# Work done by Earth's Gravity

Figure 1: As an object moves along the path $$\vec{R}(t)$$ from an initial height of $$y_i$$ to a final height of $$y_f$$, the Earth's gravitational force $$-m\vec{g}$$ does work on the object.

In this lesson, let's try to solve the following problem: if an object is at an initial height of $$y_i$$ (which could be zero if the object is initially on a surface like the Earth's surface) and I move it along any arbitrary trajectory $$\vec{R}(t)$$ to a final height of $$y_f$$, what is the work $$W$$ done by the force of gravity $$\vec{F}_g$$? We’ll assume that the object’s change in height, $$Δy=y_f-y_i$$, is very small in comparison to the Earth’s radius; this means that it is a reasonable approximation to assume that the magnitude of the force of gravity, $$F_g$$, will stay constant as it acts on the object at each point along the object’s trajectory. We’ll also assume that the disp̨lacement $$Δx$$ of the object along the x-direction is small enough so that the surface of the Earth beneath it does not curve. Thus, it is also a good approximation to assume that the direction of $$\vec{F}_g$$ does not change either. Thus, given these two approximations, we can assume that $$\vec{F}_g$$ stays constant at each point along the object’s trajectory. Also, given these assumptions, Newton’s law of gravity can be used to determine that $$F_g$$ is given by

$$F_g=-mg.\tag{1}$$

Although the force vector $$\vec{F}_g$$ does not change from point to point, since the direction of the object can, in general, change, the displacement vector $$d\vec{R}$$ will in general change direction from point to point. Thus, we cannot use the simplified expression for the work which is given by $$W=FΔR$$; instead, we must use the general definition of work which is given by

$$W≡\int_c\vec{F}_g·d\vec{R}\tag{2}$$

The displacement vector $$d\vec{R}$$ can be decomposed with respect to the x-, y-, and z-components to give

$$d\vec{R}=d\vec{x}+d\vec{y}+d\vec{z}.\tag{3}$$

Substituting Equation (3) into (2) and doing some algebraic simplification, we have

$$W=\int_c\vec{F}_g·(d\vec{x}+d\vec{y}+d\vec{z})=\int{\vec{F}_g·d\vec{x}}+\int{\vec{F}_g·d\vec{y}}+\int{\vec{F}_g·d\vec{z}}.$$

Since the direction of $$\vec{F}_g$$ is always pointing down in the y-direction, $$\vec{F}_g$$ has no x- or z-components. Thus, $$\vec{F}_g$$ is always perpendicular to $$d\vec{x}$$ and $$d\vec{z}$$ and Equation (4) can be simplified to

$$W=\int{\vec{F}_g·d\vec{y}}.\tag{5}$$

Since $$\vec{F}_g$$ is always parallel (or anti-parallel) to $$d\vec{y}$$, we know that the dot product in Equation (5) will always simplify to either $$\vec{F}_g·d\vec{y}=F_gdy$$ or

$$\vec{F}_g·d\vec{y}=-F_gdy$$. If the object's total change in height $$Δy$$ is positive, then the dot product simplifies to $$\vec{F}_g·d\vec{y}=-F_gdy$$ and Equation (5) becomes

$$W=-\int{F_gdy}.\tag{6}$$

Equation (6) is thus the work done on an object moving along any trajectory $$\vec{R}(t)$$ such that $$Δy>0$$ (in other words, such that the object is ultimately moved to a greater height). Substituting Equation (1) into (6), we have

$$W=-\int{mgdy}.\tag{7}$$

Since both the mass $$m$$ of the object and that object's gravitational acceleration $$g$$ are constant, it follows that

$$W=-mg\int{dy}.\tag{8}$$

Finally, $$\int{dy}=Δy$$ and Equation (8) becomes

$$W=-mgΔy.\tag{9}$$

Whenever an object moves across the Earth's surface (such that $$Δy≪R_E$$ and $$Δx≪R_E$$), we can use Equation (9) to determine the work $$W$$ done on that object by the force $$\vec{F}_g$$. If the object is moved to a lower height ($$Δy<0$$), we would have followed the same exact steps except we would substitute $$\vec{F}_g·d\vec{y}=-F_gdy$$ for $$\vec{F}_g·d\vec{y}=F_gdy$$ leading to the following expression for $$W$$: $$W=mgΔy$$. In other words, there's just a change in sign.