Page 62 - Satellite Communications, Fourth Edition
P. 62
42 Chapter Two
From Eq. (2.11):
nK 1
K
2
2 2
a (1 e )
6.544 deg/day
From Eq. (2.12):
d
K cos i
dt
0.981 deg/day
From Eq. (2.13):
d 2
K(2 2.5 sin i)
dt
2.904 deg/day
Example 2.6 Calculate, for the satellite in Example 2.5, the new values for w and
Ω one period after epoch.
Solution From Table 2.1:
NN 14.23304826 day ; w 0 113.5534°; Ω 0 251.5324°
−1
The anomalistic period is
1
P A
NN
0.070259 day
This is also the time difference (t t 0 ) since the satellite has completed one rev-
olution from perigee to perigee. Hence:
d
0 (t t 0 )
dt
251.5324 0.981(0.070259)
251.601°
d
0 (t t 0 )
dt
113.5534 ( 2.903)(0.070259)
113.349°
