Page 73 - Satellite Communications, Fourth Edition
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Orbits and Launching Methods  53

                                Example 2.14 For satellite no. 14452 the NASA prediction bulletin for a certain
                                epoch gives the eccentricity as 9.5981 × 10 and the mean anomaly as 204.9779°.
                                                                −3
                                The mean motion is 14.2171404 rev/day. Calculate the true anomaly and the mag-
                                nitude of the radius vector 5 s after epoch. The semimajor axis is known to be
                                7194.9 km.
                                Solution The rotation in radians per second is

                                                         14.2171404   2
                                                     n
                                                             86400
                                                       >  0.001034 rad/s


                                The mean anomaly of 204.9779°, in radians is 3.57754, and 5 s after epoch the
                                mean anomaly becomes
                                           M 5 3.57754   0.001034   5

                                             > 3.5827 rad
                                              >  3.5827   2   9.5981   10  3     sin 3.5827

                                                 5              3 2
                                                    (9.5981   10 )    sin (2   3.5827)
                                                 4
                                               3.5746 rad (  204.81°)

                                Applying Eq. (2.23) gives r as
                                                                    2
                                                    7194.9   (1 9.5981 )   10  6
                                                r
                                                   1   9.5981   10  3    cos 204.81
                                                    7257.5 km
                                The magnitude r of the position vector r may be calculated by either
                              Eq. (2.23) or Eq. (2.30). It may be expressed in vector form in the perifocal
                              coordinate system. Here, the orbital plane is the fundamental plane, and
                              the origin is at the center of the earth (only earth-orbiting satellites are
                              being considered). The positive x axis lies in the orbital plane and passes
                              through the perigee. Unit vector P points along the positive x axis as
                              shown in Fig. 2.8. The positive y axis is rotated 90° from the x axis in the
                              orbital plane, in the direction of satellite motion, and the unit vector is
                              shown as Q. The positive z axis is normal to the orbital plane such that
                              coordinates xyz form a right-hand set, and the unit vector is shown as W.
                                The subscript w is used to distinguish the xyz coordinates in this
                              system, as shown in Fig. 2.8. The position vector in this coordinate
                              system, which will be referred to as the PQW frame, is given by

                                                   r   (r cos  )P   (r sin  )Q           (2.32)
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