The Coriolis “Force” Part 2:
Reading Pond and Pickard Section 6.3 & 8.1-8.5
Ocean Circulation Sections 3.2 , 3.3, 3.4
Once again I recommend checking out these web-sites.
http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html
http://satftp.soest.hawaii.edu/ocn620/coriolis/
Last lecture ended with the introduction of the fictitious force the Coriolis force. Let’s
begin by looking at the magnitude of the acceleration that a particle on the equator feels
due to the earth’s rotation. Centrifugal acceleration is equal to v2/R where v is velocity
and R is the radius of curvature. First I will deviate from a this with a Question: Why
doesn’t this term v2/R appear in the momentum equations that we’ve written before?
Answer: Because they were written in Cartesian coordinates. If we were to transform the
advective terms in the momentum equation to polar coordinates the v2/R term would
appear. So the advective terms in the momentum equation that we’ve been writing in
class and the centrifugal acceleration are the same. The both involve a changing velocity
vector (Figure 1) and both are nonlinear—they involve velocity squared.
v2
R
u
u
v
x
y
v
v
u v
x
y
u
R
Figure 1. Schematic of eddy to demonstrate
that advective term in the momentum equation
in Cartesian coordinates is similar to
centrifugal acceleration.
For now we need not worry about the complexity of a fluid and its advective term. How
big is the centrifugal acceleration? On the equator the earth spins at 1000 miles/hour or
446 m/s. The earths radius is ~6380 km, so the centrifugal acceleration is (v2/R) is =.03
m/s2. This can be figured out a different way. The angular velocity of a spinning body
can be written as R , where is the rotation rate is radians per second (for example if
= 2 the body would rotate once time around – or 2 radians each second). The earth
rotates 2 radians every 23 hours 56 minutes and 4 seconds so  7.3 x 10-5 s-1 (the
Sidereal Day). So the centrifugal acceleration v2/R = 2 R . On the equator this
acceleration is in the same direction of gravity—but only about 0.3 percent as strong as
gravity. Thus for this effect alone a 200 lb person on the pole would with 200.6 lbs on
the equator (however there are other process that alter the effective gravity. Fortunately
all of them are small and as oceanographers we can live with a constant gravity worldwide ~ 9.8 m2/s).
But that deals with the vertical acceleration. In oceanography and meteorology most of
the motion is in the horizontal direction. To discuss how this impacts particles in motion
relative to the rotating frame consider the example found in Pond and Pickard (page 42).
Rotation rate 
B’
otion
Apparent M
A
x=vt
B
Displacement of
particle during time
t is xt=vt2.
A particle begins moving from point A towards point B at velocity v—so the distance x
equals vt. However because it is on a rotating plane it ends up at point B’. The distance
from B to B’ is the distance that the rotation has deflected the particle. This is equal to the
angular velocity (which is equal to x=vt ) times the time t. So the distance that it
moved is vt2. The speed of this deflection is the time-derivative of this distance
=2vt, and the acceleration of this acceleration is the time-derivative of speed =2v.
Thus the effect of rotation is to deflect a particle to the right (or left depending on the sign
of the rotation) at a rate equal to the particle velocity times twice the rotation rate—this is
the enigma that the Coriolis frequency (2is twice the rotational frequency . At the
pole this rotation rate is  7.29 10-5 radians/sec. Thus the Coriolis effect accelerates
particles at the North pole to the right at a rate of 2times the speed of the particle
relative to the rotating earth.
In summary the acceleration = the Coriolis frequency times the normal velocity. Often
the letter f is used to symbolize the Coriolis frequency and so this can be written as:
u
 fv
u
v
  fu
t
(1)
(2)
If you forget which equation has the negative sign on the right hand remember that in the
northern hemisphere the flow turns to the right. So northward velocity (v positive)
accelerates the flow to the east (u positive). In contrast a eastward velocity (u positive)
accelerates the flow to the south (v negative). In the northern hemisphere f is postive—so
equations 1 and 2 describe circular motion in a clockwise direction. Draw this out and
see that the flow is always turning to the right and thus flows clockwise. This motion is
also called cum-sole meaning with motion moves in the same direction that the sun
moves across the sky or anti-cyclonic for it is in the opposite direction of the flow around
a cyclone.
The magnitude of Coriolis parameter (which has the units of frequency) changes with
latitude. The Coriolis frequency equals sin(), where is the latitude in degrees and
T where T is the number of seconds in a Sidereal Day (86164 s). On the north
pole f=1.45 x 10 –4, and the inertial period—the time for a particle governed by the
momentum balance described by 1 & 2 to complete one circle, is 11.96 hours. At our
latitude f= 9.37 x 10 –5 and the inertial period is 18.6 hours. In the tropics at 18 degrees
north f=4.5 37 x 10 –5 and the inertial period is 36 hours. At the equator f=0 and the
inertial period is infinity. At 40 degrees south f=- 9.37 x 10 –5 and the Coriolis effect
accelerates a moving particle to the left.
If you’re willing to accept at face value these arguments then all you need to know is that
f=2sin (lat) and that it enters our momentum equation relative to a rotating earth as:
u
 fv 
forces x
t

v
 fu 
forces y
t

Note that if we multiply equation 1 by u and equation 2 by v (without the sum of the
forces term) and add them up we obtain


 (u 2  v 2 )
0
t
Or in other words despite the changing motion the kinetic energy remains the same.
A second example is in the
atmosphere where we have high
pressures and low pressure systems
and therefore pressure gradients. In
large scale weather systems (100’s1000’s of km) most of the
momentum balance is between the
Coriolis force and the pressure
gradient (figure 3). In this case
winds flow along isobars rather
than across them. This type of
momentum balance can be written
as:
1 P
 fv
 x
1 P

 fu
 y

Figure 3. Atmospheric pressure in milibars (mb) at a height of 500
mb showing higher pressure to the south. Wind speed is indicated
by number of sticks on the wind vector. Note the strongest winds
occur where the pressure gradient is largest off the NW coast.
and is called geostrophic flow. It is the first order momentum balance that occurs in much
of the atmosphere and the ocean. It is why winds flow clockwise around a high (in the
northern hemisphere) and counter-clockwise around a low.
So does the toilet flush the way it does because of the Coriolis effect? Let’s assume that
the flow in the toilet is 1 m/s. The Coriolis effect would accelerate the flow at a rate of 1
m/s * 9.37 x 10 –5 m/s2. Thus in the 10 seconds it takes to flush the toilet the fluid
moving down the slope of the bowl would acquire a flow to the right at a speed of * 9.37
x 10 –4 m/s2 or less than 1 mm/s. Compare this to the centrifugal accelerations in the toilet
(v2/R) (assume R = 20 cm) and we find that the centrifugal accelerations are 5 m/s2—or
40,000 times larger then the Coriols effect. Note that you can compare these two terms
v2
R  v
fv
fR
(3)
Where R is the radius of the toilet. Several things to note about equation 3. First for our
toilet bowl example it equals over 50,000 -- indicating that effects of the earth’s rotation
are over 50,000 times smaller than those associated with centrifugal acceleration.
Secondly note that 3 is dimensionless -- so we’ve come up with another non-dimensional
number. This one is known as the Rossby
Number- after Carl-Gustaf Rossby, (who
appeared on the cover of TIME magazine in
1956) More generally this can be written as
v
R
(4)
fL
Where L is a characteristic length scale. It is a
ratio of the fluid’s inertia to effects of the
earth’s rotation. When the Rossby number is
near 1 or less effects of rotation are important.
(like all non-dimensional numbers it’s value
should be viewed as an order of magnitude
estimate). Consider a weather system with a
length scale of 500 km, winds 20 miles per hour
(~ 10 meters/sec) at 40 degrees north. Here the
Rossby number is ~ 0.1 and the effects of the
earth’s rotation are clearly important. For a Hurricane v=100 miles/hour =44 m/ s The
length scale of a hurricane is ~ 200 km. At a latitude of 30 N the Rossby number is about
3—indicating that effects rotation are important but that effects of the winds inertia also
play an important role in the dynamics.
In the ocean if currents are 50 cm/s and for f=10-4 any feature with a length scale of less
than 20 km will have a Rossby number less than 1. Note that in the ocean because
velocities are smaller than in the atmosphere that in the ocean the earth’s rotation effects
smaller scale features than in the atmosphere. Subsequently the eddying motion in the
ocean is of a much smaller scale than that in the atmosphere. This is a challenge to ocean
modelers because to capture all the eddying motion in the ocean requires a much finer
resolution than atmospheric modelers require.
In both the atmosphere and the ocean the first order balance is what we call
“Geostrophic” in which the pressure gradient is balanced by the Corolis acceleration. In
this balance the flow is steady—and thus if you know the pressure gradient you would
know the velocity. This has been useful to Oceanographers who can measure the density
field in the ocean and calculate the pressure gradient and thus estimate the Geostrophic
velocity of the ocean. However, this requires an assumption that the pressure gradient
goes to zero at some depth. This can happen if isopycnals tilt in the opposite direction of
the barotropic pressure gradient and produce a lower layer with no pressure gradient and
thus no geostrophic velocities occurs this level. As such we call this the level the level of
no motion. Since we have velocity above the level—and no velocity below—we have
vertical shear. Note that this requires a tilting of the isopycnals and this corresponds to a
horizontal density gradient.
Thermal Wind Equation
Writing the momentum balance once again for geostrophic flow:
1 P
 fv
 x
(5 & 6)
1 P
  fu
 y
Multiplying 5 by and taking the vertical gradient we get
 P   fv

z x
z
(7)
If we change the order if differentiation we get
 P  ( fv)

x z
z
(8)
Now recall the equation for hydrostatic pressure:


P( z )   g z '
(9)
z
So vertical gradient in Pressure is simply:
P
  g
z
(10)
By using equation 10 in 8 we get
g
  ( fv)

x
z
(11)
In the ocean the change in density in the vertical is small with respect to changes in
density (this is consistent with the Boussinesq approximation) and since f does not
change with height 11 can be rewritten as:
g

v
 f
x
z
(12)
Equations 11 and 12 are known as the thermal wind equations and demonstrate that in a
rotating system a horizontal density gradient generates a vertical shear in the flow.
As a simple example imagine if we were able to run our gravitational adjustment problem
is a large rotating tank. The surface slope—from the light fluid to the dense fluid would
drive the fluid to the right—this can be determined from a geostropic flow for a tilting
sea-surface

g
 fv
x

g
 fu
y
However, below this level there is a
horizontal density gradient and in the
case of this set up it produces a total
pressure gradient at depth that is equal to
the surface pressure gradient- but in the
opposite direction. In the non-rotating
case this produced a current in the upper
layer that flowed in one direction and an
equal and opposite current in the lower
layer. However, if we were able to
rotate the tank the upper layer would
flow into the page and the lower layer
flow out of the page (Figure 4)
Consequently this drives a flow in the
opposite direction producing what we
call “Thermal Wind Shear” for it’s initial
use was to determine atmospheric winds
based on temperature measurements.
Non Rotating release of a gravity current
Rotating release of a gravity current
Figure 4. Gravity current release (upper
panel) in non-rotating system. Lower panel
shows steady state (for infinitely wide tank!)
for rotating system
But we’ve neglected Friction! One place that a level of no motion can be assumed is on
the bottom because flow goes to zero at the bottom because of friction—not due to a
vanishing pressure gradient. After the exam we will talk about friction, how it is
generated and its influence on geophysical flows. Next lecture will be a review.