Consider the Earth as a homogenous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is `T= 2pi sqrt((R)/(g))`

Explanation

Earth is assumed to be a homogeneous sphere .
Its centre is at O and Radius = R
The hole is bored straight through the centre along its diameter.
The acceleration due to gravity at the surface of the earth = g
Mass of the body dropped inside the hole = m
After time t, the depth it reached (inside the earth )=d
The value of 'g' decreases with deportation
So acceleration due to gravity at deportation = g'
i.e., `g'=g(1-d//R)=g((R-d)/R)" "...(1)`
Let y be distance from the centre of the earth
Then y = Radius - distance = R - d
Substitute y in (1)
`g'=gy//R`
Now, force on the body of mass m due to this new acceleration g' will be
`F=mg'=mgv//R`
and this force is directed towards the mean position O.
The body dropped in the hole will execute S.H.M spring factor k = mg/Radius
`T=2pisqrt(("Inertial factor")/("spring factor"))=2pisqrt((m)/(mg//R))=2pisqrt((R)/(g))`