Question

The height above the surface of the earth at which the acceleration due to gravity becomes $\frac{g}{9}$ in terms of radius of earth $R$ is ($g$ is acceleration due to gravity at the surface of the earth)}

• A. $\frac{R}{4}$
• B. $\frac{R}{3}$
• C. $\frac{R}{2}$
• D. $2R$
• E. $3R$

Hint:

If \( g \) becomes \( g/n^2 \), the distance from the center becomes \( nR \). The height above the surface is then simply \( (n-1)R \). Here \( n=3 \), so height is \( 2R \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The acceleration due to gravity decreases as we move away from the surface of the Earth. It follows the inverse square law with respect to the distance from the center of the Earth.
Key Formula or Approach:
The formula for gravity at a height \( h \) is:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]

Step 2: Detailed Explanation:
Given that \( g' = \frac{g}{9} \).
Substitute this into the formula:
\[ \frac{g}{9} = g \left( \frac{R}{R + h} \right)^2 \]
Cancel \( g \) from both sides:
\[ \frac{1}{9} = \left( \frac{R}{R + h} \right)^2 \]
Take the square root of both sides:
\[ \frac{1}{3} = \frac{R}{R + h} \]
Cross-multiply:
\[ R + h = 3R \]
\[ h = 3R - R = 2R \]

Step 3: Final Answer:
The required height is $2R$.