Connection of the Potential Energy Formula with mgh

The formula $U(r)=-\dfrac{GMm}{r}$ derived in the previous section is the EXACT expression for gravitational potential energy at any distance r from the Earth's centre. For an object resting on the Earth's surface, $r=R$, so $U_1=-\dfrac{GMm}{R}$; lifted to height h above the surface, $r=R+h$, so $U_2=-\dfrac{GMm}{R+h}$. The increase in potential energy on lifting the object is therefore $\Delta U=U_2-U_1=-\dfrac{GMm}{R+h}+\dfrac{GMm}{R}=GMm\left(\dfrac{1}{R}-\dfrac{1}{R+h}\right)=\dfrac{GMmh}{R(R+h)}.$

Using $g=\dfrac{GM}{R^2}$, i.e. $GM=gR^2$, this can be rewritten as $\Delta U=\dfrac{gR^2mh}{R(R+h)}=\dfrac{mgh}{1+h/R}=\dfrac{mgh\,R}{R+h}$. Only in the SPECIAL CASE where the height h is negligibly small compared to the Earth's radius R (so that $R+h\approx R$) does this reduce to the familiar $\Delta U\approx mgh.$ …