g(h)= GM/(R+h)^2 which decreases with h increases
g(h) / g = R^2/(R+h)^2
g(h) = g * (1+ (h/R))^-2
now we know that for the fall if we consider energy conservation
1/2 mV^2 = mg(h)h
V^2 = 2*g(h)*h
V^2 = 2*g*h * (1+ (h/R))^-2
V = sqrt(2gh) * (1+(h/R))^-1
now if i expand...
Suppose an object is dropped from height h, where h < R but gravity is not constant). Show that the speed with which it hits the ground, neglecting friction, is approximately given by:
v = sqrt(2gh)(1 - (h/2r))
where g is the acceleration due to gravity on the surface of the Earth.
HINT...