Angular momentum conservation — prove Kepler's second law

Question

Using the principle of conservation of angular momentum, prove Kepler’s second law — that a planet sweeps out equal areas in equal time intervals.

(JEE Advanced 2022, Paper 1 — this derivation connects two major topics)

Solution — Step by Step

The gravitational force on a planet of mass $m$ due to the Sun (mass $M$ ) is:

\vec{F} = -\frac{GMm}{r^2}\hat{r}

This force is directed along $\hat{r}$ (the line joining the Sun and planet). A force that always points toward (or away from) a fixed point is called a central force.

Torque about the Sun:

\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \left(-\frac{GMm}{r^2}\hat{r}\right) = -\frac{GMm}{r^2}(\vec{r} \times \hat{r}) = 0

Since $\vec{r}$ and $\hat{r}$ are parallel, their cross product is zero. Therefore, $\vec{\tau} = 0$ and angular momentum $\vec{L}$ is conserved.

In a small time $dt$ , the planet moves by $d\vec{r}$ . The area swept out by the radius vector is:

dA = \frac{1}{2}|\vec{r} \times d\vec{r}| = \frac{1}{2}|\vec{r} \times \vec{v}|\,dt

But $\vec{L} = m(\vec{r} \times \vec{v})$ , so $|\vec{r} \times \vec{v}| = L/m$ .

\frac{dA}{dt} = \frac{L}{2m}

Since $L$ and $m$ are both constants, the areal velocity $dA/dt = L/(2m)$ is constant.

This means the planet sweeps equal areas in equal time intervals — which is precisely Kepler’s second law.

\boxed{\frac{dA}{dt} = \frac{L}{2m} = \text{constant}}

Why This Works

The proof rests on one physical fact: gravity is a central force. Central forces produce zero torque about the centre, which conserves angular momentum. Angular momentum conservation then directly implies constant areal velocity.

Notice that we never used the inverse-square nature of gravity — only that it’s central. This means Kepler’s second law holds for any central force, not just gravity. The inverse-square law is needed for the elliptical orbit shape (Kepler’s first law) and the $T^2 \propto a^3$ relation (third law), but equal areas in equal times follows from centrality alone.

Alternative Method

Using polar coordinates, $\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$ . Then:

L = m|\vec{r} \times \vec{v}| = mr^2\dot{\theta}

Area swept: $dA = \frac{1}{2}r^2\,d\theta$ , so $\frac{dA}{dt} = \frac{1}{2}r^2\dot{\theta} = \frac{L}{2m}$ .

In JEE Advanced, this derivation often appears as part of a larger problem — for example, “A comet’s speed at perihelion is $v_1$ at distance $r_1$ . Find its speed at aphelion distance $r_2$ .” The answer uses $L = mr_1v_1 = mr_2v_2$ , which is just angular momentum conservation — the same principle behind Kepler’s second law.

Common Mistake

Students sometimes try to “prove” Kepler’s second law using the formula for the area of an ellipse. That approach is circular — Kepler’s law is about the rate of sweeping area, not the total area. The correct approach requires calculus: compute $dA/dt$ and show it’s constant using angular momentum conservation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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