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.

Explanation

Consider the situation shown in ure. Suppose at an instant t the particle in the tunnel is at a distance x from the centre of the earth. Let us draw a sphere of erradius x with its centre at the centre of the earth. Only the part of the earth within this sphere will exert a net attraction on the particle. Mass of this part is
`M'=(4/3pix^3)/(4/3piR^3)M=x^3/R^3M`
The force of attraction is therefore
`F=(G(x^3/r^3)Mm)/^2=(GMm)/R^3x`
This force acts towards the centre of the earth. Thus, the resultant force on the particle is opposite to the displacement from the centre of the earth and is proportioN/Al to it. The particle therefore executes a simple harmonic motion in the tunnel with the centre of the earth as the mean position.
The force constant is `k=(GMm)/R^3` so that the time period is
`T=2pisqrt(m/k)=2pisqrt(R^3/(GM)`