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the Earth (
r
=
d
), this expression becomes
2
d
3
1
3
I
33
P
2
(cosθ)
GM
d
−
3
G
1
3
I
22
2
V
=−
3
I
11
+
−
2
2
3
I
13
P
2
(cosθ)cosφ
+
3
I
23
P
2
(cosθ)sinφ
(4.52)
+
1
6
(
I
11
−
1
I
22
)
P
2
(cosθ)cos2φ
−
3
I
12
P
2
(cosθ) sin 2φ
−
+···
.
Since the solid spherical harmonics in expressions (4.41) and (4.52) are linearly
independent, equating
V
2
to
k
2
U
2
following equation (4.34) gives five equations in
the six components of the inertia tensor. A sixth equation is
Tr(
I
)
=
I
11
+
I
22
+
I
33
.
(4.53)
Solution of these equations for the components of the inertia tensor yields
Tr(
I
)
k
2
d
5
3
G
ω
i
ω
j
k
2
d
5
2
1
3
ω
i
I
ij
=
+
−
δ
j
.
(4.54)
3
G
2
d
2
The termω
/3 in the deforming potential (4.41) will cause a purely radial expan-
sion, contributing only equal amounts to
I
11
,
I
22
,
I
33
, thus altering only Tr(
I
)
(Rochester and Smylie, 1974). The contributions to the inertia tensor,
r
ij
, caused by
deformation under the changed centrifugal force, neglecting squares and products
of the small quantities
m
1
,
m
2
,
m
3
,arethen
⎭
2
k
2
d
5
9
G
Ω
1
3
Tr(
R
)
2
m
3
r
11
=
−
2
k
2
d
5
9
G
Ω
1
3
Tr(
R
)
2
m
3
(4.55)
r
22
=
−
4
k
2
d
5
9
G
Ω
1
3
Tr(
R
)
2
m
3
,
r
33
=
+
where Tr(
R
)
=
r
11
+
r
22
+
r
33
is the contribution to the trace of
the inertia
tensor, and
k
2
d
5
3
G
Ω
k
2
d
5
3
G
Ω
2
m
2
,
r
13
2
m
1
.
r
12
=
0,
r
23
=
=
(4.56)
Thus, the complex phasor,
k
2
d
5
3
G
Ω
2
m
,
r
=
r
13
+
ir
23
=
(4.57)
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