Download Chapter 4: Fundamental Forces Newton`s Second Law: F=ma In

Chapter 4: Fundamental Forces
Newton’s Second Law: F=ma
In atmospheric science is typical to consider the force per unit mass acting
on the atmosphere:
Force
a
mass
In order to understand atmospheric motion (accelerations) we need to
know what forces act on the atmosphere.

Body and Surface Forces
Body (or volume) force – a force that acts on a volume of the atmosphere
Surface force – a force that acts locally upon a part of a fluid
Forces in an Inertial Reference Frame
Inertial reference frame – a non-accelerating frame of reference
Gravity – a force that arises from the mutual attraction between two objects
GM r 
g *   2  
 r r 
G – gravitational constant (= 6.673x10-11 N m2 kg-2)
M – product of the mass of the two objects being considered
r – distance between 
objects
For a unit mass of air at the surface of the Earth:
M = 5.9742x1024 kg (mass of Earth)
r = 6.37x106 m (radius of Earth)
g0* = 9.83 m s-2

Geopotential (): work required to raise a unit mass to height z
(z) 
z
 g* z dz
0

Note that g * varies with z since it depends on the distance from the center
of the Earth.
Geopotential height (Z):

 1
Z *  *
g0 g0
z
 g* dz
0
Pressure Gradient Force

Example: Real-time weather map
The force exerted on the left face of this air parcel due to pressure is:
pA  pyz
The force exerted on the right face of this air parcel due to pressure is:

 p 
p  pyz  p  xyz
 x 

The net force exerted by pressure on this air parcel is:

p
xyz
x
By dividing by the mass of the air parcel ( xyz ) we get the force per unit
mass due to changes in pressure (i.e. the pressure gradient force):

P
g i  
1 p
 x
We can write all three components of the pressure gradient force as:

1 p
p
p 
Pg    i 
j  k 
 x
y
z 
In what direction does this force act relative to locations with high and low
pressure?

Example: Calculation of the pressure gradient force from a weather map
Viscous Force
Each layer in this fluid is experiencing a shear stress due to differences in
the speed of motion of adjacent layers of fluid.
  i  
u
z
 - coefficient of shear viscosity (depends on fluid being considered)
 change between adjacent layers of the fluid?
How does the shear stress
What does this imply about the net force acting on a layer of this fluid?
For a net force to arise there must be a gradient in the shear stress.
The force that arises due to a gradient in the shear stress is called a
viscous force and can be represented as:
  2 u  2 u  2 u 
Fr  i   2  2  2 
 x y z 
Kinematic viscosity coefficient:  

(~ 1.5x10-5 m2 s-1 in the atmosphere)


In three dimensions this force is written as:

 2 u  2 u  2 u  
 2  2  2 i 
y z  
x
 2


v  2 v  2 v 
Fr   2  2  2 j  
y z  
x
 2
2
2  

w

w

w 



k
2

y 2 z 2  
x


Hydrostatic Balance
For an atmosphere that is at rest, what forces act in the vertical direction?



1 dp
 g*
 dz
dp
 g *
dz
This equation is known as the hydrostatic equation.
balance between the vertical pressure gradient
Hydrostatic balance – a
force and the gravitational force
Combining the hydrostatic equation with the ideal gas law gives:
 p  z g*
ln    
dz
p0  0 RdT
Using a layer average temperature, T , we can write:


where H 

p  p0 e
z H
Rd T
g*

This equation is known as the hypsometric equation and relates pressure
and height.
Using the definition of geopotential we can also write:
dp
g*

dz
p
Rd T
d  Rd T
dp
p
p
  Rd  T d ln p
p0

R
 Z   *d
g0
p
 T d ln p
p0
Thickness (Z) – the difference in geopotential height (Z) between two
pressure levels

What causes the thickness to increase (or decrease)?
Example: Thickness and atmospheric soundings
Forces in a Rotating Reference Frame
Non-inertial frame of reference – a frame of reference which is undergoing
an acceleration
Why is a frame of reference fixed to the Earth a non-inertial frame of
reference?
Angular velocity of rotation of the Earth:

2
2

 7.292 10 5 s 1
1 sidereal day 86164 sec
Sidereal day – the time required for the earth to make one complete
revolution in an absolute (fixed) coordinate system, that is, with respect to

the stars.
Centrifugal Force
Rotating table experiment:
What happens to the object when it is placed on the rotating table?
What force is required to keep the object stationary on the rotating table?
mr  2
Centrifugal force – outward radial force
Centripetal force – inward radial force

Is this object experiencing an acceleration when viewed in a frame of
reference rotating with the table?
Is this object experiencing an acceleration when viewed from a fixed frame
of reference?
Which forces are acting in each frame of reference?
The centrifugal force arises only in observations taken in a rotating frame of
reference and is due to the acceleration of the frame of reference.
The centrifugal force and the Earth:
What is the direction of the centrifugal force for an object on the Earth?
The centrifugal force is a body force and can be combined with the gravity
force to give an effective gravity.
Effective gravity: g = 9.81 m s-2 at sea level
For atmospheric science applications we use effective gravity (g) rather
than gravity (g*). Therefore effective gravity should be used in place of
gravity in all of the previous equations in this chapter.
The addition of the centrifugal force to the gravity force results in the
effective gravity force not being directed towards the center of the Earth.
In reality the Earth is not a perfect sphere and the effective gravity is
exactly normal to the surface of the Earth at all locations (neglecting
topography). Therefore, the effective gravity force acts only in the zdirection.
This discussion has only considered an object at rest in a rotating frame of
reference.
What happens when an object is in motion relative to a rotating frame of
reference?
Coriolis Force
For an object in motion relative to a rotating frame of reference we need to
consider:
- an additional centrifugal force
- changes in the relative angular momentum of the object
What is the centrifugal force acting on an object moving towards the east?
CFtotal
2

u 
 m   R
R 

2mu
u2
 m R 
R  m 2 R
R
R
2
The first term on the RHS of this equation is the same as the centrifugal
force for an object at rest, and is incorporated into the effective gravity.

The last term on the RHS of this equation is assumed to be small since
R  u .
The remaining term on the RHS of this equation can be written as:

CFmotion  2musin j  2mucosk
and is the centrifugal force due to the motion of the object that we need to
consider in our study of the atmosphere.

This component of the centrifugal force arises only for zonal (east/west)
motion.
Conservation of angular momentum
For a rotating object the angular momentum of the object is given by:
I  12 mR2 
The angular momentum (I) of this object will remain constant.

What happens to the terms in this equation if the object is moving towards
the North Pole (v>0)?
1
2
2 
u 
mR2   12 mR  R  

R


R



R   R   2R R 
2

2
R2u
R  R
As the object moves towards the pole its zonal velocity (u) increases.
Noting that R vsin t we get:
u  2vsin t
Ast  0 we then have:


du
 2vsin 
dt
Conservation of angular momentum results in a zonal acceleration for
objects moving in a meridional (north/south) direction.

Similarly, conservation of angular momentum also results in objects moving
in the vertical direction experiencing a zonal acceleration equal to:
du
 2w cos
dt

Conservation of angular momentum does not lead to an acceleration for
objects moving in a purely zonal direction.
The accelerations due to the additional component of the centrifugal force
and due to conservation of angular momentum for an object in motion in a
rotating frame of reference can be combined to form a single force known
as the Coriolis force.
FCoriolis
 2vsin   2w cos  i  2usin j  2ucosk
m
Using vector notation, and noting that  cosj  sin k the Coriolis
force can also be expressed as:

FCoriolis
 m  2 u
In what direction does the Coriolis force act relative to the motion of the air?
Does this differ in theNorthern and Southern hemispheres?
Example: The Coriolis force on the roof of Duane Physics
The Navier-Stokes Equation
The sum of the forces acting in each direction gives the acceleration
experienced by an air parcel.
 2 u  2 u  2 u 
1 p
   2  2  2  2vsin   2w cos  
 x x
y z 
Du

Dt

Dv

Dt
 2 v  2 v  2 v 
1 p

   

 2u sin  
 y x 2 y 2 z 2 
 2 w  2 w  2 w 
Dw
1 p
 g 
   2  2  2  2u cos  
Dt
 z
y
z 
x
What does each term in these equations represent?

These equations are known as the Navier-Stokes equations.
We can rewrite the Navier-Stokes equations in an Eulerian framework as:
 2 u  2 u  2 u 
1 p
   2  2  2  2vsin   2w cos  
 x
y z 
x
u
u u
u
u v w 
t
x
y
z

v
v v
v
u v w 
t
x y
z
 2 v  2 v  2 v 
1 p

   

 2u sin  
 y x 2 y 2 z 2 
 2 w  2 w  2 w 
w
w w
w
1 p
u
v
w
 g 
   2  2  2  2u cos  
t
x
y
z
 z
y
z 
x
Perturbation Pressure

We can rewrite the vertical momentum equation by defining a reference
pressure, p0, that is in hydrostatic balance with a reference density, 0, as:
dp0
  0 g
dz
The total pressure is then given by:

p  po z   pd
Substituting this definition into the vertical component of the Navier-Stokes
equation gives:

 2 w  2 w  2 w 
Dw


 g  p0 z   pd     2  2  2   2u cos  
Dt
z
y
z 
x
 2 w  2 w  2 w 
p
 g   0 g  d    2  2  2  2u cos  
z
y
z 
x
   0  1 pd  2 w  2 w  2 w 
Dw

g
   2  2  2  2u cos  
Dt

 z
y
z 
x
