Why eccentric orbits give uneven transits

The problem

Two planets share an orbital period, a star and a planet-to-star size ratio. One is on a circular orbit and the other on an eccentric one. Do their transits (the dip in starlight while the planet crosses the star's disc) last the same amount of time? It is tempting to say yes, since "same period, same sizes" sounds like it should fix everything else. It does not.

The idea

A planet on an eccentric orbit does not move at constant angular speed. It moves fastest near periapsis (closest approach) and slowest near apoapsis (farthest point). That is Kepler's second law: equal areas are swept in equal times. If a transit happens near periapsis, the planet is moving quickly through that part of its orbit and crosses the star's disc quickly. Near apoapsis it moves slowly, and the same crossing takes longer.

The sandbox below assumes the simplest case, a transit aligned with periapsis. One eccentricity value drives both the orbit shape and the transit light curve underneath it. Push eccentricity up and the dip narrows.

Full proof

Setup. Mean anomaly $M$ increases uniformly in time by definition: $M = nt$, where $n = 2\pi/T$ is the mean motion and $T$ is the orbital period. Kepler's equation relates $M$ to the eccentric anomaly $E$:

$M = E - e\sin E \implies \frac{dM}{dE} = 1 - e\cos E$

Modeling convention. Treat a transit as the planet crossing a fixed window of eccentric anomaly $[-\Delta E, \Delta E]$ centered on periapsis ($E = 0$). This is a simplification. A real transit is a fixed window of true anomaly (the star's actual angular size as seen from the planet), and the two agree only to leading order for small $\Delta E$, which is the regime a distant star's disc puts us in anyway. This is exactly the convention the interactive above computes.

Duration. Because $M$ is uniform in time, the time spent between $E=-\Delta E$ and $E=\Delta E$ is:

$\Delta t = \frac{\Delta M}{n} = \frac{1}{n}\int_{-\Delta E}^{\Delta E} (1 - e\cos E)\, dE$

For small $\Delta E$, $\cos E \approx 1$ across the whole window, giving:

$\Delta t \approx \frac{2\Delta E(1-e)}{n} = \frac{\Delta E (1-e)\, T}{\pi} \qquad \blacksquare$

Consequence. For a periapsis-aligned transit, duration shrinks linearly toward zero as $e \to 1$, because the planet is moving fastest exactly where the transit happens. Flip the alignment to apoapsis and the sign of the effect flips too: eccentricity would lengthen the transit instead.

Where it breaks in practice. The error this editorial is filed under, assuming constant orbital speed, is really two errors that compound: (1) forgetting that speed varies around an eccentric orbit at all, and (2) even granting that, assuming the transit necessarily lands at periapsis. Real transiting systems can have any argument of periapsis, and duration depends on both. This sandbox fixes the alignment to periapsis to isolate the eccentricity effect. Don't reach for the formula above on a system where periapsis alignment isn't given.