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AOSS 321, Winter 2009
Earth System Dynamics
Lecture 6 & 7
1/27/2009
1/29/2009
Christiane Jablonowski
cjablono@umich.edu
734-763-6238
Eric Hetland
ehetland@umich.edu
734-615-3177
What are the fundamental forces in
the Earth’s system?
• Surface forces:
– Pressure gradient force
– Viscous force
• Body force: Gravitational force
• Apparent forces: Centrifugal and Coriolis
• Electromagnetic force (important for charged
particles, e.g. in the ionosphere, not important
for the neutral atmosphere)
• Total Force is the sum of all of these forces.
Viscous force(1): Friction
• In a fluid there is friction which resists the flow.
• It is dissipative, and if the fluid is not otherwise
forced, it will slow the fluid and bring it to rest.
• Away from boundaries in the atmosphere this
frictional force is often small, and it is often
ignored.
• Close to the boundaries, however, we have to
consider friction.
Viscous force (2)
velocity ≡ u
in m/s
Ocean
Land
Biosphere
velocity must be 0 at surface
Viscous force (3)
velocity ≡ u
in m/s
Ocean
Land
Biosphere
Velocity is zero at the surface; hence,
there is some velocity profile.
Viscous force (4)
(How do we think about this?)
z
Moving plate with velocity u0
u(h) = u0
h
u(z)
u(0) = 0
u0  u(0) 
u(z)  
z
 h

Linear velocity profile!

Viscous force (5)
(How do we think about this?)
The drag on the moving plate is the same of as the force required
to keep the plate moving. It is proportional to the area (A),
proportional to the velocity u0 of the plate, and inversely
proportional to the distance h between the plates; hence,
u(h) = u0
h
u(z)
u0
F  A
h
u(0) = 0
Proportional usually means we assume linear relationship. This is a
model based on observation, and it is an approximation. The constant of
proportionality assumes some physical units. What are they?

Viscous force (6)
(How do we think about this?)
Recognize the u0/h can be represented by
∂u/∂z
u(h) = u0
h
u(z)
u(0) = 0
u
F  A
z
Force per unit area is F/A is defined as shearing
stress (t). Like pressure the shearing stress is
proportional to area. 
Viscous force (7) & Shearing stress
tzx = μ(∂u/∂z) is the component of the shearing
stress in the x direction due to the vertical (z)
shear of the x velocity component (velocity u)
u(h) = u0
h
u(z)
u(0) = 0
Fzx
u
   t zx
A
z
Fzx/A: Force per unit area with units kg m-1 s-2:
hence, like pressure

Viscous force (8) & Shearing stress
Remember Newton’s third law:
"For every action, there is an equal and opposite reaction."
tzx = -(tzx0 + (∂tzx/∂z)Dz/2)
Stress above
a surface
tzx = tzx0 + (∂tzx/∂z)Dz/2
z
• x
tzx0
Stress below
a surface
Front view of 3D box
tzx = -(tzx0 - (∂tzx/∂z)Dz/2)
tzx = tzx0 - (∂tzx/∂z)Dz/2
The opposing force introduces a ‘-’ sign!
We are interested in the net force inside the volume.
Viscous force (9) & Shearing stress
(Do the same thing we did for pressure)
stress below the upper
boundary (inside the volume):
t zx Dz
t zx  t zx0 
z 2
mid point:
(x0, y0, z0)
.
Dz
Dy
stress above the lower
 boundary (inside the volume):
Dx
tzx0 = stress at (x0, y0, z0)

t zx Dz 
t zx  t zx0 


z 2 
Viscous force (10) & Shearing stress
Viscous force = shearing stress * area
Area of side C: DyDx
Viscous force below the upper
boundary:
C
Dz
.
Dy
D
Dx

Area of side D: DyDx
FCzx

t zx Dz 
 t zx0 
DyDx

z 2 
Viscous force above the lower
boundary:

t zx Dz 
FDzx  t zx0 
DyDx

z 2 
Viscous force (11)
Net viscous force (inside a volume) in the
x direction due to the vertical (z) shear of the
x velocity component (u):
Sum of the forces that act below the upper
boundary C and above the lower boundary D
on the fluid F  F  F
zx
Czx
Dzx
t zx 
 
DzDyDx
 z 
We want the viscous
force per unit mass:

Fzx 1 t zx 
 

m   z 
Viscous force (12)
(using definition of t)
Recall the definition of the shearing stress:
u
t zx  
z
It follows:

Fzx 1 t zx

m  z
1  u

( )
 z z
Viscous force (13)
μ is also called dynamic viscosity coefficient.
Assume that μ is constant, then:
Fzx   u
u

 2
2
m  z
z
2
2
 ≡ /kinematic viscosity coefficient
Viscous force (14)
Do the same for the shearing stress in x direction
due to the horizontal (x and y) shear of the
x velocity component (u): txx and tyx
The total frictional (viscous) force Fr in the x direction
is called Frx .
It is the sum of all viscous forces acting in x:
Frx
u u u
 ( 2  2  2 )
m
x y z
2
2
2
Viscous force (15)
Do same for other directions of force
N-S (y) direction:
 v  v  v
 ( 2  2  2 )
m
x y z
Fry
2
velocity vector:
2
Frz
 w  w  w
 ( 2  2  2 )
m
x
y
z
2
vertical direction:
2
v  (u,v,w)
2
2
Viscous force (16)
Viscous force per unit mass written in vector form:
F
2
  (v )
m
Kinematic viscosity coefficient
Where Laplacian is operating on the velocity vector:

v  u,v,w 
Our surface forces
dv
1
2
  p   (v ) 
dt

other forces
Here, we use the text’s convention that the velocity is
v  u,v,w 
Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
Summary: Surface forces
• Pressure gradient force and viscous force are
examples of surface forces.
• They were proportional to the area of the
surface of our particle of atmosphere.
• They are independent of the mass of the
particle of atmosphere.
• They depend on characteristics of the particle
of atmosphere; characteristics of the flow.
Body forces
• Body forces act on the center of mass of the
parcel of fluid.
• Magnitude of the force is proportional to the
mass of the parcel.
• The body force of interest to dynamic
meteorology is gravity.
Newton’s Law of Gravitation
Gm1m2 r 
F
 
2
r 
r
Newton’s Law of Gravitation: Any two elements of
mass in the universe attract each other with a force
proportional to their masses m1 and m2 and
inversely
proportional to the square of the distance r

between the centers of the mass.
The force acts along the line joining the particles and
has the magnitude proportional to G, the universal
gravitational constant.
Gravitational Force
Gravitational force for dynamic
meteorology
GMm r 
F   2  
r r 
Newton’s Law of Gravitation:
M = mass of Earth
m = mass of air parcel
r = distance from center (of mass) of Earth to parcel
force directed down, towards Earth, hence ‘-’ sign
G = gravitational constant (6.673 × 10-11 m3 kg-1 s-2 )

Gravitational force per unit mass
Force per unit mass exerted on the atmosphere
by the gravitational attraction of the earth:
F
GM r 
*
 g   2  
m
r r 
r

m
M
g* = gravitational acceleration

Adaptation to dynamical meteorology
GM r 
* r 
* r 
g   2    g0   g0  
r 
r 
a r 
r  a z
*
0
This is the gravitational acceleration at mean sea level.
a: radius of the Earth
z: height above the mean sea level
For
z  a  g  g
*
*
0
Gravitational force per unit mass
F
GM a r 

2
2  
m
a r r 
2
r 
a
*
 g0
2  
(a  z) r 
r 
1
*
 g0
2  
(1 z /a) r 
2
Our momentum equation so far
2
dv
1
a
r
2
*
  p   (v )  g0
2
dt

(a  z) r
+ other forces
Material
derivative of
the velocity
vector v
Pressure
gradient
force
Viscous
force
Gravitational
force
Apparent forces
• Due to the fact that we live on a rotating
planet
• Centrifugal force
• Coriolis force
Back to Basics:
Newton’s Laws of Motion
• Law 1: Bodies in motion remain in motion with
the same velocity, and bodies at rest remain
at rest, unless acted upon by unbalanced
forces.
• Law 2: The rate of change of momentum of a
body with time is equal to the vector sum of all
forces acting upon the body and is the same
direction.
• Law 3: For every action (force) there is and
equal and opposite reaction.
Back to basics:
A couple of definitions
• Newton’s laws assume we have an “inertial”
coordinate system; that is, and absolute frame
of reference – fixed, absolutely, in space.
• Any motion relative to the coordinate system
fixed in space is known as inertial motion.
• Velocity is the change in position of a particle
(or parcel). It is a vector and can vary either
by a change in magnitude (speed) or
direction.
Non-inertial coordinate systems
Unit 6,frames 16-21:
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf
• The Earth rotates, it is an accelerating frame of
reference, hence:
• Our coordinate system ‘north, south, east, west’
is a non-inertial, non-absolute coordinate
system.
• The ‘north’ direction in Ann Arbor is different
from the ‘north’ direction in San Francisco or
Sydney, Australia.
Two coordinate systems
Can describe the velocity
and forces (acceleration)
in either coordinate
system.
dx dx '
z
y
x
dt
or
dt
One coordinate system related to another by:
x'  Ax (t ) x  Bx (t ) y  C x (t ) z
y '  Ay (t ) x  By (t ) y  C y (t ) z
z '  Az (t ) x  Bz (t ) y  C z (t ) z
Example: Velocity (x’ direction)
x' Ax (t)x  Bx (t)y  Cx (t)z
dx'
dx
dy
dz
 Ax (t)  Bx (t)  Cx (t) 
dt
dt
dt
dt
d(Ax (t))
d(Bx (t))
d(Cx (t))
x
y
z
dt
dt
dt
So we have the velocity relative to the coordinate
system and the velocity of one coordinate system
relative to the other.

This velocity of one coordinate system relative to the other
leads to apparent forces. They are real, observable
forces to the observer in the moving coordinate system.
Two coordinate systems
z’ axis is the same as z,
and there is rotation of
the x’ and y’ axis
z’
z
y
y’
x
x’

Example:
One coordinate system related to another by
x' x cos(t)  y sin( t)
y' x sin( t)  y cos(t)
z' z
2
2
2
with  


 7.292 105 s1
T sidereal day 86164 s

We also use the symbol  for .
 is the angular speed of rotation of the earth.
T is time needed to complete rotation.
On Earth it takes a sidereal day = 86164 s.
Acceleration (force) in rotating coordinate system
d 2 x' dx 2
dy 2

cos(t) 
sin( t)
2
dt
dt
dt
dx

dy
 2 sin( t)  cos(t)
dt

dt
  2 (x cos(t)  y sin( t))
The apparent forces that are proportional to rotation and
the velocities in the inertial system (x,y,z) are called the
Coriolis forces.

The apparent forces that are proportional to the square of
the rotation and position are called centrifugal forces.
Acceleration (force) in rotating coordinate system
d 2 x' dx 2
dy 2

cos(t) 
sin( t)
2
dt
dt
dt
dx

dy
 2 sin( t)  cos(t)   2 (x cos(t)  y sin( t))
dt

dt
d 2 y'
dx 2
dy 2

sin( t) 
cos(t)
2
dt
dt
dt
dx

dy
 2 cos(t)  sin( t)   2 (x sin( t)  y cos(t))
dt

dt
d 2 z' d 2 z
 2
2
dt
dt
Coriolis forces
Centrifugal forces