A linearly increasing radiative forcing produces a fast adjustment over about four years superposed on a slow deep-ocean response over about four hundred years. Once forcing stops increasing at year 70, surface warming continues and approaches the final $3\mathrm{K}$ equilibrium. The right panel of solution gives a model response normalized to the supplied $2\mathrm{K}$ endpoint and $3\mathrm{K}$ equilibrium.

Write $u=t-70$ after stabilization and $3-T_s=A_f{e}^{-u/4}+A_s{e}^{-u/400}$, with $A_f+A_s=1\mathrm{K}$. For a 70-year forcing ramp, the normalized step response is $1-b_f{e}^{-t/4}-b_s{e}^{-t/400}$, $b_f+b_s=1$. The ramp endpoint impliesThis gives $b_s\simeq 0.322$, $A_f\simeq 0.116\mathrm{K}$ and $A_s\simeq 0.884\mathrm{K}$. Thusat $u=10,15,20$ years respectively. The usual order estimate after the fast tail has largely disappeared is $\boxed{{\dot{T}}_s\sim (3-2)/400\simeq 0.0025\mathrm{K}\mathrm{y}{\mathrm{r}}^{-1}}$, with a still visible fast contribution near year 10.

To hold $T_s=T_{\ast }$ fixed, the deep layer obeys $T_s-T_d\propto e^{-\gamma u/C_d}$ and the surface equation requires $F=\lambda T_{\ast }+\gamma (T_{\ast }-T_d)$. Hence $\dot{F}=-{\gamma }^2(T_{\ast }-T_d)/C_d$. Using $b_f\simeq \lambda /(\lambda +\gamma )$ gives $\gamma /\lambda \simeq 0.475$ and $\gamma /C_d\simeq 1/(400b_f)\simeq 0.00369\mathrm{y}{\mathrm{r}}^{-1}$. With $T_{\ast }=2\mathrm{K}$ and initial forcing $F_0=3\lambda$, the initial fractional decline isThe rate declines slowly as the ocean contrast disappears. A coarser estimate treating the post-fast warming as an effective equilibrium change gives $0.0025/3\simeq 0.08\%$ per year; distinguishing the instantaneous surface feedback from the equilibrium feedback explains that difference.

Wording qualification: the printed text says a “1% increase … over 70 years”. The conventional transient climate response experiment uses $1\%$ per year, which nearly doubles concentration in 70 years. A literal total $1\%$ increase would not define the usual equilibrium climate sensitivity or transient climate response. The supplied $3\mathrm{K}$ and $2\mathrm{K}$ are used here as the endpoint normalizations, and a logarithmically uniform concentration growth gives the assumed linear forcing ramp.

Set ${\lambda }^{\prime }=0$. A decaying normal mode $e^{-kt}$ obeysFor $C_d\gg C_s$, its sum and product of roots giveInitially $T_s=T_d=T_0$, so $a_1+a_2=1$ and $a_1{k}_1+a_2{k}_2=\lambda /C_s$. To leading order,On the slow normal mode, the first equation gives $T_d/T_s=(\lambda +\gamma -C_s{k}_2)/\gamma \simeq (\lambda +\gamma )/\gamma$, verifying proportionality after the fast transient has decayed. The small parameter must also make $k_2/k_1$ small; the supplied positive coefficients and $C_d\gg C_s$ satisfy this.

$F$ is the imposed radiative forcing; $-\lambda T_s$ is the restoring change in outgoing radiation; and $-\gamma (T_s-T_d)$ transfers heat from the surface layer to the deep ocean, whose equal and opposite gain appears in its own energy balance. The additional $-{\lambda }^{\prime }(T_s-T_d)$ is a radiative adjustment correlated with the surface–deep-ocean temperature contrast, representing a pattern-dependent effective feedback rather than another conserved interlayer heat flux.

For constant forcing, final equilibrium has $T_s=T_d=F/\lambda$. During transient warming $T_d<T_s$, so the radiative response inferred from a recent global energy budget is ${\lambda }_{\mathrm{eff}}=\lambda +{\lambda }^{\prime }(1-T_d/T_s)>\lambda$. If that response were mistaken for the final feedback, the inferred equilibrium climate sensitivity would be too small. Summing the two equations makes the distinction explicit: $d(C_s{T}_s+C_d{T}_d)/dt=F-\lambda T_s-{\lambda }^{\prime }(T_s-T_d)$.

Use a convective layer $T(z)=T_s-\Gamma z$ capped at an isothermal minimum $T_t$ at height $z_t$. Let the initial optical depth be $\tau (z)={\tau }_s{e}^{-z/H_{\tau }}$ and ${\tau }_t={\tau }_s{e}^{-z_t/H_{\tau }}$. Within the lapse-rate layer, $d{B}_{\lambda }/d\tau =B^{\prime }\Gamma H_{\tau }/\tau$; above it the derivative vanishes. The supplied Schwarzschild equation therefore givesAssume the additional carbon dioxide multiplies the relevant opacity by $a=1.1$, leaving temperatures, the lapse rate and $H_{\tau }$ initially fixed. All material surfaces then acquire $a$ times their previous optical depth. HenceFor ${\tau }_t\ll 1$ and ${\tau }_s\gg 10$, the specified approximation givesFor illustration, $\Gamma =6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ and $H_{\tau }=7\mathrm{km}$ give $-4.34\mathrm{K}$; these are illustrative, not supplied data. The inequalities put the effective emitting level $\tau \sim 1$ inside the decreasing-temperature layer, well above an opaque surface and below the thin isothermal cap. Increasing opacity raises that emitting level by $H_{\tau }\ln 1.1$, so it is colder. In a spectral window or above a temperature inversion the approximation and possibly the sign change. Opacity proportional to mixing ratio is also an assumption, not exact for every absorption mechanism.

Downward-from-space optical depth at wavelength $\lambda$ iswhere ${\kappa }_{\lambda }$ is the mass absorption coefficient. A nearly isothermal, hydrostatic equilibrium, compositionally well-mixed atmosphere with height-independent ${\kappa }_{\lambda }$ has $\rho \propto e^{-z/H}$ and $\tau \propto e^{-z/H_{\tau }}$ with $H_{\tau }=H=R_{a}T/g$. Thus temperature, mean molecular mass and gravity set the basic scale height. Changes in the absorber mixing ratio, pressure broadening, temperature dependence of line strengths, collision-induced absorption and departures from hydrostatic equilibrium modify it. For example if local absorption scales as ${\rho }^2$ in an isothermal atmosphere, $H_{\tau }=H/2$.

The usual estimate adds a grey, infrared-absorbing radiative equilibrium upper atmosphere to the convective lower atmosphere. Consider a very optically thin upper layer: it receives upwelling infrared flux approximately $F_0$ from below, absorbs a small fraction $d\tau$, and emits $\sigma T^{4}d\tau$ into each hemisphere. Radiative balance givesfor $F_0=300\mathrm{W}{\mathrm{m}}^{-2}$. The corresponding effective emission temperature is $(F_0/\sigma {)}^{1/4}\simeq 270\mathrm{K}$. Without shortwave absorption, the model has no solar-heated stratospheric inversion.

This skin estimate requires the grey absorption/emission and angular approximations just stated. Horizontal mixing and no shortwave absorption alone do not determine an exact minimum for a spectrally selective nitrogen–carbon-dioxide atmosphere: absorption coefficients and spectral windows can change its upper-atmosphere radiative equilibrium.

For a unit mass of ideal gas, $h=u+P/\rho =u+R_{a}T$. Since $u$ depends only on temperature, $c_p=dh/dT=c_v+R_a$, giving $\boxed{R_a=c_p-c_v}$. For a dry parcel in adiabatic, quasistatic ascent, the first law of thermodynamics is $c_{p}dT={\rho }^{-1}dP$. Environmental hydrostatic equilibrium gives $dP/dz=-\rho g$, so a convectively well-mixed layer has the dry adiabatic lapse rateThis assumes negligible condensation, approximately constant $g$, well-mixed composition and efficient convection; a radiative or stably stratified layer need not follow it. For mole fractions $x_N,x_C$ and molar masses $M_N,M_C$, $R_a=R/(x_N{M}_N+x_C{M}_C)$ and $c_p=[x_N{C}_{p,N}+x_C{C}_{p,C}]/(x_N{M}_N+x_C{M}_C)$. Molecular vibrational excitation can make $c_p$ depend on temperature; then ${\Gamma }_d=g/c_p(T)$. Nitrogen dominance near terrestrial temperatures gives the familiar order $10\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$.

Three physically distinct contributions are unseen dark matter, hot diffuse intracluster gas omitted from the summed stellar mass, and faint stars or stellar remnants missed by transferring the Milky Way mass-to-light ratio to the cluster. The first two are actual additional components; the third changes the inferred stellar budget. Departures from virial equilibrium, velocity anisotropy or extra nonthermal pressure can also bias the dynamical or X-ray estimates, although they need not remove the discrepancy.

Assume an isotropic, stationary spherical galaxy cluster. Its three-dimensional velocity dispersion obeys $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$. For a uniform sphere, the virial theorem gives $3M{\sigma }_{\mathrm{los}}^2=3G{M}^2/(5R)$, henceA different density profile changes the coefficient, so $10^{15}{M}_{\odot }$ is the meaningful scale. For isothermal X-ray gas in hydrostatic equilibrium,Using $\mu =0.6$ and a representative slope $-d\ln {\rho }_g/d\ln r=2$–$3$ at $1\mathrm{Mpc}$ gives $M_X\simeq (1.3$–$1.9)\times 10^{15}{M}_{\odot }$. The temperature and extent alone cannot fix this coefficient without a density profile. The total luminosity is $5\times 10^{12}{L}_{\odot }$, so the previously inferred mass-to-light ratio gives

Treat the gravitational field as approximately spherical. The circular orbit condition givesIf every star had a representative mass $m$ and obeyed $L/L_{\odot }=(m/M_{\odot }{)}^3$, its mass-to-light ratio would be $(m/M_{\odot }{)}^{-2}$, giving $\boxed{m\simeq 0.47M_{\odot }}$. This is a one-mass estimate, not an actual average inferred uniquely from the data: a galactic stellar mass function, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the galactic disk also changes the spherical conversion by an order-one factor.

The Milky Way has a thin stellar galactic disk, radius roughly $15$–$20\mathrm{kpc}$ and scale height about $0.3\mathrm{kpc}$, containing gas, dust, young and old stars, spiral arms and most current star formation. Its stellar mass is a few $10^{10}{M}_{\odot }$. A thicker, older galactic disk has scale height around $1\mathrm{kpc}$ and stellar mass of order $10^{10}{M}_{\odot }$, with little cold gas.

The central galactic bulge and bar extend a few thousand parsecs and contain about $10^{10}{M}_{\odot }$, largely older stars. The central supermassive black hole has mass about $4\times 10^6{M}_{\odot }$ and is negligible in the total galactic budget. A diffuse old metal-poor stellar halo, extending tens to over $100\mathrm{kpc}$, contains globular clusters and roughly $10^9{M}_{\odot }$ in stars. An extended dark matter halo reaches roughly $200\mathrm{kpc}$ and dominates the total mass, of order $10^{12}{M}_{\odot }$. These are approximate component scales; the gaseous galactic halo also extends beyond the stellar galactic disk.

The right panel of the figure in solution has luminosity increasing upwards and effective temperature decreasing to the right. An old globular cluster contains a low-mass main sequence, a main-sequence turnoff, subgiants, a red giant branch, a horizontal branch, an Asymptotic giant branch and a faint hot white dwarf sequence. Its high-mass main sequence is absent because those stars have evolved.

If a fraction $f_c=0.1$ of the star participates in hydrogen burning, with initial hydrogen mass fraction $X=0.7$, the available nuclear binding energy is $0.007f_{c}XM{c}^2$. Assuming a roughly constant luminosity and coeval stars,Taking the quoted core fraction as hydrogen fuel itself, $f_{c}X=0.1$, gives $1.98\times 10^{10}\mathrm{yr}$. Stating the fuel convention matters: burning the whole stellar mass overestimates the age by about an order of magnitude. Evolving luminosity and detailed stellar structure are needed for an accurate age.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

Assume a circular orbit and an orbital inclination $i\simeq 90^{\circ }$, as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,Thus $a_1=0.1103\mathrm{AU}$, $a_2=0.2206\mathrm{AU}$ and Kepler's third law givesThe orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$; the numerical answer requires the circular assumption.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

Quasineutrality gives $n_e=4\times 10^{26}{\mathrm{m}}^{-3}$. Using the supplied $90^{\circ }$ collision-time expression and $\ln \Lambda =10$ givesThe printed absorption coefficient then yields $\kappa =1.62\times 10^2{\mathrm{m}}^{-1}$. Since the unabsorbed fraction is $e^{-\kappa L}$,This uses exactly the coefficient specified by the paper and neglects reflection and changing plasma conditions. With $\nu =1/{\tau }_{ei}$ instead, the corresponding thickness is $2.14\mathrm{mm}$, illustrating the convention dependence discussed in part (c).

Introduce the electron momentum-relaxation rate $\nu$ through $m_e(\dot{\mathbf{v}}+\nu \mathbf{v})=-e\mathbf{E}$. For a harmonic field $\mathbf{E}=\mathrm{Re}({\mathbf{E}}_0{e}^{-i\omega t})$, the real part of the electrical conductivity is $\mathrm{Re}\sigma =n_e{e}^2\nu /[m_e({\omega }^2+{\nu }^2)]$. At weak damping, the mean heating is $Q=\frac{1}{2}\mathrm{Re}\sigma |E_0{|}^2$ and the beam intensity is $I=\frac{1}{2}{ϵ}_{0}cn|E_0{|}^2$, where $n=\sqrt{1-{\omega }_p^2/{\omega }^2}$. ThereforeWith the relaxation convention $\nu =1/(2{\tau }_{ei})$, this becomes the printed resultConvention qualification: a $90^{\circ }$ deflection time does not by itself specify a momentum-relaxation rate without a collision model. If instead ${\tau }_{ei}$ is defined by $\nu =1/{\tau }_{ei}$, the intensity absorption coefficient is twice the printed one. Amplitude and intensity damping also differ by two. The result above states the extra convention needed rather than silently identifying these quantities. It assumes $\omega >{\omega }_p$, $\nu \ll \omega$ and sufficiently weak damping away from the cutoff; the apparent divergence near cutoff is outside that approximation.

Use quasineutrality, $n_e=Z{n}_i$. Multiplying the supplied collision time by the electron plasma frequency and using the preceding definition givesThus ${\tau }_{ei}\gg 2\pi /{\omega }_p$ requires $N_D\gg (2\pi /9.6)Z\ln \Lambda$; for ordinary $Z$ and $\ln \Lambda \sim 1$–$30$, this is a large plasma parameter. Many particles then collectively screen a charge, fluctuations in a Debye sphere are relatively small, and mean-field electrostatics descriptions become meaningful. A large plasma parameter is also associated with weak Coulomb coupling, well-separated screening and close-collision scales, and collective plasma oscillations surviving many oscillations before collisionally relaxing. Roughly, relative counting fluctuations scale as $N_D^{-1/2}$ and the ratio of interaction to thermal energy decreases as $N_D$ increases. Small $N_D$ signals that strong correlations can invalidate the simple weakly coupled plasma description.