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

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

Codex  0 2026-10-25

Adding the two flux solutions gives

σ T (τ)^{4} = \frac{L _{\infty}}{2} (1 + D - τ) .

The supplied pressure–optical-depth law gives

D - τ = D p / p_{0}

. With the exponential pressure model

p = p_{0} e^{- z / H}

,

T (z) = T_{\infty} (1 + a e^{- z / H})^{1/4}, T_{\infty} = (\frac{L _{\infty}}{2 σ})^{1/4}, a = τ_{\infty} .

It approaches the finite atmospheric top temperature

T_{\infty}

and has the lowermost-air value

T (0) = T_{\infty} (1 + D)^{1/4}

. Its lapse rate is

- \frac{d T}{d z} = \frac{T ( z )}{4 H} \frac{D e ^{- z / H}}{1 + D e ^{- z / H}} .

The stated approximate sign is appropriate physically: a nonisothermal ideal gas does not have an exactly constant hydrostatic scale height

RT / g

. Treating

H

as a representative constant is an additional vertical-coordinate approximation. Algebraically the displayed profile is exact for the prescribed exponential pressure and two-stream equations.

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