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The geometric albedo $A_g$ compares the planet's reflected brightness at full phase with that of a perfectly reflecting Lambertian surface in the form of a flat disc of the same projected area, illuminated by the same incident flux. The Bond albedo $A_B$ is the fraction of total incident stellar power reflected in all directions and over all wavelengths. They need not be equal: the planetary phase integral and the incident spectral weighting connect them.

For a blackbody star, the incident flux at orbital distance $a$ isAssume negligible intrinsic luminosity, thermal emissivity one, and efficient day-night heat redistribution giving uniform emission over the whole planet. The absorbed and emitted powers areUsing the Stefan–Boltzmann law gives the planetary equilibrium temperatureFor a representative emitting area encoded by a stellar flux redistribution factor $f$, the equivalent expression iswith $f=1/4$ for global uniform emission and $1/2$ for uniform dayside-only emission. The Bond albedo, rather than the geometric albedo, belongs in the energy budget. Internal heat, wavelength-dependent emissivity and imperfect redistribution require corresponding modifications.

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.

In the proper-eccentricity annulus, averaging over random proper phases cancels the proper vector in the first-order angular variation. The mean line density is proportional to $1-e_f\cos \theta$, while the full ring width is independent of angle. Thus at corresponding radial locationsThere is a forced-apocentre enhancement in number surface density of a disk, together with the proper-excursion edge profile.

For a narrow aligned eccentric ring with locally constant $e_f$, its width has the same first-order factor $1-e_f\cos \theta$ as its line density. Their ratio is constant: the constant-eccentricity aligned ring has no first-order surface-density contrast. It still contains more particles per unit longitude at apoapsis, spread across a wider area.

The more general surface density contrast in an eccentric ring depends on the eccentricity gradient. With a smooth number distribution per unit semi-major axis and aligned apsides,Hence decreasing $e_f(a)$ can also make the aligned ring apocentre-dense; for $e_f\propto a^{-1}$ it has the same first-order contrast as the proper-phase case. A positive gradient gives a pericentre enhancement. The problem leaves this profile unspecified, so there is no universal observational distinction based solely on the two apse densities.

Resolved ring width and a measured forced-eccentricity profile can distinguish the constant-width and similar-ellipse constructions. Interpreting observed brightness additionally requires the grain temperature, illumination, emissivity, and viewing geometry; brightness is not itself particle number density. Inferences from one contrast alone can therefore be degenerate.

A planetary surface energy balance equates absorbed short-wave radiation and heat supplied from below with reflected short-wave radiation and emitted long-wave radiation. With incident flux $F_r$, Bond albedo $\alpha$, conductive upward flux $F_c$, and emissivity $ϵ$,

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

The view factor $F_{i\to j}$ is the fraction of uniformly distributed, diffuse outgoing radiant flux from surface $i$ that directly reaches surface $j$. It depends on geometry, including obstruction, rather than on temperature or emissivity. All destinations in a closed enclosure have fractions summing to one. For a convex body fully enclosed by walls, $F_{\mathrm{body}\to \mathrm{walls}}=1$; the reverse view factor is generally smaller. Reciprocity gives $A_i{F}_{i\to j}=A_j{F}_{j\to i}$. Two unobstructed spheres with outer radius ten times the inner radius have $F_{\mathrm{inner}\to \mathrm{outer}}=1$ and $F_{\mathrm{outer}\to \mathrm{inner}}=1/100$. Do not multiply both directions by an assumed identical view factor.^[ref]

Consider an idealized planet that is a sphere of radius $R$, illuminated by parallel sunlight of irradiance $S$, measured on a plane perpendicular to the rays. Its intercepting projected area is $\pi R^2$, whereas its total surface area is $4\pi R^2$. Therefore the incoming power averaged per unit total surface area iswhere $\alpha$ is the broadband Bond albedo. The factor four is a sphere geometry result, not an atmospheric transmission factor. Local sunlight still varies over the illuminated hemisphere and vanishes on the dark side.

Now stipulate a single spatially uniform temperature $T$, no atmosphere, no internal heating, and emissivity one for outgoing thermal radiation. This is a thermal blackbody model using the Stefan–Boltzmann law; the prescribed solar-band reflection is treated separately, so the body is not a perfect absorber at all wavelengths. Let $C>0$ be the effective heat capacity per unit surface area, in $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Conservation of energy gives the ordinary differential equationHere $F$ is net heating per unit surface area, in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma$ is the Stefan-Boltzmann constant. Use $\sigma =5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, rounded from the exact constant reported by NIST. ^[ref] The uniform temperature is a model assumption: generally the surface average of $T^4$ differs from the fourth power of the surface-average temperature.

For fixed Bond albedo $\alpha$, radiative equilibrium requires $F(T_{\ast })=0$, givingWith the illustrative solar constant $S=1360\mathrm{W}{\mathrm{m}}^{-2}$, the global incoming average is $340\mathrm{W}{\mathrm{m}}^{-2}$:

Doubling albedo here lowers the equilibrium temperature by $33.231\mathrm{K}$, rather than halving it: the absorbed fraction changes from $0.7$ to $0.4$, and thermal radiation scales as $T^4$. These stipulated Bond albedos are whole-planet examples, not measurements of bare water and ice. The model predicts neither Earth's modern surface temperature nor its response to actual climate change; it omits the greenhouse effect, spatial heat transfer and seasons.

Each fixed-albedo equilibrium is a stable equilibrium: for $T_{\ast }>0$, the derivative is $F^{\prime }(T_{\ast })=-4\sigma T_{\ast }^3<0$. A small temperature increase produces net cooling, and a small decrease produces net heating. This is the simplest expression of the stabilizing Planck feedback. Allowing albedo to fall with temperature adds a competing term, as in ice-ocean albedo feedback.