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AN INTRODUCTION TO DYNAMIC METEOROLOGY: THIRD EDITION J. R. Holton PROBLEM SOLUTIONS 1.1 Neglecting the latitudinal variation in the radius of the carth, calculate the angle between the gravitarional force and gravity veciors at the surface of the earth as à function of latitude. Solution: Ler à be latimde and o the angle berween g”and g. Then from the law of sines: dl. SRI, But jri =acosó. sing sina cosd 22 asinZé . snasa=9 asinÊ lgl 28 1.2 Calculate the altitude at which an artificial satellite orbiting in the equatorial plane can be a synchronous satellite (i.e. can reimain above the same spot on the surface of the earth). M Solution: From (1.7) 2+ NR =0, but g*=— (5) (CA) and R=r for q=0. . 25 - -(Se = 42,2000 km, 2 = r-a = 36,000 km. 1.3 An artificial satellite às placed into a natural synchronous orbit above the equator and is attached to the sarth below by a wire. A second sateilite is attached to the first by a wire of the same length and is placed in orbit directly above the first at the same angular velocity. Assuming that the wires have zero mass, calculate the tension in ths wires per unit mass of satellite. Could this tension be nsed to lift objects into obit with no additional expenditure of energy? . : : GM : : Solution: Tensionfunitmass = T = 0% where r, is the distance of . Ti the outer satellite from the center of the carth (r1 = 78,000 km). T =0.349N kg'l. This tension could be used once to lift a mass less than Ê = 3.56% of the mass of the outer satellite into orbit — but only by dispiacing the satellites and wire further into space. (ng = Tm, whersm, and marc the mass to be lifted and the tnass Of the outer satellite, respectively.) 1.4 A train is moning smoothly along a curved track at the rate of 50 m 5). A passenger standing on a ser of scales observes that his weight is 10% greater than when the train is at rest. The wack is banked so that the force acting on the passenger is normal to the floor of the main. What às the radius of curvarure of the track? * Solution: From the force diagram 2 2 (LI mg)? = (m, (2) rs É = 556m. ) (mg) F [0.214z) llmg Fig. 1,4 AAA ADADADDAAARABADARO: AARAAARAARASARARNSAA. 1.8 1.9 1.10 Find the horizontal displacement of a body dropped from a fixed platform at a height 4 at th equator nsglecting the eifecis of air resistance. What is ih numerical value of the displacement for h = 5 km? di A Solution: From (1.11) we have at the equator (5) = 28%w. But c ==g [= 2 Vel so that w=—g1. But E = w so = 0% (2) isthe dt 2 g total time of the fall from height A. From these we get du dx É = 20gr th =? =O. dy O CtEs us HS dr nie e (68) (608 A bullet is fired vertically upward with initial speed wg at latitude à. Neglecing air resistance, by what distance will it be displaced horizontally when it returns to the ground? (Neglect 28 cos compared to g in the vertical equation.) Solution: Now e =-g sothat w=wo-gr=0 attopof trajectory. Thus, total time of flight (upward + downward) is ig = 2wo, 8 Fon (ID É. sgcsdw. dt go u4u=-20 caso om sé) and since 2) e I& 2 3 =u, x=-20 coso(MMÉ) EE, awe But y = 2wa, So x= Q coro [Se8, g 3g A biock of mass M = 1 kg is suspended from the end of a weighiless string. The other end of the string is passed through a small hole in a horizontal platform and a ball of mass m = 10 kg is amached. At what angular velocity must the ball rotate on the horizontal platform to balance the weight of the block if the horizontal distance of the ball from the hole is 1 m ? While the ball is rotanng, the block is pulled down 10 cm. What is the new angular velocity of the ball? How much work is done in pulling down the block? AAAADAAASANSANAABAADABAABARADAADSAADANA A DADA tm | & Solution; Force balance for equilibrium: Mg =mm,a,, where to, is angular velocity of the ball,a, = im. Mg Vó (98y% E = (2) - (55) = 15, If the block M is pulled down ma 10 10 em then the distance of the ball from the hole is 0.9 m. By conservation of angular momentum mal,= 22, where Wa ar are final radii and angular velocity, respectively. 2 e Thus 0, = o(5) = 1.22257!. The total work done is given by f the sum of the changes in kinetc and potential energy: AW = AK + AP. Now AK = (Etoosos) — (0:04) =115], AP=Mgdh where 4h =-10cmheightchange. . AP=0.98, W=1.15-0.98=0.173. [rá 2 muça 1)» M pa Fig. 1.10 1.11 A paricle is free to slide or a horizontal frictioniess plane located at a latitude 4 on the earth. Find the equation goveming the path of the particle if it is given an impulsive eastward velocity u=ug at £=0. Give the solution for the position of the parúcle as a function of times. WWW UT UT IM =7. 1.14 Show thar a hoihogencous atmosphere (densiry independent of height) has a fínite height which depends Gniy on the temperature at the lower boundary. Compute the heighr of a homogeneous atmosphere with surface temperature To = 273 K and surface pressure 100 kPa. (Use the ideal gas law and hydrostatic balance.) Solution: Now e =-pg so for p = constant. Jo = “oe | d and Po = PogzH- Bu: po = poRTo so 29 = Po. = Bh - CDE cogim Po8 8 (9,81) 1.15 For the condinons of Problem 14 compute the variation of the temperatare with respect to height. This is referred 10 as an autoconvective lapse rate. Solution: From problem (1.14) p=po-— pgz when po = surface pressure. But T=P -tocer) r=t0 (Bion (E), COR) pR pro AR R . - g 34ºC temperature decreases with height at rate R = strongly super - adiabanc! r . . ar 1.16 Show íhar for an atmosphere with uniform lapse rare ny = E. the n/A r geopotential height at pressure level p, is given by Z = (5) | — (22) Y A where To and po are the seá level temperature and pressure, respectively. Solution: From hydrostaric equation and ideal gas law: aln = (EJae tu, Th : ? (E BE om P z 2 di Thus, dnp=-+ a fuera [rs Po o 5 R (2) (8 o(s 2): o «(2º Po Ry hn» Po Er So solving for z: 7, h 2 «f- v; Fal 1.17 Calculate the 100-50-kPa (1000-500 mb) thickness for a constant lapse rate atmosphere with y= 6.5 K km"! and To = 273K. Compare your results with the results in Problem 12. Solution: From the result of problem 1.16: Y a 6.5 x 107 1.18 Derive an expression for the variation of density with respect to height in a constant lapse rate atmosphere. Solution: Now p= + Lering 9 be density at height 7; Po Po To be surface values (z = 0) we have RR “ÉS Po R [E =) du Po = , - = 1-L UE RTy Po E] pol E) e a (from 1.16) 1.1y Denve an expression ior Le aitiwde variaúon vi nt pressiue change op which occurs when an atmosphere with constant lapse Tate is subjected to a height independent temperanwe change ô7 while surface pressure remains constant. At what height is the magnitude of the pressure change a maximum? EA 287(6.5 x 107?) pr=D (2) Es 1-2 58 l=si8gm. 1949999999 4 449 VAI VEILLTTIAIIITIIIIISIIIIIIIDIA Chapter 2 2. A ship is steaming northward at a rate of 10 km h”!. The surface pressure increases towards the northwest at the rate of 5 Pa km. What is thé pressure tendency recorded at a nearby island starion if the pressure aboard the ship decreases at a rate of 100 FE h? Soution: P.DP y.y Dr... ram E t Pp but Dt ( 3 . V-Vp=|VI |Vplcosa,. Ea = 33.33 — do(s) 5) =-687 pah”, 2mb or =———— 3h Yp Fig. 2.1 E The temperature at a point 50 km north of a station is 3'C cooler than at the station. If the wind is blowing from the northeast at 20 m s-| and the air is being heated by radiation ar the rate of 1ºC h-1, what is the local temperature change at the station? Solution: ELI g=1"Chi, 97. DT. V-vr, Dr à Dr “ 3ºC | . 5 x10m|: =4º€ V-VT=(0 =847 x 10", Ca x 5 OT 1ºC 305€ 8) . A o Q PERRRRRARAR Ls DA A ! SETERPEICOCCEECEC TEIA AAA AAA AAA AAA AAA 2.3 q Fac ar * vTr Fig. 2.2 Derive the relationship O x(2 xr) = —22R which was used in Eq. (2.7). Solution: |Rf=lrlsing and R is perpendicular to Q, so by definition of cross producer Q xr = Rxr =I9QHRIn. where hn is unit vecior pointing into paper perpendicular to plane oi 2 and R. Then Qx(2xr) = Qxn IQUIRI, bu Qxn = ar Qx(Qxr)=-22R. Fig. 2.3 Derive the expression given in Eq. (2.13) for the rate of change of k following he motion. solution: 2H cu2E , v2K now [SEI jm (12, Di do. % % Now [ok j5%|= 2 and ôK is direcied paralletto j, 1] a ok =d “q E | | | DT For constant temperature parcel Pr” =0. 10) w kg” Dz = 2 = fio )w ke = 00102ms! or | +L02cm st Dig 9.8ms 2.7) Denve an expression for the density p that results when an air parcel initiaily at ” pressure p, and density p, expands adiabatically to pressure p. R Solution: For adiabaric expansion 9 = (20º is conserved. Thus p DA x 8=T,= r(ee)". or subsatuting T = E ES “(8) º , p Ro pp, plp CDs xp= (2) where we have used c, = c, — R. 's 23 An air parcel that has a temperature of 20ºC at the 1000 mb level is lifted dy o adiabatically. What is its density when it reaches the 500 mb level? Solution: From problem 7, Pp = jp » - RT, 10º Pa (or 4 s00 mb) =| CE (1) -g72sk or P(S0O mb) fere] 2 em (2.9) Suppose an air parcel starts from rest at the 80 kPa (800-mb) level and rises — verrtically to 50 kPa ( 500 mb) while maintaining a constant 1*C temperature excess over the environment. Assuming that the mean temperature of the 80-50-kPa layer is 260 K, compute the energy released due to the work of the buoyancy force. Assuming that all the released energy is realized as kinetic energy of the parcel, what will thé vertical velocity of the parcel be at 50-kPa? Solution: Using (2.44) to express r.h.s. of (2.51) in terms of T we get (T-T buoyancy force / unitmass = g but energy reteased/unit mass = force x distance. | - Distance = Az = (E) In (a) where p; = 80-kPa, 8 ) ipa Pp; = 50-kPa. em [PE enem) = 135] kg, If all 135 Jkg-1 goes into kinetic energy, then sé = 135] kg, w=1643mslar 50 kPat (2.10) (2.10) Show that for an atmosphere with an adiabatic lapse rate (i.e., constant potential VE temperature) the geopotential height is given by Z = vm) — (po, p J | + Where po Po . [A . - is the pressure at Z = 0 and H6 = E is the total geopotential height of the atmosphere. Solution: Inproblem (1.16) let y A (lapse rate of adiabaric armosphere Ed and note that for po = 1000mb Ty = 8. Then immediately z= eo) - (27) Po) In the isentropic coordinare system potential temperature is sed as the vertical coordinate. Since in adiabatic flow potential temperature is conserved following the motion, isentropic coordinates are useful for tracing the actual paths of travel of individual air parcels. Show that the transformation of the horizontal pressure gradient force from z tofê coordinates is given by (3) Vep=VoM, where M=cT + istheMontgomery smreamfunction. a Solution: In equaúon (1.22) let s= 6 (2) = = (25), + ( ). cias Boom an(D (8) (2/2) (8), RSRASA AAA SRA SANA SA ASSAR SANA RRASRAANDASADAAA 16 Chapter 3 3.1 An aircraft flying a heading of 60" (i.e., 60º east of north) ar air speed 200 m s! moves relative to the ground due east (90º) at 225 m s 1. If the plane is flying at constant pressure, what is its rate of change in altitude in meters per kilometer of horizontal distance assuming a steady pressure ficid, geostrophic winds, and f=1051, Solution: F i ! E. E i i * For geostrophic motion (5) “2. Thus, assuming that the wind p is geostrophic we need the y component to compute the changein height in x-direction. Now | Vayl=200ms]; | Vprouna | = 200 1 57). From diagram Vg =j: Vwind=-100m si. =) o! . (&) - (1025 (-100 ms Do, mo à), 9.8ms* km ground Fig. 31 3.2 - The actual wind is directed 30" to the right of the geostrophic wind. If he é geostrophic wind is 20 m s-!, what is the rate of change of wind speed? Let f=i0451 CALDA REDPPERORDE PA ERRES RRR Solution: From (3.9) =. = (é. Ea is component of Vp Jo V. = 2 Vol sin E. Bu |[Vp|= pflVgl, " I as a = =-f vel sinã. = =(10*sN(20 ms) =-102ms*. Oem mms uma assa ISA CIC CI CA CEESIECEERERECECEZRERERREREPIRPRRERERRr: ipa. 17 é Pç+5P O : v vp Fig. 3.2 A tornado rotates with constant angular velocity wo. Show that the surface pressure 2,2 É o at the center of the tornado is given by P = Po es/ SÉ), where pq is the surface pressure at a distance ry from the center and T is the temperature (assume constant). If the temperature is 288 K, pressure at 100 mi from the center is 102 kPa, and wind speed at 100 m from center is 100 m s”!, what is the central pressure? Solution: Letting r be the distance form the axis of rotation of the tornado, equation (3.14) can be rewritten as 2 - Lol gm Bu pl=2T E rp p o 52 rá 2,2 dinp = o rdr, sn EE: RT Po (QRT) Pa To or p= exp| — ny ” 42 P=pPoexp GRT) | “o 2,2 2,2 Since 2% <<1l,p=Pp 1 - 2% = 94 kPa. 2RT 2RT Calculare the geostrophic wind speed in meters per second for a pressure gradient of Ti and compare with alí possible gradient wind speeds for the same pressure gradient anda radius of curvature of +500 km. Let p= 1 knm andf=10*s71. 19 mera dé pn « «— sx > Fig. 3.6 3.7” Determine the radii of curvature for the trajecrories of air parceis located 500 km to the east, north, south, and west of the center cf a circular low-pressure system, - respectively. The system is moving castward at 15 m sl. Assume geostrophic flow with a uniform tangential wind speed of 15 ms. ecosy Solution: From (3.24) R, = Ri : c=ISms!— speed of storm V = 15ms”! is wind speed. R, = 500 km. North of center Ro = e =250km (y= 7): West of center R=R,=500km (1-5), South of center Ro c(y=0); East of center R = R, = 500km (7 = 3, 1 tá f Shows y due east of storm center i Fig. 3.7 AAA AAA AAA AAA 20 3.8 Determine the gradient wind speeds for the four air parceis in Problem 7 and compare these speeds with the geostophic speed. (Let f= 104 st) Solution: Gradient wind speed for normal low: 202 VA R, Voad E (5) + (Se + 4) Vanda = 10.5ms”! North of center, - Voaa = 12. 1ms West and East of center, Voa = 15m s1 South of center. [Note: forcase R, — o (South of center) we get from equation GIN V=VA. 3,9 Show that as the pressure gradient approaches zero the gradient wind reduces to the geostrophic wind for a normal anúcycione and to thé inerna circle for an anomalous antcyclone. Solution: For pressure grad > 0 V, 0. put lg = (2) e (E) + Sa A IR 2v, Vo > O Vad = +( [-ra(i a] Positive root gives Voa = Vo. Negative root gives Vyad=—/R. (See equarion 3.12.) Q.1 0: The mean temperature in the layer berween 75 and 50 kPa decreases eastwárd by 3*C per 100 km. If the 75-kPa geosmrophic wind is from the southeast at 20 m s”!, what is the geostrophic' wind speed and direction at 50 kPa? Let /= 10455, Solution: From (3.33) us = uys + ur = —14.1 ms + ur, vg = vs + vp = +41 ms! RY(oT 75 e 0, =[8(D) a D) = -3asme. meo dr SS nl) dóms + vp, romeo nmmnoo a TRLADAAAA ARAL ADOOS DOR ER EEEDRA SERES RARA OEREENDO À 
