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The component γ λ is zero because U does not contain λ . This is also evident
from the rotational symmetry.
Performing the partial differentiations, we find
ω 2 a 2 E
u 2 + E 2
q
q 0 2 sin 2 β
6
GM
u 2 + E 2 +
1
ω 2 u cos 2 β,
u =
(2-132)
β =
q 0 + ω 2 u 2 + E 2 sin β cos β,
ω 2 a 2
u 2 + E 2
q
where we have set
du =3 1+ u 2
E 2 1
u 2 + E 2
E
dq
u
E
tan 1 E
u
q =
1 .
(2-133)
Note that q does not mean dq/du ; this notation has been borrowed from Hir-
vonen (1960), where q is the derivative with respect to another independent
variable which we are not using here.
For the level ellipsoid S 0 itself, we have u = b and get
γ β, 0 =0 .
(2-134)
(Note that we will often mark quantities referred to S 0 by the subscript
0.) This is also evident because on S 0 the gravity vector is normal to the
level surface S 0 . Hence, in addition to the λ -component, the β -component
is also zero on the reference ellipsoid u = b . Note that the other coordinate
ellipsoids u = constant are not equipotential surfaces U = constant, so that
the β -component will not in general be zero.
Thus, the total gravity on the ellipsoid S 0 , which we simply denote by
γ ,isgivenby
GM
a a 2 sin 2 β + b 2 cos 2 β ·
γ =
|
γ u, 0 |
=
(2-135)
1+ ω 2 a 2 E
GM
cos 2 β ,
q 0
q 0 2
6
ω 2 a 2 b
GM
1
sin 2 β
·
since on S 0 we get the relations
u 2 + E 2 = b 2 + E 2 = a,
b 2 + E 2 sin 2 β = 1
a
a 2 sin 2 β + b 2 cos 2 β.
(2-136)
w 0 = 1
a
Now we introduce the abbreviation
m = ω 2 a 2 b
GM
(2-137)
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