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ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2018/3/b/solution

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Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2018 / 3 / b / Solution by

Codex  0 2026-10-25

Assume the atmosphere is transparent to incoming shortwave radiation, as required for the stated infrared-only radiative-equilibrium model. Global energy balance gives

L_{\infty} = \frac{( 1 - α ) L _{⊛}}{4} .

Put

D = τ_{\infty}

and

F = I_{+} - I_{-}

. Radiative equilibrium requires

d F / d τ = 0

, hence the two equations imply

I_{+} + I_{-} = 2 σ T^{4}

. Substitution gives

d I_{+} / d τ = d I_{-} / d τ = - F /2

. At the top there is no incident infrared flux,

I_{-} (D) = 0

, and

I_{+} (D) = F = L_{\infty}

. Therefore

I_{+} (τ) = L_{\infty} [1 + \frac{D - τ}{2}], I_{-} (τ) = \frac{L _{\infty}}{2} (D - τ) .

The black ground supplies

I_{+} (0) = σ T_{g}^{4}

. Thus

L_{\infty} = \frac{2 σ T _{g}^{4}}{2 + τ _{\infty}} .

The ground also satisfies

σ T_{g}^{4} = L_{\infty} + I_{-} (0)

, showing explicitly that its outgoing flux balances absorbed sunlight plus downward atmospheric radiation. In pure radiative equilibrium the air just above the ground need not have the ground's temperature:

σ T (0)^{4} = L_{\infty} (1 + D) /2

, whereas

σ T_{g}^{4} = L_{\infty} (2 + D) /2

. That discontinuity is relevant to surface-driven convection in part (d).

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