Environmental Engineering Reference
In-Depth Information
the deflection is very small, we can use the small angle approximation to write
ω
2
R
2
g
0
.
1
◦
α
max
≈
≈
and we have substituted for the angular velocity of the Earth: ω
=
2
π/(
1day
)
10
−
5
s
−
1
.
≈
7
.
3
×
Let us now turn our attention to the case of an object moving with velocity
v
on the Earth's surface. The Coriolis acceleration is given by
1
m
F
Cor
=−
2
ω
×
v
e
1
e
2
e
3
0
ω
cos
λω
sin
λ
v
1
=−
2
v
2
v
3
=
2
ω(v
2
sin
λ
−
v
3
cos
λ)
e
1
−
2
ωv
1
sin
λ
e
2
+
2
ωv
1
cos
λ
e
3
.
(8.29)
From now on we shall ignore the
e
3
component since it is negligible compared to
the acceleration due to gravity, which acts also in this direction. Let us now use
Eq. (8.29) to consider two different types of motion close to the surface of the
Earth.
First we take a look at the case of an object dropped downwards. Initially, the
object has velocity
v
0
. Our goal is to deduce its velocity sometime later. The
precise motion is rather complicated because it is non-linear, i.e. once the particle
starts to move the size of the Coriolis and centrifugal forces changes with time, but
if we are happy to neglect terms in
ω
2
(and higher powers of
ω
) then the situation
is much simpler. For a start, we can therefore neglect the centrifugal acceleration.
Moreover, the motion in the
e
3
direction is dominated by the force of gravity, i.e.
v
3
=
≈−
gt.
All that remains is to consider the
v
1
and
v
2
components of the velocity. We obtain
these by integrating the accelerations, i.e.
t
2
ω
t
0
(F
Cor
)
1
m
v
1
≈
d
t
=
(v
2
sin
λ
+
gt
cos
λ)
d
t
0
t
2
ω
t
0
(F
Cor
)
2
m
v
2
=
d
t
=−
(v
1
sin
λ)
d
t.
(8.30)
0
These are coupled equations and as such we cannot go ahead and solve them for
v
1
and
v
2
. However, we have not made full use of the assumption that we can neglect
terms which are proportional to
ω
2
. Immediately we see that in this approximation
v
2
=
0 for all
t
, since the first of the two equations (8.30) tells us that
v
1
∝
ω
and
