The first term describes collision-coalescence: a collector sweeps area $\pi R^2$, overtakes small droplets at relative terminal speed $u_R-u_r$, encounters liquid mass per volume $L$, and retains a fraction $E$ of that intercepted liquid. The second term describes diffusion-limited condensational growth, with vapour supersaturation $S-1$ and combined thermal and vapour-diffusion resistances $F_k+F_d$. Condensation is positive for $S>1$ and reverses to evaporation for $S<1$. The liquid-water content $L$ here is a local mass-per-volume quantity, not a vertically integrated liquid-water path.

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 $0.20$–$0.15\mathrm{m}$.

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.

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 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.

Pressure broadening is proportional to collision frequency and therefore becomes small in the upper atmosphere. Doppler broadening remains finite at finite temperature, with a characteristic width of order ${\nu }_0\sqrt{k_{B}T/(m{c}^2)}$ up to the stated width convention. Where that exceeds the collisional width, the line core is Doppler dominated. A Voigt profile combines the two mechanisms; natural broadening can contribute at still lower pressures. Thus a Lorentz pressure-broadened profile cannot hold all the way to $p=0$.

At line centre, replacing the fictitious $1/p$ opacity by a finite Doppler-core coefficient removes the divergent column optical depth. The emitting pressure is then finite rather than arbitrarily close to zero. If that level still lies in the isothermal stratosphere and the core is opaque, the central brightness temperature stays approximately $T_{\mathrm{strat}}$, so the flux floor changes little. If it lies below the stratosphere, or the Doppler-core column becomes thin, warmer layers or the surface contribute and the core brightens. The core-to-wing transition changes shape and width, but Lorentz wings formed mainly at higher pressures remain approximately as in part (c). The size of the correction needs the missing spectroscopic widths, but no unique numerical correction is requested.

For a vertical ray, the formal Schwarzschild equation solution isIn the wing model $\tau \propto p^2$, the contribution per interval $d\ln p$ is proportional to $2\tau e^{-\tau }$ when the source varies slowly, and is concentrated near $\tau =1$. This motivateslimited to $p_s$ in transparent spectral regions. Rays contributing to the hemispheric flux replace this by $p_s\sqrt{\mu }d/w$, an order-one angular refinement.

For $T_s>T_{\mathrm{strat}}$, put $p_t=p_s(T_{\mathrm{strat}}/T_s{)}^{1/x}$ and $d_t=w{p}_t/p_s$. With the vertical optical-depth-one estimate, the spectral outgoing longwave radiation is approximatelyThe cold flat central portion comes from the isothermal stratosphere; intermediate wings sample progressively warmer, deeper tropospheric levels; far outside the line the warm surface is seen. The transitions are smooth in the exact transfer solution. If $T_s=T_{\mathrm{strat}}$, the entire column and ground are isothermal and there is no line contrast. If $T_s<T_{\mathrm{strat}}$, the given max-profile has no cooler troposphere: the atmosphere can instead produce an emission feature above the colder surface background. Assuming a narrow isolated line permits the local Planck function background to be treated as nearly constant across its width. The original schematic below also compares the optical-depth-one estimate with angular and source-function integration for illustrative, explicitly chosen parameters.

For a vertical ray, the transmittance from $p_2$ to space is ${\mathcal{T}}_{\nu }(p_2)=e^{-{\tau }_{\nu }(p_2)}$. Direct integration of the full pressure-dependent Lorentz line shape gives, for $d>0$,The exact Lorentz-model transmittance and its strong-line limit are thereforeThe approximation is justified at the radiating level by $C\gg 1$. At zenith-angle cosine $\mu$, replace $\tau$ by $\tau /\mu$; if a hemispheric flux transmission is desired, its normalized value is $2{\int }_0^1\mu e^{-\tau /\mu }d\mu =2E_3(\tau )$. The pure Lorentz model formally gives infinite line-centre optical depth from every nonzero $p_2$ to $p=0$, because ${\kappa }_{{\nu }_0}\propto 1/p$. This is an unphysical central-core singularity, addressed in part (d), rather than a divergence of the integrated outgoing flux.

Take $S$ as the integrated line mass absorption coefficient, and $q$ as the absorber mass fraction, assumed constant; for a pure absorbing gas set $q=1$. All spectral intervals below use the paper's wavenumber coordinate. The normalized Lorentz line shape isIf instead an absorption coefficient per length is wanted, multiply by absorber density. Neglecting temperature dependence, pressure broadening gives $\Delta (p)={\Delta }_{s}p/p_s$. In the line wings, with $d=|\nu -{\nu }_0|\gg \Delta (p)$,Define the line-opacity parameter and characteristic wavenumber widthAt an emitting level ${\tau }_{\nu }\sim 1$, the wing estimate implies $d/\Delta (p)\sim \sqrt{C}$. Thus the strong-line limit relevant to outgoing radiation isIt ensures that the opaque-to-transparent transition lies in the Lorentz wings, rather than the narrow central core, and the characteristic absorption width is much larger than ${\Delta }_s$. At a grazing ray angle the criterion is even easier to satisfy because of the larger optical depth. This is a criterion for the radiatively important wings, not a claim that collision broadening remains physical at arbitrarily small pressure.

For ordinary condensibles, $L/(R_{B}T)$ is typically of order ten or more near low atmospheric temperatures. At the approximate triple-point temperatures, the supplied substances give $\beta$ about $20$ for water and $10$–$12$ for carbon dioxide, methane and nitrogen. This is consistent with the supplied large ratios of latent heat to sensible heating: for molecular gases $c_p/R$ is only a few. The relevant $x=R_A/c_{pA}$ is typically a few tenths.

The integral is concentrated where ${\beta }_{s}x\zeta =O(1)$. There $x\zeta =O({\beta }_s^{-1})\ll 1$, so $e^{x\zeta }-1\simeq x\zeta$ givesThis is an Laplace's method, rather than an assumption that vapour is uniformly mixed. The effective contributing depth in log-pressure is $\Delta \zeta \sim 1/(x{\beta }_s)$, explaining why the near-surface atmosphere dominates.

An exact representation within the same dilute model makes the error explicit. Set $y={\beta }_s{e}^{x\zeta }$ and define the exponential integral $E_1(b)={\int }_b^{\infty }{e}^{-y}dy/y$. ThenThe saturated vapor column on a dry adiabat estimate overstates the exact dilute-model column by roughly $1/{\beta }_s$, about five to ten per cent for those illustrative values. It is not accurate for a warm, nondilute steam atmosphere.

Interpret the requested mass path as the vapour column; no condensed-water loading is prescribed. Hydrostatic equilibrium gives $dp=-\rho gdz$, so if $q_B={\rho }_B/\rho$ is vapour mass fraction,For the mixture $q_B=ϵe_s/[p+(ϵ-1)e_s]$. Under the dilute conditions of part (a), $q_B\simeq ϵe_s/p$. Introduce $\zeta =-\ln (p/p_s)$, so $T=T_s{e}^{-x\zeta }$ and $-dp=pd\zeta$. Integrating the constant-$L$ Clausius-Clapeyron equation relative to the surface givesThereforeThe explicit $p_s$ cancels: decreasing vapour mass fraction in a denser dry column compensates its greater total mass. This cancellation belongs to the dilute, dry-adiabatic approximation, and is not a general property of moist atmospheres. With suspended condensate, an additional and unspecified condensate mass path would have to be added.

Let $R_A,R_B$ be the specific specific gas constants and $c_{pA},c_{pB}$ the specific heat capacities of the noncondensible gas and vapour. Write $e_s(T)$ for the saturation vapor pressure, $r$ for vapour mass per unit dry-gas mass, and $ϵ=R_A/R_B$. For a dilute ideal gas mixture,The second equation is the Clausius-Clapeyron equation, neglecting condensed-phase volume and taking constant latent heat $L$. The first law of thermodynamics for saturated ascent, with condensate removed and negligible condensate heat capacity, is to leading orderSince $dr=r[LdT/(R_B{T}^2)-d\ln p]$, the resulting moist adiabat has logarithmic slopeConsequently a sufficient dry adiabat criterion isThe first two conditions justify neglecting the vapour's effects on the gas law and sensible heat capacity; the last two suppress latent heating. With $x=R_A/c_{pA}$ and $\beta =L/(R_{B}T)$, the latent conditions are $\beta e_s/p\ll 1$ and $x{\beta }^2{e}_s/p\ll 1$. Usually $x\beta \gg 1$, so the squared-latent term gives the stronger condition. At sufficiently low surface temperature, the exponential smallness of $e_s(T_s)$ makes these conditions hold andThese are local conditions along the column. They can be checked at the surface when $x{\beta }_s>1+2x$: on this dry profile the strongest latent parameter is proportional to $\exp [(1+2x)\zeta -{\beta }_s(e^{x\zeta }-1)]$ and decreases upwards. Without that restriction a surface-only test need not bound every altitude. If suspended condensate is retained, its own heat capacity and loading must also be negligible; saturation alone does not specify those quantities.

Assuming the mean wind remains $10\mathrm{m}{\mathrm{s}}^{-1}$, the weakly stratified upper layer hasFor the inverse-width estimate $|k|\sim 10^{-4}{\mathrm{m}}^{-1}$, it is near the turning threshold $m=0$, with negligible upward group-velocity component. Components of larger $|k|$ are evanescent aloft and are substantially reflected; only horizontal wavelengths exceeding $2\pi U/N_{\mathrm{upper}}\simeq 62.8\mathrm{km}$ can propagate vertically there. For the alternative $10\mathrm{km}$ sinusoidal wavelength, $m_{\mathrm{upper}}^2<0$ and the evanescent decay length is about $1.6\mathrm{km}$.

The sharp inversion and change of stability can therefore cap the upward-radiating disturbance. Reflection between the ground and cap may trap energy in the lower layer and allow downstream lee-wave trains or resonant guided modes, with permitted wavelengths depending on the interface and lower-layer structure. Reflection, leakage and possible resonance are distinct: a specified $5\mathrm{km}$ height does not by itself fix a unique mode amplitude or wavelength.

Wave clouds may become confined below the cap or organized into persistent lee bands if there is enough moisture. Inversion-limited vertical mixing can also retain low cloud, haze or pollutants; sufficiently strong trapped motions can promote rotors or turbulence. These are possible changes, not inevitable weather outcomes: humidity, wave amplitude, detailed stability and wind shear are unspecified. The new structure suppresses upward energy transmission for the dominant short horizontal scales even though the lower atmosphere can still support waves.

A mountain width is not an exactly specified sinusoidal wavelength. Taking its characteristic horizontal inverse length as $|k|\sim 1/a$ with $a=10\mathrm{km}$ gives $|k|\sim 10^{-4}{\mathrm{m}}^{-1}$. Since $N/U=10^{-3}{\mathrm{m}}^{-1}$, the stationary vertical wavenumber isThis is close to the long-horizontal-scale/hydrostatic estimate $2\pi U/N=6.28\mathrm{km}$. If instead the quoted width is identified with a sinusoidal wavelength ${\lambda }_x=10\mathrm{km}$, then $|k|=2\pi /{\lambda }_x$ and ${\lambda }_z\simeq 8.1\mathrm{km}$. Thus several kilometres, roughly $6$–$8\mathrm{km}$ under these two common width conventions, is the justified estimate; the terrain profile would fix its actual forcing spectrum.

For the inverse-width choice, both $k$ and $m$ are negative for upward energy radiation. The stationary phase lines $kx+mz=\text{constant}$ tilt upstream with increasing height, while energy propagates upward and slightly downstream. Numerically $\widehat{{\mathbf{c}}_g}\simeq (0.10,0.995)\mathrm{m}{\mathrm{s}}^{-1}$, with ray inclination about $84.3^{\circ }$. Intrinsic phase propagation points upstream and downwards; the ground pattern itself remains fixed. The original sketch below uses this explicit convention.

A ground observer may see stationary lenticular or wave clouds where ascent cools sufficiently moist air, with continuing air flow through a cloud whose position remains almost fixed. Descending regions can clear or warm. An aircraft crossing the pattern encounters alternating updraughts and downdraughts; gliders can exploit the lift. Strong amplitudes, shear or rotors may cause turbulence, but a linear gravity wave can also be smooth, and clear-air waves need not produce any visible cloud. Mountain height, humidity and nonlinear stability are not supplied, so neither a vertical-wind amplitude nor a definite atmospheric precipitation forecast follows.

Put $K=(k^2+m^2{)}^{1/2}$ and $P=p^{\prime }/{\rho }_0$. Acting on the Fourier mode, $D_0=-i(\hat{\omega }-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The buoyancy equation gives $g{\rho }^{\prime }/{\rho }_0=i{N}^{2}w/\omega$. Substitution into vertical momentum yieldsThe two signs describe the two frequency branches; when the intrinsic frequency is required positive, $\omega =N|k|/K$. For a stationary disturbance $\hat{\omega }=0$, so $\omega =-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat{\omega }=Uk-Nk/K$. The stationary dispersion condition isDifferentiate the unconstrained dispersion before imposing stationarity. The stationary mountain-wave group velocity is obtained from the ground-frame dispersion; before imposing stationarity it isOn the stationary locus this becomesDifferentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.

For upward energy propagation from the lower boundary, choose $m<0$, since $k<0$ and ${\hat{c}}_{gz}>0$. It requires real nonzero $m$, henceThe downstream ray inclination above horizontal isIt is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.

Frame qualification for the requested phase angle: the normal intrinsic phase velocity is ${\mathbf{c}}_p=\omega (k,m)/K^2$. Because $\omega =-Uk$, the boxed stationary group velocity is exactly $-{\mathbf{c}}_p$: its angle to intrinsic phase propagation is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic group velocity ${\mathbf{c}}_g=\widehat{{\mathbf{c}}_g}-U\hat{\mathbf{x}}$ satisfies ${\mathbf{c}}_g\cdot (k,m)=0$, so its angle to intrinsic phase propagation is $\pi /2$. In the ground frame the phase velocity of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{{\mathbf{c}}_g}=(0.36U,0.48U)$ and ${\mathbf{c}}_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.

Linearize about a uniform horizontal flow $U$ and a hydrostatic, vertically stratified basic state. The basic-flow material derivative is $D_0={\partial }_t+U{\partial }_x$. The first two equations are the horizontal and vertical perturbation momentum balances: pressure gradients accelerate the disturbance, while the vertical equation retains the buoyancy force $-g{\rho }^{\prime }/{\rho }_0$ after the basic hydrostatic balance is subtracted.

The Boussinesq approximation neglects fractional density variations in inertia and enforces incompressibility, $u_x+w_z=0$, but retains them in buoyancy. Advecting the stable background density gradient gives $D_0{\rho }^{\prime }+wd\bar{\rho }/dz=0$. With $N^2=-(g/{\rho }_0)d\bar{\rho }/dz>0$, this is the fourth supplied equation. A displaced parcel then has a restoring buoyancy tendency.

Small amplitude allows products of perturbations to be discarded; two-dimensional motion removes $y$ dependence and transverse flow. Rotation, viscosity, diffusion, diabatic heating and mean shear are neglected. ${\rho }_0$ is the reference inertial density, whereas the small background density gradient still sets $N$. In a compressible real atmosphere the appropriate stability measure involves potential temperature; this Boussinesq system is a local idealization that filters acoustic waves rather than a full compressible atmospheric model.

Per unit horizontal area, the linear shallow-water kinetic and gravitational potential energies areFor the Kelvin mode, $u=0$ and $v=(c/H)\eta$ with $c^2=gH$. Therefore $H{v}^2=g{\eta }^2$, and the two energies are actually equal pointwise at this order. For a harmonic wave $\eta =A_0{e}^{-r/L_R}\cos (ky-\omega t)$,The averaging supplies $\overline{{\cos }^2}=1/2$. These energies refer to the perturbation relative to the flat resting state; background ocean potential energy is not included.

For an eastward tidal signal entering the English Channel from the Atlantic, the northern coast of France lies on the right of propagation, with land to the south. It is therefore the trapping coast of the corresponding northern-hemisphere Kelvin mode. The south coast of England is across the ocean from that boundary and receives a smaller amplitude through the offshore exponential decay. For a uniform channel of width $W$, this ideal mode would have $A_{\mathrm{England}}/A_{\mathrm{France}}=e^{-W/L_R}$.

Thus the mode assumption predicts the greater tidal range on the French side. It does not mean all local tides are a single Kelvin wave: channel depth and shape, friction, reflections, resonances and interference also influence the actual tidal range. The explanation concerns the stated eastward incoming tidal-mode idealization, not a universal comparison at every location or tidal constituent.