# I ν = λ 2 I λ. λ<0.35 µm F λ = µm <λ<1.00 µm F λ =0.2 Wm 2 µm 1. λ>1.00 µm F λ =0. F λi 4λ i. i 1

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1 Chapter Remote sensing in the microwave part of the spectrum relies on radiation emitted by oxygen molecules at frequencies near 55 ghz. Calculate the wavelength and wavenumber of this radiation. From (4.2) λ = c eν = ms =5450µm s The spectrum of monochromatic intensity can be defined either in terms of wavelength λ or wavenumber ν such that the area under the spectrum, plotted as a linear function of λ or ν is proportional to intensity. Show that I ν = λ 2 I λ. di = I λ dλ = I ν dν From (4.1) ν = 1 λ from which it follows that dν = dλ λ 2 Substituting for dλ in the first expression, cancelling the common factor dν, and ignoring the minus sign, which is taken into account by reversing the direction of the integration, we obtain I ν = λ 2 I λ 4.14 A body is emitting radiation with the following idealized spectrum of monochromatic flux density λ<0.35 µm F λ = µm <λ<0.50 µm F λ =1.0 Wm 2 µm µm <λ<0.70 µm F λ =0.5 Wm 2 µm µm <λ<1.00 µm F λ =0.2 Wm 2 µm 1 λ>1.00 µm F λ =0 Calculate the flux density of the radiation. F = Z F λ dλ = NX F λi 4λ i i 1 = = = 0.31 Wm An opaque surface with the following absorption spectrum is subjected to the radiation described in the previous exercise. 54

2 λ<0.70 µm A λ =0 λ>0.70 µm A λ =1 How much of the radiation is absorbed? How much is reflected? F (absorbed) = F (reflected) = Z A λ F λ dλ = NX A λi F λi 4λ i i 1 = = = 0.25 Wm 2 Z NX (1 A λ )F λ dλ = (1 A λi )F λi 4λ i = = 0.06 Wm Calculate the ratios of the incident solar radiation at noon on north and south facing 5 slopes (relative to the horizon) in seasons in which the solar zenith angle is (a) 30 and (b) 60. For the 30 solar zenith angle the ratio r = F north facing slope F south facing slope is I cos 35 r = I cos 25 =0.84 and for the 60 zenith angle the ratio is r = i 1 I cos 65 I cos 55 = Compute the daily insolation at the North Pole at the time of the summer solstice when the earth-sun distance is km. The tilt of the earth saxisis23.5. Compare this value with the minimum value that occurs in association with the Earth s orbital cycles described in Answer: 46.4 MJ m 2 day 1 versus xxx MJ m 2 day 1. At the time of the summer solstice the incident solar radiation is independent of time of day and the solar zenoth angle is ( )=66.5. Hence the flux density is equal to 1368 Wm 2 µ km cos 66.5 = km 55

3 4.18 Compute the daily insolation at the top of the atmosphere at the equator at the time of the equinox (a) by integrating the flux density over a 24 hour period and (b) by simple geometric considerations. Compare your result with the value in the previous exercise and with Fig Answer: 38.0 MJ m 2 day What fraction of the flux of energy emitted by the sun does the earth intercept? Answer Show that for small variations in the earth s radiation balance δt E = 1 δf E T E 4 F E where T E is the planet s equivalent blackbody temperature and F E is the flux of radiation emitted from the top of its atmosphere. Use this relationship to estimate the change in effective temperature that would occur in response to (a) the seasonal variations in the sun-earth distance due to the eccentricity of the earth s orbit (presently 3%),(b) an increase in the earth s albedo from 0.30 to From (4.12) F = σt 4 Taking the log yields ln F =4lnT Taking the differential yields δf F =4δT T and dividing both sides by 4 yields δt E = 1 δf E (1) T E 4 F E (a) From the inverse square law F E = const d 2 where d is the earth-sun distance. Taking the log yields ln F E =lnconst 2lnd Taking the differential yields δf E F E = 2 δd d (2) Combining (1) and (2) yields δt E = 1 δd T E 2 d (1) Substituting δd/d =0.03 yields δt E /T E = If T E = 255 K, then δt E =1.5 K. 56

4 (b) F E = const (1 A) where A is the eplanetary albedo. Taking the log yields ln F E =lnconst ln(1 A) Taking the differential yields δf E δ(1 A) = F E (1 A) (2) Combining (1) and (2) yields δt E = 1 δ(1 A) T E 4 (1 A) (3) Substituting δd/d =0.01/0.70 yields δt E /T E = If T E = 255 K, then δt E =0.91 K Show that the emission the flux density of incident solar radiation on any planet in our solar system is 1366 W m 2 d 2,whered is the Planet-Sun distance, expressed in astronomical units (A.U., multiples of the Earth- Sun distance). Method 1: The flux density is equal to the intensity of solar radiation I s (the same for all planets) times the arc of solid angle δω subtended by the sun, as viewed from the planet, i.e., where F = I s δω δω =4π µ 2 Rs d where R s is the radius of the sun and d isthedistancebetweentheplanet and the sun. Hence, F =(4πI s R 2 s)d 2 where the factor in parentheses is the same for all planets. For Earth, 1368 = (...)d 2 E (1) and for any other planet Dividing (2) by (1) yields F p =(...)d 2 p (2) µ 2 F p 1368 = dp (3) d E 57

6 where ψ is the angle between the zenith and the limb of the Earth (ie., a straight line passing through the satellite and tangent to the circle that defines the Earth). It is readily verified that ψ =sin 1 R E d where R E is the radius of the Earth and d is the distance between the center of the satellite and the center of the Earth. Substituting values yields ψ = sin = and 4ω 2.21 sr. Fig Geometric setting for Exercises If the Earth radiates as a black body at an equivalent blackbody temperature T E = 255 K, calculate the radiative equilibrium temperature of the satellite when it is in the Earth s shadow. Let dq be the amount of heat imparted to the satellite by the flux density de received within the infinitesimal element of solid angle dω. Then, dq = πr 2 Idω where r is the radius of the satellite and I is the intensity of the isotropic blackbody radiation emitted by the Earth, which is given by I = σt E 4 π Hence, dq = r 2 σtedω 4 Integrating the above expression over the arc of solid angle subtended by the Earth, as computed in the previous exercise, yields the total energy absorbed by the satellite per unit time Q =2.21r 2 σt 4 E 59

7 For radiative equilibrium Q =2.21r 2 σt 4 E =4πr 2 σt 4 s where T s is the equivalent blackbody temperature of the satellite. Solving, we obtain µ 1/ T s = T E 4π and substituting T E = 255 KyieldsT s = 166 K Show that the approach in Exercise 4.5 in the text, when applied to the previous exercise, yields a temperature of T s = T E " 1 4 µ # 2 1/ = 158 K 8370 Applying the inverse square law, the flux density of Earth s radiation in the orbit of the satellite is µ 2 RE F = σt E (1) d where d is the distance between the center of the satellite and the center of the Earth. In analogy with Exercise 4.5, with the Earth s radiation treated as parallel beam, the radiation balance of the satellite is Combining (1) and (2) yields πr 2 F E =4πr 2 σt 4 s (2) T s = T E " µ1 4 µ # 2 1/4 RE (3) d This approach underestimates the temperature of the satellite because it treats the radiation that impinges on the satellite as parallel beam. The answer obtained in the previous exercise converges to this limiting value as R E /d. In this limiting case, the angle ψ in Exercise 4.30 becomes very small and converges to sin 1 (R E /d) =R E /d. Using the half angle formula r 1 cos x sin (x/2) = 2 the arc of solid angle 4ω =2π(1 cos ψ) subtended by the Earth in the "sky" of the satellite in Exercise 4.30 may be written as 4ω = π sin 2 2ψ = π sin 2 (2R E /d) = 4π (R E /d) 2 60

8 From here on, we can proceed as in Method 1 of Exercise 4.21, replacing the (sun by the Earth) to show that the flux density of Earth radiation reaching the orbit of the satellite is given by the inverse square law F E = σt 4 E (R E /d) 2 as assumed in Eq. (1). It follows that in this limiting case, the equivalent blackbody temperature of the satellite is the same regardless of whether it is computed using the equations in Exercise 4.5 or Calculate the radiative equilibrium temperature of the satellite immediately after it emerges from the earth s shadow (i.e., when the satellite is sunlit but the earth, as viewed from the satellite, is still entirely in shadow). Answer 289 K The satellite has a mass of 100 kg, a radius of 1 m and a specific heat of 10 3 Jkg 1 K 1. Calculate the rate at which the satellite heats up immediately after it (instantaneously) emerges from the earth s shadow. Answer Cs (a) Extend the proof in the previous exercise to the case in which absorptivity and emissivity are wavelength dependent. Let one of the walls be black, as in the previous exercise and let the other wall also be black, except within a very narrow wavelength range of width δλ, centered at λ = λ 1 where a λ1 < 1. [Hint: Since blackbody radiation is isotropic, it follows the blackbody flux in the interval δλ is πb (λ 1,T) δλ. Using this relationship, consider the energy balance as in the previous exercise and proceed to show that α λ1 = ε λ1.] (b) Indicate how this result could be extended to prove that ε λ = α λ (a) If δλ is of infinitesimal width, then in the region outside this band, the proof in the previous exercise should still be valid. Within the band H = α λ1 πb (λ 1,T) δλ ε λ1 πb (λ 1,T) δλ = (α λ1 ε λ1 ) πb (λ 1,T) δλ Based on the same line of reasoning as in the previous exercise, it follows that α λ1 = ε λ1. (b) This proof could be extended by proving that ε λ2 = α λ2 within a second narrow wavelength range centered at λ = λ 2, and then within a third at λ = λ 3,andsoon,toconstructaspectrumin which ε λ and α λ vary continuously, but are equal within any arbitrarily narrow wavelength range. 61

9 4.37 Consider a closed spherical cavity in which the walls are opaque and all at the same temperature. The surfaces on the top hemisphere are black and the surfaces on the bottom hemisphere reflect all the incident radiation at all angles. Prove that in all directions I λ = B λ. Downward directed radiation is blackbody radiation for which I λ = B λ by definition. Since the bottom hemisphere is a perfect reflector, along each ray path, I λ = I λ and since the lower hemisphere is at the same temperature as the upper hemisphere, I λ = I λ = B λ (a) Consider the situation described in Exercise 4.35, except that both plates are gray, one with absorptivity α 1 and the other with absorptivity α 2. Prove that F1 0 = F 2 0 α 1 α 2 where F1 0 and F2 0 are the flux densities of the radiation emitted from the two plates. Make use of the fact that the two plates are in radiative equilibrium at the same temperature but do not make use of Kirchhoff s Law. [Hint: Consider the total flux densities F 1 from plate 1 to plate 2andF 2 from plate 2 to plate 1. The problem can be worked without explicitly dealing with the multiple reflections between the plates.] For each plate, the total radiation equals the emitted radiation plus the reflected radiation. For plate 1 F 1 = F F 2 (1 a 1 ) and for plate 2 F 2 = F2 0 + F 1 (1 a 2 ) If the plates are in radiative equilibrium, F 1 = F 2, so we can write or, since F 1 = F 2, or F F 2 (1 a 1 )=F F 1 (1 a 2 ) F F 1(1 a 1 )=F F 2(1 a 2 ) F 0 1 a 1 F 1 = F 0 2 a 2 F 2 which may be written as µ µ F 0 1 F 0 F 1 a 1 = 2 F 2 a 2 a 1 a 2 Since F 1 = F 2, but a 1 need not be equal to a 2, this relationship can be generally satisfied only if F1 0 = F 2 0 α 1 α 2 62

10 4.39 Consider the radiation balance of an atmosphere with a large number of isothermal layers, each of which is transparent to solar radiation and absorbs the fraction α of the longwave radiation incident on it from above or below. (a) Show that the flux density of the radiation emitted by the top-most layer is αf/(2 α) where F is the flux density of the planetary radiation emitted to space. By applying the Stefan-Boltzmann law (4.12) to an infinitesimally thin topmost layer, show that the radiative equilibrium temperature at the top of the atmosphere, sometimes referred to as the skin temperature, isgivenby µ 1/4 1 T = T E 2 (Were it not for the presence of stratospheric ozone, the temperature of the km layer in the earth s atmosphere would be close to the skin temperature.) Consider the radiation balance for the topmost layer of the atmosphere. Since this layer is isothermal and since its absorptivity a equals its emissivity ε, it follows that it must emit flux density F 1 in both the upward and downward directions. Hence, it loses energy at a rate 2F 1. The loss due to emission is balanced by the absorption of upwelling radiation from below. The upwelling radiation from below is (F + F 1 ) and the fraction of it that is absorbed in the topmost layer is a. Hence 2F 1 = a(f + F 1 ) (1) Solving, we obtain F 1 = αf (2 α) But from the Stefan Boltzmann law and Kirchhoff s law (2) F 1 = aσt 4 1 (3) where T 1 is the equivalent blackbody temperature of the topmost layer. Combining Eqs. (2) and (3) yields In the limit, as a 0, Solving, we obtain aσt 4 1 = αf (2 α) σt 4 1 = F 2 = σt 4 E 2 T 1 =(0.5) 1/4 T E For the Earth;s atmosphere, T E = 255 KandT 1,theso-calledskintemperature, is 214 K, which is quite comparable to the mean temperature of the lower stratosphere. 63

11 4.40 Consider an idealized aerosol consisting of spherical particles of radius r with a refractive index of 1.5. Using Fig. 4.13, estimate the smallest radius for which the particles would impart a bluish cast to transmitted white light, as in the rarely observed "blue moon". To impart a bluish (as opposed to a reddish) cast to transmitted light, the particles would need to exhibit a range of sizes for which the scattering efficiency K λ decreases with increasing values of the size parameter, the reverse of the situation for Rayleigh scattering. It is evident from Fig. 4.13, that the smallest range of values of x for which dk λ /dx < 0 occurs around x = 2πr λ 5 For visible light with λ 0.5 µm 2πr µm =2.5 µm r 0.4 µm 4.41 Consider an idealized cloud consisting of spherical droplets with a uniform radius of 20 µm and concentrations of 1 cm 3. How long a path through such a cloud would be required to deplete a beam of visible radiation by a factor of e due to scattering alone? (Assume that none of the scattered radiation is subsequently scattered back into the path of the beam.) From Eq. (4.16), di λ = I λ K λ Nσds Thesizeparameterx =2πr/λ =2π 20 µm 0.5 µm > 50. Hence, if absorption is assumed to be negligible, Fig yields a value of K λ =2. The number density N = 1 cm 3 = 10 6 m 3 and the cross-sectional area σ = π = Hence, the volume scattering coefficient K λ Nσ = m 1. The incident radiation is depleted by a factor of e over a path in which the optical depth τ λ = R K λ Nσds is equal to unity. If we assume that conditions are uniform along the ray path, the path length is given by s =(K λ Nσ) 1 = m m 4.43 Consider radiation with wavelength λ and zero zenith angle passing through a gas with an volume absorption coefficient k λ of 0.01 m 2 kg 1.Whatfraction of the beam is absorbed in passing through a layer containing 1 kg m 2 of the gas? What mass of gas would the layer have to contain on order to absorb half the incident radiation? The absorptivity of the layer is given by A =1 e R k λ u 64

12 where u is the density-weighted path length. For a layer containing 1 kg m 2 of gas, the optical depth k λ u =0.01. Since e x = 1+x for x<<1, it follows that A = 1 (1 0.01) = 0.01 or 1% For a layer thick enough to absorb exactly half the incident radiation, Taking the natural logarithm yields T λ = e τ λ =0.5 τ λ =ln0.5 =0.694 Hence, k λ u =0.694 and the mass per unit area along the path is u = =69.4 kg k λ 4.45 For incident parallel beam solar radiation in an atmosphere in which k λ is independent of height, (a) show that optical depth is linearly proportional to pressure. (b) Show that the absorption per unit mass (and consequently the heating rate) is strongest, not at the level of unit optical depth but near the top of the atmosphere, where the incident radiation is virtually undepleted. (a) If k λ and r are assumed to be independent of height, optical depth is given by τ λ = rk λ sec θ = rk λ sec θ g µ rkλ sec θ = g Z z Z z p ρdz ρgdz (b) The absorption per unit pass is equal to di λ /ρ where ρ is the air density. In a hydrostatic atmosphere, the mass per unit height in a column. From Eq. (4.16) di λ = I λ ρrk λ sec θdz from which it follows that di λ ρ = I λrk λ sec θdp/g Provided that r, the density of the absorbing constitruent and k λ are not changing with height, it follows that the absorption per unit mass is strongest at the top of the atmosphere, where the monochromatic intensity of the incident radiation is undepleted. 65

13 4.46 Consider a hypothetical planetary atmosphere comprised entirely of the gas in Exercise The atmospheric pressure at the surface of the planet is 1000 hpa, the lapse-rate is isothermal, the scale height is 10 km and the gravitational acceleration is 10 m s 2. Estimate the height and pressure of the level of unit normal optical depth. At the level of unit normal optical depth Z τ λ = k λ ρdz =1 Hence, k λ g Z z z ρgdz = k λp g =1 It follows that p = g = 10 k λ 0.01 =103 Pa From the hypsometric equation z = µ p0 H ln p = 10 ln 100 = 46 km 4.47 (a) What percentage of the incident monochromatic intensity with wavelength λ and zero zenith angle is absorbed in passing through the layer of the atmosphere extending from an optical depth τ λ = 0.2 to τ λ = 4.0? (b) What percentage of the outgoing monochromatic intensity to space with wavelength λ and zero zenith angle is emitted from the layer of the atmosphere extending from an optical depth τ λ = 0.2 to τ λ = 4.0? (c) In an isothermal atmosphere, through how many scale heights would the layer in (a) and (b) extend? (a) The fraction of the incident radiation that is absorbed in the layer extending from an optical depth τ λ = 0.2 to τ λ = 4.0 is given by or 80%. e 0.2 e 4.0 = = (b) Of the outgoing monochromatic intensity to space with wavelength λ and zero zenith angle, the fraction e 0.2 is emitted from the top of the layer without absorption, and the fraction e 4.0 is emitted from the layer below without absorption. Hence, the fraction emitted from the layer is as given in (a). 66

14 (c)atthetopofthelayer Hence, and k λ g Z τ λ = k λ ρdz =0.2 Z z z ρgdz = k λp T g =0.2 p T =0.2 g k p Similarly, at the bottom of the layer, p B =4.0 g k p The depth of the layer is µ pb 4z = H ln p T µ 4.0 = H ln 0.2 = 3.00H 4.48 For the atmosphere in Exercise 4.46, estimate the levels and pressures of unit (slant path) optical depth for downward parallel beam radiation with zenith angles of 30 and 60. At the level of unit optical depth τ λ = k λ sec θ Z z ρdz =1 Hence, g p = k λ sec θ =0.866 and Pa From the hypsometric equation z = µ p0 H ln p = 10 ln 115 and 10 ln 200 = 47.4 and 53.0 km 4.49 Prove that the optical thickness of a layer is equal to minus the natural logarithm of the transmissivity of the layer. 67

15 By definition T λ = e τ λ Taking the natural log of both sides yields τ λ = ln T λ 4.50 Prove that the fraction of the flux density of overhead solar radiation that is backscattered to space in its first encounter with a particle in the atmosphere is given by b = 1 g 2 where g is the asymmetry factor defined in (4.35). [Hint: The intensity of the scattered radiation must be integrated over zenith angle.] 4.52 Prove that the ratio of the flux absorptivity to the intensity absorptivity at zero zenith angle approaches a limiting value of 2 as the optical thickness of a layer approaches zero. The flux transmissivity, as defined in (4.45) is given by T f ν =2 Z 1 0 e τ ν/µ µdµ where µ =cosθ. In the limit as x 0, e x = 1+x. Hence, as the optical thickness approaches zero, T f ν = 2 = 2 Z 1 0 Z 1 0 = 1 2τ ν µ 1 τ ν µ µdµ 2 µdµ Z 1 The corresponding flux absorptivity is given by 0 τ ν dµ α f ν =1 T f ν =2τ ν and the intensity absorptivity is given by α ν = 1 T ν = 1 e τ ν = τ ν Hence, α f ν =2α ν 68

16 4.54 A thin, isothermal layer of air in thermal equilibrium at temperature T 0 is perturbed about that equilibrium value (e.g., by absorption of a burst of ultraviolet radiation emitted by the sun during a short lived solar flare) by the temperature increment δt. Using the cooling to space approximation (4.56) show that µ dt δ = α ν δt (4.65) dt ν where α ν = πk λr c p e τ ν/µ µ µ dbν (4.66) dt T 0 This formulation, in which cooling to space acts to bring the temperature back toward radiative equilibrium is known as Newtonian cooling or radiative relaxation. It is widely used in parameterizing the effects of longwave radiative transfer in the middle atmosphere. The cooling to space approximation [Eq. 4.59)] is µ dt = π k ν rb ν (z) e τ ν/µ dt ν c p µ If the layer is in radiative equilibrium at temperature T 0, then there must be a heat source that exactly balances (dt/dt) ν. Let us assume that this heat source remains fixed, while the temperature of the layer increases from T 0 to T 0 +δt. The only term in (4.59) that changes in response to the warming of the layer is B ν which changes by the increment (db ν /dt ) T0 δt in accordance with Planck s law. Hence, the imbalance in the heating rate is given by µ dt δ = πk νr e τ µ ν/µ dbν δt dt ν c p µ dt T 0 which is equivalent to Eqs. (4.65) and (4.66) Prove that weighting function w i used in remote sensing, as defined in (4.59) can also be expressed as the vertical derivative of the transmissivity of the overlying layer. By definition where τ i = T i = e τ i Z z k i ρrdz Differentiating with respect to z, we obtain dt i dz = e τ i dτ i = e τ i k i ρrdz 69

17 Hence, the weighting function in Eq. (4.59) can be expressed as w i = dt i dz 4.56 The annual mean surface air temperature ranges from roughly 23 Cin the tropics to 25 C in the polar cap regions. On the basis of the Stefan- Boltzmann law, estimate the ratio of the flux density of the emitted longwave radiation in the tropics to that in the polar cap region. From the Stefan Boltzmann law, F tropics F polar = σt 4 tropics σtpolar 4 µ = µ = 250 = Consider the simplified model of the short wave energy balance shown in Fig The model atmosphere consists of an upper layer with transmissivity T 1, a partial cloud cloud layer with fractional coverage f c and reflectivity in both directions R c, and a lower layer with transmissivity T 2. The planet s surface has an average reflectivity R s. Assume that no absorption takes place within the cloud layer and no scattering takes place except in the cloud layer. 4 Fig Geometric setting for Exercise (a) Show that the total short wave radiation reaching the surface of the planet divided by the solar radiation incident on the top of the atmosphere is given by F s = [(1 f c)+f c (1 R c )]T 1 T 2 1 T2 2f cr c R s 70

18 The fraction of the incident radiation reaching the surface of the planet, taking into account the radiation reflected upward from the surface and downward again from the cloud layer is F s = {[(1 f c )+f c (1 R c )] T 1 T 2 }[1+T2 2 f c R c R s + T f c R c R s T ] 2 f c R c R s Summing over the series, we obtain F s = [(1 f c)+f c (1 R c )]T 1 T 2 1 T 2 2 f cr c R s (b) Show that the planetary albedo is given by A = f c R c T F s R s [(1 f c )+f c (1 R c )] The fraction of ther incident radiation that is reflected off the cloud tops is f c R c T 2 1 and the fraction refected off the surface of the planet is (c) For the following values of model parameters calculate the planetary albedo: f c =0.5, R c =0.5, T 1 =0.95, T 2 =0.90, R s = Answer (d) Use the model to estimate the albedo of a cloud free and cloud covered earth. Answers 0.09 and

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