Published on METEO 300: Fundamentals of Atmospheric Science (https://www.e-education.psu.edu/meteo300)

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Lesson 10: Dynamics – Forces

Overview

Overview

To forecast the weather, we use numerical weather prediction models that are based on mathematical expressions for conservation of energy, mass, and momentum. Climate prediction models are based on the same conservation laws. Conservation simply means that the amount of a quantity such as total energy, mass, or momentum remains constant even though the forms of that quantity may change. Conservation of energy is described by the 1st Law of Thermodynamics, which was discussed in Lesson 2; conservation of mass and conservation of momentum are discussed in this lesson. The total mass of an air parcel is constant, but density and volume may change. The conservation of momentum is based on Newton's 2nd Law and involves forces that can change momentum. The conservation of momentum is relatively simple when cast in an inertial (non-accelerating) reference frame because there are only three real forces that really matter for atmospheric motion: gravity, the pressure gradient force, and friction. But when cast on a rotating Earth, we need to add apparent forces to this equation in order to compensate for the fact that an air parcel on Earth is always accelerating as the Earth rotates. We end up adding the apparent forces—Coriolis force and centrifugal force—to the real forces to get an equation of motion whose predictions we can readily match with our observations.

Some atmospheric motion occurs with air masses and waves that are thousands of kilometers across, while other motion, such as tornadoes, is at most a few kilometers across. Sometimes air flows in a straight line; sometimes it flows around ridges and troughs. In all of these different cases, the most important forces are different, allowing the momentum conservation equation to be simplified in different ways. We will discuss these different conditions and show how you can determine the wind velocity from knowledge of the balance of the most important forces and thus determine the impact of air motion at upper levels on the air motion near Earth’s surface. Finally, we will describe why midlatitude winds are westerly.

Learning Objectives

By the end of this lesson, you should be able to:

  • explain mass conservation physically, recognize the mass conservation equation, and memorize its form when density is constant
  • state the three main conservation laws in atmospheric science: the conservation of mass, the conservation of momentum, and the conservation of energy
  • name and explain the three fundamental (real) forces in the atmosphere (gravity, pressure gradient, and friction)
  • name and explain the two new (apparent) forces that emerge when momentum conservation is written in the rotating reference frame
  • draw the balance of forces for geostrophic flow, gradient flow, geostrophic flow with friction, and cyclostrophic flow
  • explain why midlatitude winds are westerly

Questions?

If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.

10.1 Why We Like Conservation

10.1 Why We Like Conservation

Scientists like things that are conserved. There are good reasons for this. First, if something is conserved, that means we can always count on it being the same no matter what happens. Second, when we write down the equation for the conserved quantity, we can use that equation to understand how the equation’s variables will change with differing conditions. For example, in Lesson 2, we were able to use the First Law of Thermodynamics (a.k.a., conservation of energy) along with the Ideal Gas Law to derive the equation for potential temperature, which is very useful for understanding and calculating the vertical motion of air parcels.

In atmospheric dynamics, we like three conservation laws:

  1. conservation of energy (The 1st Law of Thermodynamics [1])
  2. conservation of mass
  3. conservation of momentum (Newton’s Second Law, but really three equations—one in each direction)

Conservation of Mass

So, let’s step back and look at the mass of an air parcel, which equals the density times the volume of the parcel:

m=ρV MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaad2gacqGH9aqpcaGGGcGaeqyWdiNaamOvaaaa@3897@

[10.1]

In a parcel, the mass is conserved, and since m = ρV,

D Dt ( m )= D Dt ( ρV )=0 MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaamyBaaGaayjkaiaawMcaaiabg2da9maalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaeqyWdiNaamOvaaGaayjkaiaawMcaaiabg2da9iaaicdaaaa@4235@

[10.2]

Apply the product rule to Equation [10.2]:

V Dρ Dt +ρ DV Dt =0 MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadAfadaWcaaWdaeaapeGaamiraiabeg8aYbWdaeaapeGaamiraiaadshaaaGaey4kaSIaeqyWdi3aaSaaa8aabaWdbiaadseacaWGwbaapaqaa8qacaWGebGaamiDaaaacqGH9aqpcaaIWaaaaa@406A@

Divide both sides by ρV:

1 ρ Dρ Dt + 1 V DV dt =0

Recall that the specific rate of change in parcel volume is equal to the divergence (Equation 9.4) and so we can write:

1 V DV dt = ∇ → • U →

Rearranging the equation gives us an expression for the conservation of mass:

1 ρ Dρ Dt + ∇ → • U → =0

[10.3]

This equation is for the conservation of mass in a continuous fluid (i.e., the fluid particles are so small that the air parcel behaves like a fluid). It is also called the Equation of Continuity. Physically, this equation means that if the flow is converging ( ∇ → • U → <0 MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaaFiaabaGaey4bIenacaGLxdcacaGGIaYaa8HaaeaacaWGvbaacaGLxdcacqGH8aapcaaIWaaaaa@3B2C@ ), then the density must increase ( Dρ Dt MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebGaeqyWdihapaqaa8qacaWGebGaamiDaaaaaaa@3779@ >0). Note that in Lesson 9.5 [3]we said that density doesn’t change much at any fixed pressure level, which is how we were able to relate horizontal divergence/convergence with vertical ascent/descent. What did change was the vertical size of the air parcel as the horizontal size increased or decreased. The total mass, however, remained the same.

Isaac Newton
"Isaac Newton woodcut, frontispiece to Mach" by an unknown artist after a portrait by Kneller, ca. 1689.
Credit: Licensed under Public Domain via Wikimedia Commons [4]

Conservation of Momentum

Newton’s 2nd Law, F = ma, applies to a mass with respect to the inertial coordinate system of space. But we are interested in motion with respect to the rotating Earth. So, to apply Newton’s 2nd Law to Earth’s atmosphere, our mathematics will need to account for the forces of Earth’s rotating coordinate system:

F → =m a → = ∑ ​ (real forces)+ ∑ ​ (apparent forces)=m D U → Dt MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@60E8@

[10.4]

where the first set of forces are real forces and the second set are apparent (or effective) forces that will be used to correct for using a coordinate system attached to a rotating Earth.

When we use the word “specific” as an adjective describing a noun in science, we mean that noun divided by mass. So, specific force is F/m = a, acceleration. In what follows, we will use the terms “force” and “acceleration” interchangeably, assuming that if we say “force,” we mean “force/mass,” which is acceleration. At this point, you should be able to check the units—if there is no “kg,” then obviously we are talking about accelerations.

a → = F → m = D U → Dt

Extra Credit Reminder!

Here is another chance to earn one point of extra credit: Picture of the Week!

  1. You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
  2. Add a short description of the processes that you think are causing your observation. A Word file is a good format for submission.
  3. Use your name as the name of the file. Upload it to the Picture of the Week Dropbox in this week's lesson module. To be eligible for the week, your picture must be submitted by 23:59 UT on Sunday of each week.
  4. I will be the sole judge of the weekly winners. A student can win up to three times.
  5. There will be a Picture of the Week dropbox each week through Lesson 11. Keep submitting entries!

10.2 What are the important real forces?

10.2 What are the important real forces?

There are three real forces important for atmospheric motion:

  1. Gravitational Force
  2. Pressure Gradient Force (PGF)
  3. Friction

Hence we can sum these real forces:

∑ F a → = F g → + F p → + F f →

We put the subscript "a" on these forces to indicate "absolute" because they are true in an inertial reference frame. Thus, in the absolute reference frame,

D a U → a Dt = ∑ ​ F → a m MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaalaaapaqaa8qacaWGebWdamaaBaaaleaapeGaamyyaaWdaeqaaOWdbiqadwfapaGbaSaadaWgaaWcbaWdbiaadggaa8aabeaaaOqaa8qacaWGebGaamiDaaaacqGH9aqpdaWcaaWdaeaadaqfGaqabSqabeaacaaMb8oaneaapeGaeyyeIuoaaOGabmOra8aagaWca8qadaWgaaWcbaGaamyyaaqabaaak8aabaWdbiaad2gaaaaaaa@41E5@

Let's examine each of these real forces in more detail.

1. Gravitational Force

Recall that the gravitational force on a mass m is simply the weight of the mass, which is given by:

F → g =m g → * where g → *=− GM r 2 ( r → r ) MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaafaqaaeWabaaabaaeeaaaaaaaaigBHn2Aa8qaceWGgbWdayaalaWaaSbaaSqaa8qacaWGNbaapaqabaGcpeGaeyypa0JaamyBaiqadEgapaGbaSaapeGaaiOkaaWdaeaapeGaae4DaiaabIgacaqGLbGaaeOCaiaabwgaa8aabaWdbiqadEgapaGbaSaapeGaaiOkaiabg2da9iabgkHiTmaalaaapaqaa8qacaWGhbGaamytaaWdaeaapeGaamOCa8aadaahaaWcbeqaa8qacaaIYaaaaaaakmaabmaabaWaaSaaaeaaceWGYbGbaSaaaeaacaWGYbaaaaGaayjkaiaawMcaaaaaaaa@49E1@

[10.6]

where M is Earth’s mass (5.9722 x 1024 kg), r → MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaceWGYbGbaSaaaaa@3287@  is the distance vector originating from the Earth’s center, and G is the gravitational constant (6.6741 × 10–11 m3 kg–1 s-2). Ignoring the minor effects of topography and the horizontal variation of Earth's density, the real gravitational force points directly towards Earth’s center. The gravitational force per unit mass is simply g → * MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaam4zaaGaay51GaGaaiOkaaaa@3619@ .

2. Pressure Gradient Force (PGF)

The derivation of the pressure gradient force is similar to what we have already done in Lesson 2.2 to find hydrostatic equilibrium [5], except that we will look at only the pressure forces in this case, and will serve as a quick review. Consider the x-direction first:

cube with arrows pointing to forces
Drawing of the forces involved in the pressure gradient force in the x direction.
Credit: W. Brune

F px m = p( x )A−p( x+Δx )A m = p( x )A−[ p( x )+Δp ]A m MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@57CF@

F px m = AΔx[ p( x )−p( x )−Δp ] mΔx =− V m Δp Δx =− 1 ρ Δp Δx ≈− 1 ρ ∂p ∂x MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@69B9@

Adding in the y and z directions, we get the 3-D vector form of the pressure gradient force per unit mass:

F → p m =− 1 ρ ∇ → p

Example

Let's do a quick calculation of the pressure gradient force from a map of surface pressure on 26 June 2015. Note that Pennsylvania's northern border is about 250 km from its southern border.

map of surface pressure of east coast of united states with a low over ohio and green over PA and NY. see text description
Click here for an accessible alternative to the text on the image

Reads: PA is ~250 km high. Isobars are 4 mb apart. Air density ~1.2kg/m3. From 1008 mb isobar to 1016 mb isobar is ~300 km. |PGF| = (1/1.2)* (800 Pa/(300*1000) = 2.2x10-3 m s-2 pointing due west

Credit: Penn State e-Wall and W. Brune

The following video (1:20) will explain the process:

PGF Example
Click here for a transcript of the PGF Example video.

Let's go through a quick calculation of the pressure gradient force or a low-pressure system that passed over Pennsylvania on June 26, 2015. Note that the pressure increases as x increases. But because the pressure gradient force is minus 1 over the density times the pressure gradient force-- really the pressure gradient acceleration-- is negative. This makes sense since the pressure gradient force would move air from high pressure to low pressure which is to the west in this case. To find the pressure gradient we note that the height of Pennsylvania is about 250 kilometers, which is slightly smaller than the distance between the 1,008 millibar and the 1,016 millibar isobars, which is about 300 kilometers as a distance. So the air density is about 1.2 kilograms per meter cubed. When we put all these numbers together-- that is one over the density times the change in pressure over the change in the distance-- we get that the pressure gradient force, in this case, is 2.2 times 10 to the minus 3 meters per second squared and is directed to 180 degrees, or due west.

3. Friction

We can think of friction as being processes that impede the airflow. There are two different types of friction that meteorologists are concerned with: molecular friction and turbulent friction. Molecular friction is a real force that appears in the conservation of momentum equation whereas turbulent friction is an additional term that arises out of averaging the conservation of momentum equation.

Molecular friction results from the random movement of molecules. Imagine two air parcels moving towards the east. One air parcel is just to the north of the other and is moving a bit faster than the other. Due to random molecular motion, the two parcels exchange air molecules that carry the momentum of their respective air parcels. When molecules collide, some of their momentum is transferred, resulting in the faster parcel (the one to the north) slowing down and the slower parcel (the one to the south) speeding up. There is thus a transfer of momentum from the faster parcel to the slower parcel. This transfer is proportional to the velocity difference between the air parcels and a quantity called the viscosity. The viscosity depends on the fluid in question (air in this case) and the temperature. Fluids with a relatively high resistance to motion, like honey, have relatively high viscosities. Think about air near the Earth's surface. The air right at the surface is stationary due to electromagnetic forces between the air and the surface. Due to molecular friction, the air near the surface will slow down the air just above it, just as that air slows the air a little bit higher. We show without derivation that the molecular friction force (sometimes called the viscous force) per unit mass is to a very good approximation given by:

F → f m =ν  ∇ 2 U → a MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaalaaapaqaa8qaceWGgbWdayaalaWaaSbaaSqaa8qacaWGMbaapaqabaaakeaapeGaamyBaaaacqGH9aqpcqaH9oGBcaaMc8Uaey4bIe9damaaCaaaleqabaWdbiaaikdaaaGcceWGvbWdayaalaWdbmaaBaaaleaacaWGHbaabeaaaaa@3FFA@

where ν is the kinematic viscosity, ∇ 2 = ∇ → · ∇ → = ∂ 2 ∂ x 2 + ∂ 2 ∂ y 2 + ∂ 2 ∂ z 2 is called the Laplace operator or the Laplacian, and U → is the air parcel velocity. The viscous force is important for resisting flow and dissipating air flow on small scales, such as for an individual raindrop, but it is not an important force on larger scales when compared to other forces such as gravity and the pressure gradient force (as will be demonstrated in Lesson 10.5).

Turbulent friction is important for larger-scale atmospheric motion, even synoptic-scale motion. The flow in the atmosphere's lowest kilometer or two, called the atmospheric boundary layer is often turbulent, with chaotic large and small swirls of air that, when taken together, have momentum in all directions. During the day, turbulence is generated by convection. During both day and night, turbulence is also generated by wind shear throughout the boundary layer. No matter how turbulence is generated, it provides a drag on the horizontal flow throughout the boundary layer because upward moving air with low horizontal momentum collides with air aloft with high horizontal momentum, slowing it down. This turbulent drag is often referred to as friction, even though the word "friction" really applies only to molecular-scale interactions.

Turbulent friction is not a fundamental force; it is represented in the conservation of momentum equation only after the equation has been averaged over time, space, or both. New terms representing turbulent friction arise from the averaging of the advective derivative, which we will discuss in more detail in Lesson 11. For now, we take the momentum conservation equation and average it so that all of the quantities that we are predicting—like velocity, pressure, and density—really reflect average quantities that vary gradually over space and time. For example, the wind velocity averaged over an hour and over the southeastern quarter of Pennsylvania would be a good example of a quantity one could predict from the averaged momentum conservation equation. On the other hand, a wind gust measured by an anemometer on top of a building would not be a good example of such a quantity.

For a turbulent boundary layer, the turbulent friction per unit mass is a function of four quantities: the dimensionless drag coefficient Cd, the planetary boundary layer height h, the magnitude of the horizontal velocity | V → a | MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaaemaapaqaa8qaceWGwbWdayaalaWdbmaaBaaaleaacaWGHbaabeaaaOGaay5bSlaawIa7aaaa@3908@ , and the horizontal velocity itself:

− C d h | V → a | V → a MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiabgkHiTmaalaaapaqaa8qacaWGdbWdamaaBaaaleaapeGaamizaaWdaeqaaaGcbaWdbiaadIgaaaWaaqWaa8aabaWdbiqadAfapaGbaSaapeWaaSbaaSqaaiaadggaaeqaaaGccaGLhWUaayjcSdGabmOva8aagaWca8qadaWgaaWcbaGaamyyaaqabaaaaa@3F54@

[10.8]

Even though this turbulent drag is not really friction, it is an important resistance to the average horizontal flow on large scales in the boundary layer and so we will keep it, and not molecular friction, as the friction term in the averaged momentum equation. Note that the turbulent drag is greatest within the boundary layer and becomes much smaller above the boundary layer, where it is assumed that the drag coefficient becomes very small.

Inertial (Real) Force Summary

The real forces can be summarized in the following two equations. The first equation represents how the instantaneous velocity of an individual air parcel varies with time. The second equation, which is an average of the first equation, represents how the average velocity of an air mass varies with time. Both equations include acceleration, gravity, and the pressure gradient force. The first equation includes molecular friction and the second equation includes turbulent friction. The first equation is more accurate but the second equation is more practical for applications in weather and climate.

D a U → a Dt =− 1 ρ ∇ → p+ g → *+ν ∇ 2 U → a MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaalaaapaqaa8qacaWGebWaaSbaaSqaaiaadggaaeqaaOGabmyva8aagaWcamaaBaaaleaapeGaamyyaaWdaeqaaaGcbaWdbiaadseacaWG0baaaiabg2da9iabgkHiTmaalaaapaqaa8qacaaIXaaapaqaa8qacqaHbpGCaaGafy4bIe9dayaalaWdbiaadchacqGHRaWkceWGNbWdayaalaWdbiaacQcacqGHRaWkcqaH9oGBcqGHhis0daahaaWcbeqaaiaaikdaaaGcceWGvbGbaSaadaWgaaWcbaGaamyyaaqabaaaaa@4AE1@

D a U → a Dt =− 1 ρ ∇ → p+ g → *− C d h | V → a | V → a MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@4F85@

[10.9]

10.3 Effects of Earth’s Rotation: Apparent Forces

10.3 Effects of Earth’s Rotation: Apparent Forces

Newton’s Second Law applies in an inertial reference frame, which means that the reference frame is not accelerating. A point on the rotating Earth is not following a straight line through space, but instead follows a roughly circular path and hence is constantly accelerating towards the axis of rotation. Therefore, Earth does not provide an inertial reference frame. For an astronaut in distant space observing the weather on Earth, the air motions obey Newton’s Second Law perfectly, but for an Earth-bound observer, Newton’s Second Law fails to capture the observed motion. To account for this crazy behavior, the Earth-bound observer needs to add some apparent forces to the real forces for the math to explain the observed motion from the point of view of someone standing on the rotating Earth.

circle labeled to depict earth's rotation
Earth’s rotation is to the east. The symbol ⨂ indicates that the vector U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ is going into the page with a magnitude RΩ.
Credit: W. Brune

Suppose we have an air parcel moving through space with a velocity U → a MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadggaaeqaaaaa@366B@, which we will call the absolute velocity. We want to relate this absolute velocity to U → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51Gaaaaa@3559@, the velocity observed with respect to the Earth reference frame. Let U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ be the velocity of the Earth. Here we only consider the velocity of the Earth due to rotation about its axis (the motion around the Sun is much less important), so U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ is always eastward, greatest at the equator and zero at the poles. The absolute velocity of an air parcel is simply the velocity of the air parcel with respect to the Earth plus the velocity of the Earth itself: U → a = U → + U → e MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiqadwfapaGbaSaadaWgaaWcbaWdbiaadggaa8aabeaakiabg2da98qaceWGvbWdayaalaGaey4kaSYdbiqadwfapaGbaSaadaWgaaWcbaWdbiaadwgaa8aabeaaaaa@3B08@

What is the velocity of the Earth? Consider a specific point on the Earth. Let Ω → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaeuyQdCfacaGLxdcaaaa@360D@ be Earth’s angular velocity vector, r → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOCaaGaay51Gaaaaa@3576@ be the position vector from the Earth’s center to the point in question, and R → be the shortest distance vector from the axis of rotation to the point in question (as in the figure above). The magnitude of Ω → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaeuyQdCfacaGLxdcaaaa@360D@ is  | Ω → |= 2π 23.934 hr 1 hr 3600 s =7.292× 10 −5 s −1 MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaaemaapaqaa8qacuqHPoWvpaGbaSaaa8qacaGLhWUaayjcSdGaeyypa0ZaaSaaa8aabaWdbiaaikdacqaHapaCa8aabaWdbiaaikdacaaIZaGaaiOlaiaaiMdacaaIZaGaaGinaiaabccacaqGObGaaeOCaaaadaWcaaWdaeaapeGaaGymaiaabccacaqGObGaaeOCaaWdaeaapeGaaG4maiaaiAdacaaIWaGaaGimaiaabccacaqGZbaaaiabg2da9iaaiEdacaGGUaGaaGOmaiaaiMdacaaIYaGaey41aqRaaGymaiaaicdapaWaaWbaaSqabeaapeGaeyOeI0IaaGynaaaakiaabohapaWaaWbaaSqabeaapeGaeyOeI0Iaaeymaaaaaaa@5893@ and the direction of Ω → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaeuyQdCfacaGLxdcaaaa@360D@  is determined by the right hand rule (the direction of your thumb when you curl the fingers of your right hand in the direction of rotation and point your thumb towards the North Star). To determine the angular velocity of the Earth, note that we have used the sidereal day length, 23.934 hr, which is the day length when Earth's rotation is measured with respect to the fixed stars (the inertial reference frame).

The following video (:51) will demonstrate the right hand rule:

UE Right Hand Rule
Click here for transcript of the UE Right Hand Rule video.

u sub e is the eastward velocity of the Earth. It is pointed into the page. We know the u sub e equals r-- which is the shortest distance vector between Earth's rotation axis and the point on the surface-- times omega, which is Earth's rotation vector. The units of omega are seconds to the minus 1, which makes it a frequency. Note that u sub e equals r times omega, which is also equal to omega cross r. We can see this if you take your right hand with your fingers pointed in the omega direction and the palm in the r direction. And you fold your fingers into the palm. Your thumb will point into the page, which is the direction of u sub e and is in the positive x direction.

The magnitude of U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ is RΩ, but we need to write U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ as a vector. Note that  U → e MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadwgaaeqaaaaa@366F@ is pointed into the page in the figure above, which is a direction perpendicular to both  Ω → and  R → . Hence we can use the cross product equation to write an expression for Earth's velocity:  U → e = Ω → × R → = Ω → × r →, since the component of r → perpendicular to Ω → is R →. So:

U → a = U → + Ω → × r → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadggaaeqaaOGaeyypa0Zaa8HaaeaacaWGwbaacaGLxdcacqGHRaWkdaWhcaqaaiabfM6axbGaay51GaGaey41aq7aa8HaaeaacaWGYbaacaGLxdcaaaa@42F0@

[10.10]

We have thus related velocity in the absolute reference frame to velocity in the rotating reference frame.

Now we can consider acceleration. Since U → = D r → Dt MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaGaeyypa0ZaaSaaaeaacaWGebWaa8HaaeaacaWGYbaacaGLxdcaaeaacaWGebGaamiDaaaaaaa@3BA6@ and U → a = D a r → Dt we can write:

D a r → Dt = D r → Dt + Ω → × r → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamiramaaBaaaleaacaWGHbaabeaakmaaFiaabaGaamOCaaGaay51GaaabaGaamiraiaadshaaaGaeyypa0ZaaSaaaeaacaWGebWaa8HaaeaacaWGYbaacaGLxdcaaeaacaWGebGaamiDaaaacqGHRaWkdaWhcaqaaiabfM6axbGaay51GaGaey41aq7aa8HaaeaacaWGYbaacaGLxdcaaaa@485E@

[10.11]

This equation describes the change in a position of an air parcel with time observed from an inertial reference frame (the derivative on the left) to the change with time observed from Earth’s reference frame (the derivative on the right). Equation [10.11] is general and applies not only to r → but also to any other vector.

Let’s replace  r → on the left hand side with  U → a MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiqadwfapaGbaSaapeWaaSbaaSqaaiaadggaaeqaaaaa@35BC@ and r → on the right hand side with U → + Ω → × r → since these two expressions equal each other in Equation [10.10]. By making these substitutions, we can relate acceleration in the absolute frame to acceleration in the rotating frame:

D a U → a Dt = D U → Dt + D Ω → Dt × r → + Ω → × D r → Dt + Ω → × U → + Ω → ×( Ω → × r → ) MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@60CA@

We can simplify this equation and then we can make sense of it physically. First,  Ω → is not changing significantly with time, so D Ω → Dt can be set to zero. Second,  ( Ω → × r → ) MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacIcadaWhcaqaaiabfM6axbGaay51GaGaey41aq7aa8HaaeaacaWGYbaacaGLxdcacaGGPaaaaa@3C27@ has the magnitude of ΩR MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfM6axjaadkfaaaa@352F@ and points to the east (by the right hand rule) and thus Ω → ×( Ω → × r → ) MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaeuyQdCfacaGLxdcacqGHxdaTcaGGOaWaa8HaaeaacqqHPoWvaiaawEniaiabgEna0oaaFiaabaGaamOCaaGaay51GaGaaiykaaaa@4180@ has the magnitude Ω 2 R MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfM6axnaaCaaaleqabaGaaGOmaaaakiaadkfaaaa@3623@ and points toward – R → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOuaaGaay51Gaaaaa@3556@. Finally, noting that U → = D r → Dt, we end up with the equation:

D a U → a Dt = D U → Dt +2 Ω → × U → − Ω 2 R → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamiramaaBaaaleaacaWGHbaabeaakmaaFiaabaGaamOvaaGaay51GaWaaSbaaSqaaiaadggaaeqaaaGcbaGaamiraiaadshaaaGaeyypa0ZaaSaaaeaacaWGebWaa8HaaeaacaWGwbaacaGLxdcaaeaacaWGebGaamiDaaaacqGHRaWkcaaIYaWaa8HaaeaacqqHPoWvaiaawEniaiabgEna0oaaFiaabaGaamOvaaGaay51GaGaeyOeI0IaeuyQdC1aaWbaaSqabeaacaaIYaaaaOWaa8HaaeaacaWGsbaacaGLxdcaaaa@4FDB@

[10.13]

The term on the left is the acceleration in the absolute inertial reference frame. The first term on the right is the acceleration in the Earth reference frame. The remaining terms are the apparent accelerations. The first one is the Coriolis acceleration and the second one is the centripetal acceleration.

We can now combine Equation [10.13] with the version of Equation [10.9] that is averaged to get:

− 1 ρ ∇ → p+ g → *− C d h | V → | V → = D U → Dt +2 Ω → × U → − Ω 2 R →

and then rearrange this equation to get:

D U → Dt =− 1 ρ ∇ → p+ g → *− C d h | V → | V → −2 Ω → × U → + Ω 2 R → MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbmaalaaapaqaa8qacaWGebGabmyva8aagaWcaaqaa8qacaWGebGaamiDaaaacqGH9aqpcqGHsisldaWcaaWdaeaapeGaaGymaaWdaeaapeGaeqyWdihaaiqbgEGir=aagaWca8qacaWGWbGaey4kaSIabm4za8aagaWca8qacaGGQaGaeyOeI0YaaSaaa8aabaWdbiaadoeapaWaaSbaaSqaa8qacaWGKbaapaqabaaakeaapeGaamiAaaaadaabdaWdaeaapeGabmOva8aagaWcaaWdbiaawEa7caGLiWoaceWGwbWdayaalaWdbiabgkHiTiaaikdacuqHPoWvpaGbaSaapeGaey41aqRabmyva8aagaWca8qacqGHRaWkcqqHPoWvpaWaaWbaaSqabeaapeGaaGOmaaaakiqadkfapaGbaSaaaaa@5604@

[10.14]
The first three terms on the right hand side of Equation [10.14] are the real forces. The fourth and fifth terms on the right hand side are the apparent forces: the Coriolis force and the centrifugal force, respectively.

Centrifugal Force

The centrifugal force is directed away from the Earth's axis of rotation and is the same type of force that you feel when you are in a car going around a sharp curve. Over its long history, all the material that makes up the Earth has adjusted to the real gravitational force, g → *, which is directed to Earth’s center, and the apparent centrifugal force that is directed away from Earth’s rotation axis (see figure below).

circle illustrating gravity as described in the text
The gravity we feel is the sum of gravity pointed toward Earth’s center and the outward centrifugal force. The effect is greatly exaggerated to show the vectors. The gravity we feel, g → , is perpendicular to Earth's flat surfaces at rest (i.e., oceans).
Credit: W. Brune

The resulting gravity that the Earth and everything on it feels is the vector sum of this real and this apparent force:

g → = g* → + Ω 2 R → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaam4zaaGaay51GaGaeyypa0Zaa8HaaeaacaWGNbGaaiOkaaGaay51GaGaey4kaSIaeuyQdC1aaWbaaSqabeaacaaIYaaaaOWaa8HaaeaacaWGsbaacaGLxdcaaaa@3FAC@

[10.15]

Since the centrifugal force depends on R → , it is greatest at the equator and zero at the poles. As a result of the centrifugal force, the Earth has become slightly oblate, with an equatorial radius of 6378.1 km that is 0.34% greater than the polar radius of 6356.8 km. Note that g → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaam4zaaGaay51Gaaaaa@356A@ is always perpendicular to Earth’s surface, which is very useful because the vertical coordinate is always chosen to be perpendicular to Earth's surface, so that  g → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaam4zaaGaay51Gaaaaa@356A@ is only in the z direction and, as the figure above indicates, does not point towards the center of the Earth (except at the poles and the equator). The value of g at the equator is 9.780 m s–2, which is 0.052 m s–2 smaller than the value of g at the poles, which is 9.832 m s–2. The centrifugal force at the equator is Ω2R = (7.27 x 10–5 s–1)2 (6.378 x 106 m) = 0.033 m s–2, and hence accounts for about 2/3 of the difference in g between the equator and the poles. The rest of the difference is due to the difference in g*, which is overestimated by the difference in equatorial and polar radii--the problem is more complicated than it might appear because Newton's law of gravitation only applies to point masses. At any rate, the difference between g at the poles and equator is small enough for a constant value of g = 9.8 m s–2 to be suitable for most applications in atmospheric dynamics.

Combining Equations [10.14] and [10.15] yields a more useful form of the averaged momentum conservation in the rotating reference frame:

D U → Dt =− 1 ρ ∇ → p+ g → − C d h | V → | V → −2 Ω → × U →

[10.16]
We will now move on to a discussion of the Coriolis Force. The following video (3:05) gives a basic introduction.
Coriolis Effect
Click here for transcript of the Coriolis Effect video.

[MUSIC PLAYING]

What's going on here? Every time I throw the ball straight, it seems to bend to the side. No matter what I throw, no matter how straight I throw it, the ball seems to be curving.
Maybe it has something to do with this: That whole time I was spinning, which is why it looks like I'm about to puke.
That curve is created by the Coriolis effect. You may have heard that it makes water in the bathtub spiral down the drain a certain way or that it determines the way a toilet flushes. That's actually wrong. Those are not caused by the Coriolis effect. But the Coriolis effect does create hurricanes and is the reason why Jupiter’s Great Red Spot is spinning the way it is.
So, what is the Coriolis effect? Well, it's what happens when objects moving in a straight line appear to curve because you are rotating. And it affects all kinds of things. It bends the paths of missiles and sniper shots. But how does it work, and how does it create hurricanes?
 
Right now, it seems like the ball's curving. But let's take a look at what's really happening. My friends and I put together this little experiment using a sort of spinning seesaw. Here, the camera's standing still; it's just hanging from the ceiling. And watch what happens. From the moment the ball leaves my hand, it's just going straight. Let's watch that again, frame by frame. Keep your eye on the ball. You can see that it travels in a perfectly straight line. But what if you're spinning with the seesaw? Look at what happens when you rotate the footage so the seesaw stays sideways and everything moves around it? Now let's follow the path of the ball again but from this new perspective. The trajectory looks completely different. Now let's watch this one frame by frame. From this rotating perspective, it totally looks like the ball is curving. That's crazy because you already saw that exact same throw. Watch it again on the left. The ball is going straight, but on the right, when you're rotating with the seesaw, the ball really looks like it's curving.
You might be thinking, why should I care? I don't spend a lot of time on a spinning seesaw or on a merry-go-round. But you kind of are on a merry-go-round. The earth is constantly spinning. Have you ever seen someone in a movie making fine adjustments on their gun? Well, one effect they're correcting for is the Coriolis effect. The longest sniper shot, which was over 3,000 yards, would have had to correct for a one-foot deviation due to the Coriolis effect.
 
So, how does the Coriolis effect create hurricanes? Hurricanes form when air rushes from all directions into a low-pressure region. So, imagine there's a low-pressure region between the two of us. Air is going to rush toward the center. Let's see what that looks like from space, from our camera that's hanging from the ceiling. It's not spinning, so the air--or in this case, the balls--are clearly going in a straight line. But if you're rotating with the Earth, or with the seesaw, you'll see the air bend to the right. In the northern hemisphere, this creates hurricanes with counterclockwise spirals. In the southern hemisphere it does the opposite; it creates hurricanes with clockwise spirals. This storm on Jupiter, which is actually bigger than the Earth itself, has been spinning for around 200 years.
So, when you're in a rotating frame, the Coriolis effect seems to exert a very real force on objects. But there is no force; what you observe is just a result of your perspective. And you might be wondering, if this is simply a product of your frame of reference, then could other forces, like gravity or electromagnetism, also just be products of your frame of reference? Of where we live in the universe? Or in the multiverse? For more on that, check out our video "Do We Live in the Multiverse?" There are talking fish. And if you want to find out when new episodes come, click subscribe.
Credit: NOVA PBS [6]

Coriolis Force

The Coriolis force, −2 Ω → × U →, acts on an air parcel (or any other object) only when it is moving with respect to the Earth. It acts perpendicular to Earth's angular velocity vector and the air parcel's velocity vector. The explanation for the Coriolis force is usually broken into an explanation in the zonal (constant latitude) direction and the meridional (constant longitude) direction.

Zonal Flow (East–West Wind Velocity)

Consider an air parcel that is initially at rest and in hydrostatic balance but is impulsively accelerated to a velocity u to the east (see left side of figure below). Initially when it was at rest, it had the same acceleration as the Earth below it. However, after it accelerated to velocity u, it suddenly had more acceleration than it had before, throwing it out of hydrostatic balance. Look at the change in acceleration that comes from the air parcel suddenly acquiring a velocity to the east, which is just the acceleration after the velocity changes minus the acceleration before the velocity changes and equals (Ω+u/R)2R – Ω2R. To a very good approximation, this change equals the Coriolis force, 2Ωu. There is a vertical component that points up, but there is also a horizontal component of force that points to the right of the motion in the Northern Hemisphere and to the left in the Southern Hemisphere.

Now consider an air parcel that is initially at rest but is impulsively accelerated to a velocity u to the west (see right side of figure below). The air parcel suddenly has less angular momentum than it had before and experiences a decreased centrifugal force. This decrease in angular momentum, to a very good approximation, equals the Coriolis force, 2Ωu, but is pointed toward Earth's axis of rotation. There is a vertical component that points down, but the horizontal component of force that points to the right of the motion in the Northern Hemisphere and to the left in the Southern Hemisphere.

Coriolis force on a particle traveling zonally to the east with speed u as described in the text
Coriolis force on an air parcel traveling zonally (left) to the east, where ⨂ indicates flow into the page, and (right) to the west, where the red dot in the circle indicates flow out of the page, for the Northern Hemisphere. The air parcel speed is u and the latitude is Φ.
Credit: W. Brune

We can write down the accelerations in the y and z directions due to the air moving to the east with speed u:

Coriolis acceleration in the y direction= Dv Dt =−2Ωusinϕ MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiaaboeacaqGVbGaaeOCaiaabMgacaqGVbGaaeiBaiaabMgacaqGZbGaaeiOaiaabggacaqGJbGaae4yaiaabwgacaqGSbGaaeyzaiaabkhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabckacaqGPbGaaeOBaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaamyEaiaabckacaqGKbGaaeyAaiaabkhacaqGLbGaae4yaiaabshacaqGPbGaae4Baiaab6gacqGH9aqpdaWcaaWdaeaapeGaamiraiaadAhaa8aabaWdbiaadseacaWG0baaaiabg2da9iabgkHiTiaaikdacqqHPoWvcaWG1bGaci4CaiaacMgacaGGUbGaeqy1dygaaa@686A@

Coriolis acceleration in the z direction= Dw Dt =2Ωucosϕ MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiaaboeacaqGVbGaaeOCaiaabMgacaqGVbGaaeiBaiaabMgacaqGZbGaaeiOaiaabggacaqGJbGaae4yaiaabwgacaqGSbGaaeyzaiaabkhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabckacaqGPbGaaeOBaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaamOEaiaabckacaqGKbGaaeyAaiaabkhacaqGLbGaae4yaiaabshacaqGPbGaae4Baiaab6gacqGH9aqpdaWcaaWdaeaapeGaamiraiaadEhaa8aabaWdbiaadseacaWG0baaaiabg2da9iaaikdacqqHPoWvcaWG1bGaci4yaiaac+gacaGGZbGaeqy1dygaaa@677A@

Meridional Flow (North–South Velocity)

What about an air parcel traveling to the north at a constant altitude? Note that the air parcel moving north starts at a greater distance from Earth’s axis and comes closer to Earth’s axis if it moves at the same height above the surface. Its angular momentum is conserved, so it is moving faster to the east than the Earth beneath it. As a result, it appears to move to the right or the east.

If the same air parcel moves to the south at the same height above Earth’s surface, then it moves to a greater distance from Earth’s rotation axis. Its angular momentum becomes less than that of Earth, it slows down relative to Earth, and it veers to the right of south or to the west.

In both the zonal and meridional flow cases, the air parcel's velocity with respect to the Earth causes the air parcel to have a different angular momentum from the Earth below it. Conservation of angular momentum during that motion requires that the apparent Coriolis force be added in order to describe the observed motion. See the video below (2:11) for further explanation:

Coriolis Explanation
Click here for transcript of the Coriolis Explanation video.

Coriolis force is an apparent force that accounts for motion on a rotating sphere, such as Earth. We can break the explanation of the Coriolis force in two cases-- zonal flow, which is east west, and Meridional flow, which is north south. The explanation for both cases relies on conservation of angular momentum. For zonal flow, imagine an air parcel moving to the east with velocity, u. Angular acceleration is just the angular velocity squared times the radius of rotation. If the parcel is moving at a velocity of u relative to Earth's surface, then it has some extra angular momentum, which is u divided by r. To find the total angular acceleration that the moving air parcel has, we need to square the angular momentum of the air parcel, which is omega plus u divided by r, and then multiply it by r. We then subtract the Earth's acceleration, which is just omega squared r. The difference, to good approximation, is 2 omega times u, which is just the Coriolis force, and, in the case of eastward motion, is pointed away from Earth's axis in the northern hemisphere. Thus, the Coriolis force turns the air parcel to the right for zonal flow. If the air parcel moves to the west, then by the same argument the Coriolis force points towards Earth's rotation axis in the northern hemisphere, which again turns the air parcel to the right. The explanation for the Meridional flow is simpler. An air parcel initially has the angular momentum of the Earth at its latitude. If it moves north at the same height, then it has more angular momentum than the Earth below it. And so it goes faster than the Earth and appears to move to the right. If it moves south at the height, then it has less angular momentum than Earth and appears to slow down relative to Earth and thus appears to move to the right.

Finding the Magnitude and Direction of the Coriolis Force

The magnitude of the horizontal Coriolis force is simply 2Ω| V → |sinφ,  where φ is the latitude. This magnitude applies to both the Northern and Southern Hemispheres. The direction of the Coriolis force is 90 degrees to the right of the horizontal velocity vector in the Northern Hemisphere and 90 degrees to the left of the horizontal velocity vector in the Southern Hemisphere.

Quiz 10-1: All about forces.

  1. Find Practice Quiz 10-1 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 10-1. You will be allowed to take this quiz only once. Good luck!

10.4 Equations of Motion in Spherical Coordinates

10.4 Equations of Motion in Spherical Coordinates

The three variables used in spherical coordinates are:

  • longitude (denoted by λ)
  • latitude (denoted by φ)
  • vertical distance (denoted by r from Earth’s center and by z from Earth’s surface, where z = r – a and a is Earth’s radius)
circle showing Conversion between spherical and Cartesian coordinates as described in the text
Conversion between spherical and Cartesian coordinates.
Credit: W. Brune

Note that the unit vectors in spherical coordinates change with position. For example, for an air parcel at the equator, the meridional unit vector, j → , is parallel to the Earth’s rotation axis, whereas for an air parcel near one of the poles, j → is nearly perpendicular to the Earth’s rotation axis. In spherical coordinates, the velocity vector and its components are given by:

U → =u i → +v j → +w k → u=rcosϕ Dλ Dt ,      v=r Dϕ Dt ,      w= Dz Dt MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6067@

[10.17]

where u, v, and w are the eastward, northward, and upward components of the velocity, respectively. These velocities are derived from the changes in eastward, northward, and upward distances, which are given by:

dx=rcosϕdλ=change in eastward distance≅acosϕdλ dy=rdϕ=change in northward distance≅adϕ dz=dr=change in upward distance

Let’s now write the averaged momentum conservation equation [10.16] in component form in spherical coordinates. We will just show you how this conversion is done without actually taking you through all the steps. Note that we need to take the total derivatives of the unit vectors as well as the velocities:

D U → Dt = D Dt ( i → u+ j → v+ k → w )= i → Du Dt +u D i → Dt + j → Dv Dt +v D j → Dt + k → Dw Dt +w D k → Dt

[10.18]

The terms containing derivatives of the unit vectors are called “metric terms.” They depend on the Earth being a sphere. In Cartesian coordinates, they equal zero.

Consider just one of these metric terms:

D j → Dt = ∂ j → ∂t +u ∂ j → ∂x +v ∂ j → ∂y +w ∂ j → ∂z =0+u ∂ j → ∂x +v ∂ j → ∂y +0

Since for any location, j → is constant with time and j → does not change as a function of altitude, that leaves j → MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOAaaGaay51Gaaaaa@356E@ dependent only on latitude and longitude.

Look at ∂ j → ∂y first. Set y = 0 at the equator, and y = a (Earth’s radius) near the pole. As noted above, at the equator, j → is parallel to Earth’s rotation axis, but near the pole, it is almost perpendicular to it. Thus the change in j → going from south to north (increasing y) must be pointing down to Earth’s center and so j → changes by − k → times a small angular change while y changes by a times the same small angular change. The net result is that:

∂ j → ∂y = − k → a

Using the same approach, we can show that:

∂ j → ∂x = −tanϕ  a i →

The total derivatives of all three unit vectors are:

D i → Dt = u acosϕ ( j → sinϕ− k → cosϕ ) D j → Dt =− utanϕ a i → − v a k → D k → Dt = u a i → + v a j →

[10.19]

Putting this all together:

D U → Dt =( Du Dt − uvtanϕ a + uw a ) i → +    ( Dv Dt + u 2 tanϕ a + vw a ) j → +    ( Dw Dt − u 2 + v 2 a ) k →

[10.20]

Similar analysis can be done for the other terms in the averaged momentum equation.

Coriolis force:

−2 Ω → × U → =2Ω( vsinϕ−wcosϕ ) i → −( 2Ωusinϕ ) j → +( 2Ωucosϕ ) k → MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiabgkHiTiaaikdadaWhcaqaaiabfM6axbGaay51GaGaey41aq7aa8HaaeaacaWGvbaacaGLxdcacqGH9aqpcaaIYaGaeuyQdC1aaeWaa8aabaWdbiaadAhaciGGZbGaaiyAaiaac6gacqaHvpGzcqGHsislcaWG3bGaci4yaiaac+gacaGGZbGaeqy1dygacaGLOaGaayzkaaGabmyAa8aagaWca8qacqGHsisldaqadaqaaiaaikdacqqHPoWvcaWG1bGaci4CaiaacMgacaGGUbGaeqy1dygacaGLOaGaayzkaaGabmOAa8aagaWca8qacqGHRaWkdaqadaqaaiaaikdacqqHPoWvcaWG1bGaci4yaiaac+gacaGGZbGaeqy1dygacaGLOaGaayzkaaGabm4Aa8aagaWcaaaa@663A@

[10.21]

Gravity:

g → =−g k →

[10.22]

Pressure Gradient Force (PGF):

− 1 ρ ∇ → p=− 1 ρ ∂p ∂x i → − 1 ρ ∂p ∂y j → − 1 ρ ∂p ∂z k →

[10.23]

Turbulent friction (in the boundary layer only):

− C d h | V → | V → =− C d h | V → |u i → − C d h | V → |v j →

[10.24]
Adding together all of the forces, the averaged momentum equations in spherical coordinates in the zonal, meridional, and vertical directions are, respectively:

Du Dt − uvtanϕ a + uw a =− 1 ρ ∂p ∂x +2Ωv sinϕ−2Ωw cosϕ− C d h | V → |u Dv Dt − u 2 tanϕ a + vw a =− 1 ρ ∂p ∂y −2Ωu sinϕ− C d h | V → |v Dw Dt − u 2 + v 2 a =− 1 ρ ∂p ∂z −g+2Ωu cosϕ

10.5 Are all the terms in these equations equally important? Let's use scale analysis.

10.5 Are all the terms in these equations equally important? Let's use scale analysis.

Our equations of motion have several terms in them, and the question is, “Which ones are the largest and, thus, the most important?” The answer is, “It depends on the situation.” You can follow the following steps to determine which terms to keep and which you can safely ignore when you are trying to calculate the force balance for a specific atmospheric phenomenon. This process is called scale analysis and it can be applied to any conservation equation you would like to simplify.

  1. Decide on the phenomenon of interest (e.g., cyclone, front, hurricane, tornado).
  2. Determine the characteristic (i.e., typical) lengths and times over which the phenomenon occurs.
  3. Determine the range of fluctuations of equation variables in space and time during the phenomenon.
  4. Approximate derivatives (i.e., ∂p ∂x ∼ Δp Δx MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaamiCaaqaaiabgkGi2kaadIhaaaGaeSipIOZaaSaaaeaacqqHuoarcaWGWbaabaGaeuiLdqKaamiEaaaaaaa@3D90@ ).
  5. Compare the magnitudes of the terms in the equation.
  6. Keep only the relatively large terms (say, the top two orders of magnitude) and neglect the smaller terms.

The Coriolis Parameter

Define the Coriolis parameter as f≡2Ωsinϕ MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiaadAgacqGHHjIUcaaIYaGaeuyQdCLaci4CaiaacMgacaGGUbGaeqy1dygaaa@3D3D@ . At 45o N,   f=( 2 )( 7.27× 10 −5 s −1 )sin 45 o ∼ 10 −4 s −1 MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiaadAgacqGH9aqpdaqadaWdaeaapeGaaGOmaaGaayjkaiaawMcaamaabmaapaqaa8qacaaI3aGaaiOlaiaaikdacaaI3aGaey41aqRaaGymaiaaicdapaWaaWbaaSqabeaapeGaeyOeI0IaaGynaaaakiaabohapaWaaWbaaSqabeaapeGaeyOeI0IaaGymaaaaaOGaayjkaiaawMcaaiaabohacaqGPbGaaeOBaiaaisdacaaI1aWdamaaCaaaleqabaWdbiaab+gaaaGccqWI8iIocaaIXaGaaGima8aadaahaaWcbeqaa8qacqGHsislcaaI0aaaaOGaae4Ca8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaaa@522F@ . We will use this parameter in the scale analysis in the next section and then throughout the rest of the lesson.

Example: Scale analysis of the averaged x-momentum equation for mid-latitude synoptic-scale flow in the free troposphere.

1. Phenomenon: mid-latitude synoptic-scale flow in the free troposphere.

2. L∼1000 km= 10 6  m; H∼10 km= 10 4  m; (in boundary layer only,  C d ∼ 10 −2  and h∼1000 m); t∼? MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@807B@

3.

U∼u; v∼10  m s −1 ; Δp∼10 hPa= 10 3  Pa ( in the horizontal ) t∼L/u=( 10 6  m )/( 10  m s −1 )= 10 5  s∼1 day f o  ∼2Ωsin ϕ o ∼2Ωcos ϕ o ∼ 10 −4   s −1 w∼H/t= 10 −1   m s −1 ; a∼ 10 7  m; ν∼ 10 −5   m 2  s –1 ; ρ∼1  kg m −3 MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D5C8@

4 and 5.

term magnitude (variables) magnitude (m s-2)
Du Dt = MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamiraiaadwhaaeaacaWGebGaamiDaaaacqGH9aqpaaa@3766@ U t MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamyvaaqaaiaadshaaaaaaa@34AE@ 10 10 5 = 10 −4 MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaGymaiaaicdaaeaacaaIXaGaaGimamaaCaaaleqabaGaaGynaaaaaaGccqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGinaaaaaaa@3B0E@
2Ωvsinϕ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacqqHPoWvcaWG2bGaci4CaiaacMgacaGGUbGaeqy1dygaaa@3AAF@ f o U MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAgadaWgaaWcbaGaam4BaaqabaGccaWGvbaaaa@35B9@ 10 −4 10= 10 −3 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaOGaaGymaiaaicdacqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG4maaaaaaa@3BE8@
−2Ωwcosϕ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaaikdacqqHPoWvcaWG3bGaci4yaiaac+gacaGGZbGaeqy1dygaaa@3B98@ f o W MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAgadaWgaaWcbaGaam4BaaqabaGccaWGxbaaaa@35BB@ 10 −4 10 −1 = 10 −5 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaOGaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdaaaGccqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaaa@3DC9@
uw a MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamyDaiaadEhaaeaacaWGHbaaaaaa@35B6@ UW a MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamyvaiaadEfaaeaacaWGHbaaaaaa@3576@ ( 10 )( 10 −1 ) 10 7 = 10 −7 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaWaaeWaaeaacaaIXaGaaGimaaGaayjkaiaawMcaamaabmaabaGaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdaaaaakiaawIcacaGLPaaaaeaacaaIXaGaaGimamaaCaaaleqabaGaaG4naaaaaaGccqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG4naaaaaaa@4178@
uvtanϕ a MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamyDaiaadAhaciGG0bGaaiyyaiaac6gacqaHvpGzaeaacaWGHbaaaaaa@3A4E@ U 2 a MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamyvamaaCaaaleqabaGaaGOmaaaaaOqaaiaadggaaaaaaa@358D@ 10 2 10 7 = 10 −5 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaGymaiaaicdadaahaaWcbeqaaiaaikdaaaaakeaacaaIXaGaaGimamaaCaaaleqabaGaaG4naaaaaaGccqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaaa@3C03@
− 1 ρ ∂p ∂x MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaaGymaaqaaiabeg8aYbaadaWcaaqaaiabgkGi2kaadchaaeaacqGHciITcaWG4baaaaaa@3B10@ Δp ρL MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeuiLdqKaamiCaaqaaiabeg8aYjaadYeaaaaaaa@37C6@ 10 3 ( 1 )( 10 6 ) = 10 −3 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaGymaiaaicdadaahaaWcbeqaaiaaiodaaaaakeaadaqadaqaaiaaigdaaiaawIcacaGLPaaadaqadaqaaiaaigdacaaIWaWaaWbaaSqabeaacaaI2aaaaaGccaGLOaGaayzkaaaaaiabg2da9iaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIZaaaaaaa@3FCE@
− C d h | V → | V → C d U 2 h 10 −2 10 2 10 3 = 10 −3
ν ∇ 2 u MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqqaaaaaaaaGySf2yRbWdbiabe27aUjabgEGir=aadaahaaWcbeqaa8qacaaIYaaaaOGaamyDaaaa@38E9@ νU H 2 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamODaiaadwfaaeaacaWGibWaaWbaaSqabeaacaaIYaaaaaaaaaa@3665@ ( 10 −5 )( 10 ) 10 8 = 10 −12 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaWaaeWaaeaacaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaOGaayjkaiaawMcaamaabmaabaGaaGymaiaaicdaaiaawIcacaGLPaaaaeaacaaIXaGaaGimamaaCaaaleqabaGaaGioaaaaaaGccqGH9aqpcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaikdaaaaaaa@4233@

6. Using just the most important (i.e., biggest) terms and knowing that the turbulent drag term goes to zero above the atmospheric boundary layer, we can write a simplified x-momentum balance as:

0=− 1 ρ ∂p ∂x +fv
[10.24]

Using the same scale analysis with the y-momentum equation, we can write a simplified y-momentum balance as:

0=− 1 ρ ∂p ∂y −fu
[10.25]

Only two terms remain in both equations. One term is the pressure gradient force and the other is the Coriolis force. Since Du Dt MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamiraiaadwhaaeaacaWGebGaamiDaaaaaaa@3660@ is much smaller than either the pressure gradient force or the Coriolis force, these two forces must be about in balance. We call this balance the Geostrophic Balance. It is very important for understanding atmospheric dynamics and we will talk about its consequences in more detail later.

The molecular friction term is the smallest of all the terms for the case of large-scale flow in the atmosphere. This term is almost always very small for most meteorological phenomena, which is why we had eliminated it from the averaged momentum conservation equation earlier.

We ignored the acceleration term, Du/Dt, because it is an order of magnitude smaller than the other two terms. We often must keep all terms that are within an order of magnitude of each other because our approximations may bias our results one way or the other. For instance, if we say that the velocity is 10 m s–1, the spatial scale is 100 km, and the pressure change is 10 hPa when more accurate numbers are more like 20 m s–1, 50 km, and 5 hPa, then we would be off almost an order of magnitude in our value for the centrifugal force, but get the same order of magnitude for the pressure gradient force. So, terms that are two orders of magnitude smaller than the rest you can easily neglect, but think carefully about terms that are only one order of magnitude different. For example, for very intense low-pressure systems, Du Dt MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamiraiaadwhaaeaacaWGebGaamiDaaaaaaa@3660@ must be considered because it can become about as large as the pressure gradient force and the Coriolis force.

When scale analysis is applied to the z-momentum equation for mid-latitude synoptic-scale flow, the result is the simplified z-momentum balance:

0=− 1 ρ ∂p ∂z −g
[10.26]

If you rearrange Equation [10.26], you will get the hydrostatic equilibrium equation [5], Equation [2.18]. Previously we derived it using a balance of forces on a slab of air, but here it comes naturally out of the z-momentum equation.

The following video (4:19) provides further explanation on how to complete the above example:

Scale Analysis video
Click here for transcript of the scale analysis video.

Scale analysis is very important. Because it tells us which terms in any equation are the most important and which terms we can ignore. In scale analysis you do not need to know the exact values for the variables. But instead you need to only know their order of magnitude. The process is straightforward. First, determine the phenomenon of interest whether it be cyclone, front, hurricane, tornado, synoptic-scale, winter weather. Determine the characteristic-- that is typical lengths and times-- over which phenomenon occurs. Determine the range of fluctuations of equation variables in space and time during the phenomenon. Approximate derivatives, that is the partial of p with respect to x would become delta p over delta x where they're roughly estimated. Compare the magnitudes of terms in the equation. And then keep only the relatively large terms-- say the top two orders of magnitude-- and neglect the much smaller terms. Let's look at this example of the x momentum equation for mid-latitude synoptic-scale flow. So in this case it's mid-latitude synoptic-scale flow. The lake is about 1,000 kilometers, which is 10 and 6 meters. The height is about 10 kilometers, which is 10 to the 4 meters. And if we were in the boundary layer only, we would find that the friction drag coefficients is 10 to the minus 2. And the height of the boundary layer is about 1,000 meters. Now we know that u is about 10 meters a second, roughly. It could be a lot less and a lot more. But it's that order of magnitude. Delta p is about 10 millibar over the length of interest. We see that the time then is equal to the scale of the synoptic-scale flow divided by the velocity, which is 10 to the 6 divided by 10, or 10 to the 5 seconds which is about a day. And we see that the Coriolis parameter is about 10 to the minus 4 per second. And we can estimate other factors, such as the w velocity which is height divided by time. So that's about 10 to the minus 1 meters per second. And so on. We continue on looking at derivatives and other terms. And so, for instance, the acceleration in the u direction is about 10 meters per second divided by 10 to the 5, which is about 10 to the minus 4. And so that's the size of that term. We see that the Coriolis term is about 10 to the minus 3. We see that other apparent terms are 10 to the minus 5 to 10 to the minus 7. They're quite a bit smaller. The pressure gradient force we see is 1 over the density, which is about 1 kilogram per meter cubed times the pressure difference which is about 10 to the 3 pascals divided by the distance, which is 10 to the 6 meters. So it's about 10 to minus 3. And we see that if we were in the boundary layer that the aerodynamic drag which causes friction is acting as friction. It's about 10 to the minus 3. So in the boundary layer we would need to consider this term because it's the same order of magnitude as the pressure gradient term and one of the larger terms. When we're not in the boundary layer then c sub d is actually very, very small. And this term is very small. We can ignore it. The last term is viscosity which is true friction. And we can see that for the case of viscosity is tiny. And therefore we can always ignore it for synoptic-scale flow. So when we look at the terms we have we see that we have away from the boundary layer we have two terms the count. That is we have the Coriolis term. And we have the pressure gradient. And those are the only two terms that we need to keep.

10.6 Why do weather maps use pressure surfaces instead of height surfaces?

10.6 Why do weather maps use pressure surfaces instead of height surfaces?

You can’t find a weather map of winds at 5000 m, but you can find one for 500 mb, which is about the same altitude as 5000 m (see figure below).

weather map - white with black lines
500 mb winds for 22 February 2015. Numbers are height above sea level in decameters (dam, tens of meters). For example, the height of the 500 mb surface in the high is 5940 m = 594 dam.
Credit: NOAA

We will learn why weather maps use pressure as the vertical coordinate, but for now, we will show that higher altitudes on a constant-pressure surface correspond to higher pressures on an constant-altitude surface.

If we look down on the Earth, then we can plot the isobars as a function of latitude (y) and longitude (x). We can make a second plot of height surfaces on a constant-pressure surface (see figure below). Generally the pressure on average is greater at the equator on a given height surface than it is at the poles. This tilt makes sense if you think about the hydrostatic equilibrium equation because the temperature is greater at the equator than at the poles. Therefore, the scale height is greater, and so pressure decreases with height more gradually at the equator than it does at the poles.

N. pole 2 equator on one axis, Z on other. solid, diagonal lines (490, 500, 510 hPa) intersect horizontal dashed lines (5550, 5400, 5250m)
A schematic of constant-pressure surfaces intersecting with constant-height surfaces from the pole to the equator.
Credit: W. Brune

We can arbitrarily choose one height surface and see how the pressure changes as a function of latitude. We see that it increases from pole to equator (see figure below).

isobars. 3 vertical lines rising from the y axis
A schematic of isobars (constant pressure) on an x–y horizontal map at a constant height surface of 5400 m, as in the figure above.
Credit: W. Brune

If we now arbitrarily choose a constant-pressure surface of, say, 500 mb, then the change in the height on an x–y horizontal plot on the pressure surface also shows an increase from pole to equator (see figure below).

4 vertical lines coming out of y axis, labeled 5250m, 5400m, & 5550m all at pressure = 500hPa
A schematic of lines of constant height on an x–y horizontal map at a constant pressure surface of 500 hPa, as in the figure above.
Credit: W. Brune

Thus, low pressure on constant-height surfaces is related to low heights on constant-pressure surfaces. As a result of the hydrostatic approximation, for every height there is a unique pressure, so we can replace z with p as the vertical coordinate. We can then look at changes in variables as a function of x and y, but instead of doing this on a constant-height surface, we can do it on a constant-pressure surface.

We can now show how the equations of motion change when the vertical coordinate is switched from the height, z, to the pressure, p.

Consider first the pressure gradient force (PGF). The figure below provides a schematic of the math.

parallel diagonal lines. top labeled P0, bottom P0 +delta p, triangle made by delta x and delta z between lines increasing z, increasing p
A cross section of pressure change on a x–z surface. Note that Δx, Δz, and Δp are all positive and Δp is the same in the horizontal and vertical.
Credit: W. Brune (after R. Najjar)

The slope of the isobar is just the change in z divided by the change in x on an isobar:

Δz Δx → ( ∂z ∂x ) p MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeuiLdqKaamOEaaqaaiabfs5aejaadIhaaaGaeyOKH46aaeWaaeaadaWcaaqaaiabgkGi2kaadQhaaeaacqGHciITcaWG4baaaaGaayjkaiaawMcaamaaBaaaleaacaWGWbaabeaaaaa@4111@

where the subscript “p” means “constant pressure.” Since Δp is the same in the vertical and the horizontal:

( ∂p ∂x ) z Δx=− ( ∂p ∂z ) x Δz ( ∂p ∂x ) z =− ( ∂p ∂z ) x ( ∂z ∂x ) p MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@613C@

where the subscripts "x" and "z" mean constant eastward distance and constant height, respectively.

Multiplying both sides by 1/ρ and using the hydrostatic equilibrium equation:

− 1 ρ ( ∂p ∂x ) z = 1 ρ ( ∂p ∂z ) x ( ∂z ∂x ) p =−g ( ∂z ∂x ) p or − 1 ρ ( ∂p ∂x ) z =−g ( ∂z ∂x ) p MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@723A@

[10.27]

Same for the y direction:

− 1 ρ ( ∂p ∂y ) z =−g ( ∂z ∂y ) p MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaaGymaaqaaiabeg8aYbaadaqadaqaamaalaaabaGaeyOaIyRaamiCaaqaaiabgkGi2kaadMhaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadQhaaeqaaOGaeyypa0JaeyOeI0Iaam4zamaabmaabaWaaSaaaeaacqGHciITcaWG6baabaGaeyOaIyRaamyEaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamiCaaqabaaaaa@4831@

[10.28]

So, the geostrophic balance (Equation [10.24], [10.25]) in pressure coordinates becomes:

x-momentum equation:      0=−g ∂z ∂x +fv y-momentum equation:      0=−g ∂z ∂y −fu MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7694@

[10.29]

These equations can be rearranged to give the horizontal velocity on the pressure surface and can be rewritten as one in vector form:

V g → = g f k → × ∇ → p z MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamODamaaBaaaleaacaWGNbaabeaaaOGaay51GaGaeyypa0ZaaSaaaeaacaWGNbaabaGaamOzaaaadaWhcaqaaiaadUgaaiaawEniaiabgEna0oaaFiaabaGaey4bIenacaGLxdcadaWgaaWcbaGaamiCaaqabaGccaWG6baaaa@43A7@

[10.30]

where V g → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamODamaaBaaaleaacaWGNbaabeaaaOGaay51Gaaaaa@369B@ is designated geostrophic velocity and ∇ → p z= i → ∂z ∂x + j → ∂z ∂y MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaapeGafy4bIe9dayaalaWaaSbaaSqaa8qacaWGWbaapaqabaGcpeGaamOEaiabg2da9iqadMgapaGbaSaapeWaaSaaa8aabaWdbiabgkGi2kaadQhaa8aabaWdbiabgkGi2kaadIhaaaGaey4kaSIabmOAa8aagaWca8qadaWcaaWdaeaapeGaeyOaIyRaamOEaaWdaeaapeGaeyOaIyRaamyEaaaaaaa@453E@ is on a constant pressure surface.

Geopotential

We can write these equations a little differently by using the concept of geopotential. Geopotential, Φ, is the potential energy per unit mass of an air parcel at a height z, where zero potential energy is defined at the surface (z = 0).

Φ= ∫ o z gdz MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiabfA6agjabg2da9maawahabeWcpaqaa8qacaWGVbaapaqaaiaadQhaa0qaa8qacqGHRiI8aaGccaWGNbGaamizaiaadQhaaaa@3CE0@

[10.31]

dΦ=gdz MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaabsgacqqHMoGrcqGH9aqpcaWGNbGaamizaiaadQhaaaa@391B@

[10.32]

In vector form, the velocities become:

V → g = 1 f k → × ∇ → p Φ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamODaaGaay51GaWaaSbaaSqaaiaadEgaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaamOzaaaadaWhcaqaaiaadUgaaiaawEniaiabgEna0oaaFiaabaGaey4bIenacaGLxdcadaWgaaWcbaGaamiCaaqabaGccqqHMoGraaa@43F1@

[10.33]
dot with 4 labeled arrows. PGF going left, coriolis going right, Vg(NH) goes up, Vg(SH) goes down. phi and phi+delta phi on either side
Balance between Coriolis and pressure gradient forces and the resulting velocities of the wind in the Northern Hemisphere (NH) and Southern Hemisphere (SH).
Credit: W. Brune

A major advantage of using pressure coordinates is that the gradient of z or Φ is proportional to V g → MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamODamaaBaaaleaacaWGNbaabeaaaOGaay51Gaaaaa@369B@ for all pressure levels. This statement is not true for pressure gradients on height levels because you must know the density, ρ, as in Equations [10.24] and [10.25] [7], which varies dramatically with height.

The following video (1:19) provides a good overview of pressure surfaces:

Pressure Surfaces
Click here for a transcript of the Pressure Surfaces video.

This schematic shows the relationship between height surfaces and pressure surfaces. Typically, pressure surfaces slope downward in height from the equator, where it is warmer, to the pole, where it is colder. You were able to show this in Activity 2.2, and you saw it again in Lesson 2.4 on thickness. Note that on the constant height surface, from the equator to the pole, the pressure surface is decreased with latitude. Now it also, on a constant pressure surface from the equator to the pole, the height surface is decreased on the constant pressure surface. Thus, changes in pressure are proportional to changes in height. After a little math, we can show that 1/rho-- that is, one over the density-- times the change in pressure with respect to x or y on a height surface is equal to g times the change in z with respect to x or y on a constant pressure surface. Finally, we note that gdz is just a differential of the geopotential phi, which has units of meters squared per second squared, which are the same units as energy divided by mass. So changes in height on a constant pressure surface are the same as changes in geopotential on a constant pressure surface.

10.7 Natural coordinates are better horizontal coordinates.

10.7 Natural coordinates are better horizontal coordinates.

Understanding the results of a balance of forces can often be easier if we choose a horizontal coordinate system that is aligned naturally with the air flow, and not just set up in Cartesian coordinates x and y or spherical coordinates λ and φ. We can choose one direction—let’s call it s for streamline—so that it is aligned with the streamline (and is thus always parallel with the flow) and increases in the direction of the flow (i.e., downwind). The second direction—let’s call it n for normal—increases to the left of the flow. An interesting and noteworthy feature of natural coordinates is that the wind velocity is always positive because, by the definition of natural coordinates, the velocity vector is always pointed in the positive s direction.

For the horizontal momentum equation without friction:

D V → Dt =−f  k → × V → − ∇ → p Φ

where V → = i → u+ j → v and ∇ → p MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaey4bIenacaGLxdcadaWgaaWcbaGaamiCaaqabaaaaa@3725@ is the horizontal gradient operator on a pressure surface, not a height surface.

Let’s look at each term in the equation and put it in natural coordinates.

The acceleration of the air parcel: D V → Dt

There are two ways that V → can change. It can change by changing its speed ( V=| V → | ), DV Dt , which occurs in the streamline direction s, or by changing its direction, which occurs in the normal direction, n. Recall that n points to the left of the velocity vector.

Remember your physics, which showed that the acceleration due to rotation (the centripetal acceleration) is the velocity squared divided by the radius of curvature of the rotation. In this case, the rotation is the rotation of the air parcel as it moves horizontally over the Earth (not the rotation of the Earth itself). The acceleration is directed perpendicular to the motion and is directed towards the center of curvature. R is the radius of curvature of the changing direction, and by convention, R > 0 if the curvature is in the counterclockwise direction, and R < 0 if the curvature is in the clockwise direction. Hence the sign always works out if the centripetal acceleration is V2/R.

The Coriolis force:

−f  k → × V→ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaadAgacaaMc8+aa8HaaeaacaWGRbaacaGLxdcacqGHxdaTdaWhcaqaaiaadAhaaiaawEniamaaBaaaleaacaWGibaabeaaaaa@3E90@

The n component equals −fV  MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaadAgacaaMc8UaamODamaaBaaaleaacaWGibaabeaaaaa@3821@ because when f > 0, the force is to the right of the motion (i.e., streamline) and is, thus, negative because positive n points to the left of the motion. When f < 0, the force would be to the left in the positive n direction.

The pressure gradient force (PGF):

− ∇ → p Φ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaaFiaabaGaey4bIenacaGLxdcadaWgaaWcbaGaamiCaaqabaGccqqHMoGraaa@3996@

The n component of − ∇ → p Φ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaaFiaabaGaey4bIenacaGLxdcadaWgaaWcbaGaamiCaaqabaGccqqHMoGraaa@3996@ is just − ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3901@ .

Putting these pieces all together results in the equation for the n component of the horizontal momentum equation:

V 2 R =−fV− ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaadAfapaWaaWbaaSqabeaapeGaaGOmaaaaaOWdaeaapeGaamOuaaaacqGH9aqpcqGHsislcaWGMbGaamOvaiabgkHiTmaalaaapaqaa8qacqGHciITcqqHMoGra8aabaWdbiabgkGi2kaad6gaaaaaaa@4228@

[10.34a]

This equation is called the Gradient Wind Equation. The form of the equation in z coordinates is sometimes useful:

V 2 R =−fV− 1 ρ ∂p ∂n MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaadAfapaWaaWbaaSqabeaapeGaaGOmaaaaaOWdaeaapeGaamOuaaaacqGH9aqpcqGHsislcaWGMbGaamOvaiabgkHiTmaalaaabaGaaGymaaqaaiabeg8aYbaadaWcaaWdaeaapeGaeyOaIyRaamiCaaWdaeaapeGaeyOaIyRaamOBaaaaaaa@442E@

[10.34b]

Keep in mind that V is always positive in Equations [10.34] because they are cast in natural coordinates.

Let’s rearrange Equation [10.34a] slightly by moving the centripetal acceleration over to the right hand side, where it represents a centrifugal force. Then, by considering which of the terms are the most important, we can apply this equation to different situations:

− V 2 R −fV− ∂Φ ∂n =0,     gradient balance    −fV− ∂Φ ∂n =0,     geostrophic balance − V 2 R     − ∂Φ ∂n =0,     cyclostrophic balance − V 2 R −fV      =0,     inertial balance MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C58C@

[10.35]

We can use the Rossby number to help determine which of the above balances dominates. The Rossby number is defined to be the magnitude of the acceleration divided by the magnitude of the Coriolis force. When applied to the gradient wind equation, the Rossby number is given by

R o = acceleration Coriolis force = V 2 /R fV = V fR MathType@MTEF [2]@5@5@+=faaahmart1ev3aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadkfadaWgaaWcbaGaam4BaaqabaGccqGH9aqpdaWcaaqaaiaabggacaqGJbGaae4yaiaabwgacaqGSbGaaeyzaiaabkhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaaqaaiaaboeacaqGVbGaaeOCaiaabMgacaqGVbGaaeiBaiaabMgacaqGZbGaaeiiaiaabAgacaqGVbGaaeOCaiaabogacaqGLbaaaiabg2da9maalaaabaWaaSGbaeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOuaaaaaeaacaWGMbGaamOvaaaacqGH9aqpdaWcaaqaaiaadAfaaeaacaWGMbGaamOuaaaaaaa@56D5@

If Ro << 1, then the centrifugal force is much smaller than the Coriolis force and the Coriolis force must balance the PGF (geostrophic balance).

If Ro >> 1, then the Coriolis force is much smaller than the centrifugal force and the centrifugal force must balance the PGF (cyclostrophic balance).

If Ro ~ 1, then the Coriolis force and the centrifugal force balance each other (inertial balance).

Quiz 10-2: Coordinates and scales.

  1. Find Practice Quiz 10-2 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 10-2. You will be allowed to take this quiz only once. Good luck!

10.8 A Closer Look at the Four Force Balances

10.8 A Closer Look at the Four Force Balances

Geostrophic Balance

This balance occurs often in atmospheric flow that is a straight line (R = ±∞) well above Earth’s surface, so that friction does not matter. The Rossby number, Ro, is much less than 1.

Let’s think about how this balance might occur. Assume that an air parcel is placed in the midst of a fixed horizontal pressure gradient in the Northern Hemisphere and is initially at rest (V → =0 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaFiaabaGaamOvaaGaay51GaGaeyypa0JaaGimaaaa@3719@), as shown in the figure below. The Coriolis force is thus zero and the parcel begins to move from high pressure toward low pressure. However, as the parcel accelerates and attains a velocity parallel to the pressure gradient, the Coriolis force develops perpendicular and to the right of the velocity vector and the PGF. The resulting acceleration is now the vector sum of the PGF and the Coriolis force and turns the parcel to the right. As the velocity continues to increase, the Coriolis force increases but always stays perpendicular and to the right of the velocity vector while the PGF always stays parallel to the pressure gradient. Eventually, the PGF and Coriolis force become equal and opposite and the air parcel will move perpendicular to the horizontal pressure gradient. This condition is called geostrophic balance. We have simplified somewhat the approach to geostrophic equilibrium because, in reality, air parcels would overshoot and undergo inertial oscillations (discussed below) and because the pressure field would evolve in response to the motion.

diagram of geostrophic balance. More details in the video below "Into Geostrophic Balance"
How geostrophic balance is achieved for an air parcel starting at rest. The PGF is always there, but the Coriolis force is zero until the air parcel acquires some velocity. In the figure, vg is used to represent the geostrophic velocity.
Credit: H.N. Shirer

−f V − ∂Φ ∂n =0          geostrophic wind balance MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaadAgacaWG2bWaaSbaaSqaaiaadIeaaeqaaOGaeyOeI0YaaSaaaeaacqGHciITcqqHMoGraeaacqGHciITcaWGUbaaaiabg2da9iaaicdacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae4zaiaabwgacaqGVbGaae4CaiaabshacaqGYbGaae4BaiaabchacaqGObGaaeyAaiaabogacaqGGaGaae4DaiaabMgacaqGUbGaaeizaiaabccacaqGIbGaaeyyaiaabYgacaqGHbGaaeOBaiaabogacaqGLbaaaa@5A9B@

[10.36]

Let the Coriolis force per unit mass be designated as Co=−f V MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadoeacaWGVbGaeyypa0JaeyOeI0IaamOzaiaadAhadaWgaaWcbaGaamisaaqabaaaaa@3958@ and the pressure gradient force per unit mass as P=− ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfacqGH9aqpcqGHsisldaWcaaqaaiabgkGi2kabfA6agbqaaiabgkGi2kaad6gaaaaaaa@3ADB@ . Then the force balances are shown in the figure below.

Geostrophic force balance in (a) the Northern Hemisphere and (b) the Southern Hemispherezes!
Geostrophic force balance in (a) the Northern Hemisphere and (b) the Southern Hemisphere shown in natural coordinates. Note that the n direction is always to the left of the velocity when looking downwind. In the figure, Vg is used to represent the geostrophic velocity.
Credit: W. Brune (after R. Najjar and An Introduction to Dynamic Meteorology, Fifth Edition, J. R. Holton and G. J. Hakim, 2013)

Note that the Coriolis force is always to the right of the velocity vector in the Northern Hemisphere. It is always to the left of the velocity vector in the Southern Hemisphere. When the pressure gradient force and Coriolis force are in balance, the PGF is to the left of the velocity vector and the Coriolis force is to the right in the Northern Hemisphere. Watch the video below (1:10) for further explanation:

Into Geostrophic Balance
Click here for transcript of the into geostrophic balance video.

Let's see how an air partial initially at rest achieves geostrophic balance. At rest, the air parcel velocity equals 0. And the only horizontal force acting on the parcel is the pressure gradient force, which has a constant magnitude and direction as long as the pressure gradient remains the same. As soon as the parcel has some velocity, the Coriolis force starts, perpendicular and to the right of velocity in the northern hemisphere. The Coriolis force begins to move the parcel to the right because the sum of forces on the parcel now has a y component. Note that the PGF is still always perpendicular to the pressure gradient, and the Coriolis force is always perpendicular to the velocity. Eventually, the PGF and the Coriolis force come into opposition with the velocity in between and Coriolis to the right of the velocity. In the end, the y component of the forces is 0 again so that the air parcel remains at the geostrophic velocity.

Inertial Balance

In this case, the pressure gradient force is minimal and the centrifugal and Coriolis forces are in balance.

− V 2 R −fV=0          inertial balance MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaamODamaaBaaaleaacaWGibaabeaakmaaCaaaleqabaGaaGOmaaaaaOqaaiaadkfaaaGaeyOeI0IaamOzaiaadAhadaWgaaWcbaGaamisaaqabaGccqGH9aqpcaaIWaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabMgacaqGUbGaaeyzaiaabkhacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabkgacaqGHbGaaeiBaiaabggacaqGUbGaae4yaiaabwgaaaa@51F1@

[10.37]

Let the centrifugal force be designated by Ce=− V 2 R MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadoeacaWGLbGaeyypa0JaeyOeI0YaaSaaaeaacaWG2bWaaSbaaSqaaiaadIeaaeqaaOWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOuaaaaaaa@3A47@ .

two circles depicting Inertial balance in (a) the Northern Hemisphere and (b) the Southern Hemisphere as described in the text
Inertial balance in (a) the Northern Hemisphere and (b) the Southern Hemisphere shown in natural coordinates. Note that the n direction is always to the left of the velocity when looking downwind.
Credit: W. Brune (after R. Najjar)

We can manipulate Equation [10.37] to find the radius of the circle:

R=− V f MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadkfacqGH9aqpcqGHsisldaWcaaqaaiaadAhadaWgaaWcbaGaamisaaqabaaakeaacaWGMbaaaaaa@388D@

For f = 10–4 s–1 and V = 10 m s–1, R = –100 km. Inertial balance is not a major balance in the atmosphere because there is almost always a significant pressure gradient, but it can be important in oceans.

Cyclostrophic Balance

The balance in this case is between the pressure gradient force and the centrifugal force.

− V 2 R  − ∂Φ ∂n =0 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaamODamaaBaaaleaacaWGibaabeaakmaaCaaaleqabaGaaGOmaaaaaOqaaiaadkfaaaGaaGPaVlabgkHiTmaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaacqGH9aqpcaaIWaaaaa@4110@

[10.38]
two circles depicting Cyclostrophic balance in (a) the Northern Hemisphere and (b) the Southern Hemisphere as described in the text.
Cyclostrophic balance for (a) cyclonic flow and (b) anticyclonic flow in the Northern Hemisphere shown in natural coordinates. Note that the n direction is always to the left of the velocity when looking downwind. In the Southern Hemisphere, (b) is cyclonic and (a) is anticyclonic.
Credit: W. Brune (after R. Najjar and An Introduction to Dynamic Meteorology, Fifth Edition, J. R. Holton and G. J. Hakim, 2013)

In this case, the scale of the motion is so small that Coriolis acceleration is not important. The Rossby number, Ro = centrifugal acceleration / Coriolis >> 1.

Examples of motion in cyclostrophic balance are tornadoes, dust devils, water spouts, and other small atmospheric circulations, such as the vortex you sometimes see when leaves get swept off the ground. These can be either cyclonic or anticyclonic and, in fact, a few percent of tornadoes in the Northern Hemisphere are anticyclonic. Another common example of cyclostrophic balance is the vortex seen when a bathtub or sink is draining.

water spout over water next to boat
NOAA Ship NANCY FOSTER dwarfed by waterspout. Gulf of Mexico. Summer, 2007.
Credit: NOAA Photo Library [8] via flickr

Gradient Balance

In gradient balance, the pressure gradient force, Coriolis force, and horizontal centrifugal force are all important. This balance occurs as wind in a pressure gradient field goes around a curve. There are many examples of this type of flow on any weather map—any synoptic-scale pressure gradient for which the isobars curve is an example of gradient flow.

− V 2 R −fV− ∂Φ ∂n =0  MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaalaaabaGaamODamaaBaaaleaacaWGibaabeaakmaaCaaaleqabaGaaGOmaaaaaOqaaiaadkfaaaGaeyOeI0IaamOzaiaadAhadaWgaaWcbaGaamisaaqabaGccqGHsisldaWcaaqaaiabgkGi2kabfA6agbqaaiabgkGi2kaad6gaaaGaeyypa0JaaGimaiaabccaaaa@43FE@

[10.39]

To solve this equation for velocity, we can use the quadratic equation:

a x 2 +bx+c=0 → x= −b± b 2 −4ac 2a V 2 R +f V + ∂Φ ∂n =0 →  V = −f± f 2 −4( 1 R ) ∂Φ ∂n 2 R MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7725@

V=− f R 2 ± f 2 R 2 4 −R ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaamisaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaadAgacaaMi8UaamOuaaqaaiaaikdaaaGaeyySae7aaOaaaeaadaWcaaqaaiaadAgadaahaaWcbeqaaiaaikdaaaGccaWGsbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGinaaaacqGHsislcaWGsbWaaSaaaeaacqGHciITcqqHMoGraeaacqGHciITcaWGUbaaaaWcbeaaaaa@4866@

[10.40]

f 2 R 2 4 MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamOzamaaCaaaleqabaGaaGOmaaaakiaadkfadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaaaaa@3740@       is always positive, so, for a given hemisphere (say, the Northern Hemisphere) there are eight possibilities because R can be either positive or negative, ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@   can be positive or negative, and we have the ± sign in between the two terms on the right-hand side of Equation [10.40].

Gradient balance velocity solutions
Northern Hemisphere R > 0 R < 0
∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@ > 0 no roots are physical only positive root is physical
∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@ < 0 only positive root is physical both roots are physical

The table above gives the results for the Northern Hemisphere (f > 0). We are looking for whether positive or negative values of R and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@   give non-negative and real values for V because only non-negative and real values for V are physically possible. The reason why real negative values of V are not possible is because the gradient wind balance has been written down in natural coordinates.

For R > 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@    > 0, V is always negative, so there are no physical solutions.

For R > 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@   < 0, only the plus sign gives a positive V and thus a physical solution.

For R < 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@   > 0, only the plus sign gives a positive V and thus a physical solution.

For R < 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@   < 0, both roots give positive V and thus physical solutions.

So there are four physical solutions. However, there is one more constraint. This additional constraint is that the absolute angular momentum about the axis of rotation at the latitude of the air parcel should be positive in the Northern Hemisphere (and negative in the Southern Hemisphere). Without proof, we state that only two of the four physically possible cases meet this criterion of positive absolute angular momentum in the Northern Hemisphere. They are:

  1. Regular low: R > 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@  < 0 and V =− f R 2 + f 2 R 2 4 −R ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaamisaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaadAgacaaMi8UaamOuaaqaaiaaikdaaaGaey4kaSYaaOaaaeaadaWcaaqaaiaadAgadaahaaWcbeqaaiaaikdaaaGccaWGsbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGinaaaacqGHsislcaWGsbWaaSaaaeaacqGHciITcqqHMoGraeaacqGHciITcaWGUbaaaaWcbeaaaaa@475A@
  2. Regular high: R < 0 and ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@  < 0 and V =− f R 2 − f 2 R 2 4 −R ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaamisaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaadAgacaaMi8UaamOuaaqaaiaaikdaaaGaeyOeI0YaaOaaaeaadaWcaaqaaiaadAgadaahaaWcbeqaaiaaikdaaaGccaWGsbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGinaaaacqGHsislcaWGsbWaaSaaaeaacqGHciITcqqHMoGraeaacqGHciITcaWGUbaaaaWcbeaaaaa@4765@

These two cases are depicted in the second and third panels of the figure below.

Gradient balance in Northern Hemisphere as described in the text and video below
Gradient balance in Northern Hemisphere. left: Geostrophic balance; center: regular low balance; right: regular high balance. Note that the PGF is independent of velocity but both the Coriolis force and the centrifugal force are dependent on velocity. In the figure, vg is used to represent the geostrophic velocity (only the PGF and Coriolis force are important) and vgr is used to represent the gradient wind velocity (the PGF, Coriolis force, and centrifugal force are all important).
Credit: H.N. Shirer

The video below ( 3:22) explains these four force balances in more detail:

Four Force Balances
Click here for transcript of the Four Force Balances video.

Let's look at four force balances. Let's start which geostrophic balance, which occurs in straight line flow and [INAUDIBLE] troposphere. In geostrophic flow only the pressure gradient force and the Coriolis force are important. Pressure gradient points to low pressure on the height surface or low height and thus low geopotential on a constant pressure surface is opposed by the Coriolis force which is to the right of the velocity vector in the northern hemisphere and to the left of the velocity vector in the southern hemisphere. Note that we can find the geostrophic velocity if we know the pressure gradient on a constant height surface, or the geopotential or height gradient on a constant pressure surface. For inertial balance the Coriolis force is balanced by the horizontal centrifugal force with the Coriolis force to the right of the velocity vector in the northern hemisphere and to the left in the southern hemisphere. This balance is rarely seen in the atmosphere. Because there's almost always a pressure gradient force of the same magnitude as the centrifugal force and the Coriolis force. In cyclostrophic balance the pressure gradient force is balanced by the centrifugal force. In this case, the velocity vector can be either to the right or to the left of the centrifugal force in both hemispheres. And the Coriolis force is much smaller. This balance is seen in tornadoes and other small vortices. In the northern hemisphere tornadoes are mostly cyclonic with only a few percent anticyclonic. While smaller vortices are about as often anticyclonic as they are cyclonic. For gradient wind balance the pressure gradient force, Coriolis force, and horizontal centrifugal force are all about equal. To two physical cases are shown for the northern hemisphere in the figure along with the geostrophic balance. With a cyclonic gradient-- that is curvature around the low pressure center-- the PGF points to the low and is constant as long as the pressure gradient is constant. In this case, the PGF is opposed by both the Coriolis force, which depends on the velocity, and the centrifugal force, which depends on the velocity squared. Since the PGF is constant then the sum of the centrifugal and Coriolis force must be equal to it. And since they both depend on velocity the velocity must be less than in the geostrophic case in order for them to be in force balance. This velocity is called subgeostrophic because it is less than the geostrophic velocity. For the anticyclonic gradient-- which is flow around a high-- the PGF points away from the high and is joined by the centrifugal force, which means that the Coriolis force must be stronger than in the geostrophic case because it must balance both the PGF-- which is the same in the geostrophic case-- and the centrifugal force. The Coriolis force can only be greater if the velocity is greater. Thus this velocity is called super geostrophic. Because it is greater than the geostrophic velocity.

10.9 See how the gradient wind has a role in weather.

10.9 See how the gradient wind has a role in weather.

Note that the wind speed for gradient flow differs from the wind speed of geostrophic flow. Let’s see why. Start with the geostrophic balance (Equation [10.36]) and rearrange the equation to get an expression for the geostrophic wind speed:

V g =− 1 f ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaam4zaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaWGMbaaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3DD9@

[10.40]

Replacing the pressure gradient force ( ∂Φ ∂n MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamOBaaaaaaa@3813@ ) with –fVg in the gradient balance equation results in an equation that relates these gradient velocities to the geostrophic velocity:

V 2 R +fV−f V g =0 or  V g V =1+ V f R MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaamODamaaBaaaleaacaWGibaabeaakmaaCaaaleqabaGaaGOmaaaaaOqaaiaadkfaaaGaey4kaSIaamOzaiaadAhadaWgaaWcbaGaamisaaqabaGccqGHsislcaWGMbGaamODamaaBaaaleaacaWGNbaabeaakiabg2da9iaaicdacaaMf8Uaam4BaiaadkhacaaMf8+aaSaaaeaacaWG2bWaaSbaaSqaaiaadEgaaeqaaaGcbaGaamODamaaBaaaleaacaWGibaabeaaaaGccqGH9aqpcaaIXaGaey4kaSYaaSaaaeaacaWG2bWaaSbaaSqaaiaadIeaaeqaaaGcbaGaamOzaiaayIW7caWGsbaaaaaa@5158@

In a regular low (middle, figure below), R > 0 so that Vg > V. The velocity in a curve around a low-pressure area is subgeostrophic.

In a regular high (right, figure below), R < 0 so that Vg < V. The velocity in a curve around a high-pressure area is supergeostrophic.

Gradient balance in Northern Hemisphere as described in the text
Gradient balance in Northern Hemisphere. left: Geostrophic balance; center: regular low balance; right: regular high balance. Note that the PGF is independent of velocity but both the Coriolis force and the centrifugal acceleration are dependent on velocity. In the figure the geostrophic velocity is represented by vg and the gradient wind velocity is represented by vgr.
Click here for an accessible alternative to the figure above

Balance of Forces (Northern Hemisphere)

*Represented as boxes with arrows*

Geostrophic:

PGF arrow pointing up, COR arrow pointing down (about the same size as PGF), vg arrow pointing right

Subgeostrophic in a trough: Low pressure in a trough with a green arrow underneath pointing right

Supergeostrophic; in a ridge: high pressure below a hill with a green arrow above pointing right

Cyclonic Gradient:

PGF up, COR down (smaller than PGF), CENT down (smaller than COR), vgr right (smaller than geostrophic vg)

Anticyclonic Gradient:

PGF up, CENT up (smaller than PGF), COR down (larger than PGF), vgr ( larger than geostrophic vg)

Credit: H.N. Shirer

Think of it this way. The pressure gradient force is independent of velocity and so is always there for a given geopotential gradient. In a regular low, the centrifugal and Coriolis forces, both dependent on velocity, sum together to equal the pressure gradient force, whereas for geostrophic flow, only the Coriolis force does. Thus, the velocity in the gradient balance case must be less than the geostrophic velocity for the same geopotential gradient.

So how do subgeostrophic and supergeostrophic flow affect weather?

Supergeostrophic flow around ridges and subgeostrophic flow around troughs helps to explain the convergence and divergence patterns aloft that are linked to vertical motions.

Look at the figure below, starting on the left. Going from geostrophic flow in the straight section to supergeostrophic flow at the ridge’s peak causes divergence aloft. This divergence causes an upward vertical velocity, which causes a low pressure area and convergence at the surface. As the air rounds the ridge’s peak, it slows down to become geostrophic, and then continues to slow down even more as the flow becomes subgeostrophic around the trough, thus causing convergence aloft. This convergence aloft causes a downward velocity, which causes high pressure and divergence at the surface.

Subgeostrophic and supergeostrophic velocities in flow aloft in the Northern Hemisphere as described in the text and video below
Subgeostrophic and supergeostrophic velocities in flow aloft in the Northern Hemisphere.
Click here for an accessible alternative to the figure above

Gradient Wing & Divergence Patterns Aloft, Surface Low and High Pressure

Diagram of a wave shape. It starts as a low-pressure zone with a constant geostrophic wind (vg) moving up along the wave, it is also labeled with divergence, at the ridge of the wave super geostrophic gradient wind vgr is horizontal. The ridge also has anticyclonic curvature. The wave then moves downward, in a high-pressure zone, labeled with convergence until it reaches a trough with cyclonic curvature and subgeostropic gradient wing vgr which is represented by a horizontal arrow. The wave then repeats and vg goes upward toward the ridge from a low-pressure zone that is diverging.

Credit: H.N. Shirer

So downwind of a trough is the favored location for divergence aloft, upward motion, and a surface low [9]. Downwind of a ridge is the favored location for convergence aloft, downward motion, and a surface high. Since ridges form around high pressure aloft and troughs form around low pressure aloft, we see that the high aloft is offset relative to the surface low and the low aloft is offset relative to the surface high.

Thus subgeostrophic flow and supergeostrophic flow aloft are directly related to the formation of weather at the surface. Other factors like vorticity are also very important. The video below (1:09) describes how the gradient wind flow aloft can affect surface weather.

Trough Aloft Surface Low Video
Click here for a transcript of the Trough Aloft Surface Low video.

Let's see how the gradient wind flow aloft can affect surface weather. Look at how the velocity changes as air flows around the ridge and then a trough aloft. Initially, the velocity is about geostrophic and straight-line flow. As it rounds the ridge, it speeds up. And then it slows down again to geostrophic in the straight section. As it goes through the trough, around the low-pressure loft, it slows down to subgeostrophic and then speeds up to geostrophic in the next straight section. The speeding up causes divergence aloft. And the slowing down causes convergence aloft, just as you learned in lesson nine. You also saw how convergence aloft can lead to divergence at the surface. This contributes to a surface high. And how divergence aloft can lead to convergence at the surface, which contributes to a surface low. Thus, gradient flow contributes to surface weather. We often see a surface low forming on the downwind side of a trough.

10.10 What does turbulent drag do to horizontal boundary layer flow?

10.10 What does turbulent drag do to horizontal boundary layer flow?

The turbulent drag force on the horizontal air flow within the atmospheric boundary layer can approach the size of the other terms in the horizontal equation of motion. Note that this turbulent drag force acts to reduce the air velocity and therefore is opposite the air flow velocity vector. 

Let’s look at the force balance when the turbulent drag force is included. We will call this turbulent drag "friction" because it is commonly called that, but it is really quite different.

shows a box between high & low pressure with PGF arrow going up, velocity arrow, friction arrow opposite velocity & coriolis force arrow
Horizontal map showing the geostrophic wind with friction in the Northern Hemisphere.
Credit: W. Brune

Note that the turbulent drag (friction) force is parallel to the velocity vector and is opposite in direction.

In the x-direction (along the isobars) the balance of forces is:

friction (x-component) = Coriolis (x-component).

In the y-direction (perpendicular to the isobars), the balance of opposing forces is:

friction (y-component) + Coriolis (y-component) = PGF

Because of the turbulent drag force and the velocity dependence of the Coriolis force, the parcel velocity points toward the lower pressure and the air parcel will tend to move across isobars toward the low pressure. Typical cross-isobaric flow in the boundary layer makes an angle of 30o with the isobars. Thus, surface air moves toward low pressure and away from high pressure.

Above the atmospheric boundary layer, however, turbulent drag is not generally important and geostrophic and gradient flows are good approximations.

The effects of turbulent drag are very important for weather. When divergence aloft causes upward-moving air below, it becomes associated with a low-pressure region near Earth's surface. A pressure gradient force is created, but the air moving toward the low pressure is turned to travel counterclockwise around the low by the Coriolis force. However, turbulent drag slows the wind and turns it to cross isobars toward the low pressure, creating convergence, which causes uplift, clouds, and perhaps precipitation. The opposite is true for surface high-pressure regions occurring under regions of convergence aloft, which causes descent. The anticyclonic winds around the surface high are slowed and turn outward, causing divergence near the surface, leading to descending air and clear skies. The following video (1:43) provides further discussion of friction:

Friction
Click here for transcript of the Friction video.

When the flow is in the upper part of the atmospheric boundary layer, which is a kilometer or two above the surface, the turbulence in the boundary layer acts to impede the horizontal wind. This resistance to flow is not really friction. But it does act to slow the wind down, no matter the wind's direction. As a result, we can assume that this turbulent drag is a force that opposes the wind velocity. Look at what adding a turbulent resistant turn does to the force balance for straight line flow in the upper boundary layer. The PGF force is perpendicular to the pressure gradient as usual. However, the friction opposes the velocity, and slows it down. At the same time, the Coriolis force is always perpendicular to the velocity and to the right. And because the velocity is slowed down, the Coriolis force is less. The velocity vector gets turned toward the PGF vector, and thus toward the low pressure. Note that the x direction of the frictional force must balance and be opposite to the x direction of the Coriolis force. And the y directions of the friction and Coriolis forces must be opposite to and balance the PGF in the y direction in the diagram. Boundary layer turbulent drag turns to velocity across isobars for low pressure, which causes convergence, while boundary layer turbulent drag turns to velocity across isobars away from high pressure, which causes divergence. These friction effects tend to amplify the convergence into surface low pressure and divergence from surface high pressure.

Quiz 10-3: Balance of forces and motion.

  1. Find Practice Quiz 10-3 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 10-3. You will be allowed to take this quiz only once. Good luck!

10.11 Why are midlatitude winds mostly westerly (i.e., eastward)?

10.11 Why are midlatitude winds mostly westerly (i.e., eastward)?

The answer seems simple. More solar energy is deposited in the tropics than near the poles and as a result, the air is warmer in the tropics than the poles, where there is net radiative cooling. According to the hydrostatic equilibrium equation, the fall-off in pressure with altitude is less in the tropics than at higher latitudes, and as a result, for any height surface, the pressure is greater in the tropics than near the poles, setting up a pressure gradient force on each height surface that drives the wind poleward. As the air moves poleward, atmospheric processes such as the Coriolis force, which we saw occurs because of conservation of angular momentum, and synoptic-scale disturbances at higher latitudes cause the air to veer to the right in the Northern Hemisphere and to the left in the Southern Hemisphere.

Thermal Wind

With this broad concept in mind, we can consider the idea of the thermal wind, which is not really a wind, but instead is a difference in winds at two heights. We will see that the thermal wind is proportional to the horizontal temperature gradient. To show this relationship mathematically, we start with the geostrophic balance equation and apply the Ideal Gas Law and the hydrostatic equilibrium equation.

Look at the x and y components of Equation [10.33] for geostrophic winds:

v g = 1 f ∂Φ ∂x    u g =− 1 f ∂Φ ∂y MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaam4zaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaWGMbaaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamiEaaaacaaMf8UaaGzbVlaadwhadaWgaaWcbaGaam4zaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaWGMbaaamaalaaabaGaeyOaIyRaeuOPdyeabaGaeyOaIyRaamyEaaaaaaa@4B2B@

[10.41]

Use the hydrostatic equation and the Ideal Gas Law to relate T to Φ MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfA6agbaa@3444@ :

∂p ∂z =−ρg →  ∂p g∂z =−ρ →  ∂p ∂Φ =−ρ →  ∂Φ ∂p =− 1 ρ =− RT p MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIyRaamiCaaqaaiabgkGi2kaadQhaaaGaeyypa0JaeyOeI0IaeqyWdiNaam4zaiaaywW7cqGHsgIRcaaMf8+aaSaaaeaacqGHciITcaWGWbaabaGaam4zaiabgkGi2kaadQhaaaGaeyypa0JaeyOeI0IaeqyWdiNaaGzbVlabgkziUkaaywW7daWcaaqaaiabgkGi2kaadchaaeaacqGHciITcqqHMoGraaGaeyypa0JaeyOeI0IaeqyWdiNaaGzbVlabgkziUkaaywW7daWcaaqaaiabgkGi2kabfA6agbqaaiabgkGi2kaadchaaaGaeyypa0JaeyOeI0YaaSaaaeaacaaIXaaabaGaeqyWdihaaiabg2da9iabgkHiTmaalaaabaGaamOuaiaadsfaaeaacaWGWbaaaaaa@6C33@

Take ∂ ∂p MathType@MTEF [2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeyOaIylabaGaeyOaIyRaamiCaaaaaaa@369C@ of equations in [10.41], starting with the equation for v g MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaaWcbaGaam4zaaqabaaaaa@34DD@ :

∂ v g ∂p = 1 f ∂ ∂p ( ∂Φ ∂x )= 1 f ∂ ∂x ( ∂Φ ∂p )= 1 f ∂ ∂x ( − RT p ) p ∂ v g ∂p =− R f ∂T ∂x  and p ∂ v g ∂p = ∂ v g ∂lnp MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8444@

Taking the same approach with the equation for u g MathType@MTEF [2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwhadaWgaaWcbaGaam4zaaqabaaaaa@34DC@ gives the Thermal Wind Equations:

∂ v g ∂lnp =− R f ( ∂T ∂x ) p     and     ∂ u g ∂lnp = R f ( ∂T ∂y ) p ∂ V → g ∂lnp =− R f k → × ∇ → p T   in vector form MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8862@

[10.42]

These equations are very powerful because they reveal that measurements of temperature only (which are relatively easy to make) make it possible to determine the changes in the horizontal wind with height, assuming that the geostrophic and hydrostatic approximations are valid (which is the case for large-scale flow in the free atmosphere). The thermal wind velocity is defined as the change in horizontal geostrophic velocity between two layers.

V → T ≡ V → g ( p 1 )− V → g ( p 2 ),   p 1 < p 2 MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaceWGwbGbaSaadaWgaaWcbaaeeaaaaaaaaigBHn2Aa8qacaWGubaapaqabaGcpeGaeyyyIO7daiqadAfagaWcamaaBaaaleaapeGaam4zaaWdaeqaaOWdbiaacIcacaWGWbWdamaaBaaaleaapeGaaGymaaWdaeqaaOWdbiaacMcacqGHsislpaGabmOvayaalaWaaSbaaSqaa8qacaWGNbaapaqabaGcpeGaaiikaiaadchapaWaaSbaaSqaa8qacaaIWaaapaqabaGcpeGaaiykaiaacYcacaaMf8UaaGzbVlaadchapaWaaSbaaSqaa8qacaaIXaaapaqabaGcpeGaeyipaWJaamiCa8aadaWgaaWcbaWdbiaaicdaa8aabeaaaaa@4D00@

[10.43]

So the thermal wind velocity equals the horizontal velocity at the upper level minus the horizontal velocity at the lower level. Equation [10.43] can be drawn as a two-dimensional vector subtraction as if we were looking down on them from above (see figure below).

3 Thermal wind vectors, 2 vectors moving north east, labeled p1 and p2. vector #3 labeled vT moves northwest connecting the tips of p1 & p2
The thermal wind is the geostrophic wind vector at pressure level (p1) higher in the atmosphere minus the geostrophic wind vector at a pressure level (p2) lower in the atmosphere, where p1 < p2.
Credit: W. Brune

We can integrate Equation [10.42] and if we let <T> be the average temperature between pressure surfaces p1 and p2 (where p1 < p2), yielding the expressions for the thermal wind vectors in the x and y directions:

V → T = u T i → + v T j → = R f [ j → ( ∂〈T〉 ∂x ) p − i → ( ∂〈T〉 ∂y ) p ]ln( p 2 p 1 ) V → T = R f ln( p 2 p 1 ) k → × ∇ → p 〈T〉 u T =− R f ( ∂〈T〉 ∂y ) p ln( p 2 p 1 )   v T = R f ( ∂〈T〉 ∂x ) p ln( p 2 p 1 ) MathType@MTEF [2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaafaWaaeGabaaaeaqabeaaqqaaaaaaaaGySf2yRbWdbiqadAfapaGbaSaadaWgaaWcbaWdbiaadsfaa8aabeaak8qacqGH9aqpcaWG1bWdamaaBaaaleaapeGaamivaaWdaeqaaOWdbiqadMgapaGbaSaapeGaey4kaSIaamODa8aadaWgaaWcbaWdbiaadsfaa8aabeaak8qaceWGQbWdayaalaWdbiabg2da9maalaaapaqaa8qacaWGsbaapaqaa8qacaWGMbaaamaadmaapaqaa8qaceWGQbWdayaalaWdbmaabmaapaqaa8qadaWcaaWdaeaapeGaeyOaIy7enf2yl9MCLfMBaeHbbbhCSvgitrei7QET5cB71gpiov2BGaBRhhTU8wB5dZlaiuaacaWFiaJaamivaiaa=Lama8aabaWdbiabgkGi2kaadIhaaaaacaGLOaGaayzkaaWdamaaBaaaleaapeGaamiCaaWdaeqaaOWdbiabgkHiTiqadMgapaGbaSaapeWaaeWaa8aabaWdbmaalaaapaqaa8qacqGHciITcaWFiaJaamivaiaa=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Lama8aabaWdbiabgkGi2kaadMhaaaaacaGLOaGaayzkaaWdamaaBaaaleaapeGaamiCaaWdaeqaaOWdbiGacYgacaGGUbWaaeWaa8aabaWdbmaalaaapaqaa8qacaWGWbWdamaaBaaaleaapeGaaGOmaaWdaeqaaaGcbaWdbiaadchapaWaaSbaaSqaa8qacaaIXaaapaqabaaaaaGcpeGaayjkaiaawMcaaiaaywW7caaMf8UaamODa8aadaWgaaWcbaWdbiaadsfaa8aabeaak8qacqGH9aqpdaWcaaWdaeaapeGaamOuaaWdaeaapeGaamOzaaaadaqadaWdaeaapeWaaSaaa8aabaWdbiabgkGi2kaa=HamcaWGubGaa8xcWaWdaeaapeGaeyOaIyRaamiEaaaaaiaawIcacaGLPaaapaWaaSbaaSqaa8qacaWGWbaapaqabaGcpeGaciiBaiaac6gadaqadaWdaeaapeWaaSaaa8aabaWdbiaadchapaWaaSbaaSqaa8qacaaIYaaapaqabaaakeaapeGaamiCa8aadaWgaaWcbaWdbiaaigdaa8aabeaaaaaak8qacaGLOaGaayzkaaaaaaaa@BF00@

[10.44]

The vertical change in geostrophic wind is called the geostrophic vertical shear. Since the geostrophic vertical shear is directly proportional to the horizontal temperature gradient, it is also called the Thermal Wind.

2 graphs: 2 vectors moving from cold to warm connected by vector VT & 2 vectos moving from warm to cold connected by vector VT
Thermal wind is the vector difference of the geostrophic wind at the upper-level pressure (p1) minus the lower-level pressure (p0, which should be written as p2 to be consistent with the previous figure and the text). These figures show the view looking down on the x-y plane. Cold air is on the left of VT. Left: Backing (flow turning counterclockwise with height) indicates cold advection. Right: Veering (flow turning clockwise with height) indicates warm air advection.
Credit: W. Brune

We can learn a lot from the thermal wind (seen in the figure above).

  • Thermal wind blows with cold air to the left (on average) and with the warm air on the right in the Northern Hemisphere (remember “air is light on the right”), and with warm air on the left in the Southern Hemisphere.
  • If the geostrophic wind vector rotates counterclockwise with height (called backing), then there is cold air advection. For the Northern Hemisphere, remember “CCC: counterclockwise cold.”
  • If the geostrophic wind vector rotates clockwise with height (called veering), then there is warm air advection.
  • In the Northern Hemisphere, because low thickness means lower layer temperature and higher thickness means higher layer temperature, the thermal wind blows parallel to the lines of constant thickness [10] with the low thickness on the left.
  • This statement is an analogy to “The geostrophic wind blows parallel to the height contours on a pressure surface (pressure contours on a height surface) with the low height (pressure) to the left."

Quiz 10-4: Feeling the thermal wind.

  1. Find Practice Quiz 10-4 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 10-4. You will be allowed to take this quiz only once. Good luck!

Summary and Final Tasks

Summary

Understanding atmospheric dynamics is built upon three conservation laws: energy (the 1st Law of Thermodynamics), mass, and momentum. When we use conservation of momentum on the rotating Earth, we need to consider not only the real forces of gravity, pressure gradient force, and turbulent drag in the lower troposphere, but also apparent forces—centrifugal and Coriolis. With these terms, we can use conservation of momentum to write down the equations of motion in the Earth's reference frame and then show how they can be transformed into spherical coordinates or even pressure coordinates in the vertical and natural coordinates in the horizontal.

Using natural coordinates simplifies the equation of motion for geostrophic flow (the balance of Coriolis and pressure gradient forces), cyclostrophic flow (the balance of centrifugal and pressure gradient forces), inertial flow (the balance between centrifugal and Coriolis forces), and gradient flow (the balance among the pressure gradient force, Coriolis force, and horizontal centrifugal force). Upper air motion around high and low pressure causes upper-air convergence and divergence, which leads to high and low pressure at the surface.

Finally, the temperature decrease at each pressure level from tropics to the poles leads to a pressure gradient force that drives air toward the poles. The Coriolis force turns the air toward the east, creating westerlies observed in the midlatitudes of both hemispheres. This connection between the latitudinal temperature gradient and wind is expressed in the thermal wind equation. The thermal wind vector, which is the difference between the geostrophic winds at two different pressure levels, is parallel to isotherms, with cold air on the left in the Northern Hemisphere. If the geostrophic wind vector turns counterclockwise with height in the Northern Hemisphere, cold air advection is occurring in that layer of air. This relationship is a handy way to figure out if the advection is cold or warm.

Reminder - Complete all of the Lesson 10 tasks!

You have reached the end of Lesson 10! Double-check that you have completed all of the activities before you begin Lesson 11.


Source URL: https://www.e-education.psu.edu/meteo300/node/696

Links
[1] https://www.e-education.psu.edu/meteo300/8
[2] mailto:MathType@MTEF
[3] https://www.e-education.psu.edu/meteo300/node/726
[4] http://commons.wikimedia.org/wiki/File:Isaac_Newton_woodcut,_frontispiece_to_Mach.jpg#/media/File:Isaac_Newton_woodcut,_frontispiece_to_Mach.jpg
[5] https://www.e-education.psu.edu/meteo300/node/7
[6] https://www.youtube.com/channel/UCjHz5SVHeMT0AViCYZvsGDA
[7] https://www.e-education.psu.edu/meteo300/732
[8] https://www.flickr.com/photos/noaaphotolib/5033385343/in/photolist-8EMq6V-8EQBmq-9W83TQ-fF8Ugj-fEU41n-fF8Uef-ivdUBU-fERkeK-fEU43z-rdAmRY-7d61Wf-8GHPRS-8EMrfT-5sDJmC-nEsBE5-p7iVA-6JPMWs-fDhBfW-46nJo-2VYUeU-5ivcYq-7G866b-4tFTc8-7akywQ-8fJgoz-8tYfiv-8tYff8-dp5iaq-9zjYaJ-ePi3dG-FB6zY-2Vxn8F-9zgYBg-8tYfnH-8u2km1-8tY5qe-8tY3Zn-8tY1ZF-9zjXLQ-e3Wjs-2Vxcjr-ePj5ah-8sctt-FB7u2-7MdgbH-a4JYTA-BpnYU-ePzp3w-ePsyey-ePiKJU
[9] https://www.e-education.psu.edu/meteo300/726
[10] https://www.e-education.psu.edu/meteo300/node/667