# Two expressions for potential energy in the gravitational field of the earth

Let $M$ be the mass of the earth, considered as a point mass, then the potential energy of a point with distance $r$ away from the center (assume $r > \textrm{radius of earth})$ is $$U(r) = -\frac{G M m} r = -\frac{gm}{r}$$ with $g := G\cdot M$. Now, as it is written in the textbooks, the potential energy of a body of height $h$ on earth is approximately $$U(h) = mgh.$$ Now for me both expression contradict each other, one is negative, the other positive, one decreases as $r \to \infty$, the other increases as $h \to \infty$ (and the height is directly related to the distance/radius $r$ I think).

So whats my misconception, what have I missed that I came to conclusion that both expression contradict each other? Thanks for any help!

EDIT: Changed $$U(r) = -\frac{GM}{r} = -\frac{g}{r}$$ to $$U(r) = -\frac{GMm}{r} = \frac{gm}{r}.$$

• Where did you get the first formula? – HDE 226868 Aug 28 '14 at 21:39
• for example here: en.wikipedia.org/wiki/… – StefanH Aug 28 '14 at 21:43
• The last part of that first expression is invalid. GM/r has units of velocity squared, while g/r has units of inverse time squared. – David Hammen Aug 28 '14 at 21:46
• @Stefan Try this link: en.wikipedia.org/wiki/Gravitational_potential That could be the source of confusion. – HDE 226868 Aug 28 '14 at 21:47
• That first expression is still invalid. $GM$ has dimensions of length$^3$/time$^2$ but $g$ has dimensions of length/time$^2$. Typically the product $GM$ is denoted as $\mu$ and is called the standard gravitational parameter. (Note that this conflicts with use of $\mu$ for reduced mass. C'est la vie. There are only so many English, Greek, and German letters around. There are bound to be conflicts.) – David Hammen Aug 28 '14 at 22:03

The first the expression $U(r) = -\frac {GM} r$ is a potential, but not potential energy. The units are velocity2. This is a widely used potential in solar system astronomy, geology, geophysics, and in aerospace engineering. For example, see http://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-201-essentials-of-geophysics-fall-2004/lecture-notes/ch2.pdf beginning at the bottom of the third page.
This becomes energy when you multiple by the mass $m$ of the test particle that is subject to the mass $M$: $U(r) = -\frac {GMm}r$. The second expression, $U=mgh$, is linear approximation to the former.
A linear approximation of some function $f(x)$ about some point $x=x_0$ can be constructed by using the first two terms of the Taylor expansion of $f(x)$. This yields $f(x) \approx f(x_0) + \frac{df(x)}{dx}(x-x_0)$. Applying that concept to the potential $U(r)$ results in $$U(r) \approx U(r_0) + \left.\frac {dU(r)}{dr}\right|_{r=r_0}(r-r_0)$$ Upon choosing $r_0$ as the Earth's radius, $r-r_0$ becomes height $h$. The derivative $\frac {dU(r)}{dr}$ is $\frac {GM}{r^2}$, and at the surface of the Earth this becomes $g$. Finally, potential energy involves an arbitrary additive constant. We can get rid of that constant term $U(r_0)$ without no loss of generality. The final result is $$U(r) \approx mgh$$
• You mean $U(r) = -\frac{GMm}{r}$, not $U(r) = \frac{GMm}{r}$? – StefanH Aug 28 '14 at 22:05
• If you use positive potential the other formula would read $U(r) \approx -mgh$, right? – StefanH Aug 28 '14 at 22:31
• That can be corrected by negating that potential and multiplying by mass to yield a potential energy. Being able to yield $mgh$ is not why people use that potential, however. It is instead a first step towards hairy models that make your hair stand straight up and then fall out. – David Hammen Aug 28 '14 at 23:22
Notice that $h$ and $r$ are related in the following way: \begin{align} r = R + h \end{align} where $R$ is the radius of the Earth (the distance from the center to the surface) and $h$ is the height above the surface. Then notice that \begin{align} U = -\frac{GmM}{r} = -\frac{GMm}{R+h} = -\frac{GMm}{R}\left[1 -\frac{h}{R} + O(\frac{h}{R})^2\right]. \end{align} Now $g$ is defined as the magnitude of the acceleration due to gravity at the surface of the Earth, namely \begin{align} g = \frac{GM}{R^2} \end{align} so we can write \begin{align} U = -mgR\left[1 -\frac{h}{R} + O(\frac{h}{R})^2\right] \end{align} Now we simply note that the potential energy is only defined up to an additive constant, so we can add $mgR$ to this expression to obtain \begin{align} U = mgR\left[\frac{h}{R} + O(\frac{h}{R})^2\right]. \end{align} In other words, the formula $mgh$ is the approximate form of the more precise formula $-GmM/r$ for heights $h$ above the surface of the Earth that are small compared to the radius R$of the Earth since when that ratio is small, the higher order terms in brackets on the right hand side can be neglected to good approximation. • Thanks for your answer, but how do you derived$1/(R+h) = -1/R\left[ 1 - \frac{h}{R} + O(\frac{h}{R})^2\right]$? – StefanH Aug 28 '14 at 22:42 • @Stefan There should be no negative sign on the right hand side of what you wrote, but in any case, the Taylor expansion of$1/(1+x)$about$x=0$is$1-x + O(x^2)$. Now simply notice that$1/(R+h) = \frac{1}{R}\frac{1}{1+\frac{h}{R}}$and apply that Taylor expansion to the second factor. – joshphysics Aug 28 '14 at 22:52 • In your last step, in going from$U = -mgR\left[ 1 - \frac{h}{R} + O(\frac{h}{R})^2\right]$to$U = mgh + O(\frac{h}{R})^2$you use$-mgR \cdot O(\frac{h}{R})^2 = O(\frac{h}{R})^2$? Why is this equality true? – StefanH Aug 29 '14 at 11:02 • @Stefan It's just notation. Since$mgR$is a constant, it doesn't affect the fact that the higher order terms behave like squares of$h/R$and higher. Take, for example,$x^2$and$1000 x^2$, both of these are$O(x^2)$. I could have also left the$mgR$in there. It wouldn't have changed the result, namely that$mgh$is the fist non-vanishing order when you Taylor expand the potential energy. – joshphysics Aug 29 '14 at 14:40 • @Stefan Sure. If you're viewing the potential energy as a function of the parameter$h$, then that's true, but in physics it's usually only useful to discuss certain quantities being "small" or "large" when they are dimensionless because then smallness and largeness doesn't depend on one's choice of units, so that's why I would prefer to refer to the size of the dimensionless ration$h/R\$. I'm going to edit my answer at the end to make all of these issues less confusing; let me know if it helps. – joshphysics Aug 29 '14 at 22:49