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ID: past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2010/a/1/iii/solution

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Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / A / 1 / iii / Solution by

Codex  0 2026-10-31

Use dimensional quantities

T_{d}, t_{d}

and set

T_{d} = T_{0} T

t_{d} = t_{0} t

. Dividing the energy balance by

σγ T_{0}^{4}

and the albedo equation by its time scale gives

ε T_{t} = 2 (1 - a) - T^{4}, a_{t} = a^{*} (T) - a, a^{*} (T) = \frac{2 - tanh [ μ ( T ^{4} - 1 )]}{4}, ε = \frac{ρ c _{p} d}{t _{0} σγ T _{0}^{3}} = \frac{8 ρ c _{p} d T _{0}}{Q t _{0}} .

(1)

Thus

ε

is the ratio of the atmospheric thermal response time to the much longer albedo response time. Converting kilometres to metres and years to seconds gives

C = 1 0^{7} J m^{- 2} K^{- 1}

T_{0} ≃ 266.3 K

t_{0} ≃ 3.156 \times 1 0^{10} s

, and

ε ≃ 4.93 \times 1 0^{- 4} .

(2)

The thermal time is about half a year, compared with a millennium for albedo adjustment. This is a singular perturbation: setting

ε = 0

removes the independent initial condition for temperature.

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