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 is
$$
F=\int_{\rm hemisphere}B(T)\cos\theta\,d\Omega
=\pi B(T)=\sigma T^4,
$$
where the last equality is bolometric. Multiplication by the emitting area gives the <luminosity of a spherical blackbody>
$$
\boxed{L_p=4\pi R_p^2\sigma T^4}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 315 / 4 / c / Solution

An isotropic blackbody surface has outward flux $F=\pi\int B_\nu\,d\nu=\sigma T^4$. A sphere of radius $R$ therefore has luminosity
$$
L=4\pi R^2\sigma T^4.
$$


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 Incoming links: Luminosity of a spherical blackbody