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Peroxyacetyl nitrate forms reversibly from a peroxyacetyl radical and NO${}_2$. It stores reactive nitrogen while intact, then thermal dissociation can release it again. It can therefore transport nitrogen through colder free-tropospheric air and recycle it more rapidly in a warmer planetary boundary layer. Deposition or further chemistry can compete, but reversible storage is its reservoir role.

Dinitrogen pentoxide is likewise a reservoir because $\mathrm{N}{\mathrm{O}}_2+\mathrm{N}{\mathrm{O}}_3+\mathrm{M}\rightleftharpoons {\mathrm{N}}_2{\mathrm{O}}_5+\mathrm{M}$ can return reactive species. Conversion to nitric acid makes it a sink from the active nitrogen oxides family, especially through heterogeneous uptake on wet aerosol or cloud particles, followed by deposition or washout. Water vapour alone and wet-particle uptake are different mechanisms; they should not be assigned the same rate constant. Temperature, particle surface area and removal are often very different between the boundary layer and the free troposphere, so a reservoir can be transported in one environment and chemically removed in another.

For the stated total number density, multiply the number atmospheric mixing ratio by $[\mathrm{M}]$. With NO${}_x$ defined as NO plus NO${}_2$,The water partial pressure is $30/1000=0.030$ of the total pressure. It is a 3% number fraction, not 30 ppm.

Activate the oxime hydroxyl oxygen with thionyl chloride to give a good N–O leaving group. In the Beckmann rearrangement branch, the C–C bond anti to that leaving group migrates onto nitrogen as the N–O bond breaks. Here the migrating carbon is the spiro centre. Inserting N between that centre and the former oxime carbon expands the five-membered oxime ring to a six-membered nitrilium ion ring. Water attacks the electrophilic nitrilium carbon, followed by proton transfers and tautomerization to the observed spiro lactam. The lactam carbonyl oxygen is introduced in the aqueous work-up.

The competing Beckmann fragmentation cleaves the same anti C–C bond instead of completing migration onto N. Formation of the C≡N bond gives an open-chain nitrile and a tertiary carbocation at the former spiro centre. Loss of a proton from the adjacent chain carbon then gives the exocyclic alkene attached to the retained cyclopentane. The key distinction is N insertion and ring expansion for the lactam, versus C–C cleavage followed by elimination for the unsaturated nitrile. Stabilization of the resulting tertiary carbocation makes fragmentation plausible.

Methane reaching the stratosphere is oxidized and is an important source of stratospheric water vapor. That water can feed HOx radical family chemistry and alter radiative cooling and polar-cloud conditions. Methane also converts chlorine atoms into hydrogen chloride, opposing some chlorine-driven ozone loss. These effects compete: increasing methane does not have one universal ozone sign at every altitude and latitude.

A humidity assumption is necessary: warming does not by itself specify how much water is available. At fixed relative humidity, the Clausius-Clapeyron equation gives the first-order vapor pressure responseIf water abundance is approximated as proportional to $e_s$, the usual first-order estimate is a $26.7\%$ increase in hydroxyl radical production. Integrating the constant-latent-heat expression instead of linearizing gives $e_s(289)/e_s(285)=\exp [(45050/8.314)(1/285-1/289)]=1.3010$.

There is a small but explicit concentration correction when the square brackets are literal molecules per volume. Since $[{\mathrm{H}}_2\mathrm{O}]=\varphi e_s/(k_{B}T)$ and $P_{\mathrm{OH}}=2k[\mathrm{O}({}^{1}D)][{\mathrm{H}}_2\mathrm{O}]$,The corresponding finite-change concentration result is $(285/289)(1.3010)-1=0.2830$. The $25.3\%$ linear concentration estimate and the $26.7\%$ vapor pressure approximation express different stated approximations; neither follows without a humidity condition.

Extra water increases the fraction of oxygen singlet D atoms reacting to form hydroxyl radicals rather than being collisionally quenched and recycled into ozone. In clean, low-nitrogen oxides air this tends to increase ozone loss. In nitrogen oxides-rich air, increased hydroxyl radical can accelerate precursor oxidation and ozone production, while radical termination also changes. Greater hydroxyl radical production therefore does not imply a universally positive or negative ozone response.

The lost ozone-column photolysis shielding is $\Delta N=20(2.69\times 10^{16})=5.38\times 10^{17}\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-2}$, so its optical-depth decrement is $\sigma \Delta N=0.0538$. Under the stipulated effective-cross-section Beer-Lambert model, the transmitted photodissociation flux increases by $\exp (0.0538)$.

The short-lived oxygen singlet D atom satisfies $[\mathrm{O}({}^{1}D)]\simeq J_{{\mathrm{O}}_3}[{\mathrm{O}}_3]/(k_q[M])$ when quenching dominates. At fixed local ozone, water and air density, hydroxyl radical production is proportional to $J_{{\mathrm{O}}_3}$. ConsequentlyA linear optical-depth approximation would give $5.38\%$. A realistic calculation integrates the wavelength-dependent absorption cross section, quantum yield and actinic flux; clouds and multiple scattering modify the simple transmitted-beam model.

Ozone change can also alter the radiative temperature structure, water vapor and circulation, thereby changing the water reaction channel and the supply of ozone/precursors to the troposphere. If the held-fixed tropospheric ozone assumption is relaxed, altered ozone itself changes the photochemical source of oxygen singlet D atom. These are additional feedbacks, distinct from the isolated UV-transmission calculation above.

For the tropospheric OH budget, let $P$ be hydroxyl radical production per unit volume and $K$ its total first-order loss frequency. The present hydroxyl radical steady state hasHere one year is $365.25$ days; using it consistently makes the reported lifetimes in years independent of that conversion convention. Methane's chemical lifetime is ${\tau }_0=1/(k_2[\mathrm{OH}{]}_0)$, givingThe current combined loss frequency already includes existing hydrogen, so add only the incremental hydrogen sink:Holding production and the other losses fixed then gives $[\mathrm{OH}{]}_1=5.107\times 10^5\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-3}$ andThe rapid hydroxyl radical adjustment time, $1/K$ of order a second, supports its steady-state approximation in this box model. Subsequent methane growth would itself change hydroxyl radical consumption, so the computed increase is not a fully coupled long-term atmospheric prediction. Detailed molecular hydrogen modelling treats changes in methane, ozone and stratospheric water alongside reductions in other precursor emissions.^[ref]

Carbon dioxide is relatively well mixed, its abundance is measured accurately, and its spectroscopy and large-scale radiative effect are well constrained. Aerosols are short lived and spatially heterogeneous; uncertain emissions, removal, size, composition, water uptake and height change both scattering and absorption. Their aerosol-cloud interaction effects and subsequent cloud adjustments add uncertainty, and the unobserved preindustrial aerosol baseline complicates anthropogenic attribution. Thus total aerosol forcing combines poorly constrained physical pathways and a less certain counterfactual, rather than simply an uncertain molecular absorption coefficient.^[ref]

Absorption depends on an allowed transition with a nonzero transition dipole moment and appreciable line strength at wavelengths carrying terrestrial thermal radiation. Molecular vibration and molecular rotation, level populations and pressure/temperature-dependent line shapes determine the spectrum. A band in an atmospheric window can be more effective incrementally than one already strongly absorbed by water or carbon dioxide. Symmetry can make a vibration infrared inactive even when the molecule has many normal modes. Atmospheric lifetime affects the persistence of a released gas, not the intrinsic absorption strength of one molecule.

First, weakly absorbing sulfate scatters shortwave electromagnetic radiation and tends to cool the surface/planetary energy budget, with magnitude and sign influenced by cloud background and particle optics. Second, sulfate can supply cloud condensation nuclei: at fixed cloud water, more activated droplets give smaller droplets and greater cloud albedo through the Twomey effect; atmospheric precipitation and water-content adjustments can modify that idealized cooling. Third, sulfate particles change atmospheric composition through uptake, aqueous chemistry and heterogeneous surface reactions, with their acidity and available reactants controlling what is converted. Particle surfaces alone do not guarantee polar-style stratospheric chlorine activation in an ordinary tropospheric cloud.^[ref]

Convert the molar mass to $0.0288\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ before substitution:The often-used environmental value near $-6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ is less steep because rising moist air can condense water vapor, releasing latent heat and offsetting some expansion cooling. A saturated parcel therefore follows a smaller-magnitude moist adiabatic lapse rate, which varies with temperature, pressure and water content. Real environmental profiles also reflect radiation, mixing and weather systems; $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ is a representative profile value, not a universal parcel law.

Lower winter temperatures favour formation and persistence of polar stratospheric clouds, providing particle surfaces on which reservoir chlorine can be activated, for exampleThese are heterogeneous surface processes; cooling changes particle abundance, adsorption, phase and reaction uptake, so a simple gas-phase Arrhenius argument is insufficient. Removal of nitric acid in sedimenting particles can also delay conversion of active chlorine back into chlorine nitrate. The relevant condensation thresholds depend on pressure and water/nitric-acid abundances, rather than a single universal temperature.

Once sunlight is present, $\mathrm{C}{\mathrm{l}}_2+\mathrm{h}\nu \to 2Cl$ provides active chlorine. In the chlorine monoxide dimer cycle,The net reaction is $2{\mathrm{O}}_3\to 3{\mathrm{O}}_2$, with chlorine regenerated. Low temperature favours formation of the ClOOCl dimer and suppresses its thermal dissociation, making photodissociation rather than thermal return of two ClO radicals more competitive. Laboratory flow experiments formed the dimer at 220–240 K and distinguished its infrared and ultraviolet signatures, supporting the chemical identity needed for this mechanism.^[ref]

Colder polar conditions can prepare and sustain greater springtime ozone loss, but darkness alone does not drive the photolytic cycle. In polar night reactive precursors can accumulate; sunlight is required for rapid dimer-cycle ozone destruction. This differs from the middle-stratospheric gas-phase cooling response derived above.^[ref]

Excited $\mathrm{O}({}^{1}D)$ has two competing fates. Collisional deactivation returns ground-state oxygen, which reforms ozone rapidly; reaction with water instead gives net ozone loss and two OH radicals. The chemical branching fraction for the latter isThus only about 4.92% of these ozone photolyses directly lead to the specified net chemical loss; treating every photodissociation as permanent removal would overestimate the sink. The corresponding first-order rate is $J_1{f}_w=9.8361\times 10^{-7}{\mathrm{s}}^{-1}$. Converting the given hydroperoxyl mixing ratio to a concentration gives $[\mathrm{H}{\mathrm{O}}_2]=20\times 10^{-12}[\mathrm{M}]=4.4\times 10^8\mathrm{c}{\mathrm{m}}^{-3}$, so $k_4[\mathrm{H}{\mathrm{O}}_2]=9.6800\times 10^{-7}{\mathrm{s}}^{-1}$.

For a well-mixed marine atmospheric boundary layer of depth $H=2\times 10^5\mathrm{cm}$, the downward surface flux is $v_{\mathrm{dep}}[{\mathrm{O}}_3]$. Dividing by the layer volume per unit area gives a first-order dry deposition rate $v_{\mathrm{dep}}/H=7.5\times 10^{-7}{\mathrm{s}}^{-1}$. The air density cancels on expressing the entire balance in ozone mixing ratio $\chi$:At steady state, with $P_{\chi }=0.25/3600\mathrm{ppb}{\mathrm{s}}^{-1}$,The effective removal time is about 4.28 days. This calculation assumes a uniform layer of the diagnosed depth, fixed chemical conditions and a local time-independent balance.

For a liquid cloud at fixed water content and geometrical thickness, more cloud condensation nuclei can produce more, smaller droplets. The increased total extinction area raises optical depth and reflected sunlight: the Twomey effect gives negative shortwave radiative forcing. It is an indirect aerosol effect mediated by cloud microphysics, rather than direct aerosol scattering.^[ref]

Smaller droplets can also delay collision-coalescence and drizzle, potentially increasing liquid-water retention, cloud lifetime or coverage; that adjustment would usually strengthen cooling for low reflective clouds. It is not universal: evaporation and mixing with dry air can reduce water content or cloud cover, and dynamical changes can oppose the simple lifetime argument. Absorbing particles can heat an atmospheric layer, sometimes evaporating cloud and producing a positive semi-direct forcing, but their vertical position can instead stabilize a cloud layer and reverse the response. Aerosols can additionally alter ice nucleation and mixed-phase precipitation, with no single forcing sign independent of cloud type. The net response combines shortwave reflection and longwave thermal effects, and depends on cloud height, optical depth and adjustments. Thus the fixed-water Twomey sign is robust within its assumptions; adjustment and ice-cloud signs require context.^[ref]

Activation as a cloud condensation nucleus depends primarily on dry particle size, soluble composition and hygroscopicity, together with wetting and surface-tension effects. In Köhler theory, droplet curvature raises the equilibrium vapor pressure, while dissolved solute lowers water activity. Each particle therefore has a critical supersaturation; larger or more hygroscopic particles generally activate at a lower value. A size distribution cannot be replaced by an undifferentiated total aerosol count.^[ref]

Actual droplet number also depends on the supersaturation attained in the cloud updraught. Adding particles increases competition for condensing water vapor and can lower the peak supersaturation, leaving some particles unactivated. Updraught speed, entrainment and the aerosol size distribution therefore make the number of activated droplets a nonlinear function of the available nuclei. Suitable nuclei need sufficiently low activation thresholds, and additional aerosol does not guarantee one additional droplet per particle.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.

After onset, protrusions reach more weakly saline liquid and advance faster, while grooves accumulate rejected brine. The planar interface develops cells and, as growth becomes sufficiently strong, crystal dendrites with concentrated liquid between them. Nonlinear competition changes the selected spacing; linear rates alone cannot fix the final geometry. The interdendritic liquid forms a mushy layer, concentrates toward the eutectic composition, and ultimately freezes into a salt-bearing eutectic product when it reaches $T_E$. This replaces the pure planar salt-rejection picture with trapping and freezing of brine between crystals.

Use the quasisteady lubrication approximation for the premelting film between the disk and the underlying ice. The ice is anchored to the lower plane; new ice formed from water supplied through this upper film raises the disk. Let $U$ be its upward separation rate, $d(r)$ the premelted thickness, $\mu$ the liquid dynamic viscosity, $\rho$ a common ice/water mass density, $L$ the specific latent heat of fusion, and $T_m$ the absolute bulk melting temperature. As usual in this leading model, neglect the ice–water density difference, pressure-dependent changes in $T_m$ due to a common phase pressure, inertia and edge corrections. The imposed thermal field is taken as maintained by an external heat supply/removal system, so no additional heat-conduction bottleneck is introduced.

Write $\Delta =T_m-T_0>0$, $K_T=\rho L/T_m$ and $\Delta (r)=\Delta (1-r^2/a^2)$. The Clapeyron pressure relation for equal-density phases makes the pressure difference across the solid–liquid interfaceThis thermomolecular pressure is the repulsive stress transmitted through the film. To calculate a definite rate one also needs a film-thickness constitutive law. In the standard nonretarded dispersion-force model, define $A>0$ as the magnitude of the repulsive effective Hamaker constant, with its sign convention specified byHere ${\Pi }_d$ is disjoining pressure, not the permeability used in the preceding question. The sign of an effective Hamaker coefficient depends on the three materials; the positive parameter here denotes an interaction that stabilizes a premelted film. The problem does not specify $A$ numerically, so it must remain a defined material parameter.

Take the reservoir pressure at the rim to be $p_{\infty }$. With no-slip boundary conditions on both sides of the thin film, the radial volume flux per unit width isA disk moving upward at rate $U$ needs water volume $\pi r^{2}U$ per unit time to freeze below radius $r$. Regularity on the axis and mass conservation therefore give $2\pi r{q}_r=-\pi r^{2}U$, or $q_r=-Ur/2$. ThusDefine $B=A/(6\pi K_T)$, so $d^3=B/[\Delta (1-r^2/a^2)]$. Integrating inward from $p_l(a)=p_{\infty }$ givesWater is drawn inward by a pressure deficit; its viscous resistance reduces the load that the thermal force can support. The net normal stress on the disk is $p_l-p_{\infty }+P_T$. Let $W$ be the downward apparent load after subtracting ordinary buoyancy. For an ideal thin disk with negligible buoyant volume, $W=Mg$, where $g$ is gravitational acceleration. If a finite displaced volume is known, use its submerged weight instead; the stated mass and radius alone do not determine that volume.

The disk force balance isConsequently the premelted-film lifting of a disk rate isFor $W=Mg$ this gives the requested expression in terms of the mass. Every additional symbol in this formula has been defined above. The growing-ice lifting branch exists only when the bracket is positive.

At the threshold the required water flux and viscous pressure deficit vanish. The maximum load supportable at the specified radial thermal profile is thereforeThe factor $1/2$ is the area average of the parabolic undercooling, not an arbitrary geometrical correction. At equality the lift rate is zero; strictly positive lift requires $W<W_{\max }$. There is no radius-only, temperature-independent maximum load in this ideal model: one must specify the imposed undercooling or a physically admissible range of it.

For a fixed positive load define ${\Delta }_c=2W{T}_m/(\pi a^2\rho L)$ and $U_{\infty }=A/(6\pi \mu a^2)$. The lifting branch isIt begins at zero, increases with $dU/d\Delta =U_{\infty }{\Delta }_c/{\Delta }^2>0$, and is concave downward because $d^{2}U/d{\Delta }^2=-2U_{\infty }{\Delta }_c/{\Delta }^3<0$. Below threshold no positive steady heave is possible: the thermomolecular force is too weak to support the load. The mathematical continuation to negative $U$ is not a prediction of upward ice-driven separation there.

Increasing undercooling increases the available thermal pressure, but it also thins the premelting film. In the inverse-cube disjoining pressure model, $d^3\propto {\Delta }^{-1}$, so the hydraulic conductance decreases in exact inverse proportion to the driving scale. The remaining load correction tends to zero and the rate approaches a plateau:For positive load this is approached from below, not attained at finite undercooling. With zero load the reduced model gives the same constant rate wherever its thin-film assumptions hold.

Two limits of the calculation are worth making explicit. First, the formal $d(r)$ diverges at the warm rim, so a narrow outer connection to bulk water replaces the thin-film solution there; this leading model assumes that its contribution to the integrals is negligible. Second, very large undercooling produces a molecularly thin film and eventually invalidates continuum lubrication theory and the simple dispersion-force law. Thus the plateau is the maximum within the stated ideal constitutive model, not an unrestricted extrapolation for real water.

If a different molecular interaction is intended, the result is not universal. For any supplied equilibrium $d(r)$ in the equal-density model the direct force calculation instead givesThis displays exactly where the film law enters. The closed formula and plateau above use the conventional repulsive inverse-cube law; premelting alone, without that material law, would not determine a unique rate curve.

Interpret the quoted band amplitude as an absolute transit-depth increment $\Delta D=0.0003$, or $300$ parts per million. For a thin atmospheric annulus of extra height $\Delta z$, the atmospheric spectral-feature amplitude isLet the water band span $N_H$ atmospheric scale heights, so $\Delta z=N_{H}H$. For a nearly isothermal atmosphere with constant gravity and mean molecular weight $\mu$, $H=k_{B}T/(\mu m_{u}g)$. The temperature from a transmission-band amplitude is thereforeHere $T_0,H_0,{\mu }_0,g_0$ refer to the chosen Jupiter reference layer. Assuming $g=g_J\simeq 25\mathrm{m}{\mathrm{s}}^{-2}$, an H/He exoplanet atmosphere with $\mu \simeq 2.3$, and a clear strong band with $N_H\simeq 5$, gives $H\simeq 210\mathrm{km}$ and $T\simeq 1.4\times 10^3\mathrm{K}$. Equivalently, a Jovian reference with $T_0\simeq 165\mathrm{K}$ has $H_0\simeq 24\mathrm{km}$, giving $H/H_0\simeq 8.7$.

The assumptions are hydrostatic, approximately isothermal limb gas, constant gravity, negligible cloud truncation, H/He composition, and an effective opacity contrast yielding about five scale heights. In an ideal clear limb, $N_H\simeq \log ({\kappa }_{\mathrm{band}}/{\kappa }_{\mathrm{cont}})$ follows from the slant optical depth of an isothermal atmosphere. The data fix $N_{H}H$, not $H$ separately: the temperature is conditional on the band opacity contrast and composition. For $N_H=3$–$7$ with the same composition and gravity, the estimate is about $1000$–$2400\mathrm{K}$.

The continuum exoplanet transit photometry depth is approximately the opaque projected area ratio, assuming a corrected or negligible limb darkening contribution:For a Sun-sized host, $R_p\simeq 6.96\times 10^7\mathrm{m}\simeq R_J\simeq 11R_{\oplus }$. The natural size analogue is Jupiter, not Earth.

The expected bulk constituent is molecular hydrogen, with helium next most abundant for a retained primary planetary atmosphere. The water band detects a strong trace absorber; it does not imply that water is the dominant gas. This interpretation assumes a conventional H/He gas giant; the measured radius by itself is not a measurement of mass or composition.

The mass-radius curve of solar-composition substellar objects reflects the transition from weak compression to pressure ionization and electron degeneracy pressure, followed by sustained hydrogen burning. A schematic joining representative object classes is:

The illustration is not an age-specific numerical evolutionary model. In particular, ice giants contain much more heavy material than a solar-composition giant, so a single uniform-composition equation of state does not describe the entire joined curve.

Planet/brown-dwarf naming conventions and deuterium burning do not define a universal discontinuity in the mass-radius relation. Composition, age and irradiation move the curves; the hydrogen-burning transition changes the long-term energy source more fundamentally.