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## Building a Planet: A First Simple Model of Temperature

I expect that a recurring theme in my writing here will be to outline and describe simple models of climate. In doing so, I wish to highlight the assumptions/flaws in those models, as well as the properties of those models which appear robust to more complex calculations (i.e., those properties which emerge in a full hierarchy of model complexity, and are thus more likely to yield meaningful insight in the real world).

Keeping this goal in mind, one could argue that a reasonable starting point for understanding a planet is to simply know how much energy it receives and how that might translate into a temperature if you stood somewhere on the planet. By “energy it receives,” we are accustomed to thinking about the energy from the host star like our sun. Earth receives virtually 100% of its energy from the sun, although it is quite possible to receive significant amounts of energy from sources other than a star (this is the case for several of the gaseous planets in our own solar system). I will ignore those terms in this post, and will return to them in a later article.

As a starting point, I like to envision a spherical rock in space- essentially a “planet” that has been stripped of any atmosphere and which gains energy by absorbing a certain fraction of stellar energy. Let $L_{*}$ be the power output of the star, called the stellar luminosity (measured in Watts for example, with dimensions of energy per unit time). The luminosity for our own sun is approximately 3.85 x 1026 W. The instantaneous stellar flux, $Q$ reaching a planet that orbits its star at a distance $d$ is,

$\displaystyle Q = \frac{L_{*}}{4 \pi d^{2}}$

(more generally, a factor of $\sqrt{1-e^2}$ appears in the denominator for planets that orbit their star in a non-circular orbit, where the factor $e$ is called the eccentricity. Annual-mean insolation (INcoming SOLar radiATION) thus increases for higher eccentricity. $e$ is close to zero (circular orbit) for most solar system planets, though indications from extrasolar planetary systems suggests this is not a general rule).

Imagine now a solid, square surface that stood in space perpendicular to the incoming star light. On Earth, we call the solar constant $S$ the amount of energy that impinges on such a surface at the mean Earth-sun distance. Measured at Earth, $S = Q_{earth} = 1365 W/m^{2}$. The units “Watts per square meter” are helpful because in calculating temperature it turns out to be more useful than just knowing the total power the entire planet receives.

When starlight impinges upon the planet, not all of it is absorbed at the surface. Surfaces typically reflect some fraction of that energy away, depending on the nature of the surface and the wavelengths of the electromagnetic spectrum that strike the surface. Let us define a parameter $\alpha$ to be called the planetary albedo, the bulk fraction of reflected stellar energy (across all wavelengths). It follows that $1-\alpha$ is the fraction absorbed. Earth reflects about 30% of all the energy it receives from the sun right back to space, though most of this is actually due to clouds, but in this simple no-atmosphere model the planetary albedo is the same as the surface albedo. Although we typically think of albedo as a property of the planet, it depends also on the spectrum of starlight received, since atmospheric scattering processes and surface reflectivities are a function of wavelength on the electromagentic spectrum. If Earth were placed around a different type of star, with other climate conditions held constant, one would expect it to have a different planetary albedo. Moreover, albedo can change as climate does, possibly amplifying or dampening the magnitude of change that occurs. Such a feedback process will be discussed in following articles.

We next need to consider the spherical geometry of the planet, and how the starlight is distributed over the sphere. In this simple model, we do something that might appear silly- take the 1370 W/m2 solar energy impinging on a perfectly perpendicular plate facing the sun and distribute it over the whole planet evenly, such that every square meter of the planet is receiving equal amounts of sunlight. What value then should we use in this average for the whole planet?

We can think of a shadow that is cast on a plane sitting “behind” the Earth as the cross-section of a sphere (a circle). Imagine holding a basketball between a projector and the white screen on a board. The image displayed on the screen will be the shadow of a circle, and the area of that circle is the appropriate term to multiply $S$ by in order to obtain the total power received by Earth (or whatever value of $Q$ is appropriate for a planet in consideration). For a planet with radius $r$, the total incoming stellar energy (in Watts) is thus,

$\displaystyle I= Q (1-\alpha) \pi r^{2}$

Outgoing Energy

Obviously our planet does not continue to just soak up sunlight forever without doing anything else. All objects with a temperature also emit radiation, and it turns out that the amount of radiation being emitted is a function of temperature. The emission of such radiation is how the planet sheds the heat it receives from the sun back into space. I will have more to say about the nature of this emitted planetary radiation in many future posts, but for now suffice to say that

– The amount of energy radiated by a body is proportional to the fourth power of its absolute temperature. The relevant constant of proportionality is called the Stefan-Boltzmann constant, often represented by $\sigma$, which is a constant that emerges from quantum physics and reflects the influence of the microscopic world on the planetary scale. Its value is 5.67 x 10-8 W/(m2 K4).

– On Earth, this outgoing radiation is emitted from a planet much colder than the sun, and is thus in the infrared portion of the electromagnetic spectrum (which is why we don’t “see” energy emitted from objects other than very hot surfaces such as a stovetop). I will return to the nature of radiation in a later post to elaborate on this.

-In order to characterize the temperature of the planet, we can assume the planet to be in steady state, such that it receives an equal amount of starlight as it emits energy back into space. This near-perfect balancing act is assured to us by the temperature dependence of the outgoing radiation. If a planet were to suddenly receive more stellar energy, its temperature would rise, and by the Stefan-Boltzmann law, emit more energy until the radiative fluxes balanced again.

– We can assume in this simple model a spherically uniform temperature, such that all points on the planet are characterized by an equilibrium temperature $T_{eq}$ and the energy is emitted over the whole sphere (of area $4 \pi r^{2}$).

Energy balance then assures that:

$\displaystyle Q (1-\alpha) \pi r^{2} = 4 \pi r^{2} \sigma T_{eq}^4$

The planetary radius cancels and becomes irrelevant. Re-arranging terms,

$\displaystyle T_{eq}^{4} = \frac{L_{*} (1- \alpha)}{16 \pi \sigma d^2}$

or,

$\displaystyle T_{eq} = (\frac{Q (1-\alpha)}{4 \sigma})^{0.25}$

Note that temperature goes as the fourth root of the radiation flux, but from equation 1 the radiation flux goes down as the squared distance from the star, so that we expect planetary temperature to drop fairly slowly (as the square root) as you move away from the star.

Limitations/Things to Come

Finally, some notes on the limitations of this energy-conserving zero-dimensional model, much of which will provide fodder for later discussions:

1) The “uniform planet” model is not valid for a bare rock, since one would expect rapid changes in the temperature between day and night. However, it works by coincidence for Earth, Venus, and other bodies that in reality have an atmosphere (or oceans) to transport heat around. Since we are measuring temperature in Kelvins, the absolute value fluctuates by only a few percent between day and night, or equator to pole. However, the approximation is not useful for Mercury, the moon, or other airless bodies. What is relevant then may be the hemispheric-averaged temperature, or the instantaneous temperature at a point assuming radiative equilibrium, which affects the term one ends up using in the geometrical considerations (e.g., for the local, instantaneous temperature at the equator, at noon, on a bare rock one would ignore the division by 4).

2) We have implicitly ignored any temperature dependence on albedo. This, in general, is not a good assumption since we expect the characteristics of a planet (like cloud or ice cover) to change with temperature. In considering a broad range of climate-space, this temperature dependence of albedo gives rise to a family of solutions for what equilibrium climate one ends up with for a given stellar luminosity, and can introduce dependence on the history of what path a planet took to get to a given state. This gives rise to bifurcation structures and hysteresis, which will be explored in the next post. This is perhaps an unsatisfactory limitation of the simple model, in that the albedo is largely determined by the intrinsic properties of a planet, so one cannot obtain a unique solution for temperature on some distant exoplanet by only knowing the distance to the star.

3) The calculated equilibrium temperature depends only on the absorbed starlight. However, a fundamental characteristic of many planets is its atmosphere, and all known bodies with a significant atmosphere in some way alter the flow of energy into and/or out of the planet. On Earth, $T_{eq}=255 K$, below freezing, yet the true global mean temperature is closer to 290 K. This is due to a greenhouse effect generated by the atmosphere which inhibits radiative heat loss to space. Venus’ true temperature is a hellish 735 K due to its atmosphere, even though its calculated $T_{eq}$ is just over 230 K (colder than Earth because its high albedo over-compensates for the closer distance to the sun). Moreover, there are no feedbacks in this model: in reality, if one made the Earth that cold, high-albedo ice would begin to cover the oceans and reflect more sunlight forcing $T_{eq}$ to be much less than 255 K. Similarly, Venus would not retain its high albedo (~75%) if it actually had no atmosphere, yet it would still be much colder than currently observed. It should be kept in mind however that the parameters one uses in the energy balance calculation are not always self-consistent with the assumptions used. In a future post, we will see that $T_{eq}$ will no longer correspond to the surface temperature when we include an atmosphere, but rather a temperature encountered at a certain height above the surface. This, and building on other assumptions, will also be developed in a later post.

4) Once we cover the basics in detail, I will discuss in a later post that for a planet with liquid oceans, $T_{eq}$ must reside below a critical value, typically 275 K or so, in order for that ocean to exist without vaporizing completely into the atmosphere (this value increases for higher values of the gravitational acceleration of a planet). This fact is crucial to the question of habitability on other worlds. 255 K is well below that limit, and in fact Venus resides even below this (~235 K). It is a necessary but not sufficient condition, however, since Venus and Mars both meet that criteria yet neither have liquid water.

### 6 Responses

1. on June 12, 2012 at 9:51 am | Reply ozzieostrich

Hi Chris,

Quick question. What would happen if you started with a molten Earth sized rock 4.5 billion years ago, with a radiogenic core?

Would it be possible that by now, the net result of cooling, internal radioactive heat generation, insolation, meteorite impact etc., would result in a temperature of around 290K?

Alternatively, when do you estimate the average global surface temperature dropped to about 255K?

Is there any evidence that the Earth cooled to 255K (average, so parts would have been much colder) when CO2 was at a minimum in the atmosphere?

Thanks.

Mike Flynn.

• Mike- I don’t really know how to answer this, but I know of a few studies who estimate the current geothermal heat flux to be on the order of ~0.1 W/m2 (compared to the order 300 W/m2 from the sun), of which a large fraction of this is from radioactive decay, see e.g., http://www.nature.com/ngeo/journal/v4/n9/full/ngeo1205.html so just from present-day calculations it doesn’t work as a mechanism to heat the surface on a global-scale.

There’s some literature on the impact collision between the Earth and “moon” that happened early in Earth’s history, and eventually resulted in the formation of the moon we know today. The cooling timescale of such an event is on the order of a few thousand years, and possibly a few million years for a magma ocean to cool off at the surface, depending on how efficiently the atmosphere allows heat to escape to space.

2. on June 13, 2012 at 12:02 am | Reply ozzieostrich

Chris,

You may be missing my point. The fact that the Earth is still losing heat from the core, as you have said, shows that the core is cooling.

If the surface started out in a molten state, it has now cooled to the point where the crustal skin is, as you say, around 290K or thereabouts.

At the base of the crust (somewhere between 5km and 45 km give or take), mantle temperatures are sufficiently high that the outer mantle’s viscosity can drop to low levels. You have no doubt seen pictures of liquid lava flowing extremely quickly due to its low viscosity. Other lava is much more viscous, at a lower temperature, and flows much more slowly. Obviously this is a much simplified picture, but demonstrates that not far below the surface of the Earth, it is very hot.

So it is logical to assume that the Earth is still cooling, and indeed, geophysicists have written papers quantifying the rate in degrees per year. I assume you are aware of these.

You can see why I am wondering why you would assume that the Earth’s surface is at 255K for the purpose of climate modelling, when it appears it will not reach that temperature for some millions of years.

It might give some perspective to realise that the thickness of the crust on this molten blob we call the Earth is about as thin as the skin on an apple, in proportional terms.

I trust I have clarified my question. What evidence do you have that the Earth’s surface temperature (or skin temperature, if you wish), was ever 255K?

I am unaware of any.

The Earth seems to have cooled to its present temperature, and appears to be continuing to cool, according to the geophysicists.

Thanks,

Mike Flynn.

• I don’t know what point you’re trying to make, but I’m talking about climate in this post. The rate of interior cooling is a distraction.

You seem to be imagining a situation in which the whole climate system starts off in some molten, warm state and gradually cools to some threshold temperature over billions of years that is finally determined by sunlight (it’s actually easy to show that the Earth would have been unstable to a runaway greenhouse in such a case and would have lost all its liquid water over history, a subject for a future post).

Of course, this view isn’t at all close to reality, and I doubt there are any geophysicists that would disagree. The geothermal flux doesn’t propagate onto the surface temperature at all, except perhaps in localized hot springs/geysers, etc.. For the purposes of this topic, you don’t need any knowledge of what is happening in the interior of the Earth, and you don’t need a geologic time-history of radioactive decay rates.

(For the purposes of maintaining moderation in the early goings of this blog, I’m leaving it at that. I don’t want to clutter up comments with science fiction).

3. “Energy balance then assures that:

\displaystyle Q (1-\alpha) \pi r^{2} = 4 \pi r^{2} \sigma T_{eq}^4 ”

This equation assumes the planet radiates as a perfect blackbody, which isn’t quite true. In reality, the emission of a solid body approximates that of a blackbody with a small correction factor, the emissivity $\latex \varepsilon$. Therefore, the actual equation of radiative equilibrium or energy balance is:

\displaystyle Q (1-\alpha) \pi r^{2} = 4 \pi r^{2} \sigma \varepsilon T_{eq}^4

In practice this isn’t very important because, in the thermal infrared, $\latex \varepsilon$ is close to unity for most common natural materials. For ice, snow, water, vegetation, soil and rock, $\latex \varepsilon$ is between 0.95 and 0.995 in the infrared (however, artificial materials can have much lower emissivities). Still it should be included for completeness.

4. on June 17, 2012 at 11:08 am | Reply ozzieostrich

Hi Chris,

May I suggest that you read a few papers relating to the geology of the Hadean. “Hellish” for good reasons. Obviously, the water was not lost, in spite of conditions.

Anyway, good luck.

Live well and prosper,

Mike Flynn.