Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2019 / 3

Assume an ideal gas dominated by carbon dioxide, constant mean molar heat capacity $C_p$, and a dry convective adiabat connecting the stated emitting layer to the ground. Conservation of entropy gives $C_{p}d\ln T=Rd\ln p$, where $R$ is the universal molar gas constant. ThusThe estimate extrapolates across a pressure ratio of $450$. It neglects radiatively stable layers, variable heat capacity, departures from an ideal gas and changes in composition; these approximations limit its accuracy. A brightness temperature observed from space cannot generally be equated to a unique material level, but this calculation uses the layer interpretation explicitly supplied by the question.

For a non-scattering atmosphere above a blackbody surface, the difference between the two curves is the transmitted surface term $B_{\lambda }(T_s){\mathcal{T}}_{\lambda ,s}$. The atmospheric-only curve is an integral of local thermal emission weighted by transmission to space. Molecular absorption blocks some wavelengths and permits narrow windows at others. In a transparent window, adding the hot surface substantially increases the emerging radiance; in an opaque interval, the two curves nearly coincide because the surface contribution is extinguished.

The original graph shows the clearest surface-dominated window around $0.99$–$1.04\mu \mathrm{m}$, with the blue curve near $5\mu \mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mu {\mathrm{m}}^{-1}$ at its maximum and the atmospheric-only contribution small. Weaker possibilities are around $0.85$–$0.92\mu \mathrm{m}$ and near $0.8\mu \mathrm{m}$, with less available signal. The $1.17$–$1.20\mu \mathrm{m}$ peak contains a measurable surface contribution but a larger atmospheric background. The near-equality around $1.27$–$1.29\mu \mathrm{m}$ makes that peak a poor surface thermometer despite its appreciable total radiance. These are approximate readings, not precisely tabulated band edges.

At $T_s\simeq 766\mathrm{K}$, the Wien displacement law puts the thermal wavelength peak near $3.8\mu \mathrm{m}$. A $1\mu \mathrm{m}$ window is still in the Wien approximation regime, but its radiance is exponentially larger than for an Earth-like $300\mathrm{K}$ or colder Martian surface. For example, at $1\mu \mathrm{m}$ the Planck function ratio $B_{\lambda }(766)/B_{\lambda }(300)$ is of order $10^{12}$. A modest hot-spot temperature increase likewise gives a relatively large fractional radiance change. Inverting this signal for surface temperature requires an atmospheric transmission and emission model, not just identifying a peak.

Assume a circular orbit, nadir viewing, negligible planetary rotation and a fixed camera relative to the orbital ground track. If $R_V$ is the planet's radius and $h$ the orbital altitude, the orbital angular speed isThe nadir footprint moves across the surface at $v_g=n{R}_V\simeq 6.88\times 10^3\mathrm{m}{\mathrm{s}}^{-1}$, rather than the slightly larger satellite speed $n(R_V+h)$. HenceThis is an upper integration time for approximately one $200\mathrm{m}$ along-track sample without motion compensation. Using satellite speed directly changes it by only about four per cent; the expression above keeps the projection onto the planetary surface.

The etendue is the area–solid-angle acceptance of a ray bundle, more generally $\mathcal{G}=n_{mopt}^2\int \cos \theta dAd\Omega$. In an ideal lossless optical system it is conserved. Increasing pupil area increases accepted power only together with the appropriate field and detector cone; passive optics cannot arbitrarily increase radiance. For a small nadir pixel here, $A\Omega$ denotes entrance-pupil area times the solid angle of its surface footprint.

Choose $1.00$–$1.03\mu \mathrm{m}$ within the surface-sensitive window. A representative band-mean total radiance read from the original blue curve is ${\bar{L}}_{\lambda }\simeq 4\mu \mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mu {\mathrm{m}}^{-1}$, uncertain at roughly the tens-of-per-cent level. Its band integral isThe provided SNR formula depends on $L\Delta \nu$ only through this band integral. If $\nu$ is wavenumber, the selected band spans about $291\mathrm{c}{\mathrm{m}}^{-1}$ and $L_{\nu }d\nu =L_{\lambda }|d\lambda |$; numerical radiance densities in the two coordinates must not be mixed. Convert the specific detectivity to SI: $D^{\ast }=5\times 10^9\mathrm{m}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$ and $\sqrt{a}=2\times 10^{-4}\mathrm{m}$. Using the footprint dwell time from part (c),A square $200\mathrm{m}$ footprint at altitude $259\mathrm{km}$ subtends $\Omega \simeq (200/259000{)}^2=5.96\times 10^{-7}\mathrm{sr}$. Thus the collecting area and circular aperture diameter areThe Rayleigh criterion independently requires only $D\gtrsim 1.22\lambda /(200/h)\simeq 1.6\mathrm{mm}$ at $1.015\mu \mathrm{m}$, so the noise requirement dominates diffraction. The $200\mu \mathrm{m}$ detector size corresponds to focal length about $0.259\mathrm{m}$ for this pixel field. This assumes one detector footprint per stated feature size; stricter spatial sampling would need smaller footprints and a revised photon budget.

The estimate assumes a clear, non-scattering atmosphere as in the graph, a night-side or otherwise isolated thermal signal, negligible extra stray-light and read-noise penalties beyond the supplied detectivity, and transmission $0.7$ for the instrument alone. Atmospheric attenuation is already included in the plotted incoming radiance. The selected band has little atmospheric-only signal, but actual hot-spot detection concerns contrast relative to the background: SNR $50$ on total radiance is not a guarantee for an arbitrarily small temperature anomaly. An illustrative radiance range $3$–$5$ in the plotted units would give diameters about $0.20$–$0.15\mathrm{m}$.