Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 315 / 3 / d

Solution

The stellar luminosity is

L_{⋆} = 4 π R_{⋆}^{2} σ_{SB} T_{⋆}^{4}

. At separation

a

, the planet intercepts a disk of area

π R_{p}^{2}

A_{B}

and uniform emission over its full surface, energy balance gives

(1 - A_{B}) \frac{L _{⋆}}{4 π a ^{2}} π R_{p}^{2} = 4 π R_{p}^{2} σ_{SB} T_{e}^{4}, T_{e} = T_{⋆} (\frac{R _{⋆}}{2 a})^{1/2} (1 - A_{B})^{1/4} .

This planetary equilibrium temperature neglects intrinsic heat. Dayside-only or nonuniform emission changes the redistribution factor.

For the single-layer greenhouse model, assume a black infrared surface, an atmosphere transparent to incoming stellar radiation, and infrared absorptivity/emissivity

α

. Let

T_{s}

T_{a}

layer temperature. Kirchhoff's law makes the layer's emission in each direction

α σ_{SB} T_{a}^{4}

. Its balance is

α σ_{SB} T_{s}^{4} = 2 α σ_{SB} T_{a}^{4},

T_{a}^{4} = T_{s}^{4} /2

for

α > 0

. Surface balance includes the downward layer emission:

σ_{SB} T_{e}^{4} + α σ_{SB} T_{a}^{4} = σ_{SB} T_{s}^{4} .

Therefore

T_{s} = T_{e} (1 - α /2)^{- 1/4} .

The no-atmosphere limit is

T_{s} = T_{e}

, and a completely IR-absorbing single layer gives $T_{s} = 2^{1/4} T_{e}$ . The outgoing flux

(1 - α) σ T_{s}^{4} + α σ T_{a}^{4}

remains

σ T_{e}^{4}

. The layer delays escape by absorption and re-emission; it does not permanently retain a fraction

α

of every emitted photon, which would incorrectly predict a divergent temperature at

α = 1

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