Question

A pendulum on the earth has a time period of 1 sec. It's time period in a box orbiting around the earth at a distance R from the surface of the earth will be? (R=radius of earth)

Answer 1

Final Answer :2s

Steps:
1) Time period of simple pendulum at surface of earth = T(1) = 1s

2) Then when we place that pendulum in box at distance 2R from centre of earth .
Here, Radial force to orbit is provided by Gravitational force of Earth.
m =mass of Bob in pendulum
M. =Mass of Earth.

F(radial ) = F(Gravitational )
=> ma(r) = GMm/(2R) ^2
=> a(r) = GM/4R^2
=> a(r) = g/4 , since g = GM/R^2 )

3) We also know that, Time period of simple pendulum is proportional to 1/√g(eff) )
(As length is constant)
Therefore,
$\frac{t(1)}{t(2)} = \frac{ \sqrt{ \frac{g}{4} } }{ \sqrt{g} } \\ = > t(2) = 2t(1) = 2 \times 1 = 2s$
Therefore, Time Period of pendulum when it is orbiting the Earth at distance R from surface of planet .