For balanced global error $\epsilon$, a method second order in both variables can take $h\sim {\epsilon }^{1/2}$ and $k\sim {\epsilon }^{1/2}$. With $O(1/h)$ work per time step in one spatial dimension, its cost scales as ${\epsilon }^{-1}$. A method second order in space but first order in time instead needs $k\sim \epsilon$, so its cost is $\boxed{O({\epsilon }^{-3/2})}$ under the same per-step cost assumption. Equivalently at fixed $h$, using $k\sim h^2$ rather than $k\sim h$ needs about $1/h$ more steps. A restrictive stability condition or different linear-solver cost can dominate these accuracy-based estimates.

Write $h=\Delta x$, $k=\Delta t$ and ${\delta }_{xx}{u}_i=(u_{i+1}-2u_i+u_{i-1})/h^2$. Insert a smooth exact heat equation solution into the Crank-Nicolson method. Expanding about the midpoint $(x_i,t_n+k/2)$ givesThus the residual, defined as left side minus right side, isSince $u_t=D{u}_{xx}$ implies $u_{ttt}=D{u}_{xxtt}$,This proves second-order numerical consistency in both space and time. About the old time level, uncombined Taylor remainders can include $O(k^3)$; the midpoint expression displays the symmetric leading errors most cleanly. Multiplying by $k$ gives the one-step defect $O(k^3+k{h}^2)$. Stable evolution with suitable initial and boundary data then gives global error $O(k^2+h^2)$; consistency alone does not prove global convergence.

Round-off error arises from representing numbers and carrying out arithmetic at finite precision: storing $1/3$ or subtracting nearly equal numbers can lose information even when the mathematical algorithm is exact. Truncation error arises from replacing a continuous or infinite mathematical operation by a finite approximation, such as discarding higher Taylor series terms in a finite difference. Refining a grid usually reduces truncation error but may increase accumulated round-off error; the two have different origins.

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.

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.

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 $T=100\mathrm{K}$ with $\mu =2.3$, $c_s=\sqrt{k_{B}T/(\mu m_p)}\simeq 599\mathrm{m}{\mathrm{s}}^{-1}$. A solar-mass Keplerian disk has $v_K\simeq 29.8\mathrm{km}{\mathrm{s}}^{-1}$ at $1\mathrm{AU}$, so $\boxed{h/r\simeq 0.020}$: the thin-disk approximation is consistent.

For fixed central mass, $h/r\propto T^{1/2}{r}^{1/2}$. If $T\propto r^{-q}$, thenA flared outer disk intercepts more stellar radiation than a flat or self-shadowed disk, affecting its temperature, chemistry, vertical structure and observed spectral energy distribution. Heating changes the sound speed and therefore transport and dispersal; it can also enhance irradiation-driven mass loss. The actual coupled evolution depends on optical properties and transport, so flaring alone does not fix a unique lifetime. “Mid-planet” in the source is read as mid-plane.

Neglect disk self-gravity and take $|z|\ll r$ around a central point mass $M_{\star }$. Its vertical gravitational acceleration isVertical hydrostatic equilibrium is $dP/dz=\rho g_z$. For a vertically isothermal ideal gas, $P=\rho c_s^2$ and $c_s$ is independent of $z$, givingHere $v_K=r\Omega =\sqrt{G{M}_{\star }/r}$, $c_s^2=k_{B}T/(\mu m_p)$ is the isothermal sound speed squared, and $h$ is the Gaussian scale height. A different convention writing $e^{-z^2/h^2}$ changes the quoted height by $\sqrt{2}$ but not the physics.

A cold, self-gravitating region in a molecular cloud can collapse when gravity overcomes thermal, turbulent and magnetic support. Increasing central density produces a protostar, while radiation removes part of the released gravitational energy. The cloud initially carries angular momentum; material that cannot fall radially to the centre settles into a rotating protoplanetary disk. Collisions and cooling reduce random vertical motion, while angular-momentum transport allows some material to accrete and some to move outwards. Bipolar outflows and magnetic stresses can remove additional angular momentum.

The schematic sequence is $\text{rotating cloud}\to \text{protostar plus disk}\to \text{accreting solids and gas}\to \text{planetary system plus cleared disk}$. In the cool disk, grains condense and collide; settling and concentration can produce planetesimals, whose gravitational growth leads to planetary embryos. Massive cores may capture gas before the disk disperses; sufficiently massive unstable disks can also form bound condensations. Migration and interactions modify the final orbits. Radiation, accretion and outflows eventually disperse the remaining gas. Thus planet formation uses the same rotating reservoir produced by ordinary star formation, although the efficiency and final architecture are not universal and individual growth stages remain active research topics.

The habitable zone is the range of orbital distances where an appropriate planetary atmosphere could sustain liquid water at the surface; it is not a guarantee of inhabited or even habitable conditions for every planet. At a fixed habitable incident flux, $a_{\mathrm{HZ}}\propto L_{\star }^{1/2}$. With the supplied $L_{\star }\propto M_{\star }^3$, Kepler's third law givesTaking Earth as a rough central reference, a $0.1M_{\odot }$ star has $a_{\mathrm{HZ}}\sim 0.0316\mathrm{AU}$ and $P_{\mathrm{HZ}}\sim 6.5$ days. Its six-day, Earth-sized companion receives about $1.11$ times terrestrial irradiance and could therefore lie in the habitable zone in this simplified model. Atmosphere, stellar activity, tidal locking and volatile retention remain unknown. The six-day companion around the Sun-like star receives about $239$ times terrestrial irradiance and is approximately Jupiter-sized; it is not a plausible surface-water Earth analogue. The stellar spectral dependence of actual habitable-zone edges is omitted by the simple scaling.

To distinguish the companion, measure the host's radial velocity reflex motion to infer a mass; a low-mass star generally produces a much larger signal than a planet. Also test for a detectable secondary eclipse, colour-dependent dilution, doubled spectral lines and out-of-eclipse ellipsoidal variation. High angular-resolution imaging and shifts of the image centroid can identify a blended background eclipsing binary. A Jupiter-sized radius alone is not decisive because low-mass stars and brown dwarfs can have comparable radii. The three labelled effects on the transit are treated individually below.

The common one-per-cent exoplanet transit depth gives $R_p=0.1R_{\star }$. Apply the supplied main-sequence mass-radius relation, $R_{\star }/R_{\odot }=M_{\star }/M_{\odot }$, and Kepler's third law, neglecting the planet mass. For $P=6$ days,The companions are approximately Jupiter-sized and Earth-sized respectively; these are sizes, not independently determined masses. The duration uses the small-planet chord approximation. Counting the finite planet radius would increase first-to-fourth-contact durations by about ten per cent for central transits.

For equal distance, passband, cadence and noise dominated by stellar photon counting, the supplied mass-luminosity relation makes the solar-mass host about $10^3$ times more luminous, with a transit about $4.64$ times longer. At equal fractional exoplanet transit depth, signal-to-noise per event scales as $\delta \sqrt{{\dot{N}}_{\gamma }{t}_{\mathrm{tr}}}$, favouring the solar-mass host by about $\sqrt{1000\times 4.64}\simeq 68$. Equal apparent brightness would remove the luminosity advantage but retain the duration advantage. Without distances and noise properties the relative detectability is conditional. Transit probability is also larger for the solar-mass case, about $7.2\%$ versus $1.55\%$.

A exoplanet transit occurs when a planet crosses the projected stellar disk and blocks some stellar flux. For a circular orbit with isotropically distributed angular-momentum directions, $\cos i$ is uniform on $[0,1]$. A transit requires $a|\cos i|<R_{\star }+R_p$, soFor a dark planet, a uniformly bright stellar disk and negligible dilution, the transit exoplanet transit depth is $\boxed{\delta \simeq (R_p/R_{\star }{)}^2}$. For $R_{\star }/a\ll 1$, a small planet and impact parameter $b=a\cos i/R_{\star }$,The central-transit result sets $b=0$. A first-to-fourth-contact duration retaining planet size replaces $\sqrt{1-b^2}$ by $\sqrt{(1+R_p/R_{\star }{)}^2-b^2}$ and includes the exact projected-speed geometry when needed. Eccentricity, limb darkening and grazing geometry modify these expressions.

With $\gamma =2$ and $T(t+s)=T(t)+\dot{T}s$, the marginal damage is $2\kappa D_{0}W(t)e^{gs}[T+\dot{T}s]$. Integrating the constant and linear terms givesThe prefactor is $2(0.5\times 10^{-12})(0.003)(90\times 10^{12})=0.27$ in the corresponding dollar–temperature–time units. For the historical scenario with $r=0.03$ and $g=0.02$, the bracket is $1/0.01+0.02/0.01^2=300$, so $\boxed{S=81\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$. For $r=0.07$, $r-g=0.05$, giving global $S=0.27(20+8)=7.56$ dollars per tonne. Restricting impacts to ten per cent gives $\boxed{S_{\mathrm{domestic}}=0.756\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$.

The factor of about 107 reduction comes from both the higher discount rate and the geographical restriction; it is not a change in climate physics. The administration descriptions are those of the 2019 question, not claims about present policy. With a net discount rate of only one per cent, a century-scale horizon contributes substantially, making the linear warming and exponential income extrapolations consequential assumptions.

Equating marginal mitigation cost to the previous marginal damage givesUsing $\dot{T}=\kappa E$ from the cumulative-emissions approximation,Thus $T\propto t^{1/\gamma }$ at late times and $E\propto t^{-(\gamma -1)/\gamma }$. Warming slows but never stops, and cumulative emissions remain unbounded. This follows from a cost model in which zero emissions is infinitely costly and from a marginal-damage rule with constant positive $r-g$.

The approximation eventually fails if cumulative emissions leave the range of constant TCRE, if large damages suppress economic growth, if technological change alters $M_0$, or if fossil resources and mitigation options impose different limits. The slowly varying-temperature approximation also needs $\dot{T}/[T(r-g)]\ll 1$. Thresholds, adaptation, uncertainty and unequal impacts are omitted; the formula is a property of this model rather than a universal policy trajectory.

An additional tonne changes eventual temperature by approximately $\delta T=\kappa$. Linearizing the damage function gives $\delta L(t^{\prime })=\kappa \gamma D_{0}W(t^{\prime })T(t^{\prime }{)}^{\gamma -1}$. Put $s=t^{\prime }-t$, $W(t+s)=W(t)e^{gs}$, and hold $T$ approximately fixed across the effective discount horizon. The discounted integral issoIf $r\le g$, this infinite-horizon approximation diverges; a finite result then needs different growth, damages or discounting assumptions. It also treats the marginal warming from the extra tonne as effectively persistent over the relevant horizon.

The transient climate response to cumulative carbon emissions is $\boxed{\kappa \simeq \Delta T/\mathcal{E}}$, where $\mathcal{E}$ is cumulative emitted carbon dioxide in a stated mass unit. Here the unit is tonnes of $\mathrm{C}{\mathrm{O}}_2$, rather than tonnes of carbon. Increasing concentration weakens the incremental radiative forcing per added atmospheric tonne because forcing is approximately logarithmic. Over a substantial range this is partly offset by a weakening of carbon sinks and by the changing ratio between transient warming and forcing as ocean heat uptake evolves. These approximate compensations make warming nearly proportional to cumulative emissions; no exact cancellation follows from the logarithmic forcing law alone. The approximation is range- and pathway-dependent, and is not a theorem valid for arbitrary emissions histories.