Assume that as narrow tunnel is dug between two diametrically opposite points of the earth. Treat the earth as a solid sphere of uniform density. Show that if a particle is released in this tunnel, it will execute a simple harmonic motion. Calculate the time period of this motion.

Options

A. `2pisqrt((R^(3))/(Cm))`
B. `2pisqrt((R^(3))/(Gm))`
C. `2pisqrt((m)/(g))`
D. `2pisqrt((R)/(mg))`

Answer

B

Explanation

At centre of each force on particle is Zero. So it is the mean position.
At x distance from centre , force is to wards centre
`therefore F=-mg'=-m(GM)/(R^(3))x Rightarrow a=-(GM)(R^(3))x`
`therfore omega^(2)=(GM)(R^(3))x Rightarrow T=(2pi)/(omega)=2pisqrt((R^(3))/(GM))`