Imagine a frictionless tunnel along a chord of earth having length equal to radius of earth R. Particle is thrown from surface of earth inside it along its length at a velocity `sqrt((3gR)/(4))` where is acceleration due to gravity on surface of earth. Find the time (in sec) taken by particle to cross the tunnel. Take `g = 10 m//s^(2), pi = 3.14` and `R = 6400 km`.

Answer

`837.33 sec`

Explanation

The particle will perform SHM `(w=sqrt((GM)/(R^(3)))=sqrt((g)/(R )))`
`=w sqrt(A^(2)-(R^(2))/(4))=sqrt((3gR)/(4))`
`= sqrt((g)/(R )) sqrt(A^(2)-(R^(2))/(4))=sqrt((3gR)/(4))`
`= A^(2) - (R^(2))/(4) = (3)/(4) R^(2)`
`A^(2)=R^(2)`
`A = R`
`t = (pi)/(3w) = (pi)/(3) sqrt((R )/(g ))`
`= (3.14)/(3)xx800 rArr 837.33 sec`.