Geology Reference
In-Depth Information
after replacing f and f by expressions (5.169) and (5.170) in the first-order terms,
respectively, and by e and e in the second-order terms. To first order, equation
(5.181) reduces to
5
2 m ,
de +
2 e =
(5.182)
which, on squaring, yields the second-order relation
25
d 2 e 2
4 dee +
4 e 2
4 m 2
+
=
,
(5.183)
and, on multiplication through by e , yields another second-order relation
5
2 me .
dee +
2 e 2
=
(5.184)
These permit the reduction of (5.181) to the condition
5
2 m
10
6
7 me
4
de +
21 m 2
7 e 2
2 e
+
=
0.
(5.185)
Similarly, replacing f and f by e and e in second-order terms, (5.161) gives
d 2 e 2
2 dee
+
16κ =
0,
(5.186)
which reduces to
5
4 me
25
d κ +
16 m 2
+
=
0,
(5.187)
after substitution from (5.183) and (5.184).
There are then two ways of solving for e . We can take the homogeneous first-
order form of (5.172) and integrate it to find the solution that is regular at the
geocentre and satisfies the first-order condition (5.182) at the surface. This gives a
first-order estimate of e ,su
cient for calculation of the second-order non-
homogeneous terms in (5.172). The full equation can be integrated to find the solu-
tion that is regular at the geocentre and satisfies the full second-order condition
(5.185). Alternatively, we can use the integral equation for η that resulted from the
Radau transformation. First, ηis calculated assuming F (η)
=
1. Then, the first-order
equation (5.173) is integrated to give
e (0) exp r 0
0
r 0 dr 0
e ( r 0 )
=
.
(5.188)
It is shown later that η( r 0 ) is linear in r 0 for small r 0 , so the integral in the expo-
nent is proper. Substituting e η for de , the surface condition (5.185) becomes the
quadratic equation
1
4 (14
35
8 m
5
e 2
6 m 2
+
+
6 m ) e
+
+
=
0.
(5.189)
 
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