MadSci Network: Astronomy

Re: Habitable zones, how are they calculated (formulas, not conjecture)?

Date: Wed Jul 12 13:58:03 2006
Posted By: Bryan Dunne, Instructor, Astronomy, University of Illinois
Area of science: Astronomy
ID: 1152650303.As

First, we need to define the inner and outer edges of a habitable zone. Life needs liquid water, so lets define the "HZ" as the distances over which the average surface temperature of the planet will be between the temperatures at which water boils (100 C, or 373 K) and freezes (0 C, or 273 K). Now lets see how to compute the average surface temperature of the planet.

The planet's surface is heated by solar radiation. The amount of solar energy the planet will receive depends 1) the luminosity (total energy output) of its sun, 2) how big the planet is, 3) how reflective the planet is, and 4) how far it is from its sun.

Lets call the luminosity of the sun L*. The radius of the planet is R, and so the planet will collect solar energy over an area πR2 (the area of a circle). The reflectivity of the planet, called its "albedo" (A in our formula), is the fraction of solar radiation that is reflected back to space; Earth's albedo is about 0.3. We are interested in 1-A, how much of the solar radiation goes into heating the planet. The distance from the planet to the sun is d. The sun's total energy (luminosity) will be spread out over a sphere with area 4πd2.

Solar energy warming the planet = L*πR2(1-A)(4πd2)

In order to stay at a steady surface temperature, the planet must radiate away an amount of energy equal to the amount it gets from its sun. This energy will be emitted as thermal or "blackbody" radiation. Assuming the energy is emitted evenly from all over the planet's surface,

Energy emitted by the planet = 4πR2σT4, where R is again the radius of the planet, T is the average surface temperature of the planet (in Kelvins), and σ is a physical constant known as Boltzmann's constant (equal to 1.380650310-23 m2 kg s-2 K-1).

Therefore, 4πR2σT4 = L*πR2(1-A)(4πd2)

You can then solve this equation for d. Finally, given the luminosity of the planet's sun and assuming an albedo (as I mentioned above, the Earth's albedo is about 0.3), plug-in T=273 K (0 C) and T=373 K (100 C) to find the outer and inner limits of the habitable zone, respectively.

Now, this is a very simplified calculation. We have only found an average surface temperature for the planet. Obviously, the surface temperature of a planet will vary, especially with latitude. It is possible that a planet with an average surface temperature at or even slightly below freezing could have regions warm enough for liquid water at its equator.

We have also neglected the effects of the greenhouse gases. The strength of a planet's greenhouse effect depends on the thickness and composition of a planet's atmosphere. For example, the greenhouse effect warms Earth's surface by an average of 32 degrees C (or K). Venus' atmosphere is 100 times the thickness of Earth's and is 96% CO2 (as opposed to less than 1% for Earth); the greenhouse effect on Venus is over 300 degrees C! On the other hand Mars' thin atmosphere only has a greenhouse effect of 5 degrees C.

Another simplication we have made is to ignore the feedback loop between temperature and albedo. As a planet grows colder, its polar regions would get more highly reflective ice, increasing albedo. Conversely, as a planet grows warmer, ice melts, and the albedo decreases. For example, if the Earth were covered in ice, its albedo would be 0.84 instead of 0.30, but if the Earth had no ice caps, its albedo would be about 0.14.

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