Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2021 / 4

The optical telescope supplies collecting area and spatial selection, the spectral selector supplies a passband, and the photodetector converts received radiant flux into a measured signal. For an extended source, a useful schematic photon count iswhere $A\Omega$ is the etendue, $\tau$ the integration time, $\eta$ the net transmission and quantum efficiency, and $L_{\lambda }$ the spectral radiance. An illustrative independent-noise model givesHere $N_b$ and $N_d$ denote background and dark counts, and ${\sigma }_{\mathrm{read}}$ is readout standard deviation. This makes the roles of the three components explicit.

A larger optical telescope collects more photons at fixed angular pixel size and improves the diffraction-limited resolution. Its limits include optical aberrations, alignment, launch size, stray light and pointing. Conservation of etendue constrains simultaneous collecting area and angular acceptance. Increasing spatial resolution by narrowing a pixel's solid angle reduces its photons.

A narrower spectral passband or higher spectral resolving power separates diagnostic features and rejects unwanted background, but usually reduces photons per channel and hence signal-to-noise ratio. Selector transmission, leakage, slit width, detector format and the required time cadence limit the trade. A wider band helps only insofar as it adds useful signal rather than mostly background.

The photodetector's quantum efficiency, dark current, read noise, dynamic range and response time determine conversion losses and added noise. Cooling can suppress thermal background and dark current, but cannot remove photon shot noise. Longer integration improves a stationary signal's signal-to-noise ratio until motion, saturation, drift or required cadence becomes limiting.

A geostationary orbit provides continuous viewing of approximately the same terrestrial region, favouring fast cloud evolution, storm tracking and the diurnal cycle. The large range makes fine spatial resolution demanding, high latitudes have poor viewing geometry, and a single platform has no global coverage. A polar orbit gives much smaller range and broad latitude coverage as Earth rotates beneath it, favouring detailed global mapping; any one location is sampled intermittently. A Sun-synchronous orbit can hold approximately fixed local solar time for comparing observations, but that sampling can miss much of the diurnal evolution. The science therefore sets a trade between spatial coverage, spatial resolution and temporal sampling.

The Moon's very slow, tidally locked rotation makes a lunar stationary orbit fundamentally problematic. Let its orbital angular velocity around Earth be $n$, approximately also its spin rate. For lunar mass $m$, terrestrial mass $M$, and separation $a$, the two-body estimates giveThus $r_{\mathrm{sync}}\simeq 3^{1/3}{r}_H$ lies outside the lunar Hill sphere. Earth's tidal perturbations prevent the desired simple stable, lunar-centred stationary orbit. This is stronger than merely saying that the required orbit is high; specially maintained three-body orbits are a different construction.

For a prograde equatorial synchronous circular orbit, equate gravitational acceleration and centripetal acceleration. With planetary radius $R_M$, $GM=g_s{R}_M^2$, and rotation period $P$,The tabulated $g_s$ is a gravitational acceleration; its printed unit must be read as $\mathrm{m}{\mathrm{s}}^{-2}$, rather than $\mathrm{m}{\mathrm{s}}^{-1}$. Using $P=88642\mathrm{s}$ givesThe orbital altitude is measured from the surface, rather than the centre.

At the visible horizon, the line of sight is tangent to the spherical surface. The limiting central angle ${\theta }_h$ satisfies $\cos {\theta }_h=R_M/r_o$. The visible spherical cap has area $2\pi R_M^2(1-\cos {\theta }_h)$, givingAbout $41.7\%$ of the whole surface is geometrically visible, with ${\theta }_h\simeq 80.4^{\circ }$. This is a surface-area fraction, not an apparent-disc area ratio. Useful imaging near the horizon is further limited by foreshortening and the long atmospheric path.

Interpret the stated detector width $d=20\mu \mathrm{m}$ as one square spatial sample. At the point directly below the spacecraft, the small-angle approximation gives pixel angular width $d/f\simeq \Delta x/h$. ThusThere is a definition error in the source: the usual focal ratio, or f-number, divides focal length by the entrance-aperture diameter $D$, whereas the printed parenthesis divides by detector width $d$. Reporting both avoids silently changing the question:The actual optical design is therefore approximately f/1.36, a fast optical telescope, not an optically slow f/34000 instrument. Such a focal ratio is demanding but possible in principle: optical aberrations, obscuration, field coverage, stray light and alignment need careful design. The single-sample angular width is $2.94\times 10^{-5}\mathrm{rad}$. Spatial sampling alone does not guarantee true resolving power; diffraction, pointing and signal-to-noise ratio also matter. At large viewing angles, the same angular pixel covers a larger, distorted surface footprint. If the detector is literally one element, scanning is additionally needed to form a map.

A calibrated multispectral filter radiometer is a natural choice for rapid thermal mapping: a few selected bands provide higher photon throughput and faster imaging than spreading the light over many spectrometer channels. Several bands can help separate temperature, dust emissivity and atmospheric effects. A spectrometer would be justified if identifying composition or resolving spectral structure were a central additional objective; one broadband radiance alone cannot generally determine both temperature and unknown emissivity.

Thermal infrared radiation is preferable to reflected visible light for measuring temperature. The Rayleigh criterion setsA band near $10\mu \mathrm{m}$ is therefore a reasonable compromise between useful planetary thermal emission and the required diffraction-limited resolution. At that wavelength, the nominal surface Rayleigh criterion separation is about $415\mathrm{m}$; at $15\mu \mathrm{m}$ it is about $623\mathrm{m}$ and fails the target. The exact bands must avoid or model strong atmospheric absorption and account for the dust spectrum. The $500\mathrm{m}$ sampling pitch is still not a guarantee of faithful reconstruction of arbitrary $500\mathrm{m}$ features: detector sampling and the point spread function must be considered together.