Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to reach centre of earth if it is projected into the tunnel from surface of earth with speed needed for it to escape the gravitational field to earth.

Answer

`t=sin^(-1)((1)/(sqrt(3)))sqrt((R_(e))/(g))`

Explanation

let x= distance of the particle form the surface
acceleration
`(vdv)/(dx)=(GM)/(R^(3))(R-x)`
`implies_(v_(e))^(v)vdv=int_(0)^(x)g(1-(x)/(R))dx`
`implies v=sqrt(v_(e)^(2)+2g(x-(x^(2))/(2R)))=(dx)/(dt)`
`impliesint_(0)^(t)dt=int_(R)^(0)(dx)/(sqrt(2g[R+x-(x^(2))/(2R)]))`
`impliest=int_(R)^(0)(dx)/(sqrt((g)/(R)[3R^(2)-(x-R)^(2)]))`
let x-R=`sqrt(3)Rsintheta`
`dx=sqrt(3)Rcosthetad theta`
`thereforet=sqrt((R)/(g))intd theta=sqrt((R)/(g))sin^(-1)((x-R)/(sqrt(3)R))_(0)^(R)`
`impliest=sqrt((R)/(g))sin^(-1)((1)/(sqrt(3)))`