Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S30 / 3

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.

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.

At maximum projected separation, an Earth analogue at $10\mathrm{pc}$ is separated from its star by $\vartheta =a/D_{\mathrm{sys}}=0.1\mathrm{arcsec}=4.85\times 10^{-7}\mathrm{rad}$. The Rayleigh criterion for a circular pupil givesThe relevant quantity is the star–planet separation, not the tiny angular diameter of the planet itself. Simple angular optical resolution is insufficient: the planet lies in the star's diffraction pattern at reflected contrast of order $10^{-10}$, or thermal contrast of order $10^{-6}$. Residual speckles, wavefront errors, scattered light, detector/background noise and, in the infrared, strong telescope and sky thermal emission obstruct detection. A coronagraph or other stellar suppression method generally has an inner working angle of several $\lambda /D$, making practical aperture requirements larger than the bare Rayleigh criterion. The question's instrumentation claim is interpreted in its 2019 context; the physical requirements remain the basis of the estimate. Maximum reflected brightness near full phase also coincides with small projected separation, so maximum brightness and maximum resolvable separation cannot generally be obtained simultaneously.