# Newton’s second law for a parcel of air in an inertial coordinate system (a coordinate system in which the coordinate axes do not change direction and.

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Newton’s second law for a parcel of air in an inertial coordinate system (a coordinate system in which the coordinate axes do not change direction and are not accelerated) In the case of more than one force:

y y y x x x z (toward you) z (toward you) z (toward you) Earth not rotating An inertial coordinate system and the meaning of the vector and Inertial coordinate system: is with respect to a distant star Let’s first take a look at the term in the yellow box…

Earth-based (non-inertial) “spherical” coordinate system (favored by meteorologists) x = positive toward east y = positive toward north z = positive toward zenith wind component toward east wind component toward north wind component toward zenith = longitude  = latitude

Earth-based (non-inertial) “spherical” coordinate system (favored by meteorologists) x = positive toward east y = positive toward north z = positive toward zenith Where unit vectors are a function of position on the earth (not Cartesian)

We need to determine what these are Since the unit vectors are not constant… Let’s start with What is the relationship between and

At a point (constant x,y,z) none of the unit vectors change with time so As one moves north or south the i direction experiences no change so As one moves up or down the i direction experiences no change so So:

From figure on right looking down at north pole at latitude  : This gives us the magnitude, but not the direction. Note that is pointed toward the center of the earth at the original point

Looking at the figure at the right: We see that has components in the and directions The unit vector describing the direction of has two components: So:

We still need to determine what these are Since the unit vectors are not constant… Let’s do next REMEMBER THIS SLIDE?

does not change with time or elevation, but does change in the x and y directions

Lets first figure out Look at light gray triangle in (a): Dark gray triangle is shown in a different view in (b) or So: Direction of is the -x direction

Now lets figure out Direction of is the direction So:

We still need to determine what this is Since the unit vectors are not constant… Let’s do next REMEMBER THIS SLIDE?

does not change with time or elevation, but does change in the x and y directions

Let’s do first Direction of is the positive direction

Let’s do next Direction of is the positive direction

Since the unit vectors are not constant… REMEMBER THIS SLIDE? or

What are the correction terms and what do they mean? Consider third equation: Air moving in a straight line at a constant speed initially southward. u = 0, v = -10 m/s, w = 0 Air will accelerate upward! The earth curves away from the path of the air parcel.

Newton’s second law in an inertial coordinate system Newton’s second law in a spherical coordinate system Now let’s look at the right side of the equation!

Forces and the governing equations Three Fundamental Forces in the lower atmosphere Pressure Gradient Force Gravity Friction Two Apparent Forces in the lower atmosphere due to the rotation of the earth Centrifugal Force Coriolis Force

1. Pressure gradient force: Directed from high pressure toward low pressure. A wall with higher pressure on its left side (indicated by greater molecular density on the left side) Simple illustration of a pressure gradient acting on a wall (left) and a molecule (right) An air molecule with higher air pressure on its left (indicated by greater molecular density on the left side)

Volume of the fluid element Mass of the fluid element  = density (g m -3 ) Pressure at center of the fluid element Definition of pressure: Pressure is equal to the force applied per unit area

Find Pressure at A and B Use Taylor expansion:

Find Force at B and A Force = Pressure  Area

Find net x-direction force per unit mass

In a similar way: The vector form of the Pressure Gradient Force can therefore be expressed as

2. Gravity: Directed toward center of earth Newton’s law of universal gravitation G = 6.673  10 -11 N m 2 kg -2 M = mass of Earth m = mass of fluid element r = vector from Earth center r = magnitude of r vector

Newton’s law of universal gravitation… …expressed as a force per unit mass Over the depth of the troposphere the change in the force of gravity is insignificant and we can approximate r as the earth radius, a

3. Friction: Directed opposite the flow Friction is manifested as: A drag force in a very thin layer (a few mm) near the surface Turbulent mixing of blobs of faster and slower air at altitudes above the surface.

Although largely irrelevant to the atmosphere, we will consider the drag force first and then make a simple analogy to formulate friction for the rest of the atmosphere Not moving moving

Expect that drag force on upper plate will be proportional to: Speed of the plate (u 0 ) Area of the plate (A) And inversely proportional to the depth of the fluid (l) Where  is the proportionality constant called the dynamic viscosity coefficient

Next let’s define a “Shearing Stress” , as the Force/Unit Area on the plate Let’s also assume the lower plate is moving at an arbitrary speed so that the speed difference is  u Let’s also assume that the distance between the plates is arbitrary and that that difference is  z Then:andas

Force per unit area acting in x direction due to shear in z direction Now let’s consider how that force acts on a fluid element of volume  x  y  z Illustration of x component of shearing stress (F/A) on a fluid element

Stress acting across the upper boundary on the fluid below it Use Taylor expansion Stress acting across the lower boundary on the fluid below it Using Newton’s third law: Stress across the lower boundary on the fluid above it must be: To find the net stress, we want to sum the forces that act on fluid within the fluid element

Dividing by the mass of the fluid element we have the viscous force per unit mass arising from the vertical shear in the x direction on the fluid element

Assuming  is constant, this can be written where  is called the kinematic viscosity coefficient

We have figured out one component of the frictional force per unit mass: Considering how shear can act in each direction to create a frictional force in each direction, there are nine components

Unfortunately, all of the preceding theory applies to the air in a layer a few mm thick above the earth’s surface For blobs of air mixing, an analogous “eddy viscosity coefficient”, K, is defined that is calculated based on the average length that an eddy can travel before mixing out its momentum. In general, vertical wind shear is much stronger than horizontal shear for synoptic scale flows so this reduces to:

Acceleration pressure gradient force gravity friction Force balance in the atmosphere for a non-rotating earth expressed in spherical coordinates Correction for spherical Coordinate system

Same statement expressed as 3 orthogonal scalar equations Acceleration Correction terms for earth based coordinate system Pressure gradient Force Friction (East-West equation) (North-South equation) (Up-Down equation) Gravity

Apparent forces on a rotating earth – the Centrifugal Force We will derive the force terms twice First from a more conceptual approach The second time from a more formal vector approach

Apparent forces on a rotating earth – the Centrifugal Force Consider an object at rest on the rotating earth Equator Speed = 465 m/s 40 N: Speed = 356 m/s 60 N: Speed = 232 m/s Actual speed due to rotation of earth:

A ball whirling on a string experiences a CENTRIPETAL acceleration toward the axis of rotation The acceleration is equal to where  is the angular rotation rate

A person on the ball whirling on a string experiences a CENTRIFUGAL acceleration toward the axis of rotation The acceleration is equal to where  is the angular rotation rate

Standing still on the rotating earth Outward acceleration away from Earth’s axis is where is the Earth’s angular rotation rate Where 86156.09 s is the time it takes for the Earth to rotate once with respect to a fixed star (a sidereal day) Vectors not to true size!

Earth has distorted its shape into an oblate spheroid in response to this force. The distortion is such that the combined gravitational force and centrifugal force (g*) act exactly perpendicular to the Earth’s (flat) surface. We combine the centrifugal force with gravity and forget about it!!

Apparent forces on a rotating earth – the Coriolis Force In the absence of a twisting force called a torque, air in motion across the earth must conserve its angular momentum (M  V  R) where M is the parcel mass, V is velocity about the axis of rotation and R is the distance from the axis of rotation. A simple interpretation of the Coriolis effect: Air moving across the earth’s surface will try to come to equilibrium at a latitude/altitude where its angular momentum equals that of the earth so that the parcel has no relative motion.

Motion on a rotating earth – the Coriolis Force Consider air moving: Eastward: The air has greater angular momentum than the earth beneath it. It will experience an “outward” centrifugal acceleration equatorward Westward: The air has less angular momentum Than the earth beneath it. It will experience an “inward” centripetal acceleration poleward. Poleward: The air will progressively move over points on the earth with less angular momentum. Air will accelerate eastward relative to the earth below it. Equatorward: The air will progressively move over points on the earth with greater angular momentum. Air will accelerate westward relative to the earth below it.

Suppose an object is moving eastward on the rotating earth at speed u The total centrifugal force it will experience will be its angular velocity squared  distance to axis of rotation Centrifugal force (combined with gravity) Magnitude of 2 nd term on RHSMagnitude of 3 rd term on RHS

Suppose an object is moving eastward on the rotating earth at speed u The total centrifugal force it will experience will be its angular velocity squared  distance to axis of rotation Centrifugal force (combined with gravity)  ignore Coriolis Force

Suppose an object is moving equatorward on the rotating earth at speed -v Apply principle of conservation of momentum Expand this expression Since and are small, neglect terms with their product

(From previous slide) If we assume that << this expression can be approximated as: or

For a displacement in the –y direction, from Figure: Dividing by and taking the limit as

For a displacement in the z direction, from Figure: Dividing by and taking the limit as

Combining the results for east-west and north-south movement of air Where: and f is the Coriolis parameter

In a reference frame on the earth the Coriolis effect appears as a force acting on an air parcel. Motion on a rotating earth – the Coriolis Force The Coriolis Force Causes air to deviate to the right of its direction of motion in the Northern Hemisphere (and to the left in the Southern Hemisphere); Affects the direction an object will move across the earth’s surface, but has no effect on its speed; Is strongest for fast-moving objects and zero for Stationary objects; and Has no horizontal component at the equator and has a maximum horizontal component at the poles

In our previous discussion of spherical coordinates, the coordinates were fixed in position. A At Point A, for example, the x, y, and z axes always point toward distant stars. At point B, the axes point at different stars, but always in the same direction. B A formal way of deriving the Coriolis and Centrifugal Forces When the earth rotates, the axes at point A find themselves at a later time at point B. The direction the axes at a specific location point is a function of time

To account for the local movement of the coordinate system mathematically, we must consider the behavior of a vector in a rotating coordinate system Consider a vector in a stationary coordinate system The components of the vector in the rotating coordinate system are: where,, and are fixed at local points

The time derivative of the vector in a stationary coordinate system would be: in the rotating coordinate system are:

Stationary Rotating We can write the equation for the rotating frame of reference as: The derivatives,, and, represent the rate of change of the unit vectors,, and, that arise because the coordinate system is rotating

Let’s determine and expression for Rotation vector,, points upward from north pole By similar triangles The magnitude of is given by:The direction of is given by:

By similar triangles The magnitude of is given by:The direction of is given by: Let’s determine and expression for Rotation vector,, points upward from north pole

By similar triangles The magnitude of is given by:The direction of is given by: Let’s determine and expression for Rotation vector,, points upward from north pole

Our previous equation was: and we can write then

For a vector A, the coordinate transformation from a non-rotating to a rotating system is given by: Let’s apply this to two vectors: Let be a position vector, perpendicular to the axis of rotation of the earth with magnitude equal to the distance from the surface of the Earth to the axis of rotation By definition: The absolute velocity of an object on a rotating earth is equal to its velocity relative to the rotating earth + the velocity of the Earth’s rotation Also apply vector transform to The absolute acceleration of an object on a rotating earth is equal to its acceleration relative to the rotating earth + acceleration due to the Earth’s rotation

(1) (2) Put (2) into (1) Coriolis Force Centrifugal Force

Equations of motion on non-rotating earth Acceleration Correction terms for earth based coordinate system Pressure gradient Force Friction (East-West equation) (North-South equation) (Up-Down equation) Gravity

Equations of motion on rotating earth Answer We must add terms to account for the acceleration required for air to conserve its angular momentum. In scalar form the equations of motion for each direction become: (East-West equation) (North-South equation) (Up-Down equation) Acceleration Correction terms for spherical coordinate system Pressure gradient Force Friction Effective Gravity Coriolis Force

The complete momentum equations on a spherical rotating earth FOR SYNOPTIC SCALE MOTIONS….. WHICH TERMS ARE LARGE AND IMPORTANT? WHICH TERMS ARE SMALL AND INSIGNIFICANT? See Scale Analyses: Table 3.1 P. 60, Table 3.2 P. 64

Scale Analysis of the horizontal momentum equations *numbers in yellow boxes apply to free atmosphere above boundary layer

Scale Analysis of the vertical momentum equations

        ABOVE THE BOUNDARY LAYER        WITHIN THE BOUNDARY LAYER 

ABOVE THE BOUNDARY LAYER, ALL HORIZONTAL PARCEL ACCELERATIONS CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE PRESSURE GRADIENT AND CORIOLIS FORCES WITHIN THE BOUNDARY LAYER, ALL HORIZONTAL PARCEL ACCELERATIONS CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE PRESSURE GRADIENT, CORIOLIS AND FRICTIONAL FORCES The Hidden Simplicity of Atmospheric Dynamics: THE ATMOSPHERE IS IN HYDROSTATIC BALANCE – VERTICAL PGF BALANCES GRAVITY – ON SYNOPTIC SCALES

Note the Total Derivative: Note that: The rate of change of a property (temp) at a fixed point (x,y,z) The rate of change of the property Following a parcel as it moves to the point (x,y,z) The advection of the property from upstream to the point (x, y, z) =

Simplified u Momentum Equation expressed in terms of local time derivatives The rate of change of west-east wind at a fixed point (x,y,z) PGFCOR The advection of the west-east wind Component from upstream to the point (x, y, z)

Simplified v Momentum Equation expressed in terms of local time derivatives The rate of change of south-north wind at a fixed point (x,y,z) PGFCOR The advection of the south-north wind Component from upstream to the point (x, y, z)

Pressure as a vertical coordinate Pressure decreases monotonically with altitude, and therefore can be used as a vertical coordinate instead of altitude We will find this coordinate system more convenient mathematically and so our task will be to recast the momentum equations in pressure coordinates

The simplified horizontal momentum equations in height coordinates are given by: or more succinctly: In order to cast these equations in pressure coordinates, we need to convert the PGF term into an equivalent expression in isobaric coordinates……..

Consider the differential on a constant pressure surface Expanding… (Subscripts indicate differentiation carried out holding subscripted variable constant) Now expand as a function of x and y

Rearrange: For this statement to be true for all and, the statements in each square bracket must equal zero

Use hydrostatic equation Divide both sides by or where is called the geopotential height

We will now drop the use of subscripts and write the momentum equations in p coordinates

In pressure coordinates, the horizontal momentum equations (above the boundary layer) become: Pressure coordinates where , the geopotential height, is given by  = gz and f = 2  sin  and the vertical momentum equation becomes (using the ideal gas law)

The Total Derivative in pressure coordinates Note that: Note that the vertical velocity in pressure coordinates has a different form: The term,, represents the rate of change of pressure following parcel motion as a parcel rises or descends in the atmosphere

Note the Total Derivative: Note that: The rate of change of a property (temp) at a fixed point (x,y,z) The rate of change of the property Following a parcel as it moves to the point (x,y,z) The advection of the property from upstream to the point (x, y, z) =

Simplified u Momentum Equation in pressure coordinates expressed in terms of local time derivatives The rate of change of west-east wind at a fixed point (x,y,z) PGFCOR The advection of the west-east wind Component from upstream to the point (x, y, z)

Simplified v Momentum Equation expressed in terms of local time derivatives The rate of change of south-north wind at a fixed point (x,y,z) PGFCOR The advection of the south-north wind Component from upstream to the point (x, y, z)

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