Planetary equilibrium temperature ($T_{\mathrm{eq}}$)

For a planet of Bond albedo $A_B$ orbiting a blackbody star of radius $R_{\ast }$ and temperature $T_{\ast }$, complete global redistribution and negligible internal heat giveUniform dayside-only reradiation replaces $4a^2$ by $2a^2$. The planetary equilibrium temperature describes absorbed stellar power; including an intrinsic cooling flux gives a total effective temperature $T_{\mathrm{eff}}^4=T_{\mathrm{int}}^4+T_{\mathrm{eq}}^4$, distinct from $T_{\mathrm{eq}}$. Neither quantity is necessarily the greenhouse-warmed surface temperature.

Define a planet's effective radiating temperature by its actual outgoing longwave flux, regardless of the surface temperature. In radiative equilibrium with solar irradiance $S_0$ and Bond albedo A, $\sigma T_{\mathrm{eff}}^4=(1-A)S_0/4$. A greenhouse effect changes the surface temperature needed to emit that flux. An effective emissivity relating OLR to surface temperature is a diagnostic of the atmosphere–surface system, not necessarily a material emissivity. Away from equilibrium, the outgoing-flux definition and the temperature inferred from absorbed sunlight need not agree.

The ratio of top-of-atmosphere outgoing thermal flux to the blackbody flux at a chosen representative surface temperature. In global equilibrium with a uniform surface, ${ϵ}_{\mathrm{eff}}=(T_{\mathrm{eq}}/T_s{)}^4$. It parametrizes the relation between surface and radiating temperatures and need not equal the material emissivity of the surface or an atmospheric layer. An effective value below one is compatible with a black surface when colder atmospheric layers contribute to radiation escaping to space.

A sphere intercepts a parallel stellar beam over projected area $\pi R_p^2$ and radiates over surface area $4\pi R_p^2$. Dividing total absorbed power by total surface area gives the factor one-quarter, independent of planet radius. Under uniform-temperature reradiation this implies $T=T_{\ast }\sqrt{R_{\ast }/(2a)}(1-A{)}^{1/4}$. Without heat redistribution it constrains the area mean of $T^4$, not the arithmetic mean of $T$.