Planetary equilibrium temperature

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The planetary equilibrium temperature is a theoretical temperature that the planet would be at when considered simply as if it were a black body being heated only by its parent star. In this model, the presence or absence of an atmosphere (and therefore any greenhouse effect) is not considered, and one treats the theoretical black body temperature as if it came from an idealized surface of the planet.

Other authors use different names for this concept, such as equivalent blackbody temperature of a planet,[1] or the effective radiation emission temperature of the planet.[2] Similar concepts include the global mean temperature, Global radiative equilibrium, global-mean surface air temperature,[3] which includes the effects of Global warming.

Theoretical Model[edit]

Consider a spherical star and a spherical planet. The star and the planet are considered to be perfect black bodies. The planet has an albedo and only absorbs a fraction of radiation, depending on its surface characteristics. The star emits radiation isotropically according to the Stefan–Boltzmann law which travels a distance equal to the orbital distance of the planet. The planet absorbs the radiation that isn't reflected by the albedo of the surface, and heats up. Since the planet is also a black body which emits radiation according to the Stefan–Boltzmann law, it will emit radiation and lose energy. Thermal equilibrium exists when the power supplied by the star is equal to the power emitted by the planet. The temperature at which this balance occurs is the planetary equilibrium temperature and is equal to:

{ T }_{ eq }={ T }_{ \bigodot  }{ \left( 1-a \right)  }^{ 1/4 }\sqrt { \frac { { R }_{ \bigodot  } }{ 2D }  }

The equilibrium temperature is neither an upper nor lower bound on actual temperatures on a planet. Because of the greenhouse effect, planets with atmospheres will have temperatures higher than the equilibrium temperature. For example, Venus has an equilibrium temperature of 260 K, but a surface temperature of 740 K.[4] The Moon has a black body temperature of 271 K,[5] but can have temperatures of 373 K in the daytime and 100 K at night.[6] This is due to the relatively slow rotation of the moon compared to its size, so that the entire surface doesn't heat evenly. Orbiting bodies can also be heated by Tidal heating,[7] Geothermal energy which is driven by radioactive decay in the core of the planet,[8] or accretional heating.[9]

Detailed derivation of the planetary equilibrium temperature[edit]

{ P }_{ in }={ P }_{ out } The power absorbed by the planet from the star is equal to the power emitted by the planet.

{ P }_{ in }={ L }_{ \bigodot  }\left( 1-a \right) \left( \frac { \pi { { R }_{ p } }^{ 2 } }{ 4\pi { D }^{ 2 } }  \right) [4] The power input to the planet is equal to the luminosity (i.e. power emitted) of the star, times the ratio absorbed by the planet (1 minus the albedo), times the area of the planet illuminated by the star, divided by the area of the sphere that all of the star's radiation is cast on at the distance of the planet.

P=\sigma A { T }^{ 4 } Any incoming power to a black body is radiated as heat according to the Stefan–Boltzmann law, where P is the incoming power, σ is the Stefan–Boltzmann constant, A is the surface area of the black body, and T is the equilibrium temperature.

{ L }_{ \bigodot  }=\left( \sigma { { T }_{ \bigodot  } }^{ 4 } \right) \left( 4\pi { { R }_{ \bigodot  } }^{ 2 } \right) The luminosity of the star is equal the Stefan-Boltzmann constant, times the area of the star, times the fourth power of the temperature of the star.

{ P }_{ out }=\left( \sigma { { T }_{ eq } }^{ 4 } \right) \left( 4\pi { { R }_{ p } }^{ 2 } \right) The power emitted by the planet

Rearranging, it can be shown that: { T }_{ eq }={ T }_{ \bigodot  }{ \left( 1-a \right)  }^{ 1/4 }\sqrt { \frac { { R }_{ \bigodot  } }{ 2D }  }

It is interesting to note that the equilibrium temperature does not depend on the size of the planet, because both the incoming radiation and outgoing radiation depend on the area of the planet.

Calculation for extrasolar planets[edit]

For extrasolar planets the temperature of the star can be calculated from the color of the star using Planck's law. The calculated temperature of the star can be used with the Hertzsprung–Russell diagram to determine the absolute magnitude of the star, which can then be used with observational data to determine the distance to the star and finally the size of the star. Orbital simulations are used to determine what orbital parameters (including orbital distance) produce the observations seen by astronomers.[10] Astronomers use a hypothesized albedo [11] and can then estimate the equilibrium temperature.

References[edit]

  1. ^ Wallace, J.M., Hobbs, P.V. (2006). Atmospheric Science. An Introductory Survey, second edition, Elsevier, Amsterdam, ISBN 978-0-12-732951-2. Section 4.3.3, pp. 119–120.
  2. ^ Stull, R. (2000). Meteorology For Scientists and Engineers. A technical companion book with Ahrens' Meteorology Today, Brooks/Cole, Belmont CA, ISBN 978-0-534-37214-9., p. 400.
  3. ^ Wallace, J.M., Hobbs, P.V. (2006). Atmospheric Science. An Introductory Survey, second edition, Elsevier, Amsterdam, ISBN 978-0-12-732951-2., p.444.
  4. ^ a b "Equilibrium Temperatures of Planets". Burro.astr.cwru.edu. Retrieved 2013-08-01. 
  5. ^ "Moon Fact Sheet". Nssdc.gsfc.nasa.gov. 2013-07-01. Retrieved 2013-08-01. 
  6. ^ "What’s the Temperature on the Moon? | Lunar Temperatures". Space.com. Retrieved 2013-08-01. 
  7. ^ "Planetary Science". Astronomynotes.com. 2013-05-12. Retrieved 2013-08-01. 
  8. ^ Jul 27, 2008 (2008-07-27). "Probing Question: What heats the earth's core?". Phys.org. Retrieved 2013-08-01. 
  9. ^ "accretional heating – Dictionary definition of accretional heating | Encyclopedia.com: FREE online dictionary". Encyclopedia.com. Retrieved 2013-08-01. 
  10. ^ pages 3-4
  11. ^ page 16

External links[edit]