# How to reach $U = mgh$ Using Newton's Law of Universal Gravitation? [duplicate]

today i was curious about the potential energy, so, i started studying the Newton's Law of Universal Gravitation which its equation is $$\begin{eqnarray} U= -\frac{GMm}{r}.\end{eqnarray}$$ Well, since i gotthis equation i really know how to reach at this point ( due some experiments, also some documents help me to get until here), i was very excited about reaching this point, and i told to myself : "Alright, good job." the excitement passed away when a friend told me that i can transform this equation into:$$\begin{eqnarray} U= m.g.h.\end{eqnarray}$$ Sadly, i don't even know how to start with,i know that The potential should be an approximation for the general potential energy when r = r (of earth), but the problem is... Equation 1 Scales with radius and the other one with height.

## marked as duplicate by Qmechanic♦Feb 10 at 6:26

• This has been asked here before. Just trying to find the question – Aaron Stevens Feb 10 at 4:51
• Looking for it 1 hour ago, i'll check again, thanks. – Andrés HK Feb 10 at 4:54
• Possible duplicate of Near Earth vs Newtonian gravitational potential – Sandejo Feb 10 at 5:00
• Just approximate the gravitational force to a constant at near surface and then integrate over small displacements to get that potential. – user191954 Feb 10 at 5:09
• $U=mgh$ only applies when the gravitational field strength is constant, such as when you change an object's elevation from ground level to an altitude that is "small" relative to the magnitude of the earth's radius (e.g., a VERY few miles). For a "large" change in height, the force of gravity is not constant, and $U=-GMm/r$ applies. – David White Feb 10 at 5:11

Write $$r=R+h$$ where $$r$$ is the distance from the center of the Earth, $$R$$ is the radius of the Earth, and $$h$$ is height above the surface. Then the difference between the potential energy at height $$h$$ above the surface and the PE at the surface is

\begin{align} U(r)-U(R)&=-\frac{GMm}{r}+\frac{GMm}{R}\\ &=\frac{GMm}{R}-\frac{GMm}{R+h}\\ &=\frac{GMm}{R}\left[1-\left(1+\frac{h}{R}\right)^{-1}\right] \end{align}.

For $$h<, use the Taylor series

$$(1+x)^{-1}=1-x+O(x^2)$$

to get

\begin{align} U(r)-U(R)&=\frac{GMm}{R}\left[\frac{h}{R}+O\left(\frac{h^2}{R^2}\right)\right]\\ &\approx m\left(\frac{GM}{R^2}\right)h\\ &=mgh \end{align}.

So: Taking the PE outside the surface as $$-GMm/r$$, it is is zero at infinity and negative (namely $$-GMm/R$$) at the surface. Slightly above the surface, it is slightly less negative by $$mgh$$. (By the way, inside the Earth, $$-GMm/r$$ doesn’t apply.) In Newtonian physics it doesn’t really matter where you take the PE to be zero, so you can take it to be zero at the surface and say that it simply is $$mgh$$ for $$h<.