orbital mechanics homework6_solution

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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 a circle of radius 15 000 km to a collinear ellipse with perigee altitude of 500 km and apogee radius of 22000 km. 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 r A = r C = r E = 15000 km r B = 22000 km r D = 6878 km (a) Orbit 1: v A 1 = μ r A = 398 600 15 000 = 5.155 km s γ A1 = 0 Orbit 2: e 2 = r B r D r B + r D = 22 000 6878 22 000 + 6878 = 0.5237 h 2 = 2μ r B r D r B + r D = 2 398 600 22 000 6878 22 000 + 6878 = 64 630 km 2 s At the maneuvering point A: r A = h 2 2 μ 1 1 + e 2 cos θ A 15 000 = 64 630 2 398 600 1 1 + 0.5237 cos θ A θ A = 125.1° 15 000 km 22 000 km 6878 km 2 A B C D E γ 2 v A 1 v A 2 Δv Common apse line Earth 3 1 4

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Page 1: Orbital Mechanics Homework6_Solution

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

Page 2: Orbital Mechanics Homework6_Solution

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

Page 3: Orbital Mechanics Homework6_Solution

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°

Page 4: Orbital Mechanics Homework6_Solution

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

Page 5: Orbital Mechanics Homework6_Solution

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

Page 6: Orbital Mechanics Homework6_Solution

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)

Page 7: Orbital Mechanics Homework6_Solution

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.

Page 8: Orbital Mechanics Homework6_Solution

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

Page 9: Orbital Mechanics Homework6_Solution

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)

Page 10: Orbital Mechanics Homework6_Solution

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