How much below the earth's surface the acceleration due to gravity reduces to 64% of its value on the surface of earth ?

Explanation

`g. = (64)/(100)g`
`(g.)/(g) = (R^(2))/((R+h)^(2))`
`(64)/(100) = (R^(2))/((R+h)^(2))`
`rArr (8)/(10) = (R )/(R + d)`
`8R + 8d = 10R`
`8d = 2R rArr d = (1)/(4)R`
`d = (1)/(4)R = (1)/(4) xx 6,400 = 1,600 km`