How much above the surface of earth does the accelration due to gravity reduces by `64%` of its value on the earth. Radius of earth `= 6400 km`.

Explanation

(i) `g' = (64)/(100) g = (gR^(2))/((R + h)^(2))` or `(8)/(10) = (R )/((R + h))`
or `8 R + 8h = 10R`
or `h= (R )/(4) = (6400)/(4) = 1600 km`
(ii) `g' = 100 - 64 = 36% = (36)/(100) g = (gR^(2))/((R + h)^(2))`
`:. (6)/(10) = (R )/((R + h))`
or `h = (4R)/(6) = (2)/(3) R = (2 xx 6400)/(3) = 4266.7 km`