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We see plenty of seasonal variation on Earth, due to the large axial tilt, but the Earth follows an almost perfectly circular orbit, so we don't get much variation as a result of that.

What if we had a planet with similar axial tilt, similar conditions, etc., but in an eccentric orbit? What will that do to the climate, especially in terms of temperature variance? How much hotter/colder are the extremes going to be, and how much worse will things get when the two effects (eccentricity and axial tilt) combine forces to create even nastier extremes? For bonus points, what will temperature bands look like, given that an eccentric orbit will cause a specific latitude (somewhere between the tropic lines) to be very hot at perihelion, since the sun will be directly overhead on that line at the point of closest approach?

I'm hoping for answers with formulas that explain how one could calculate this, so that other people can build on the information and tweak it to suit their worlds. For the same reason, I'll offer a few sample orbits for people to use in providing examples.

Case A: eccentricity of 0.1, perihelion at the northern summer solstice.

Case B: eccentricity of 0.1, perihelion at the vernal (northern spring) equinox.

Case C: eccentricity of 0.2, perihelion at the northern summer solstice.

For the purposes of this question, assume that Earth is the planet following an eccentric orbit; it'll make things easier to understand. Answers that express temperature variance in proportion to what Earth faces are fine: if, for instance, case A is going to have a northern winter 20% colder than Earth would, that's pretty straightforward.

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    $\begingroup$ It's pretty complicated. Most of Earth's temperature stabilization comes from water, and ocean currents. Once oceans freeze, changes starts to be dramatic. $\endgroup$ – Mołot Oct 14 '16 at 8:24
  • $\begingroup$ @Molot I would think the increased/decreased solar radiation hitting the planet would have a significant effect on temperatures, if nothing else. $\endgroup$ – Palarran Oct 14 '16 at 14:27
  • $\begingroup$ Related: worldbuilding.stackexchange.com/q/10434/627 $\endgroup$ – HDE 226868 Oct 14 '16 at 15:18
  • $\begingroup$ Palarran, I've removed the hard-science tag after Youstay put it on, because it requires answers to be very exact. You can put it back on if you want; flag if you do so, so I can add the hard science post notice. $\endgroup$ – HDE 226868 Oct 21 '16 at 19:04
  • $\begingroup$ Also related: worldbuilding.stackexchange.com/q/55560/1971 $\endgroup$ – ApproachingDarknessFish Dec 6 '16 at 5:55
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That's a difficult question to answer. Every formula, I don't explain is from Wikipedia. I'll try to answer your first question: How would an more eccentric orbit than earth influence the temperature on a planet which orbits around a star. However, since I still deem this to be a difficult question, let me make a few simplifications: (I will later discuss what happens if you don't make this)

  1. The Planet consists completely of only one solid material x. Where $x$ has a density of $\rho_x$, a specific heat capacity of $c_x$ and a specific thermal conductivity of $\lambda_x$. No fluids, no gases, no athmosphere.
  2. The temperature on every point of the surface planet is the same, at any time.
  3. The planet has no internal heat source
  4. Only the surface changes temperature, deep inside it will always stay at the same temperature.

Calculation

What I will effectively do is calculating the temperature y meters below the surface. In other words, I'll calculate a function $f(t,d)$ of the temperature, where $t$ is the time and $d$ the depth, how many meters below the surface we calculate the temperature. Further let $D(t)$ be the distance the planet has at time $t$ from the star.

How much power does reach the planet on every square meter? You have to calculate how much light reaches the planet.

In formulas: $\frac{Lr^2(1-A)}{4 D^2}$ Where $L$ is the luminosity of the star (for our Sun about $3.845*10^{26} W$), $r$ the radius of the planet, $A$ the albedo (for our earth about $0.3$) and $D$ the distance to the star. To calculate the power per square meter you divide by the surface of the planet, which gives us $P_{in} = \frac{L(1-A)}{16 \pi D^2}$ (in $\frac{W}{m^2}$).

How much energy gets lost due to infrared radiation per square meter? $P_{out} = \epsilon \sigma T^4$, where $\epsilon$ is the emissivity of the planet (for earth approx $0.612$) and $\sigma \approx 5.67 * 10^{-8} \frac{W}{m^2K}$ the Stephen-Bolzman-constant.

Due to thermal conduction, the interior of the planet heats the surface and in the process cools itself down. The equation for the heat flow density is $q(\frac{\Delta T}{l}) = \lambda \, \frac{ \Delta T }{ l }$, where $\lambda$ is the specific thermal conductivity, $\Delta T$ the temperature difference, $l$ the length of the object (in our example the depth) and $q(\frac{\Delta T}{l})$ is the thermal conductivity as a function of the length and the temperature difference. Here I'm using 5., since the planet is smaller at a depth of $d$ than at the surface. For small $d$ this is negligible. The temperature at a depth $d$ only depends upon the temperature directly above and below. Therefore the energy flow at $d$ meters below surface can be calculated as $\frac{\mathrm{d}}{\mathrm{d}d}q(f(t,d))$. On the surface it is only half as much since noting is above the surface. Therefore, we get a power per square meter on the surface of $P_{flow} = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}d}q(f(t,d))$.

Now we have $P_{in} - P_{out} + P_{flow} = 0$, since only these three factors influence the surface temperature. Therefore we have: $$ \frac{L(1-A)}{16 \pi D^2} - \epsilon \sigma f(t,0)^4 + \left[\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}d}q(f(t,d))\right]_{d=0} = 0 $$ Inserting $q$ gives us: $$ \frac{L(1-A)}{16 \pi D^2} - \epsilon \sigma f(t,0)^4 + \left[\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}d}\lambda \, f(t,d)\right]_{d=0} = 0 $$ I don't even try to solve this differential equation exactly. However, it is possible to calculate the temperature $f(t,d)$ at the surface approximatively given all other data.

To heat $1m^3$ of $x$ up by one degree, we need an energy of $\rho_x * c_x$. $q$ is the heat flow. To calculate how the thermal energy changes, we have to differentiate $q$. Therefore the temperature change at a depth $d$ is $$ \frac{\mathrm{d}}{\mathrm{d}t}f(t,d) = \frac{\frac{\mathrm{d}}{\mathrm{d}d}q(d, f(t,d))}{\rho_x * c_x} = \frac{\lambda\frac{\mathrm{d}}{\mathrm{d}d}f(t,d)}{\rho_x * c_x} $$ This allows us to calculate how the temperature changes given that we know how it is at the moment. With computer aid you can approximatively calculate the temperature of the planet.

Example

Let us choose $x$ to be $SiO_2$. The mantle consists to a large part of $SiO_2$. We have $\rho_{SiO_2} \approx 2500\frac{kg}{m^3}$, $\lambda_{SiO_2} \approx 1.3 \frac{W}{m * K}$ and $c_{SiO_2} \approx 1050 \frac{J}{kg K}$. Let us assume at first our planet circles the sun at a distance of $1AE = 1.496 * 10^{11} m$, just like the earth, but on a perfect circle. (Remember: $L = 3.845 * 10^{26} W$, $A = 0.3$ and $\epsilon = 0.612$)

After some time the temperature will become the same everywhere. This means, we have $P_{flow} = 0$ and $P_{in} = P_{out}$. This gives us a constant temperature of $288.15K \approx 15 °C$. (That is about what is true for the earth)

Now let us assume, by magic the planet is suddenly on an circular orbit $2AE$ from the sun. That means that it gets only $\frac{1}{4}$ of the light. The planet will eventually cool down to about $203K$. However, that will take some time. How is it after say $180$ days? My calculation says, that up to 7 meters, everything is completely frozen, while below 8 meters the temperature is still about $288 K$. (Compare to permafrost where the ice never melts starting from a few meters below the surface)

This example tells us, that an eccentric orbit will only influence the surface temperature.

Generalizations

In the beginning I made a few assumptions.

  1. Point gives me a little trouble: You can easily use the same model for several different materials. (Simply assume, that $x$ is a mix of them or calculate it independent for different places on the surface) However, the athmosphere can play havoc on the calculations. The same goes for liquids like oceans. However, I think, the atmosphere would rather benefit the living creatures, since it slows the temperature changes. Oceans will likely stay the same temperature a few hundred meters below the surface, like now (even when the eccentricity is as extreme, as my example above.

  2. is easy to work a round: One can make the same calculation for different points of the surface. However, one has to consider, that different places get a different amount of sun light: On midday at the equator, you get 4 times the light than an average place.

  3. The internal doesn't has a relevant effect on the surface temperature. However, it does help to keep everything below the surface warm, a little bit.

  4. We saw in the example that this assumption holds.

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    $\begingroup$ So, you think that the cooling/warming effects at perihelion and aphelion will be more or less completely carried out over the course of a year? We won't end up going halfway to the predicted temperature and then having the temperatures turn around because the planet's out of that part of the orbit already? $\endgroup$ – Palarran Oct 24 '16 at 14:30
  • $\begingroup$ This is useful, especially the part about internal heat conduction. I've been looking for a way to analyze that for the past week and a half. Do you have any more information on the $P_{\text{flow}}$ term, especially as specifically applied to planets? $\endgroup$ – HDE 226868 Oct 24 '16 at 14:35
  • $\begingroup$ @Palarran, I didn't want to say, that only the perihelion and aphelion temperature matters. Since I was short time and would have to find a good formula for the precise orbit data, I had to decide, whether I post what I have or nothing at all. I decided to go for the first $\endgroup$ – lurch Oct 26 '16 at 16:55
  • $\begingroup$ If my calculations are correct, you can ignore that $P_{flow}$ term. It's to small to matter. However, it exists because the heat from the inside of the planet warms the surface. $\endgroup$ – lurch Oct 26 '16 at 16:57
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So this problem basically has two parts. The answers so far have tried to answer the problems with evening heat out over the surface of the planet. I am working on a circulation model to do just that, but that is complicated and not ready. However, the other part of the problem is the question of insolation. How much solar energy is your planet really receiving? That I can answer with the approach I used here.

Defining planet's orbit

This is just application of Kepler's laws. The semi-latus rectum can be defined by $$ ap = b^2,\quad b = a\sqrt{1-e^2}$$ where $p$ is the semi-latus rectum, $a$ is semi-major axis, $b$ is semi-minor axis, and $e$ is eccentricity. Assuming semi-major axis is 1 AU, those two equations combined give $$p = 1-e^2.$$

The semi-latus rectum is used to find orbital distance as a function of time by $$r = \frac{p}{1+e\cos(\theta)}.$$

Since theta goes from $0$ to $2\pi$ and we want that orbit to take 365 days, we can use python (3.5.2) code to return distance as a function of eccentricity and time (in days) as:

def f(e, t):
    p = 1 - e**2
    theta = 2 * pi * t / 365
    return p / (1 + e * cos(theta)), theta

Zero time for this function is defined at perihelion.

Plot solar energy as a function of time

The first component of solar energy is the relative solar energy due to distance. This is simple, since solar energy flux follows an inverse square law, energy = 1/r**2.

The second component is the seasons. Copying my work from that other problem I linked, I use insolation at 45 degrees north. Insolation as a percentage of maximum (maximum insolation of 1 unit being when the sun is directly overhead) is the cos(latitude - axial_tilt * sin(time)). Axial tilt is 23.5 degrees, same as earth.

Since this seasonal component is a percentage of maximum, we simply multiply it by the distance component to get total energy.

Here is the code that gives us some graphs:

def plot_energy():
    x = [i/10 for i in range(3650)]
    orbit_a = [r for r, theta in [f(0.1, t/10) for t in range(3650)]]
    orbit_c = [r for r, theta in [f(0.2, t/10) for t in range(3650)]]

    seasons_a = [cos(radians(45-23.5*sin(i/10/365*2*pi+pi/2))) for i in range(3650)]
    seasons_b = [cos(radians(45-23.5*sin(i/10/365*2*pi))) for i in range(3650)]

    energy_a = [1/r**2 * n for r, n in zip(orbit_a, seasons_a)]
    energy_b = [1/r**2 * n for r, n in zip(orbit_a, seasons_b)]
    energy_c = [1/r**2 * n for r, n in zip(orbit_c, seasons_a)]

    plt.plot(x,energy_a, 'b', x, seasons_a, 'g')
    plt.axis([0, 365, 0, 1.5])
    plt.show()

    plt.plot(x,energy_b, 'b', x, seasons_b, 'g')
    plt.axis([0, 365, 0, 1.5])
    plt.show()

    plt.plot(x,energy_c, 'b', x, seasons_a, 'g')
    plt.axis([0, 365, 0, 1.5])
    plt.show()

And here are the graphs. Blue represents your planet's insolation, and Green represents the Earth, with the seasons aligned.

Case A

enter image description here

Case B

enter image description here

Case C

enter image description here

I added the python code so you can replicate or modify if you want. If you do use it, make sure you are in python3 and import from math and matplotlib:

from math import sin, cos, pi, radians
from matplotlib import pyplot as plt

Discussion

While the seasons play a bigger role in insolation changes than the orbit does, the combination of summer at perihelion is significant: a 45% increase in solar energy. It is worthwhile pointing out that with Northern Hemisphere summer aligned with perihelion, the Northern Hemisphere on your planet gets significantly higher solar energy than the southern hemisphere. If the equinox is aligned to perihelion, then the seasons are equal in North and South.

If 1.00 is the amount of sunlight energy received in a year with the sun directly overhead at 1 AU, than on Earth 45 N gets 0.68 and the Equator gets 0.96 over the course of the year.

For your cases, the Northern hemisphere, equator, and southern Hemisphere get the following values:

  • Case A: N = 0.72, E 0.98, S = 0.67
  • Case B: N = 0.69, E 0.98, S = 0.69
  • Case C: N = 0.81, E 1.06, S = 0.69

You will also notice that you get more energy the more eccentric your orbit is. This is due to the sun's delivered energy increasing more for moving 10% inwards than for moving 10% outwards (inverse square law, and all).

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As it's been a week without any answers offered, I think I'll share the results of my own further research here and see if people think it's close to the mark.

To get a basic temperature estimate for the planet, using the calculations for effective temperature seems like a good idea. The Wikipedia calculations, however, calculate it over the course of a year; to cover an eccentric orbit, I'm going to run the calculations based on perihelion and aphelion distances, compare those to the result at the mean distance, and see how big the gap is.

Formula: $$T = (\frac{1}{4} * \frac{L(1 - a)}{4\pi\delta\epsilon D^2})^\frac{1}{4}$$ The first ratio of 1/4 indicates how incoming solar energy is spread across the surface; a quickly rotating sphere ends up with about 1/4 (the area of a disc of equal radius), but a tidally locked planet would have about 1/2. L is stellar luminosity, about $3.84 * 10^{26}$ in the case of Earth's Sun. $\delta$ is the Stefan-Boltzmann constant, around $5.67 * 10^{-8}$. D is the distance between planet and star (for Earth, this averages about 149.5 million kilometres over the course of a year). $\epsilon$ is the emissivity, approximated as 0.96 to 1; this is how much of the incoming radiation is reradiated out to space again. a is the albedo, or how much energy is reflected; this is around 0.3 for the Earth.

For anybody doing this themselves, units: L is in watts per square metre, D in metres. T is in degrees Kelvin (this is basically the Celsius scale, but 0 C is 273.15 K).

The basic formula doesn't account for the greenhouse effect, internal heating, etc.; it effectively assumes that the planet has no atmosphere, and produces a figure of about 255 K for temperature (-18 C), which is obviously far too low. For a simplistic approximation that lumps all of those extra factors into the greenhouse effect, emissivity can be altered: for the Earth, this results in an emissivity of about 0.612, mostly due to cloud cover. T then becomes around 288 K (15 C), a much more reasonable figure.

Now, to actually calculating some examples, where PT = T at perihelion, AT = T at aphelion:

Case 1: D ~ 134.5km at perihelion, and ~ 164.5km at aphelion. $$PT = (\frac{1}{4} * \frac{3.84*10^{26}(1 - 0.3)}{4\pi*5.67*10^{-8}*0.612* 134,550,000,000^2})^\frac{1}{4} = ~303.7 K (~30.59 C)$$ $$AT = (\frac{1}{4} * \frac{3.84*10^{26}(1 - 0.3)}{4\pi*5.67*10^{-8}*0.612* 164,450,000,000^2})^\frac{1}{4} = ~274.75 K (~1.60 C)$$ Yikes! Those figures don't look too promising for life; those are averages over the planet, which means that perihelion probably annihilates the ice caps and turns the equator line into a 40 C or higher deathtrap. Still, I could be wrong; maybe these predictions are only equilibrium temperatures in these cases, which means that the Earth is trying to reach them but not fully doing so (courtesy of the day going by, putting various parts under shade to avoid baking them like bread).

If anybody can offer an answer covering how the extremes might be softened, that would be appreciated. Answers or comments covering how seasons might look different (what spring might turn into if perihelion is during spring instead of summer, for instance) are also welcome.

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  • $\begingroup$ Yes I think you need to factor in sources of thermal inertia. On the flip side, it suggests how the author can make it work: provide sufficient buffers e.g. melting and forming ice. $\endgroup$ – JDługosz Oct 21 '16 at 18:47
  • $\begingroup$ A robust biosphere could theoretically modify the albedo of the planet (either by accident or purposefully), and stabilize the temperature. Think leaves falling on the ground during colder periods, showing the dark ground underneath, which absorbs more sunlight, keeping the planet above freezing. $\endgroup$ – Bloc97 Oct 22 '16 at 17:34
  • $\begingroup$ I think you missed a very important point: The planet needs time to warm up (cool down). For the circular orbit, this doesn't matter, since the temperature does not change. However in this case, it does matter. It would take about 2.6 years to cool the earth down by 16 degrees in average if the sun stops shining and it loses the same amount of heat to the space as it does now. That said, we are only interested in the surface. $\endgroup$ – lurch Oct 23 '16 at 20:30
  • $\begingroup$ @lurch I didn't miss that point: notice where I mention that the predictions could be equilibrium temperatures. The problem is that I don't know how to adequately calculate what the actual temperatures would become as a result; I can't figure out how close to those predictions the temperatures would actually get. Feel free to post an answer if you have better estimates than I do. $\endgroup$ – Palarran Oct 23 '16 at 20:44
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There's about an 20 W/m^2 difference in solar input when averaged over the curved surface and night/day but the effect of this goes largely unnoticed because perihelion aligns almost exactly with the N hemisphere winter solstice and its effect is conflated with the seasonal solar difference which when averaged over each hemisphere is about 280 W/m^2 in the N and 300 W/m^2 in the S. Notice the difference in seasonal variability in the S? In about 11K years, this will be reversed.

The planet behaves quite differently when perihelion is aligned with the N hemisphere summer because of the significant asymmetry between how the 2 hemispheres respond. This is evident in the ice core data which has local minimum and maximum temperature swings of about 2 degrees C that corresponds exactly to the periodicity of the precession of perihelion.

To put this in perspective, the solar energy increase that's equivalent to doubling atmospheric CO2 since pre-industrial times is only about 3.7 W/m^2.

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