Question

Time period of pendulum, on a satellite orbiting the earth, is

• A. 1/$\pi$
• B. zero
• C. $\pi$
• D. infinity

Answer 1

Answer: (D)

infinity

Answer 2

On an artificial satellite orbiting the earth the acceleration is given by $\frac{GM}{R^{2}}$ towards the centre of the earth. Now for a body of mass m on the satellite the graviational force due to earth is $\frac{GMm}{R^{2}}$ towards the centre of the earth. Let the reaction force on the surface of thesatellite be N, then $\frac{GMm}{R^{2}}-N=m \left(\frac{GM}{R^{2}}\right)$ $\Rightarrow\quad N=0$ That is on the satellite there is a state ofweightlessness or g = 0 $?\quad$ The time period of the simple pendulum, $T=2\pi\sqrt{\frac{l}{g}}=\infty$