# Planetary equilibrium temperature

(Redirected from Surface temperature)

The planetary equilibrium temperature is a theoretical temperature that a 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, and global-mean surface air temperature,[3] which includes the effects of global warming.

## Calculation of semi-blackbody temperature

If the incident solar radiation ("insolation") on the planet at its orbital distance from the Sun is Io, the amount of energy absorbed by the planet will depend on its albedo a and its cross-sectional area:

${\displaystyle {P}_{in}={I_{o}}\left(1-a\right)\pi {{R}_{p}}^{2}}$

Note that the albedo would be zero (${\displaystyle a=0}$) for a blackbody. However, in planetology, more useful results are obtained by accounting for a measured or assumed planetary albedo ${\displaystyle a>0}$.

The infrared power radiated by the planet as thermal radiation will depend on its emissivity and its surface area, according to the Stefan–Boltzmann equation:

${\displaystyle {P}_{out}=\epsilon \sigma A{T}^{4}}$

where Pout is the radiated power, ${\displaystyle \epsilon }$ is the emissivity, σ the Stefan–Boltzmann constant, A the surface area, and T the absolute temperature. For a spherical planet, the surface area is ${\displaystyle A=4\pi {{R}_{p}}^{2}}$.

The emissivity is typically assumed to be ${\displaystyle \epsilon =1}$, as would be the case for a perfectly emitting blackbody. This is often a good assumption, as emissivities of natural surfaces tend to be in the range of 0.9 to 1, e.g. ${\displaystyle \epsilon _{Earth(Planet)}=0.96}$.

The equilibrium temperature is then calculated by setting Pin=Pout. Thus,

${\displaystyle {T}_{eq}={\left({\frac {I_{o}\left(1-a\right)}{4\sigma }}\right)}^{1/4}}$

### Theoretical model

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, D. 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:

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

Where ${\displaystyle {T}_{\bigodot }}$ and ${\displaystyle {{R}_{\bigodot }}}$ are the temperature and radius of the star.

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 approximately 227 K, but a surface temperature of 740 K.[4][5] The Moon has a black body temperature of 271 K,[6] but can have temperatures of 373 K in the daytime and 100 K at night.[7] 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,[8] geothermal energy which is driven by radioactive decay in the core of the planet,[9] or accretional heating.[10]

### Detailed derivation of the planetary equilibrium temperature

The power absorbed by the planet from the star is equal to the power emitted by the planet: ${\displaystyle {P}_{in}={P}_{out}}$

The power input to the planet is equal to the solar irradiance (i.e. power received per unit area) of the star at the distance of the planet, Io, times the fraction absorbed by the planet (1 minus the albedo), times the area of the planet illuminated by the star: ${\displaystyle {P}_{in}={I_{o}}\left(1-a\right)\pi {{R}_{p}}^{2}}$

Io, the solar intensity at the distance of the planet from the sun, is equal to the luminosity (i.e. total power emitted) of 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. This yields: ${\displaystyle {P}_{in}={L}_{\bigodot }\left(1-a\right)\left({\frac {\pi {{R}_{p}}^{2}}{4\pi {D}^{2}}}\right)}$ [5]

Any incoming power to a black body is radiated as heat according to the Stefan–Boltzmann law ${\displaystyle P=\epsilon \sigma A{T}^{4}}$.

(The emissivity ${\displaystyle \epsilon }$ is usually expected to be very close to 1, and thus is often left out). Multiplying by the area, the power emitted by the planet is: ${\displaystyle {P}_{out}=\left(\epsilon \sigma {{T}_{eq}}^{4}\right)\left(4\pi {{R}_{p}}^{2}\right)}$

Setting these equal:

${\displaystyle {T}_{eq}={\left({\frac {{L}_{\bigodot }\left(1-a\right)}{16\epsilon \sigma \pi {D}^{2}}}\right)}^{1/4}}$

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

Inserting this into the previous equation, it can be shown that:

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

By assuming that the emissivity ${\displaystyle \epsilon }$ is equal to 1, this reproduces the equation in the previous section. 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

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.[11] Astronomers use a hypothesized albedo [12] and can then estimate the equilibrium temperature.

## References

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. ^ "Venus Fact Sheet". nssdc.gsfc.nasa.gov. Retrieved 2017-02-01.
5. ^ a b "Equilibrium Temperatures of Planets". Burro.astr.cwru.edu. Retrieved 2013-08-01.
6. ^ "Moon Fact Sheet". Nssdc.gsfc.nasa.gov. 2013-07-01. Retrieved 2013-08-01.
7. ^ "What's the Temperature on the Moon? | Lunar Temperatures". Space.com. Retrieved 2013-08-01.
8. ^ "Planetary Science". Astronomynotes.com. 2013-05-12. Retrieved 2013-08-01.
9. ^ Anuta, Joe (March 27, 2006). "Probing Question: What heats the earth's core?". Penn State.
10. ^ "accretional heating – Encyclopedia.com". Encyclopedia.com. Retrieved 2013-08-01.
11. ^ pages 3-4
12. ^ page 16