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Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$

The experimentally verified near-Earth gravitational potential is $$U ~=~ mgh.\tag{2}$$

The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance. $(1)$ scales as $\frac 1r$. So the greater the distance from the Earth, the less potential energy an object should have. But $(2)$ scales proportionally to distance. So the greater the distance from the Earth, the more potential energy an object should have.

How is this reconcilable?

Qmechanic
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Bobbie D
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5 Answers5

24

Your equation (2) is the change in potential energy when the object moves vertically by a distance $h$ i.e. when the object moves from $r$ to $r+h$. Let's use equation (1) to calculate this:

$$ \Delta U = GMm\left(\frac{1}{r}-\frac{1}{r+h}\right) $$

Subtracting the two fractions inside the bracket gives:

$$\begin{align} \Delta U &= GMm\left(\frac{r+h}{r(r+h)}\frac{r}{r(r+h)}\right) \\ &= GMm\frac{h}{r(r+h)} \end{align}$$

Since $h \ll r$ that means $r+h\approx r$ and our equation becomes:

$$\begin{align} \Delta U &\approx GMm\frac{h}{r^2} \\ &\approx \frac{GM}{r^2}mh \\ &\approx gmh \end{align}$$

Footnote:

I've just noticed that in your comment to Itachí's answer you ask if you can use a Taylor series. You can use a binomial expansion to make the aproximation more obvious. You rewrite:

$$ \Delta U = GMm\frac{h}{r(r+h)} $$

as:

$$ \Delta U = \frac{GM}{r^2}mh\left(1+\frac{h}{r}\right)^{-1} $$

then a binomial expansion gives:

$$ \Delta U = \frac{GM}{r^2}mh\left(1-\frac{h}{r} + O\left(\frac{h}{r}\right)^2 \right) $$

And as before since $h \ll r$ the term in the brackets is approximately one and we once again get:

$$ \Delta U = \frac{GM}{r^2}mh $$

John Rennie
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7

Given a force $F$, the work done on an object over a distance between two points $s_0$ and $s_f$ by that force is $$W=-\int_{s_0}^{s_f} Fds$$ In the case of gravity, $$F=\frac{GMm}{r^2},\quad ds=dr$$ Thus, in the case where $U=W$, $s_0=0$ and $s_f=r$, so $$U=\frac{GMm}{r}$$ Now, over small distances by the Earth's surface, the force is approximately constant. If we substitute in $$g\equiv\frac{GM}{r_e^2}$$ and assume that $g$ is essentially constant between our reference point and $h$, we can say that $$\Delta U=\int_0^hmgds=mg\int_0^hds=mgh$$ So $(1)$ is the actual expression for the potential energy at a point if we assume that $g$ changes; $(2)$ is an approximation if we assume that the change in $g$ is small. This is valid near Earth's surface, as John Rennie showed, but it's generally not valid over large distances.

I should note something about reference points. In the case of $(1)$, $r$ is a coordinate from the center of mass $M$; in the case of $(2)$, $h$ is a coordinate from some arbitrary reference point from the center of $M$. Generally, you could take this to be the radius of the Earth, but it's often unimportant for conservation of energy problems, and you can choose any value that makes the calculations simpler - so long as $g$ is approximately constant.

HDE 226868
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1

The potential energy of an object in an gravitational field is $$U =-\frac{GMm}{r}\tag{1}$$

If I plot the graph of $-\frac{1}{r}$,It appears like:

enter image description here

So the greater the distance from the Earth, the more potential energy an object is having.

1

Simple answer that I like the most: Because $g$ is simplified to a fixed value.

The general expression is: $$U=gmr \quad \quad\text{with}\quad g=\frac{GM}{r^2}$$ and works for any distance $r$ from the Earth centre. (Usually written $U=mgh$ when the reference point is somewhere else than the Earth centre.)

At the surface of Earth with (mean) radius $r=6371\times 10^3 \;\mathrm m$, we get:

$$g=\frac{GM}{r^2}=\frac{6.674\times 10^{-11}\;\mathrm{\frac{m^3}{kg\cdot s^2}}\;*5.972\times 10^{24}\;\mathrm{kg}}{(6371\times 10^3 \;\mathrm m)^2}\;\mathrm{m/s^2}=9.820\;\mathrm{m/s^2}$$

the wellknown value of $g$. Moving upwards changes the value but only negligibly little in the beginning:

  • For $h=1\;\mathrm{km}$: $\quad g=9.816\;\mathrm{m/s^2}$
  • For $h=2\;\mathrm{km}$: $\quad g=9.813\;\mathrm{m/s^2}$
  • For $h=3\;\mathrm{km}$: $\quad g=9.810\;\mathrm{m/s^2}$
  • For $h=4\;\mathrm{km}$: $\quad g=9.807\;\mathrm{m/s^2}$
  • For $h=5\;\mathrm{km}$: $\quad g=9.804\;\mathrm{m/s^2}$
  • $\quad\vdots$
  • For $h=10\;\mathrm{km}$: $\quad g=9.789;\mathrm{m/s^2}$
  • $\quad\vdots$
  • For $h=100\;\mathrm{km}$: $\quad g=9.518\;\mathrm{m/s^2}$

A nice plotting shows it clearer. So, smaller values - especially when staying below 1 km - makes no difference in $g$. Down here it is safe to assume a fixed value. So all in all, the expression $U=mgr$ (or $U=mgh$) is not the simplification; the fixed $g$ is the simplification.

Steeven
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1

All other answers are correct. The shortest answer can simply be found by inspection of equation (1) for the potential energy $U(r)$. If $r_2>r_1$, then the potential energy $U(r_2)$ at $r_2$ is larger (less negative) than $U(r_1)$ giving a positive potential energy difference $U(r_2)-U(r_2)>0$ which is consistent with equation (2).

Assuming $h<<r_{earth}$, you find the Taylor approximation $$\Delta U≈\frac{dU}{dr}∣_{r_{earth}}h=\frac{GMm}{r_{earth}^2}h=gmh$$ giving the linear potential energy dependence on $h$ near the earth surface according to equation (2).

freecharly
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