Question

If a simple pendulum is taken to a height equal to the radius of the earth, how many times will the period of oscillation increase?

Answer 1

The period of oscillation of a simple pendulum is given by

$$T_0=2\pi\sqrt{\frac{l_0}{g_0}}$$

The acceleration due gravity

$$g_0=G\frac{M}{R^2}$$

At the height $H=R$ above Earth's surface

$$g=G\frac{M}{(2R)^2}=\frac{1}{4}g_0$$

Hence

$$T=2\pi\sqrt{\frac{l_0}{g}}=2T_0$$

Answer: two times.