In planetary astronomy, this is the temperature a planet would have just taking into account its reflectivity and its distance from the sun (or another star, if we're talking about a planet in another system). It is also called equilibrium temperature or emission temperature.

Here's how we derive an expression for the effective temperature of a planet. We start with the sun (or another star, but we'll just use the sun for now). The Stefan-Boltzmann law as modified for a gray-body radiator is:

F = ε σ T^{4} |
(1) |

Here,

F is the flux per unit area emitted by the object in question,

ε is the emissivity (a value that can vary from 0 to 1),

σ is the Stefan-Boltzmann constant, and

T is the absolute temperature.

In the SI (Système Internationale d'unités, the mks metric system), F is measured in watts per square meter (W m^{-2}), ε is dimensionless, σ has the value 5.6704 x 10^{-8} W m^{-2} K^{-4}, and T is in degrees Kelvin (K).

The Sun has an effective temperature of 5,779 K and an emissivity indistinguishable from 1, so according to equation (1), it radiates about 63,240,000 watts per square meter. Pretty bright!

The area of a sphere is:

Area = 4 π R^{2} |
(2) |

The Sun has an average radius of 6.955 x 10^{8} meters, according to NASA, so its area is about 6.079 x 10^{18} square meters, and combining equations (1) and (2), the Sun's total luminosity is 3.844 x 10^{26} watts.

Now, all that light spreads out as it travels through space in every direction. At a given distance from the sun, it is spread out over a sphere with the radius of that distance (for instance, a million miles from the sun it's spread over a sphere a million miles in radius). This is another way of saying that light propagates by an inverse-square law.

If we use equation (2) to find the area of the sphere with a radius equal to Earth's mean distance from the sun -- the astronomical unit or AU -- we find that area to be 2.812 x 10^{23} square meters. (I assume that the Wikipedia's value of 149,597,870,691 meters for the AU is accurate). Divide the Sun's luminosity by this area and we have the "Solar constant" -- the amount of sunlight falling through one square meter perpendicular to the sun at the Earth-sun distance:

S = (R / a)^{2} σ T^{4} |
(3) |

The "a" above stands for "semimajor axis" in astronomy -- the mean Earth-sun distance. R is the sun's radius, T its effective temperature. From our figures so far, S comes out as 1,367 watts per square meter.

According to Judith Lean's solar studies, the mean value for S from 1951 to 2000 was 1,366.1 W m^{-2}, so we'll use that. (I.e., my figure for the Sun's effective temperature was a little too high.)

The amount of energy the Earth intercepts from the Sun is then equal to the solar constant times the Earth's cross-sectional area -- or it would be if the Earth were jet black. Actually, the Earth has a nonzero "bolometric Bond albedo" A -- it reflects away some of the sunlight falling on it. About 30.6%, according to NASA. So incoming solar power to the Earth's climate system is:

P_{IN} = S π R_{e}^{2} (1 - A) |
(4) |

The amount the Earth radiates is its total surface area (4 π R^{2}, not the cross-sectional area, π R^{2}) times its flux from the Stefan-Boltzmann law (equation (1)):

P_{OUT} = 4 π R_{e}^{2} σ T_{e}^{4} |
(5) |

The crucial insight from physics is that energy must be conserved. P_{IN} must equal P_{OUT}. If they weren't equal, Earth would be steadily heating up or cooling down with time. In the long run, if the Sun and the Earth's orbit and the Earth's reflectivity are stable, Earth's effective temperature should be stable as well.

Setting P_{IN} and POUT equal, the π R_{e}^{2} factors cancel and we can solve for the Earth's effective temperature:

T_{e} = [S (1 - A) / (4 σ)]^{1/4} |
(6) |

If this looks a little unwieldy, you can break it up into two stages. Define the flux absorbed by the climate system as:

F_{a} = S (1 - A) / 4 |
(7) |

The effective temperature is then:

T_{e} = (F_{a} / σ)^{1/4} |
(8) |

One last point: If we know the Solar constant at Earth's distance from the Sun and define this as "S_{0}," then a quick-and-dirty equation for the solar constant at any planet is:

S = S_{0} / a^{2} |
(9) |

as long as you express a in AUs (a = 1.00000 for Earth, for instance, but 0.72333 for Venus).

With this toolkit of equations in hand, we can now go to NASA web pages and the primary astronomy literature to pick up the latest values for a and A for each planet. That lets us calculate S, F_{a}, and T_{e}. Here are the results, neatly tabulated for your amusement and edification:

Planet | a (AUs) | A (Bond) | S (W m^{-2}) |
F_{a} (W m^{-2}) |
T_{e} (K) |
---|---|---|---|---|---|

Mercury | 0.3871 | 0.119 | 9116.7 | 2007.9 | 433.8 |

Venus | 0.7233 | 0.750 | 2611.2 | 163.2 | 231.6 |

Earth | 1.000 | 0.306 | 1366.1 | 237.0 | 254.3 |

Mars | 1.524 | 0.250 | 588.1 | 110.2 | 210.0 |

Ceres | 2.766 | 0.042 | 178.6 | 42.77 | 165.7 |

Jupiter | 5.203 | 0.343 | 50.46 | 8.289 | 110.0 |

Saturn | 9.537 | 0.342 | 15.02 | 2.471 | 81.2 |

Uranus | 19.19 | 0.300 | 3.710 | 0.6492 | 58.2 |

Neptune | 30.07 | 0.290 | 1.511 | 0.2682 | 46.6 |

Pluto | 39.48 | 0.5 | 0.8765 | 0.1096 | 37.3 |

Eris | 67.67 | 0.86 | 0.2983 | 0.01044 | 20.7 |

For good luck, here they are for some moons:

Moon | Primary |
a (AUs) | A (Bond) | S (W m^{-2}) |
F_{a} (W m^{-2}) |
T_{e} (K) |
---|---|---|---|---|---|---|

Luna | Earth |
1.000 | 0.11 | 1366.1 | 303.9 | 270.6 |

Io | Jupiter |
5.203 | 0.52 | 50.46 | 6.056 | 101.7 |

Titan | Saturn |
9.537 | 0.30 | 15.02 | 2.628 | 82.5 |

Triton | Neptune |
30.07 | 0.89 | 1.511 | 0.04155 | 29.3 |

For planets with A) no atmosphere, and B) no internal heat source, and C) surface emissivity equal to 1, effective temperature is the whole story. But not all worlds meet those criteria. For Venus, Earth, and Mars, the atmospheric greenhouse effect raises the surface temperatures to 735 K, 288 K and 214 K, respectively. For the giant planets, which have internal heat sources, the measured radiating temperature is higher than the effective temperature -- e.g. Jupiter gives off radiation equivalent to an effective temperature of 124 K, rather than the 110 K we expected. And the moon Triton, while having no atmosphere, has an internal heat source, and also has a surface emissivity which may be as low as 0.59, both of which combine to raise its surface temperature to 38 K rather than the 29 K we expect.

To predict the surface temperature of a planet, you need to take into account a lot more processes than just radiation and reflection. But start with effective temperature, because that gives you the basic facts about the planet's radiation balance. You can go on from there.

Page created: | 2010? |

Last modified: | 02/01/2011 |

Author: | BPL |