Oppgavesett kap. 4 (2 av 2) GEF2200

Oppgavesett kap. 4 (2 av 2) GEF2200 hans.brenna@geo.uio.no Oppgave 1 a) W&H uses two dierent expressions to describe how the monochromatic intensity of a beam is weakened over the distance ds. Write down the expressions and explain the symbols. Change of monochromatic intensity, di λ of a beam traversing a distance ds is di λ = I λ K λ Nσds, where K λ is saying how ecient the scattering/absorption is (dimension less), N is the volume density of the scattering/sborbing particles (numbers per unit volume, m 3 ), σ is the cross section (m 2 ) of each particle (not the Stefan-Boltzmann constant!). Note the minus sign on the right hand side, meaning that the new intensity I λ + di λ still represents a dimming of the beam of radiation. We can also express it as di λ = I λ ρrk λ ds, where ρ is the density of air (kg m 3 ), r is the mass of the absorbing gas per unit mass of air (dimensionless), while K λ is the mass absorption coecient m 2 kg 1 ) saying how likely absorption is for a beam hitting a molecule of the absorbing gas. b) Show that the monochromatic intensity of a beam traversing from s = 0 til s = s, is given as ( s ) I λ (s) = I λ (0)exp ρrk λ ds 0 1

Start from the last expression of the previous exercise and collect all factors containing intensity on one side. Integrate over ds, solve with respect to I λ (s) Iλ (s =s) I λ (s =0) di λ = I λ ρrk λ ds 1 di λ = ρrk λ ds I λ 1 s =s I λ (s ) d(i λ(s )) = ρrk λ ds s =0 ( ) Iλ (s = s) s =s ln = ρrk I λ (s λ ds = 0) s =0 ( I λ (s ) = s) s =s I λ (s = 0) = exp ρrk λ ds s =0 ( s ) I λ (s) = I λ (0) exp ρrk λ ds In the rest of this exercise we use Figure 4.10 from W&H (gur 1). There is a sign error between the text and the gure in the book (an you nd it?)! The gure is a bit misleading since in the text it says di λ = I λ ρrk λ ds (minus sign included), while weakening of the beam is expressed as I λ di λ. This gives us two minus signs meaning that a beam passing throught the layer would be strengthened. 0 Figure 1: Weakening of beams passing through an atmospheric layer of absorbing gases and/or aerosols. 2

c) Show that ds = sec θdz and explain what the following equation means di λ = I λ ρrk λ sec θdz From trigonometry we know: (draw gure) We recognize 1/ cos θ as sec θ, giving us cos θ = dz ds ds = dz 1 cos θ ds = dz sec θ Substitute for ds in the second expression from a) and we get di λ = I λ ρrk λ sec θdz This means that the beam is weakening through the layer due to the minus sign. d) Parallel beams travel from the top of the atmosphere (z = ) and down to z = z with zenith angle θ. Show that the monochromatic intensity to the beams at z = z can be expressed as ( ) I λ = I λ, exp sec θ ρrk λ dz What do we call this law? We start from the same expression as in exercise b, but integrate over z instead of s. Note that the beams are going down, which is the opposite direction to the z-axis. This means that z = is the lower integration limit and z = z is the z 3

upper limit. This results is Iλ (z =z) I λ (z = ) di λ = I λ ρrk λ sec θdz 1 di λ = ρrk λ sec θdz I λ 1 z =z I λ (z ) d(i λ(z )) = ρrk λ sec θdz z = ( ) Iλ (z = z) z =z ln = ρrk I λ (z λ sec θdz = ) z = ( I λ (z = z) = I λ (z = ) exp ( I λ = I λ, exp sec θ ( I λ = I λ, exp sec θ z sec θ z =z z = ρrk λ dz ) z ρrk λ dz ) ρrk λ dz ) This is called Beer's law (also called Beer-Bouguer-Lambert's law). e) The integral in the exponent in the above exercise has what name? What foes this quantity express and what values can it take? The integral is an expression for the optical depth, τ λ τ λ = z ρrk λ dz This is a measure of how optically thick the atmosphere is from the top down to z = z. Large optical depth mean that a beam passing through the layer becomes attenuated to a large extent through absorption or scattering. τ λ can have values from 0 and up (no upper limit), it's a dimensionless quantity. f ) The full exponential function in ex d) has which name? What does it express? All of the exponential function expresses the transmissivity, T λ, of the atmosphere 4

between the top, z =, and z = z. We have ( ) T λ = exp sec θ ρrk λ dz T λ = e τ λ sec θ The transmissivity (monochromatic) expresses how much of the original monochromatic intensity entering the layer at the top (z = ) is left at the bottom of the layer (z = z). z g) Given an isothermal atmosphere with the density prole ρ(z) = ρ(0)e z/h, where H is the constant scale height. show that the optical depth of the layer (τ λ ) over the height z is given as τ λ = Hrk λ ρ(0)e z/h Insert the density prole into the expression for optical depth τ λ : τ λ = = z z ρrk λ dz ρ(0)e z /H rk λ dz = rk λ ρ(0) e z /H dz z ( ) = Hrk λ ρ(0) lim /H z e z e z/h = Hrk λ ρ(0) ( e z/h) = Hrk λ ρ(0)e z/h h) Assume that radiation from the Sun come in at zenith angle θ = 0), also assume the following: ρ(0) = 1 kgm 3 H = 10 km rk λ = 10 3 m 2 kg 1 5

Table 1: Viser en oversikt over optisk tykkelse, τ λ, tranmissivitet, T λ, og absorbtivitet, α λ, fra høyden z og opp til TOA. Høyde, z (km) Optisk tykkelse, τ λ Transmissivitet, T λ Absorptivitet, α λ 10 3,68 0,025 0,975 20 1,35 0,259 0,741 30 0,50 0,607 0,393 40 0,18 0,835 0,165 Calculate the optical depth, transmissivity og absorbtivity to the atmosphere over 40, 30, 20 and 10 km. By using the formulas for τ λ from Ex g) and for T λ from Ex f), as well as α λ = 1 T λ og sec 0 = 1, yo get the values in Table 1. i) Between which of the two levels does the absorption change most. How is this related to the optical depth? We see that the dierence in absorptivity is largest between the layer at 20 km and the layer at 30 km. in thisregion we nd τ λ = 1. It can be shown that the change in absorptivity is largest where τ λ = 1. Oppgave 2 Oppgave 4.47a) fra boken We're asked to nd the percentage of incoming intensity at TOA, I λ,, that is absorbed in the layer between τ λ,2 = 0, 2 and τ λ,1 = 4, 0. We start from Beer's law which is: ( ) I λ = I λ, exp sec θ ρrk λ dz Recognize the integral in the exponent as the optical depth: I λ = I λ, e sec θτ λ Since the radiation is coming from straight above, we have θ = 0, giving: I λ = I λ, e τ λ z 6

(Memorize this expression!!). Insert values for the two optical depths and we get that the incoming radiation at TOA has been reduced to the following at the two levls: I λ,2 = I λ, e 0,2 I λ,1 = I λ, e 4,0 The dierence between them has been absorbed in the layer. We get: I λ, abs = I λ, e 0,2 I λ, e 4,0 = I λ, ( e 0,2 e 4,0) To nd the percentage, divide by the incoming intensity at TOA, I λ,. Fraction absorbed in the layer = I λ, (e 0,2 e 4,0 ) I λ, = e 0,2 e 4,0 = 0, 8 = 80 % Oppgave 3 Explain what Figure 4.23 in W&H illustrates. The gure illustrates how I λ changes with height. The peak of this curve z will be where the peak attenuation is located, i.e. where eht absorption is greatest. The altitude of this level depends on the variables I λ and ρ, how much incoming intensity is availables and how many absorbing particles are present. We hnow that density varies as ρ e z, but we also know that radiation decreases downwoard in the atmosphere. By optimizing we can nd that the absorption has a maximum where the optical depth equals 1. Oppgave 4 a) Explain Figure 4.20 in W&H. Tie this to optical depth. Figure 4.20 shows at what altitude dierent wavelengths of solar ultraviolet radiation penetrates to in the atmosphere. We see that this altitude varies with wavelength λ, and none of the radiation with λ < 0.3 µm reaches the ground. If the depth of penetration is high this means that the optical depth of the atmosphere for this wavelength is very high. Also indicated is which gases contribute most to absorbing at dierent λ. 7

b) Draw as line between the corresponding terms NIR and IR IR UV and VIS UV og VIS E orbit NIR E vibrasjon IR E rotasjon E vibrasjon E orbit E rotasjon c) Which of the following gases will absorb incoming solar radiation and which will absorb outgoind terrestrial radiation? Below a few important absorption bands are listed ˆ O 3 : 0.2-0.3 µm, 9.6 µm. ˆ CO 2 : 15 µm ˆ H 2 O: 2.70 µm, 6.25 µm, 12 µm and up Which gases dominates absorption of solar and terrestrial radiation in the troposphere and stratosphere? ˆ O 3 : 0.2-0.3 µm (solar radiation, UV), 9.6 µm (terrestrial radiation, IR). ˆ CO 2 : 15 µm (terrestrial radiation, IR). ˆ H 2 O: 2,70 µm (solar radiation, NIR), 6,25 µm (terrestrial radiation, IR), 12 µm and up (terrestrial radiation, IR) solar radiation: O 3 dominates in the stratosphere (ozon layer), while H 2 O dominates in the troposphere (Large consentration of H 2 O at ground level, declinse exponentially with height). terrestrial radiation: H 2 O dominates in the troposphere, CO 2 is also very absorbing here. In the stratosphere O 3 and CO 2 dominates (a lot of O 3 here, consentration of CO 2 is the same throughout the atmosphere). 8

Oppgave 5 a) Write down an equation for how the temperature of an atmospheric layer changes with time Explain the symbols and their units. Draw a sketch which explains why you get a minus sign on the right hand side of the equation. ρc p dt dt = df (z) dz ρ is air density (kgm 3 ), c p is the specic heat capacity of air under constant pressure (Jkg 1 K 1 ), dt expresses how temperature in the layer changes in time dt (Ks 1 ) and dz is the thickness of the thin layer we're considering (m). df (z) is the dierence betw net radiation up and out of the top of the layer (z 2 ) (F (z 2 ) - F (z 2 )) and net radiation from below into the layer (F (z 1 ) - F (z 1 )), where z 1 is lower in the atmosphere than z 2. The unit of df (z) is Wm 2. Figure 2: Vertical section of a simplied atmospheric layer A simplied atmospheric layer in the troposphere is shown in Figure 2. The layer absorbs and emits long wave radiation only. The emissivity for all wavelengths the layer emits is given in the gure. We assume that the emissivity/absoptivity between the layer and the ground is zero. In the following exercises you will need σ = 5, 67 10 8 Wm 2 K 4 and c p = 1004 Jkg 1 K 1. 9

b) Consider terrestrial radiation, F s. How much of this is left at z 2? Assume ε λ = α λ in the layer. This means that 40 % of the ux density entering from the bottom of the layer will be absorbed while 60 % will be transmitted through the layer. We assume no scattering which is a good approximation for long wave radiation. The following will be left at z 2 : F s (z 2 ) = (1 ɛ)σt 4 s = 0, 6 5, 67 10 8 (288 4 ) Wm 2 = 234 Wm 2 c) Show that the temperature change of the atmospheric layer is 0, 61 K/h. F (z 2 ) = F (z 2 ) F (z 2 ) = (ɛσta 4 + (1 ɛ)σts 4 ) 0 = ɛσta 4 + (1 ɛ)σts 4 F (z 1 ) = F (z 1 ) F (z 1 ) = σt 4 s ɛσt 4 a F (z) = F (z 2 ) F (z 1 ) = (ɛσt 4 a + (1 ɛ)σt 4 s ) (σt 4 s ɛσt 4 a ) = ɛσ(2t 4 a T 4 s ) Insert into expression from Ex a) Insert values: dt dt = 1 F (z) ρc p z dt dt = 1 ɛσ(2ta 4 Ts 4 ) ρc p z dt dt = 1, 7016 10 4 K/s We want heating rates in K/h så we have to multiply by 3600s/h. dt dt = 0, 61 K/h 10

d) Which two gases contributes most to the cooling of the atmosphrer in the troposphere? Draw a vertical prole of the troposphere illustrating approximately how these two gases contribute to dt. Explain why the values are what they are at the dz tropopause and just above the ground. Do you think the vertical proles would look dierent in tropical and arctic areas? How? H 2 O and CO 2 contributes most of the cooling in the troposphere. A vertical prole is shown in Figure 4.29 in W&H. look at the lowest 15 km for the troposphere. We see that the cooling is largest at the ground level because water vapor is abundant here and the layers above are colder, so the lower layers emit more radiation than they recieve. A the tropopause there is very little water vapor and very cold temperatures (see Figure 1.9 on page 10) so the cooling potential is very small. Both layer below and above are warmer and it will recieve almost as much energy as it emits. For arctic areas the cooling rates would be expected to be lower, aprticularly in the lower troposphere where there is little water vapor due to the low temperatures. In the tropics, the tropopause is very cold compared to lower layers so the cooling potential is extremely small. In fact, we often nd small warming rates happening here. e) Avarage temperature in the atmosphere are approximately constant. Which processes counteract the cooling caused by emission of long wave radiation? Vater vapor in the troposphere absorbs short wave (NIR) radiation and contributes a small warming eect. In addition we have convective processes making warm air rise and cold air sink. Together this cancels out the cooling eect of the greenhouse gases on the troposphere and temperaturesw will remain approximately constant (unless we add more greenhouse gases to the atmosphere...) Oppgave 6 a) Dene the single scattering albedo ω and explain the variables in the denition. What values can the single scattering albedo take? Single scattering albedo expresses how likely scattering is given extinction, (i.e. absorption + scattering). 11

We have ω λ = K λ (scattering) K λ (scattering) + K λ (absorption) Remember that K λ is the scattering or absorbtion eciency. Single scattering albedo can only take values between 0 and 1. ω λ = 0 means that none of the radiation that interacts with a particel is scattered (everything is absorbed). ω λ = 1 means all of the radiation interacting with a particle is scattered. b) What does it mean that the single scattering albedo of a particle is small? Small ω λ means that absorption is relatively more inportant than scattering for this particle when radiationinteracts with it. c) Approximately what value to you think the single scattering albedo is for ˆ CO 2 for radiation with wavelength 15 µm ˆ air molecules for radiation with wavalength 0,5 µm ˆ small since CO 2 absorbs a lot here. (between 0 and 0,5) ˆ large since we know that blue light is eciently scattered by air molecules through Rayleighspredning. (close to 1) Oppgave 3 a) Infrared radiation from the Earth passes through an atmospheric layer withs thickness ds. Explain what the equation di λ = I λ k λ ρrds + B λ (T )k λ ρrds expresses. Explain the terms. What is the name of this equation. The equation di λ = I λ k λ ρrds + B λ (T )k λ ρrds 12

expresses how muc the intensity is attenuated (if di λ is negative) or amplied (if di λ is positive) over the distance ds in the layer. The rst term expresses attenuation of the original intensity while the second term quanties the strengthening contribution from emission in the layer. The equation is called Schwarzschild's equation b) Show that the intensity of the radiation after passing distance s 1 through an atmospheric layer is I λ = I λ,0 e τ λ(s 1,0) + s1 0 k λ ρrb λ [T (s)]e τ λ(s 1,s) ds Here τ λ is optical path, and not optical depth. It goes along s and not z. By using Figure 4.26, the optical path between s s 1 can be dened as: τ λ (s 1, s) = s1 s k λ ρrds Use this togethert with that τ λ (s 1, s) increases in the opposite direction of s sso that dτ λ (s 1, s) = k λ ρrds Solution: di λ k λ ρrds + I λ = B λ (T ) di λ dτ λ (s 1, s) + I λ(s) = B λ (T (s)) di λ = I λ k λ ρrds + B λ (T )k λ ρrds This is a dierential equation which we can solve by multiplying the equation with an integrating factor di λ dτ λ (s 1, s) e τ(s 1,s) + I λ (s)e τ(s1,s) = B λ (T (s))e τ(s 1,s) Use the product rule in reverse to see that s=s1 s=0 d ( I λ e τ λ(s 1,s) ) = ( I λ (s 1 )e τ λ(s 1,s 1 ) I λ (0)e τ λ(s 1,0) ) = s=s1 s=0 s=s1 s=0 I λ (s 1 ) = I λ (0)e τ λ(s 1,0) + 13 B λ (T (s))e τ(s 1,s) dτ λ (s 1, s) B λ (T (s))e τ(s 1,s) dτ λ (s 1, s) s=s1 s=0 kρrb λ (T (s))e τ(s 1,s) ds

c) Explain what the terms in the equation from the previous exercise means. The rst term is the radiation entering the layer at s = 0 and has been attenuated travelling through the layer to s = s 1. We call this the sink term since you would get lower intensity at s = s 1 than at s = 0 if this was the only term. The second term is a source term representing the extra intensity added to the readiation from emission in the layer between s = 0 oand s = s 1. Oppgave 4 We consider a layer in the atmosphere with thicknes 2 km. The density is given as ρ = 0, 5 kgm 3. Net downwelling radiation ux in the layer increases with 100 Wm 2 from top to bottom. a) Will the layer warm or cool? Since net downwelling ux exiting the layer is larger than the ux entering the layer we know that more enrgy is leaving the layer than entering. This means we mus have net cooling in the layer. Alternate approach: Both F (z 1 ) and F (z 2 ) are negative (since they are positive in df (z) the reverse z direction), but F (z 1 ) is more negative. If we want to nd we dz must put F (z 2 ) F (z 1 ) in the numerator. This number will then be positive. From the expression for heating rates we'll see that dt must be negative since dt dt dt = 1 df (z) ρc p dz b) How large will the heating/cooling rate be? Give the answer in K/day. 14

Insert values: dt dt = 1 df (z) ρc p dz 1 100 Wm 2 = 0, 5 kgm 3 1004 Jkg 1 K 1 2000 m = 9, 96 10 5 Ks 1 = 8, 60 Kday 1 c) Where in the atmosphere are we and which gases contribute to the emission/absorption leading to the heating/cooling? Since the density is 0,5 kgm 3 we know we're in the troposphere. We know that the ux of downwelling solar radiation decreases deeper into the atmosphere (see gure 4.23), and since our ux increases downward, this has to be from longwave radiation. In the troposphere, H 2 O and CO 2 are the most important gases emitting and absorbing this radiation. Since we have cooling, emitted ux density must be greater than absorbed ux density. Oppgave 5 Use Figure 4.29 in W&H to answer the following questions 1. What does the thick black line indicate? 2. Where in the atopshere do we have approximate radiative equilibrium? 3. Where in the atmosphere do we not have radiative equilibrium and which processes compensates for this? 4. How can H 2 O have both a warming and a cooling eect? 5. Write down the most important absorbation bands of O 3, H 2 O and CO 2, and explain which of there absorb solar radiation and which absorb terrestrial radiation. 6. How sould the vertical prole of dt dt cloud into the troposphere? be aected for LW and SW if we pu a 15

1. The thick black line shows the net heating/cooling rate at dierent altitudes in the atmosphere due to absorption/emission of radiation. The included gases are CO 2, H 2 O and O 3. 2. Above approximately 15 km (above the tropopause) we have approximate radiative equilibrium. We see this since the net heating rate from the gasses is very close to 0 here. 3. We do not have ratiative equilibrium in the troposphere. The thick black line is far fro 0 here. Convective processes compensate for the cooling. 4. H 2 O has a cooling eect in LW since it absorbs and emits IR (cooling since emission usually is greater than absorption since it mostly recieves IR from below, but emits equally up and down). It has a heating rate in SW since it absorbs NIR. 5. ˆ O 3 : 0.2-0.3 µm (SW), 9.6µm (LW) ˆ H 2 O: 2.70 µm (SW), 6.25 µm (IR), 12 µm and up (LW) ˆ CO 2 : 15 µm (LW) 6. ˆ SW: Lower at the ground since less SW gets down there, higher at above since the cloud absorbs NIR, higher in the stratosphere because more SW is reected back up there to be absorbed by the ozone layer. ˆ LW: lower at the top of the cloud because more is emitted up. Higher at the cloud bottom since more is absorbed from below. Oppgave 6 a) What is remote sensing and what can it be used for? Remote sensing is measurement from a distance by using electromagnetic radiation. By analysing a signal that was emitted or reected from one or more objects we can learn something about them. E.g. we can nd: ozon concentrations, aerosol consentrations, aerosol size distributions, temperatures, temperature proles surface albedo, droplet sizes, etc- One must keep in mind that one is measuring the radiation coming from an object, and not the object itself! 16

b) Which wavelengths should one ise if you want to use remote sensing to establish: ˆ Suface albedo ˆ Ozone concentration ˆ Vertical temperature prole ˆ Surface albedo: Wavelengths that are not absorbed travelling through the atmosphere. ˆ Ozone concentration: Wavelengths that are absorbed by ozone, e.g. the region 0.2-0.3 µm. ˆ Vertical temperature prole: Wavelengths where the center of the absorption band is rapidly absorbed on the way from the ground into the atmosphere, e.g. 15 µm. This ensures that the intensity of radiation with 15 µm will not originate from the ground (this radiation has already been absorbed), but from emission sources from a level in the stratosphere. Wavelengths close to 15 µm will originate from emission sources a little further down, etc. by converting the recieved intensities to temperaturs (assume approximately black bodies at the given wavelengths), we can derive a temperature prole. Oppgave 7 a) What is illustrated in FigurE 4.31 in W&H? Figure 4.31 shows the black body spectra for black bodies at dierent temperatures (gray curves). The blue curve shows the emission spectrum of the Earth including the atmosphere. We see that the Eartg can be approximated as a black body with a temperature between 275 and 300 K. The jumps show where greenhouse gases emit and absorb radiation. b) Place some important absorption bands for H 2 O, CO 2 og O 3 into the gure. The large jump between 600 and 800 cm 1 is the 15 µm band of CO 2. The small jump in the atmospheric window (800-1200 cm 1 ) is the 9,6 µm band of O 3. From 17

1200 cm 1 and up (shorter waves) we have many H 2 O bands. The one centered ant wavelength 6,25 µm is one of them. From 800 cm 1 and down we have a lot of absorption from H 2 O. We call this the Rotation band of H 2 O. Note that this overlaps with the 15 µm band of CO 2. c) Where can we nd the atmospheric window? What is it? The atmospheric windoe is a wavelength interval where little radiation from the earth is absorbed by the atmosphere. We nd it between 800-1200 cm 1. Oppgave 8 a) What is illutsred in the top panel in Figur 4.33 in W&H? The gure illustrates how you can determine the vertical temperature prole of the atmosphere. A staellite pointing towards Earth measures the intensity of radiation in and around the 15 µm band of CO 2. This intensity is converted to temperatures by assuming black bodies. The channel 1 measures intensity at 15 µm. This radiation must come from the atmosphere (since the radiation from the ground and lower layers has been absorbed), The estimated temperature for channel 1 will come from high up in the atmosphere. Channel 2 measures the radiation coming from a layer a little deeper into the atmosphere and so on, while channel 6 is right at the edge of the 15 µm-band and this radiation will not be signicantly absobed through the atmosphere. This radiation must come from a layer close tho the ground. b) How can you use this gure to determine the vertical prole of the atmosphere? See a). c) Why do you get a peak at channel/band 1? Because the stratosphere is warmer than the upper troposphere(see Figure 1.9). 18

d) At which channel is the tropopause located From Figure 1.9 we know that the tropopause has the lowest temperature in the troposphere/stratosphere. Thus we can say that the radiation in channel 3 comes from the tropopause. e) What is the value of the surface temperature? The surface temperature seems to be close to 300 K (i.e. 27 C). 19