Question

The time period of a simple pendulum in a satellite is:

Answer 1

1. In a satellite orbiting the earth the acceleration is given by $\frac{\text{GM}}{{\text{R}}^2}$ towards the centre of the earth. Now for a body of mass m on the satellite, the gravitational force due to earth is $\frac{\text{GMm}}{{\text{R}}^2}$ towards the centre of the earth.
2. Let the reaction force on the surface of the satellite be N, then
   $\frac{\text{GMm}}{{\text{R}}^2}$−$\text{N}$= $\text{m}\left(\frac{\text{GM}}{{\text{R}}^2}\right)$
   $\text{⇒N=0}$
   That is on the satellite there is a state of weightlessness or $\text{g}=0$
   ∴ The time period of the simple pendulum,
   $\text{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$
3. Now, by putting the value of $\text{g}=0$ and solving it , we get

$\text{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{0}=}\infty$

Hence, The time period of a simple pendulum in a satellite is : infinity.