Kepler's Second Law: Equal Areas

Kepler's Second Law is the odd sibling of the three. The First and Third Laws are about ellipses — shapes and sizes special to the inverse-square force. The Second Law says something that sounds almost bureaucratic: the line from the Sun to a planet sweeps out equal areas in equal times. Kepler found it in 1602, before the ellipse, while fighting through Tycho Brahe's Mars data — and it turned out to be the deepest of the three, because it is not really about planets at all. In modern language the equal-area law is angular momentum conservation wearing geometric clothing. A central force points along the line to the force centre, so it exerts zero torque, so $\mathbf L$ is conserved — and the areal velocity is $L/2m$ , a constant, whatever the force law's radial profile. Swap gravity for a spring, a Yukawa force, or the attraction of a charged sphere: orbits change shape completely, but the equal-area rule survives untouched. Where the Kepler problem article uses this fact in passing — as equation (2) of its machinery — this article makes it the star: we derive $\mathrm{d}A/\mathrm{d}t=L/2m$ completely, from the swept triangle and from the vanishing torque, and then put it to work. The law's practical voice is speed variation. A planet must race through perihelion and crawl through aphelion in exact inverse proportion to its distance, which is why comets vanish into the outer dark for decades, why the northern-hemisphere summer is a week longer than winter, and why mission designers time engine burns to the fast point of an orbit.

Key takeaways

Kepler's Second Law: the line joining a planet to the Sun sweeps out equal areas in equal times — the areal velocity $\mathrm{d}A/\mathrm{d}t$ is constant.
It is angular momentum conservation in geometric form: $\mathrm{d}A/\mathrm{d}t=L/2m$ , and $L$ is conserved because a central force exerts zero torque.
The law is universal across central forces — nothing in the derivation uses the inverse-square form, unlike the First and Third Laws.
Speed and distance trade off exactly: at the apsides $v_p r_p=v_a r_a$ , so a comet with $r_a=60\,r_p$ moves 60 times slower at aphelion.
Integrated over one orbit, the law gives $T=2\pi m ab/L$ — the bridge that leads directly to Kepler's Third Law.
Key idea: Equal areas in equal times is the statement $r^2\dot\theta = L/m = \text{const}$ . It follows solely from zero torque (central force), so it is universal across all central forces and is the geometric face of angular momentum conservation.

Where this lives today

Every interplanetary mission is choreographed around the Second Law's speed–distance trade. Burns that change orbital energy are cheapest where the spacecraft moves fastest (the Oberth effect), so insertion and escape manoeuvres are scheduled at periapsis; long coasts and science campaigns are parked near apoapsis, where the craft loiters almost for free. ESA's JUICE, en route to Jupiter, threads a chain of gravity assists — including its Earth flyby of September 2026 — each one a hyperbolic orbit whose incoming and outgoing legs obey the same conserved sweep rate about the assisting planet. The law also runs the astrophysics of things we cannot visit. The star S2, orbiting the Milky Way's central black hole on a 16-year, $e=0.88$ ellipse, was clocked at $7{,}700\ \mathrm{km/s}$ at pericentre in 2018 — the equal-area speed-up, observed at the galaxy's heart and used to test general relativity in the strongest field available.

The physical picture

Picture the Sun–planet line as the hand of a strange clock: long and slow near aphelion, short and fast near perihelion. The Second Law is the statement that this hand paints area at a perfectly steady rate. Here is one month of painting at two different places on an $e=0.6$ orbit:

Figure

Equal areas in equal times — one month near perihelion covers 70° of travel, one month near aphelion only 4.6°

The two shaded wedges are computed to have identical areas, yet they could hardly look more different: near perihelion the planet covers $70^\circ$ of true anomaly in the time it takes to cover $4.6^\circ$ near aphelion — a factor of fifteen in angular rate, because the sweep rate goes as $r^2\dot\theta$ and the radius differs by a factor of four between the two ends. The law is a budget rule: the orbit is allotted a fixed area-per-second, and the planet trades speed against distance to keep paying it. Two features of this picture deserve emphasis. First, the areas are measured from the Sun — from the focus, not the centre of the ellipse; the law is about the radius vector of the force. Second, nothing in the picture depends on the curve being an ellipse. Draw any orbit produced by any central force — a precessing rosette, a hyperbolic flyby, a spiral in a modified potential — and the equal-area rule holds wedge for wedge, because it is bookkeeping for a conserved quantity, not a property of a shape.

From swept triangles to a conservation law

Start with pure geometry: how much area does the radius vector cover in an instant $\mathrm{d}t$ ? The planet moves by $\mathrm{d}\mathbf r$ , and the region swept is the sliver of triangle between the old and new position vectors. Before any dynamics enters, this is a statement about triangles:

\mathrm{d}A=\tfrac12\,r^2\,\mathrm{d}\theta \qquad (1)

where: $r$ — distance from the force centre; $\mathrm{d}\theta$ — the angle turned through in time $\mathrm{d}t$ ; $\mathrm{d}A$ — the swept area. The $\tfrac12$ is the triangle's half, and only the tangential part of the displacement matters — sliding directly toward or away from the Sun sweeps no area at all. The first derivation below builds this from the cross product.

Figure

The swept sliver — only the tangential part of the displacement paints area

Now the dynamics. The angular momentum of the orbiting body about the force centre, written in the same polar coordinates, picks out exactly the combination that appeared in the triangle:

L=m\,r^2\dot\theta \qquad (2)

where: $m$ — the orbiting mass (for a true two-body problem, the reduced mass $\mu=m_1m_2/(m_1+m_2)$ of the relative motion, as set up in the Kepler problem article); $\dot\theta$ — the angular velocity. Only the tangential velocity $r\dot\theta$ enters, for the same cross-product reason that only tangential displacement sweeps area — $L$ and $\mathrm{d}A/\mathrm{d}t$ are the same geometry, one with velocity, one with displacement. Divide (1) by $\mathrm{d}t$ and compare with (2), and the two statements fuse into the law itself — constant because a central force can exert no torque about its own centre:

\frac{\mathrm{d}A}{\mathrm{d}t}=\tfrac12 r^2\dot\theta=\frac{L}{2m}=\text{const} \qquad (3)

where: $L$ — the conserved magnitude of the angular momentum. Read it as an exchange rate: every second, the same area. When $r$ doubles, $\dot\theta$ must fall by a factor of four; when a comet is sixty times farther out, its angular progress is $3600$ times slower. The full derivation — including why $L$ is conserved and why the orbit stays in a plane — is the second derivation below. One immediate global consequence: integrate (3) over a full orbit and the left side is the total area of the ellipse, $\pi ab$ . That single line converts the sweep rate into the orbital period,

T=\frac{\pi ab}{L/2m}=\frac{2\pi m\,ab}{L} \qquad (4)

where: $a,b$ — the semi-major and semi-minor axes of the bound orbit; $T$ — the period. This is the kinematic bridge to Kepler's Third Law: feed in the inverse-square-specific relations between $b$ , $a$ and $L$ (the ellipse geometry of the First Law article) and $T^2\propto a^3$ falls out — a derivation the Third Law article completes.

Derivation AThe area swept in an instant

The goal is equation (1), the geometric half of the Second Law: the area swept by the radius vector in time $\mathrm{d}t$ is $\tfrac12 r^2\mathrm{d}\theta$ , independent of how fast the radius itself is stretching. The strategy is to write the swept sliver as a triangle with two sides $\mathbf r$ and $\mathrm{d}\mathbf r$ and use the cross product, whose magnitude is twice a triangle's area — the same tool that defines angular momentum, which is exactly why the two will match up so neatly afterwards. The point to watch is the radial component of the displacement: it contributes nothing, and seeing why it drops out is most of the understanding.

Step 1. Write the sliver as a triangle. In time $\mathrm{d}t$ the position vector goes from $\mathbf r$ to $\mathbf r+\mathrm{d}\mathbf r$ ; to first order in $\mathrm{d}t$ , the region swept is the triangle spanned by $\mathbf r$ and $\mathrm{d}\mathbf r$ , of area $\mathrm{d}A=\tfrac12\left|\mathbf r\times\mathrm{d}\mathbf r\right|$ . Why: the area between two nearly-parallel position vectors differs from the straight-sided triangle only at second order — the curved outer edge contributes $O(\mathrm{d}t^2)$ and vanishes in the limit.

Step 2. Split the displacement into polar components. In polar coordinates, $\mathrm{d}\mathbf r=\mathrm{d}r\,\hat{\mathbf r}+r\,\mathrm{d}\theta\,\hat{\boldsymbol\theta}$ — a radial stretch plus a tangential swing. Why: these are the natural components when the Sun is the origin; each will behave differently under the cross product.

Step 3. Cross with the position vector. Since $\mathbf r=r\hat{\mathbf r}$ and $\hat{\mathbf r}\times\hat{\mathbf r}=0$ , the radial part is annihilated: $\mathbf r\times\mathrm{d}\mathbf r=r\hat{\mathbf r}\times\left(r\,\mathrm{d}\theta\,\hat{\boldsymbol\theta}\right)=r^2\,\mathrm{d}\theta\,\hat{\mathbf z}$ . Why: motion along the radius sweeps no area — a body falling straight toward the Sun paints nothing. Only the sideways swing counts.

Step 4. Take the magnitude. $\mathrm{d}A=\tfrac12 r^2\,\mathrm{d}\theta$ , which is equation (1). Why: the half is the triangle's half — base $r$ , height $r\,\mathrm{d}\theta$ . *In Cartesian components $\tfrac12|\mathbf r\times\mathrm{d}\mathbf r|=\tfrac12|x\,\mathrm{d}y-y\,\mathrm{d}x|$ , the standard shoelace-formula element for the area of a polygon traversed about the origin — the same result by an independent route.*

Go deeper · Derivation B · 2 minEqual areas from zero torque

Derivation BEqual areas from zero torque

Now the dynamical half, and the heart of the article: the areal velocity $\tfrac12 r^2\dot\theta$ is constant, because the angular momentum is. The strategy is to compute $\mathrm{d}\mathbf L/\mathrm{d}t$ directly from Newton's second law and watch it vanish for any force that points along the line to the centre — the inverse-square form is never used, which is what makes the Second Law universal across central forces. Two bonuses fall out of keeping the calculation vectorial: the conservation of $\mathbf L$ 's direction proves the orbit stays in a single plane (a fact usually smuggled in as an assumption), and the final identification with the sweep rate is one line because Derivation A already did the geometry.

Step 1. Define the angular momentum about the force centre. $\mathbf L=\mathbf r\times m\mathbf v$ , with $\mathbf r$ measured from the centre of force. Why: torque and angular momentum only mean anything relative to a chosen point; choosing the force centre is what will make the torque vanish.

Step 2. Differentiate. $\dfrac{\mathrm{d}\mathbf L}{\mathrm{d}t}=\mathbf v\times m\mathbf v+\mathbf r\times m\dot{\mathbf v}$ . Why: product rule; the cross product is bilinear, so it differentiates like any product.

Step 3. Kill the first term. $\mathbf v\times\mathbf v=0$ identically. Why: any vector crossed with itself vanishes — the "velocity crossed with momentum" term never survives.

Step 4. Insert Newton's law and the central form of the force. $m\dot{\mathbf v}=\mathbf F=f(r)\,\hat{\mathbf r}$ , so $\dfrac{\mathrm{d}\mathbf L}{\mathrm{d}t}=\mathbf r\times f(r)\hat{\mathbf r}=f(r)\,r\,(\hat{\mathbf r}\times\hat{\mathbf r})=0$ . Why: this is the only place the physics enters, and all it needs is that $\mathbf F$ is parallel to $\mathbf r$ — the radial profile $f(r)$ , inverse-square or otherwise, is irrelevant. Zero torque, conserved $\mathbf L$ .

Step 5. Extract planarity from the direction. $\mathbf L$ is a fixed vector, and $\mathbf r\cdot\mathbf L=\mathbf r\cdot(\mathbf r\times m\mathbf v)=0$ always; so the position vector remains forever perpendicular to a fixed direction — the motion is confined to the plane through the force centre normal to $\mathbf L$ . Why: this is why we may use plane polar coordinates at all; the Second Law's "area" lives in this invariant plane.

Step 6. Extract the magnitude in polar form. In the orbital plane, $\mathbf v=\dot r\,\hat{\mathbf r}+r\dot\theta\,\hat{\boldsymbol\theta}$ , and crossing with $\mathbf r=r\hat{\mathbf r}$ kills the radial part: $L=m r^2\dot\theta$ — equation (2). Why: the same cross-product selection rule as in Derivation A, now acting on velocity instead of displacement.

Step 7. Identify the areal velocity. Dividing equation (1) by $\mathrm{d}t$ : $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\tfrac12 r^2\dot\theta=\dfrac{L}{2m}$ , constant because $L$ is. Kepler's Second Law, in full. Why: geometry (Derivation A) supplied $\tfrac12 r^2\dot\theta$ ; dynamics supplied its constancy — the law is precisely the meeting of the two. *In Cartesian form $L_z=m(x\dot y-y\dot x)$ and the areal velocity is $\tfrac12(x\dot y-y\dot x)$ — the ratio is $2m$ in any coordinates.*

Go deeper · Derivation C · 1 minApsidal speeds from the conserved sweep rate

Derivation CApsidal speeds from the conserved sweep rate

The goal is the Second Law's most-used practical output: the exact trade between speed and distance at the two ends of an orbit, $v_p/v_a=r_a/r_p=(1+e)/(1-e)$ . The strategy is to evaluate the conserved $L$ at perihelion and aphelion, where — and only where — the velocity is purely tangential, so the full speed enters $L=mrv$ with no component-splitting. The thing to watch is exactly that restriction: at a general point of the orbit $L=mrv_\perp$ involves only the tangential velocity component, so this clean speed-ratio trick works at the apsides and nowhere else.

Step 1. Recognize the apsides as turning points. At perihelion and aphelion the distance $r$ is at an extremum, so $\dot r=0$ and the velocity is purely tangential: $v=r\dot\theta$ there. Why: "closest" and "farthest" are exactly the moments when the planet is momentarily neither approaching nor receding.

Step 2. Write the conserved angular momentum at each apsis. $L=m\,r_p v_p=m\,r_a v_a$ . Why: $L$ is the same number all along the orbit — evaluating it at the two special points where it takes the simple form $mrv$ links them directly.

Step 3. Take the ratio. $\dfrac{v_p}{v_a}=\dfrac{r_a}{r_p}$ — speeds in exact inverse proportion to distances. Why: the mass and the constant $L$ cancel; this line is pure conservation bookkeeping.

Step 4. Express through the eccentricity. For an ellipse the apsidal distances are $r_p=a(1-e)$ and $r_a=a(1+e)$ (the geometry dictionary of the First Law article), so $\dfrac{v_p}{v_a}=\dfrac{1+e}{1-e}$ . Why: this puts the speed swing directly in terms of the orbit's shape parameter — $e=0$ gives no swing, $e\to1$ an arbitrarily violent one. *As a check: Earth, $e=0.0167$ , gives a predicted speed ratio $(1+e)/(1-e)=1.0340$ ; the observed perihelion and aphelion speeds $30.29$ and $29.29\ \mathrm{km/s}$ give $30.29/29.29=1.034$ , in exact agreement.*

Limiting cases

Case	Result	Meaning
Circular orbit	$r=\text{const}\Rightarrow\dot\theta=\text{const}$	Constant sweep rate forces uniform angular speed — the pie grows in identical slices.
Apsides	$v_p r_p=v_a r_a$ , ratio $(1+e)/(1-e)$	Velocity purely tangential at the turning points; the clean speed–distance trade.
Radial infall, $L=0$	$\mathrm{d}A/\mathrm{d}t=0$	A body falling straight in sweeps no area — the law holds, vacuously.
Unbound orbits	$\mathrm{d}A/\mathrm{d}t=L/2m$ still	Parabolic and hyperbolic paths sweep equal areas too; nothing required boundedness.
Two-body system	$\mathrm{d}A/\mathrm{d}t=L/2\mu$	For comparable masses, the relative orbit sweeps at $L/2\mu$ with $\mu$ the reduced mass.
Any central force	$\mathrm{d}A/\mathrm{d}t=\text{const}$	Springs, Yukawa, modified gravity — torque-free means equal areas, whatever $f(r)$ .

Common trapEqual areas means equal speeds

It means almost the opposite: the speed varies precisely so that the area rate can stay fixed. The conserved quantity is $\tfrac12 r^2\dot\theta$ — not the speed $v$ , not the angular velocity $\dot\theta$ , and not the kinetic energy, all of which change continuously around a non-circular orbit. Two related traps: first, $L=mrv$ is valid only at perihelion and aphelion, where the velocity is purely tangential — at a general point $L=mrv_\perp$ and using the full speed there overestimates $L$ . Second, the areas are swept about the Sun at its focus: wedges measured from the ellipse's centre are not equal in equal times. And remember which mass appears: for a genuine two-body system the sweep rate of the relative orbit is $L/2\mu$ with the reduced mass $\mu$ , as the Kepler problem article sets up.

Worked examples

1 · Earth's speed swing (confidence builder)

Earth's orbit has $e=0.0167$ and perihelion speed $v_p=30.29\ \mathrm{km/s}$ at $r_p=1.471\times10^{8}\ \mathrm{km}$ . Find the aphelion speed, and Earth's constant areal velocity.

Step 1. Aphelion distance: $r_a=r_p\dfrac{1+e}{1-e}=1.471\times10^{8}\times1.0340=1.521\times10^{8}\ \mathrm{km}$ .

Step 2. Aphelion speed from the apsidal trade (Derivation C): $v_a=v_p\dfrac{r_p}{r_a}=30.29\times\dfrac{1.471}{1.521}=\mathbf{29.29\ \mathrm{km/s}}$ — a $3.4\%$ swing, matching $(1+e)/(1-e)=1.034$ .

Step 3. Areal velocity from perihelion data: $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\tfrac12 r_p v_p=\tfrac12\times1.471\times10^{11}\times3.029\times10^{4}=\mathbf{2.23\times10^{15}\ \mathrm{m^2/s}}$ . Earth paints two thousand trillion square metres of orbital plane every second — the same number in January as in July.

2 · Halley's period from the sweep rate alone (exam level)

Halley's comet has $a=17.8\ \mathrm{AU}$ , $e=0.967$ , and perihelion speed $v_p=54.5\ \mathrm{km/s}$ (numbers from the Kepler problem article). Using only the Second Law and ellipse geometry, predict its orbital period.

Step 1. Perihelion distance: $r_p=a(1-e)=17.8\times0.033=0.587\ \mathrm{AU}=8.78\times10^{10}\ \mathrm{m}$ .

Step 2. The conserved sweep rate, evaluated at perihelion where $v$ is purely tangential: $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\tfrac12 r_p v_p=\tfrac12\times8.78\times10^{10}\times5.45\times10^{4}=2.39\times10^{15}\ \mathrm{m^2/s}$ .

Step 3. Total area to be swept in one period: $b=a\sqrt{1-e^2}=17.8\times0.2548=4.535\ \mathrm{AU}$ , so $A=\pi ab=\pi\times17.8\times4.535=253.6\ \mathrm{AU^2}=5.68\times10^{24}\ \mathrm{m^2}$ (using $1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm{m}$ ).

Step 4. Period = area ÷ rate: $T=\dfrac{5.68\times10^{24}}{2.39\times10^{15}}=2.37\times10^{9}\ \mathrm{s}=\mathbf{75.2\ \mathrm{yr}}$ — Halley's observed 75–76-year return, recovered without ever integrating an equation of motion.

Step 5. The same conservation gives the other end of the story: $v_a=v_p\,r_p/r_a$ with $r_a=a(1+e)=35.0\ \mathrm{AU}$ , so $v_a=54.5\times0.587/35.0=\mathbf{0.91\ \mathrm{km/s}}$ — a factor of $60$ slower. The comet spends decades drifting through the region it crosses in months near the Sun, which is why "equal areas" reads, observationally, as "long absences".

Origins · 1 minJohannes Kepler and how this was discovered

Johannes Kepler

1571–1630 · Prague

Kepler found the Second Law first — in 1602, three years before the ellipse — and by a route no modern course would dare teach. Working through Tycho Brahe's Mars observations, he first guessed that a planet's speed varies inversely with its distance from the Sun (true only at the apsides), then computed orbital times by summing hundreds of little distance-weighted steps — effectively integrating by hand, decades before calculus. The sums kept working better than his premise deserved, and he eventually recognized the clean invariant underneath: the radius vector sweeps equal areas in equal times. He had discovered a conservation law empirically, with no concept of angular momentum, torque, or force to explain it. That explanation came in 1687, when Newton opened the Principia with precisely this theorem — and its converse became his signature that the planetary force points at the Sun.

History

1602

— Kepler, deep in the Mars calculations, finds that the radius vector sweeps equal areas in equal times — his "area law", discovered before the ellipse.

1609

— Astronomia Nova publishes the area law alongside the First Law.

1687

— Newton's Principia makes it Book I, Proposition 1: equal areas hold for any central force — and, conversely, equal areas imply the force is central.

1918

— Noether's theorem identifies the deep root: angular momentum conservation is the signature of rotational symmetry, with the equal-area law its oldest empirical footprint.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 7 (Central Forces) · Goldstein, Classical Mechanics, 3rd ed., Ch. 3 · Landau & Lifshitz, Mechanics (Vol. 1), §14 · Taylor, Classical Mechanics, Ch. 8 Key results to keep in mind: the areal-velocity identity $\mathrm{d}A/\mathrm{d}t=L/(2m)$ follows from the triangle-area element $\mathrm{d}A=\tfrac12 r^2\mathrm{d}\theta$ combined with zero torque; the apsidal speed ratio $(1+e)/(1-e)$ is standard ellipse geometry; Earth's perihelion and aphelion speeds (≈30.29 / 29.29 km/s) with $e=0.0167$ yield a ratio ≈1.034, consistent with the formula; and $T=2\pi m\,ab/L$ reduces to Kepler's third law via the standard orbit relations.

Builds on: Central Force Fundamentals · Angular Momentum Conservation in Central Fields · The Two-Body Problem and Reduced Mass Leads to: Reduction to the Radial Equation · Kepler's Third Law: Period-Semimajor Relation · Kepler's First Law: Elliptical Orbits

Equal areas of what, exactly — measured from where?

The area swept by the position vector drawn from the force centre — the Sun, sitting at the focus. In time $\mathrm{d}t$ that line turns through $\mathrm{d}\theta$ and paints the sliver $\tfrac12 r^2\mathrm{d}\theta$ ; the law says these slivers accumulate at a constant rate. It is emphatically not arc length (the planet covers unequal path lengths in equal times — that is the whole point), and not area measured from the ellipse's centre, which grows at a non-uniform rate. If you re-anchor the sweep at any point other than the force centre, the force develops a torque about that point, the corresponding angular momentum is not conserved, and the equal-area property dies with it.

The planet speeds up and slows down — doesn't that violate conservation of momentum?

Linear momentum is simply not conserved here, and no law says it should be: the Sun's gravity is a real external force on the planet, continuously bending and re-scaling its velocity. What survives are the quantities whose conservation the geometry protects. Angular momentum survives because the force, pointing always at the Sun, has zero torque about the Sun. Energy survives because gravity is conservative — the planet trades kinetic for potential energy as it climbs from perihelion to aphelion, which is exactly the speed variation you noticed. So three bookkeeping accounts run simultaneously: $\mathbf p$ changes constantly, $E$ shuffles between kinetic and potential, and $L$ — the Second Law's account — never moves at all.

Is the angular velocity constant too, like the areal velocity?

No — and the contrast is instructive. From $L=mr^2\dot\theta$ , the angular velocity is $\dot\theta=L/(mr^2)$ : it varies as $1/r^2$ around the orbit. Halley's comet turns through true anomaly $3600$ times faster at perihelion ( $r_a/r_p=60$ , squared) than at aphelion. Constant $\dot\theta$ happens only for the circular orbit, where $r$ never changes. This $1/r^2$ variation is what killed the ancient dream of uniform circular motion: Ptolemy's equant was an attempt to find some point about which the angular rate looks uniform, and the Second Law explains both why it nearly worked (the sweep is uniform about the focus) and why it could never work exactly.

Why can I write L = mrv at perihelion but not at other points?

Because $L$ only sees the velocity component perpendicular to the radius: in general $L=mrv_\perp=mrv\sin\phi$ , where $\phi$ is the angle between $\mathbf r$ and $\mathbf v$ . At a generic point the planet is partly approaching or receding, so $v_\perp<v$ and using the full speed would overestimate $L$ . Perihelion and aphelion are the two moments when $\dot r=0$ — the planet is momentarily neither falling in nor climbing out — so the velocity is purely tangential, $\phi=90^\circ$ , and $L=mrv$ holds with the total speed. That is why the apsidal speed ratio drops out of the Second Law alone, while speeds elsewhere on the orbit need the energy equation too, as the Kepler problem article's machinery shows.

Does the second law work for hyperbolic flybys and for forces that are not gravity?

Yes to both, and that generality is its signature. The derivation used exactly one physical fact: the force points along the line to the centre, so its torque vanishes. Nothing required the force to be inverse-square, attractive, or even gravitational — a mass on a spring anchored at a point, an electron scattering off a nucleus, a star in a modified-gravity halo all sweep equal areas about their force centres. And nothing required the orbit to close: 'Oumuamua on its $e\approx1.2$ hyperbola swept equal areas about the Sun on the way in and on the way out. Contrast the First and Third Laws, which are inverse-square exclusives — that is why Newton could use the area law as a diagnostic: equal areas told him the planetary force is central, before he knew its strength law.

Where does the one-half in dA = half r squared d-theta come from?

From the triangle. The swept sliver has one side $\mathbf r$ (length $r$ ) and, to first order, the far edge contributes a height $r\,\mathrm{d}\theta$ — so its area is $\tfrac12\times\text{base}\times\text{height}=\tfrac12 r^2\mathrm{d}\theta$ , half of the $r^2\mathrm{d}\theta$ rectangle you might guess. Integrate the same element over a full circle and the halves add up correctly: $\int_0^{2\pi}\tfrac12 r^2\mathrm{d}\theta=\pi r^2$ , the full disc — a quick sanity check that the factor is right. The deeper version: $\mathrm{d}A=\tfrac12|\mathbf r\times\mathrm{d}\mathbf r|$ , and the same $\tfrac12$ separates the areal velocity $L/2m$ from the angular momentum — the two quantities are the identical cross product wearing different units.

If the sweep rate never changes, why are the seasons different lengths?

Because equal areas are not equal angles. The seasons are defined astronomically by the equinoxes and solstices — fixed directions of the Sun as seen from Earth — so each season corresponds to a quarter-turn of angle, not a quarter of the orbit's area. Earth passes perihelion in early January, so the autumn-to-spring half of the orbit (containing January) is traversed quickly, while the spring-to-autumn half near aphelion is traversed slowly: about $186.4$ days from March equinox to September equinox versus $178.9$ days back. The imbalance — a week — is the Second Law acting on Earth's mild $e=0.0167$ . On Mars, with $e=0.093$ , the same effect splits the two halves of the year by nearly 80 sols.