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The question is "Find the maximum orbital eccentricity that the Earth would be allowed to have if it stays in the “habitable zone” all the time. Be careful to define what you mean by “habitable zone"". I said that the habitable zone would be where liquid water would exist, so at temps between 275K and 335K but not sure how to go about finding the max eccentricity. Help would be much appreciated.

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    $\begingroup$ Have you determined the minimum (periapsis) and maximum (apoapsis) distances from the sun which satisfy your temperature limits? If so, the WP page on orbital eccentricity has the equations you need to answer your questions. $\endgroup$ – Russell Borogove Mar 8 '18 at 20:07
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Start by finding the maximum distance from the star before water freezes. Then find the minimum distance before water boils. This would be your "habitable zone". This zone can also be expressed as parameters for an orbit (apogee and perigee).

Orbit eccentricity of 0 means a perfect circle, meaning that apogee and perigee (the highest point in an orbit and the lowest point with respect to the object being orbited) happen at the same distance. A eccentricity between 0 and 0.999...<1 form an elliptic orbit. Eccentricity of 1 means that the orbit is no longer an orbit and escape velocity has been archived in a parabola (bye bye!).

Earth's current eccentricity is very close to a perfect circle at 0.0167. The maximum eccentricity would be such that keeps apogee and perigee (when referring to an object orbiting the Sun its perihelion and aphelion) inside the habitable zone limits while and below a eccentricity of 1. Keep in mind that not all orbits are possible.

It is necessary to defining Earth's "habitable zone", is it the zone where life is possible?, if so what life?, Humans and Space Bears have very different tolerances. Also the actual "zone" does not have defined temperature boundaries, there are fluctuations over time.

The habitable zone temperature is not just defined by the distance to the star. Consideration must also be given to local factors like the internal crust and volcanic activity, greenhouse effect, magnetic shield characteristics, ice presence, percentage of water and other gases in the atmosphere, etc.

This is a very simplistic calculation, many other factors and variables are involved but I hope this helps to shed some light in the matter.

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  • $\begingroup$ So if I look it up, and it says the Earth is in the habitable zone between (0.95-1.4)AU, so I set aphelion to 1.4 and perihelion to 0.95, and plug in to the two equations (Rp=(1-e)a , setting a to 1 au, is that how I solve the question? I get eccentricities of 0.05 and 0.4, so 0.4 would be the answer to the question, the max eccentricity $\endgroup$ – Haley Mar 9 '18 at 3:38
  • $\begingroup$ Say the limits are 1.4 and 0.95 a.u. as described above. You don't separately have 1+e=1.4 and 1-e=0.05, you just have the ratio (1+e)/(1-e)=1.4/0.95. Solving then gives e=9/47 or about 0.19. $\endgroup$ – Oscar Lanzi Mar 8 at 0:55
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There is some arbitrariness to how you define the temperature range, particularly on the high side as the boining point of water depends on pressure and not every habitable thing magically has one bar pressure.

Assume that you set temperature limits of $275$ K and $335$ K. The ratio of these temperatures is $335/275=1.22$. Now, if we ignore atmospheric effects the local light intensity from the star varies as $T^{1/4}$ by the Stefan-Boltzmann Law, and the light intensity in turn varies inversely with the square if distance.

Putting these results together we find that the apoapsis/perhaps if ratio maxesout at $(335/275)^2=4489/3025=1.48$.

This $1.48$ ratio is in turn the value of $(1+e)/(1-e)$ for the maximum eccentricity $e$, from which we then get $e=0.19$. This is suspiciuosly close to the value obtained from the other answer (see the comments), and one wonders whether a little "fudging" went on with the temperature limits to get the "right" answer.

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