Question

A simple pendulum has a time period of 2 s on Earth's surface. What is its time period at a height equal to the Earth's radius (R)? (Acceleration due to gravity at height h is \( g_h = \frac{g}{(1 + h/R)^2} \)).

• A. 2 s
• B. 2\(\sqrt{2}\) s
• C. 4 s
• D. \(\sqrt{2}\) s

Hint:

Account for variation in gravity with height using \( g_h = \frac{g}{(1 + h/R)^2} \).