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AOSS 321, Winter 2009 Earth System Dynamics
Lecture 82/3/2009
Christiane Jablonowski Eric Hetland
[email protected] [email protected]
734-763-6238 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
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swfCheck out Unit 6 (winds), frames 20 & 21:
Circle Basics
ω
θ
s = rθ
r (radius)
Arc length ≡ s = rθ
€
ds
dt= r
dθ
dt
ω ≡dθ
dt
v ≡ds
dt
€
... Magnitude
Centripetal acceleration:towards the axis of rotation
€
dr v
dt= −
r v
dθ
dt
r r
rω
€
... Magnituder (radius)
centrifugalcentripetal
€
rv + δ
r v
€
δθ
€
rv
€
δ r
v
€
rv
€
δ r
v =r v δθ
Centripetal force: for our purposes
€
dr v
dt= −
r v
dθ
dt
r r
rdθ
dt≡ ω ≡ Ω( )
r v = ωr
dr v
dt= −ω2r
r
Directed toward the axis of rotation
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?
Ω
Ω2RR
Earth
this is a vector forcedirected away from the axis of rotation
€
Ω=7.292 ×10−5 s−1
Earth’s angular speed of rotation:
axis of rotation
Magnitude of R
Ω
R
Earth
Magnitude (length) of vector RR = |R| = a cos()
Φ
a
latitude
Earth’s radius
Ω
R
Earth
Φ
a
Φ: latitude
a: Earth’s radius
Tangential coordinate system
Place a coordinate system on the surface.
x = west-east (longitude)y = south-north (latitude)z = local vertical
x
y z
Ω
R
Earth
Φ
a
Φ: latitude
a: Earth’s radius
Angle between R and axes
x
-y
z Φ
Reversed (negative)y direction
Assume magnitude of vector in direction R
Ω
R
Earth
Φ = latitude
a
B: Vector of magnitude B
Vertical component
Ω
R
Earth
Φ = latitude
a
z component = B cos()
Meridional component
Ω
R
Earth
Φ = latitude
a -y component = -B sin()
What direction does the Earth’s centrifugal force point?
Ω
Ω2RR
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.
a
What direction does the Earth’s centrifugal force point?
Ω
Ω2RR
Earth
So there is a component that is in the same coordinate direction as gravitational acceleration: ~ aΩ2cos2()
And there is a component pointing towards the equator ~ - aΩ2cos()sin()
Φ = latitude
What direction does the gravitational acceleration point?
Ω
R
Earth
a (radius)
€
F
m= −g0
* a2
(a + z)2
r
r
So we re-define gravitational acceleration g*
as gravity g
€
F
m= −g0
* a2
(a + z)2+ aΩ2 cos2(φ)
⎛
⎝ ⎜
⎞
⎠ ⎟r
r
F
m= −g
r
r
What direction does the Earth’s centrifugal force point?
Ω
Ω2RR
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.
Sphere
g*g
g: gravity
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
€
dr v
dt= −
1
ρ∇p + ν∇ 2(
r v ) +
r g + other forces
Pressure gradientforce
Viscous force
Gravity force = gravitational force + centrifugal force
€
rg ≡ −g
r k =
r g * + Ω2
r R
with g = 9.81 m s-2
Material derivative of the velocity vector
€
rv
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-go-round 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/fw/crls.rxml
Apparent forces:Coriolis force
• Observe the flying aircrafts• http://www.classzone
.com/books/earth_science/terc/content/visualizations/es1904/es1904page01.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
• 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
€
L = (Ωacosφ + u)acosφ
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf
Angular speed(circle)
€
v = ωrω
ΔθΔvr (radius)
v
v
Earth’s angular momentum (1)
Ω
R
Earth
Φ = latitude
a
What is the speed of this point due only to the rotation of the Earth?
€
v = ΩR
Earth’s angular momentum (2)
Ω
R
Earth
Φ = latitude
a
Angular momentum is
€
L = v R
Earth’s angular momentum (3)
Ω
R
Earth
Φ = latitude
a
Angular momentum due only to rotation of Earth is
€
L = v R
L = ΩR2
Earth’s angular momentum (4)
Ω
R
Earth
Φ = latitude
a
Angular momentum due only to rotation of Earth is
€
L = ΩR2
L = Ωa2 cos2(φ)
Angular momentum of parcel (1)
Ω
R
Earth
Φ = latitude
a
Assume there is some x velocity, u. Angular momentum associated with this velocity is
uRLu =
Total angular momentum
Ω
R
Earth
Φ
a
Angular momentum due both to rotation of Earth and relative velocity u is
)(
))cos()(cos(
)cos()(cos
2
22
2
R
uRL
uaaL
uaaL
uRRL
+Ω=
+Ω=+Ω=
+Ω=
φφφφ
Displace parcel south (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air south (or north). What happens? Δy y
Displace parcel south (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR (R gets longer)
y
yR −= )(sinφ
For the southward displacement we get
Displace parcel south (3)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
But if angular momentum is conserved, then u must change.
)()(
)(
2
2
RR
uuRR
R
uRL
++
+Ω+=
+Ω=y
ΔR
Displace parcel south (4)(Conservation of angular momentum)
)()()( 22
RR
uuRR
R
uR
++
+Ω+=+Ω
Expand right hand side, ignore second-order difference terms, solve for u (change in eastward velocity after southward displacement):
RR
uRu −Ω−≈ 2
Displace parcel south (5)(Conservation of angular momentum)
yR −= )(sinφ
yR
uyu +Ω≈ )(sin)(sin2 φφ
For our southward displacement
ya
uyu +Ω≈ )(sin
)(cos)(sin2 φ
φφ
Displace parcel south (6)(Conservation of angular momentum)
€
u
Δt≈ 2Ωsin(φ)
Δy
Δt+
u
acos(φ)sin(φ)
Δy
Δt
Divide by Δt:
Displace parcel south (7)(Conservation of angular momentum)
€
du
dt= 2Ωsin(φ) +
u
acos(φ)sin(φ)
⎛
⎝ ⎜
⎞
⎠ ⎟dy
dt
Take the limit Δt0:
€
du
dt= 2Ωvsin(φ) +
uv
atan(φ)
Coriolis termTotal derivative Metric term
v
Displace parcel south (8)(Conservation of angular momentum)
€
du
dt= 2Ωvsin(φ) +
uv
atan(φ)
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
Earth
Φ
a
Let’s imagine we move our parcel of air up (or down). What happens? Δz
Δz
Displace parcel up (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR (R gets longer)
zR = )(cosφ
For our upward displacementΔz
Displace parcel up (3)(Conservation of angular momentum)
)()()( 22
RR
uuRR
R
uR
++
+Ω+=+Ω
Expand right hand side, ignore second-order difference terms, solve for u (change in eastward velocity after vertical displacement):
RR
uRu −Ω−≈ 2
Do the same form of derivation (as we did for the southward displacement)
Displace parcel up (4)(Conservation of angular momentum)
Divide by Δt:
€
Δu ≈ −2ΩΔR −u
RΔR
⇔ Δu ≈ −2ΩcosφΔz −u
acosφcosφΔz
⇔ Δu ≈ −2ΩcosφΔz −u
aΔz
€
u
Δt≈ −2Ωcosφ
Δz
Δt−
u
a
Δz
Δt
w
Displace parcel up (5)(Conservation of angular momentum)
€
du
dt= −2Ωw cos(φ) −
uw
a
Remember this is ONLY FOR a VERTICAL displacement.
Take the limit Δt0:
So far we got(Conservation of angular momentum)
€
du
dt= 2Ωvsin(φ) +
uv
atan(φ) − 2Ωw cos(φ) −
uw
a
Coriolis
term
Total derivative
Metric
term
Coriolis
term
Metric
term
From N-S displacement
From upwarddisplacement
Displace parcel east (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air east (or west). What happens? Δx
Displace parcel east (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Well, there is no change of ΔR.
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
• 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.
)(2
R
uRL +Ω=
Displace parcel east (4)(Conservation of angular momentum)
Therefore: excess centrifugal force (per unit mass):
€
Fexcesscentrifugal = (Ω +
u
R)2R − Ω2R
=2Ωu
RR +
u2
R2R
Remember: centrifugal force per unit mass
€
F centrifugal = Ω2R
This is a vector forcewith two terms!
Coriolis term Metric term
Displace parcel east (5)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Vector with component in north-south and vertical direction:Split the two directions.
€
Fexcesscentrifugal =
2Ωu
RR +
u2
R2R
Displace parcel east (6)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
For the Coriolis component magnitude is 2Ωu. For the curvature (or metric) term the magnitude is u2/(a cos())
Displace parcel east (7)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
€
Fexcesscentrifugal
( ) • j = −2Ωu
acosφacosφsinφ
−u2
a2 cos2 φacosφsinφ
= −2Ωusinφ −u2
atanφ
Meridional (N-S) component
Displace parcel east (8)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
€
Fexcesscentrifugal
( ) • k =2Ωu
acosφacosφcosφ
+u2
a2 cos2 φacosφcosφ
= 2Ωucosφ +u2
a
Vertical component
Displace parcel east (9)(Conservation of angular momentum)
These forces in their appropriate component directions are
a
uu
dt
dw
a
uu
dt
d
2
2
)cos(2
)tan()sin(2v
+Ω=
−Ω−=
φ
φφ
Coriolis force and metric terms in 3-D
So let’s collect together all Coriolis forces and metric terms:
€
du
dt= 2Ωv sin(φ) +
uv
atan(φ) − 2Ωw cos(φ) −
uw
a
dv
dt= −2Ωusin(φ) −
u2
atan(φ) −
vw
a
⎛
⎝ ⎜
⎞
⎠ ⎟
dw
dt= 2Ωucos(φ) +
u2
a +
v 2
a
⎛
⎝ ⎜
⎞
⎠ ⎟
2 additional metric terms (due to more rigorous derivation,Holton 2.2, 2.3)
Coriolis force and metric terms
€
du
dt= 2Ωv sin(φ) +
uv
atan(φ)
dv
dt= −2Ωusin(φ) −
u2
atan(φ)
If the vertical velocity w is small (w close to 0 m/s), we can make the following approximation:
€
f ≡ 2Ωsin(φ)Define the Coriolis parameter f:
Coriolis force and metric terms
For synoptic scale (large-scale) motions |u| << ΩR .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
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= 2Ωv sinφ = fv
dv
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= −2Ωusinφ = − fu
Coriolis force
For synoptic scale (large-scale) motions:
€
du
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= fv
dv
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= − fu
where is the horizontal velocity vectorand is the vertical unit vector.
€
dr v
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= − fr k ×
r v ( )
Vector notation:
€
rv = (u,v,0)
€
rk = (0,0,1)
Since -k v is a vector rotated 90° to the right of vit 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
€
dr v
dt= −
1
ρ∇p + ν∇ 2(
r v ) − g
r k + fv
r i − fu
r j
+ other forces
Pressure gradientforce
Viscous force
Gravity
force
Material derivative of
€
rv
Coriolis
forces
Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
In Northern Hemisphere velocity is deflected to the right by the Coriolis force
Class exercise: Coriolis forceSuppose 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
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= 2Ωv sin(φ) − 2Ωw cos(φ)
dv
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= −2Ωusin(φ)
dw
dt
⎛
⎝ ⎜
⎞
⎠ ⎟Co
= 2Ωucos(φ)
Coriolis forces: