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 $a$ is
$$
S=\frac{L_\star}{4\pi a^2}=\sigma_{\rm SB}T_\star^4\left(\frac{R_\star}{a}\right)^2.
$$
Assume negligible intrinsic luminosity, thermal emissivity one, and efficient <day-night heat redistribution> giving uniform emission over the whole planet. The absorbed and emitted powers are
$$
\pi R_p^2S(1-A_B)=4\pi R_p^2\sigma_{\rm SB}T_e^4.
$$
Using the <Stefan–Boltzmann law> gives the <planetary equilibrium temperature>
$$
\boxed{T_e=T_\star\sqrt{\frac{R_\star}{2a}}(1-A_B)^{1/4}}.
$$
For a representative emitting area encoded by a <stellar flux redistribution factor> $f$, the equivalent expression is
$$
T_e=T_\star\sqrt{R_\star/a}\,[f(1-A_B)]^{1/4},
$$
with $f=1/4$ 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 315 / 4 / b / Solution

{title2=$q=2\int_0^\pi\Phi(\alpha)\sin\alpha\,d\alpha$}

The phase integral converts a monochromatic <geometric albedo> into the corresponding reflected power fraction. The bolometric <Bond albedo> is the incident-spectrum-weighted average of $q_\lambda A_{g,\lambda}$.

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 Incoming links: Planetary phase integral