Geography Reference
In-Depth Information
The subscript Co indicates that the acceleration is the part of the total acceleration
due only to the Coriolis force. Thus, for example, an object moving eastward in
the horizontal is deflected equatorward by the Coriolis force, whereas a westward
moving object is deflected poleward. In either case the deflection is to the right of
the direction of motion in the Northern Hemisphere and to the left in the Southern
Hemisphere. The vertical component of the Coriolis force in (1.11b) is ordinarily
much smaller than the gravitational force so that its only effect is to cause a very
minor change in the apparent weight of an object depending on whether the object
is moving eastward or westward.
The Coriolis force is negligible for motions with time scales that are very short
compared to the period of the earth's rotation (a point that is illustrated by several
problems at the end of the chapter). Thus, the Coriolis force is not important for
the dynamics of individual cumulus clouds, but is essential to the understanding of
longer time scale phenomena such as synoptic scale systems. The Coriolis force
must also be taken into account when computing long-range missile or artillery
trajectories.
As an example, suppose that a ballistic missile is fired due eastward at 43
◦
N
latitude (f
10
−
4
s
−
1
at 43
◦
N). If the missile travels 1000 km at a horizontal
=
1000 ms
−
1
, by how much is the missile deflected from its eastward
path by the Coriolis force? Integrating (1.12b) with respect to time we find that
speed u
0
=
v
=−
fu
0
t
(1.13)
where it is assumed that the deflection is sufficiently small so that we may let
f
and u
0
be constants. To find the total displacement we must integrate (1.13) with
respect to time:
t
y
0
+
δy
fu
0
t
0
vdt
=
dy
=−
tdt
0
y
0
Thus, the total displacement is
fu
0
t
2
/2
δy
=−
=−
50 km
Therefore, the missile is deflected southward by 50 km due to the Coriolis effect.
Further examples of the deflection of objects by the Coriolis force are given in some
of the problems at the end of the chapter.
The
x
and
y
components given in (1.12a) and (1.12b) can be combined in vector
form as
D
V
Dt
Co
=−
f
k
×
V
(1.14)
where
V
(u, v) is the horizontal velocity,
k
is a vertical unit vector, and the
subscript
C
o indicates that the acceleration is due solely to the Coriolis force.
Since
≡
−
k
×
V
is a vector rotated 90˚ to the right of
V
, (1.14) clearly shows the
