Download AOSS 321, Fall 2006 Earth Systems Dynamics 10/9/2006

AOSS 321, Winter 2009
Earth System Dynamics
Lecture 8
2/3/2009
Christiane Jablonowski
[email protected]
734-763-6238
Eric Hetland
[email protected]
734-615-3177
What are the fundamental forces in
the Earth’s system?
•
•
•
•
•
Pressure gradient force
Viscous force
Gravitational force
Apparent forces: Centrifugal and Coriolis
Can you think of other classical forces and
would they be important in the Earth’s system?
Electromagnetic force
• Total Force is the sum of all of these forces.
Apparent forces: A physical approach
Check out Unit 6 (winds), frames 20 & 21:
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf
Circle Basics
Arc length ≡ s = rθ
ω
r (radius)
θ
ds
d
r
dt
dt
d

dt
ds
v
dt
...
s = rθ
 
Magnitude
Centripetal acceleration:
towards the axis of rotation
dv
d r
 v
dt
dt r
ω
centripetal

centrifugal

v
r (radius)

v


v  v
v
v  v 
...
Magnitude
Centripetal force: for our purposes
Directed toward
the axis of
rotation
dv
d r
 v
dt
dt r
d
   
dt
v  r
dv
2
  r
dt
Now we are going to think about the
Earth
• The preceding was a schematic to think about the
centripetal acceleration problem.
Remember Newton’s third law:
"For every action, there is an equal and opposite reaction.”
• View from fixed system: uniform centripetal
acceleration towards the axis of rotation
• View from rotating system: centrifugal
acceleration (directed outward) equal and
opposite to the centripetal acceleration (directed
inward)
What direction does the Earth’s
centrifugal force point?
axis of rotation
Ω
R
Ω2R
this is a vector force
directed away from
the axis of rotation
Earth’s angular speed of rotation:
Earth
  7.292 105 s1
Magnitude of R
Ω
Magnitude (length) of vector R
R = |R| = a cos(f)
R
Φ
Earth
a
Earth’s radius
latitude
Tangential coordinate system
Place a coordinate
system on the surface.
Ω
y
R
a
z
x
x = west-east (longitude)
y = south-north (latitude)
z = local vertical
Φ
a: Earth’s radius
Earth
Φ: latitude
Angle between R and axes
Ω
z
R
Φ
Φ
x
a
-y
Reversed (negative)
y direction
a: Earth’s radius
Earth
Φ: latitude
Assume magnitude of vector in
direction R
Ω
B: Vector of magnitude B
R
a
Φ = latitude
Earth
Vertical component
Ω
z component = B cos(f)
R
a
Φ = latitude
Earth
Meridional component
Ω
R
a
Φ = latitude
Earth
-y component = -B sin(f)
What direction does the Earth’s
centrifugal force point?
Ω
Ω2R
R
a
Earth
So there is a component
that is in the same
coordinate direction as
gravity (and local vertical).
And there is a component
pointing towards the equator
We are now explicitly
considering a coordinate
system tangent to the
Earth’s surface.
What direction does the Earth’s
centrifugal force point?
Ω
R
So there is a component
that is in the same
coordinate direction as
gravitational acceleration:
2
ΩR
~ aΩ2cos2(f)
Φ = latitude
Earth
And there is a component
pointing towards the
equator
~ - aΩ2cos(f)sin(f)
What direction does the gravitational
acceleration point?
Ω
2
F
a
r
*
 g0
2
m
(a  z) r
R
a (radius)

Earth
So we re-define
gravitational acceleration g*
as gravity g
F
m
F
m
 * a 2
r
2
2
 g0
 a cos (f )
2
 (a  z)
r
r
 g
r
What direction does the Earth’s
centrifugal force point?
Ω
Sphere
Ω2R
R
g*
g
g: gravity
Earth
And there is a
component pointing
towards the equator.
The Earth has bulged to
compensate for the
equatorward component
(how much?)
Hence we don’t have to
consider the horizontal
component explicitly.
Centrifugal force of Earth
• Vertical component incorporated into
re-definition of gravity.
• Horizontal component does not need to be
considered when we consider a coordinate
system tangent to the Earth’s surface,
because the Earth has bulged to compensate
for this force.
• Hence, centrifugal force does not appear
EXPLICITLY in the equations.

Our momentum equation so far
dv
1
2
  p   (v )  g + other forces
dt

Material
derivative of
the velocity
vector v
Pressure
gradient
force
Viscous
force
Gravity force =
gravitational force +
centrifugal force
g  gk  g   R
*
2
with g = 9.81 m s-2
Where is the low pressure center?
How and why
does the
system rotate?
Apparent forces:
A physical approach
• Consider a dynamics field experiment in which
one student takes a position on a merry-goround and another student takes a position
above the ground in an adjacent tree.
• Merry-go-round is spinning, a ball is pushed
• On the Merry-go-round: the ball is deflected
from its path. This is due to the Coriolis force.
• http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/f
w/crls.rxml
Apparent forces:
Coriolis force
• Observe the flying aircrafts
• http://www.classzone.com/books/earth_science
/terc/content/visualizations/es1904/es1904page
01.cfm?chapter_no=visualization
• What happens?
• http://www.physics.orst.edu/~mcintyre/coriolis/
Effects of the Coriolis force
on motions on Earth
Angular momentum
• Like momentum, angular momentum is conserved in
the absence of torques (forces) which change the
angular momentum.
• The absolute angular momentum per unit mass of
atmosphere is
L  (acos f  u)acos f
• This comes from considering the conservation of
momentum of a body in constant body rotation in the
polar coordinate system.
•Coriolis force & angular momentum: Check out Unit 6,
frames 25-32
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf
Angular speed
(circle)
v  r
ω
r (radius) v
Δθ

v
Δv
Earth’s angular momentum (1)
What is the speed of this
point due only to the
rotation of the Earth?
Ω
R
a
Φ = latitude
Earth

v  R
Earth’s angular momentum (2)
Angular momentum is
Ω
L  vR
R
a
Φ = latitude

Earth
Earth’s angular momentum (3)
Angular momentum due
only to rotation of Earth
is
Ω
R
a
Φ = latitude
Earth
L  vR
L  R
2
Earth’s angular momentum (4)
Angular momentum due
only to rotation of Earth
is
Ω
R
a
Φ = latitude
Earth

L  R
2
L  a cos (f )
2
2
Angular momentum of parcel (1)
Assume there is some x
velocity, u. Angular
momentum associated
with this velocity is
Ω
R
a
Φ = latitude
Earth
Lu  uR
Total angular momentum
Angular momentum due
both to rotation of Earth
and relative velocity u is
Ω
R
a
Φ
Earth
L  R 2  uR
L  a cos (f )  ua cos(f )
L  a cos(f )(a cos(f )  u )
u
2
L  R (  )
R
2
2
Displace parcel south (1)
(Conservation of angular momentum)
Ω
y
R
a
Φ
Earth
Let’s imagine we move
our parcel of air south
(or north). What
happens? Δy
Displace parcel south (2)
(Conservation of angular momentum)
We get some change
ΔR (R gets longer)
Ω
R
a
y
For the southward
displacement we get
R   sin( f )y
Φ
Earth
Displace parcel south (3)
(Conservation of angular momentum)
Ω
ΔR
R
a
But if angular momentum is
conserved, then u must
change.
y
u
L  R (  )
R
2
u  u
 ( R  R) ( 
)
R  R
2
Φ
Earth
Displace parcel south (4)
(Conservation of angular momentum)
Expand right hand side, ignore second-order
difference terms, solve for u (change in eastward
velocity after southward displacement):
u
u  u
2
R (  )  ( R  R) ( 
)
R
R  R
2
u
u  2R  R
R
Displace parcel south (5)
(Conservation of angular momentum)
For our southward displacement R   sin( f )y
u
u  2 sin( f )y  sin( f )y
R
u
u  2 sin( f )y 
sin( f )y
a cos(f )
Displace parcel south (6)
(Conservation of angular momentum)
Divide by Δt:
u
y
u
y
 2sin( f )

sin( f )
t
t acos(f )
t


Displace parcel south (7)
(Conservation of angular momentum)
Take the limit Δt0:
v
dy
du 
u
 2sin( f ) 
sin( f )
dt 
acos(f )
 dt
du
uv
 2vsin( f )  tan( f )
dt
a
Total
derivative
Coriolis term
Metric term
Displace parcel south (8)
(Conservation of angular momentum)
du
uv
 2vsin( f )  tan( f )
dt
a
What’s this? “Curvature or metric term.” It takes
into account that y curves, it is defined on the
surface of the Earth. More later.
Remember this is ONLY FOR a NORTH-SOUTH
displacement.
Displace parcel up (1)
(Conservation of angular momentum)
Ω
R
a
Φ
Earth
Δz
Let’s imagine we move
our parcel of air up (or
down). What happens?
Δz
Displace parcel up (2)
(Conservation of angular momentum)
We get some change
ΔR (R gets longer)
Ω
R
Φ
Earth
Δz
a
For our upward
displacement
R  cos(f )z
Displace parcel up (3)
(Conservation of angular momentum)
Do the same form of derivation
(as we did for the southward displacement)
Expand right hand side, ignore second-order
difference terms, solve for u (change in eastward
velocity after vertical displacement):
u
u  u
2
R (  )  ( R  R) ( 
)
R
R  R
2
u
u  2R  R
R

Displace parcel up (4)
(Conservation of angular momentum)
u
u  2R  R
R
u
 u  2cos fz 
cos fz
acos f
u
 u  2cos fz  z
a
Divide by Δt:
u
z u z
 2cos f 
t
t a t
w
Displace parcel up (5)
(Conservation of angular momentum)
Take the limit Δt0:
du
uw
 2wcos(f ) 
dt
a

Remember this is ONLY FOR a
VERTICAL displacement.
So far we got
(Conservation of angular momentum)
From N-S
displacement
From upward
displacement
du
uv
uw
 2vsin( f )  tan( f )  2wcos(f ) 
dt
a
a
Total
derivative
Coriolis
term
Metric
term
Coriolis
term
Metric
term
Displace parcel east (1)
(Conservation of angular momentum)
Ω
R
a
Φ
Earth
Let’s imagine we move
our parcel of air east (or
west). What happens?
Δx
Displace parcel east (2)
(Conservation of angular momentum)
Well, there is no change
of ΔR.
Ω
R
a
Φ
Earth
But the parcel is now
rotating faster than the
earth:
Centrifugal force on
the object will be
increased
Displace parcel east (3)
(Conservation of angular momentum)
• Remember: Angular momentum
u
L  R (  )
R
2
• The east displacement changed u (=dx/dt).
Hence again we have a question of
conservation of angular momentum.
• We will think about this as an excess (or
deficit) of centrifugal force per unit mass
relative to that from the Earth alone.
Displace parcel east (4)
(Conservation of angular momentum)
Remember: centrifugal force per unit mass
F
centrifugal
 R
2
Therefore: excess centrifugal force (per unit mass):

u 2
2
F
 (  ) R   R
R
2
2u
u
This is a vector force

R 2 R
with two terms!
R
R
centrifugal
excess
Coriolis term
Metric term
Displace parcel east (5)
(Conservation of angular momentum)
Ω
2
2u
u
centrifugal
Fexcess

R 2 R
R
R
R
Φ
Earth
a

Vector with component in
north-south and vertical
direction:
Split the two directions.
Displace parcel east (6)
(Conservation of angular momentum)
Ω
R
Φ
Earth
a
For the Coriolis
component magnitude is
2Ωu.
For the curvature (or
metric) term the
magnitude is u2/(a cos(f))
Displace parcel east (7)
(Conservation of angular momentum)
Ω
F
2u
 j   acos f acos f sin f
centrifugal
excess
R
a
u2
 2
acos f sin f
2
a cos f
2
Φ
Earth

u
 2usin f  tan f
a
Meridional (N-S) component
Displace parcel east (8)
(Conservation of angular momentum)
Ω
F
2u
 k  acos f acos f cos f
centrifugal
excess
R
a
u2
 2
acos f cos f
2
a cos f
2
Φ
Earth

u
 2ucos f 
a
Vertical component
Displace parcel east (9)
(Conservation of angular momentum)
These forces in their appropriate component
directions are
2
dv
u
 2u sin( f )  tan( f )
dt
a
2
dw
u
 2u cos(f ) 
dt
a
Coriolis force and metric terms in 3-D
So let’s collect together all Coriolis forces and
metric terms:
du
uv
uw
 2v sin( f )  tan( f )  2w cos(f ) 
dt
a
a
2
 vw 
dv
u
 2usin( f )  tan( f ) 

 a 
dt
a
2 
2 
dw
u
v
 2ucos(f ) 


2 additional metric
dt
a  a 
terms (due to more
rigorous derivation,
Holton 2.2, 2.3)
Coriolis force and metric terms
If the vertical velocity w is small (w close to 0 m/s),
we can make the following approximation:
du
uv
 2v sin( f )  tan( f )
dt
a
2
dv
u
 2usin( f )  tan( f )
dt
a
Define the Coriolis parameter f:

f  2sin( f)
Coriolis force and metric terms
For synoptic scale (large-scale) motions |u| << R .
Then the metric terms (last terms on previous slide)
are small in comparison to the Coriolis terms. We
discuss this in more detail in our next class (scale
analysis).
This leads to the approximation of the Coriolis
force:
du 
   2v sin f  fv
dt Co
dv 
   2usin f   fu
dt Co
Coriolis force
For synoptic scale (large-scale) motions:
du 
   fv
dt Co
dv 
    fu
dt Co

Vector notation:
dv 
    f k  v
dt Co
 
where v  (u,v,0) is the horizontal velocity vector
and k  (0,0,1) is the 
vertical unit vector.
Since -k  v is a vector rotated 90° to the right of v

it
shows
that
the
Coriolis
force
deflects
and

changes only the direction of motion, not the speed.
Effects of the Coriolis force on
motions on Earth
Summary: Coriolis force
• Fictitious force only arising in a rotating frame of
reference
• Is directed 90º to the right (left) of the wind in the
northern (southern) hemisphere
• Increases in proportion to the wind speed
• Increases with latitude, is zero at the equator.
• Does not change the wind speed, only the wind
direction. Why?
Our approximated momentum
equation so far
dv
1
2
  p   (v )  gk  fvi  fuj
dt

+ other forces
Material
derivative
of v
Pressure
gradient
force
Viscous
force
Gravity
force
Coriolis
forces
Highs and Lows
In Northern
Hemisphere velocity
is deflected to the
right by the Coriolis
force
Motion initiated by pressure gradient
Opposed by viscosity
Class exercise: Coriolis force
Suppose a ballistic missile is fired due eastward at
43° N latitude (assume f ≈10-4 s-1 at 43° N ). If
the missile travels 1000 km at a horizontal speed
u0 = 1000 m/s, by how much is the missile
deflected from its eastward path by the Coriolis
force?
du 
   2v sin( f )  2w cos(f )
dt Co
dv 
Coriolis forces:
   2usin( f )
dt Co
dw 
   2ucos(f )
 dt Co