Orbital Mechanics for Engineering
ST
Students – 4th Edition
UV
SOLUTION
IA
MANUAL
_A
PP
Howard D. Curtis
RO
VE
Comprehensive Solutions Manual for Instructors
and Students
D?
© Howard D. Curtis
??
All rights reserved. Reproduction or distribution without permission is prohibited.
©STUDYSTREAM
, 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 )
ST
= 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 )
( )
SSUTT
2
= 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)
= Ax 2 + Ay 2 + Az 2
UV
But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of
UVIV
the vector A . Thus A ⋅ A = A2 .
(b)
AIIA_A
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
) ( )
_A_A
( )
= Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx ( )
or
APPPP
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)
PPR
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)
ROO
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
VVEE
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
DD?
= 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
, 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
( ) ( )
ST
( 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
SSUTT
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
UV
UVIV
But
[( A × B) × C] ⋅ D = − [C × ( A × B)] ⋅ D (1)
AIIA_A
Using the bac – cab rule on the right, yields
[( A × B) × C] ⋅ D = −[ A(C ⋅ B) − B(C ⋅ A)] ⋅ D
or
_A_A
[( A × B) × C] ⋅ D = −( A ⋅ D)(C ⋅ B) + ( B ⋅ D)(C ⋅ A) (2)
APPPP
Substituting (2) into (1) we get
[( A × B) × C] ⋅ D = ( A ⋅ C)( B ⋅ D) − ( A ⋅ D)( B ⋅ C)
PPR
Problem 1.3
Velocity analysis
ROO
From Equation 1.38,
v = v o + Ω × rrel + v rel . (1)
From the given information we have
VVEE
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)
DD?
Iˆ Jˆ ˆ
K
Ω × rrel = 0.6 −0.4 1.0 = 320Iˆ − 270
0 Jˆ − 300K
ˆ (4)
???
−150 −400 200
?
2
, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
v rel = −20iˆ + 25 ˆj + 70kˆ
(
= −20 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
ˆ )
(
+ 25 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
ˆ )
(
+ 70 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K
ˆ )
ST
so that
v rel = −53.826Iˆ − 28.115 Jˆ + 47.300K
ˆ ( m s) (5)
SSUTT
Substituting (2), (3), (4) and (5) into (1) yields
(
v = −10Iˆ + 30 Jˆ − 50K
ˆ + 320Iˆ − 270 Jˆ − 300K ) (
ˆ + −53.826Iˆ − 28.115 Jˆ + 47.300K
ˆ) ( )
UV UVIV
v = 256.17Iˆ − 268.12 Jˆ − 302.7K
ˆ
(
= 478.68 0.53516Iˆ − 0.56011Jˆ − 0.63236K ( m s ) )
AIIA_A
Acceleration analysis
From Equation 1.42,
a = a O + Ω × rrel + Ω × ( Ω × rrel ) + 2Ω × v rel + a rel (6)
_A_A
Using the given data together with (4) and (5) we obtain
ao = 25Iˆ + 40 Jˆ − 15K
ˆ (7)
APPPP
Iˆ Jˆ ˆ
K
Ω × rrel = −0.4 0.3 −1.0 = −340Iˆ + 230 Jˆ + 205K
ˆ (8)
PPR
−150 −400 200
Iˆ Jˆ ˆ
K
ROO
Ω × ( Ω × rrel ) = 0.6 −0.4 1.0 = 390IIˆ + 500 Jˆ − 34K
ˆ (9)
320 −270 −300
Iˆ Jˆ Kˆ
VVEE
2Ω × v rel = 2 0.6 −0.4 1.0 = 2 9.151Iˆ − 82.206 Jˆ − 38.399K
ˆ ( ) (10)
−53.826 −28.115 47.300
a rel = 7.5iˆ − 8.5 ˆj + 6.0kˆ
DD?
(
= 7.5 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
ˆ )
(
− 8.5 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
ˆ )
(
+ 6.0 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K̂ )
???
a rel = 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
ˆ (11)
Substituting (7), (8), (9), (10) and (11) into (6) yields
?
(
a = 25Iˆ + 40 Jˆ − 15K
ˆ + −340Iˆ + 230 Jˆ + 205K) (
ˆ + 390Iˆ + 500 Jˆ − 34K
ˆ ) ( )
( ) (
ˆ + 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
+ 2 9.151Iˆ − 82.206 Jˆ − 38.399K
ˆ )
3
ST
Students – 4th Edition
UV
SOLUTION
IA
MANUAL
_A
PP
Howard D. Curtis
RO
VE
Comprehensive Solutions Manual for Instructors
and Students
D?
© Howard D. Curtis
??
All rights reserved. Reproduction or distribution without permission is prohibited.
©STUDYSTREAM
, 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 )
ST
= 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 )
( )
SSUTT
2
= 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)
= Ax 2 + Ay 2 + Az 2
UV
But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of
UVIV
the vector A . Thus A ⋅ A = A2 .
(b)
AIIA_A
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
) ( )
_A_A
( )
= Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx ( )
or
APPPP
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)
PPR
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)
ROO
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
VVEE
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
DD?
= 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
, 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
( ) ( )
ST
( 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
SSUTT
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
UV
UVIV
But
[( A × B) × C] ⋅ D = − [C × ( A × B)] ⋅ D (1)
AIIA_A
Using the bac – cab rule on the right, yields
[( A × B) × C] ⋅ D = −[ A(C ⋅ B) − B(C ⋅ A)] ⋅ D
or
_A_A
[( A × B) × C] ⋅ D = −( A ⋅ D)(C ⋅ B) + ( B ⋅ D)(C ⋅ A) (2)
APPPP
Substituting (2) into (1) we get
[( A × B) × C] ⋅ D = ( A ⋅ C)( B ⋅ D) − ( A ⋅ D)( B ⋅ C)
PPR
Problem 1.3
Velocity analysis
ROO
From Equation 1.38,
v = v o + Ω × rrel + v rel . (1)
From the given information we have
VVEE
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)
DD?
Iˆ Jˆ ˆ
K
Ω × rrel = 0.6 −0.4 1.0 = 320Iˆ − 270
0 Jˆ − 300K
ˆ (4)
???
−150 −400 200
?
2
, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1
v rel = −20iˆ + 25 ˆj + 70kˆ
(
= −20 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
ˆ )
(
+ 25 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
ˆ )
(
+ 70 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K
ˆ )
ST
so that
v rel = −53.826Iˆ − 28.115 Jˆ + 47.300K
ˆ ( m s) (5)
SSUTT
Substituting (2), (3), (4) and (5) into (1) yields
(
v = −10Iˆ + 30 Jˆ − 50K
ˆ + 320Iˆ − 270 Jˆ − 300K ) (
ˆ + −53.826Iˆ − 28.115 Jˆ + 47.300K
ˆ) ( )
UV UVIV
v = 256.17Iˆ − 268.12 Jˆ − 302.7K
ˆ
(
= 478.68 0.53516Iˆ − 0.56011Jˆ − 0.63236K ( m s ) )
AIIA_A
Acceleration analysis
From Equation 1.42,
a = a O + Ω × rrel + Ω × ( Ω × rrel ) + 2Ω × v rel + a rel (6)
_A_A
Using the given data together with (4) and (5) we obtain
ao = 25Iˆ + 40 Jˆ − 15K
ˆ (7)
APPPP
Iˆ Jˆ ˆ
K
Ω × rrel = −0.4 0.3 −1.0 = −340Iˆ + 230 Jˆ + 205K
ˆ (8)
PPR
−150 −400 200
Iˆ Jˆ ˆ
K
ROO
Ω × ( Ω × rrel ) = 0.6 −0.4 1.0 = 390IIˆ + 500 Jˆ − 34K
ˆ (9)
320 −270 −300
Iˆ Jˆ Kˆ
VVEE
2Ω × v rel = 2 0.6 −0.4 1.0 = 2 9.151Iˆ − 82.206 Jˆ − 38.399K
ˆ ( ) (10)
−53.826 −28.115 47.300
a rel = 7.5iˆ − 8.5 ˆj + 6.0kˆ
DD?
(
= 7.5 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
ˆ )
(
− 8.5 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
ˆ )
(
+ 6.0 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K̂ )
???
a rel = 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
ˆ (11)
Substituting (7), (8), (9), (10) and (11) into (6) yields
?
(
a = 25Iˆ + 40 Jˆ − 15K
ˆ + −340Iˆ + 230 Jˆ + 205K) (
ˆ + 390Iˆ + 500 Jˆ − 34K
ˆ ) ( )
( ) (
ˆ + 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
+ 2 9.151Iˆ − 82.206 Jˆ − 38.399K
ˆ )
3
