The time period of a simple pendulum in a satellite orbiting very close to earth surface is equal to
Answer & explanation
Correct answer: option 3
Since the satellite is orbiting in a circular path, if the point of suspension P towards the centre of earth with an acceleration gp. The bob θ of the pendulum also is accelerating towards the centre of earth with an acceleration gQ
⇒ The relative acceleration $g_{rel}=g_{P}-g_{Q}=0$ because $g_P \approx g_Q=g=9.8 m / s^2$
∴ $g_{\text {effective }}=g_{\text {rel }}=0$
$\Rightarrow T=2 \pi \sqrt{\frac{\ell}{g_{\text {rel }}}}=\infty$