1}{2} = \frac{R}{R + h}. \] Rearranging: \[ R + h = 2R \quad \Rightarrow \quad h = R. \] \[ \therefore \text{The height is: } R. \]