We’ll start our exploration of potential energy with gravitational potential energy, the interaction energy associated with the gravitational interaction between two masses. The symbol for gravitational potential energy is $U_{\text{G}}$. We’ll restrict ourselves to the “flat-earth approximation” $\vec{F}{}_{\text{G}}=-mgĵ$. The gravitational potential energy of two astronomical bodies will be taken up in Chapter 13.

FIGURE 10.2 shows a ball of mass $m$ moving upward from an initial vertical position $y_{\text{i}}$ to a final vertical position $y_{\text{f}}$. The earth exerts force $\vec{F}{}_{\text{E on B}}$ on the ball and, by Newton’s third law, the ball exerts an equal-but-opposite force $\vec{F}{}_{\text{B on E}}$ on the earth.

FIGURE 10.2 The ball$+$earth system has a gravitational potential energy.

We could define the system to consist of only the ball, in which case the force of gravity is an external force that does work on the ball, changing its kinetic energy. We did exactly this in Chapter 9. Now let’s define the system to be ball $\text{+}$ earth. This brings the interaction inside the system, so (ignoring any gravitational forces from distant astronomical bodies) there’s no external work. Instead, we have an energy of interaction—the gravitational potential energy—described by Equation 10.4:

where $W_{\text{B}}$ is the work gravity does on the ball and $W_{\text{E}}$ is the work gravity does on the earth. The latter, practically speaking, is zero. $\vec{F}{}_{\text{E on B}}$ and $\vec{F}{}_{\text{B on E}}$ have equal magnitudes, by Newton’s third law, but the earth’s displacement is completely insignificant compared to the ball’s displacement. Because work is a product of force and displacement, the work done on the earth is essentially zero and we can write

You learned in Chapter 9 to compute the work of gravity on the ball: $W_{\text{B}}=$ $(F_{\text{G}}){}_y\text{Δ}y=-mg\text{Δ}y$. So if the ball changes its vertical position by $\text{Δ}y$, the gravitational potential energy changes by

Notice that increasing the ball’s height $(\text{Δ}y>0)$ increases the gravitational potential energy $(\text{Δ}{U}_{\text{G}}>0)$, as we would expect.

Our energy analysis has given us an expression for $\text{Δ}{U}_{\text{G}}$, the change in potential energy, but not an expression for $U_{\text{G}}$ itself. If we write out what the $\text{Δ}$ in Equation 10.7 means—final value minus initial value—we have

Consequently, we define the gravitational potential energy to be

Notice that gravitational potential energy is an energy of position. It depends on the object’s position but not on its speed. You should convince yourself that the units of mass times acceleration times position are joules, the unit of energy.

EXAMPLE 10.1 Launching a pebble

Rafael uses a slingshot to shoot a 25 g pebble straight up at 17 m/s. How high does the pebble go?

MODEL Let the system consist of both the earth and the pebble, which we model as a particle. Assume that air resistance is negligible. There are no external forces to do work, but the system does have gravitational potential energy.

VISUALIZE FIGURE 10.3 is a before-and-after pictorial representation. The before-and-after representation will continue to be our primary visualization tool.

FIGURE 10.3 Pictorial representation of the pebble $\text{+}$ earth system.

FIGURE 10.3 Full Alternative Text

SOLVE The energy principle for the pebble $\text{+}$ earth system is

$$\text{Δ}{E}_{\text{sys}}=\text{Δ}K+\text{Δ}{U}_{\text{G}}=W_{\text{ext}}=0$$

That is, the system energy does not change at all. Instead, kinetic energy is transformed into potential energy without loss inside the system. In principle, the kinetic energy is that of the ball plus the kinetic energy of the earth. But as we just noted, the enormous mass difference means that the earth is effectively at rest while the pebble does all the moving, so the only kinetic energy we need to consider is that of the pebble. Thus we have

$$0=\text{Δ}K+\text{Δ}{U}_{\text{G}}=\left(\frac{1}{2}m{v}_1{}^2-\frac{1}{2}m{v}_0{}^2\right)+\left(mg{y}_1-mg{y}_0\right)$$

We know that $v_1=0\text{m}\text{/}\text{s}$, and we chose a coordinate system in which $y_0=0\text{m,}$ so we’re left with

$$y_1=\frac{v_0{}^2}{2g}=\frac{{(17\text{ m/s)}}^2}{2(9.80{\text{ m/s}}^2)}=15\text{ m}$$

The answer did not depend on the pebble’s mass, which is not surprising after our earlier practice with free-fall problems.

REVIEW A height of 15 m $\approx$ 45 ft seems reasonable for a slingshot.