Precise Orbit Determination of Global Navigation Satellite System of ...

Chapter 8 Geostationary Orbit Determination And Prediction During Satellite Maneuvers
From Figure 8-1,
cosδ = cos( v+ ω) cos( Ω− v′− ω′ ) − sin( v+ ω) sin( Ω−
v′− ω′ ) cos( i−ε)
In Eq.(8-34) if only the first term is considered, Eq.(8-34) becomes
R G M r
2
� 1 3 2 2
i = i i − + cos ( v+ ω) cos ( − v′−
ω′
)
3
ri
�� 2 2 Ω
− 3 cos( v+ ω) cos( Ω− v′ − ω′ )sin( v+ ω)sin( Ω−
v′ − ω′ ) cos( i−ε)
R
i
+ + − ′− ′ − �
3 2 2 2
sin ( v ω)sin ( Ω v ω ) cos ( i ε)
(8-35)
2
��
2 GM i i a 2 2�
1 1
= ( 1−e)
�−
2 ri
ri
� 2 ( 1+
ecos v)
−
2
2
3 cos ( v + ω)
2
+
cos ( Ω − v′
− ω ′ )
2
2 ( 1+
ecos v)
+
+
− ′ − ′ − +
− ′ − ′ −
+
+
�
2
3 sin 2(
v ω)
3 sin ( v ω)
2 2
sin 2(
Ω v ω ) cos( i ε)
sin ( Ω v ω ) cos ( i ε)
2
2
� (8-36)
4 ( 1 ecos v)
2 ( 1 ecos v)
�
Inserting Eq.(8-36) into Eq.(8-1), the following differential equations can be obtained
di
dt
dΩ
dt
n GM
=
2 3
n sin( i−ε)
r
i i
i
� 2 3/ 2 � 3 sin 2(
v + ω)
2
( 1−e
) �cos(
i−ε)
�−
cos ( Ω − v′−
ω ′ )
2
��
� 2 ( 1+
ecos v)
+
+
−
− ′ − ′ − +
− ′ − ′ −
+
+
� 3 cos 2(
v ω)
3 sin 2(
v ω)
2 2
sin 2(
Ω v ω ) cos( i ε)
sin ( Ω v ω ) cos ( i ε)
2 2
�
2 ( 1 ecos v)
2 ( 1 ecos v)
�
� 2
3 cos ( v + ω)
3 sin 2(
v + ω)
−�− sin 2(
Ω− v′
− ω ′ ) −
cos 2(
Ω−
v′ − ω′ ) cos( i−ε)
2 2
� 2 ( 1+
ecos v)
2 ( 1+
ecos v)
+
+
− ′− ′ −
+
� 2
3 sin ( v ω)
�
2
sin 2(
Ω v ω ) cos ( i ε)
2
�� (8-37)
2 ( 1 ecos v)
���
n GM
=
2 3
n sin( i−ε)
r
i i
i
2 3/ 2
( 1−
e )
� 3 sin 2(
v + ω)
�
sin 2(
Ω − v′ − ω′ )sin( i−ε)
2
� 4 ( 1+
ecos v)
+
−
− ′ − ′ −
+
�
2
3 sin ( v ω)
2
sin ( Ω v ω ) sin 2(
i ε)
2
�
(8-38)
2 ( 1 ecos v)
�
Before solving Eq.(8-37) and Eq.(8-38), the following integrations should be solved first, considering
dM r
= ( )
dv a − e
2 1
2
1
te
�
ts
te
�
ts
te
�
ts
te
�
ts
2π
2π
sin 2(
v + ω) sin 2(
v + ω) sin 2(
v + ω) r 2 1
2 3/ 2 sin 2(
v + ω)
ndt =
dM
( ) dv ( 1 e )
dv
2 2 2
( 1 cos ) ( 1 cos ) ( 1 cos )
2
4
+ e v �
=
+ e v �
= −
+ e v a �
1−
e
( 1+
ecos v)
0
2π
cos 2(
v + ω) 2 3/ 2 cos 2(
v + ω)
ndt = ( 1−e)
dv
2
4
( 1+
ecos v)
� ( 1+
ecos v)
2
sin ( v + ω) 2 3/ 2 sin ( v + ω)
ndt = ( 1−e)
dv
2
4
( 1+
ecos v)
� ( 1+
ecos v)
2
0
2π
cos ( v + ω) 2 3/ 2 cos ( v + ω)
ndt = ( 1−e)
dv
2
4
( 1+
ecos v)
� ( 1+
ecos v)
0
2π
0
2
2
0
101
2π
0
(8-39)
(8-40)
(8-41)
(8-42)
Eq. (8-39) to Eq.(8-42) are difficult to integrate, but for geostationary satellite, e ≈ 0 , then Eq.(8-39) to Eq.(8-
42) become