Classical-Mechanics · Unit 15 · Video 3 · Interactive Practice

Inverse-Square Forces and the Work-Energy Theorem in Three Dimensions

IKey Formulas

Formula	Name	What you need
$dW = \vec{F}\cdot d\vec{r} = F_r\,dr$	Central-force work element	A force along $\hat{r}$ ; the sideways term $F_r\,r\,d\theta\,(\hat{r}\cdot\hat{\theta})$ dies because $\hat{r}\cdot\hat{\theta} = 0$
$W = C\left(\dfrac{1}{r_f} - \dfrac{1}{r_i}\right)$	Work of an inverse-square force	Only the two radii, plus the constant: $C = G m_s m$ for gravity, $C = -\dfrac{q_1 q_2}{4\pi\varepsilon_0}$ for Coulomb
$\int_i^f \vec{F}\cdot d\vec{r} = K_f - K_i$	Work-energy theorem in three dimensions	The force along the path and the two speeds, $K = \tfrac{1}{2}mv^2$
$P = \vec{F}\cdot\vec{v} = \dfrac{dK}{dt}$	Instantaneous power	Force and velocity at a single instant

Key Insight: Both inverse-square results are the same formula wearing a different constant — and whichever way the force points, a body that moves the way the force pushes it receives positive work. Gravity pulling a comet inward and repulsion driving two like charges apart both give $W > 0$ .

IIVisualization 1 — Which Part of the Step Does the Work

A central force can only work through the radial piece of a step; the sideways piece dots to zero.

Path Position (r) 4.00

💡 Any other path joining the same two radii gives exactly the same work — the integral remembers $r_i$ and $r_f$ and nothing else about the route.

IIIVisualization 2 — Attraction, Repulsion, and the Sign of $W$

Opposite signs or like signs, a charge that moves the way the force pushes it gains energy.

Charges Final separation (r_f) 1.60

💡 Replace $k\,q_1 q_2$ by $-G m_s m$ and this panel becomes the gravitational case: gravity is the inverse-square force whose constant can never change sign.

IVVisualization 3 — Power, and the Force That Bends Without Working

Only the component of force along the velocity changes the speed; the rest merely bends the path.

Angle φ between F and v 60° Time 0.00 s

VQuiz Questions

Problem 1 · Work Done by the Sun's Gravity

Given: A body falls along a curved orbital arc from $r_i = 6.0\times10^{11}\ \text{m}$ to $r_f = 2.0\times10^{11}\ \text{m}$ from the centre of the sun, with $G m_s m = 3.0\times10^{22}\ \text{J}\cdot\text{m}$ — find the work $W$ done by gravity.

✅ Correct! The body moves inward, gravity points inward, and the bracket

\left(\tfrac{1}{r_f} - \tfrac{1}{r_i}\right)

comes out positive — gravity feeds

1.0\times10^{11}\ \text{J}

of kinetic energy in.

❌ Check the order of the reciprocals. The bracket is

\tfrac{1}{r_f} - \tfrac{1}{r_i}

, final first. Since

r_f < r_i

, the reciprocal

\tfrac{1}{r_f}

is the larger one, so the bracket — and the work — is positive.

❌ You dropped the second term.

\tfrac{1}{r_i} = 1.67\times10^{-12}\ \text{m}^{-1}

is not negligible; subtract it from

\tfrac{1}{r_f} = 5.0\times10^{-12}\ \text{m}^{-1}

before multiplying.

❌ The force is not constant. Treating

F

as its final value

Gm_sm/r_f^2 = 0.75\ \text{N}

over a displacement of

4.0\times10^{11}\ \text{m}

overcounts badly — the force was far weaker over most of the fall, which is exactly why the integral is needed.

❌ Not quite. Use

W = G m_s m\left(\tfrac{1}{r_f} - \tfrac{1}{r_i}\right)

with the reciprocals of the two radii.

Show solution

Gravity is central, so the line integral collapses to a one-dimensional integral in $r$ with $F_r = -Gm_sm/r^2$ :

W = \int_{r_i}^{r_f}\left(-\frac{G m_s m}{r^2}\right)dr = \left.\frac{G m_s m}{r}\right|_{r_i}^{r_f} = G m_s m\left(\frac{1}{r_f} - \frac{1}{r_i}\right)

Substituting the numbers:

W = (3.0\times10^{22})\left(\frac{1}{2.0\times10^{11}} - \frac{1}{6.0\times10^{11}}\right) = (3.0\times10^{22})(3.33\times10^{-12}) = 1.0\times10^{11}\ \text{J}

Sign check: $r_f < r_i \Rightarrow \tfrac{1}{r_f} > \tfrac{1}{r_i} \Rightarrow W > 0$ . The force points inward and the body moves inward, so gravity does positive work — the same sign as a stone falling toward the earth.

Note that the shape of the curved arc never entered the calculation: only the two radii did.

Problem 2 · Two Like Charges Flying Apart

Given: $q_1 = +2.0\ \mu\text{C}$ is held fixed while $q_2 = +3.0\ \mu\text{C}$ moves from $r_i = 0.10\ \text{m}$ out to $r_f = 0.30\ \text{m}$ . Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ — find the work done by the electric force on $q_2$ . The formula carries a leading minus sign — and so does the bracket here.

✅ Correct! Two minus signs meet: the bracket is negative because

r_f > r_i

, and the formula's own leading minus flips it back — repulsion driving the charges apart does positive work.

❌ One minus sign too few (or too many). Here

q_1q_2 > 0

and

\tfrac{1}{r_f} - \tfrac{1}{r_i} = 3.33 - 10 = -6.67\ \text{m}^{-1} < 0

; the leading minus in

W = -\tfrac{q_1q_2}{4\pi\varepsilon_0}\left(\tfrac{1}{r_f}-\tfrac{1}{r_i}\right)

turns that product positive.

❌ Both reciprocals belong in the bracket. Compute

\tfrac{1}{0.30} - \tfrac{1}{0.10} = 3.33 - 10 = -6.67\ \text{m}^{-1}

— not one term alone.

❌ Not quite. Apply

W = -\dfrac{q_1q_2}{4\pi\varepsilon_0}\left(\dfrac{1}{r_f} - \dfrac{1}{r_i}\right)

with

q_1q_2 = +6.0\times10^{-12}\ \text{C}^2

Show solution

Coulomb's law gives $F_r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ along $\hat{r}$ , with no built-in minus sign — the sign lives in the charge product. The central-force collapse then gives

W = \int_{r_i}^{r_f}\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\,dr = -\frac{1}{4\pi\varepsilon_0}q_1q_2\left(\frac{1}{r_f} - \frac{1}{r_i}\right)

Step 1 — the prefactor. $q_1q_2 = (2.0\times10^{-6})(3.0\times10^{-6}) = 6.0\times10^{-12}\ \text{C}^2$ , so

\frac{q_1q_2}{4\pi\varepsilon_0} = (9.0\times10^{9})(6.0\times10^{-12}) = 0.054\ \text{J}\cdot\text{m}

Step 2 — the bracket. $\dfrac{1}{0.30} - \dfrac{1}{0.10} = 3.33 - 10 = -6.67\ \text{m}^{-1}$

Step 3 — combine.

W = -(0.054)(-6.67) = +0.36\ \text{J}

Why positive? Like charges repel, so outward is the direction the force pushes. A body moving the way the force pushes it always receives positive work — attraction moving inward and repulsion moving outward give the same verdict.

Problem 3 · Arc, Then Plunge — Work and Final Speed

Given: A $1000\ \text{kg}$ probe runs along a frictionless guide that first sweeps a quarter circle at fixed radius $r_i = 3.0\times10^{11}\ \text{m}$ about a star, then runs radially inward to $r_f = 1.0\times10^{11}\ \text{m}$ . The guide always pushes perpendicular to the motion, so it does no work. With $G m_s m = 2.4\times10^{23}\ \text{J}\cdot\text{m}$ and an initial speed $v_i = 2.0\times10^{4}\ \text{m/s}$ — find the work done by gravity and the final speed.

Total work done by gravity?

Final speed?

✅ Correct! The quarter circle contributes nothing, the plunge contributes

1.6\times10^{12}\ \text{J}

, and the work-energy theorem turns that into a final speed of

6.0\times10^{4}\ \text{m/s}

❌ Check the work. The circular arc holds

r

fixed, so

dr = 0

and it contributes zero; only the change of radius counts, through

G m_s m\left(\tfrac{1}{r_f} - \tfrac{1}{r_i}\right)

with both reciprocals.

❌ Check the speed. The theorem gives

K_f = K_i + W

— the probe was already moving, and

K = \tfrac{1}{2}mv^2

carries the factor

\tfrac{1}{2}

on both ends.

Show solution

Step 1 — the quarter circle. On it $r$ never changes, so $dr = 0$ and $dW = F_r\,dr = 0$ at every point. Gravity stays perpendicular to the motion and does no work on that leg.

Step 2 — the radial plunge.

W = G m_s m\left(\frac{1}{r_f} - \frac{1}{r_i}\right) = (2.4\times10^{23})\left(\frac{1}{1.0\times10^{11}} - \frac{1}{3.0\times10^{11}}\right)

W = (2.4\times10^{23})(6.67\times10^{-12}) = 1.6\times10^{12}\ \text{J}

Step 3 — the work-energy theorem. The guide does no work, so gravity's work is the total:

K_i = \tfrac{1}{2}(1000)(2.0\times10^{4})^2 = 2.0\times10^{11}\ \text{J}

K_f = K_i + W = 2.0\times10^{11} + 1.6\times10^{12} = 1.8\times10^{12}\ \text{J}

Step 4 — back to a speed.

v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(1.8\times10^{12})}{1000}} = \sqrt{3.6\times10^{9}} = 6.0\times10^{4}\ \text{m/s}

Dropping $K_i$ would have given $5.7\times10^{4}\ \text{m/s}$ , and forgetting the $\tfrac{1}{2}$ in $K$ would have given $4.5\times10^{4}\ \text{m/s}$ .

Problem 4 · A Force That Does Nothing At All

Given: At one instant a $2.0\ \text{kg}$ drone has velocity $\vec{v} = (6.0\,\hat{i} - 3.0\,\hat{j})\ \text{m/s}$ while the net force on it is $\vec{F} = (4.0\,\hat{i} + 8.0\,\hat{j})\ \text{N}$ — find the instantaneous power and say what is happening to the drone's speed.

Instantaneous power?

The drone's speed at this instant is…

✅ Excellent!

\vec{F}\cdot\vec{v} = 0

, so the force is perpendicular to the velocity: it bends the path hard while leaving the speed untouched.

❌ Watch the sign of the $\hat{j}$ term.

v_y = -3.0\ \text{m/s}

, so

F_yv_y = (8.0)(-3.0) = -24\ \text{W}

— it cancels the

x

term instead of doubling it.

❌ That is $|\vec{F}||\vec{v}|$ . The scalar product carries a

\cos\theta

P = |\vec{F}||\vec{v}|\cos\theta = (8.94)(6.71)\cos 90\degree = 0

. Multiplying magnitudes alone assumes the vectors are parallel.

❌ Not quite. Use components:

P = \vec{F}\cdot\vec{v} = F_xv_x + F_yv_y

❌ Not quite. The theorem governs speed through energy, not through the size of the force:

dK/dt = \vec{F}\cdot\vec{v}

, and here that rate is zero even though

\vec{F}

is large and

\vec{v}

is nonzero.

Show solution

Step 1 — the power.

P = \vec{F}\cdot\vec{v} = F_xv_x + F_yv_y = (4.0)(6.0) + (8.0)(-3.0) = 24 - 24 = 0\ \text{W}

Step 2 — read it as a rate of energy. Since $\dfrac{dK}{dt} = \vec{F}\cdot\vec{v} = P = 0$ , the kinetic energy — and therefore the speed — is momentarily unchanging.

Step 3 — what the force is doing. Neither vector is zero: $|\vec{v}| = \sqrt{36+9} = 6.71\ \text{m/s}$ and $|\vec{F}| = \sqrt{16+64} = 8.94\ \text{N}$ . Their scalar product vanishes only because they are perpendicular, so the whole force goes into turning the velocity:

a_\perp = \frac{|\vec{F}|}{m} = \frac{8.94}{2.0} = 4.5\ \text{m/s}^2

This is the punchline of the three-dimensional theorem: work governs speed, not direction. A force can bend a path all day and never transfer a joule — which is exactly why gravity does zero work on a circular orbit.

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Inverse-Square Forces and the Work-Energy Theorem in Three Dimensions

IKey Formulas

IIVisualization 1 — Which Part of the Step Does the Work

IIIVisualization 2 — Attraction, Repulsion, and the Sign of WWW

IVVisualization 3 — Power, and the Force That Bends Without Working

VQuiz Questions

IIIVisualization 2 — Attraction, Repulsion, and the Sign of $W$