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Effective temperature of a planet, calculated by first approximation
Definition:
T = Effective temperature of an object such as a star or planet. It is the temperature of an ideal
blackbody that radiates the same total energy per unit area as the given object, unit in kelvin (K).
Also known as blackbody temperature.
L = Luminosity of a star, defined as the total energy per second (i.e. power) radiated by the star, unit
in watts (W). It propagates in all directions according to the inverse-square law, i.e. the intensity
of radiation is inversely proportional to the square of distance from the source.
A = Albedo, the fraction of radiation reflected from a planetary body. An albedo of 1.0 means that all
incident radiation is reflected, an albedo of 0.0 means all radiation is absorbed. For examples,
the Earth’s average albedo is 0.30, Mars is 0.16, Jupiter is 0.51.
 = Stefan-Boltzmann constant = 5.67 x 10–8 W/m2/K4. The effective temperature of a star can be
derived from the Stefan-Boltzmann law: L = 4  R2  T4, where R is the radius of the object.
For the Sun, R = 696,000 km, L = 3.85 x 1026 W and hence T = 5780 K. The thermal radiation
emitted from a planet can also be derived from the Stefan-Boltzmann law.
Theory:
Imaginary sphere, surface area = 4  a2
Suppose a planet of radius R is orbiting at
distance a from a star which has luminosity L.
If the albedo of the planet is 0, the total power
absorbed by the planet will be
Planet,
radius R
a
Pabs = L‧  R / (4  a ) = L‧ R / (4 a )
2
2
2
2

In general, all planets have albedo A < 1. Hence
Pabs = L‧ R2 / (4 a2)‧ (1 – A)
Star, luminosity = L
The planet, as a blackbody at temperature T,
also possesses intrinsic radiation according
to Stefan-Boltzmann law. This radiation is
Prad = 4  R2  T4
Next, we assume the planet has reached a state of thermal equilibrium, i.e. Pabs = Prad
L‧ R2 / (4 a2)‧ (1 – A) = 4  R2  T4
Simplifying gives
T4 = L (1 – A) / (16   a2),
or T4 = L (1 – A) / (285 x 10–8 a2).
Note that the planet’s radius has cancelled out of the above equation. This equation is also applicable
to an exoplanet, provided its orbital radius is computed from Kepler’s law and the luminosity of the
parent star is derived from spectroscopic analysis.
Temperature_Planets.doc
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Ref:
http://en.wikipedia.org/wiki/Effective_temperature
http://en.wikipedia.org/wiki/Solar_luminosity
http://en.wikipedia.org/wiki/Black_body
http://www.universetoday.com/guide-to-space/the-sun/solar-luminosity/
http://en.wikipedia.org/wiki/51_Pegasi_b
http://www.universetoday.com/2008/07/14/observing-an-evaporating-extrasolar-planet/
Example 1:
The luminosity of the Sun is 3.85 x 1026 W. The Earth is 1.50 x 1011 m from the Sun and has an albedo
of 0.30. Its effective temperature4
T4 = L (1 – A) / (285 x 10–8 a2)
= 3.85 x 1026 (1 – 0.30) / [(285 x 10–8) x (1.50 x 1011)2]
= 2.70 x 1026 / [(285 x 10–8) x (2.25 x 1022)]
= 42.1 x 108  2554
The effective temperature of Earth is 255 K. The actual mean surface temperature is 150 C or 288 K.
The higher surface temperature is due to the greenhouse effect in the atmosphere and the internal
heat emitted by the crust. 
Example 2:
Mars is 2.28 x 1011 m from the Sun and has an albedo of 0.16. Its effective temperature4
T4 = 3.85 x 1026 (1 – 0.16) / [(285 x 10–8) x (2.28 x 1011)2]
= 3.23 x 1026 / [(285 x 10–8) x (5.20 x 1022)]
= 21.8 x 108  2164
The effective temperature of Mars is 216 K (close to mean surface temperature of –500 C or 223 K).
Example 3：
Jupiter is 7.78 x 1011 m from the Sun and has an albedo of 0.51. Its effective temperature4
T4 = 3.85 x 1026 (1 – 0.51) / [(285 x 10–8) x (7.78 x 1011)2]
= 1.89 x 1026 / [(285 x 10–8) x (60.5 x 1022)]
= 1.10 x 108  1024
The effective temperature of Jupiter is 102 K (Mean temperature at top of its atmosphere is 130 K.)
Again, the atmosphere and the internal heat emitted by Jupiter are supposed accounting for the
difference of temperatures.
Example 4：
The luminosity of star 51 Pegasi is 1.30 times the solar luminosity. It has a planet (51 Peg b) with
orbital radius of 0.053 AU and mass like Jupiter. Assuming the planet has an albedo of 0.1 (like
Mercury) and has no greenhouse or tidal effects, its effective temperature4 will be
T4 = L (1 – A) / (285 x 10–8 a2)
= 1.30 x (3.85 x 1026) (1 – 0.1) / [(285 x 10–8) x (0.053 x 1.5 x 1011)2]
= 4.50 x 1026 / [(285 x 10–8) x (6.32 x 1019)]
= 2.50 x 1012  12604
The effective temperature of 51 Peg b is 1260 K. It is so high that the planet is called a “hot-Jupiter”.
Temperature_Planets.doc
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Example 5：
This news http://www.universetoday.com/2010/01/07/second-smallest-exoplanet-found/
reports a “small” exoplanet orbiting star HD15668 (HIP84607) at 0.05 AU. What is the effective
temperature of this planet?
From stellar database http://www.geody.com/geolook.php?world=space&map=col&q=hd156668,
we know the star is at apparent magnitude of 8.43 and at parallax of 41.58 milli-arcsec (= distance of
24.05 parsec or 4.96 x 106 AU). We also know the apparent magnitude of the Sun is –26.73.
By inverse-square law, Lstar / LSun = (dstar / dsun)2‧ 2.512 (Sun’s apparent mag – Star’s apparent mag)
= (4.96 x 106 / 1)2‧ 2.512 (–26.73 – 8.43)
= 24.6 x 1012‧ 8.63 x 10–15 = 0.212
Luminosity of the star = 0.212 x (3.85 x 1026 W) = 0.816 x 1026 W.
Assuming the planet has an albedo of 0.1 (like Mercury) and has no greenhouse or tidal effects, its
effective temperature4 will be
T4 = L (1 – A) / (285 x 10–8 a2)
= 0.816 x 1026 (0.9) / [(285 x 10–8) x (0.05 x 1.5 x 1011)2]
= 4580 x 108  8204
The planet is roughly 800 K hot and is supposed not inhabitable although it is only 4 times the mass
of Earth.
-----------Remark:

A more exact calculation is to let Prad = 4  R2  T4 , where  is the emissivity (value between 1
and zero). Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by
a blackbody at the same temperature. A planet with  = 1.0 implies that the planet has little or no
atmosphere, and 100% of the radiation emitted is released to space. Further descriptions of
emissivity are given in
http://www.woodrow.org/teachers/esi/2001/Princeton/Project/wetherald/p6desemiss.htm
http://en.wikipedia.org/wiki/Climate_model
Once  is included, the equation of effective temperature becomes
T4 = L (1 – A) / (16    a2)
The Earth has average emissivity  = 0.61. So
T4 = 3.85 x 1026 (1 – 0.30) / [16  (5.67 x 10–8) x 0.61 x (1.50 x 1011)2]
= 2.70 x 1026 / [(285 x 10–8) x 0.61 x (2.25 x 1022)]
= 69.0 x 108  2884
Earth’s effective temperature = 288 K, close to the actual mean surface temperature of 150 C.
< End >
Temperature_Planets.doc
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2010 Jan