Corrected exercises: greenhouse effect, albedo and radiative forcing (203-SN3-RE)

The power of the Sun is spread over a sphere of radius $d$: $S=\dfrac{L}{4\pi d^{2}}=\dfrac{3.828\times 10^{26}}{4\pi\times(1.496\times 10^{11})^{2}}=1361$ W/m².

The Earth intercepts the beam as a DISC of radius $R$: $P=\pi R^{2}S=\pi\times(6.371\times 10^{6})^{2}\times 1361=1.74\times 10^{17}$ W.

On average over a day and over the whole globe, this power is spread over the entire SPHERE: $\dfrac{\pi R^{2}S}{4\pi R^{2}}=\dfrac{S}{4}=340$ W/m².

It is the ratio of the surface area of the sphere, $4\pi R^{2}$, to that of the shadow disc, $\pi R^{2}$. It combines two effects: half of the globe is in night, and on the lit half the rays arrive at a slant on average.

$30$ % is reflected: $(1-\alpha)\dfrac{S}{4}=0.70\times 340=238$ W/m². This is the flux that the Earth must re-emit to space in order to stay in equilibrium.

$\sigma T_{e}^{4}=(1-\alpha)\dfrac{S}{4}$, so $T_{e}=\left(\dfrac{0.70\times 1361}{4\times 5.67\times 10^{-8}}\right)^{1/4}=254.6$ K, or about $-19$ °C.

$288-254.6=33$ K. Without this extra warming, the surface would be frozen on average.

$T_{e}=\left(\dfrac{S}{4\sigma}\right)^{1/4}(1-\alpha)^{1/4}$, so $\dfrac{dT_{e}}{d\alpha}=-\dfrac{1}{4}\left(\dfrac{S}{4\sigma}\right)^{1/4}(1-\alpha)^{-3/4}=-\dfrac{T_{e}}{4(1-\alpha)}$. Numerically: $-\dfrac{254.6}{2.8}=-90.9$ K per unit of albedo.

To first order: $\Delta T_{e}\approx -90.9\times 0.01=-0.91$ K. The exact calculation gives $253.7$ K, which is also $-0.91$ K: the linearization is excellent for such small changes. One hundredth of albedo is almost a degree.

To the GREENHOUSE EFFECT: the atmosphere, transparent to most of the solar radiation, absorbs a large part of the infrared radiation emitted by the surface and sends part of it back down. The surface therefore receives more than the $238$ W/m² of solar radiation and must be warmer to re-emit it.

$\lambda_{max}=\dfrac{2.898\times 10^{-3}}{5772}=5.02\times 10^{-7}$ m $=502$ nm, in the visible.

$\lambda_{max}=\dfrac{2.898\times 10^{-3}}{255}=1.14\times 10^{-5}$ m $=11.4$ µm, in the thermal infrared.

Wien's law makes $\lambda_{max}$ inversely proportional to $T$: the ratio of the wavelengths is the ratio of the temperatures, $\dfrac{5772}{255}=22.6$. The two spectra, each about a factor of ten wide around its maximum, barely overlap: we speak of “shortwave” and “longwave” radiation.

$E=\dfrac{hc}{\lambda}=\dfrac{6.63\times 10^{-34}\times 3.00\times 10^{8}}{1.14\times 10^{-5}}=1.75\times 10^{-20}$ J $=0.109$ eV. This energy matches the gaps between VIBRATIONAL levels of molecules such as CO₂ and H₂O, whereas a visible photon, twenty times more energetic, does not excite them.

Absorption depends on wavelength. Greenhouse gases (H₂O, CO₂, CH₄), transparent in the visible, absorb strongly in the infrared, where their molecules have vibrational modes. Because the two radiations occupy separate ranges, the atmosphere can be nearly transparent to one and nearly opaque to the other.

$T_{s}=2^{1/4}T_{e}=1.189\times 254.6=302.8$ K. An opaque atmosphere overestimates the greenhouse effect by $15$ K: the real atmosphere lets part of the infrared escape through the “window” from $8$ to $13$ µm.

$\left(\dfrac{288}{254.6}\right)^{4}=\dfrac{2}{2-f}$, so $1.637=\dfrac{2}{2-f}$ and $f=2-\dfrac{2}{1.637}=0.78$. The atmosphere of the model absorbs $78$ % of the radiation from the surface.

$T_{a}=\dfrac{288}{1.189}=242$ K. The layer is colder than the surface: that is what lets it hold energy back, since it radiates less to space than the surface sends it.

$f\sigma T_{a}^{4}=0.779\times 5.67\times 10^{-8}\times 242.2^{4}=152$ W/m².

Received: $238+152=390$ W/m². Emitted: $\sigma\times 288^{4}=390$ W/m². The equality holds, and it shows the key to the mechanism: the surface receives MORE than the solar energy, because the atmosphere returns part of it.

$\Delta F=5.35\,\ln\dfrac{420}{280}=5.35\ln 1.5=2.17$ W/m².

$\Delta F_{2\times}=5.35\ln 2=3.71$ W/m². Note that it does not depend on the starting concentration: every doubling adds the same forcing.

$\dfrac{\ln 1.5}{\ln 2}=0.585$: with only half of the way covered in concentration ($+140$ ppm out of $+280$), $58.5$ % of the forcing is already there.

$\Delta F=0.036\left(\sqrt{1900}-\sqrt{722}\right)=0.036\times(43.59-26.87)=0.60$ W/m², a little over a quarter of that of CO₂ although the concentration is two hundred times smaller: molecule for molecule, methane is far more effective.

At the centre of the CO₂ absorption band near $15$ µm, the atmosphere is already SATURATED: adding gas there changes almost nothing. The additional forcing comes from the wings of the band, where absorption increases more and more slowly. Hence slow, logarithmic growth; methane, less saturated, follows a square-root law.

$\lambda_{P}=\dfrac{d}{dT}\left(\sigma T^{4}\right)=4\sigma T_{e}^{3}=4\times 5.67\times 10^{-8}\times 254.6^{3}=3.74$ W/(m²·K). Each extra kelvin sends $3.74$ W/m² more back to space.

The warming offsets the forcing when $\lambda_{P}\Delta T_{0}=\Delta F$: $\Delta T_{0}=\dfrac{3.71}{3.74}=0.99$ K. With no feedback, a doubling of CO₂ would warm the planet by about one degree.

$\Delta T=\dfrac{0.99}{1-0.60}=2.5$ K. Feedbacks multiply the basic response by $2.5$. The main one is water vapour: warmer air holds more of it, and it is itself a greenhouse gas.

$1-g=\dfrac{0.99}{4.5}$, so $g=0.78$. Going from $2.5$ K to $4.5$ K takes only a slightly larger feedback factor: this is why the uncertainty on feedbacks, especially that of clouds, dominates the uncertainty on future warming.

The denominator would vanish: the response would no longer have a finite value, and a small disturbance would trigger a runaway with no equilibrium state. This is the “runaway greenhouse effect” invoked for Venus. For the Earth today, $g<1$: feedbacks amplify, but the system remains stable.

Each square metre that changes from ice to water absorbs an extra $(0.60-0.07)\times 200=106$ W. Over $4.0\times 10^{12}$ m²: $\Delta P=4.0\times 10^{12}\times 106=4.2\times 10^{14}$ W. Watch the square kilometres: $1$ km² $=10^{6}$ m².

$\dfrac{4.24\times 10^{14}}{5.10\times 10^{14}}=0.83$ W/m².

The season lasts only a third of the year: $\dfrac{0.83}{3}=0.28$ W/m².

$\dfrac{0.28}{2.17}=0.13$: about $13$ % of the forcing of CO₂, from an area that does not even cover $1$ % of the globe. The estimate is rough (clouds, sun angle, spring ice), but the order of magnitude is right.

Because the effect reinforces its own cause: warming melts the ice, the surface becomes darker, it absorbs more energy, which warms it further and melts even more ice. A negative feedback, on the other hand, would oppose the disturbance, like the Planck response.

Two distinct phenomena. The ozone hole lets more ULTRAVIOLET reach the ground, a health problem caused by CFCs. The greenhouse effect concerns the INFRARED emitted by the surface. CFCs are also powerful greenhouse gases, which keeps the confusion alive, but the hole itself is not the cause of the warming.

Without the greenhouse effect, the mean surface temperature would be about $255$ K, or $-19$ °C, instead of $15$ °C. The natural phenomenon makes the Earth habitable; it is its STRENGTHENING by human emissions, of the order of $2$ W/m² of forcing, that is the problem.

CO₂ is almost transparent to solar radiation, which is mostly visible. It absorbs the INFRARED emitted by the surface and the atmosphere, in particular near $15$ µm, and then re-emits it in all directions, including toward the ground.

Water vapour is the MAIN greenhouse gas, by its total contribution. But its concentration is set by temperature, not by our direct emissions: it acts as a FEEDBACK that amplifies the warming caused by CO₂. Clouds have a double effect: they reflect sunlight and they hold back infrared.

A garden greenhouse heats up mainly because its glass BLOCKS CONVECTION: the air warmed at ground level cannot rise and be replaced. Wood's experiment (1909), with a pane transparent to infrared, showed this. The atmosphere has no lid: it works by absorbing and re-emitting infrared radiation. The name is an analogy, not an identity.

$0.70$ kg of octane, so $n=\dfrac{700}{114.23}=6.13$ mol. From the equation, $1$ mol of octane gives $8$ mol of CO₂: $49.0$ mol, that is $49.0\times 44.01=2.16\times 10^{3}$ g $=2.16$ kg.

15,000 km at $7.5$ L/100 km: $1125$ L, so $1125\times 2.16=2.43\times 10^{3}$ kg $=2.4$ t of CO₂ per year.

$40\times 0.45=18$ Gt remain in the air, that is $\dfrac{18}{7.8}=2.3$ ppm/yr, the order of magnitude of current measurements.

$\dfrac{560-420}{2.31}=61$ years. In practice emissions are not constant, but the calculation shows what a business-as-usual scenario implies.

The carbon of the gasoline combines with the oxygen of the AIR: each carbon atom ($12$ g/mol) leaves in a CO₂ molecule of $44$ g/mol. The mass of the dioxygen consumed is added, and conservation of mass is respected.

The flux falls off as $\dfrac{1}{d^{2}}$: $S_{V}=\dfrac{1361}{0.723^{2}}=2604$ W/m².

$T_{e}=\left(\dfrac{0.23\times 2604}{4\sigma}\right)^{1/4}=227$ K. The surface is at $737$ K: the Venusian greenhouse effect is worth more than $500$ K.

$S_{M}=\dfrac{1361}{1.524^{2}}=586$ W/m² and $T_{e}=\left(\dfrac{0.75\times 586}{4\sigma}\right)^{1/4}=210$ K, exactly the observed temperature: the very thin Martian atmosphere has practically no greenhouse effect.

$(n+1)=\left(\dfrac{737}{226.7}\right)^{4}=112$, so $n\approx 111$ layers. The atmosphere of Venus, $93$ times more massive than that of the Earth and made almost entirely of CO₂, is opaque to infrared over an enormous thickness: the figure gives an order of magnitude, not a real number of layers.

Its albedo of $0.77$: its sulfuric acid clouds send back more than three quarters of the radiation. Venus absorbs only $0.23\times\dfrac{2604}{4}=150$ W/m², less than the $238$ W/m² of the Earth. Its scorching surface comes not from the flux received but from the greenhouse effect.

$T_{e}=\left(\dfrac{0.40\times 1361}{4\times 5.67\times 10^{-8}}\right)^{1/4}=221$ K.

$\left(\dfrac{2}{1.22}\right)^{1/4}=1.1315$, so $T_{s}=221.3\times 1.1315=250$ K, or $-23$ °C.

No: at $250$ K the ice does not melt, the albedo stays at $0.60$ and the planet stays cold. The ice-covered state is STABLE. In reality, it was the slow accumulation of volcanic CO₂, no longer absorbed by rock weathering under the ice, that eventually increased $f$ and freed the planet.

We need $T_{e}=\dfrac{273}{1.1315}=241.3$ K, so $1-\alpha=\dfrac{4\sigma T_{e}^{4}}{S}=0.565$ and $\alpha=0.44$. The albedo would have to drop from $0.60$ to $0.44$, which a frozen surface cannot do on its own.

For the SAME sunlight, the system has two equilibrium states, one warm and dark, the other cold and bright: this is hysteresis. The state the system is in depends on its history, and moving from one to the other takes a strong disturbance. The ice-albedo feedback is what creates these two states.

$C=1025\times 3990\times 70=2.86\times 10^{8}$ J/(m²·K). It takes nearly three hundred megajoules to warm the water column under each square metre by one kelvin.

$\tau=\dfrac{2.863\times 10^{8}}{1.2}=2.39\times 10^{8}$ s, that is $\dfrac{2.39\times 10^{8}}{3.156\times 10^{7}}=7.6$ years. We check the solution: $T(0)=0$, and $C\dfrac{dT}{dt}=\dfrac{C\Delta F}{\lambda\tau}e^{-t/\tau}=\Delta F-\lambda T$.

When $\dfrac{dT}{dt}=0$: $T_{eq}=\dfrac{\Delta F}{\lambda}=\dfrac{3.7}{1.2}=3.1$ K.

$T(10)=3.08\times\left(1-e^{-10/7.56}\right)=3.08\times 0.734=2.26$ K.

$1-e^{-t/\tau}=\dfrac{1}{2}$, so $t=\tau\ln 2=7.56\times 0.693=5.2$ years. The climate “runs after” the forcing: part of the warming due to past emissions is not yet visible. With the deep ocean, much slower, part of it arrives over centuries.

$T=\left(\dfrac{240}{5.67\times 10^{-8}}\right)^{1/4}=255.1$ K.

The temperature drops by $288-255.1=32.9$ K between the surface and the emission altitude: $z=\dfrac{32.9}{6.5}=5.07$ km. It is near $5$ km, and not from the ground, that the Earth cools to space.

Rising by $150$ m loses $6.5\times 0.150=0.98$ K: $T=254.1$ K.

$\Delta F=\sigma\left(255.07^{4}-254.09^{4}\right)=3.65$ W/m², of the same order as the forcing of a doubling of CO₂. The linear approximation $4\sigma T^{3}\Delta T$ gives $3.67$ W/m².

The Earth still receives $240$ W/m² but now emits only $236$: it accumulates energy and warms. The whole profile shifts toward warmer temperatures until the NEW emission altitude is back at $255$ K; the surface, $5.2$ km lower, has then gained about one kelvin, before feedbacks.

$\Delta\alpha=\dfrac{3.71}{340}=0.0109$, about one hundredth.

$\dfrac{0.0109}{0.30}=3.6$ %. It would be enough to send back a little over one percent of the incident sunlight, which explains the interest in, and the risks of, projects to inject aerosols into the stratosphere.

$T_{eq}=\dfrac{-3.0}{1.2}=-2.5$ K.

$T(1)=-2.5\times\left(1-e^{-1/7.6}\right)=-2.5\times 0.123=-0.31$ K. A global cooling of the order of $0.3$ to $0.5$ K was measured: the inertia of the ocean prevents a brief forcing from having its full effect, which also explains why the cooling disappeared within two or three years.

Aerosols fall out within a year or two: they would have to be reinjected without interruption for centuries, at the risk of abrupt warming if we stop. Above all, they do not remove the CO₂: ocean acidification continues, and the greenhouse effect and the aerosol screen only cancel on average, not region by region or season by season.

Received: $161+333=494$. Lost: $396+17+80=493$. Net flux: $+1$ W/m². The budget is not zero: the surface, mainly the ocean, is slowly warming.