# Why can gravitational potential energy be expressed both as $mgh$ and $-GMm/r$? [duplicate]

In these two different equations for the same (?) thing, not only is one directly proportional to height and one is inversely proportional to height, but they contain completely different variables, and this is my confusion.

Additionally, in the second equation, we can see that the $$U$$ will decrease to 0 as the distance between the objects increases to infinity. This also confuses me, because my thinking would be that if you took two objects and separated them very far in empty space, they would come together because of gravity and they would both be moving pretty fast at the end of their journeys, but for this to happen we would need a high starting $$U$$.

$$mgh$$ is an approximate formula that can be used only close to the Earth's surface, i.e., when $$h\ll R$$. It is obtained from $$-GMm/(R+h)$$ by Taylor expansion in h.

The formula $$E_\text{pot}=mgh$$ only holds for objects which are located in the vicinity the surface of the earth where $$F_g=mg\qquad \text{with}\qquad g=\frac{Gm_\text{earth}}{r_\text{earth}^2}$$

The formula $$E_\text{pot}=-\frac{Gm_\text{earth}m}{r}$$ holds for all distances $$r$$, not just for distances $$r\approx r_\text{earth}$$. Also, the reference point where the energy is considered zero is shifted. In the first formula we put zero to $$h=0$$, that is, on the surface of earth, whereas in the second formula we put it to $$r\to\infty$$.

• Thank you, but I am still confused on why in these two equations, one exact and one approximation, the relation with R is different (direct vs inverse square). I am also confused as to how, if you put an object infinitely far away, it would have 0 potential energy, Wouldnt it speed up towards the Earth for a long time and gain a lot of kinetic energy? – cyalatergator Mar 15 at 9:15
• Are you familiar with Taylor series or at least linear approximation? Regarding your second question: Yes, exactly, it would gain negative potential energy and positive kinetic energy (thus keeping zero energy in total, as energy conservation demands). – Photon Mar 15 at 9:22

$$h$$ is a change in height, it might help to see the derivation of $$mgh$$ from a change in $$\frac{-GMm}{R}$$

The increase in potential energy, when lifting by a small height $$h$$ is

$$\frac{-GMm}{R+h} - \frac{-GMm}{R}$$

this is approximately $$GMm(\frac{1}{R}-\frac{1}{R+h})$$ = $$GMm(\frac{R+h-R}{R^2})$$

so it's $$\frac{GMmh}{R^2}$$ or $$mgh$$

and the last expression is really only useful to calculate a gain in P.E for small changes in height.