Physics - Practical physics Question with Solution | TestHub

The periodic time of a simple pendulum is given by $T=2\pi \sqrt{\frac{L}{g}}$.

On the surface of the earth, the acceleration due to gravity is $g_1=\frac{GM}{R^2}$ and the periodic time is $T_1=2\pi \sqrt{\frac{L}{g_1}}$.

At a height $h=R$ above the earth's surface, the distance from the center of the earth is $r=R+h=2R$.

The acceleration due to gravity at this height is $g_2=\frac{GM}{(2R{)}^2}=\frac{g_1}{4}$.

The periodic time at this height is $T_2=2\pi \sqrt{\frac{L}{g_2}}$.

Taking the ratio of the periodic times:

$\frac{T_2}{T_1}=\sqrt{\frac{g_1}{g_2}}=\sqrt{\frac{g_1}{g_1/4}}=\sqrt{4}=2$.

Answer: 2