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.