Imagine a shaft going all the way through the Earth from pole to pole along its rotation axis. Assuming the Earth to be a homogenous ball and neglecting the ari drag, find `:`
(a) the equation of motion of a body falling down into the shaft ,
`(b)` how long does it take the body to reach the other end of the shaft,
`(c)` the velocity of the body at the Earth's centre.

Explanation

The only force acting on the ball is the gravitational force `vec(F)`, of magnitude `gamma(4)/(3) pi rho m r`, whey `gamma` is the gravitational constant `rho`, the density of the Earth and `r` is the distance of the body from the centre of the Earth.
But, `g=gamma(4pi)/(3)rho R`, so the expression for `vec (F)` can be written as,
`vec(F)-mg ((vec(r))/(R)),` here `R` is the radius of the Earth and the equation of motion in projection form has teh form, or, `m ddot(x) + (mg )/(R)x=0`
`(b)` The equation , obtained above has the form of an equation of S.H.M. having the time period,
`T=2pi sqrt((R)/(g)),`
Hence the body wil reach the other end of the shalft in the time,
`t=(T)/(2)=pi sqrt((R)/(g))=42 mi n`
(c) From the conditions of S.H.M., the speed of the body at the centre of the Earth will be maximum, having the magnitude,
`v=Romega=Rsqrt(g//R)=sqrt(gR)=7.9km//s.`