2.8.2 The Coriolis Acceleration On The Rotating Earth
Now consider the earth. Let i^{∗},j^{∗},k^{∗}, be the usual basis vectors fixed in space with k^{∗} pointing in
the direction of the north pole from the center of the earth and let i,j,k be the unit vectors
described earlier with i pointing South, j pointing East, and k pointing away from the center of
the earth at some point of the rotating earth’s surface p. Letting R
(t)
be the position
vector of the point p, from the center of the earth, observe the coordinates of R
(t)
are constant with respect to i
(t)
,j
(t)
,k
(t)
. Also, since the earth rotates from West
to East and the speed of a point on the surface of the earth relative to an observer
fixed in space is ω
|R |
sinϕ where ω is the angular speed of the earth about an axis
through the poles and ϕ is the polar angle measured from the positive z axis down as
in spherical coordinates. It follows from the geometric definition of the cross product
that
′ ∗
R = ωk × R
Therefore, the vector of Theorem 2.8.4 is Ω = ωk^{∗} and so
◜-=◞0◟-◝
R ′′ = Ω′ × R + Ω× R ′ = Ω× (Ω × R)
since Ω does not depend on t. Formula 2.30 implies
In this formula, you can totally ignore the term Ω×
(Ω × rB)
because it is so small whenever
you are considering motion near some point on the earth’s surface. To see this, note
ω
se◜conds◞ in◟-a da◝y
(24)(3600)
= 2π, and so ω = 7.2722 × 10^{−5} in radians per second. If you are using
seconds to measure time and feet to measure distance, this term is therefore, no larger
than
( )2
7.2722× 10−5 |rB |.
Clearly this is not worth considering in the presence of the acceleration due to gravity which is
approximately 32 feet per second squared near the surface of the earth.
If the acceleration a is due to gravity, then
aB = a − Ω× (Ω× R )− 2Ω × vB =
◜-----------≡◞g◟-----------◝
GM--(R--+rB-)
− |R + rB|3 − Ω× (Ω × R)− 2Ω × vB ≡ g − 2Ω × vB.
Note that
Ω× (Ω × R) = (Ω ⋅R )Ω− |Ω |2 R
and so g, the acceleration relative to the moving coordinate system on the earth is not directed
exactly toward the center of the earth except at the poles and at the equator, although the
components of acceleration which are in other directions are very small when compared with the
acceleration due to the force of gravity and are often neglected. Therefore, if the only force acting
on an object is due to gravity, the following formula describes the acceleration relative to a
coordinate system moving with the earth’s surface.
aB = g− 2(Ω ×vB )
While the vector Ω is quite small, if the relative velocity, v_{B} is large, the Coriolis acceleration
could be significant. This is described in terms of the vectors i
(t)
,j
(t)
,k
(t)
next.
Letting
(ρ,θ,ϕ )
be the usual spherical coordinates of the point p
(t)
on the surface taken with
respect to i^{∗},j^{∗},k^{∗} the usual way with ϕ the polar angle, it follows the i^{∗},j^{∗},k^{∗} coordinates of this
point are
( )
ρsin(ϕ)cos(θ)
|( ρ sin(ϕ)sin (θ) |) .
ρcos(ϕ)
It follows,
i = cos(ϕ)cos(θ) i∗ + cos(ϕ)sin (θ)j∗ − sin(ϕ)k∗
j = − sin(θ)i∗ + cos(θ)j∗ + 0k∗
and
k = sin(ϕ)cos(θ)i∗ + sin(ϕ)sin (θ)j∗ + cos(ϕ)k∗.
It is necessary to obtain k^{∗} in terms of the vectors, i,j,k. Thus the following equation needs to
be solved for a,b,c to find k^{∗} = ai+bj+ck
∗
◜-k◞◟-◝
( 0 ) ( cos(ϕ)cos(θ) − sin (θ) sin (ϕ )cos(θ) ) ( a )
| | | | | |
( 0 ) = ( cos(ϕ)sin (θ) cos(θ) sin (ϕ)sin(θ) ) ( b) (2.32)
1 − sin (ϕ ) 0 cos(ϕ) c
(2.32)
The first column is i, the second is j and the third is k in the above matrix. The solution is
a = −sin
Remember ϕ is fixed and pertains to the fixed point, p
(t)
on the earth’s surface. Therefore, if the
acceleration a is due to gravity,
′ ′ ′ ′
aB = g − 2ω [(− y cosϕ)i+ (x cosϕ + zsinϕ)j− (y sin ϕ)k]
where g = −
GM (R+r )
-|R+rB|3B-
− Ω×
(Ω ×R )
as explained above. The term Ω×
(Ω× R )
is pretty small
and so it will be neglected. However, the Coriolis force will not be neglected.
Example 2.8.5Suppose a rock is dropped from a tall building. Where will it strike?
Assume a = −gk and the j component of a_{B} is approximately
′ ′
− 2ω (x cosϕ + zsinϕ).
The dominant term in this expression is clearly the second one because x^{′} will be small. Also, the i
and k contributions will be very small. Therefore, the following equation is descriptive of the
situation.
′
aB = − gk − 2z ωsin ϕj.
z^{′} = −gt approximately. Therefore, considering the j component, this is
2gtω sinϕ.
Two integrations give
( 3 )
ωgt ∕3
sinϕ for the j component of the relative displacement at time
t.
This shows the rock does not fall directly towards the center of the earth as expected but
slightly to the east.
Example 2.8.6In 1851 Foucault set apendulum vibrating and observed the earth rotateout from under it. It was a very long pendulum with a heavy weight at the end so that it wouldvibrate for a long time without stopping^{2}.This is what allowed him to observe the earth rotate out from under it. Clearly such apendulum will take 24 hours for the plane of vibration to appear to make one completerevolution at the north pole. It is also reasonable to expect that no such observed rotationwould take place on the equator. Is it possible to predict what will take place at variouslatitudes?
where T, the tension in the string of the pendulum, is directed towards the point at which the
pendulum is supported, and m is the mass of the pendulum bob. The pendulum can be thought of
as the position vector from
(0,0,l)
to the surface of the sphere x^{2} + y^{2} +
(z − l)
^{2} = l^{2}.
Therefore,
T = − T xi− T yj+T l−-zk
l l l
and consequently, the differential equations of relative motion are
′′ -x- ′
x = − T ml + 2ωy cosϕ
y
y′′ = − T--− 2ω(x′cosϕ+ z′sin ϕ)
ml
and
′′ l−-z ′
z = T ml − g+ 2ωy sinϕ.
If the vibrations of the pendulum are small so that for practical purposes, z^{′′} = z = 0, the last
equation may be solved for T to get
All terms of the form xy^{′} or y^{′}y can be neglected because it is assumed x and y remain small.
Also, the pendulum is assumed to be long with a heavy weight so that x^{′} and y^{′} are also small.
With these simplifying assumptions, the equations of motion become
x′′ + gx-= 2ωy ′cosϕ
l
and
y′′ + gy-= − 2ωx′cosϕ.
l
These equations are of the form
′′ 2 ′ ′′ 2 ′
x + a x = by ,y + a y = − bx (2.34)
(2.34)
where a^{2} =
g
l
and b = 2ω cosϕ. Then it is fairly tedious but routine to verify that for each
constant, c,
( ) ( )
( bt) √b2 +-4a2 ( bt) √b2 +-4a2
x = csin 2- sin ---2----t ,y = ccos 2- sin ---2----t (2.35)
(2.35)
yields a solution to 2.34 along with the initial conditions,
It is clear from experiments with the pendulum that the earth does indeed rotate out from under
it causing the plane of vibration of the pendulum to appear to rotate. The purpose of this
discussion is not to establish these self evident facts but to predict how long it takes for the plane
of vibration to make one revolution. Therefore, there will be some instant in time at which the
pendulum will be vibrating in a plane determined by k and j. (Recall k points away from the
center of the earth and j points East. ) At this instant in time, defined as t = 0, the conditions of
2.36 will hold for some value of c and so the solution to 2.34 having these initial conditions will
be those of 2.35 by uniqueness of the initial value problem. Writing these solutions
differently,
( ) ( ( ) ) ( √ -------)
x (t) sin bt2- --b2 +-4a2
y(t) = c cos (bt) sin 2 t
2
This is very interesting! The vector, c
( (bt) )
sin(2 )
cos bt2
always has magnitude equal to
|c|
but its
direction changes very slowly because b is very small. The plane of vibration is determined by this
vector and the vector k. The term sin
(√----- )
-b2+24a2t
changes relatively fast and takes
values between −1 and 1. This is what describes the actual observed vibrations of the
pendulum. Thus the plane of vibration will have made one complete revolution when t = T
for
bT-≡ 2π.
2
Therefore, the time it takes for the earth to turn out from under the pendulum is
T = ---4π-- = 2π-secϕ.
2ω cosϕ ω
Since ω is the angular speed of the rotating earth, it follows ω =
2π24
=
1π2
in radians per hour.
Therefore, the above formula implies
T = 24secϕ.
I think this is really amazing. You could actually determine latitude, not by taking readings with
instruments using the North Star but by doing an experiment with a big pendulum. You would set
it vibrating, observe T in hours, and then solve the above equation for ϕ. Also note the
pendulum would not appear to change its plane of vibration at the equator because
lim_{ϕ→π∕2} secϕ = ∞.
The Coriolis acceleration is also responsible for the phenomenon of the next example.
Example 2.8.7It is known that low pressure areas rotate counterclockwise as seen from above in the Northern hemisphere but clockwise in the Southern hemisphere. Why?
Neglect accelerations other than the Coriolis acceleration and the following acceleration which
comes from an assumption that the point p
(t)
is the location of the lowest pressure.
a = − a (rB )rB
where r_{B} = r will denote the distance from the fixed point p
(t)
on the earth’s surface which is
also the lowest pressure point. Of course the situation could be more complicated but this will
suffice to explain the above question. Then the acceleration observed by a person on the earth
relative to the apparently fixed vectors, i,k,j, is
Therefore, one obtains some differential equations from a_{B} = x^{′′}i + y^{′′}j + z^{′′}k by matching the
components. These are
′′ ′
x + a (rB)x = 2ωy cosϕ
y′′ + a (rB )y = − 2ωx ′cosϕ − 2ωz′sin (ϕ)
z′′ + a (rB )z = 2ωy′sin ϕ
Now remember, the vectors, i,j,k are fixed relative to the earth and so are constant
vectors. Therefore, from the properties of the determinant and the above differential
equations,
| | | |
|| i j k ||′ || i j k ||
(r′× rB)′ = || x′ y′ z′|| = || x′′ y′′ z′′||
B || || || ||
x y z x y z
| |
|| i j k ||
= || − a(rB)x + 2ωy′cosϕ − a (rB)y − 2ωx′cosϕ− 2ωz′sin(ϕ) − a(rB)z + 2ωy′sinϕ ||
|| x y z ||
Then the k^{th} component of this cross product equals
( 2 2)′ ′
ω cos(ϕ) y + x + 2ωxz sin (ϕ).
The first term will be negative because it is assumed p
(t)
is the location of low pressure causing
y^{2} + x^{2} to be a decreasing function. If it is assumed there is not a substantial motion in the k
direction, so that z is fairly constant and the last term can be neglected, then the k^{th} component
of
(r′B × rB)
^{′} is negative provided ϕ ∈
( )
0, π2
and positive if ϕ ∈
( )
π2,π
. Beginning with a point
at rest, this implies r_{B}^{′}× r_{B} = 0 initially and then the above implies its k^{th} component is
negative in the upper hemisphere when ϕ < π∕2 and positive in the lower hemisphere when
ϕ > π∕2. Using the right hand and the geometric definition of the cross product, this shows
clockwise rotation in the lower hemisphere and counter clockwise rotation in the upper
hemisphere.
Note also that as ϕ gets close to π∕2 near the equator, the above reasoning tends to break
down because cos
(ϕ)
becomes close to zero. Therefore, the motion towards the low pressure has
to be more pronounced in comparison with the motion in the k direction in order to draw this
conclusion.