Question

12.104. A satellite describes a circular orbit at an altitude of 19 110 km above the surface of the earth. Determine (a) the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit, (b) the decrease in speed required at point A for the satellite to enter an elliptic orbit with a minimum altitude of 6370 km, (c) the eccentricity e of the elliptic orbit. R = 6370 km 19 110 km 6370 km Fig. P12.104

Answer 1

for earth GM = R² = 9.81X6.37x108 - 398.06x102 m² EA = 6370+ 19110 = 25480 km = 25.48x10 m. - GM-1398.068182 - 3.9525x183 m V ZA 25:48X106 vesc = 120m = 12 veir = 5.589x10 m/s a) or=vecc - vae = (1 637x13 m/s) VEA 62 h2 Es = 6370 + 6370 = 12740 Rm = 12.74x100 m. I = GM + CCOSA & I = GM + Closoa OB - QA +180 C OSOA = – coso Adding a 2 + 1 = 8+2 = 26m EAEB h = 12 GMA &B 2x398.06x102 x 25.488,0 x 12.74x106 eates 38.22x100 h = 82.23x109 m/s 6) decrease in speed - DV = vce - VA I VA = h = 82.83x109 = 3.227x 103 m/s EA 25.488,06 Or = 3.9525 4,03 – 3.227x103 1725 m/s) m/s c) 2 - 1 1 - - - ccosog-CCSSOA = 20 - 12.748106 = 19.623x15 m 2EAEB 2x 25.488,0 6x 12.74x100 EA c=EA-E6 EALB = 2 x 25.48X10' 10-333 = 19.623x10x (82.23 x 10 398 06 X1012