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Home , Coriolis force

Science A-30 Section Handout 4 March 4-5, 2008

CONTENTS: I. /Land Breeze II. III. Angular and Force IV. Geostrophic Flow

Relevant formulas and conversions: Scale Height: H = kT/(mairg ) = R'T/g -z/H Barometric Law: P(z) = P(zo)e P = ρ(k/mair)T = ρR'T m ΔP Pressure gradient force: F = × pg ρ Δx 1 day = 86400 seconds Circle: 360 degrees = 2π radians Angular : ω = 2π/t : L = m r2 ω : Fc = 2 m Ω v sin(λ) Coriolis deflection: Δy = [ Ω (Δx)2 / v ] sin(λ)

I. Sea/Land Breeze

Onshore sea breeze Offshore land breeze Warm land, cool Cool land, warm water

ƒ During the day, land heats up faster than the sea. This makes the scale height, H larger over the land than over sea. Pressure at some altitude over land is greater than pressure over sea at same altitude, which induces flow from high pressure (land) to low pressure (sea).

ƒ Now there is less over land than over the sea. This makes the pressure at the surface higher over the sea. At the ground, air flows from high pressure (sea) to low pressure (land).

ƒ Conservation of mass completes the circulation.

ƒ During nighttime, land cools off faster than the sea. The flow reverses. At the ground, air flows from inland towards the sea.

1 II. Pressure Gradient Force Pressure gradient force (Fpg) is the force due to a pressure difference (ΔP) across a horizontal distance (Δx) (directed from high pressure to low pressure). ΔP Pressure gradient = (units: N m-2 / m = N m-3 Æ force on 1 m3 of air) Δx

m ΔP Pressure gradient force = F = × (units: kg (m3 kg-1) (N m-3) Æ N Æ force) pg ρ Δx

III. Coriolis Force Coriolis force is an apparent force caused by the of the . It causes moving objects to appear to follow curved trajectories when viewed from a such as Earth. Consequently, freely moving objects on the surface of the Earth appear to be deflected to the right in the and to the left in the .

Angular velocity (ω) is the amount of angle an object rotates per unit , or simply one rotation (2π radians) divided by the time it takes to make one rotation: 2π ω = , where T is the time required for one rotation (units: radians sec-1 or sec-1) T Stationary objects on the Earth’s surface will make one rotation per day, so their is the same as that of the Earth (represented by Ω): 2π Ω = = 7.3×10−5 radians sec-1 (units: radians sec-1 or sec-1) 86,400 sec

Angular momentum (L) describes the relationship between the distance to the axis of rotation, r and the angular velocity, ω. Ignoring external such as , angular momentum is conserved. L = m r 2 ω (units: kg m2 sec-1)

A way to visualize the Coriolis Effect is to think about it in terms of angular momentum. Conservation of angular momentum dictates that freely moving objects will preserve its value of L. Consider an object traveling from the to the North Pole. Its r decreases. To keep L constant, the object’s angular velocity must increase, causing the object to rotate faster than the ground underneath it therefore appearing to veer eastward. Conversely for an object moving from the North Pole to the equator, r will increase, so its angular velocity must decrease to keep L constant. Therefore, the object will rotate slower than the land underneath it, appearing to veer westward.

Magnitude of the force (Fc) and (ac) due to Coriolis: -1 -1 -2 Fc = 2 m Ω v sin()λ (units: kg (s ) (m s ) Æ kg m s Æ N Æ force) F a = c = 2 Ω v sin()λ (units: (s-1) (m s-1) Æ m s-2 Æ acceleration) c m • λ = (at the equator λ = 0 and at the North Pole λ = 90º); • m = mass of the object on which Coriolis force is acting; • v = velocity of the moving object.

Deflection due to Coriolis Force: 2 Δx Ω sin(λ) 2 -1 -1 -1 Δy = (units: (m ) (s ) (m s ) Æ m Æ distance) v

2 IV. Geostrophic Flow Geostrophic Balance: The Coriolis force and the pressure gradient force on an air parcel balance each other, meaning they have the same magnitude but opposite signs. Northern Hemisphere Geostrophic Balance: The Coriolis force and the pressure gradient force on an air parcel balance each other (Note: below we F PGF cancel mass on both sides and equate acceleration). Coriolis

1 ΔP 2 Ω v sin(λ) = H L ρ Δx

When in geostrophic balance, the air will always flow parallel to the isobars (contours of constant pressure). • The closer together the isobars are (change in pressure per unit distance), F the stronger the pressure gradient force and the faster the will blow. Coriolis PGF • In Northern Hemisphere, wind will blow with higher pressure to its right, H L lower pressure to its left (opposite in Southern Hemisphere).

Pressure Gradient + Coriolis + Friction: In reality, below an altitude of 1 km will be subjected to a frictional force with the earth’s surface. • This is because the presence of objects on the ground, whether mountains, trees, or buildings, will slow the wind down. • Friction decreases the magnitude of the velocity and therefore the Coriolis force. • Friction does not change the magnitude of the pressure gradient force. • Therefore, wind blowing close to the ground surface is deflected towards low pressure, as the pressure gradient force is larger. Winds around highs and lows

Convergence: air is moving toward low pressure at the surface: • Once it reaches the center, the air will rise • As the air rises, it will cool resulting in condensation, clouds and rain.

Divergence: air is moving away from high pressure at the surface: • Mass will move out of the center of the high towards lower pressure. • Air will descend from above to replace the loss of mass. • As the air sinks, it will warm up and its relative humidity will decrease -- sunny and fair .

Note: In order to close the circulation lows and highs at the surface are associated with the reversed flows aloft, highs and lows, respectively.

3 Example 1: Sea/land breeze The temperature on a beach is 25ºC and 21ºC over the approximately 4 km away. The surface pressure over the beach and over the ocean is 1010 mb. a) Calculate the mass densities (kg m-3) of the surface air over land and over the ocean.

P ρ = R'T Over the beach (T = 25o C) : 1010 mb 100 Pa N m-2 J ρ = × × × = 1.179 kg m-3 (287.5 J kg-1 K -1) × (273 + 25) K 1 mb Pa N m Over the ocean (T = 21o C) : 1010 mb 100 Pa N m-2 J ρ = × × × = 1.195 kg m-3 (287.5 J kg-1 K -1) × (273 + 21) K 1 mb Pa N m

b) Calculate the local scale heights (km) over the beach and over the ocean. Assume the given temperatures are representative of the air column overlying these locations.

R'T Scale height = H = g Over the beach (T = 298 K) (287.5 J kg-1 K-1) × (298 K) kg m2 s-2 H = × = 8733 m = 8.733 km 9.81 m s-2 J Over the ocean (T = 294 K) (287.5 J kg-1 K-1) × (294 K) kg m2 s-2 H = × = 8616 m = 8.616 km 9.81 m s-2 J c) Calculate the pressure at 3 km altitude over both locations in mb.

Use the barometric law and the scale height from above to calculate the pressure at 3 km height : Over the beach (H = 8.733 km) : ⎛ z ⎞ ⎛ 3 km ⎞ P(3 km) = P(0)exp⎜- ⎟ = 1010 mb× exp⎜- ⎟ = 716 mb ⎝ H ⎠ ⎝ 8.733 km ⎠ Over the ocean (H = 8.616 km) : ⎛ z ⎞ ⎛ 3 km ⎞ P(3 km) = P(0)exp⎜- ⎟ = 1010 mb× exp⎜- ⎟ = 713 mb ⎝ H ⎠ ⎝ 8.616 km ⎠ d) Calculate the pressure gradient (mb km-1) at 3 km elevation from the coast to 4 km offshore. What is the wind direction at 3 km altitude? How about at the surface?

ΔP (Pressure over beach) - (Pressure over ocean) (716 − 713) mb Pressure gradient = = = = 0.75 mb km-1 Δx Distance between locations 4 km

4 Example 2: Coriolis Force a) Where is the Coriolis effect zero? b) Where is the Coriolis effect the greatest?

Example 3: Angular Momentum A circular station in outer space (no friction) with a radius of 40 m is rotating three per minute to induce artificial for its occupants. a) In order to decrease the gravity felt by its passengers, should the space station increase or decrease its angular velocity?

The angular velocity would need to be decreased for the gravity to decrease. The radius of the space station would need to be increased for the gravity to decrease. b) Calculate the final angular velocity if the space station magically increases its radius to 55 m. Express your answer in units of per minute and radians per minute.

Conservation of angular momentum :

Linitial = Lfinal 2 2 m rinitial ωinitial = m rfinal ωfinal 2 2 rinitial (40 m) ωfinal = ωinitial × 2 = (3 rotations per minute)× 2 = 1.6 rotations per minute rfinal (55 m) 1 rotation = 2π radians 1.6 rotations 2π × = 10 radians per minute minute 1 rotation

5 Example 4: Pressure Gradient Force, Coriolis Force and Geostrophic Flow The geostrophic velocity of a 1.0 kg mass of imaginary air parcel with a mass density of 0.801 kg m-3 at 45°N latitude is 12 m s-1. a) What is the pressure gradient force acting on this air parcel?

Since we are assuming geostrophic balance, pressure gradient force equals the coriolis force (calculate Coriolis Force) : −5 -1 o Fpg = Fc = 2 m Ω v sin()λ = 2× (1.0 kg) × (7.3×10 radians per second)× (12 m s ) × sin(45 ) = 0.00124 N

b) If the air parcel was instead over Boston (42°N), what would be its velocity assuming the same pressure gradient force from above was acting on it.

Fc = 2 m Ω v sin()λ

Use the value for Fc above, to solve for v : F 0.00124 N v = c = = 12.7 m s-1 2 m Ω sin(λ) 2×1.0 kg × 7.3×10-5 radians sec-1 sin(42o ) c) Near the ground, friction force plays a significant role. Does friction increase, decrease or have no impact on the pressure gradient force?

Friction does not affect the pressure gradient or pressure gradient force. d) Does friction increase, decrease or have no impact on the Coriolis Force?

Friction decreases the velocity of a moving object, therefore decreases the Coriolis Force.

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