Kepler found from Tycho Brahe's measurements that a planet's line to the Sun sweeps out equal areas in equal times. Angular Momentum of a Particle shows why, using only one fact about gravity: it points straight at the Sun. See it in the Orbit Lab. It builds on Satellites in Orbit.
Why it works
Gravity on a planet points along $-\vec{r}$, from the planet to the Sun. Its torque about the Sun is $\vec{r}\times\vec{F} = 0$, because the two vectors are parallel. So $\vec{L} = m\,\vec{r}\times\vec{v}$ is constant. In a short time $dt$ the line from the Sun sweeps a thin triangle with sides $\vec{r}$ and $\vec{v}\,dt$, of area $$dA = \tfrac{1}{2}|\vec{r}\times\vec{v}\,dt| = \dfrac{L}{2m}\,dt.$$ So $\dfrac{dA}{dt} = \dfrac{L}{2m}$ is constant: equal areas in equal times.
Key idea
Any central force conserves angular momentum, which is Kepler's second law: $\dfrac{dA}{dt} = \dfrac{L}{2m}$ is constant. At the nearest and farthest points the velocity is perpendicular to $\vec{r}$, so $$r_pv_p = r_av_a:$$ a planet moves fastest at perihelion and slowest at aphelion.
Worked example: Earth's speed through the year
Earth passes perihelion ($1.471 \times 10^{11}\ \text{m}$) at $30.29\ \text{km/s}$ in early January. Its aphelion distance is $1.521 \times 10^{11}\ \text{m}$. How fast does it move in early July?
- No torque about the Sun: $r_pv_p = r_av_a$.
- $v_a = 30.29 \times \dfrac{1.471}{1.521} = 29.29\ \text{km/s}$.
Answer: About $29.3\ \text{km/s}$, about 3% slower.
Worked example: A comet
A comet's closest distance to the Sun is $0.5\ \text{AU}$ and its farthest is $10\ \text{AU}$. How do its speeds at those points compare?
- $\dfrac{v_p}{v_a} = \dfrac{r_a}{r_p} = \dfrac{10}{0.5} = 20$.
- It whips around the Sun 20 times as fast as it creeps along at the far end, so it spends most of its time far away.
Answer: 20 times faster at its closest point.
Common mistake
Thinking a planet moves at the same speed all around its orbit. That holds only for a perfect circle. On an ellipse the distance changes, and to keep $rv\sin\phi$ constant the speed must rise as the distance falls. Check at the two ends: $v_pr_p = v_ar_a$ with $r_p < r_a$ forces $v_p > v_a$.
Nothing here used the inverse-square law: equal areas hold for any force that points toward a fixed center. The inverse square is what makes the orbit an ellipse and fixes the period, as in Kepler's Third Law from Newton.