It was observed that the Galileo space craft reached the
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, 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 h = rovo = (2820 x 103)(15 x 103) = 42.3x 109 m2/s 1 GM i = 2 (1 + ε ) r0 h
(42.3 × 10 9 ) 2 h2 = = 106.55 r0 GM i (2820 × 10 3 )(5.955 × 1012 ) ε = 106.55 − 1 = 105.55
1+ ε =
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 = 14.24 × 1013 m 2 3
( 2)(14 .24 × 1013 ) = 3749 .34 s 75 .96 × 10 9
1.0414 h
6
rA=1778.6 km =1.7786 x 106 m; rB = 3589 km =3.589 x 10 m
(2)(4.896 × 1012 )(1.7786 × 10 6 )(3.589 × 10 6 ) 5.3676 × 10 3.412 × 10
6 6
9
3.589 × 10 6
= 950.68 m/s
9
2
= 3.412x10 m /s
3589 / 1730.5 − 1 = 0.444 3590 / 1730.5 + cos 70° (4.896 × 1012 )(0.556)(3.589 × 10 6 ) = 3.126 x 109 m2/s 3.126 × 10 9 3.589 × 10 6
= 871 m/s
=871-950.68 = -79.68 m/s
79.68 m/s
6
rC= 6370 + 1490 = 7860 km = 7.86 x 10 m
1.2339
1.2339 − 1 = 0.683 1.2339 + cos153.016°
(398.06 × 1012 )(0.317)(7.86 × 10 6 ) = 31.493x109 m2/s 31.493 × 10 9 = 4006 m/s 7.86 × 10 6 0.683
PROBLEM 12.135 A space shuttle is describing a circular orbit at an altitude of 563 km above the surface of the earth. As it passes through point A, it fires its 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 121 km. (Hint. Point A is the apogee of the elliptic descent orbit.)
SOLUTION
(
GM = gR 2 = ( 9.81) 6.37 × 106
)
2
= 398.06 × 1012 m3/s 2
rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m For the circular orbit through point A, GM = rA
vcirc =
398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106
For the descent trajectory, v A = vcirc + ∆v = 7.5773 × 103 − 152 = 7.4253 × 103 m/s
(
)(
)
h = rAv A = 6.933 × 106 7.4253 × 103 = 51.4795 × 109 m 2 /s 1 GM = 2 (1 + ε cosθ ) r h At point A,
θ = 180°,
r = rA 1 GM = 2 (1 − ε ) rA h
(
)
2
51.4795 × 109 h2 1−ε = = = 0.96028 GM rA 398.06 × 1012 6.933 × 106
(
)(
)
ε = 0.03972 1 GM = 2 (1 + ε cosθ B ) rB h
(
)
2
51.4795 × 109 h2 1 + ε cosθ B = = = 1.02567 GM rB 398.06 × 1012 6.491 × 106
(
cosθ B =
θ B = 49.7°
1.02567 − 1
ε
)(
)
= 0.6463
AOB = 180° − θ B = 130.3°
AOB = 130.3° W
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, 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 h = rovo = (2820 x 103)(15 x 103) = 42.3x 109 m2/s 1 GM i = 2 (1 + ε ) r0 h
(42.3 × 10 9 ) 2 h2 = = 106.55 r0 GM i (2820 × 10 3 )(5.955 × 1012 ) ε = 106.55 − 1 = 105.55
1+ ε =
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 = 14.24 × 1013 m 2 3
( 2)(14 .24 × 1013 ) = 3749 .34 s 75 .96 × 10 9
1.0414 h
6
rA=1778.6 km =1.7786 x 106 m; rB = 3589 km =3.589 x 10 m
(2)(4.896 × 1012 )(1.7786 × 10 6 )(3.589 × 10 6 ) 5.3676 × 10 3.412 × 10
6 6
9
3.589 × 10 6
= 950.68 m/s
9
2
= 3.412x10 m /s
3589 / 1730.5 − 1 = 0.444 3590 / 1730.5 + cos 70° (4.896 × 1012 )(0.556)(3.589 × 10 6 ) = 3.126 x 109 m2/s 3.126 × 10 9 3.589 × 10 6
= 871 m/s
=871-950.68 = -79.68 m/s
79.68 m/s
6
rC= 6370 + 1490 = 7860 km = 7.86 x 10 m
1.2339
1.2339 − 1 = 0.683 1.2339 + cos153.016°
(398.06 × 1012 )(0.317)(7.86 × 10 6 ) = 31.493x109 m2/s 31.493 × 10 9 = 4006 m/s 7.86 × 10 6 0.683
PROBLEM 12.135 A space shuttle is describing a circular orbit at an altitude of 563 km above the surface of the earth. As it passes through point A, it fires its 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 121 km. (Hint. Point A is the apogee of the elliptic descent orbit.)
SOLUTION
(
GM = gR 2 = ( 9.81) 6.37 × 106
)
2
= 398.06 × 1012 m3/s 2
rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m For the circular orbit through point A, GM = rA
vcirc =
398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106
For the descent trajectory, v A = vcirc + ∆v = 7.5773 × 103 − 152 = 7.4253 × 103 m/s
(
)(
)
h = rAv A = 6.933 × 106 7.4253 × 103 = 51.4795 × 109 m 2 /s 1 GM = 2 (1 + ε cosθ ) r h At point A,
θ = 180°,
r = rA 1 GM = 2 (1 − ε ) rA h
(
)
2
51.4795 × 109 h2 1−ε = = = 0.96028 GM rA 398.06 × 1012 6.933 × 106
(
)(
)
ε = 0.03972 1 GM = 2 (1 + ε cosθ B ) rB h
(
)
2
51.4795 × 109 h2 1 + ε cosθ B = = = 1.02567 GM rB 398.06 × 1012 6.491 × 106
(
cosθ B =
θ B = 49.7°
1.02567 − 1
ε
)(
)
= 0.6463
AOB = 180° − θ B = 130.3°
AOB = 130.3° W
