Planetary Phenomena - 2D Seasons Lab: Axial Tilt, Eccentric Orbits and the Insolation Map

Presets

Live readouts

Graphs

Daily insolation over one orbit at three latitudes

Day length against time of year at the chosen latitude

Annual mean insolation against latitude (line = global mean)

Experiments

Compare Earth with "Earth with no tilt" (no seasons, hot equator, cold poles), then Uranus (the poles get more annual sunlight than the equator) and Venus (tilt 177 degrees: like Earth upside down). Raise eccentricity to see perihelion summers.

Numerical checks

running...

How to use

Top left: the orbit seen from the north with the equinox and solstice directions. Top right: the planet seen from the side with sunlight from the left; the yellow dot is the subsolar latitude (the solar declination) and the blue line is your latitude of interest. Bottom: the heat map of daily insolation (or day length) for every latitude through one orbit, with a white time cursor. Choose a preset or change tilt, eccentricity and distance. Space pauses, R rewinds.

The science

Declination. With axial tilt epsilon and ecliptic longitude of the planet lambda (0 at the northern spring equinox), the Sun's latitude on the planet is sin(delta) = sin(epsilon) sin(lambda) so the subsolar point oscillates between +epsilon and -epsilon; for epsilon above 90 degrees the planet spins backwards and the same formula holds.

Daily insolation. With h0 = acos(-tan phi tan delta) the half-day angle (0 in polar night, pi in polar day) and flux S = 1361 W/m2 / r^2 (r in AU), the daily mean over a rotation is Q = (S / pi) (h0 sin phi sin delta + cos phi cos delta sin h0) Day length is 24 h x h0 / pi. The animation assumes the planet rotates many times per orbit, true for all planets shown.

Elliptical orbit. Position follows Kepler's equation M = E - e sin E solved by Newton iteration, true anomaly nu = 2 atan2(sqrt(1+e) sin(E/2), sqrt(1-e) cos(E/2)) and distance r = a (1 - e cos E). Eccentric planets spend more of the year far from the Sun (Kepler's second law): Mars has a long cold southern summer... and a short hot one near perihelion. The longitude of perihelion decides which season is hot; defaults are the real values for Earth and Mars and illustrative for the others.

What the map shows. The annual mean at the equator beats the pole for small tilt, but above 54 degrees the poles receive more sunlight per year than the equator (Uranus at 98 degrees). The global mean of daily insolation is always S / 4 (times 1/sqrt(1 - e^2) for the orbit average), whatever the tilt.

Honest simplifications. No atmosphere, ocean or ice: the map is sunlight, not temperature. No precession of the axis or perihelion over time, no changes of eccentricity, no refraction at sunrise, a point-like Sun; the tilt and eccentricity values are the real ones but only roughly (rounded).

Checks run in the page. The area-weighted global mean of Q against S/4; a direct integral of the daily insolation over a rotation against the closed form; the orbital time between true anomalies 0 and 90 degrees from the integral of Kepler's second law against Kepler's equation; the Kepler equation residual; 12-hour days at an equinox.