For a small additional carbon mass, logarithmic forcing has derivative inversely proportional to the background atmospheric carbon inventory. With other conversion factors fixed, $b_0$ therefore decreases as background concentration rises. Solve the temperature equation using an integrating factor and the pulse burden:At $b_1=2k$ the second term is its continuous limit $t{e}^{-b_{1}t}$. For $0<k\ll b_1$, temperature initially rises over $b_1^{-1}$ toward roughly $b_0/b_1$, then relaxes over $(2k{)}^{-1}$ toward half that value. This produces an overshoot, rather than permanent growth. The two numerical limits requested separately below also follow directly from the differential equation.

The second box represents a reservoir exchanging carbon with the atmosphere, with equal capacity in the chosen concentration units. For a unit atmospheric pulse, $C(0^{+})=1$, $C_1(0^{+})=0$ and no subsequent emissions, total $C+C_1=1$ is conserved while $d(C-C_1)/dt=-2k(C-C_1)$. ThusThe atmospheric burden falls from one to one half with time constant $(2k{)}^{-1}$, not $k^{-1}$. The model has a permanent airborne fraction because neither box contains an external sink. The carbon-response panel of the original figure in S12 shows this explicitly.

Albedo reduction increases absorbed short-wave flux by $S_0(a_{1}T+a_2{T}^2)/4$. Define $A={\lambda }_P-1.5-S_0{a}_1/4$ and $B=S_0{a}_2/4$. The perturbation energy balance is $N(T)=F-AT+B{T}^2$, where $F=3.7$ W m${}^{-2}$ for doubling. With the preceding constants, $A\simeq 1.220$ W m${}^{-2}$ K${}^{-1}$ and $B=0.068$ W m${}^{-2}$ K${}^{-2}$. The two roots areStability follows from $N^{\prime }(T)=-A+2BT<0$ at the lower root. They merge at $T=A/(2B)\simeq 8.97$ K and $F_{\max }=A^2/(4B)\simeq 5.47$ W m${}^{-2}$; at equality the equilibrium is marginal, not linearly stable.

Assuming the usual logarithmic forcing extension $F=3.7\log (C/280)/\log 2$ gives a stable branch only for $\boxed{C<C_{\mathrm{crit}}\simeq 7.8\times 10^2\mathrm{ppm}}$. This “maximum” is the supremum of stable concentrations in the stated quadratic, linearized model. It is not a physical prediction of a real-world runaway threshold; extrapolating its feedback laws far beyond their small-change regime is an additional assumption.

At a fixed surface warming, a smaller lapse rate makes the upper troposphere warmer than in an unchanged-lapse profile. Its emission to space therefore increases, giving a negative feedback. Water vapor has the opposite effect under the given coefficients. The net restoring coefficient is ${\lambda }_P-2+0.5\simeq 2.24$ W m${}^{-2}$ K${}^{-1}$, so $\boxed{\Delta T\simeq 1.65\mathrm{K}}$. This calculation uses the same $S_0$ and linearized Planck model as the preceding part.

Effective emission temperature satisfies $\sigma T_e^4=S_0(1-\alpha )/4$: absorbed sunlight over the planetary disc balances radiation over the whole sphere. Thus $T_e=[S_0(1-\alpha )/(4\sigma ){]}^{1/4}$. Taking $S_0=1360$ W m${}^{-2}$ gives about 254.5 K. If only a uniform atmospheric/surface temperature shift is allowed, the blackbody response coefficient is ${\lambda }_P=4\sigma T_e^3\simeq 3.74$ W m${}^{-2}$ K${}^{-1}$. HenceThe surface and troposphere shift together under this fixed-profile approximation; their absolute temperatures need not equal $T_e$. A more detailed spectral Planck feedback requires additional information not supplied by the simple model.

For dry ascent, $c_{p}dT=Vdp$ and hydrostatic pressure change is $dp/dz=-\rho g=-g/V$, so $\boxed{{\Gamma }_d=-dT/dz=g/c_p}$. During condensation $dq<0$, and $-Ldq>0$ offsets expansion cooling, making the saturated ascent less steep.

For a saturated parcel with $q=q_s(T,p)$, use $dq=q_{T}dT+q_{p}dp$ to obtainWith dilute vapor, $q_p\simeq -q_s/p$ and $q_T\simeq L{q}_s/(R_v{T}^2)$, this becomes ${\Gamma }_d[1+L{q}_s/(R_{a}T)]/[1+L^2{q}_s/(c_p{R}_v{T}^2)]$. In terrestrial conditions the denominator's latent-heating correction dominates. Warming increases saturation humidity strongly, so latent heat increasingly compensates expansion cooling and the moist lapse rate falls. This is a statement about saturated convective adjustment, not every instantaneous environmental temperature profile.

Let ${\tau }_t$ be the tropopause optical depth and linearize $B$ with positive derivative $B_T$. In the troposphere $T=T_t+\Gamma H_{\tau }\log (\tau /{\tau }_t)$, so the requested schematic isIt is constant in the isothermal stratosphere and rises linearly with $\log \tau$ below it. At unchanged temperatures and a multiplicative abundance change $a$, optical depth at each height changes to $a\tau$, hence $\boxed{B_a(\tau )=B_{\mathrm{old}}(\tau /a)}$. Doubling moves the tropopause kink to $2{\tau }_t$ and gives colder emission temperatures at a fixed optical depth; halving does the opposite.

Hydrostatic balance and the ideal gas law give $dp/dz=-pg/(R_{a}T)$, hence approximately $p=p_0{e}^{-z/H}$ with $H=R_{a}T/g$ for weak temperature variation. For well-mixed absorber abundance, ${\rho }_{\mathrm{abs}}\propto p/T$, while pressure broadening gives $k\propto p$. Thus $k{\rho }_{\mathrm{abs}}\propto e^{-2z/H}$ and its integral above height $z$ has the same scale. ThereforeThe factor two depends on both pressure dependence of absorption and constant mixing ratio; it is not the ordinary density scale height.

Multiplying $I^{\prime }-I=-\pi B$ by $e^{-\tau }$ gives $(I{e}^{-\tau }{)}^{\prime }=-\pi B{e}^{-\tau }$. Integrating to depth $T$ yields $I(0)=I(T)e^{-T}+\pi {\int }_0^{T}B(t)e^{-t}dt$. If the first term vanishes as $T\to \infty$,Each layer emits in proportion to $Bd\tau$ and is attenuated by its overlying optical depth. The normalized exponential weighting emphasizes optical depths of order unity, while retaining contributions from a range of heights.

Optical depth increases downward because $d\tau =-k\rho dz$. Along an upward ray the beam loses irradiance by absorption, as in Beer-Lambert attenuation, and gains it from local thermal emission. Kirchhoff's law equates emissivity and absorptivity in local thermodynamic equilibrium, giving source term $\pi B$. Thus $-dI/d\tau =-I+\pi B$, equivalently the given equation. The factor $\pi$ converts isotropic radiance into hemispheric irradiance in the stipulated diffuse approximation. The following two limits hold for the same downward optical-depth coordinate.

The matter rest-energy density is ${\rho }_m{c}^2\simeq 2.25\times 10^{-10}$ J m${}^{-3}$, whereas the given photon density is $8.01\times 10^{-14}$ J m${}^{-3}$. Their present ratio is about $2.8\times 10^3$. Matter scales as $(1+z{)}^3$ and freely redshifting radiation as $(1+z{)}^4$, soThis compares matter specifically with the supplied CMB photons. Adding relativistic neutrinos changes the total radiation equality redshift; it is not included in these data.

Set the spatial-curvature constant to zero in the stipulated matter-only Friedmann equation to get ${\rho }_c=3H^2/(8\pi G)$. Conservation of pressureless matter with $R(t_0)=1$ gives $\rho ={\rho }_0/R^3$, and thereforeChoosing the Big Bang as $t=0$ yields $R=(t/t_0{)}^{2/3}$ and $H=2/(3t)$. Consequently $\boxed{t_0=2/(3H_0)\simeq 9.3\mathrm{Gyr}}$ for the supplied $H_0$. This is the age of the flat pressureless model without a cosmological constant, not a determination in the current multi-component cosmological model.

At a fixed cosmological time, homogeneous expansion has $\mathbf{v}=H(t)\mathbf{r}$. A comoving galaxy has $\mathbf{r}=R(t)\mathbf{x}$ with constant $\mathbf{x}$, hence $\mathbf{v}=\dot{R}\mathbf{x}=(\dot{R}/R)\mathbf{r}$. Thus $\boxed{H=\dot{R}/R}$. Peculiar velocities must be removed before applying this ideal relation to individual galaxies.

A continuum peak gives a color temperature through Wien's law, provided the continuum is approximately Planckian and extinction is corrected. Broad-band colors or a fitted spectral energy distribution compare flux ratios with model atmospheres; they require a reddening estimate, abundance and gravity assumptions, and calibrated passbands. Spectroscopic line ratios or excitation/ionization balances provide another thermometer, requiring atomic data and a suitable atmosphere model, often local thermodynamic equilibrium.

For this cool giant, line blanketing and molecular absorption substantially distort a simple continuum peak, so Wien's law is the least reliable of these approximations. A detailed spectroscopic analysis, checked against a multi-band atmosphere fit, would generally be the strongest of the three if good data and model parameters are available. There is no unconditional accuracy ranking without measurement errors and model systematics; an accurately measured angular diameter plus bolometric flux would offer a more direct effective-temperature determination, if obtainable.

The parallax gives $d=20$ pc and the distance modulus gives $M_{\mathrm{bol}}=-0.641-5{\log }_{10}2=-2.146$. Therefore $L/L_{\odot }=10^{(4.72-M_{\mathrm{bol}})/2.5}\simeq 558$. Taking $T_{\odot }\simeq 5772$ K,This is a cool luminous giant, in the red-giant region of the Hertzsprung-Russell diagram. On the usual first-ascent interpretation, luminosity comes predominantly from hydrogen shell burning around an inert helium core, with gravitational adjustment also contributing. The two measured quantities alone do not uniquely determine evolutionary history: a later giant phase can overlap this region, so they do not prove the precise core-burning state.

The effective temperature is defined by $\boxed{L=4\pi R_{\ast }^2\sigma T_{\mathrm{eff}}^4}$, with bolometric luminosity $L$, photospheric radius $R_{\ast }$ and Stefan-Boltzmann constant $\sigma$. It is the blackbody temperature giving the same total surface flux, not a claim that each spectral line forms at that temperature.