At what depth from the surface of the earth, the acceleration due to gravity will be `(1)/(4)` th the value of g on the surface of the eath ?

Explanation

As `g'=g[1-(d)/(R_(e))]`
According to question
`g'=(g)/(4)`
`therefore (g)/(4)=g[1-(d)/(R_(e))]`
`therefore (1)/(4)=1-(d)/(R_(e))`
`therefore (d)/(R_(e))=1-(1)/(4)`
`therefore d=0.75 R`.