Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 3

LetThe enclosed mass isThis is continuous at the core boundary. Hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$, with $P(R_p)=0$. DefineSince $F^{\prime }(r)=Gm(r)(A-br)/r^2$, the mantle pressure isInside the uniform core, the shell theorem gives $m(r)=4\pi {\rho }_c{r}^3/3$, so

The mantle density decreases outward when ${\rho }_m\ge 0$. Requiring a nonnegative surface density gives the immediate physical boundThe measured planetary mass also fixesEquivalently,and a viable model requires this value to obey the boxed bound.

Assume a thin isothermal atmosphere, so $g\simeq G{M}_p/R_p^2$ and the surface area are constant through it. Integrating hydrostatic equilibrium from base pressure $P_0$ to negligible top pressure gives atmospheric column mass $P_0/g$. ThereforeEquivalently, $M_{\mathrm{atm}}\sim 4\pi R_p^2{\rho }_{0}H$, with ${\rho }_0=P_0\mu m_H/(k_B{T}_0)$ and $H=k_B{T}_0/(\mu m_{H}g)$; the explicit temperature dependence cancels.

The location between pure-silicate and pure-water planetary mass-radius curves does not determine a unique composition. Possibilities include a rocky iron-silicate interior with a deep water layer, a rock-water mixture, or a mostly rocky planet whose small hydrogen-helium envelope inflates its radius. The atmosphere could accordingly be hydrogen-helium, steam-rich, carbon-dioxide-rich, or secondary gas released from the interior. Residence in the circumstellar habitable zone constrains stellar flux but does not by itself distinguish these models or guarantee surface liquid water.

For an atmosphere spanning fixed base and photospheric pressures,Thus two compositions haveAt equal temperature, a hydrogen-helium atmosphere with $\mu \simeq 2.3$ is abouttimes thicker than a steam atmosphere with $\mu \simeq 18$. This large difference is one reason atmospheric composition contributes strongly to the exoplanet interior-composition degeneracy.

The stellar flux at the orbit isLet $ϵ$ be the fraction of the planet's total absorbed power transported to the night side, and let $A_B$ be its Bond albedo. Equating transported power to nightside blackbody emission givesHence the nightside equilibrium temperature isPerfect global day-night heat redistribution has $ϵ=1/2$ and gives $T_n=T_s\sqrt{R_s/a}[(1-A_B)/4{]}^{1/4}$; without redistribution, stellar heating alone gives $T_n=0$ in this idealization. Efficient atmospheric or oceanic transport prevents volatile cold trapping and broadens the habitable region of a tidally locked planet, whereas a thin atmosphere can leave a frozen night side even when the day side is temperate.