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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 869950.56
869.44950.56=81.12 m/s
81.56
=871950.68 = 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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 Fall '10
 Prof.V.H.Chu

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