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A passive disk is heated predominantly by intercepted stellar radiation, rather than by its own viscous dissipation. A rough luminosity criterion, ignoring geometric factors, isThe intrinsic accretion disk luminosity is of order $GM\dot{M}/(2R_{\ast })$, with additional accretion energy potentially released at the stellar surface. The local test is more precise: absorbed irradiation must dominate the viscous flux at the radius being considered. For an ordinary steady Keplerian accretion disk, sufficiently far outside its inner edge,where $\alpha$ is the irradiation grazing angle of a disk. Reflection, very small incidence angles and shadowing can invalidate a simple total-luminosity comparison. The following calculation assumes an illuminated, absorbing surface.

At a point $(R,z_s)$ on the upper surface, the outward unit normal in the meridional plane is proportional to $(-z_s^{\prime },1)$, and a light ray from the point source is directed along $(R,z_s)$. Its flux through the upper face is the incident inverse-square flux times minus their scalar product:For a thin disk with small surface slope this becomesThe tangent slope and ray slope differ by $\alpha \simeq z_s^{\prime }-z_s/R$. Positive incidence requires disk flaring, $d(z_s/R)/dR>0$, and an unobstructed ray from the star. A negative value of the formula would mean that direct illumination of this face is absent, not that it receives a negative heating flux. The hierarchy $z_s\gg R_{\ast }$ makes the finite angular size of the star negligible here.

Let $a=\mathcal{R}/\mu$, so the ideal gas equation of state gives the isothermal sound speed $c_s^2=a{T}_c$. In the stellar vertical gravitational potential, hydrostatic equilibrium has vertical acceleration $-{\Omega }_K^{2}z$, where ${\Omega }_K^2=GM/R^3$. Integration of $c_s^2{\partial }_z\rho =-\rho {\Omega }_K^{2}z$ gives a Gaussian atmosphere with disk scale heightConsequently $q=z_s/R=fH/R$ satisfies $T_c=GM{q}^2/(a{f}^{2}R)$. Substitute this into the stipulated Stefan–Boltzmann law $F_d=\sigma T_c^4$ and the irradiation flux:Here $f$ specifies surface height in units of disk scale height and is unrelated to the wind-energy fraction in the preceding solution. This separable differential equation integrates exactly toIf the thin-disc solution is formally matched to $q(R_M)=1$, then $B=1-R_0^2/R_M^2$, givingIt follows that the flaring passive disk asymptotic isThis applies in the thin intermediate region $R\ll \min (R_0,R_M)$, in addition to the point-source and stellar-gravity assumptions. The corrections are of relative order $R^2/R_0^2+R^2/R_M^2$. In particular, the printed hierarchy $R\ll R_M$ must still be accompanied by the initial thin-disc condition $q\ll 1$: if $R_M\gg R_0$, radii larger than $R_0$ cannot satisfy that approximation. The matching at $q=1$ is an extrapolation used to set an integration constant; the derivation is not a valid treatment of a thick surface at $R_M$ itself.

Identify the stellar surface temperature with its effective radiating temperature, so that luminosity of a spherical blackbody gives $L_{\ast }=4\pi R_{\ast }^2\sigma T_{\ast }^4$. The given stellar surface scale height has $H_{\ast }^2=a{T}_{\ast }{R}_{\ast }^3/(GM)$, hence $GM/(a{T}_{\ast }{R}_{\ast })=(R_{\ast }/H_{\ast }{)}^2$. Substituting into $R_0$ yieldsThus a small stellar surface scale height leaves a wide interval above $R_{\ast }$ and below $R_0$ in which the thin disk flaring law can apply. The point-source approximation imposes the stronger lower cutoff obtained from $z_s\gg R_{\ast }$:This cutoff is still much smaller than $R_0$ when $R_0\gg R_{\ast }$. The full interval is therefore $R_{\ast }^{7/9}{R}_0^{2/9}\ll R\ll \min (R_0,R_M)$, whenever these bounds overlap. An assumed thickening radius must lie above the lower cutoff for all of the original approximations to hold simultaneously. The stellar luminosity-temperature relation is essential to the final comparison; independent arbitrary values of $L_{\ast }$ and $T_{\ast }$ would not imply it.

Take an unresolved spherical planet at Euclidean distance $d\gg R$, with uniform surface temperature $T$, stationary isotropic emission, and no intervening extinction. Isothermality does not by itself imply blackbody radiation; first derive the geometric result for a uniform emergent specific intensity $I_{\lambda }$.

The observer sees the projected disc with apparent solid angle ${\Omega }_p\simeq \pi R^2/d^2$. Integrating its specific intensity over that angle gives the observed spectral radiative fluxwhere ${\theta }_{\mathrm{obs}}$ is the small angle to the detector normal. For a blackbody, $I_{\lambda }=B_{\lambda }(T)$, and the Stefan–Boltzmann law gives ${\int }_0^{\infty }\pi B_{\lambda }(T)d\lambda ={\sigma }_{\mathrm{SB}}{T}^4$. HenceEquivalently, the surface radiative flux is ${\sigma }_{\mathrm{SB}}{T}^4$ and the luminosity of a spherical blackbody is $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}^4$, leading to the same result by inverse-square dilution. A uniform wavelength-dependent emissivity ${ϵ}_{\lambda }$ multiplies the spectral formula; an atmospheric spectrum can require a full angular radiative transfer integral rather than a single Planck function.

In vacuum, a narrow ray bundle conserves both power and geometrical etendue $dA\cos \theta d\Omega$. Their ratio, the specific intensity, is therefore independent of source-receiver distance. Equivalently, geometric dilution reduces received power and apparent solid angle by the same inverse-square factor.

For an isotropically emitting isothermal blackbody atmosphere, the outward surface flux iswhere the last equality is bolometric. Multiplication by the emitting area gives the luminosity of a spherical blackbody