Gravity, like electricity, is a long-range force. Much as we defined the electric field $\vec{E}=\vec{F}{}_{\text{on }q}\text{/}q$, we can also define a gravitational field—the agent that exerts gravitational forces on masses—as $\vec{F}{}_{\text{on }m}\text{/}m$. But $\vec{F}{}_{\text{on }m}=m\vec{g}$ near the earth’s surface; thus the familiar $\vec{g}=(9.80\text{ N/kg}\text{,}\text{ down})$ is really the gravitational field! Notice how we’ve written the units of $\vec{g}$ as $\text{N}\text{/}\text{kg}$, but you can easily show that $\text{N}\text{/}\text{kg}=\text{m}\text{/}{\text{s}}^2$. The gravitational field near the earth’s surface is a uniform field in the downward direction.

FIGURE 25.2 shows a particle of mass m falling in the gravitational field. The gravitational force is in the same direction as the particle’s displacement, so the gravitational field does a positive amount of work on the particle. The gravitational force is constant, hence the work done by the gravitational field is

We have to be careful with signs because $\text{Δ}r$, the magnitude of the displacement vector, must be a positive number.

FIGURE 25.2 Potential energy is transformed into kinetic energy as a particle moves in a gravitational field.

Now we can see how the definition of $\text{Δ}U$ in Equation 25.2 makes sense. The change in gravitational potential energy is

Comparing the initial and final terms on the two sides of the equation, we see that the gravitational potential energy near the earth is the familiar quantity

We could add a constant to $U_{\text{G}}$, but we’ve made the customary choice that $U_{\text{G}}=0$ at $y=0$.