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Use spectral radiant exitance per unit wavelength. The Planck law isA hotter blackbody radiates more at every wavelength, its wavelength peak moves to shorter wavelengths according to Wien's displacement law, and its integrated flux rises as $\sigma T^4$ by the Stefan–Boltzmann law. The wavelength convention matters: a spectrum plotted per unit frequency has a different peak transformation.

For planetary equilibrium temperature, assume an isotropic blackbody star, a circular orbit, negligible internal heating, unit surface thermal emissivity and efficient global heat redistribution so the surface has one temperature. The received stellar irradiance is $F_{\ast }=\sigma T_{\ast }^4(R_{\ast }/a{)}^2$. A planet of radius $R_p$ intercepts $\pi R_p^2{F}_{\ast }$ but radiates over $4\pi R_p^2$, giving the projected-area factor in planetary energy balanceWithout the uniform-temperature assumption, this calculation gives $⟨T^4{⟩}^{1/4}$, not a unique arithmetic mean of surface temperature.

With Bond albedo $A=0.05$, the absorbed fraction is $1-A$. Hencea decrease of approximately $3.87\mathrm{K}$. This albedo weights incident stellar light and does not imply thermal emissivity $0.95$.

Define the planetary effective emissivity by outgoing top-of-atmosphere flux $F_{\mathrm{out}}={ϵ}_{\mathrm{eff}}\sigma T_s^4$. Maintaining the same absorbed stellar power with $T_s=312\mathrm{K}$ givesThe greenhouse effect allows thermal radiation to escape from colder layers, so the outgoing flux is less than a black surface at $T_s$ would emit. Downward atmospheric emission contributes to the surface energy balance; this does not violate the top-of-atmosphere energy balance.^[ref]

A single-layer greenhouse model makes the distinction quantitative. Take a shortwave-transparent, isothermal layer with infrared absorptivity and emissivity ${ϵ}_a$, above a surface black in the thermal infrared. Writing $Q=(1-A)F_{\ast }/4$, its surface and atmospheric balances areFor ${ϵ}_a>0$, $T_a^4=T_s^4/2$. Radiation escaping to space is $(1-{ϵ}_a)\sigma T_s^4+{ϵ}_a\sigma T_a^4$, soThus $0.849$ is the planet's effective emissivity; $0.301$ is the layer's infrared emissivity within this particular model. They are not interchangeable. Real atmospheres require wavelength-dependent absorption, a vertical temperature profile and nonradiative heat transport; the measured temperatures alone do not uniquely retrieve an atmospheric optical profile.

The stellar luminosity is $L_{\star }=4\pi R_{\star }^2{\sigma }_{\mathrm{SB}}{T}_{\star }^4$. At separation $a$, the planet intercepts a disk of area $\pi R_p^2$. For Bond albedo $A_B$ and uniform emission over its full surface, energy balance givesThis planetary equilibrium temperature neglects intrinsic heat. Dayside-only or nonuniform emission changes the redistribution factor.

For the single-layer greenhouse model, assume a black infrared surface, an atmosphere transparent to incoming stellar radiation, and infrared absorptivity/emissivity $\alpha$. Let $T_s$ be surface temperature and $T_a$ layer temperature. Kirchhoff's law makes the layer's emission in each direction $\alpha {\sigma }_{\mathrm{SB}}{T}_a^4$. Its balance isso $T_a^4=T_s^4/2$ for $\alpha >0$. Surface balance includes the downward layer emission:ThereforeThe no-atmosphere limit is $T_s=T_e$, and a completely IR-absorbing single layer gives $T_s=2^{1/4}{T}_e$. The outgoing flux $(1-\alpha )\sigma T_s^4+\alpha \sigma T_a^4$ remains $\sigma T_e^4$. The layer delays escape by absorption and re-emission; it does not permanently retain a fraction $\alpha$ of every emitted photon, which would incorrectly predict a divergent temperature at $\alpha =1$.

Use the single-layer greenhouse model with a blackbody surface and one isothermal atmospheric layer. The layer is transparent to incoming stellar radiation, absorbs a fraction $\alpha$ of the surface's thermal radiation, and has the same thermal emissivity $\alpha$ by Kirchhoff's law of thermal radiation. Here $0\le \alpha \le 1$, and $T_e$ is the planetary equilibrium temperature defined by the absorbed global stellar flux ${\sigma }_{\mathrm{SB}}{T}_e^4$. Neglect convection, latent heat transport and intrinsic heat.

The layer absorbs $\alpha {\sigma }_{\mathrm{SB}}{T}_s^4$ and emits $\alpha {\sigma }_{\mathrm{SB}}{T}_a^4$ upward and downward. For $\alpha >0$, its energy balance givesThe surface receives the absorbed stellar flux plus downward atmospheric emission, soEliminating $T_a$ yieldsThe outgoing top-of-atmosphere flux is $(1-\alpha ){\sigma }_{\mathrm{SB}}{T}_s^4+\alpha {\sigma }_{\mathrm{SB}}{T}_a^4={\sigma }_{\mathrm{SB}}{T}_e^4$, confirming overall energy conservation. At $\alpha =0$, $T_s=T_e$ and the decoupled layer's temperature is not determined; at $\alpha =1$, $T_s=2^{1/4}{T}_e$. This warming is the restriction of thermal escape by the absorbing and emitting layer.

Let $F_{\mathrm{abs}}={\sigma }_{\mathrm{SB}}{T}_{\mathrm{eq}}^4$ be the globally averaged absorbed stellar flux. The single-layer greenhouse model has an atmosphere transparent to incoming visible light and with infrared absorptivity and emissivity $\alpha$. Treat the surface as a blackbody and neglect internal heating. The layer emits $\alpha {\sigma }_{\mathrm{SB}}{T}_a^4$ both upward and downward.

Atmospheric radiative balance givesSurface radiative balance givesEliminating $T_a$ yieldsThe transparent limit has $T_s=T_{\mathrm{eq}}$, and a fully infrared-opaque layer has $T_s=2^{1/4}{T}_{\mathrm{eq}}$. The model omits convection, wavelength-dependent opacity, and atmospheric absorption of starlight, so it captures the simplest greenhouse energy balance rather than a full surface-temperature prediction.

Use a single-layer greenhouse model whose atmosphere is transparent to incident stellar light and has infrared emissivity $ϵ$. Let $T_s$ and $T_a$ be the surface and atmospheric temperatures. Atmospheric balance givesAt the top of the atmosphere, the escaping flux is the directly transmitted surface radiation plus upward atmospheric emission:ThereforeFor a perfectly infrared-opaque one-layer atmosphere, $T_s=2^{1/4}{T}_{\mathrm{eq}}$; for $ϵ=0$, $T_s=T_{\mathrm{eq}}$.

In an infrared-opaque single-layer greenhouse model, the atmospheric layer obeys $2\sigma T_a^4=\sigma T_s^4$. The outgoing planetary flux is $\sigma T_a^4$, while the globally averaged absorbed stellar flux is $(1-A_B)L/(16\pi a^2)$. Radiative equilibrium therefore givesHence the orbit at which the prescribed surface temperature can be maintained isThis assumes uniform redistribution, constant Bond albedo, unit longwave emissivity, a transparent atmosphere to starlight, and no internal heat. Without the greenhouse layer, replace $8\pi$ by $16\pi$.

Assume uniform global temperatures, all visible light not scattered back to space is absorbed by the surface, and use Kirchhoff's law of thermal radiation so that the atmospheric infrared emissivity is $\alpha$. If $F_{\ast }=\sigma T_{\ast }^4(R_{\ast }/a{)}^2$, the globally averaged absorbed stellar flux isThe atmosphere absorbs $\alpha \sigma T_{\mathrm{surf}}^4$ from the surface and emits from both faces, soSurface balance isTherefore the single-layer greenhouse model givesVisible scattering cools the surface, whereas infrared absorption and downward re-emission warm it.

Let absorbed solar flux per unit surface area be $F_{\odot }=\sigma T_0^4$. The single-layer greenhouse model emits infrared upward and downward. Top-of-atmosphere balance gives $\sigma T_a^4=F_{\odot }$, hence $T_a=T_0$. Surface balance is $\sigma T_s^4=F_{\odot }+\sigma T_a^4$, soThe layer simultaneously satisfies $\sigma T_s^4=2\sigma T_a^4$. This ideal radiative model excludes convection, spectral windows and atmospheric temperature structure; it is not a quantitative prediction for the real atmosphere.

A single-layer greenhouse model replaces the vertical structure of an exoplanet atmosphere by one layer transparent to incoming stellar light and having thermal emissivity $ϵ$. Let the surface be a uniform blackbody receiving absorbed stellar irradiance $Q$. In radiative equilibrium, the layer absorbs $ϵ\sigma T_s^4$ and emits $ϵ\sigma T_a^4$ in each direction, so for $ϵ>0$,The second equation follows from surface balance, $\sigma T_s^4=Q+ϵ\sigma T_a^4$. It demonstrates how absorption and downward reemission warm the surface at fixed absorbed power. The limiting $ϵ=1$ surface is $2^{1/4}$ times the planetary equilibrium temperature. The model omits convection, spectral variation and a continuous atmospheric pressure-temperature profile, so the fitted emissivity is not a unique atmospheric composition.