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Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 282 Copyright © 2010, Elsevier, Inc.
Problem 6.24 (a) With a single delta-v maneuver, the earth orbit of a satellite is to be changed from acircle of radius 15 000 km to a collinear ellipse with perigee altitude of 500 km and apogee radius of 22000km. Calculate the magnitude of the required delta-v and the change in the flight path angle Δγ .(b) What is the minimum total delta-v if the orbit change is accomplished instead by a Hohmann transfer?
Solution
�
rA = rC = rE = 15000 km rB = 22000 km rD = 6878 km
(a)
Orbit 1:
vA1= µ
rA=
398 60015000
= 5.155 km s
γ A1 = 0
Orbit 2:
e2 =rB − rDrB + rD
=22000 − 687822000 + 6878
= 0.5237
h2 = 2µ rBrDrB + rD
= 2 ⋅398 600 22000 ⋅687822000 + 6878
= 64 630 km2 s
At the maneuvering point A:
rA =h2
2
µ1
1 + e2 cosθA
15000 =64 6302
398 6001
1 + 0.5237 cosθA ⇒ θA = 125.1°
15 000 km
22 000 km6878 km
2A
B C D E
γ2vA 1
vA2
Δv
Common apseline
Earth
3
1
4
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 283 Copyright © 2010, Elsevier, Inc.
vA2 )⊥ =
h2rA
=64 63015000
= 4.309 km s
vA 2 )r = µ
h2e2 sinθA =
398 60064 630
0.5237 sin125.1° = 2.641km s
vA2
= vA2 )⊥2 + vA2 )r
2 = 4.3092 + 2.6412 = 5.054 km s
γ A 2= tan−1
vA 2 )rvA 2 )⊥
= tan−1 2.6414.309
= 0.5499 ⇒ γ A 2= 31.51°
Δγ A = γ A 2
− γ A1= 31.51° − 0 = 31.51°
ΔvA = vA1
2 + vA 22 − 2vA1
vA 2cosΔγ A = 5.1552 + 5.0542 − 25.1555.054 cosΔγ A = 2.773 km s T
(b)Try Hohmann transfer (orbit 3) from point E on orbit 1 to point B on orbit 2.
h3 = 2µ rErB
rE + rB= 2 ⋅398 600 15000 ⋅ 22000
15000 + 22000= 84 320 km2 s
vE1= vA1
= 5.155 km s
vE3=
h3rE
=84 32015000
= 5.621km s
vB3=
h3rB
=84 32022000
= 3.833 km s
vB 2=
h2rB
=64 63022000
= 2.938 km s
Δvtotal = vE3− vE1
+ vB 2− vB3
= 0.4665 + 0.985 = 1.362 km s
Try Hohmann transfer (orbit 4) from point C on orbit 1 to point D on orbit 2.
�
h4 = 2 µrCrD
rC + rD= 2 ⋅398600
15000 ⋅687815000 + 6878 = 61310 km2 s
vC1= vA1
= 5.155 km s
vC 4=
h4rC
=6131015000
= 4.088 km s
vD 4=
h4rD
=613106878
= 8.914 km s
vD 4=
h2rD
=64 6306878
= 9.397 km s
Δvtotal = vC 4− vC1
+ vD 2− vD 4
= 1.067 + 0.4824 = 1.55 km sThis is larger than the total computed above; thus for minimum Hohmann transfer
Δv = 1.362 km s
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 284 Copyright © 2010, Elsevier, Inc.
Problem 6.25 An earth satellite has a perigee altitude of 1270 km and a perigee speed of 9 km/s. It isrequired to change its orbital eccentricity to 0.4, without rotating the apse line, by a delta-v maneuver at θ = 100° . Calculate the magnitude of the required Δv and the change in flight path angle Δγ .
Solution
Orbit 1:
�
rperigee1= 6378 + 1270 = 7648 km
vperigee1= 9 km s
h1 = rperigee1vperigee1 = 7648 ⋅9 = 68 832 km2 s
rperigee1=
h12
µ1
1 + e1
7648 =68 8322
398 6001
1 + e1 ⇒ e1 = 0.5542
At the maneuver point,
�
θ = 100° .
�
r =h1
2
µ1
1 + e1 cosθ =688322
3986001
1 + 0.5542 cos100°= 13150 km
�
v1⊥ = h1r
=68 83213150
= 5.234 km s
v1 r =µh1
e1 sinθ =39860068 832 0.5542 sin100° = 3.16 km s
v1 = v1⊥2 + v1r
2 = 5.2342 + 3.162 = 6.114 km s
γ1 = tan −1 v1rv1⊥
= tan −1 3.165.234 = 31.13
Orbit 2:
�
e2 = 0.4
r =h2
2
µ1
1 + e2 cosθ
13150 =h2
2
398 6001
1 + 0.4 cos100° ⇒ h2 = 69 840 km2 s
�
v2⊥ = h2r
=69 84013150
= 5.311 km s
v2 r =µh2
e2 sinθ =39860069 840 0.4 sin100° = 2.248 km s
v2 = v2⊥2 + v2 r
2 = 5.3112 + 2.2482 = 5.767 km s
γ2 = tan −1 v2 rv2⊥
= tan −1 2.2485.767 = 22.94°
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 285 Copyright © 2010, Elsevier, Inc.
Δγ = γ 2 − γ1 = 22.94° − 31.13° = −8.181°
Δv = v12 + v2
2 − 2v1v2 cosΔγ = 6.1142 + 5.7672 − 2 ⋅6.114 ⋅5.767 cos −8.181( ) = 0.9155 km s
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 291 Copyright © 2010, Elsevier, Inc.
Problem 6.29 At point A on its earth orbit, the radius, speed and flight path angle of a satellite are
rA = 12 756 km , vA = 6.5992 km/s and γ A = 20° . At point B, at which the true anomaly is 150°, animpulsive maneuver causes Δv⊥ = +0.75820 km/s and Δvr = 0 .
(a) What is the time of flight from A to B?(b) What is the rotation of the apse line as a result of this maneuver?
Solution
Orbit 1:
vA⊥= vA cosγ A = 6.5992cos 20° = 6.20122 km s
∴h1 = rAvA⊥= 12756 ⋅6.20122 = 79102.8 km2 s
vAr= vA sin γ A = 6.5992 ⋅sin 20° = 2.25706 km s
vAr= µ
h1e1 sinθA
2.25706= 398 60079102.8
e1 sinθA ⇒ e1 sinθA = 0.447917
rA =h1
2
µ1
1 + e1 cosθA
12756 =79102.82
398 6001
1 + e1 cosθA ⇒ e1 cosθA = 0.230641
∴ e12 sin2 θA + cos2 θA( ) = 0.4479172 + 0.2306412
e12 = 0.253825 ⇒ e1 = 0.50381
∴sinθA = 0.447917
0.50381= 0.889058 ⇒ θA = 62.755 2° or θA = 117.235°
Since
�
cosθA > 0 ,
�
θA = 62.7552° .
B
150°
1
2
η
θΒ2
P1
P2A
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 292 Copyright © 2010, Elsevier, Inc.
The period of orbit 1 is
T1 = 2πµ2
h1
1 − e12
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 2π
398 600279102.8
1 − 0.503812
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 30 368.2 s
The eccentric anomaly at point A of orbit 1 is
EA = 2 tan−1 1 − e1
1 + e1tan θA
2⎛
⎝⎜
⎞
⎠⎟ = 2 tan−1 1 − 0.50381
1 + 0.50381tan 62.7552°
2⎛
⎝⎜
⎞
⎠⎟ = 0.67392
From Kepler’s equation, the corresponding mean anomaly is
MA = EA − e1 sin EA = 0.67392 − 0.50381sin 0.67392 = 0.35951
Therefore, the time since perigee passage is
tA =
MA2π
T1 = 0.359512π
30 368 = 1737.6 s
At point B the eccentric anomaly is
EB = 2 tan−1 1 − e1
1 + e1tan θB
2⎛
⎝⎜
⎞
⎠⎟ = 2 tan−1 1 − 0.50381
1 + 0.50381tan 150°
2⎛
⎝⎜
⎞
⎠⎟ = 2.2687
Thus
MB = EB − e1 sin EB = 2.2687 − 0.50381sin 2.2687 = 1.8826
and
tB =
MB2π
T1 = 1.88262π
30 368 = 9099.2 s
It follows that the time of flight from A to B is
tof = tB − tA = 9099.2 − 1737.6 = 7361.6 s = 2.045 hr
(b)
rB =
h12
µ1
1 + e1 cosθB1
=79102.82
398 6001
1 + 0.50381cos150°= 27 848.9 km (1)
vB⊥ )1 =
h1rB
=79102.827 848.9
= 2.84043 km s (2)
vBr )1 = µ
h1e1 sinθB1
=398 60079102.8
⋅0.50381 ⋅sin150° = 1.26945 km s (3)
ΔvB⊥ = 0.75820 km s (4)
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 293 Copyright © 2010, Elsevier, Inc.
ΔvBr = 0 (5)
∴vBr )2 = vBr )1 + ΔvBr = 1.26945 + 0 = 1.26945 km s (6)
According to Equation 6.18b
tan θB2( ) =vB⊥ )1 + ΔvB⊥⎡⎣⎢
⎤⎦⎥
vBr )1 + ΔvBr⎡⎣⎢
⎤⎦⎥
vB⊥ )1 + ΔvB⊥⎡⎣⎢
⎤⎦⎥
2e1 cosθB1
+ 2 vB⊥ )1 + ΔvB⊥⎡⎣⎢
⎤⎦⎥
vB⊥ )12
µ rB
Substituting Equations (1) through (5) above yields
tan θB2( ) = 2.8404 + 0.7582( ) 1.2694 + 0( )2.8404 + 0.7582( )2 ⋅0.50381 ⋅cos150° + 2 ⋅ 2.8404 + 0.7582( )
2.84042
398 600 27 849
tan θB2( ) = −3.3521 ⇒ θB2
= −73.389° or 106.612°
According to Equation (6) above, the spacecraft is flying away from perigee on orbit 2, so 0 ≤ θB 2
≤ 180° .Therfore,
θB 2
= 106.612°
This means
η = 150 − 106.612° = 43.3877°
That is, the apse line is rotated 43.387 7° ccw from that of orbit 1.
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 309 Copyright © 2010, Elsevier, Inc.
Problem 6.40 With a single impulsive maneuver, an earth satellite changes from a 400 km circular orbitinclined at 60° to an elliptical orbit of eccentricity e = 0.5 with an inclination of 40°. Calculate theminimum required delta-v.
Solution
For the circular orbit
v1 = µ
r=
398 6006778
= 7.668 km s
Assume the maneuver is done at apogee of the ellipse (orbit 2).
r = h22
µ1
1 − e2
6778 =h2
2
398 6001
1 − 0.5 ⇒ h2 = 36750 km2 s
Then
�
rperigee =h2
2
µ1
1 + e2=
367502
3986001
1 + 0.5 = 2259 km
which is inside the earth. So the maneuver cannot occur at apogee. Assume it occurs at perigee.
�
r =h2
2
µ1
1 + e2
6778 =h2
2
398 6001
1 + 0.5 ⇒ h2 = 63 660 km2 s
�
v2 =h2r =
636606778 = 9.392 km s
Δv = v12 + v2
2 − 2v1v2 cosδ = 7.6682 + 9.3922 − 2 ⋅7.668 ⋅9.392cosδ = 3.414 km s
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 314 Copyright © 2010, Elsevier, Inc.
Problem 6.44 A spacecraft is in a 300 km circular parking orbit. It is desired to increase the altitude to600 km and change the inclination by 20°. Find the total delta-v required if
(a) The plane change is made after insertion into the 600 km orbit (so that there are a total of threedelta-v burns).
(b) If the plane change and insertion into the 600 km orbit are accomplished simultaneously (sothat the total number of delta-v burns is two).
(c) The plane change is made upon departing the lower orbit (so that the total number of delta-vburns is two).
Solution
The initial and target orbits are “1” and “2”, respectively, and “3” is the transfer orbit.
r1 = 6678 km
v1 = µ
r1=
398 6006678
= 7.726 km s
�
r2 = 6978 km
v2 = µ
r2=
398 6006978
= 7.558 km s
�
a3 =r1 + r2
2 =6678 + 6978
2 = 6828 km
vperigee3
= µ 2r1
− 1a3
⎛⎝⎜
⎞⎠⎟= 398 600 2
6678− 1
6828⎛⎝⎜
⎞⎠⎟ = 7.810 km s
vapogee3
= µ 2r2
− 1a3
⎛⎝⎜
⎞⎠⎟= 398 600 2
6978− 1
6828⎛⎝⎜
⎞⎠⎟ = 7.474 km s
(a)
Δv = vperigee3− v1( ) + v2 − vapogee3( ) + 2 ⋅ v2 sin Δi
2
= 7.810 − 7.726( ) + 7.558 − 7.474( ) + 2 ⋅7.558sin 20°2
= 0.0844 + 0.083 48 + 2.625 = 2.793 km s
(b)
Δv = vperigee 3− v1( ) + vapogee 3
2 + v22 − 2vapogee 3
v2 cosΔi
= 7.810 − 7.726( ) + 7.4742 + 7.5882 − 2 ⋅7.474 ⋅7.558cos 20°= 0.0844 + 2.612 = 2.696 km s
(c)
Solutions Manual Orbital Mechanics for Engineering Students Second Edition Chapter 6
Howard D. Curtis 315 Copyright © 2010, Elsevier, Inc.
Δv = vperigee32 + v1
2 − 2vperigee3v1 cosΔi + v2 − vapogee3( )
= 7.812 + 7.7262 − 2 ⋅7.81 ⋅7.726cos 20° + 7.558 − 7.474( )
= 2.699 + 0.083 48 = 2.783 km s