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× [ A 1 (1 − E 2 ) 2 + E 2 A 2 sin 2 i sin 2 α 0 ][ A 1 +( A 2 cos 2 i − A 1 )sin 2 α 0 ]sin α 0 +
+2[ A 1 +( A 2 cos 2 i
A 1 )sin 2 α 0 ] E 2 A 2 sin 2 i sin α 0 cos 2 α 0 +
A 1 )sin α 0 cos 2 α 0 .
+[ A 1 (1
E 2 ) 2 + E 2 A 2 sin 2 i sin 2 α 0 ]( A 2 cos 2 i
Box 16.5 (Longitude L ( α ) as a function of meta-longitude α ).
Power series expansion Δl = N
r =1 l r Δα r up to order N =2:
A 1 A 2 cos i
l 1 := +
A 1 cos 2 α 0 A 2 cos 2 i sin 2 α 0 ,
(16.58)
2 A 1 A 2 cos i ( A 2 cos 2 i
A 1 )sin α 0 cos α 0
( A 1 cos 2 α 0 + A 2 cos 2 i sin 2 α 0 ) 2
2 l 2 :=
.
The relations ( 16.44 ) together with the relations ( 16.50 ) lead to the relations ( 16.51 ). Let us prove
the other central relations here.
Proof (( 16.47 ): b 1 ).
A 2
sin α
A 1 +( A 2 cos 2 i
tan B =
1 − E 2 sin i
A 1 )sin 2 α ,
A 1 cos α
[ A 1 +( A 1 cos 2 i
A 2
dtan B
d α
1
cos 2 B
d B
d α =
=
E 2 sin i
A 1 )sin 2 α ] 3 / 2 ,
1
d B
d α =cos 2 B dtan B
1
1+tan 2 B
dtan B
d α
=
=
(16.59)
d α
=[ A 2 A 1 (1
E 2 )sin i cos α ][ A 1 +( A 2 cos 2 i
A 1 )sin 2 α ] 1 / 2
×
× [ A 1 (1 − E 2 ) 2 + E 2 A 2 sin 2 i sin 2 α ] 1
b 1 := d B
d α ( α 0 ) .
End of Proof.
 
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