Atmosphere, Ocean and Climate Dynamics Answers to Chapter 3

Transcription

1 Atmosphere, Ocean and Climate Dynamics Answers to Chapter 3 1. Use the hydrostatic equation to show that the mass of a vertical column of air of unit cross-section, extendin from the round to reat heiht, is ps,wherep s is the surface pressure. Insert numbers to estimate the mass on a column of air of area 1m 2. Use your answer to estimate the total mass of the atmosphere. From the hydrostatic relationship we can write (see Eq.3.4): p s = ρd surface Thus the mass of a column of air of area 1m 2 is the atmosphere is p s = =104 k. The total mass of the atmosphere is M a = 4πa2 p s = 4π ( ) = k Usin the hydrostatic equation, derive an expression for the pressure at the center of a planet in terms of its surface ravity, radius a and density ρ, assumin that the latter does not vary with depth. Insert values appropriate for the earth and evaluate the central pressure. [Hint: the ravity at radius r is (r) = Gm(r) where m(r) is the mass inside a radius r and G = k 1 m 3 s 2 is the ravitational constant. r 2 You may assume the density of rock is 2000 k m 3.] The hydrostatic equation is = ρ (r) (r) r where the acceleration due to ravity (r) at distance r is (r) = Gm (r) r 2 where m (r) is the mass inside radius r. If the planet has a uniform density m (r) = 4 3 πr3 ρ r = ρg4 3 πr3 ρ r 2 1

2 and so Zp o dp = 4 Z 0 3 πρ2 G rdr 0 a ivin p 0 = 2 3 πgρ2 a 2 Notin that M = 4 3 πa3 ρ isthemassoftheplanetand a = MG is the a 2 surface ravity, then p 0 = 1 2 aρa. Insertin numbers for the earth we find that: p 0 = = ' 10 6 atmospheres. 3. Consider a horiontally uniform atmosphere in hydrostatic balance. The atmosphere is isothermal, with temperature of 10 o C. Surface pressure is 1000hPa. (a) Consider the level that divides the atmosphere into two equal parts by mass (i.e., one-half of the atmospheric mass is above this level). What is the altitude, pressure, density and potential temperature at this level? For an isothermal atmosphere in hydrostatic balance, whence = ρ = p RT ³ p() =p(0) exp H where H = RT/. With R =287Jk 1 K 1, T = 10 C = 263 K, H =7.69 km. Mass above level is M() where M() = Z ρd= Z The level at which M() = M(0)/2 has d = p(). p() =p(0)/2 =500hPa = Pa. 2

3 At this level, the altitude is µ p() = H ln = H ln 2 = 7.69 ln 2 = 5.33 km. p(0) Temperature is 263K, so density is ρ = p/rt = / ( ) = k m 3. (b) Repeat the calculation of part (a) for the level below which lies 90% of the atmospheric mass. The level which has 90% of the mass below it has 10% above, so M() =0.1M(0), whence The altitude is Here, density is p() =0.1p(0) = 100hPa = Pa. = H ln µ p() =7.69 ln 10 = km. p(0) ρ = p()/rt =10 4 / ( ) = k m Derive an expression for the hydrostatic atmospheric pressure at heiht above the surface in terms of the surface pressure p s and the surface temperature T s for: (i) an isothermal atmosphere at temperature T s and (ii) an atmosphere with constant lapse rate of temperature Γ = dt. d Express your results in terms of the dry adiabatic lapse rate Γ d = (i) In interatin the hydrostatic equation over a depth of 10km in the atmosphere, can be taken as constant. Thus we have: ivin = p RT p() =p s e ( RTs ) for an isothermal atmosphere c p. 3

4 and (ii) µ γ Ts Γ p() =p s T s γ 1 Γcp where γ = c p c v for an atmosphere with constant lapse rate. 5. Spectroscopic measurements show that a mass of water vapor of more than 3 km 2 in a column of atmosphere is opaque to the terrestrial waveband. Given that water vapor typically has a density of 10 2 k m 3 at sea level (see Fi.3.1) and decays in the vertical like e ( b),where is the heiht above the surface and b 3 km,estimate at what heiht the atmosphere becomes transparent to terrestrial radiation. By inspection of the observed vertical temperature profile showninfi.3.3, deduce the temperature of the atmosphere at this heiht. How does it compare to the emission temperature of the Earth, T e =255K,discussed in Chapter 2? Comment on your answer. Writin the density of water vapor ρ v = ρ vsurface e ( b) then the atmosphere becomes transparent to terrestrial radiation at a heiht, say, such that ρ v d = ρ ( b) e vsurface d = bρvsurface e b =3km 2 This ives =7kmif b =3km. The temperature at this heiht is 250 K, rouhly the same as the emission temperature of the planet. This provides excellent independent support to the idea that convective systems transport heat and moisture up to a heiht (and hence temperature) at which water vapor is the principle re-radiator of terrestrial wavelenths. 6. Make use of your answer to Q.1 of Chapter 1 to estimate the error incurred in p at 100 km throuh use of Eq.(3.11) if a constant value of is assumed. 4

5 Assumin an isothermal atmosphere, Eq.(3.11) reduces to p = p s exp( /H). Thus, iven that = p H s exp( /H) then δp = p H 2 H 2 s exp( /H)δH. Since H = RT this implies that: δp p = δ H Puttin in numbers we find, notin that δ 1), δp = 100 p =3%(from Q.1 of Chapter 3 =0.41 which is a lare error. When pressures are bein computed at many scale heihts above the surface, lare errors are incurred if variations in are not accounted for. 5