Geoscience Reference
In-Depth Information
leads, after simple manipulations, to
q
=2
1
3
...
.
E
u
3
E
u
5
E
u
7
2
3
−
+
−
(2-156)
·
5
5
·
7
7
·
9
More concisely, we have
E
u
2
n
+1
+
∞
tan
−
1
E
u
=
E
u
1
2
n
+1
1)
n
(
−
,
n
=1
(2-157)
E
u
2
n
+1
∞
2
n
(2
n
+ 1)(2
n
+3)
1)
n
q
=
−
(
−
.
n
=1
By inserting these relations into (2-151) we obtain
E
u
2
n
+1
∞
V
=
GM
u
+
GM
E
1
2
n
+1
1)
n
(
−
n
=1
E
u
2
n
+1
∞
ω
2
a
2
3
q
0
2
n
(2
n
+ 1)(2
n
+3)
1)
n
−
(
−
P
2
(sin
β
)
.
n
=1
(2-158)
Introducing
m
, defined by (2-137), and the second eccentricity
e
=
E/b
,we
find
E
u
2
n
+1
+
∞
V
=
GM
u
GM
(2
n
+1)
E
1)
n
(
−
·
n
=1
1
2
n
+3
P
2
(sin
β
)
.
me
3
q
0
2
n
·
−
(2-159)
We expand the potential
V
into a series of spherical harmonics. Because
of the rotational symmetry, there will be only zonal terms, and because of
the symmetry with respect to the equatorial plane, there will be only even
zonal harmonics. The zonal harmonics of odd degree change sign for negative
latitudes and must, therefore, be absent. Accordingly, the series has the form
V
=
GM
r
+
A
2
P
2
(cos
ϑ
)
r
3
+
A
4
P
4
(cos
ϑ
)
r
5
+
···
.
(2-160)
We next have to determine the coecients
A
2
,A
4
, ...
. For this purpose, we
consider a point on the axis of rotation, outside the ellipsoid. For this point,
we have
β
=90
◦
,ϑ
=0
◦
, and, by (2-153),
u
=
r
. Then (2-159) becomes
1
1
r
2
n
+1
,
+
∞
1)
n
GM E
2
n
2
n
+1
me
3
q
0
V
=
GM
r
2
n
2
n
+3
(
−
−
(2-161)
n
=1
