Blackbody temperature To find the effective (blackbody) temperature of a
planet, it can be calculated by equating the power received by the planet to the known power emitted by a blackbody of temperature . Take the case of a planet at a distance from the star, of
luminosity . Assuming the star radiates isotropically and that the planet is a long way from the star, the power absorbed by the planet is given by treating the planet as a disc of radius , which intercepts some of the power which is spread over the surface of a sphere of radius (the distance of the planet from the star). The calculation assumes the planet reflects some of the incoming radiation by incorporating a parameter called the
albedo (a). An albedo of 1 means that all the radiation is reflected, an albedo of 0 means all of it is absorbed. The expression for absorbed power is then: :P_{\rm abs} = \frac {L r^2 (1-a)}{4 D^2} The next assumption we can make is that the entire planet is at the same temperature , and that the planet radiates as a blackbody. The
Stefan–Boltzmann law gives an expression for the power radiated by the planet: :P_{\rm rad} = 4 \pi r^2 \sigma T^4 Equating these two expressions and rearranging gives an expression for the effective temperature: :T = \sqrt[4]{\frac{L (1-a)}{16 \pi \sigma D^2}} Where \sigma is the Stefan–Boltzmann constant. Note that the planet's radius has cancelled out of the final expression. The effective temperature for
Jupiter from this calculation is 88 K and
51 Pegasi b (Dimidium) is 1,258 K. A better estimate of effective temperature for some planets, such as Jupiter, would need to include the
internal heating as a power input. The actual temperature depends on
albedo and
atmosphere effects. The actual temperature from
spectroscopic analysis for
HD 209458 b (Osiris) is 1,130 K, but the effective temperature is 1,359 K. The internal heating within Jupiter raises the effective temperature to about 152 K.
Surface temperature of a planet The surface temperature of a planet can be estimated by modifying the effective-temperature calculation to account for emissivity and temperature variation. The area of the planet that absorbs the power from the star is which is some fraction of the total surface area , where is the radius of the planet. This area intercepts some of the power which is spread over the surface of a sphere of radius . We also allow the planet to reflect some of the incoming radiation by incorporating a parameter called the
albedo. An albedo of 1 means that all the radiation is reflected, an albedo of 0 means all of it is absorbed. The expression for absorbed power is then: :P_{\rm abs} = \frac {L A_{\rm abs} (1-a)}{4 \pi D^2} The next assumption we can make is that although the entire planet is not at the same temperature, it will radiate as if it had a temperature over an area which is again some fraction of the total area of the planet. There is also a factor , which is the
emissivity and represents atmospheric effects. ranges from 1 to 0 with 1 meaning the planet is a perfect blackbody and emits all the incident power. The
Stefan–Boltzmann law gives an expression for the power radiated by the planet: :P_{\rm rad} = A_{\rm rad} \varepsilon \sigma T^4 Equating these two expressions and rearranging gives an expression for the surface temperature: :T = \sqrt[4]{\frac{A_{\rm abs}}{A_{\rm rad}} \frac{L (1-a)}{4 \pi \sigma \varepsilon D^2} } Note the ratio of the two areas. Common assumptions for this ratio are for a rapidly rotating body and for a slowly rotating body, or a tidally locked body on the sunlit side. This ratio would be 1 for the
subsolar point, the point on the planet directly below the sun and gives the maximum temperature of the planet — a factor of (1.414) greater than the effective temperature of a rapidly rotating planet. Also note here that this equation does not take into account any effects from internal heating of the planet, which can arise directly from sources such as
radioactive decay and also be produced from frictions resulting from
tidal forces.
Earth effective temperature Earth has an albedo of about 0.306 and a
solar irradiance () of at its mean orbital radius of 1.5×108 km. The calculation with ε=1 and remaining physical constants then gives an Earth effective temperature of . The actual temperature of Earth's surface is an average as of 2020. The difference between the two values is called the
greenhouse effect. The greenhouse effect results from materials in the atmosphere (
greenhouse gases and clouds) absorbing thermal radiation and reducing emissions to space, i.e., reducing the planet's emissivity of thermal radiation from its surface into space. Substituting the surface temperature into the equation and solving for ε gives an
effective emissivity of about 0.61 for a 288 K Earth. Furthermore, these values calculate an outgoing thermal radiation flux of (with ε=0.61 as viewed from space) versus a surface thermal radiation flux of (with ε≈1 at the surface). Both fluxes are near the confidence ranges reported by the
IPCC. == See also ==