ight of h from the surface of the earth, the gravitational FORCE on an object of mass m is F = GMm/(R+h)^2 Here (R + h) is the distance between the object and the centre of earth. Say at that height h, the gravitational acceleration is g1. So we can write, mg1 = GMm / (R+h)^2 => g1 = GM/(R+h)^2 _________________ (5) Now we KNOW on the surface of earth, it is  g =  GM / R^2 TAKING ratio of these 2, g1/g = R^2 /(R+h)^2 = 1/(1 + h/R)^2 = (1 + h/R)^(-2) = (1 – 2h/R) => g1/g = (1 – 2h/R) => g1 = g (1 – 2h/R) ___________________ (6) So as altitude h increases, the value of acceleration due to gravity falls.     Let’s say, a body of mass m is resting at point A , where A is at a depth of h from the earth’s surface. Distance of point A from the centre of the earth = R – h,  where R is the radius of the earth. Mass of inner sphere = (4/3). Pi. (R-h)^3. p Here p is the density. Now at point A, the gravitational force on the object of mass m is F = G M m/ (R-h)^2 = G. [(4/3). Pi. (R-h)^3. p] m/(R-h)^2 = G. (4/3). Pi. (R-h). p. m Again at point A, the acceleration due to gravity (say g2) = F/m = G. (4/3). Pi. (R-h). p  _________________ (7) Now we know at earth’s surface, acceleration due to gravity = g = (4/3) Pi R p G Taking the ratio, again, g2/g = [G. (4/3). Pi. (R-h). p ]/ [(4/3) Pi R p G] = (R-h) / R = 1 – h/R. => g2 = g (1 – h/R)  _________________________ (8) So as depth h increases, the value of acceleration due to gravity falls.HOPE IT HELPS YOU