Visual walkthrough — Kepler's second law — equal areas in equal times, from angular momentum conservation

This is the picture-companion to the parent note. Same result, but here the figures carry the proof.

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Step 1 — What an arrow from the Sun even means

WHAT. Draw the Sun as a dot. The planet is another dot. The straight arrow from the Sun to the planet is called the radius vector, written $\vec{r}$. The little arrow-hat on top just means "this is a thing with a length AND a direction," not just a number.

WHY. Everything in this law is about the area painted by this one arrow as the planet moves. So before area, before speed, we must be crystal-clear on what arrow we are watching. Its length $r$ is the Sun–planet distance; its direction is "which way the planet lies."

PICTURE. In the figure, the burnt-orange arrow is $\vec{r}$. The dotted circle is not the orbit — it is just there to show that the arrow can point any direction. The planet sits at the arrowhead.

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Step 2 — In a tiny instant, the arrow sweeps a thin triangle

WHAT. Let a very short time $dt$ pass ($d$ = "a tiny bit of"). The planet nudges forward, so the radius vector becomes $\vec{r}+d\vec{r}$. Here $d\vec{r}$ is the tiny displacement arrow — the little step the planet took. The region the arrow swept between "before" and "after" is a skinny triangle.

WHY. Kepler's law is about area swept. The smallest possible swept region is this one thin triangle. If we can find its area, we can add up triangles to get any swept area. So we zoom all the way in.

PICTURE. Two orange arrows fan out from the Sun: the old $\vec{r}$ and the new $\vec{r}+d\vec{r}$. The teal arrow between their tips is $d\vec{r}$. The shaded plum sliver is the swept area $dA$. Notice it is a genuine triangle: two straight edges from the Sun, one short edge $d\vec{r}$ closing it.

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Step 3 — The cross product: a machine that outputs triangle area

WHAT. For any two arrows $\vec{a}$ and $\vec{b}$ starting at the same point, there is an operation written $\vec{a}\times \vec{b}$ (say "a cross b") whose size equals the area of the parallelogram they span. Half of that parallelogram is the triangle. So:

$dA=\frac{1}{2}|\underset{\text{one edge}}{\underbrace{\vec{r}}}\times \underset{\text{other edge}}{\underbrace{d\vec{r}}}|$

The vertical bars $||$ mean "the length/size of," throwing away direction and keeping the positive area number. The $\frac{1}{2}$ turns the parallelogram into our triangle.

WHAT — the direction, too. A cross product is not just a size; it is itself an arrow, and we must say which way it points. The rule: $\vec{a}\times \vec{b}$ points perpendicular to the flat sheet containing $\vec{a}$ and $\vec{b}$ — straight out of the page or straight into it. Which of the two? The right-hand rule: point your right hand's fingers along the first arrow $\vec{a}$, then curl them toward the second arrow $\vec{b}$; your thumb then points along $\vec{a}\times \vec{b}$. Sweeping counter-clockwise (as $\vec{r}$ swings toward $d\vec{r}$ for a normal orbit) gives a thumb pointing out of the page — that is why the angular-momentum arrow in Step 5 sticks up out of the orbital plane. For area alone we only use the size $|\vec{a}\times \vec{b}|$; the direction becomes important the moment we call it $\vec{L}$.

WHY this tool and not just base×height? Because base×height forces us to find the perpendicular height each instant — annoying as the arrows tilt. The cross product does it automatically: $|\vec{a}\times \vec{b}|=ab\sin \varphi$, where $\varphi$ is the angle between the arrows and $\sin \varphi$ already extracts the perpendicular part. One formula, every angle, every quadrant. That is exactly why we reach for it. (More at Cross Product and Area.)

PICTURE. The full parallelogram is outlined faintly; the plum triangle is the half we want. The right-angle mark shows the height $b\sin \varphi$ that the cross product secretly computes. The small circled dot marks $\vec{a}\times \vec{b}$ coming out of the page by the right-hand rule.

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Step 4 — Divide by time to get area per second

WHAT. Kepler cares about area per second, so divide the tiny triangle by the tiny time:

$\frac{dA}{dt}=\frac{1}{2}\left|\vec{r}\times \frac{d\vec{r}}{dt}\right|=\frac{1}{2}|\vec{r}\times \vec{v}|$

The piece $\frac{d\vec{r}}{dt}$ is "tiny step divided by tiny time" — that is exactly the velocity $\vec{v}$: how fast and in which direction the planet moves. We slid the $1/dt$ inside the cross product because the cross product treats each arrow linearly (scaling one arrow scales the answer).

WHY. This converts a static triangle into a rate — the areal velocity, the quantity Kepler claims is constant. Now the fixed pieces are $\vec{r}$ and $\vec{v}$, both real physical arrows we can track.

PICTURE. The teal velocity arrow $\vec{v}$ is drawn tangent to the orbit at the planet. The swept-per-second triangle is $\frac{1}{2}|\vec{r}\times \vec{v}|$ — same shape as before, now measured against the clock.

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Step 5 — Rename the cross product: angular momentum enters

WHAT. Physicists already have a name for $m$ (the planet's mass) times $\vec{r}\times \vec{v}$. It is the angular momentum, a vector written $\vec{L}$:

$\vec{L}=\vec{r}\times m\vec{v}⟹\vec{r}\times \vec{v}=\frac{\vec{L}}{m}.$

Our area formula from Step 4 uses the size $|\vec{r}\times \vec{v}|$, so we take the size of both sides. Write $L\equiv |\vec{L}|$ for that length (a plain positive number — the magnitude of the vector $\vec{L}$). Then $|\vec{r}\times \vec{v}|=L/m$, and substituting into Step 4:

$\boxed{\frac{dA}{dt}=\frac{1}{2}\cdot \frac{L}{m}=\frac{L}{2m}}$

Term by term: $L=|\vec{L}|$ = the length of the spin-vector (a scalar), $m$ = planet mass (a fixed number), so $L/2m$ is "half the spin per mass." Only lengths appear here, so everything on the right is an ordinary number.

WHY this rename? Because $\vec{L}$ is famous for a reason we exploit next: under the right condition it cannot change. If $L=|\vec{L}|$ is frozen and $m$ is fixed, then $L/2m$ is frozen — and that is Kepler's second law. We have converted a geometry question into a physics conservation question. (See Angular Momentum Conservation.)

PICTURE. $\vec{L}$ is drawn as a plum arrow pointing out of the orbital plane — that "out of page" direction is exactly the right-hand-rule output from Step 3, applied to $\vec{r}$ curling toward $\vec{v}$. Its length is $L=mrv\sin \varphi$.

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Step 6 — The central force twists nothing (torque = 0)

WHAT. Ask how $\vec{L}$ changes in time. We differentiate the product $\vec{r}\times m\vec{v}$ (the product rule works for cross products too, keeping the order):

$\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times m\vec{v})=\underset{=\vec{v}\times m\vec{v}=0}{\underbrace{\frac{d\vec{r}}{dt}\times m\vec{v}}}+\vec{r}\times m\frac{d\vec{v}}{dt}=\vec{r}\times m\vec{a}=\vec{r}\times \vec{F}$

Symbol by symbol: $\frac{d\vec{r}}{dt}=\vec{v}$ (Step 4's definition), so the first piece is $\vec{v}\times m\vec{v}$; and $\frac{d\vec{v}}{dt}=\vec{a}$ is the acceleration just defined, so the second piece is $\vec{r}\times m\vec{a}$. The first term dies because $\vec{v}\times \vec{v}=0$ (an arrow crossed with itself spans zero area — angle $\varphi =0$, $\sin 0=0$).

Bringing in force. Newton's second law says the net force equals mass times acceleration, $\vec{F}=m\vec{a}$ — so wherever we see $m\vec{a}$ we may write $\vec{F}$. That is the single substitution that turns $\vec{r}\times m\vec{a}$ into $\vec{r}\times \vec{F}$, our torque, a vector we write $\vec{\tau }$ ("tau"):

$\vec{\tau }\equiv \vec{r}\times \vec{F}(\text{the “twisting” a force applies about the Sun}).$

So the box above says simply $\frac{d\vec{L}}{dt}=\vec{\tau }$: angular momentum changes only when something twists it.

Now gravity's force is $\vec{F}=-\frac{GMm}{r^2}\hat{r}$, pointing straight back along $\vec{r}$ toward the Sun ($\hat{r}$ = unit arrow along $\vec{r}$; the minus = "inward"). Two parallel arrows span no area, so

$\vec{\tau }=\vec{r}\times \vec{F}=0⟹\frac{d\vec{L}}{dt}=0⟹\vec{L}=\text{const}\Rightarrow L=|\vec{L}|=\text{const}.$

WHY. This is the whole secret. A central force (one aimed at the center) can never twist the planet about that center, so it can never bleed away angular momentum. No leak $\Rightarrow$ $\vec{L}$ frozen $\Rightarrow$ area rate frozen. (See Central Forces and Torque.)

PICTURE. The orange $\vec{r}$ and the red gravity arrow $\vec{F}$ lie along the same line (opposite directions). The angle between them is $180^{\circ }$; $\sin 180^{\circ }=0$; so the torque $\vec{\tau }$ — their cross product — is a zero-length arrow, drawn as a tiny crossed-out dot.

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Step 7 — Naming the swept angle, then every case

WHAT — first, define the angle. Instead of tracking $\vec{r}$ and $\vec{v}$ separately, mark the direction $\vec{r}$ points as a plain angle from a fixed reference line. Call it $\theta$ (the polar angle): it is the planet's angular position, like the hour-hand reading on a clock centred on the Sun. As the planet orbits, $\theta$ grows.

In a tiny time the arrow swings by a tiny angle $d\theta$, and the thin triangle of Step 2 becomes a slice of a circle of radius $r$: base $\approx rd\theta$ (arc), height $r$, so

$dA=\frac{1}{2}\underset{\text{height}}{\underbrace{r}}\cdot \underset{\text{arc base}}{\underbrace{rd\theta }}=\frac{1}{2}{r}^{2}d\theta .$

Why is the base $rd\theta$? The true swept region is bounded by the slightly curved arc, but over an infinitesimally small angle $d\theta$ the arc and the straight chord differ only at second order in $d\theta$ (the gap shrinks like $d{\theta }^2$, far faster than $d\theta$ itself). To leading order they are identical, so we may use the clean arc length $rd\theta$ — the exact result appears once we integrate. This is the same "zoom in until curves look straight" move that makes all of calculus work.

Now write $\dot{\theta }\equiv \frac{d\theta }{dt}$ — the little dot on top means "rate of change in time," so $\dot{\theta }$ is how fast the angle sweeps (angular speed). Dividing $dA$ by $dt$:

$\frac{dA}{dt}=\frac{1}{2}{r}^2\dot{\theta }.$

Why does this match Step 5? The part of the velocity that is perpendicular to $\vec{r}$ (the only part that sweeps area) is exactly $v\sin \varphi$, and geometrically that perpendicular speed equals the arc rate $r\dot{\theta }$. So $v\sin \varphi =r\dot{\theta }$, and multiplying by $mr$:

$L=mrv\sin \varphi =m{r}^2\dot{\theta }.$

Both forms of $L$ agree — one written with the angle-between-arrows $\varphi$, the other with the sweep-rate $\dot{\theta }$.

WHY (all cases). With $\frac{1}{2}{r}^2\dot{\theta }=L/2m$ fixed everywhere, we must check it survives every orbit position:

• Perihelion (closest): $r$ small, so $\dot{\theta }$ must be large — a short, fat triangle. Fastest here.
• Aphelion (farthest): $r$ large, so $\dot{\theta }$ small — a long, thin triangle. Slowest here.
• At the apsides, the velocity is exactly perpendicular to $\vec{r}$ ($\varphi =90^{\circ }$, $\sin \varphi =1$), so $L=mvr$ cleanly — the only place you may drop the $\sin \varphi$. Hence $v_p{r}_p=v_a{r}_a$.
• Anywhere else $\varphi \ne 90^{\circ }$, so you MUST keep $L=mrv\sin \varphi =m{r}^2\dot{\theta }$. Using $L=mvr$ off-apsis is the classic error.
• Degenerate case — radial fall ($L=0$): if the planet were dropped straight at the Sun, $\vec{v}\parallel \vec{r}$, so $\vec{r}\times \vec{v}=0$, $\dot{\theta }=0$, and $dA/dt=0$. The law still holds — it sweeps zero area per second, consistently, forever. No contradiction; just the boundary value.

PICTURE. Three plum slices drawn on one ellipse — fat near the Sun, medium at the side, long-and-thin far away — all labeled equal area. The angle $\theta$ of $\vec{r}$ from the reference line and the arrow's angle $\varphi$ to $\vec{v}$ are both marked; $\varphi =90^{\circ }$ only at the two apsides.

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The one-picture summary

Everything above, compressed: a central force $\Rightarrow$ zero torque $\Rightarrow$ constant $\vec{L}$ $\Rightarrow$ constant $dA/dt=L/2m$ $\Rightarrow$ equal areas in equal times.

Recall Feynman retelling of the whole walkthrough

Picture a rubber band from the Sun to a planet. Every second, that band paints a little triangle of area. In Step 1 we named the band $\vec{r}$. In Step 2 we watched it paint one thin triangle in a blink of time. In Step 3 we found a magic ruler — the cross product — that measures any triangle's area no matter how it's tilted, and points its answer straight out of the page by the right-hand rule. In Step 4 we asked "how much area per second?" and that pulled in the planet's speed. In Step 5 we noticed that "area per second" is really just the length of the planet's spin-vector $\vec{L}$ in disguise, divided by twice its mass. Then the punchline, Step 6: gravity always yanks the planet straight toward the Sun (Newton's $\vec{F}=m\vec{a}$ points inward), and a straight-toward pull can never twist it about the Sun — the torque is zero — so the spin-amount is locked. Step 7 gave the sweep a name, the angle $\theta$: fat slices when it's close and whirling ($\dot{\theta }$ big), skinny slices when it's far and dawdling ($\dot{\theta }$ small) — but always, always the same amount of pie. That's the whole law, and it never needed to know that gravity was inverse-square; it only needed gravity to point at the Sun.

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Connections

• Angular Momentum Conservation — the conserved quantity doing all the work
• Central Forces and Torque — why $\vec{r}\times \vec{F}=0$
• Cross Product and Area — the triangle-area machine of Step 3
• Kepler's First Law (Ellipse) — where inverse-square does matter
• Kepler's Third Law (T² ∝ a³) — uses $\pi ab=(L/2m)T$
• Vis-viva Equation — the speed at any radius