 Solution

At full phase, reflected-light contrast iswhere $A_g$ is the geometric albedo, and other phases multiply this by a phase function normalized to one at full phase. Thermal contrast for uniform blackbody emitters isThe corresponding bolometric thermal ratio is $(R_p/R_{\star }{)}^2(T_p/T_{\mathrm{eff}}{)}^4$. If zero internal heating and equilibrium are assumed, energy balance also gives a bolometric reflected ratio $\alpha (R_p/d{)}^2/4=(\alpha ϵ/(1-\alpha ))(R_p/R_{\star }{)}^2(T_p/T_{\mathrm{eff}}{)}^4$ for the luminosity ratio; it is not a directional full-phase contrast.

Earth's thermal emission peaks around $10\mu \mathrm{m}$, and solar emission around $0.5\mu \mathrm{m}$. At $0.5\mu \mathrm{m}$ the thermal contrast for $T_p=288\mathrm{K}$ is negligible, about $5\times 10^{-46}$; reflected light dominates. Taking illustrative visible $A_g\simeq 0.3$ gives $\boxed{C_{0.5}\sim 5\times 10^{-10}}$. A Bond albedo does not uniquely determine a spectral geometric albedo; a Lambert sphere with Bond albedo $0.3$ would instead have $A_g=0.2$, giving $3.6\times 10^{-10}$. At $20\mu \mathrm{m}$, the Planck function gives $\boxed{C_{20}\simeq 1.0\times 10^{-6}}$ for $288\mathrm{K}$; using $255\mathrm{K}$ gives about $7\times 10^{-7}$. These are order estimates, since the terrestrial spectrum is not a uniform blackbody.