Kepler's second law — equal areas in equal times, from angular momentum conservation: Worked examples

This page is a drill. The parent note proved the law once. Here we hit it from every angle — every geometric case, every degenerate input, every trap an exam can set. Before each answer you get a Forecast line: cover the solution, guess, then check yourself.

Two tools from the parent that we lean on constantly. Both are just labels for pictures:

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

Look at the figure: the coral slice is the swept triangle; its area uses only $v_{⊥} = v sin ϕ$ (the mint arrow). The lavender component $v cos ϕ$ slides along $r$ and paints nothing.

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below tag which cell they clear.

Cell	Case class	What's tricky	Example
A	Apsides only ( $ϕ = 90°$ )	clean $L = m v r$	Ex 1
B	General point ( $ϕ \neq = 90°$ )	must use $sin ϕ$	Ex 2
C	Two general points compared	$r v sin ϕ$ conserved	Ex 3
D	Areal velocity as a rate (time from area)	$A = \frac{L}{2 m} t$	Ex 4
E	Degenerate: radial "fall" ( $ϕ = 0$ )	$L = 0$ , law trivial	Ex 5
F	Limiting: circular orbit	constant speed, still equal areas	Ex 6
G	Non-gravity central force (spring)	law still holds	Ex 7
H	Real-world word problem	translate story → symbols	Ex 8
I	Exam twist: fraction of area / time	ratio reasoning	Ex 9

Ex 1 — Cell A: apsides, the clean case

Worked example Perihelion vs aphelion speed

A planet has perihelion distance $r_{p} = 4.0 \times 1 0^{10} m$ with speed $v_{p} = 60 km/s$ . Its aphelion distance is $r_{a} = 6.0 \times 1 0^{11} m$ . Find $v_{a}$ .

Forecast: aphelion is 15× farther. Will $v_{a}$ be bigger or smaller, and by roughly what factor?

Step 1 — Recognise both points are apsides. At perihelion and aphelion the velocity is perpendicular to the radius, so $ϕ = 90°$ and $sin ϕ = 1$ . Why this step? Only then does $L = m v r$ hold with no angle factor — the whole reason apsides are the "easy" points.

Step 2 — Set the two $L$ values equal. $m v_{p} r_{p} = m v_{a} r_{a} \Rightarrow v_{p} r_{p} = v_{a} r_{a} .$ Why this step? $L$ is conserved (central force, zero torque). Mass cancels.

Step 3 — Solve. $v_{a} = v_{p} \frac{r _{p}}{r _{a}} = 60 \times \frac{4.0 \times 1 0 ^{10}}{6.0 \times 1 0 ^{11}} = 60 \times \frac{1}{15} = 4.0 km/s .$

Verify: ratio came out $1/15$ exactly, matching the distance ratio — as forecast, slower and by the inverse factor. Units: $(km/s) \times$ dimensionless $=$ km/s. ✓

Ex 2 — Cell B: a general point, $ϕ \neq = 90°$

Worked example Only the perpendicular part counts

At some point on its orbit a comet is $r = 2.0 \times 1 0^{11} m$ from the Sun, moving at $v = 25 km/s$ , with the velocity making an angle $ϕ = 30°$ with the radius vector. Find $L / m$ (the specific angular momentum) and the areal velocity $d A / d t$ .

Forecast: since $ϕ$ is only $30°$ , will $L$ be near its maximum $r v$ or well below it?

Step 1 — Use the full formula $L = m r v sin ϕ$ . $\frac{L}{m} = r v sin ϕ = (2.0 \times 1 0^{11}) (2.5 \times 1 0^{4}) sin 30°.$ Why this step? The velocity is not perpendicular to $r$ , so $L = m v r$ would over-count. Only $v sin ϕ = v_{⊥}$ sweeps area (mint arrow in the figure).

Step 2 — Plug $sin 30° = 0.5$ . $\frac{L}{m} = (2.0 \times 1 0^{11}) (2.5 \times 1 0^{4}) (0.5) = 2.5 \times 1 0^{15} m^{2} / s .$

Step 3 — Halve it for areal velocity. $\frac{d A}{d t} = \frac{L}{2 m} = 1.25 \times 1 0^{15} m^{2} / s .$ Why this step? $d A / d t = \frac{1}{2} ∣ r \times v ∣ = \frac{L}{2 m}$ — the swept-area rate is always half the specific angular momentum.

Verify: with $ϕ = 30°$ we got half of the $r v = 5.0 \times 1 0^{15}$ max — well below maximum, as forecast. Units: $m \cdot (m/s) = m^{2} / s$ . ✓

Ex 3 — Cell C: compare two general points

Worked example Conservation with two angle factors

At point 1: $r_{1} = 1.0 \times 1 0^{11} m$ , $v_{1} = 40 km/s$ , $ϕ_{1} = 90°$ . At point 2: $r_{2} = 2.0 \times 1 0^{11} m$ , $ϕ_{2} = 30°$ . Find $v_{2}$ .

Forecast: point 2 is farther AND its velocity is tilted — do these push $v_{2}$ up or down? (Careful, they fight.)

Step 1 — Write $L / m$ conserved between the two points. $r_{1} v_{1} sin ϕ_{1} = r_{2} v_{2} sin ϕ_{2} .$ Why this step? $L$ is the same everywhere; only its factored form changes point to point.

Step 2 — Solve for $v_{2}$ . $v_{2} = \frac{r _{1} v _{1} s i n ϕ _{1}}{r _{2} s i n ϕ _{2}} = \frac{( 1.0 \times 1 0 ^{11} ) ( 40 ) ( 1 )}{( 2.0 \times 1 0 ^{11} ) ( 0.5 )} .$ Why this step? Isolate the unknown; keep every factor honest.

Step 3 — Evaluate. $v_{2} = \frac{40}{1.0} = 40 km/s .$

Verify: the "farther by 2×" would halve $v_{2}$ , but the tilt $sin 30° = 0.5$ doubles it back — they exactly cancel, giving $v_{2} = v_{1}$ . The two effects fought to a draw, as forecast. ✓

Ex 4 — Cell D: time from swept area

Ex 5 — Cell E: the degenerate radial drop ( $ϕ = 0$ )

Worked example Falling straight at the Sun

An object is released from rest far away and falls straight toward the Sun, so its velocity always points along $r$ ( $ϕ = 0°$ ). What is its angular momentum and areal velocity?

Forecast: if it never sweeps sideways, what area does its radius vector paint?

Step 1 — Plug $ϕ = 0$ into $L = m r v sin ϕ$ . $L = m r v sin 0° = 0.$ Why this step? $sin 0° = 0$ : motion purely along $r$ has no perpendicular component, so nothing sweeps.

Step 2 — Areal velocity follows. $\frac{d A}{d t} = \frac{L}{2 m} = 0.$ Why this step? Zero angular momentum ⇒ zero swept area. The radius vector just shrinks along a fixed line, painting a degenerate triangle of zero width.

Verify: the "orbit" is a straight line through the Sun — a degenerate ellipse with $b = 0$ . Its area $π ab = 0$ , consistent with $d A / d t = 0$ . The equal-areas law still holds: every second sweeps the same area, namely zero. ✓

Ex 6 — Cell F: the circular limit

Worked example Equal areas when speed is already constant

A satellite is in a circular orbit of radius $r = 7.0 \times 1 0^{6} m$ at constant speed $v = 7.5 km/s$ . Find $d A / d t$ , and confirm the "equal areas" statement is not vacuous.

Forecast: in a circle the speed is constant everywhere — is the equal-areas law then trivially satisfied?

Step 1 — In a circle $v ⊥ r$ everywhere, so $ϕ = 90°$ . $\frac{d A}{d t} = \frac{1}{2} r v sin 90° = \frac{1}{2} r v .$ Why this step? A circular orbit's velocity is always tangent, hence perpendicular to the radius — the clean case at every point.

Step 2 — Evaluate. $\frac{d A}{d t} = \frac{1}{2} (7.0 \times 1 0^{6}) (7.5 \times 1 0^{3}) = 2.625 \times 1 0^{10} m^{2} / s .$

Step 3 — Sanity vs full circle. Full circle area $π r^{2} = π (7.0 \times 1 0^{6})^{2} = 1.539 \times 1 0^{14} m^{2}$ ; period $T = 2 π r / v = 5.864 \times 1 0^{3} s$ ; so $π r^{2} / T = 2.625 \times 1 0^{10} m^{2} / s$ . ✓

Verify: both routes agree. Equal areas is not vacuous — it's automatically true because a circle has constant $r$ AND constant $v$ ; the interesting content of the law only shows for eccentric orbits (Ex 1, 3). ✓

Ex 7 — Cell G: a non-gravity central force

Worked example A spring-tethered bead — law still holds

A bead moves on a frictionless plane, attached to the origin by an ideal spring with force $F = - k r$ (a central force, but $F \propto r$ , not $1/ r^{2}$ ). At one instant $r_{1} = 0.20 m$ , $v_{1} = 3.0 m/s$ , $ϕ_{1} = 90°$ . Later $r_{2} = 0.50 m$ . Find the perpendicular speed $v_{2 ⊥}$ .

Forecast: the force is a spring, not gravity. Does Kepler's second law still apply?

Step 1 — Check the only requirement: is the force central? $F = - k r$ is antiparallel to $r$ , so $τ = r \times F = 0$ . See Central Forces and Torque. Why this step? Kepler's 2nd law needs only zero torque, not inverse-square. Angular Momentum Conservation then guarantees constant $L$ .

Step 2 — Conserve $L$ : $r_{1} v_{1} = r_{2} v_{2 ⊥}$ (both use perpendicular speed). $v_{2 ⊥} = \frac{r _{1} v _{1}}{r _{2}} = \frac{( 0.20 ) ( 3.0 )}{0.50} = 1.2 m/s .$ Why this step? $L / m = r v_{⊥}$ is conserved regardless of the force law; farther out means smaller perpendicular speed.

Verify: $r_{2} v_{2 ⊥} = 0.50 \times 1.2 = 0.60 = r_{1} v_{1} = 0.20 \times 3.0$ . ✓ The orbit here is an ellipse centred on the origin (not focus) — different shape from gravity, but areas still equal. This is exactly the parent's "any central force" mistake-buster. ✓

Ex 8 — Cell H: real-world word problem

Worked example Comet in winter vs summer

Halley-type comet: at perihelion $r_{p} = 8.8 \times 1 0^{10} m$ , speed $v_{p} = 54.6 km/s$ . At aphelion $r_{a} = 5.3 \times 1 0^{12} m$ . (a) Find $v_{a}$ . (b) Find its specific angular momentum $L / m$ . (c) How many times faster is it at perihelion?

Forecast: aphelion is ~60× farther — expect the comet nearly frozen out there.

Step 1 — (a) Apsides, so $v_{p} r_{p} = v_{a} r_{a}$ . $v_{a} = v_{p} \frac{r _{p}}{r _{a}} = 54.6 \times \frac{8.8 \times 1 0 ^{10}}{5.3 \times 1 0 ^{12}} = 54.6 \times 0.016604 = 0.9066 km/s .$ Why this step? Both are apsides ( $ϕ = 90°$ ), the clean case again.

Step 2 — (b) $L / m = r_{p} v_{p}$ (evaluate at perihelion). $\frac{L}{m} = (8.8 \times 1 0^{10}) (5.46 \times 1 0^{4}) = 4.805 \times 1 0^{15} m^{2} / s .$ Why this step? $L$ is the same everywhere; perihelion is convenient because $ϕ = 90°$ .

Step 3 — (c) Speed ratio. $\frac{v _{p}}{v _{a}} = \frac{r _{a}}{r _{p}} = \frac{5.3 \times 1 0 ^{12}}{8.8 \times 1 0 ^{10}} \approx 60.2.$ Why this step? From $v_{p} r_{p} = v_{a} r_{a}$ , the speed ratio is just the inverse distance ratio.

Verify: check (b) at aphelion: $r_{a} v_{a} = (5.3 \times 1 0^{12}) (906.6) = 4.805 \times 1 0^{15} m^{2} / s$ — matches Step 2. ✓ Comet crawls at ~0.9 km/s far out, as forecast. ✓

Ex 9 — Cell I: exam twist (fraction of period from area)

Worked example What fraction of a year to sweep a quarter of the ellipse?

Because $d A / d t$ is constant, prove that sweeping any fraction $f$ of the total ellipse area takes fraction $f$ of the period $T$ . Then: Earth's orbital area is $π ab$ ; how long ( $f$ and days) to sweep a quarter of that area? Use $T = 365.25$ days.

Forecast: is "quarter of the area" the same as "quarter of the time"? Same as "quarter of the angle"? (Two of these are equal, one is not.)

Step 1 — Time is proportional to area swept. $t = \frac{A _{swept}}{d A / d t}, T = \frac{π ab}{d A / d t} .$ Why this step? Both use the same constant $d A / d t$ . Dividing kills it.

Step 2 — Take the ratio. $\frac{t}{T} = \frac{A _{swept}}{π ab} = f .$ Why this step? The $d A / d t$ cancels — area fraction equals time fraction, always. This is the exam's whole point.

Step 3 — Quarter of the area, $f = 1/4$ . $t = \frac{1}{4} \times 365.25 = 91.31 days .$

Verify: area-fraction $=$ time-fraction ✓, but NOT angle-fraction — a quarter-area slice near perihelion spans a bigger angle than one near aphelion (see Kepler's First Law (Ellipse)). The forecast trap: area ↔ time equal; angle differs. ✓

Connections

Parent: Kepler's Second Law
Angular Momentum Conservation — what stays fixed in every example
Central Forces and Torque — why the spring (Ex 7) still obeys the law
Cross Product and Area — the $sin ϕ$ and the swept triangle
Kepler's First Law (Ellipse) — why angle-fraction ≠ area-fraction (Ex 9)
Kepler's Third Law (T² ∝ a³) — uses $π ab = (L /2 m) T$
Vis-viva Equation — get $v$ at any $r$ when $ϕ$ is unknown