Can a planet with 2/3 Orbital eccentricity sustain life?

The eccentricity of an orbit is defined as $$e=\frac{r_a-r_p}{r_a+r_p}$$ where $r_a$ and $r_p$ are the distances to the farthest and closest points of the planet's orbit from the star. If $e=2/3$, then $r_a=5r_p$, which is quite significant. It seems that the planet will not remain in the habitable zone for its entire orbit.

However, it turns out that it doesn't have to. The flux on a planet's surface scales like $$F\propto a^{-2}(1-e^2{)}^{-1/2}$$ where $a$ is the semi-major axis, with $a=(r_a+r_p)/2=3r_p$. Given that $e=2/3$, the eccentricity factor is only $$(1-(2/3{)}^2{)}^{-1/2}=(5/9{)}^{-1/2}\approx 1.34$$ which actually isn't that much. In other words, the mean flux is not substantially different from the flux incident on a planet orbiting at $3r_p$ with $e=0$.

In winter everything freezes over, at about -200ºC surface temperature.

The effective temperature at a distance $r$ from the star is proportional to $r^{-1/2}$. Therefore, at $r=r_a=5r_p$, $T(r_a)\approx 0.45T(r_p)$. At $r=a=3r_p$, $T(a)\approx 0.58T(r_p)$. In other words, ignoring things like thermal inertia and atmospheric currents, the temperature at aphelion should not be substantially different from the temperature of the planet in the habitable zone. So even if we ignore thermal inertia, such dramatic temperature swings seem a little unlikely.

Could animal and plan life sustain this period by implementing something similar to existing hibernation/dormancy?

We did see that at the closest approach, temperatures will be significantly higher than at the habitable zone, and as planets move quicker when they're closer to the star, it seems likely that there will indeed be hibernation of some sort - just when it's hot. They'll need to either hibernate for much of a year or else just adapt to the conditions. But the environment there is pretty hostile to life. For the other half of the orbit, though, things seem more conducive to life.

posted about 10 years ago

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