A5 9th Edition Solution

A5 9th Edition Solution - 12.101 PROBLEM 12.102 It was...

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Unformatted text preview: 12.101 PROBLEM 12.102 It was observed that the Galileo space craft reached the point of its trajectory 2800+km 2800*km 2800*km closet to Io, a moon of the planet Jupiter, it was at a distance of 2820 km from the center of Io and had a velocity of 15km/s. Knowing that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory of the spacecraft as it approached Io. SOLUTION For earth, GMe = gR2 = (9.81)(6.37 x 106)2 = 398.06 x 1012 m3/s2 For Io, GMi = 0.01496 x GMe = 0.01496 x 398.06 x 1012 m3/s2 = 5.955 x 1012 m3/s2 42 h = rovo = (2820 x 103)(15 x 103) = 42.3x 109 m2/s 2800 1 r0 1 GM i h2 (1 h2 r0 GM i ) 42 (42.3 10 9 ) 2 (2800 10 3 )(5.955 1012 ) (2820 104.79 106.55 1 105.55 105.79 105.79 106.55 10.4 7307 6 7.307 km 7.307 x 10 m 10.4 km/s =10400 m/s 9 = 75.96 x 10 m/s 8 (7.307 10 6 ) 2 3 ( 2)(14 .24 1013 ) 75 .96 10 9 3749 .34 s 14.24 1013 m 2 1.0414 h 12.115 12.115 R=1737 km 1790 km 3600 km 6 1790 km 1.790 3600 3.6 rA=1778.6km =1.7786 x 106 m; rB = 3589 km =3.589 x 10 m 3.6 (2)(4.896 1012 )(11.790 10 6 )(3.589 10 6 ) .7786 5.3676 10 5.39 3.422 10 3.412 B 6 9 3.6 3.589 10 6 950.56 m/s = 950.68 m/s 6 9 2 = 3.422 3.412x10 m /s 12.115 3600/1737 3589 / 1730.5 1 = 0.444 3590 / 1730.5 cos 70 3600/1737 3.60 3.130 (4.896 1012 )(0.556)(3.589 10 6 ) = 3.126 x 109 m2/s 3.130 10 9 3.126 3.60 3.589 10 6 869.44 m/s = 869 m/s 871 869-950.56 869.44-950.56=-81.12 m/s -81.56 =871-950.68 = -79.68 m/s 81.56 81.12 79.68 m/s 6 rC= 6370 + 1490 = 7860 km = 7.86 x 10 m 7860 5928.5 km 6370 km 153.338 0.93069 1.2339 53.325 o 153.338 o 153.338 1.2339 1.2339 1 = 0.687 0.683 153.338 1.2339 cos153.016 0.313 (398.06 1012 )(0.317)(7.86 10 6 ) = 31.294 9 m2/s 31.493x10 31.294 31.493 10 9 = 3981 m/s 4006 7.86 10 6 3981 m/s 0.683 0.687 560 12.117 12.117 PROBLEM 12.135 A space shuttle is describing a circular orbit at an altitude of 560 km 563 above the surface of the earth. As it passes through point A, it fires its 150 engine for a short interval of time to reduce its speed by 152 m/s and begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at point B is 120 km. (Hint. Point A is the apogee of 121 the elliptic descent orbit.) SOLUTION gR 2 GM 9.81 6.37 106 2 398.06 1012 m3/s 2 rA 6370 560 563 6930 km 6933 6.930 6.933 106 m rB 6370 120 121 6490 km 6491 6.490 6.491 106 m For the circular orbit through point A, 398.06 1012 6.930 6.933 106 GM rA vcirc 7.579 7.5773 103 m/s For the descent trajectory, vA h vcirc rAv A 7.579 7.5773 103 v 6.930 7.429 6.933 106 7.4253 103 1 r At point A, r 180 , 7.429 7.4253 103 m/s 150 152 GM 1 h2 51.483 51.4795 109 m 2 /s cos rA 1 rA 51.483 51.4795 109 h2 GM rA 1 GM 1 h2 2 6.930 398.06 1012 6.933 106 0.9608 0.96028 0.0392 0.03972 1 rB 1 cos B B 48.5 49.7 cos B 51.483 51.4795 109 h2 GM rB cos GM 1 h2 2 398.06 1012 6.490 106 6.491 1.0259 1.02567 B 7 AOB 180 1 1.0259 1.02567 7 0.6625 0.6463 B 131.5 130.3 AOB 131.5 130.3 ...
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