Geology Reference
In-Depth Information
4.3.4 Geometry of free polar motion, wobble and sway
Neglecting the small dissipation reflected in the high value of
Q
, free polar motion
is governed by equation (4.60) with σ
0
real. The solution to this equation has
the form
e
i
(σ
0
t
+
δ)
tanβ,
m
=
(4.78)
with δ an arbitrary phase constant and β the angle the rotation axis makes with the
reference axis
e
3
. The instantaneous pole of rotation describes a circle of radius
d
tanβ about the reference pole with a period of 435.8 days in the prograde sense.
The axis of rotation cuts out a body cone (the polhode) in the Earth every complete
revolution. Changes in the trace of the inertia tensor are second order in small
quantities. Correct to first order in small quantities, the components of the total
angular momentum are
A
2
k
2
d
5
3
G
Ω
L
1
=
A
Ω
m
1
+
r
13
Ω=
+
Ω
m
1
,
(4.79)
A
2
k
2
d
5
3
G
Ω
L
2
=
A
Ω
m
2
+
r
23
Ω=
+
Ω
m
2
,
(4.80)
A
+
2
C
−
A
−
2
k
2
d
5
3
G
Ω
k
2
d
5
3
G
Ω
L
3
=
C
Ω=
Ω+
Ω
.
(4.81)
Thus, using (4.61), the total angular momentum vector is
A
1
2
(
ω
k
2
d
5
3
GA
Ω
L
=
+
+
σ
0
e
3
).
(4.82)
Since the torques have been assumed to vanish,
L
is an invariable axis in space.
Further, because
L
is a linear combination of vectors in the directions of
ω
and
e
3
,
the vectors
ω
,
e
3
and
L
are coplanar.
Let γ be the angle between the rotation axis and the invariable axis of angular
momentum. Then, the angle between the reference axis and the invariable axis is
β
−
γ, as shown in Figure 4.10. Since
ω
=Ω
m
1
e
1
+Ω
m
2
e
2
+Ω
e
3
,
(4.83)
we have that
|
ω
|=Ω
secβ.
(4.84)
Writing
A
1
2
k
2
d
5
3
GA
Ω
A
=
+
,
(4.85)
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