Calculate g at the bottom of a mine 8 km deep and at an altitude 32 km above the earth's surface . Radius of earth = `6.4 xx 10^(4)` m and g on earth's surface = `9.8 m//s^(2)` .

Explanation

Given `R_(E) = 6.4 xx 10^(6) m , g = 9.8 ms^(-2)`
(i) `g^(1) = g(1- (h)/(R))`
i.e. , `g^(1) = 9.8 (1 - (8)/(6400)) , h 8 ` km , R = `6400` km i.e `g^(1) = (9.8 xx 799)/(800) = 9.787 ms^(-2)`
acceleration due to gravity at a depth of 8 km is `9.787 ms^(-2)`
(ii) `g^(1) = g (1 - (2h)/(R))`
i.e. `g^(1) = 9.8 (1 - (2 xx 32)/(6400)) i.e. g^(1) = (9.8 xx 99)/(100) = 9.702 ms^(-2)` .