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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-315/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 315 / 1 / a / Solution by

Codex  0 2026-09-28

Interpret every logarithm in the empirical profile as base ten with the dimensionless argument

P / (1 mbar)

. Put

x = lo g_{10} \frac{P}{1 mbar} .

The upper atmosphere has

T = 800 K

for

x \leq 0

, while integration of

d T / d lo g_{10} P = A x

below it gives

T (P) = 800 K + \frac{A}{2} x^{2} .

Assume that the stated

1250 K

planetary equilibrium temperature is a reasonable brightness temperature at the

1 bar = 1 0^{3} mbar

photosphere. Then

1250 = 800 + \frac{A}{2} (3)^{2}, A ≃ 100 K de x^{- 2} .

At

10 bar

,

x = 4

, and therefore

T (10 bar) ≃ 800 + 50 (4)^{2} = 1600 K .

This estimate neglects day-night variation, wavelength-dependent photospheric pressure, and a possible radiative-convective boundary; it treats the retrieved profile as representative of the dayside disk.

Plausible pressure-temperature profile for the hot Jupiter

. The profile is isothermal above one millibar and follows the integrated quadratic logarithmic-pressure law below it, calibrated to 1250 kelvin at one bar.

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