It was observed that the Galileo space craft reached the point of its ...
12.101 PROBLEM 12.102
It was observed that the Galileo space craft reached the point of its trajectory 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 3 3 42 109 m2/s h = rovo = (2820 2800 x 10 )(15 x 10 ) = 42.3x
1 r0
1
GM i h2
(1
h2 r0 GM i
)
42.3 10 9 ) 2 (42
(2800 (2820 10 3 )(5.955 1012 ) 106 .55 1 105 .55 105.79 104.79
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 R=1737 km
1790 km
3600 km
6
1790 km 1.790 x 106 m; rB = 3589 3600 km =3.589 3.6 x 10 m rA=1778.6 km =1.7786
(2)(4.896 1012 )(11.790 .7786 10 6 )(33.6 .589 10 6 ) 55.39 .3676 10 3.422 3.412 10 B
6
9
33.6 .589 10 6
950.56mm/s = 950.68 /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 9 2 3.130 x 10 m /s (4.896 1012 )(0.556)(33.60 .589 10 6 ) = 3.126
3.130 3.126 10 9
33.60 .589 10 6
869 = 869.44 871 m/sm/s
869-950.56 = -79.68 869.44-950.56=-81.12 -81.56 m/s m/s =871-950.68
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
9 2 (398.06 1012 )(00.313 .317)(7.86 10 6 ) = 31.294 31.493x10 m /s
31.294 31 .493 10 9 = 3981 4006 m/s 7.86 10 6
3981 m/s
0.683 0.687
560
12.117 PROBLEM 12.135 A space shuttle is describing a circular orbit at an altitude of 560 563 km above the surface of the earth. As it passes through point A, it fires its 150 m/s and engine for a short interval of time to reduce its speed by 152 begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at point B is 120 121 km. (Hint. Point A is the apogee of 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 6933 km
6.930 6.933
106 m
rB
6370
120 121
6490 6491 km
6.490 6.491
106 m
For the circular orbit through point A, 398.06 1012 6.930 106 6.933
GM rA
vcirc
7.579 7.5773 103 m/s
For the descent trajectory, vA h
vcirc
rAv A
7.579 103 7.5773
v
6.930 106 7.4253 7.429 103 6.933
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
6.930 106 398.06 1012 6.933
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 6.491 106 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
It was observed that the Galileo space craft reached the point of its trajectory 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 3 3 42 109 m2/s h = rovo = (2820 2800 x 10 )(15 x 10 ) = 42.3x
1 r0
1
GM i h2
(1
h2 r0 GM i
)
42.3 10 9 ) 2 (42
(2800 (2820 10 3 )(5.955 1012 ) 106 .55 1 105 .55 105.79 104.79
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 R=1737 km
1790 km
3600 km
6
1790 km 1.790 x 106 m; rB = 3589 3600 km =3.589 3.6 x 10 m rA=1778.6 km =1.7786
(2)(4.896 1012 )(11.790 .7786 10 6 )(33.6 .589 10 6 ) 55.39 .3676 10 3.422 3.412 10 B
6
9
33.6 .589 10 6
950.56mm/s = 950.68 /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 9 2 3.130 x 10 m /s (4.896 1012 )(0.556)(33.60 .589 10 6 ) = 3.126
3.130 3.126 10 9
33.60 .589 10 6
869 = 869.44 871 m/sm/s
869-950.56 = -79.68 869.44-950.56=-81.12 -81.56 m/s m/s =871-950.68
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
9 2 (398.06 1012 )(00.313 .317)(7.86 10 6 ) = 31.294 31.493x10 m /s
31.294 31 .493 10 9 = 3981 4006 m/s 7.86 10 6
3981 m/s
0.683 0.687
560
12.117 PROBLEM 12.135 A space shuttle is describing a circular orbit at an altitude of 560 563 km above the surface of the earth. As it passes through point A, it fires its 150 m/s and engine for a short interval of time to reduce its speed by 152 begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at point B is 120 121 km. (Hint. Point A is the apogee of 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 6933 km
6.930 6.933
106 m
rB
6370
120 121
6490 6491 km
6.490 6.491
106 m
For the circular orbit through point A, 398.06 1012 6.930 106 6.933
GM rA
vcirc
7.579 7.5773 103 m/s
For the descent trajectory, vA h
vcirc
rAv A
7.579 103 7.5773
v
6.930 106 7.4253 7.429 103 6.933
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
6.930 106 398.06 1012 6.933
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 6.491 106 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
