# CORIOLIS by liuhongmei

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• pg 1
```									The Coriolis “Force”
Reading Knauss pages 87-95.

Some useful web sites

http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html

http://satftp.soest.hawaii.edu/ocn620/coriolis/

Something funny but not useful and propagates a myth about the Coriolis force.
The Simpson‘s—‗Bart vs Australia‖

What is the speed of a rotating object? Suppose a particle was on a disk that rotated
once each second. The speed of the object would be v=2r, where r is the radius,. In this
example the rotation rate is 2 radians per second. Often we write the rotation rate as,
which can be called the angular velocity and has units of frequency (2*pi/time).

How long does it take for the earth to rotate once? You’d probably guess 24 hours.
But in fact it rotates once every 23 hours 56 minutes and 4 seconds. So the earth actually
spins more that ones in 24 hours but once and 1/(365.25)th of a rotation. This is because
as the earth spins it also moves around the sun and thus requires a little more rotation to
have the same orientation with the sun as it did 24 hours earlier. Since on earth the Sun is
our main time-piece we call 24 hours one day—the amount of time from noon to noon.
But if we’re interested in the rotation of the sun in fixed space —the inertial reference
frame—we use the Sidereal Day. This is why the stars in the sky sift slightly each night.
In fact the stars along the earth’s ecliptic- where the 12 Zodiacs lie- appear 3 minutes and
56 seconds earlier each day due to the difference between the sidereal day and our solar
day.

Now suppose there is a frictionless particle on the earth. This particle, if it is to stay at the
same location on this rotating frame, must accelerate in such a way that it moves with the
underlying earth. Before I get into how this happens –lets look at what the accelerations
are like and how they vary with latitude.

A decent derivation of the force appears on 89 of Knauss and a more complete one
follows on page 90 in box 5.3—but that one requires vector calculus so in class I‘ll go
over the more simpler one on page 89.
The vertical component of this accelerations impacts gravity.

How big is the centrifugal acceleration? On the equator the earth spins at 1000
miles/hour or 464 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. So the
centrifugal acceleration v2/R = 2 R .

In addition to this gravity is reduced further at the equator due to the equatorial
bulge (the radius of the earth at the equator is 22 km larger than at the pole). Together
these contribute to a gravitational difference of 0.5% between the pole, where it is 9.83
m/s2 and the equator where it is 9.78 m/s2.

If Santa weighs 250 lbs at home (at the pole) what would he weigh on the equator?

Would gravity be different on a ship traveling 10 m/s eastward on the equator from a ship
heading 10m/s westward on the equator?

The earth’s gravitational field is further complicated by the earth’s structure—
which is not flat nor smooth. The example in figure 5.8 (page 96) shows how increased
gravity over a ridge results in a rise in the sea-surface, while reduced gravity over a
trench results in reduction of sea-surface. The variations in gravity occur because the
gravitational vector is pointing to local regions of enhanced mass—and since the earth’s
crust is heavier than the ocean’s waters it tends to point away from the trenches and
towards the ridges. This can result in sea level rising 10’s of meters with slopes on
the order of 1 meter per kilometer. Yet this does not drive any motion. Why?

For particle purposes, at least as far as ocean circulation is concerned, we can forget
about the earth’s gravitational anomalies, and set g=9.81 m/s2.

While we can invoke slight variations in gravity to deal with the vertical
acceleration how do we get the horizontal accelerations? Lets use the pole as an
example where the rotation can be thought of as that of a turntable. If we started rotating
a frictionless fluid that was at rest on a turntable it would never turn but rather remain
motionless relative to fixed space while the turntable spun underneath it. Relative to the
turntable the fluid would complete a circle once each rotation in a sense opposite to that
of the rotation of the turntable. However, no matter how hard we try, we have yet to find
a frictionless material outside of a physics book, so the fluid will eventually begin to spin
and at some time later will be in what is called solid body rotation and will have no
motion relative to the turntable. In this state the fluid is now constantly accelerating
inwards. This is achieved by a pressure gradient being set up in the disk with higher
elevations on the sides and a low in the middle. Thus the fluid accelerates down hill at
exactly the same rate that the table turns. A similar thing happens in the ocean and
atmosphere whereby a pressure gradient has been set up in such a way that allows the
fluid to accelerate at the same rate that the earth rotates. This inward acceleration is called
centrifugal acceleration and is equal to v2/R where v is the speed that the tank is moving
around and R is the radius of the tank. This can also be written as 2Rsin( where is
the rate of rotation and is the latitude..

As far as I nobody has proposed a clear and convincing physical interpretation for
the Coriolis force!

To quote Henry Stommel from his book with Dennis Moore ―The Coriolis force‖

“All professional meteorologists and oceanographers encounter this mathematical
demonstration at an early stage in their education. Clutching the teacher’s hand , they
are carefully guided across a narrow gangplnak over the yawing gap between the resting
frame and the uniformly rotating frame. Fearful of looking down into the cold black
water between the dock and the ship, many are glad, once safely aboard, to accept the
idea of a Coriolis force, more or less with a blind faith, confident that it has been derived
rigorously. And some prefer never to look over the side again.
So, when they are asked for an explanation by a landlubber standing on the dock,
who has never been carefully guided across the perilous plank, they find themselves
singularly unable to explain the curious force. Incomplete explanations abound in
popular books and magazines.”

SO despite the relatively complex math required to describe the effect of moving about
on a rotating earth and non-intuitive result—mathematically this result can be
incorporated into the momentum equation.

u
 forcesx
1
 fv                                                      (1)
t        
      
v
 forcesy
1                                                 (2)
 fu 
t          



Think about the size of the term fv. At 45 degrees latitude f is 10-4 so a particle moving
to the north at 1 m/s will accelerate to the east at 10-4 m/s2—indeed a very small number.
After 100 seconds it will have obtained a speed of l mm/s to the east—corresponding to a
change in direction of about 1/20th of a degree. So if you‘re walking northward at 1 m/s
the Coriolis force along would have you walking 1/20th of a degree to the east of north
after one minute. HOwever you never get to feel this because other forces – such as
friction and the forces that you put into walking are probably 1000‘s of time larger than
this.



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
shown above in equations 1 and 2.

Two simple cases are presented—that will be discussed in more detail later in the
semester. The first what oceanographers call inertial motion because it‘s motion that
occurs on the earth solely through it‘s own inertia (note that Newton‘s famous law – a
particle remains in motion at a constant speed and direction unless it is acted on by
another force – is for a fixed coordinate frame and not a rotating frame). Inertial motion
is shown in examples 4 and 5 of the animations of the rotating disk. Note that if there
are no forces (other than this apparent force we call the Coriolis force) and 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 but
the direction of the motion changes. Later we‘ll look more into this

A second example is a Hurricane. Here the Coriolis force balances the pressure gradient
and the winds flow along isobars rather than cross them. This type of momentum balance
can be written as:

1 P
       fv
 x
1 P
       fu
 y

and is called geostrophic flow.
In large scale weather systems (100‘s-
1000‘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. It is why
winds flow clockwise around a high
(in the northern hemisphere) and
counter-clockwise around a low.
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.
When is the Coriolis force
important?

One way to determine if the Coriolis force is important is to compare it to other terms in
the momentum equation. A particularly relevant force to compare it to is the centrifugal
acceleration – because it sort of like the Coriolis acceleration. The centrifugal force is
equal to v2/R where v is velocity and R is the radius of curvature. So if both forces are at
work they can be written as:

v2/R + fv=v(v/R +f)

so the importance of the Coriolis force can be determined by comparing f to V/R.

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                                               u
u
v
u
x    y
R                                                 v    v
u v
x    y
R

Figure 1. Schematic of eddy to demonstrate
that advective term in the momentum equation
in Cartesian coordinates is similar to
centrifugal acceleration.
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                               (3)
fv   fR

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