Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S25 / 2 / 3 / Solution

The quadratic coefficient must have units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$; the printed ${\mathrm{K}}^{-1}$ makes its term dimensionally incompatible with a radiative flux. Interpret its numerical value as ${\lambda }_2=0.05\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$, with ${\lambda }_1=1.5\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. The total imbalance isA quadrupling gives $F_{\mathrm{ext}}=2(3.75)=7.5\mathrm{W}{\mathrm{m}}^{-2}$. The two equilibrium roots areFor positive heat capacity, stability requires $\partial N/\partial T=-{\lambda }_1+2{\lambda }_2\Delta T<0$. Thus only the lower root is stable. A saddle-node bifurcation occurs atThe logarithmic forcing relation gives $C_{\mathrm{crit}}=2802^{11.25/3.75}=2240\mathrm{ppm}$. Strict linear stability holds below this concentration on the lower branch; the equality is marginal and is the supremum, not another strictly stable equilibrium. Sustained forcing above it causes warming without a balancing equilibrium in this quadratic model. At such large changes its local feedback expansion can fail, and other physics can intervene; it does not prove unlimited real-planet warming or locate a measured Earth runaway threshold.