View File - University of Engineering and Technology, Taxila

whereg ¼ acceleration due to gravity at the surface of the earth¼ 9:8087m=sec 2 .R e ¼ radius of the earth. The value varies with location. For example,R e at the equator ¼ 6378.39 km (6378 km).R e at the pole ¼ 6356.91 km (6357 km).Consequently, for a satellite in a stable circular orbit around the earth,F ¼ F gIn view of (2.7) and (2.8) in (2.9),r 3 ¼ g R2 eo 2ð2:9Þð2:10ÞThe period of the orbit, t s , that is, the time taken for one complete revolution(360 or 2p radians), can be expressed assffiffiffiffit s ¼ 2p o ¼ 2p r 3secð2:11ÞR e gIf we assume a spherical homogeneous earth, a satellite will have an orbitalvelocity represented byrffiffiffigv ¼ R e m=secð2:12ÞrFor elliptical orbits, Eqs. (2.11) and (2.12) are also valid by equating theellipse semimajor axis, a with the orbit radius r (i.e., r ¼ a). In terms of theorbit parameters, r is replaced with the average of apogee to focus and perigeeto focus. By definition, a perigee is the lowest altitude point of the orbit,whereas an apogee is the highest altitude point of the orbit. In a circular orbit,with variable altitude and upon substitution of empirical values in (2.11) and(2.12), Fig. 2.4, which relates period and velocity for circular orbits, assuminga spherical homogeneous earth, is plotted.For a circular orbit at an altitude of 35,784 km, Fig. 2.4 shows that ageosatellite orbit takes a period of rotation of the earth relative to the fixed star(called sidereal day) in 86163.9001 sec, or 23 h, 56 min, and 4 sec. In somebooks and papers, an approximate value of 36,000 km is frequently cited forthe altitude of the satellite in geosynchronous orbit. The geosynchronous orbitin the equatorial plane is called the geostationary orbit. Although a satellite inCopyright © 2002 by Marcel Dekker, Inc. All Rights Reserved.