Question

At what height does gravitational acceleration become 1/9th of gravity at the surface of a planet, if the radius of the planet is \( R \)?

• A. \( 4R/3 \)
• B. \( 2R \)
• C. \( 2\sqrt{2}R \)
• D. \( 2\sqrt{3}R \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The acceleration due to gravity (\( g \)) varies inversely with the square of the distance from the center of the planet. As we move to a height \( h \) above the surface, the distance becomes \( R + h \).

Step 2: Key Formula or Approach:
\[ g_h = g \left( \frac{R}{R+h} \right)^2 \] Where \( g_h \) is the gravity at height \( h \), and \( g \) is gravity at the surface.

Step 3: Detailed Explanation:
Given \( g_h = \frac{g}{9} \): \[ \frac{g}{9} = g \left( \frac{R}{R+h} \right)^2 \] \[ \frac{1}{9} = \left( \frac{R}{R+h} \right)^2 \] Taking the square root on both sides: \[ \frac{1}{3} = \frac{R}{R+h} \] Cross-multiplying: \[ R + h = 3R \] \[ h = 2R \]

Step 4: Final Answer:
The gravitational acceleration becomes 1/9th at a height of 2R.