# e=0. What Does That Mean for the Seasons?

Currently, Earth's eccentricity (orbital shape) is 0.0167086. Zero is a perfect circle whereas One is parabolic escape orbit and any greater becomes a hyperbola. And in the theory of the Milankovitch cycles, Earth's eccentricity varies between 0.000055 and 0.0679 over a period of 100,000 years. This affects the seasonality of planet Earth, as an extreme ellipsis can result in longer seasons.

Let us assume that Earth's eccentricity today does not exist, leaving it instead in a perfectly circular orbit. What would a circular Earth orbit mean for the seasons and the climate?

• I'm not sure why people are down voting this question. Could those with reasons for doing so state them in comments so John can improve, rather than just leaving a downvote with no indication as to why. – Lio Elbammalf Feb 25 '17 at 12:00
• – Mołot Feb 25 '17 at 13:22
• I DV'd because this question shows no research effort. Simply googling "what if earth had a perfect orbit" resulted in me finding this passage on the second link: "Difference between farthest point i.e perihelion and nearest point I.e.aphelion is hardly 3.3%. This means that it is almost circular orbit."This difference is very mall to cause any major change. Note: Seasons on earth are due to axis of rotation and not due to elliptical orbit." In other words, a succinct answer could be found to the question within 15 seconds of me googling the actual question. I also found... 1/2 – Aify Feb 25 '17 at 17:08
• 2/2 this link which explains that the non-perfect orbital shape of the Earth around the sun does not cause the seasons. – Aify Feb 25 '17 at 17:10
• – JohnWDailey Feb 25 '17 at 20:37

Eccentricity matters little (at least for Earth).

The eccentricity of Earth (e~0.0167086) means that the planet orbits in an ellipsoidal path that is 147.1 million kilometers away from the sun at its closest and 152.1 million kilometers away at its furthest (the maximum is 3.4% larger than the minimum). The energy output of the sun (as a blackbody radiator) is proportional to the Sun's temperature of approximately $5800K-6000K$ at the surface, which results in a power output of approximately $4.0×10^{26}\ W$. If we want to know how much solar power hits earth we typically look at the Watts-per-square-Meter Earth recieves. At a certain distance $r$ from the sun, the power per surface area is proportional to: $$P_{total}/A_{total}=P_{total}/(4\pi r^2)$$ If you plug in the numbers for $P_{sun}$ and both $r_{min}$ and $r_{max}$ you'll get the following values: $$1471.04\ W/m^2 \ (at \ minimum \ distance)\\ 1375.92\ W/m^2 \ (at \ maximum \ distance)$$ and notice that that is a total change of less than 7%, which doesn't seem like nearly enough difference to cause sweltering summers and freezing winters in so many different places on Earth.

Furthermore, if summer were caused by being closer to the Sun, then every continent should experience summer at the same time. However, the Northern hemisphere experiences winter in December while the Southern hemisphere has summer in December. So what is going on?

Obliquity matters a lot.

The Earth is tilted on its axis by $23.5^{\circ}$. The axis points toward Polaris (the North Star) throughout the year, which means that during part of our orbit the Northern hemisphere will be tilted a little more toward the Sun, and during the other half of the year the Southern hemisphere will be tilted more toward the Sun. However, we already know that distance changes don't change the $W/m^{2}$ received by much at all when we look at orbit changes, and orbital tilt is only going to change distances by a much smaller amount that eccentricity, so what is going on?

Direct sunlight is the key. Sunlight bounces off Earth's atmosphere very similarly to how it bounces off the surface of a lake or ocean. If you look straight down into the water very little light is reflected off the surface and you can see into the water. But if you look out across the water a lot of sunlight is reflected and you mostly see reflections rather than seeing into the water. This is exactly what happens to sunlight hitting the Earth's atmosphere. In December, the tilt of the Earth places the Southern hemisphere into the most direct sunlight and very little is reflected off the atmosphere (just like very little reflects of the surface of a lake when we look straight down). However, the Northern hemisphere is tilted away and which means the angle of the sunlight relative to the surface of the atmosphere is much less direct. Therefore, just like more sunlight is reflected when we look out across a surface of water, more sunlight will be reflected by Earth's atmosphere when the Sunlight is less and less direct.

What does that all mean?

It means that fictional spacefarers or worldbuilders will have to be careful in choosing (or designing) their new Earth-like homes.

•    A planet with no tilt and no eccentricity will have no seasons (barring other factors). Such a planet will be warmest at the equator where the sunlight is most direct, and will get colder and colder as latitude increases (as more sunlight is reflected at less-direct angles).
If the planet's sun delivers a similar $W/m^{2}$ as Earth, then the equator will be absolutely sweltering (even in comparison to Earth's). The Arctic will be colder, never getting a warm season, and so permafrost will extend to lower latitudes than we have on Earth. This means that the amount of land with comfortable living conditions will be centered mostly on the middle-latitude bands. (Assuming variety of landmass/oceans is similar to Earth and not some perfectly-designed masterpiece extending habitable zones by ocean currents, etc.)    Following similar logic a low-$W/m^{2}$ planet would push the habitable zones closer to the equator, and a high-$W/m^{2}$ planet would push the habitable zones closer to the poles.

•    A planet with no tilt and high eccentricity could have seasons based on orbital $W/m^{2}$ fluctuations, but the orbits would have to be quite eccentric to create large temperature differences since the $W/m^{2}$ changes proportional to $(r_{max}/r_{min})^2$.
It is also notable that Winter/Summer cycles wouldn't be very symmetric in this sort of situation. Very eccentric orbits would lead to very short hot summers and long cold winters.

•    A planet with Earth-like tilt and no eccentricity would be very Earth-like.

•    A planet with Earth-like tilt and larger eccentricity could be very Earth-like, depending on which way the tilt is pointing. (In other words, tilting-in at the minimal radius could make summers really scorching hot and the long cold winters colder for one hemisphere, but tilting-in during fall/spring would be pretty good.)

•    Any world-building contractors can make their worlds have seasons by tailoring the tilt as they see fit. Furthermore, why design spherical worlds when a cylinder-shape will tilt everyone to the same angle at the same angle at the same time (no more Southern/Northern hemisphere seasonal differences)

• If eccentricity doesn't matter, then why did Dr. Milankovitch include it in his ice age cycle theory? – JohnWDailey Feb 25 '17 at 15:08
• I'm not saying eccentricity doesn't matter "at all" just that obliquity is the "primary" factor effecting the seasons as we percieve them on Earth. If Earth's eccentricity were magically gone tomorrow farmers would still plant in the spring and harvest in the fall; if Earth's obliquity were magically gone tomorrow the seasons as we know them would cease. – fenix d.Anconia Feb 25 '17 at 17:09
• Milankovitch's work looks at obliquity, eccentricity, and precession to account for the changing of the seasons across timescales imperceptible to those of us "living in it". By carefully tracking all three factors Milankovitch shows how we can cycle through ice ages and warm periods over many tens of thousands of years. Were he to have forgotten to account for eccentricity, then the other two factors wouldn't have been enough to successfully account for the entire magnitude of the temperature changes. – fenix d.Anconia Feb 25 '17 at 17:09
• Changing the eccentricity of Earth will contribute to changing the period and severity of ice age cycles, but it won't effect the short-term seasonal weather or climate in the way that any settler or farmer will percieve. – fenix d.Anconia Feb 25 '17 at 17:09
• That's pretty much what I'm after. – JohnWDailey Feb 25 '17 at 20:37

In the northern hemisphere, the winters would be a bit colder and the summers a bit hotter. In the southern hemisphere, it would be the other way around. It wouldn't be a big effect, as you can tell from the fact that seasons work pretty much the same in the northern and southern hemispheres at present despite the Earth being closer to the sun in the northern winter and southern summer.

• This may be true, but it doesn’t follow the rules for a hard-science Answer! – JDługosz Feb 25 '17 at 13:26

Assumptions:

• When we remove eccentricity we move to an orbit with radius of 1 AU.
• The current aphelion (maximum distance) and perihelion (minimum distance) are constants (though in fact they can vary). I make this assumption because the rate of change is of the order of thousands of years.
• Times of seasons depend only on Earth's tilt on it's axis.

Earth's orbital figures:

$r_{av}=149.60\times 10^{9}m$

$r_{pe}=147.09\times 10^{9}m$

$r_{ap}=152.10\times 10^{9}m$

Difference in energy reaching us:

Using the relationship that $E \propto \frac{1}{r^{2}}$ we can compare the energy change at these different points.

$$\frac{E_{av}}{E_{pe}}=\frac{r_{pe}^{2}}{r_{av}^{2}}=\frac{147.09^{2}}{149.60^{2}}=0.9667$$

$$\frac{E_{av}}{E_{ap}}=\frac{r_{ap}^{2}}{r_{av}^{2}}=\frac{152.10^{2}}{149.60^{2}}=1.0337$$

This gives us an idea of the temperature difference between the circular and elliptical orbits. The earth's core provides some of our heat so this isn't a direct relationship, we need to add a constant for the Earth's sunless-temperature.

Gathering this sunless temperature, however, is rather difficult and would involve a lot of assumptions. We have that the power the Earth provides is $47TW$, however this heat transfer to the surface doesn't translate easily into a temperature.

Summary:

• Northern hemisphere has hotter summers (increase of $\approx 1.0337$ times the amount of radiation currently), and cooler winters (decrease of $\approx 0.9667$ times the amount of radiation).
• Southern hemisphere has cooler summers (decrease of $\approx 0.9667$ times the amount of radiation currently), and warmer winters (increase of $\approx 1.0337$ times the amount of radiation).

An explicit answer to this question would need to take into account and model the differences in cloud cover and the changes in water cycle in general

• This assumes that the only source of heat is the sun, that is that the sun is heating earth from absolute zero to where it is. – Theraot Feb 25 '17 at 12:25
• @Theraot Thank you! I knew I had done something silly somewhere. I'll make an edit to take this into account. – Lio Elbammalf Feb 25 '17 at 12:37
• There are a lot of different influences. Places at the same latitude can have totally different climates. – ths Feb 25 '17 at 12:41
• @ths I know, I'm beginning to see that a hard-science answer requires a lot more research. – Lio Elbammalf Feb 25 '17 at 12:50
• The sun is the main heat source and space is basically absolute 0. – Donald Hobson Feb 25 '17 at 14:11