 Solution

The planetary equilibrium temperature is the effective planetary temperature at which absorbed starlight balances emitted thermal radiation, neglecting internal power. With uniform reradiation over the full surface,Cancelling areas givesUsing the supplied Bond albedos and taking the thermal emissivity $ϵ=1$ yields $\boxed{T_{\mathrm{eq},\oplus }\simeq 255\mathrm{K},T_{\mathrm{eq},\mathrm{J}}\simeq 103\mathrm{K}}$. “Perfect black bodies” must here mean perfect infrared emitters, since a perfect absorber at all wavelengths would instead have $\alpha =0$, inconsistent with retaining the supplied reflected fractions. If both albedos were set to zero, the corresponding values would be about $279$ and $122\mathrm{K}$.

Earth's mean surface temperature of about $288\mathrm{K}$ exceeds its emission temperature mainly because of the greenhouse effect. Jupiter also radiates substantial internal energy from cooling and contraction, including interior differentiation processes, so absorbed sunlight alone underpredicts its emission. Jupiter has no solid surface; the quoted $150\mathrm{K}$ is a representative atmospheric level and is not identical to an effective radiating temperature.