The Equations of Motion in a Rotating Coordinate System. Chapter 3

Transcription

1 The Equations of Motion in a Rotating Coordinate System Chapter 3

2 Since the earth is rotating about its axis and since it is convenient to adopt a frame of reference fixed in the earth, we need to study the equations of motion in a rotating coordinate system. Before proceeding to the formal derivation, we consider briefly two concepts which arise therein: Effective gravity and Coriolis force

3 Effective Gravity g is everywhere normal to the earth s surface R g* g Ω 2 R g* R g Ω 2 R effective gravity g on a spherical earth effective gravity on an earth with a slight equatorial bulge

4 Effective Gravity If the earth were a perfect sphere and not rotating, the only gravitational component g* would be radial. Because the earth has a bulge and is rotating, the effective gravitational force g is the vector sum of the normal gravity to the mass distribution g*, together with a centrifugal force Ω 2 R, and this has no tangential component at the earth s surface. g = g * + Ω 2 R

5 The Coriolis force A line that rotates with the roundabout Ω A line at rest in an inertial system Apparent trajectory of the ball in a rotating coordinate system

6 Mathematical derivation of the Coriolis force Representation of an arbitrary vector A(t) k inertial reference system Ω k j A(t) = A 1 (t)i + A 2 (t)j + A 3 (t)k A(t) = A 1 (t)i + A 2 (t)j + A 3 (t)k rotating reference system j i i

7 The time derivative of an arbitrary vector A(t) The derivative of A(t) with respect to time da A da da da = i + j + k dt dt dt dt the subscript a denotes the derivative in an inertial reference frame In the rotating frame of reference daa da1 di = i + A dt dt dt da1 A 1 ( Ω i )... = i + + dt d = + Ω dt ( A ) 1i+...

8 Let a be any vector rotating with angular velocity Ω Ω a + da a b θ Unit vector perpendicular to Ω and a da dt = a Ω sinθ b = Ω a

9 Position vector r(t) O Want to calculate u a = d ar dt the absolute velocity of an air parcel The relative velocity in a rotating frame of reference is dr dr1 dr2 dr3 u = = i + j + k dt dt dt dt and ua = u + Ω r

10 Example ua = u + Ω r Ω V ΩR e Earth s radius This air parcel starts relative to the earth with a poleward velocity V. It begins relative to space with an additional eastwards velocity component ΩR e.

11 We need to calculate the absolute acceleration if we wish to apply Newton's second law dau a dt Measurements on the earth give only the relative velocity u and therefore the relative acceleration With A = u a dau dt a d u dt du dt du = + Ω u dt a a a du = + 2Ω u+ Ω ( Ω r) dt a ua = u + Ω r

12 absolute acceleration dau dt a du = + 2Ω u+ Ω ( Ω r) dt relative acceleration Coriolisacceleration Centripetal acceleration

13 Centripetal acceleration Position vector r is split up r = ( r. ΩΩ ) + R Ω Ω Ω R R Ω ( Ω r) = Ω ( Ω R) = Ω 2 R r Ω Ω r r The Centripetal acceleration is directed inwards towards the axis of rotation and has magnitude Ω 2 R.

14 Newton s second law in a rotating frame of reference In an inertial frame: ρ d au dt a = F In a rotating frame: density force per unit mass du ρ[ + 2Ω u + Ω ( Ω r )] = F dt

15 Alternative form du ρ = F 2 ρω u ρω ( Ω r) dt Coriolis force Centrifugal force

16 Let the total force F = g* + F be split up du ρ = F + g* 2 ρω u ρω ( Ω r) dt With g = g * + Ω 2 R Ω ( Ω r) = Ω 2 R d ρ u = F + ρ g 2 ρω u dt The centrifugal force associated with the earth s rotation no longer appears explicitly in the equation; it is contained in the effective gravity.

17 When frictional forces can be neglected, F is the pressure gradient force total pressure F = p T per unit volume d ρ u = F + ρ g 2 ρω u dt d u 1 = p + T g 2Ω dt ρ u per unit mass This is the Euler equation in a rotating frame of reference.

18 The Coriolis force does no work Ω the Coriolis force acts normal to the rotation vector and normal to the velocity. u is directly proportional to the magnitude of u and Ω. 2Ω u Note: the Coriolis force does no work because u ( 2Ω u) 0

19 Perturbation pressure, buoyancy force Define p = p ( T z ) + 0 p where dp dz 0 = gρ 0 p 0 (z) and ρ 0 (z) are the reference pressure and density fields p is the perturbation pressure Important: p 0 (z) and ρ 0 (z) are not uniquely defined g = (0, 0, g) Euler s equation becomes Du Dt ρ + 2Ω u = p + g ρ ρ ρ 1 0 the buoyancy force

20 Important: the perturbation pressure gradient and buoyancy force ρ ρ g ρ 0 are not uniquely defined. 1 ρ ρ0 But the total force p + g is uniquely defined. ρ ρ 1 ρ p Indeed 1 ρ ρ 1 ρ ρ ρ 0 p+ g = pt + g

21 A mathematical demonstration Du 1 = p ' + bˆ k Dt ρ u = 0 Momentum equation Continuity equation The divergence of the momentum equation gives: p' = ρ ρ 2 ˆ ( u u) ( bk) This is a diagnostic equation!

22 Newton s 2nd law mass acceleration = force ρ Dw Dt = p z g ρ

23 buoyancy form Put p = p ( z) + p' o ρ = ρ ( o z ) + ρ' where dp o dz = gρ o Then Dw Dt 1 p' = + ρ z b where b g ρ ρ = o ρ

24 buoyancy force is NOT unique b g ρ ρ = o ρ it depends on choice of reference density ρ o (z) but 1 p 1 p' g= + b ρ z ρ z is unique

25 Buoyancy force in a hurricane ρ () z o ρ () z o

26 z Initiation of a thunderstorm θ = constant Τ + ΔΤ T U(z)

27 tropopause negative buoyancy outflow original heated air θ = constant positive buoyancy LFC LCL inflow negative buoyancy

28 Some questions How does the flow evolve after the original thermal has reached the upper troposphere? What drives the updraught at low levels? Observation in severe thunderstorms: the updraught at cloud base is negatively buoyant! Answer: - the perturbation pressure gradient

29 positive buoyancy outflow HI p' original heated air LFC LCL LO HI inflow HI HI negative buoyancy

30 Scale analysis Assume a homogeneous fluid ρ = constant. Euler s equation becomes: Du Dt 1 + 2Ω u = p ρ scales: U L 2 2ΩU ΔP ρl Then 2 Du / Dt U / L U ~ = = 2Ω u 2ΩU 2ΩL Ro Rossby number

31 Extratropical cyclone 2Ω = 10 4 s 1 L = 106 m L U = 10 ms 1 Ro = 10 U 1 10 = = 2 Ω L

32 Tropical cyclone U = 50 m/sec 2Ω = 5 X 10 5 L = 5 x 10 5 m Ro = U 2 Ω L = = 2

33 Dust devil L L = m U = 10 ms 1 2Ω = 10 4 s 1 Ro U 10 = = 4 2Ω L 10 ( 10, 100) 3 4 = 10 10

34 Waterspout L = 100 m U = 50 ms 1 2Ω = 10 4 s 1 Ro = L

35 Aeroplane wing L = 10 m U = 200 m s 1 2Ω = 10 4 s 1 Ro =

36 The Rossby number Flow system L U m s 1 Ro Ocean circulation km 1 (or less) Extra-tropical cyclone 10 3 km Tropical cyclone 500 km 50 (or >) 1 Tornado 100 m Dust devil m Cumulonimbus cloud 1 km Aerodynamic 1-10 m Bath tub vortex 1 m

37 Summary (i) Large scale meteorological and oceanic flows are strongly constrained by rotation (Ro << 1), except possibly in equatorial regions. (ii)tropical cyclones are always cyclonic and appear to derive their rotation from the background rotation of the earth. They never occur within 5 deg. of the equator where the normal component of the earth's rotation is small. (iii) Most tornadoes are cyclonic, but why? (iv) Dust devils do not have a preferred sense of rotation as expected. (v) In aerodynamic flows, and in the bath (!), the effect of the earth's rotation may be ignored.

38 Coordinate systems and the earth's sphericity Many of the flows we shall consider have horizontal dimensions which are small compared with the earth's radius. In studying these, it is both legitimate and a great simplification to assume that the earth is locally flat and to use a rectangular coordinate system with z pointing vertically upwards. Holton ( 2.3, pp31-35) shows the precise circumstances under which such an approximation is valid. In general, the use of spherical coordinates merely refines the theory, but does not lead to a deeper understanding of the phenomena.

39 Take rectangular coordinates fixed relative to the earth and centred at a point on the surface at latitude. Ω j φ Φ j k i k i equator Äquator

40 Ω φ Ω Ω cos φ j Ω sin φ k φ f-plane or β-plane Ω = Ω cos φ j + Ω sin φ k 2 Ω u = Ω cos φ j u + Ω sin φ k u

41 In component form 2Ωv sin φ+ 2Ωw 2Ω u = 2 Ω u sin φ 2 Ω u cosφ cosφ I will show that for middle latitude, synoptic-scale weather systems such as extra-tropical cyclones, the terms involving cos φ may be neglected. 2Ω u = 2 Ω cosφj u + 2 Ω sinφk u The important term for large-scale motions

42 To a good approximation 2Ω u = 2 Ω sin φ k u = f u f = 2 Ω sin φ f = f k Coriolis parameter

43 Scale analysis of the equations for middle latitude synoptic systems Much of the significant weather in middle latitudes is associated with extra-tropical cyclones, or depressions. We shall base our scaling on such systems. Let L, H, T, U, W, P and R be scales for the horizontal size, vertical extent, time, u h, w, perturbation pressure, and density in an extra-tropical cyclone, say at 45 latitude, where f (= 2Ω sin φ ) and 2Ω cos φ are both of order and U = 10 ms ; W = 10 ms ; L= 10 m ( 10 km); H= 10 m ( 10 km); T= L/ U ~ 10 s (~ 1day); δp= 10 Pa ( 10 mb) R = 1kg m

44 horizontal momentum equations Du Dt 2Ωv sin φ+ 2Ωw cosφ= 1 ρ p x Dv Dt + 2Ωu sin φ = 1 ρ p y scales U 2 /L 2 ΩU sin φ 2ΩW cos φ δp/ρl orders Du Dt h 1 + fk uh = ρ h p

45 vertical momentum equations Dw Dt 1 pt 2Ω u cosφ = ρ z g scales UW/L 2ΩU cosφ δp T /ρη g orders negligible the atmosphere is strongly hydrostatic on the synoptic scale. But are the disturbances themselves hydrostatic?

46 Question: when we subtract the reference pressure p 0 from p T, is it still legitimate to neglect Dw/Dt etc.? vertical momentum equations Dw 1 p 2Ω u cosφ= + b Dt ρ z scales UW/L 2ΩU cosφ δp/ρη gδt/t 0 orders still negligible assume that H* H assume that δt 3 o K/300 km H* = height scale for a perturbation pressure difference δp of 10 mb

47 In summary orders Dw 1 p 2Ω u cosφ= + b Dt ρ z p 0= + b ρ z In synoptic-scale disturbances, the perturbations are in close hydrostatic balance Remember: it is small departures from this equation which drive the weak vertical motion in systems of this scale.

48 The hydrostatic approximation The hydrostatic approximation permits enormous simplifications in dynamical studies of large-scale motions in the atmosphere and oceans.

49 End of Chapter 3