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

For the full derivation see the parent note.

True or false — justify

True or false: "Equal areas in equal times" means the planet moves at equal speed everywhere.

False. Only the area swept per second is constant; near the Sun the planet moves fast over a short-and-fat triangle, far away slow over a long-and-thin one — same area, very different speed.

True or false: Kepler's second law requires an inverse-square force.

False. Any central force (

F ∥ r

) gives zero torque and hence constant areal velocity; inverse-square is only needed to make the orbit an ellipse (the first law).

True or false: A planet on a perfectly circular orbit still obeys the second law.

True, trivially — with

r

constant the areal rate

\frac{1}{2} r^{2} \dot{θ}

stays fixed because

\dot{θ}

is also constant, so it sweeps equal areas as equal angles pass.

True or false: The angular momentum

L

of the planet is exactly zero because gravity pulls it straight in.

False. Gravity gives zero torque, not zero angular momentum;

L

is a fixed nonzero vector unless the planet is falling dead-radially into the Sun.

True or false: If the Sun could suddenly double its mass, the areal velocity would double.

False.

d A / d t = L /2 m

depends on the planet's angular momentum, not directly on

M

; changing

M

changes the orbit's shape and the future

L

, but at the instant of change

L

(and so

d A / d t

) is unchanged.

True or false: Areal velocity is a scalar, but it comes from a vector cross product.

True.

d A / d t = \frac{1}{2} ∣ r \times v ∣

— the cross product is a vector, but we take its magnitude to get the (scalar) rate of area swept.

True or false: For a comet on a highly elongated orbit,

d A / d t

is large near the Sun and small far away.

False. That's the whole point —

d A / d t

is constant; the comet's wild speed changes exactly compensate the radius changes to keep the swept area rate fixed.

True or false: The second law would fail if there were a small tangential drag force.

True. A tangential force is not parallel to

r

, so it produces torque,

L

drifts, and areal velocity slowly changes — orbits decay, areas shrink.

Spot the error

Error: "

L = m v r

, so at any point I can find

v

from

r

L = m v r

only when

v ⊥ r

(apsides or circles). In general

L = m r v sin ϕ

, where

ϕ

is the angle between

r

and

v

; away from the apsides the

sin ϕ

factor cannot be dropped.

Error: "Torque is zero because the force is small."

Torque is zero because the force is parallel to $r$ (

r \times F = 0

), not because it's weak — even an enormous central force gives zero torque.

Error: "

d A = \frac{1}{2} r d θ

for the pizza slice."

The slice area is

d A = \frac{1}{2} r^{2} d θ

(area of a thin circular sector), not

\frac{1}{2} r d θ

— you need

r

squared, one factor for the radial length and one for the arc.

Error: "Since areas are equal, arc lengths are equal too."

Arc length is not conserved. Near the Sun a short fat wedge and far away a long thin wedge have equal area but very different arc lengths; the planet covers a longer arc near perihelion.

Error: "

d L / d t = v \times m v + r \times F

, and the first term isn't obviously zero."

v \times v = 0

always — any vector crossed with itself vanishes because the "parallelogram" it spans has zero area — so only

r \times F

survives.

Error: "Areal velocity equals the orbital speed divided by two."

No —

d A / d t = \frac{1}{2} ∣ r \times v ∣

mixes both position and velocity; it is not simply

v /2

. Only at an apsis does it reduce to

\frac{1}{2} r v

Edge cases

Edge case: What is the areal velocity of a mass falling straight toward the Sun on a radial line?

It is zero —

v

is parallel to

r

, so

r \times v = 0

; the degenerate "orbit" is a line and sweeps no area (consistent with

L = 0

Edge case: Does the law hold during a slingshot / gravity assist past a planet?

Not about the Sun during the encounter — the planet's gravity is a second central force about a different center, adding torque about the Sun;

L

about the Sun changes, which is exactly how the spacecraft gains energy.

Edge case: For a two-body problem where the Sun also moves, whose

r

do we use?

Use the vector from the center of mass (or the relative separation), not from the moving Sun's center; the conserved

L

is that of the reduced-mass motion, and areal velocity is constant in that frame.

Edge case: If

\dot{θ}

momentarily equals zero somewhere on the orbit, is the law violated?

On a real bound orbit

\dot{θ}

is never zero (that would need

L = 0

), so this never happens for an actual ellipse — a zero would signal a radial fall, not an orbit.

Edge case: Over half an orbit from perihelion to aphelion, is the swept area exactly half the ellipse?

Yes — but it takes more than half the period if you start at perihelion, because equal areas take equal times only in balanced halves; peri-to-aphelion (measured over equal time) covers less angular arc despite equal area rate. (The peri↔apo half specifically does split area in half and, by symmetry, time in half.)

Recall One-line summary of every trap

Almost every mistake here comes from forgetting one word in the chain Central → Torque-zero → L-constant → Area-rate-constant, or from confusing area rate (constant) with speed / arc length / angle (all changing).

Connections

Angular Momentum Conservation — the conserved quantity behind every trap
Central Forces and Torque — why the chain starts with zero torque
Kepler's First Law (Ellipse) — what actually needs inverse-square
Kepler's Third Law (T² ∝ a³) — uses $π ab = (L /2 m) T$
Vis-viva Equation — the changing speed that doesn't break constancy
Cross Product and Area — the geometry of $\frac{1}{2} ∣ r \times v ∣$

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

True or false — justify

Spot the error

Why questions

Edge cases

Connections