Global Positioning System Reference
In-Depth Information
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
With (3.31), (3.34), and (3.25) the geocentric satellite distance becomes
q 1 +
r
=
q 2
=
a( 1
e cos E)
(3.35)
Differentiating Equations (3.35) and (3.26) gives
dr
=
ae sin EdE
(3.36)
r 2 e
a 1
e 2 sin fdf
dr
=
(3.37)
Eq uating (3.37) and (3.36) and using (3.24), (3.25), (3.34), and (3.7) and multiplying
th e resulting equation by r gives
[60
r 2 df
rb dE
=
(3.38)
Lin
3.8
——
Nor
PgE
Su bstituting (3.25) for b and (3.35) for r , replacing df by dt using (3.15), using (3.28)
fo r h , and then integrating, we obtain
E
a 3
t
( 1
e cos E) dE
=
dt
(3.39)
E
= 0
t 0
Integrating both sides gives
[60
E
e sin E
=
M
(3.40)
M
=
n (t
t 0 )
(3.41)
a 3
=
n
(3.42)
Equation (3.42) is Kepler's third law. Equation (3.40) is called the Kepler equation.
The symbol n denotes the mean motion, M is the mean anomaly, and t 0 denotes the
time of perigee passage of the satellite. The mean anomaly M should not be confused
w ith the same symbol used for the mass of the central body in (3.1). Let P denote the
or bital period, i.e., the time required for one complete revolution; then
2
n
P
=
(3.43)
The mean motion n equals the average angular velocity of the satellite. Equation
(3.42) shows that the semimajor axis completely determines the mean motion and
thus the period of the orbit.
With the Kepler laws in place, one can identify alternative sets of Kepler elements,
such as
{Ω
,
ω
,i,a,e,M
}
or
{Ω
,
ω
,i,a,e,E
}
. Often the orbit is not specified by
 
Search WWH ::




Custom Search