# Chapter 2. The global energy balance. 2.1 Planetary emission temperature

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1 Chapter 2 The global energy balance We consider now the general problem of the radiative equilibrium temperature of the Earth. The Earth is bathed in solar radiation and absorbs much of that incident upon it. To maintain equilibrium it must warm up and radiate energy away at the same rate as it is received, as depicted in Fig.2.1. We will see that the emission temperature of the Earth is 255 K and that a body at this temperature radiates energy primarily in the infrared (IR). But the atmosphere is strongly absorbing at these wavelengths due to the presence of trace gases principally the triatomic molecules H 2 OandCO 2 which absorb and emit in the infrared, this raising the surface temperature above that of the emission temperature, a mechanism that has become known as the greenhouse effect. 2.1 Planetary emission temperature The Earth receives almost all of its energy from the Sun. At the present time in its evolution the Sun emits energy at a rate of Q = W. The flux of solar energy at the Earth called the solar constant depends on the distance of the Earth from the Sun, r, and is given by the inverse Q 4πr 2 square law: S 0 =. Of course, because of variations in the Earth s orbit (see Sections and ) the solar constant is not really constant; the terrestrial value S 0 = 1367 W m 2 set out in Table 2.1, along with that for other planets, is an average corresponding to the average distance of Earth from the Sun, r = m. The way in which radiation interacts with an atmosphere depends on 37

2 38 CHAPTER 2. THE GLOBAL ENERGY BALANCE Figure 2.1: The Earth radiates energy away at the same rate as it is received from the Sun. The Earth s emission temperature is 255 K; that of the Sun, 6000 K. The outgoing terrestrial radiation peaks in the infrared; the incoming solar radiation peaks at shorter wavelengths, in the visible. r S 0 α p T e T m 10 9 m W m 2 K K Venus Earth Mars Jupiter T s K τ Earth days Table 2.1: Properties of some of the planets. S 0 is the solar constant at a distance r from the Sun, α p is the planetary albedo, T e is the emission temperature computed from Eq.(2.4), T m is the measured emission temperature and T s is the global mean surface temperature. The rotation period, τ, is given in Earth days.

3 2.1. PLANETARY EMISSION TEMPERATURE 39 Figure 2.2: The energy emitted from the sun plotted against wavelength based on a black body curve with T = T Sun. Most of the energy is in the visible and 95% of the total energy lies between 0.25 and 2.5 µm (10 6 m). the wavelength as well as the intensity of the radiative flux. The relation between the energy flux and wavelength the spectrum is plotted in Fig.2.2. The Sun emits radiation that is primarily in the visible part of the spectrum, corresponding to the colors of the rainbow red, orange, yellow, green, blue, indigo and violet with the energy flux decreasing toward longer (infrared, IR) and shorter (ultraviolet, UV) wavelengths. Why does the spectrum have this pattern? Such behavior is characteristic of the radiation emitted by incandescent material, as can be observed, for example, in a coal fire. The hottest parts of the fire are almost white and emit the most intense radiation, with a wavelength that is shorter than that coming from the warm parts of the fire, which glow red. The coldest parts of the fire do not seem to be radiating at all, but are, in fact, radiating in the infrared. Experiment and theory show that the wavelength at which the intensity of radiation is a maximum, and the flux of emitted radiation, depend only on the temperature of the source. The theoretical spectrum,

4 40 CHAPTER 2. THE GLOBAL ENERGY BALANCE Figure 2.3: Theenergyemittedatdifferent wavelengths for black bodies at several temperatures. The function B λ (T ), Eq.(13.1) is plotted. one of the jewels of physics, was worked out by Planck 1,andisknownas the Planck or blackbody spectrum. (A brief theoretical background to the Planck spectrum is given in Appendix ). It is plotted as a function of temperature in Fig.2.3. Note that the hotter the radiating body, the more energy it emits at shorter wavelengths. If the observed radiation spectrum of the Sun is fitted to the black body curve by using T as a free parameter, we deduce that the blackbody temperature of the sun is about 6000 K. Let us consider the energy balance of the Earth as in Fig.2.5, which shows the Earth intercepting the solar energy flux and radiating terrestrial energy away. If at the location of the (mean) Earth orbit, the incoming solar energy 1 In 1900 Max Planck ( ) combined the formulae of Wien and Rayleigh describing the distribution of energy as a function of wavelength of the radiation in a cavity at temperature T, to arrive at what is now known as Planck s radiation curve. He went on to a complete theoretical deduction, introduced quanta of energy and set the scene for the development of Quantum Mechanics.

5 2.1. PLANETARY EMISSION TEMPERATURE 41 Type of surface Albedo (%) Ocean 2 10 Forest 6 18 Cities Grass 7 25 Soil Grassland Desert (sand) Ice Cloud (thin, thick stratus) 30, Snow (old) Snow (fresh) Table 2.2: Albedos for different surfaces. Note that the albedo of clouds is highly variable and depend on the type and form. See also the horizontal map of albedo showninfig.2.4. flux is S 0 =1367W m 2, then, given that the cross-sectional area of the Earth intercepting the solar energy flux is πa 2 where a is the radius of the Earth (Fig. 2.5), Solar power incident on the Earth = S 0 πa 2 = W using the data in Table 1.1. Not all of this radiation is absorbed by the Earth; a significant fraction is reflected. The ratio of reflected to incident solar energy is called the albedo, α. As set out in Table 2.2 and the map of surface albedo shown in Fig.2.4, α depends on the nature of the reflecting surface and is large for clouds, light surfaces such as deserts and (especially) snow and ice. Under the present terrestrial conditions of cloudiness and snow and ice cover, on average a fraction α p ' 0.30 of the incoming solar radiation at the Earth is reflected back to space; α p is known as the planetary albedo (see table 2.1). Thus Solar radiation absorbed by the Earth =(1 α p )S 0 πa 2 = W. (2.1) In equilibrium, the total terrestrial flux radiated to space must balance the solar radiation absorbed by the Earth. If, in total, the spinning Earth radiates

6 42 CHAPTER 2. THE GLOBAL ENERGY BALANCE Figure 2.4: The albedo of the earth s surface. Over the ocean the albedo is small (2-10%). It is larger over the land (typically 35-45% over desert regions) and is particularly high over snow and ice (80% or so): see Table 2.2.

7 2.1. PLANETARY EMISSION TEMPERATURE 43 Figure 2.5: The spinning Earth is imagined to intercept solar energy over a disk of radius a and radiate terrestrial energy away isotropically from the sphere. Modified from Hartmann, in all directions like a blackbody of uniform temperature T e (known as the effective planetary temperature, or emission temperature of the Earth) the Stefan-Boltzmann law gives: Emitted radiation per unit area = σt 4 e (2.2) where σ = W m 2 K 4 is the Stefan-Boltzmann constant. So Emitted terrestrial radiation =4πa 2 σt 4 e. (2.3) Note that Eq.(2.3) is a definition of emission temperature T e itisthe temperature one would infer by looking back at Earth if a black body curve were fitted to the measured spectrum of outgoing radiation. Equating Eq.(2.1) with Eq.(2.3) gives 1 S0 (1 α p ) 4 T e =. (2.4) 4σ Note that the radius of the Earth has cancelled out: T e depends only on the planetary albedo and the distance of the Earth from the Sun. Putting in numbers we find that the Earth has an emission temperature of 255 K.

8 44 CHAPTER 2. THE GLOBAL ENERGY BALANCE Table 2.1 lists the various parameters for some of the planets and compares approximate measured values, T m,witht e computed from Eq.(2.4). The agreement is very good, except for Jupiter where it is thought that about one-half of the energy input comes from the gravitational collapse of the planet. However, as can be seen from Table 2.1, the emission temperature of Earth is nearly 40 K cooler than the globally averaged observed surface temperature which is T s =288K. AsweshalldiscussinSection2.3,T s 6= T e because 1) radiation is absorbed within the atmosphere, principally by its water vapor blanket and 2) fluid motions air currents carry heat both vertically and horizontally. 2.2 The atmospheric absorption spectrum A property of the black body radiation curve is that the wavelength of maximum energy emission, λ m,satisfies λ m T = constant. (2.5) This is known as Wien s displacement law. Since the solar emission temperature is about 6000 K and the maximum of the solar spectrum is (see Fig.2.2) at about 0.6 µm i.e., in the visible and we have determined T e =255K for the Earth, it follows that the peak of the terrestrial spectrum is at λ earth m =0.6 µm 6000 ' 14 µm. 255 Thus the Earth radiates to space primarily in the infrared. Normalized (see Eq of the Appendix) blackbody spectra for the Sun and Earth are shown in Fig.2.6. The two spectra hardly overlap at all, which greatly simplifies thinking about radiative transfer. Also shown in Fig.2.6 is the atmospheric absorption spectrum; this is the fraction of radiation at each wavelength that is absorbed on a single vertical path through the atmosphere. From it we see that: the atmosphere is almost completely transparent in the visible, at the peak of the solar spectrum. the atmosphere is very opaque in the UV.

9 2.2. THE ATMOSPHERIC ABSORPTION SPECTRUM 45 Figure 2.6: (a) The normalized blackbody emission spectra, T 4 λb λ, for the Sun (T =6000K)andEarth(T =255K) as a function of lnλ (top) where B λ is the black body function (see Eq.(13.2)) and λ is the wavelength (see the Appendix for further discussion.) (b) The fraction of radiation absorbed while passing from the ground to the top of the atmosphere as a function of wavelength. (c) The fraction of radiation absorbed from the tropopause (typically at a height of 11 km) to the top of the atmosphere as a function of wavelength. The atmospheric molecules contributing the important absorption features at each frequency are also indicated. After Goody and Yung (1989).

11 2.3. THE GREENHOUSE EFFECT 47 Figure 2.7: The simplest greenhouse model, comprising a surface at temperature T s, and an atmospheric layer at temperature T a, subject to incoming solar radiation S o 4. The terrestrial radiation upwelling from the ground is assumed to be completely absorbed by the atmospheric layer A simple greenhouse model Consider Fig.2.7. Since the atmosphere is thin, let us simplify things by considering a planar geometry, in which the incoming radiation per unit area is equal to the average flux per unit area striking the Earth. This average incoming solar energy per unit area of the Earth s surface is average solar energy flux = intercepted incoming radiation Earth s surface area = S 0πa 2 4πa 2 = S 0 4. (2.6) We will represent the atmosphere by a single layer of temperature T a,and,in this first calculation, assume 1) that it is completely transparent to shortwave solar radiation, and 2) that it is completely opaque to IR (i.e., itabsorbsall the IR radiating up from the ground) so that the layer that is emitting to space is also seen by the ground. Now, since the whole Earth-atmosphere system must be in equilibrium (on average), the net flux into the system must vanish. The average net solar flux per unit area is, from Eq.(2.6), even though plastic (unlike glass) does not have significant absorption bands in the IR. The greenhouse works because its windows allow energy in and its walls prevent the warm air from rising or blowing away.

12 48 CHAPTER 2. THE GLOBAL ENERGY BALANCE and allowing for reflection, 1 4 (1 α p) S 0, while the terrestrial radiation being emitted to space per unit area is, using Eq.(2.2): Equating them, we find: A = σt 4 a. σt 4 a = 1 4 (1 α p) S 0 = σt 4 e, (2.7) using the definition of T e, Eq.(2.4). We see that the atmosphere is at the emission temperature (naturally, because it is this region that is emitting to space). At the surface, the average incoming shortwave flux is also 1 4 (1 α p) S 0, but there is also a downwelling flux emitted by the atmosphere, A = σt 4 a = σt 4 e. The flux radiating upward from the ground is S = σt 4 s, where T s is the surface temperature. Since, in equilibrium, the net flux at the ground must be zero, S = 1 4 (1 α p) S 0 + A, whence σt 4 s = 1 4 (1 α p) S 0 + σt 4 e =2σT 4 e, (2.8) wherewehaveusedeq.(2.7).therefore T s =2 1 4 Te. (2.9) So the presence of an absorbing atmosphere, as depicted here, increases the surface temperature by a factor =1.19. This arises as a direct consequence of absorption of terrestrial radiation by the atmosphere, which, in turn, reradiates IR back down to the surface, thus increasing the net downward radiative flux at the surface. Note that A isofthesameorder infact, in this simple model, equal to the solar radiation that strikes the ground. This is true of more complex models and indeed observations show that the

13 2.3. THE GREENHOUSE EFFECT 49 Figure 2.8: A leaky greenhouse. In contrast to Fig.2.7, the atmosphere now absorbs only a fraction, ε, of the terrestrial radiation upwelling from the ground. downwelled radiation from the atmosphere can exceed that due to the direct solar flux. Applying this factor to our calculated value T e =255K,wepredictT s = = 303 K. This is closer to the actual mean surface temperature of 288 K but is now an overestimate! The model we have discussed is clearly an oversimplification: For one thing, not all the solar flux incident on the top of the atmosphere reaches the surface typically, some 20-25% is absorbed within the atmosphere (including by clouds). For another, we saw in Section (2.2) that IR absorption by the atmosphere is incomplete. The greenhouse effect is actually less strong than in the model assumed above and so T s will be less than the value implied by Eq.(2.9). We shall analyze this by modifying Fig.2.7 to permit partial transmission of IR through the atmosphere a leaky greenhouse model.

14 50 CHAPTER 2. THE GLOBAL ENERGY BALANCE A leaky greenhouse Consider Fig.2.8. We suppose the atmosphere has absorptivity, i.e., afraction of the IR upwelling from the surface is absorbed within the atmosphere (so the case of Fig.2.7 corresponds to =1). Now, again insisting that, in equilibrium, the net flux at the top of the atmosphere vanishes gives 1 4 (1 α p)s 0 = A +(1 ) S. (2.10) Zero net flux at the surface gives 1 4 (1 α p) S 0 + A = S. (2.11) Since at equilibrium, A = A, wehave S = σt 4 s = 1 2(2 ) (1 α p) S 0 = 2 (2 ) σt 4 e. (2.12) Therefore, µ T s = Te. (2.13) 2 So in the limit 0 (transparent atmosphere), T s = T e,andfor 1 (opaque atmosphere), T s =2 1 4 T e, as found in section In general, when 0 < <1, T e <T s < T e. So, of course, partial transparency of the atmosphere to IR radiation a leaky greenhouse reduces the warming effect we found in Eq.(2.9). To find the atmospheric temperature, we need to invoke Kirchhoff s law 3, viz., that the emissivity of the atmosphere is equal to its absorptivity. Thus, A = A = σt 4 a. (2.14) We can now use Eqs. (2.14), (2.10), (2.11) and (2.12) to find µ 1 µ T a = Te = Ts. 2 2 So the atmosphere is, for <1,coolerthanT e (since the emission is then only partly from the atmosphere). Note, however, that T a <T s :theatmosphere is always cooler than the ground. 3 Kirchhoff s law states that the emittance of a body the ratio of the actual emitted flux to the flux that would be emitted by a black body at the same temperature equals its absorptance.

15 2.3. THE GREENHOUSE EFFECT 51 Figure 2.9: An opaque greenhouse made up of two layers of atmosphere. Each layer completely absorbs the IR radiation impinging on it A more opaque greenhouse Above we considered a leaky greenhouse. To take the other extreme, suppose thattheatmosphereissoopaquethatevenashallowlayerwillabsorball the IR passing through it. Now the assumption implicit in Fig.2.7 that space and the surface both see the same atmospheric layer is wrong. We can elaborate our model to include a second, totally absorbing, layer in the atmosphere, as illustrated in Fig.2.9. Of course, to do the calculation correctly (rather than just to illustrate the principles) we would divide the atmosphere into an infinite number of infinitesimally thin layers, allow for the presence of cloud, treat each wavelength in Fig.2.6 separately, allow for atmospheric absorption layer-by-layer which depends on the vertical distribution of absorbers, particularly H 2 O, CO 2 and O 3 (see section 3.1.2) and do the required budgets for each layer and at the surface (we are not going to do this). An incomplete schematic of how this might look for a rather opaque atmosphere is shown in Fig The resulting profile which would be the actual mean atmospheric temperature profile if heat transport in the atmosphere occurred only through radiative transfer is known as the radiative equilibrium temperature

16 52 CHAPTER 2. THE GLOBAL ENERGY BALANCE Figure 2.10: Schematic of radiative transfer model with many layers. Figure 2.11: The radiative equilibrium profile of the atmosphere obtained by carrying out the calculation schematized in Fig The absorbers are H 2 O, O 3 and CO 2.Theeffects of both terrestrial radiation and solar radiation are included. Note the discontinuity at the surface. Modified from Wells (1997).

18 54 CHAPTER 2. THE GLOBAL ENERGY BALANCE about a quarter of a degree. This is rather small when one notes that a 1W m 2 change in surface forcing demands a change in solar forcing of about 6W m 2 on taking in to account geometrical and albedo effects see Q7 at the end of the Chapter. A powerful positive climate feedback results from the temperature dependence of saturated water vapor pressure, e s,ont ; see Eq.(1.4). If the temperature increases, the amount of water that can be held at saturation increases. Since H 2 O is the main greenhouse gas, this further raises surface temperature. From Eq.(1.4) we find that: de s e e = βdt and so, given that β = C 1, a 1 C change is temperature leads to a full 7% change in saturated specific humidity. The observed relative humidity of the atmosphere (that is the ratio of actual to the saturated specific humidity see Section 5.3) does not vary significantly, even during the seasonal cycle when air temperatures vary markedly. One consequence of the presence of this blanket of H 2 O is that the emission of terrestrial radiation from the surface depends much more weakly on T s than suggested by the Stefan-Boltzmann law. When Stefan-Boltzmann and water vapor feedbacks are combined, calculations show that the climate sensitivity is: T s Q BB and H2 O K =0.5 W m, 2 twice that of Eq.(2.15). The albedo of ice and clouds also play a very important role in climate sensitivity. The primary effect of ice cover is its high albedo relative to typical land surfaces or the ocean see Table 2.2 and Fig.2.4. If the surface area of sea-ice, for example, were to expand in to low albedo regions, the amount of solar energy absorbed at the surface would be reduced, so causing further cooling and enhancing the expansion of ice. Clouds, due to their high reflectivity, typically double the albedo of the earth from 15 to 30% and so have a major impact on the radiative balance of the planet. However it is not known to what extent the amount or type of cloud (both of which are important for climate, as we will see in Chapter 4) is sensitive to the state oftheclimateorhowtheymightchangeastheclimateevolvesovertime.

19 2.4. FURTHER READING 55 Unfortunately our understanding of cloud/radiative feedbacks is one of the greatest uncertainties in climate science. 2.4 Further reading More advanced treatments of radiative transfer theory can be found in the texts of Houghton (1986) and Andrews (2000). Hartmann (1994) has a thorough discussion of greenhouse models, radiative/convective processes and their role in climate and climate feedbacks. 2.5 Problems 1. At present the emission temperature of the Earth is 255 K, andits albedo is 30%.Howwouldtheemissiontemperaturechangeif: (a) the albedo were reduced to 10% (and all else were held fixed); (b) the infrared absorptivity of the atmosphere in Fig.2.8 were doubled, but albedo remains fixed at 30%. 2. Suppose that the Earth is, after all, flat. Specifically, consider it to be a thin circular disk (of radius 6370 km), orbiting the Sun at the same distance as the Earth; the planetary albedo is 30%. The vector normal to one face of this disk always points directly towards the Sun, and the disk is made of perfectly conducting material, so both faces of the disk are at the same temperature. Calculate the emission temperature of this disk, and compare with Eq.(2.4) for a spherical Earth. 3. Consider the thermal balance of Jupiter. (a) Assuming a balance between incoming and outgoing radiation, and given the data in Table 2.1, calculate the emission temperature for Jupiter. (b) In fact, Jupiter has an internal heat source resulting from its gravitational collapse. The measured emission temperature T e defined by σt 4 e = (outgoing flux of planetary radiation per unit surface area)

21 2.5. PROBLEMS 57 Figure 2.12: An atmosphere made up of N slabs each which is completely absorbing in the IR. where T n is the temperature of the n th layer, for n>1. Hence argue that the equilibrium surface temperature is T s =(N +1) 1 4 Te, where T e is the planetary emission temperature. [Hint: Use your answer to part (a); determine T 1 and use Eq.(2.16) to get a relationship for temperature differences between adjacent layers.] 6. Determine the emission temperature of the planet Venus. You may assume the following: the mean radius of Venus orbit is 0.72 times that of the Earth s orbit; the solar flux S o decreases like the square of the distance from the sun and has a value of 1367W m 2 at the mean Earth orbit; Venus planetary albedo =0.77. The observed mean surface temperature of the planet Venus is about 750 K see Table 2.1. How many layers of the N layer model considered in Question 5 would be required to achieve this degree of warming? Comment. 7. Climate feedback due to Stefan-Boltzmann.

22 58 CHAPTER 2. THE GLOBAL ENERGY BALANCE (a) Show that the globally-averaged incident solar flux at the ground is 1(1 α 4 p)s 0. (b) If the outgoing longwave radiation from the earth s surface were governed by the Stefan-Boltzmann law, then we showed in Eq.(2.15) that for every 1W m 2 increase in the forcing of the surface energy balance, the surface temperature will increase by about a quarter of a degree. Use your answer to (a) to estimate by how much one wouldhavetoincreasethesolarconstanttoachievea1 C increase in surface temperature? You may assume that the albedo of earth is 0.3 and does not change.

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