 Solution

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.