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

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