At what depth from the surface of earth, the value of acceleration due to gravity is reduced by `40%` of its value on the surface of earth. Radius of the earth `= 6.4 xx 10^(6) m`.

Explanation

Here, `d = ? , g' = g -(40)/(100) g = (3)/(5) g`
As `g' = g(1 - (d)/(R ))`
or `(3)/(5) g = g(1 - (d)/(R ))` or `(3)/(5) = 1 - (d)/(R )`
or `d = (1 - (3)/(5)) R = (2)/(5) R = (2 xx 6.4 xx 10^(6))/(5)`
`= 2.56 xx 10^(6) m`