Page 61 - Satellite Communications, Fourth Edition

Orbits and Launching Methods 41

hence no rotation takes place. Use is made of this fact in the orbit chosen
for the Russian Molniya satellites (see Probs. 2.23 and 2.24).
Denoting the epoch time by t , the right ascension of the ascending
0
node by Ω , and the argument of perigee by w at epoch gives the new
0
0
values for Ω and w at time t as
d
0
0 (t t ) (2.14)
dt
d
(t t ) (2.15)
0
0
dt
Keep in mind that the orbit is not a physical entity, and it is the forces
resulting from an oblate earth, which act on the satellite to produce the
changes in the orbital parameters. Thus, rather than follow a closed
elliptical path in a fixed plane, the satellite drifts as a result of the
regression of the nodes, and the latitude of the point of closest approach
(the perigee) changes as a result of the rotation of the line of apsides.
With this in mind, it is permissible to visualize the satellite as follow-
ing a closed elliptical orbit but with the orbit itself moving relative to
the earth as a result of the changes in Ω and w. Thus, as stated earlier,
the period P is the time required to go around the orbital path from
A
perigee to perigee, even though the perigee has moved relative to the
earth.
Suppose, for example, that the inclination is 90° so that the regres-
sion of the nodes is zero (from Eq. 2.12), and the rate of rotation of
the line of apsides is −K/2 (from Eq. 2.13), and further, imagine the
situation where the perigee at the start of observations is exactly over
the ascending node. One period later the perigee would be at an angle
−KP /2 relative to the ascending node or, in other words, would be
A
south of the equator. The time between crossings at the ascending
node would be P (1 K/2n), which would be the period observed from
A
the earth. Recall that K will have the same units as n, for example,
rad/s.

Example 2.5 Determine the rate of regression of the nodes and the rate of rota-
tion of the line of apsides for the satellite parameters specified in Table 2.1. The
value for a obtained in Example 2.2 may be used.

Solution From Table 2.1 and Example 2.2 i 98.6328°; e 0.0011501; NN
14.23304826 day −1 ; a 7192.335 km, and the known constant: K 1 66063.1704 km 2

Converting n to rad/s:
n 2 NN
0.00104 rad/s