A particle takes a time `t_(1)` to move down a straight tunnel from the surface of earth to its centre. If gravity were to remain constant this time would be `t_(2)` calculate the ratio `(t_(1))/(t_(2))`

Answer

`(pi)/(2sqrt(2))`

Explanation

`F=(GMmx)/(R^(3))=momega^(2)x`
particle will perform a oscillation
with angular speed `omega^(2)=(GM)/(R^(3))`
`T_(i)=2pisqrt((R^(3))/(GM))impliest_(1)=(pi)/(2)sqrt((R^(3))/(GM))`
if acceleration is constant
`g=(GM)/(R^(2)):S=(1)/(2)at^(2)impliesR=(1)/(2)(GM)/(R^(2))t^(2)`
`impliest^(2)=(2R^(3))/(GM)impliest_(2)=sqrt((2R^(3))/(GM)),(t_(1))/(t_(2))=(pi)/(2sqrt(2))`