Gravitational Work Along Prescribed Paths

Gravity Acts Along a Path, Not Just Through a Distance

When an object moves under gravity, its work depends on how the gravitational force points relative to each small part of its motion. A displacement in the direction of gravity contributes positive work; a displacement against gravity contributes negative work; and a displacement perpendicular to gravity contributes zero work.

For a prescribed trajectory, the calculation is therefore not simply “force times distance.” It is a sum of force components in the direction of motion. Parametrizing the path makes that sum an integral that can account for vertical motion, curved motion, and gravitational fields that vary with position.

Definition: If a particle follows a piecewise-smooth path $\mathbf r(t)$, $a\le t\le b$, in a force field $\mathbf F$, the work done by that field along the path is $$ W=\int_a^b \mathbf F(\mathbf r(t))\cdot \mathbf r'(t)\,dt. $$ The path must stay within the region where the force field is defined. Work is measured in joules when force is in newtons and position in metres.

Why the Parametrized Work Formula Works

Over a short time interval, the particle’s displacement is approximately $\Delta\mathbf r$, and the force is approximately constant at its value $\mathbf F$ during that interval. The work over this small part of the motion is approximately the force dotted with the displacement:

\Delta W\approx \mathbf F\cdot\Delta\mathbf r.

For a parametrized path, a short change $\Delta t$ produces a displacement approximately $\mathbf r'(t)\Delta t$. Substituting this into the small-work expression gives

\Delta W\approx \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,\Delta t.

Adding these contributions over the whole trajectory and taking the limit as the time intervals become arbitrarily small gives the work integral. The dot product is essential: it selects the component of force along the instantaneous direction of motion.

Useful rule: To find gravitational work along a specified trajectory, substitute the position into the force field, calculate the path derivative, take their dot product, and integrate over the parameter interval.

Uniform Gravity Near Earth

For motion over a region where gravitational acceleration can be treated as constant, choose the vertical coordinate $z$ to increase upward. Let $m$ be the object’s mass and $g>0$ the magnitude of gravitational acceleration. The force is constant and points downward:

\mathbf F=-mg\,\mathbf e_z,

where $\mathbf e_z$ is the unit vector in the upward vertical direction. If $\mathbf r(t)=(x(t),y(t),z(t))$, then $\mathbf r'(t)=(x'(t),y'(t),z'(t))$, so

\mathbf F\cdot\mathbf r'(t)=-mg\,z'(t).

Thus only the vertical component of the motion contributes to work in this uniform field. Horizontal motion can be part of the trajectory, but it does not change the work done by gravity.

Worked Example: A Vertical Descent

A $2$-kg object descends vertically from $z=5$ m to $z=2$ m. Use $g=10$ m/s$^2$, and parametrize the descent by $\mathbf r(t)=(0,0,5-3t)$ m for $0\le t\le1$.

The gravitational force is $\mathbf F=(0,0,-mg)=(0,0,-2\cdot10)=(0,0,-20)$ N. The velocity with respect to the parameter is $\mathbf r'(t)=(0,0,-3)$ m. Therefore,

\mathbf F\cdot\mathbf r'(t) =(0)(0)+(0)(0)+(-20)(-3) =60\ \text{N}\cdot\text{m}.

Integrating over the parameter interval gives

W=\int_0^1 60\,dt =60[t]_0^1 =60(1-0) =60\ \text{J}.

Check the sign and magnitude using the vertical displacement: the object moves down $3$ m, in the direction of the $20$-N gravitational force. The work is $20\text{ N}\cdot3\text{ m}=60\text{ J}$, matching the integral.

Curved Paths in a Uniform Gravitational Field

The path can curve without changing the force law: in the uniform near-Earth model, gravity still acts vertically downward at every point. The path derivative may have several nonzero components, but the dot product removes those perpendicular to gravity. The next example makes this explicit.

Worked Example: A Curved Path in a Vertical Plane

A $2$-kg object follows the curve $\mathbf r(t)=(t,t^2)$ m in the $xz$-plane for $0\le t\le2$, with the second component representing height $z$. Let $g=10$ m/s$^2$. Find the work done by gravity.

The force and path derivative are $\mathbf F=(0,-mg)=(0,-20)$ N and $\mathbf r'(t)=(1,2t)$ m. Their dot product is

\mathbf F\cdot\mathbf r'(t) =(0)(1)+(-20)(2t) =-40t.

Consequently,

W=\int_0^2 -40t\,dt =-20[t^2]_0^2 =-20(2^2-0^2) =-80\ \text{J}.

To check, the starting height is $0$ m and the ending height is $2^2=4$ m. The object rises $4$ m against a downward force of $20$ N, so the work must be negative and equals $-20\text{ N}\cdot4\text{ m}=-80\text{ J}$. This agrees with the integral.

Newton’s Inverse-Square Gravitational Field

When the gravitational force varies substantially with position, the uniform-field approximation may not be appropriate. For a point source of mass $M$ at the origin and a test mass $m$ at position $\mathbf r\ne\mathbf 0$, Newton’s gravitational force is

\mathbf F(\mathbf r)=-\frac{GMm}{\|\mathbf r\|^3}\mathbf r,

where $G$ is the gravitational constant and $\|\mathbf r\|$ is the distance from the source. This formula points toward the source, and its magnitude is $GMm/\|\mathbf r\|^2$. A trajectory used with this field must not pass through the origin, where the point-source formula is undefined.

Along a path $\mathbf r(t)$, substitution into the work formula gives

W=-GMm\int_a^b \frac{\mathbf r(t)\cdot\mathbf r'(t)} {\|\mathbf r(t)\|^3}\,dt.

The dot product and denominator both vary along the path. It is important to evaluate them using the actual parametrization rather than treating the force as constant.

Worked Example: A Three-Dimensional Curved Trajectory

A test mass moves around a point source along $\mathbf r(t)=R(\cos t,\sin t,t)$ for $0\le t\le1$, where $R>0$ is measured in metres. Suppose the masses and $R$ are chosen so that $GMm/R=10$ J. Find the work done by the source’s gravitational force.

First calculate the squared distance and the derivative of the path:

\|\mathbf r(t)\|^2 =R^2\cos^2t+R^2\sin^2t+R^2t^2 =R^2(1+t^2),

\mathbf r'(t)=R(-\sin t,\cos t,1).

The dot product in the work integrand is

\begin{aligned} \mathbf r(t)\cdot\mathbf r'(t) &=R^2\bigl(-\cos t\sin t+\sin t\cos t+t\bigr)\\ &=R^2t. \end{aligned}

Since $\|\mathbf r(t)\|^3=R^3(1+t^2)^{3/2}$, the integrand becomes

-\frac{GMm\,R^2t}{R^3(1+t^2)^{3/2}} =-\frac{GMm}{R}\frac{t}{(1+t^2)^{3/2}}.

Therefore,

\begin{aligned} W &=-10\int_0^1\frac{t}{(1+t^2)^{3/2}}\,dt\\ &=-10\left[-\frac{1}{\sqrt{1+t^2}}\right]_0^1\\ &=-10\left(-\frac{1}{\sqrt2}+1\right)\\ &=10\left(\frac{1}{\sqrt2}-1\right)\text{ J}. \end{aligned}

For a check, differentiate the antiderivative: $\frac{d}{dt}\bigl[-(1+t^2)^{-1/2}\bigr]=t(1+t^2)^{-3/2}$, so the integration step is correct. The final result is negative because the trajectory’s distance from the source increases from $R$ to $R\sqrt2$; the gravitational force points inward. The computed value is approximately $-2.93$ J, with the approximation rounded to two decimal places.

Common Pitfall: Multiplying Force Magnitude by Path Length

A frequent error is to multiply the magnitude of gravity by the total distance travelled and assign a sign afterward. That ignores the angle between the force and the motion. Work uses a dot product, not the product of magnitudes unless the force and displacement are parallel throughout the motion.

For a concrete example, let an object move on the horizontal circular arc $\mathbf r(t)=(R\cos t,R\sin t,0)$, $0\le t\le\pi/2$, in the uniform gravitational field $\mathbf F=(0,0,-mg)$. The tempting but incorrect calculation treats the force magnitude $mg$ as acting along the arc length $R\,dt$:

W_{\text{wrong}}=-mg\int_0^{\pi/2}R\,dt =-\frac{\pi}{2}mgR.

But the actual path derivative is $\mathbf r'(t)=(-R\sin t,R\cos t,0)$, and the force is vertical. Thus $\mathbf F\cdot\mathbf r'(t)=(0)(-R\sin t)+(0)(R\cos t)+(-mg)(0)=0$, giving $W=\int_0^{\pi/2}0\,dt=0$. The correction is to use the component of force along the motion, not the force magnitude alone.

Check Your Understanding

For each question, identify the force and path information that determines the work before calculating.

In uniform gravity with upward $z$-coordinate, what component of a parametrized path contributes to the work integral?
A $3$-kg object moves downward $2$ m in a uniform field with $g=10$ m/s$^2$. What is the work done by gravity, including its sign?
Why does a horizontal path have zero gravitational work in the uniform near-Earth model, even if the path is curved?
For a point source at the origin, what condition must a trajectory satisfy for the inverse-square force field to remain defined along it?
What is wrong with using force magnitude times total path length as a general formula for work?