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SOLUTIONS MANUAL
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ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
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Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
Problem 1.1
(a)
( )(
A ⋅ A = Ax iˆ + Ay ˆj + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ )
( ) ( )
= Ax iˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ + Ay ˆj ⋅ Ax iˆ + Ay ˆj + Az kˆ + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ ( )
( ) ( ) ( )
= Ax 2 ( iˆ ⋅ iˆ ) + Ax Ay iˆ ⋅ ˆj + Ax Az ( iˆ ⋅ kˆ ) + Ay Ax ˆj ⋅ iˆ + Ay 2 ˆj ⋅ ˆj + Ay Az ˆj ⋅ kˆ
( )
( )
+ Az Ax ( kˆ ⋅ iˆ ) + Az Ay kˆ ⋅ ˆj + Az 2 ( kˆ ⋅ kˆ )
= Ax (1) + Ax Ay ( 0 ) + Ax Az ( 0 ) + Ay Ax ( 0 ) + Ay 2 (1) + Ay Az ( 0 ) + Az Ax ( 0 ) + Az Ay ( 0 ) + Az 2 (1)
2
= Ax 2 + Ay 2 + Az 2
But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of
the vector A . Thus A ⋅ A = A2 .
(b)
iˆ ˆj kˆ
A ⋅ ( B × C ) = A ⋅ Bx By Bz
Cx Cy Cz
( ) (
= Ax iˆ + Ay ˆj + Az kˆ ⋅ iˆ By Cz − Bz C y − ˆj ( Bx Cz − Bz Cx ) + kˆ Bx C y − By Cx
) ( )
( )
= Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx ( )
or
A ⋅ ( B × C) = AxBy C z + Ay Bz C x + Az BxC y − AxBz C y − Ay BxC z − Az By C x (1)
Note that ( A × B) ⋅ C = C ⋅ ( A × B) , and according to (1)
C ⋅ ( A × B) = C x Ay Bz + C y Az Bx + C z AxBy − C x Az By − C y AxBz − C z Ay Bx (2)
The right hand sides of (1) and (2) are identical. Hence A ⋅ ( B × C) = ( A × B) ⋅ C .
(c)
iˆ ˆj kˆ iˆ ˆj kˆ
(
A × ( B × C ) = Ax iˆ + Ay ˆj + Az kˆ × Bx ) By Bz = Ax Ay Az
Cx Cy Cz By Cz − Bz C y Bz Cx − Bx C y Bx C y − By Cx
= Ay Bx C y − By Cx − Az ( Bz Cx − Bx Cz ) î + Az By Cz − Bz C y − Ax Bx C y − By Cx ˆj
( ) ( ) ( )
)
+ Ax ( Bz Cx − Bx Cz ) − Ay By Cz − Bz C y k̂
(
= Ay Bx C y + Az Bx Cz − Ay By Cx − Az Bz Cx iˆ + Ax By Cx + Az By Cz − Ax Bx C y − Az Bz C y ˆj
( ) ( )
+ ( Ax Bz Cx + Ay Bz C y − Ax Bx Cz − Ay By Cz ) k̂
= Bx ( Ay C y + Az Cz ) − Cx ( Ay By + Az Bz ) î + By ( Ax Cx + Az Cz ) − C y ( Ax Bx + Az Bz ) ĵ
+ Bz ( Ax Cx + Ay C y ) − Cz ( Ax Bx + Ay By ) k̂
Add and subtract the underlined terms to get
1
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Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
( ) (
A × ( B × C ) = Bx Ay C y + Az Cz + Ax Cx − Cx Ay By + Az Bz + Ax Bx î
)
( ) (
+ By Ax Cx + Az Cz + Ay C y − C y Ax Bx + Az Bz + Ay By ĵ
)
+ Bz Ax Cx + Ay C y + Az Cz − Cz Ax Bx + Ay By + Az Bz kˆ
( ) ( )
( ˆ ˆ ˆ
)( ˆ
) (ˆ
)(
= Bx i + By j + Bz k Ax Cx + Ay C y + Az Cz − Cx i + C y j + Cz kˆ Ax Bx + Ay By + Az Bz )
or
A × ( B × C) = B( A ⋅ C) − C( A ⋅ B)
Problem 1.2 Using the interchange of Dot and Cross we get
( A × B) ⋅ (C × D) = [( A × B) × C] ⋅ D
But
[( A × B) × C] ⋅ D = − [C × ( A × B)] ⋅ D (1)
Using the bac – cab rule on the right, yields
[( A × B) × C] ⋅ D = −[ A(C ⋅ B) − B(C ⋅ A)] ⋅ D
or
[( A × B) × C] ⋅ D = −( A ⋅ D)(C ⋅ B) + ( B ⋅ D)(C ⋅ A) (2)
Substituting (2) into (1) we get
[( A × B) × C] ⋅ D = ( A ⋅ C)( B ⋅ D) − ( A ⋅ D)( B ⋅ C)
Problem 1.3
Velocity analysis
From Equation 1.38,
v = v o + Ω × rrel + v rel . (1)
From the given information we have
v o = −10Iˆ + 30 Jˆ − 50K
ˆ (2)
(
rrel = r − ro = 150Iˆ − 200 Jˆ + 300K ) (
ˆ − 300Iˆ + 200 Jˆ + 100K )
ˆ = −150Iˆ − 400 Jˆ + 200K
ˆ (3)
Iˆ Jˆ ˆ
K
Ω × rrel = 0.6 −0.4 1.0 = 320Iˆ − 270
0 Jˆ − 300K
ˆ (4)
−150 −400 200
2
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1 of 201MANUAL for Orbital Mechanics for Engineering
SOLUTION Students
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By Howard
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2 of 201MANUAL for Orbital Mechanics for Engineering
SOLUTION Students
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Howard Curtis.pdf-2026/2027
for Engineering Students
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4th Edition
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By Howard
SELLER-
Curtis
SOLUTIONS MANUAL
to accompany
ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
Page
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2 of 201 freshness top STUVIA SELLER-SOLUTION MANUAL for Orbital Mechanics for Engineering Students 4th Edition By Howard Curtis
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3 of 201MANUAL for Orbital Mechanics for Engineering
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By Howard
SELLER-
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Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
Problem 1.1
(a)
( )(
A ⋅ A = Ax iˆ + Ay ˆj + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ )
( ) ( )
= Ax iˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ + Ay ˆj ⋅ Ax iˆ + Ay ˆj + Az kˆ + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ ( )
( ) ( ) ( )
= Ax 2 ( iˆ ⋅ iˆ ) + Ax Ay iˆ ⋅ ˆj + Ax Az ( iˆ ⋅ kˆ ) + Ay Ax ˆj ⋅ iˆ + Ay 2 ˆj ⋅ ˆj + Ay Az ˆj ⋅ kˆ
( )
( )
+ Az Ax ( kˆ ⋅ iˆ ) + Az Ay kˆ ⋅ ˆj + Az 2 ( kˆ ⋅ kˆ )
= Ax (1) + Ax Ay ( 0 ) + Ax Az ( 0 ) + Ay Ax ( 0 ) + Ay 2 (1) + Ay Az ( 0 ) + Az Ax ( 0 ) + Az Ay ( 0 ) + Az 2 (1)
2
= Ax 2 + Ay 2 + Az 2
But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of
the vector A . Thus A ⋅ A = A2 .
(b)
iˆ ˆj kˆ
A ⋅ ( B × C ) = A ⋅ Bx By Bz
Cx Cy Cz
( ) (
= Ax iˆ + Ay ˆj + Az kˆ ⋅ iˆ By Cz − Bz C y − ˆj ( Bx Cz − Bz Cx ) + kˆ Bx C y − By Cx
) ( )
( )
= Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx ( )
or
A ⋅ ( B × C) = AxBy C z + Ay Bz C x + Az BxC y − AxBz C y − Ay BxC z − Az By C x (1)
Note that ( A × B) ⋅ C = C ⋅ ( A × B) , and according to (1)
C ⋅ ( A × B) = C x Ay Bz + C y Az Bx + C z AxBy − C x Az By − C y AxBz − C z Ay Bx (2)
The right hand sides of (1) and (2) are identical. Hence A ⋅ ( B × C) = ( A × B) ⋅ C .
(c)
iˆ ˆj kˆ iˆ ˆj kˆ
(
A × ( B × C ) = Ax iˆ + Ay ˆj + Az kˆ × Bx ) By Bz = Ax Ay Az
Cx Cy Cz By Cz − Bz C y Bz Cx − Bx C y Bx C y − By Cx
= Ay Bx C y − By Cx − Az ( Bz Cx − Bx Cz ) î + Az By Cz − Bz C y − Ax Bx C y − By Cx ˆj
( ) ( ) ( )
)
+ Ax ( Bz Cx − Bx Cz ) − Ay By Cz − Bz C y k̂
(
= Ay Bx C y + Az Bx Cz − Ay By Cx − Az Bz Cx iˆ + Ax By Cx + Az By Cz − Ax Bx C y − Az Bz C y ˆj
( ) ( )
+ ( Ax Bz Cx + Ay Bz C y − Ax Bx Cz − Ay By Cz ) k̂
= Bx ( Ay C y + Az Cz ) − Cx ( Ay By + Az Bz ) î + By ( Ax Cx + Az Cz ) − C y ( Ax Bx + Az Bz ) ĵ
+ Bz ( Ax Cx + Ay C y ) − Cz ( Ax Bx + Ay By ) k̂
Add and subtract the underlined terms to get
1
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Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
( ) (
A × ( B × C ) = Bx Ay C y + Az Cz + Ax Cx − Cx Ay By + Az Bz + Ax Bx î
)
( ) (
+ By Ax Cx + Az Cz + Ay C y − C y Ax Bx + Az Bz + Ay By ĵ
)
+ Bz Ax Cx + Ay C y + Az Cz − Cz Ax Bx + Ay By + Az Bz kˆ
( ) ( )
( ˆ ˆ ˆ
)( ˆ
) (ˆ
)(
= Bx i + By j + Bz k Ax Cx + Ay C y + Az Cz − Cx i + C y j + Cz kˆ Ax Bx + Ay By + Az Bz )
or
A × ( B × C) = B( A ⋅ C) − C( A ⋅ B)
Problem 1.2 Using the interchange of Dot and Cross we get
( A × B) ⋅ (C × D) = [( A × B) × C] ⋅ D
But
[( A × B) × C] ⋅ D = − [C × ( A × B)] ⋅ D (1)
Using the bac – cab rule on the right, yields
[( A × B) × C] ⋅ D = −[ A(C ⋅ B) − B(C ⋅ A)] ⋅ D
or
[( A × B) × C] ⋅ D = −( A ⋅ D)(C ⋅ B) + ( B ⋅ D)(C ⋅ A) (2)
Substituting (2) into (1) we get
[( A × B) × C] ⋅ D = ( A ⋅ C)( B ⋅ D) − ( A ⋅ D)( B ⋅ C)
Problem 1.3
Velocity analysis
From Equation 1.38,
v = v o + Ω × rrel + v rel . (1)
From the given information we have
v o = −10Iˆ + 30 Jˆ − 50K
ˆ (2)
(
rrel = r − ro = 150Iˆ − 200 Jˆ + 300K ) (
ˆ − 300Iˆ + 200 Jˆ + 100K )
ˆ = −150Iˆ − 400 Jˆ + 200K
ˆ (3)
Iˆ Jˆ ˆ
K
Ω × rrel = 0.6 −0.4 1.0 = 320Iˆ − 270
0 Jˆ − 300K
ˆ (4)
−150 −400 200
2
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