Multivariable-Calculus · Unit 6 · Video 4 · Interactive Practice

Equal Areas in Equal Times: Kepler's Second Law and the Length of $\mathbf{r} \times \mathbf{v}$

IKey Formulas

Formula	Step	Where it comes from
$\Delta A \approx \tfrac{1}{2}\,\|\mathbf{r} \times \Delta\mathbf{r}\|$	Wedge $\approx$ triangle	The triangle with sides $\mathbf{r}$ and $\Delta\mathbf{r}$ is half the parallelogram, whose area is $\|\mathbf{r} \times \Delta\mathbf{r}\|$
$\Delta A \approx \tfrac{1}{2}\,\|\mathbf{r} \times \mathbf{v}\|\,\Delta t$	Bring in velocity	$\Delta\mathbf{r} \approx \mathbf{v}\,\Delta t$ , and the positive scalar $\Delta t$ pulls out of the cross product and the length
$\dfrac{dA}{dt} = \tfrac{1}{2}\,\|\mathbf{r} \times \mathbf{v}\|$	Area rate	Divide by $\Delta t$ and let $\Delta t \to 0$
$\|\mathbf{r} \times \mathbf{v}\| = \|\mathbf{r}\|\,\|\mathbf{v}\|\sin\theta = \|\mathbf{r}\|\,v_\perp$	Near fast, far slow	$v_\perp = \|\mathbf{v}\|\sin\theta$ is the part of $\mathbf{v}$ perpendicular to $\mathbf{r}$

Key Insight: Equal areas in equal times $\iff$ $|\mathbf{r} \times \mathbf{v}|$ is the same at every point of the orbit, even though $\mathbf{r}$ and $\mathbf{v}$ themselves keep changing.

IIVisualization 1 — Equal Times, Equal Areas

The sun–planet line sweeps area at a constant rate, so wedges between equally spaced times all have equal area.

Time t 3.50

IIIVisualization 2 — Wedge, Triangle, Half a Parallelogram

As $\Delta t$ shrinks, the swept wedge approaches the triangle with sides $\mathbf{r}$ and $\Delta\mathbf{r}$ , of area $\tfrac{1}{2}|\mathbf{r} \times \Delta\mathbf{r}|$ .

Δt 1.00 Start

IVVisualization 3 — Short $|\mathbf{r}|$ , Fast Sideways

Around the orbit $|\mathbf{r}|$ and the perpendicular part $v_\perp$ trade off, while their product $|\mathbf{r} \times \mathbf{v}|$ never changes.

Time t 1.00

💡 This is the area half of Kepler's second law, and it fixes only the length of $\mathbf{r} \times \mathbf{v}$ . The other half, motion in a plane, fixes its direction.

VQuiz Questions

Problem 1 · The Area Rate

Given: at some instant a planet has position $\mathbf{r} = \langle 3, 0, 0\rangle$ and velocity $\mathbf{v} = \langle 1, 2, 0\rangle$ , with the sun at the origin — find the rate $\dfrac{dA}{dt}$ at which the sun–planet line sweeps out area.

✅ Correct!

\mathbf{r} \times \mathbf{v} = \langle 0, 0, 6\rangle

, and the swept region is the triangle, half the parallelogram:

\tfrac{1}{2} \cdot 6 = 3

❌ The factor of $\tfrac{1}{2}$ got lost.

|\mathbf{r} \times \mathbf{v}| = 6

is the parallelogram rate; the swept wedge is only the triangle, half of it.

❌ That is $\tfrac{1}{2}|\mathbf{r}|\,|\mathbf{v}|$ with no $\sin\theta$ .

\mathbf{r}

and

\mathbf{v}

are not perpendicular here, so only the perpendicular part of

\mathbf{v}

sweeps area.

❌ That uses the dot product.

\tfrac{1}{2}\,\mathbf{r} \cdot \mathbf{v} = \tfrac{3}{2}

measures the motion along

\mathbf{r}

, which sweeps no area; the area comes from the cross product.

Show solution

Compute the cross product:

\mathbf{r} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 0 \\ 1 & 2 & 0 \end{vmatrix} = \langle 0\cdot 0 - 0\cdot 2,\ 0\cdot 1 - 3\cdot 0,\ 3\cdot 2 - 0\cdot 1\rangle = \langle 0, 0, 6\rangle

Its length is $6$ , the area rate of the parallelogram spanned by $\mathbf{r}$ and $\mathbf{v}$ . The swept triangle is half of that:

\frac{dA}{dt} = \tfrac{1}{2}\,|\mathbf{r} \times \mathbf{v}| = \tfrac{1}{2} \cdot 6 = 3

Problem 2 · Perihelion to Aphelion

Given: a comet passes closest to the sun at distance $1$ AU with speed $40$ km/s; at its farthest point it is $5$ AU away. At both points its velocity is perpendicular to $\mathbf{r}$ — find its speed at the farthest point.

✅ Correct!

|\mathbf{r}|\,v_\perp

is fixed:

1 \cdot 40 = 5 \cdot v

, so

v = 8

km/s — five times as far, one fifth as fast.

❌ That divides by $5^2$ . The conserved quantity is

|\mathbf{r}|\,v_\perp

, linear in

|\mathbf{r}|

: five times farther means one fifth of the perpendicular speed, not one twenty-fifth.

❌ The area rate is constant, not the speed. Far from the sun the wedge is long and thin, so the planet covers less orbit in the same time.

❌ Backwards. Keeping

|\mathbf{r}|\,v_\perp

fixed while

|\mathbf{r}|

grows fivefold forces

v_\perp

to shrink fivefold.

Show solution

Equal areas in equal times means $|\mathbf{r} \times \mathbf{v}| = |\mathbf{r}|\,|\mathbf{v}|\sin\theta$ is the same everywhere. At both points $\theta = 90\degree$ , so $\sin\theta = 1$ :

|\mathbf{r}_1|\,|\mathbf{v}_1| = |\mathbf{r}_2|\,|\mathbf{v}_2| \quad\Longrightarrow\quad 1 \cdot 40 = 5\,|\mathbf{v}_2|

|\mathbf{v}_2| = \frac{40}{5} = 8 \text{ km/s}

(Mixing AU with km/s is harmless: the distance units cancel in the ratio $1/5$ .)

Problem 3 · From Two Positions to the Rate

Given: a planet is at $\mathbf{r} = \langle 2, 0\rangle$ and, a short time $\Delta t = 0.1$ later, at $\langle 1.9,\ 0.3\rangle$ (sun at the origin) — estimate the area swept in that time, then the area rate $\dfrac{dA}{dt}$ .

Area swept in $\Delta t = 0.1$

Area rate

✅ Correct!

\Delta\mathbf{r} = \langle -0.1, 0.3\rangle

, the triangle has area

\tfrac{1}{2}|\mathbf{r} \times \Delta\mathbf{r}| = 0.3

, and dividing by

\Delta t = 0.1

gives a rate of

3

❌ That is the whole parallelogram.

|\mathbf{r} \times \Delta\mathbf{r}| = 0.6

; the swept wedge is close to the triangle, half of it.

❌ That is $\tfrac{1}{2}|\mathbf{r}|\,|\Delta\mathbf{r}|$ . It ignores the angle between

\mathbf{r}

and

\Delta\mathbf{r}

; the cross product keeps only the part of

\Delta\mathbf{r}

perpendicular to

\mathbf{r}

❌ That comes from the dot product.

\tfrac{1}{2}|\mathbf{r} \cdot \Delta\mathbf{r}| = 0.1

measures motion along

\mathbf{r}

; area needs

\mathbf{r} \times \Delta\mathbf{r}

❌ Check the swept area. Find

\Delta\mathbf{r}

first, then take half the length of

\mathbf{r} \times \Delta\mathbf{r}

❌ Multiplied instead of divided.

\Delta A \approx \tfrac{dA}{dt}\,\Delta t

, so the rate is

\Delta A / \Delta t = 0.3 / 0.1

❌ That is $|\mathbf{r} \times \mathbf{v}|$ , twice the area rate. The rate is

\tfrac{1}{2}|\mathbf{r} \times \mathbf{v}| = \Delta A / \Delta t = 0.3 / 0.1

❌ That divides by $\Delta t$ twice.

0.3 / 0.1^2 = 30

; the rate is the swept area over the elapsed time once,

\Delta A / \Delta t = 0.3 / 0.1

❌ Check the rate. Divide the swept area by the elapsed time:

\Delta A / \Delta t

Show solution

Step 1: the change in position is

\Delta\mathbf{r} = \langle 1.9 - 2,\ 0.3 - 0\rangle = \langle -0.1,\ 0.3\rangle

Step 2: for plane vectors the cross product has only a $\mathbf{k}$ component, $x_1 y_2 - y_1 x_2$ :

\mathbf{r} \times \Delta\mathbf{r} = \big(2 \cdot 0.3 - 0 \cdot (-0.1)\big)\,\mathbf{k} = 0.6\,\mathbf{k}, \qquad \Delta A \approx \tfrac{1}{2}\,|\mathbf{r} \times \Delta\mathbf{r}| = 0.3

Step 3: divide by the elapsed time:

\frac{dA}{dt} \approx \frac{\Delta A}{\Delta t} = \frac{0.3}{0.1} = 3

Equivalently, $\mathbf{v} \approx \Delta\mathbf{r}/\Delta t = \langle -1, 3\rangle$ and $\tfrac{1}{2}|\mathbf{r} \times \mathbf{v}| = \tfrac{1}{2}\,|2 \cdot 3 - 0 \cdot (-1)| = 3$ .

Problem 4 · When v Is Not Perpendicular

Given: at one point of its orbit a planet is $|\mathbf{r}| = 6$ from the sun, moving with speed $|\mathbf{v}| = 2$ at an angle $\theta = 30\degree$ to $\mathbf{r}$ . At its closest point, $|\mathbf{r}| = 1.5$ and $\mathbf{v} \perp \mathbf{r}$ — find its speed there.

✅ Correct!

|\mathbf{r} \times \mathbf{v}| = 6 \cdot 2 \cdot \sin 30\degree = 6

at the first point, so at the closest point

1.5\,|\mathbf{v}| = 6

and

|\mathbf{v}| = 4

❌ The $\sin\theta$ is missing. Only the perpendicular part of the velocity,

2 \sin 30\degree = 1

, counts toward

|\mathbf{r} \times \mathbf{v}|

❌ That uses $\cos\theta$ .

|\mathbf{v}|\cos\theta

is the part of

\mathbf{v}

along

\mathbf{r}

; the cross product needs the perpendicular part,

|\mathbf{v}|\sin\theta

❌ Speed is not proportional to distance. The product

|\mathbf{r}|\,v_\perp

is constant, so a closer planet moves sideways faster.

Show solution

At the first point:

|\mathbf{r} \times \mathbf{v}| = |\mathbf{r}|\,|\mathbf{v}|\sin\theta = 6 \cdot 2 \cdot \sin 30\degree = 6 \cdot 2 \cdot \tfrac{1}{2} = 6

By Kepler's second law this length is the same at the closest point, where $\theta = 90\degree$ :

1.5 \cdot |\mathbf{v}| \cdot 1 = 6 \quad\Longrightarrow\quad |\mathbf{v}| = 4

Note that $v_\perp$ went from $1$ to $4$ as $|\mathbf{r}|$ went from $6$ to $1.5$ : a factor of $4$ each way.

Solved: 0 / 4

Equal Areas in Equal Times: Kepler's Second Law and the Length of r×v\mathbf{r} \times \mathbf{v}r×v

IKey Formulas

IIVisualization 1 — Equal Times, Equal Areas

IIIVisualization 2 — Wedge, Triangle, Half a Parallelogram

IVVisualization 3 — Short ∣r∣|\mathbf{r}|∣r∣, Fast Sideways

VQuiz Questions

Equal Areas in Equal Times: Kepler's Second Law and the Length of $\mathbf{r} \times \mathbf{v}$

IVVisualization 3 — Short $|\mathbf{r}|$ , Fast Sideways