Our expression for the gravitational potential energy $U_{\text{G}}=mgy$ seems straightforward. But you might notice, on further reflection, that the value of $U_{\text{G}}$ depends on where you choose to put the origin of your coordinate system. Consider FIGURE 10.4, where Amber and Carlos are attempting to determine the potential energy when a 1 kg rock is 1 m above the ground. Amber chooses to put the origin of her coordinate system on the ground, measures $y_{\text{rock}}=1\text{m,}$ and quickly computes $U_{\text{G}}=mgy=9.8\text{J}$. Carlos, on the other hand, read Chapter 1 very carefully and recalls that it is entirely up to him where to locate the origin of his coordinate system. So he places his origin next to the rock, measures $y_{\text{rock}}=0\text{m,}$ and declares that $U_{\text{G}}=mgy=0\text{J}$!

FIGURE 10.4 Amber and Carlos use different coordinate systems to determine the gravitational potential energy.

How can the potential energy have two different values? The source of this apparent difficulty comes from our interpretation of Equation 10.7. Our energy analysis found that the potential energy changes by $\text{Δ}{U}_{\text{G}}=mg(y_{\text{f}}-y_{\text{i}})$. Our claim that $U_{\text{G}}=mgy$ is consistent with this finding, but so also would be a claim that $U_{\text{G}}=mgy+C$, where $C$ is any constant.

In other words, potential energy does not have a uniquely defined value. Adding or subtracting the same constant from all potential energies in a problem has no physical consequences because our analysis uses only $\text{Δ}{U}_{\text{G}}$, the change in the potential energy, never the actual value of $U_{\text{G}}$. In practice, we work with potential energies by setting a reference point or reference level where $U_{\text{G}}=0$. This is the zero of potential energy. Where you place the reference point is entirely up to you; it makes no difference as long as every potential energy in the problem uses the same reference point. For gravitational potential energy, we choose the reference level by placing the origin of the $y$-axis at that point. Where $y=0,U_{\text{G}}=0$. In Figure 10.4, Amber has placed her zero of potential energy at the ground, whereas Carlos has set a reference level 1 m above the ground. Either is perfectly acceptable as long as Amber and Carlos use their reference levels consistently.

But what happens when the rock falls? When it gets to the ground, Amber measures $y=0\text{m}$ and computes $U_{\text{G}}=0\text{J}$. No problem. But Carlos measures $y=-1\text{m}$ and thus computes $U_{\text{G}}=-9.8\text{J}$. A negative potential energy may seem surprising, but it’s not wrong; it simply means that the potential energy is less than at the reference point. The potential energy with the rock on the ground is certainly less than when the rock was 1 m above the ground, so for Carlos—with an elevated reference level—the potential energy is negative. The important point is that both Amber and Carlos agree that the gravitational potential energy changes by $\text{Δ}{U}_{\text{G}}=-9.8\text{J}$ as the rock falls.