Ass. 5 fall solutions

Ass. 5 fall solutions - PROBLEM 12.102 It was observed that...

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PROBLEM 12.102 It was observed that the Galileo space craft reached the point of its trajectory 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, G M e = gR 2 = (9.81)(6.37 x 10 6 ) 2 = 398.06 x 10 12 m 3 /s 2 For Io, GM i = 0.01496 x GM e = 0.01496 x 398.06 x 10 12 m 3 /s 2 = 5.955 x 10 12 m 3 /s 2 h = r o v o = (2820 x 10 3 )(15 x 10 3 ) = 42.3x 10 9 m 2 /s ) 1 ( 1 2 0 ε + = h GM r i 55 . 106 ) 10 955 . 5 )( 10 2820 ( ) 10 3 . 42 ( 1 12 3 2 9 0 2 = × × × = = + i GM r h 55 . 105 1 55 . 106 = =
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7.307 x 10 6 m 7.307 km 10.4 km/s =10400 m/s = 75.96 x 10 9 m/s 2 13 2 6 10 24 . 14 ) 10 307 . 7 ( 3 8 m × = × s 34 . 3749 10 96 . 75 ) 10 24 . 14 )( 2 ( 9 13 = × × 1.0414 h
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r A =1778.6 km =1.7786 x 10 6 m; r B = 3589 km =3.589 x 10 6 m 6 6 6 12 10 3676 . 5 ) 10 589 . 3 )( 10 7786 . 1 )( 10 896 . 4 )( 2 ( × × × × = 3.412x10 9 m 2 /s 6 6 9 10 589 . 3 10 412 . 3 × × = 950.68 m/s
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° + 70 cos 5 . 1730 / 3590 1 5 . 1730 / 3589 = 0.444 ) 10 589 . 3 )( 556 . 0 )( 10 896 . 4 ( 6 12 × × = 3.126 x 10 9 m 2 /s 6 9 10 589 . 3 10 126 . 3 × × = 871 m/s =871-950.68 = -79.68 m/s 79.68 m/s
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r C = 6370 + 1490 = 7860 km = 7.86 x 10 6 m 0.683 1.2339 ° + 016 . 153 cos 2339 . 1 1 2339 . 1 = 0.683 ) 10 86 . 7 )( 317 . 0 )( 10
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This note was uploaded on 12/17/2011 for the course CIVE 281 taught by Professor Prof.v.h.chu during the Spring '10 term at McGill.

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Ass. 5 fall solutions - PROBLEM 12.102 It was observed that...

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