Objective: To introduce the equations of motion and describe the forces that act upon the Atmosphere

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Roland Glenn
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1 Objective: To introuce the equations of motion an escribe the forces that act upon the Atmosphere Reaing: Rea pp 18 6 in Chapter 1 of Houghton & Hakim Problems: Work 1.1, 1.8, an 1.9 on pp. 6 & 7 at the en of Chapter 1 of Holton & Hakim. They are ue in class Wenesay 08SEP17 Motions in the Atmosphere are governe by: Newton s secon law, F = ma, force is equal to mass times acceleration. First law of thermoynamics, conservation of energy Mass conservation for various substances Ieal gas law relating air ensity, temperature, an pressure The first two relations are first orer partial ifferential equations in Initial Value Form (IVF) with the first orer, time erivative of a single epenant variable on the left an values or spatial partial erivatives on the right In IVF, the equations are straightforwar to integrate numerically in time. The mass conservation laws an gas law are iagnostic relations inasmuch as they o not contain time erivatives. Newton s Secon Law in meteorological coorinates: Du Forces in the x irection Dv Forces in the y irection Dv Forces in the z irection Here D()/ are Lagrangian erivatives that follow moving parcels of air; u, v, an w are velocity components in the x, y, an z irections; t is time an ρ is the ensity of air. For now, interpret D( )/ as being the same as ( )/t, but it will take on another meaning. We represent istances an velocities as vectors, for example x xi ˆyj ˆzkˆ or v ui ˆvj ˆwkˆ the carets ( hats ) over i, j an k inicate that they are unit vectors that point in the x, y, an z irections. What about the forces on the right sie of Newton s secon law? 1. Gravity gkˆ, where g = 9.8 m s 1. Gravity always acts in the z irection.. Pressure Graient Force: Consier a cubic parcel of air with each sie = δ. The sketch at the right shows how the ifference in pressure between the right an left sies of the parcel result is a net force. The force on the left sie is p(x)δ an that on the right is is p(x+δ)δ. The mass of the parcel is ρδ 3, so that the net force in the x irection ivie by the 1

2 mass is δ [p(x+δ) p(x)]/ρδ 3, or [p(x+δ) p(x)]/ρδ = ρ 1 p/ x. Applying the same argument in the y 1 1 an z irections yiels the pressure graient acceleration ˆi p ˆj p kˆ p p x y z where p = p(x, y, z,t), / x is a partial erivative that implies ifferentiation with respect to x only, an similarly for / y an / z. 3. Coriolis Force: ˆi( sin ) vˆj( sin ) u + small (with respect to gravity) vertical terms. A game of catch between Alice an Bob, both of whom in a rotating reference frame illustrates how this apparent force arises. After Alice throws the ball towar Bob, both players rotate to positions Alice an Bob. From the perspective of Carlos, who is in a non rotating reference frame the ball flies straight from Alice s original position to Bob s original position. From the perspectives of the players the ball misses Bob because of an apparent force that eflects it to the right of its intene position. Except at the poles of the spherical planet the Earth s rotation will have a horizontal component, which gives rise to vertical apparent forces that are << g. Similarly, since w << (u, v), we neglect the Coriolis terms inuce by vertical motions. Thus only the component of the Earth s rotation aroun the local vertical is significant. As the figure at the left shows, it is f = Ωsin φ, where φ is the latitue. We call f the Coriolis parameter. x 10 4 s 1. At the pole, the Coriolis parameter is Ω = (π/86400 s) = At 0 o latitue the Coriolis parameter is Ω sin 0 o = x 10 4 s 1 5 x 10 5 s 1 At 45 o latitue the Coriolis parameter is Ω sin 45 o = 1.08 x 10 4 s 1 1 x 10 4 s 1 The Coriolis force acts perpenicular to the horizontal velocity = fuj ˆ fviˆ We will analyze the Coriolis force in more etail later on. 4. Friction: Written in various ways a. ( ui ˆ vj ˆ), where μ is the coefficient of friction

3 b. c. C u v ( uiˆ vj ˆ), where C D is the rag coefficient D ˆ u u u ˆ v w K i j... k..., where K is the kinematic x y z z z viscosity. Friction is important only near the surface, where the z erivatives of the horizontal win components preominate. So the components of Newton s secon law for the atmosphere, calle the Momentum Equations, can be written: Du 1 p u fv K... x z Dv 1 p v fu K... y z Dw 1 p g z Let s look at one more kin of motion: Circular motion. Imagine an object rotating in a circular path such that. x xiˆˆircost y yj ˆ ˆjr sint r rrˆ xi ˆyj ˆ,an r x y Here bol face x an y are the vector components of the object s position. Each is the prouct of scalar magnitues, x an y with unit vectors ˆi an ˆjpointing the positive irections along the abscissa an orinate, respectively. The position vector r, with magnitue r points along the time varying ˆr irection. It is the vector sum of x an y. The time erivative of r is: r x y x ˆ y i ˆj uiˆvj ˆv t t t t t ( rsin t) ˆi ( rcos t) ˆj ri ( ˆsintˆj cos t) Note that the motion is perpenicular to the position vector an its magnitue is proportional to the prouct of the position vector s magnitue with the rotation frequency. 3

4 v u v u ˆ v i ˆj a t t t t t ˆ ( rcos t) i ( rsin t) ˆj ˆ ˆ ri ( costjsin t) r ( v / rr ) ˆ Here, again the acceleration s magnitue is proportional to the prouct of the rotation frequency with the magnitue of the velocity vector. Since the velocity vector is perpenicular to the position vector, the acceleration vector points towar the center of rotation, in the opposite irection from the position vector. This is a reasonably rigorous erivation of the Centripetal Acceleration require for an object to move aroun a circular path. It is also an example of a vector prouct, sometimes calle the cross prouct. For two vectors A an B that make an angle θ with each other, their vector prouct C is formally efine as. C AB AB sin C is oriente perpenicular to the plane efie by A an B. Its irection is etermine by using the right han rule. If you place your right han on the plane efiene by the vectors such that your inex finger points in the A irection an your mile finger points in the B irection, your thumb will point in the C irection. = ω r. If we efine ω ˆk where k is the unit vector perpenicular to the plane efine by ˆi an ˆj the v = ω x r an a = ω x ω x r Let s exten this argument to a vector that may be changing with time in a reference frame that rotates with the Earth at a constant angular velocity Ω = π/4h. In such a reference frame, any vector obeys: An the secon erivative is: A A ΩA t t FIX ROT A A Ω ΩA t t FIX ROT t ROT A A Ω Ω Ω A t t ROT ROT But Ω Ω A the Centripetal Force ue to rotation of the reference frame. As we showe above, it is always perpenicular to the axis of rotation an nearly constant because A is primarily mae up of R E cos ф. The centripetal force is << g an the vector sum of the centripetal force an gravity is 4

5 everywhere perpenicular to the surface efine by sea level. Thus, we can write g' gωωabut we will rop the primes henceforth. With these changes Newton s secon law emerges as: Dv 1 Ωv pgk As before, we place a Cartesian coorinate system tangent to the Earth s surface. Thus u, v, an w become spherical components of the velocity in a shallow layer above sea level. The non Cartesian, geometrical terms in the equations are more or less negligible for large scale meteorological motions. Expaning the Coriolis acceleration term, Ωv [( wcos vsin ) ˆi ( usin ) ˆj( ucos ) kˆ ] The vertical component is << that g, an (as shown subsequently) w << (u, v) so that, Ωvsin ( vi ˆuj ˆ) f kˆ v Where, as before, f = Ωsinф = *(π/86400 s)*sinф is the Coriolis Parameter. Its units are s 1. Hyrostatic Law Let s look at the vertical momentum equation again. Dw 1 p g x For synoptic scale, mile latitue motions the vertical velocity, w, is 1 10 cm s 1. Synoptic scale motions require about a ay (86400 s) to change appreciably. Thus, the vertical acceleration is about 0.1 m s 1 /(86400 s) ~ 10 6 m s << 9.8 m s, an the two terms on the right must be equal. 1 p g,or z p g (Hyrostatic law) z The gas law for ry air, p/ p RT relates pressure, p, temperature, T, an ensity, ρ,or specific volume, α, such tha α = ρ 1. R is the gas constant for ry air = 87 J kg 1 K 1. More about the gas law later. p gp 1 p g g,or lnp z R T p z z R T For the special case where T is constant with height, we can integrate the equation easily to get p(z), 5

6 At the surface, where z = 0, p = p 0 Subtracting the secon equation from the first: gz ln p const. RT ln p0 0 const. p ln p 0 gz R T Taking the exponential of both sies, gz pz () p0 exp. pe 0 RT We call the quantity H the scale height. It is the height ifference over which the pressure (or ensity) in an isothermal atmosphere ecreases by a factor of e 1 = A typical vertically average temperature in the troposphere is 55K, so z/ H 1 1 RT (87m s K )(55K) H m 7.5km 1 g 9.8m s This is the same vertical scale that we talke about in Lecture 0 6

Problem Solving 4 Solutions: Magnetic Force, Torque, and Magnetic Moments

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 004 Problem Solving 4 Solutions: Magnetic Force, Torque, an Magnetic Moments OJECTIVES 1. To start with the magnetic force on a moving