Imagine the Earth had no axial tilt but had seasons due to a very elliptical orbit. How elliptical would the Earth's orbit have to be in order to have about the same seasons as it has now (just with the difference that they'd be in both hemispheres the same)? Mercury has a very elliptical orbit. If we scaled Mercury's orbit to a semi-major axis at 1 au, would we have the very same seasons? Mercury's surface temperature in its summer is ~430 °C and in winter ~280 °C (values for daytime). I guess this would translate into a too high difference if the Earth's orbit was as elliptical as Mercury's; it would have to be a bit less elliptical for the same seasons as now.

How far would the Earth have to be from the Sun at its perihelion to have about 30 °C on the midlatitudes and at its aphelion to have about 0 °C?


1 Answer 1


TL;DR: about 5 times current eccentricity for a deviation by 5 degrees from the mean... but the devil is in the celestial mechanics and climate model details.

It would not quite work, because the length of seasons would be uneven. Currently Earth's north and south hemispheres get nearly exactly as much or little sunlight as the other. But on the eccentric Earth the hot "summer" period would be short while the cool "winter" period would be long since planets spend more time in the outer part of an elliptic orbit.

There is another complication: absorbed sunlight heats up air, water and soil over time causing a lag: the warmest part of summer and the coldest part of winter comes after the time of the highest solar influx.

To add to the modelling woes, the actual temperature of the planet is strongly affected by the greenhouse effect: while CO2 is not going to do anything weird, as the entire planet cools water vapour (a potent greenhouse gas) will decline as rain and snow, and the albedo (how much light is reflected by clouds and ice) will increase.

That said, here is a simplistic model ignoring the atmosphere inertia: The distance to the star varies as $$r(\theta)=\frac{a(1-e^2)}{1+e\cos \theta}$$ where $a$ is the semimajor axis, $e$ the eccentricity and $\theta$ the "true anomaly" (annual winner for worst name of parameter since 1609). To convert it to and from time you need to solve Kepler's equation numerically.

The temperature in a zero-dimensional atmospheric model is $$T=\left( \frac{ I_0(1-a)}{4 \sigma \epsilon}\right )^{1/4}$$ where $I_0$ is the solar constant at the top of the atmosphere, $a=0.3$ the albedo, $\epsilon=0.78$ the effective emissivity in IR. If we let $I_0(\theta)=I_0/r(\theta)^2$ (measuring orbits in AU) we get a temperature dependency scaling like $T(\theta)=T_0/\sqrt{r(\theta)}$.

So, at perihelion the temperature is $$T(0)=T_0\sqrt{\frac{1+e}{1-e^2}}$$ and the aphelion temperature $$T(\pi)=T_0\sqrt{\frac{1-e}{1-e^2}}.$$ So given a desired $T_{high}$ and $T_{low}$ one can now solve for $e$ to get $$e=\frac{(T_{high}/T_{low})^2-1}{(T_{high}/T_{low})^2}.$$

If we want a 5 degree difference from a mean $T_0=288$ K this means $e\approx 0.067$, about 5 times the current eccentricity. This is probably small enough that the time asymmetry issues will be small enough to ignore (they matter more for more eccentric orbits).

  • $\begingroup$ The 288 K are 15 °C, just clarifying. So there is no way seasons like on Earth would work on an elliptical path? $\endgroup$
    – Greenhorn
    Commented Dec 29, 2020 at 17:05
  • $\begingroup$ @Greenhorn - Depends on what you mean by normal seasons. But given that some part of seasonal weather also depends on solar angle changes (affecting relative heating of hillsides and plains, light levels...) and that the different hemispheres have different insolation, I suspect it is no way to exactly fake seasons of our kind. $\endgroup$ Commented Dec 30, 2020 at 9:07
  • $\begingroup$ If you say the winter period would be longer since the planet is slower during its aphelion, it couldn't be the same as per tilt. Can you estimate how long summer, winter and spring/fall would last on the 5 times more eccentric orbit? $\endgroup$
    – Greenhorn
    Commented Dec 30, 2020 at 10:00
  • 1
    $\begingroup$ @Greenhorn See here for an interactive graph. $\endgroup$
    – PM 2Ring
    Commented Dec 30, 2020 at 15:24

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