Geology Reference
In-Depth Information
Then, with A 1,0
A 3,0 , and substituting from (3.207) for A 6,0 , the second and
fourth equations are found to be identical. Solving the first three equations in terms
of A 4,1 yields
=
1
μ
A 1,2 =−
A 3,2 +
A 4,1 ,
(3.210)
A 2,1
=−
q 1 (1) A 3,2
+
q 2 (1) A 4,1 ,
(3.211)
p 1 (1) ρ 0 + ω
A 1,0
+ ρ 0 A 6,0 . (3.212)
1
2
2
A 3,2
=
+
2
Ω
2 m ω Ω
+
p 2 (1) A 4,1
Now A 1,0 , A 6,0 , A 4,1 may be regarded as the three free constants determining the
three regular solutions at the geocentre for n
1. No further eigenvalues are
encountered in the recursions for higher terms in the power series expansions.
For the regular solution generated by the free constant A 1,0 , from equations
(3.210) through (3.212), we have
=
A 1,2
A 2,1
A 3,2
A 4,1
1
p 1 (1) + ω
= ρ 0
q 1 (1)
1
0
2
2
+
2
Ω
2 m ω Ω
A 1,0 .
(3.213)
From (3.207), A 5,1
0. For numerical integration, the y-
variables are again replaced by a system of z -variables for n
=
G ρ 0 A 1,0 for A 6,0
=
1. For the regular
solutions generated by the free constants A 1,0 and A 6,0 the new z -variables are just
those defined in (3.187) for α =
=
1. For the regular solution generated by
the free constant A 4,1 they are those defined in (3.187) for α =
n
2
=−
1. The power
series expansions for the regular solution generated by the free constant A 1,0 then
have the forms
n
=
A 1,2 r 2
A 1,4 r 4
z 1 ( r )
= y 1 ( r )
=
A 1,0
+
+
+··· ,
(3.214)
A 2,1 r 2
A 2,3 r 4
z 2 ( r )
=
r y 2 ( r )
=
+
+··· ,
(3.215)
A 3,2 r 2
A 3,4 r 4
z 3 ( r )
= y 3 ( r )
=
A 1,0
+
+
+··· ,
(3.216)
A 4,3 r 4
z 4 ( r )
=
r y 4 ( r )
=
+··· ,
(3.217)
A 5,3 r 2
A 5,5 r 4
z 5 ( r )
= y 5 ( r )/ r
=
G ρ 0 A 1,0
+
+
+··· ,
(3.218)
A 6,2 r 2
A 6,4 r 4
z 6 ( r )
= y 6 ( r )
=
+
+··· .
(3.219)
For the regular solution generated by the free constant A 6,0 , from equations
(3.210) through (3.212), we have
A 1,2
A 2,1
A 3,2
A 4,1
ρ 0 / p 1 (1)
ρ 0 q 1 (1)/ p 1 (1)
ρ 0 / p 1 (1)
0
=
A 6,0 .
(3.220)
 
Search WWH ::




Custom Search