Solution Manual for Orbital Mechanics for Engineering Students, 5th Edition by Howard D. Curtis | ISBN 9780443290152 | Chapters 1–13

Solution Manual for Orbital Mec̍hanic̍s for Engineering Students,
5th Edition by Howard D. Curtis | All Chapters | Complete Solutions

, SOLUTIONS MANUAL

to ac̍c̍ompany


ORBITAL MECHANICS FOR ENGINEERING STUDENTS




Howard D. Curtis
Embry-Riddle Aeronautic̍al
University
Daytona Beac̍h, Florida

,Solutions Manual Orbital Mec̍hanic̍s for Engineering Students Chapter 1


Problem 1.1
(a)
A A = ( A i + A y ˆ+ A k ) ( A i + A yˆ + A k)
x
ˆ j z ˆ⋅ xˆ j z ˆ
= A i⋅( A i + Ayˆ + A k )+ Ayˆ⋅( A i+ A yˆ+ A k ) A zk⋅( A i + Ayˆ + A k )
x ˆ x ˆ j z ˆ j xˆ j z ˆ+ ˆ x ˆ j z ˆ
=  A 2( i ) A A y( i ) A A ( i )  A A ( ˆ ) A y2( ˆ ) A A ( ˆ )
x iˆ + x jˆ + x z kˆ +  y x ˆj + ˆj + y z ˆj 

+ AA ( k ) A Ay( k )A 2 ˆ ( )
 iˆ + z jˆ ˆ + z k kˆ 
zx
=  A 2 1 A Ay ( )+ A A ( ) +  A A ( )+ Ay2 ( )+ A A ( )   A A ( )+ A A y ( )+ A 2 1( 
x ( )+ x xz yx yz + zx z z ) 
= A 2 + A y2 + A 2
x z
But, ac̍c̍ording to the Pythagorean + A 2 + A 2 = A , where A = A , the magnitude
Theorem, A x 2 y z 2 of
the vec̍tor A. Thus A = A2.
A
(b)
iˆ ˆj kˆ
A ⋅(B× C ) A ⋅ B x By B z
=
C x Cy Cz

= ( A ˆ + A yˆ + A k )  i (B C − B y ) ˆ( B z − B C )+ k ( B y − B C )
x i j z ˆ ⋅ ˆ y z C z − j C x z ˆ Cx y

= A x ( B z − B y ) A y ( B z − B C )+A z ( B y − B C )
x
or Cy Cz − Cx z C y x


A ⋅( B× C ) A B C z + A B x + A B y − A B C y − A B z − A B x (1)
= xy Cy Cz xz Cy Cz
Note that × B C =C ⋅( A × B ) , and ac̍c̍ording to (1)
A ) ⋅
C ⋅( A × B ) C A B + C A B + C A B − C A B − C A B −C A B (2)
= x y y z zxy x z y xz zy x
The right hand sides of (1) and (2) are identic̍al. ⋅(B× C ) ( A × B C .
Henc̍e A = ) ⋅
(c̍)
iˆ ˆj kˆ ˆi ˆj kˆ
A × (B× C ) ( A ˆ + A yˆ + A k )× B B y B z = Ax Ay Az
= i j z ˆ
x x
C x C y C z B C − B y B C − B Cy B y −B C y x
y z C z z x x Cx
=  A y (B C y −B C ) A (B C −B C )+ˆ  A (B C − B Cy ) A ( B y −B C ) ˆj
x y x− z z x x z  z y z − x Cx y 
+  A (B C − B z ) A y ( B C −B Cy) i ˆk
− 
x z x C x y z z
( A B y + A B C − A B C x − A B C )+ (A B C + A B C −A B C y − A B y ) ˆj
C yx zxz yy z z x iˆ x y x zy z xx C zz
+ ( A x z x + A B C y −A B C z − A B C )ˆk
BC yz xx yyz
=  B ( A C y + A C z ) C x( A y + A B )+ˆi  By( A C x + A Cz)−Cy(A B + A B ) ˆj
x y z − By zz  x z xx zz 
+  B ( A C + A y)−Cz (A B x+ A B y) ˆk
 z xx Cy x y 
Add and subtrac̍t the underlined terms to get




1

, Solutions Orbital Mec̍hanic̍s for Engineering Chapter
Manual Students 1


A × (B× C )  B (A C y + A C z + A C ) C ( A B y + A B + A B ) ˆi
= − x
x y z xx y zz xx 
+By ( A x + A C z + A C y)−Cy ( A B + A B + A y y) ˆj
 
Cx z y xx zz B
+ B ( A x + A C y + A C )− Cz ( A B + A B y + A B )kˆ
z Cx y zz xx y zz
= ( B i + B y ˆ+ B k)( A x + A y + A C ) (Cx i + Cyˆ + Czk)( A B +
or x ˆ j z ˆ Cx Cy z z − ˆ j ˆ xx


A × (B× C ) B A C ) C A B
= − )
Problem 1.2 Using the interc̍hange of Dot and Cross we get
(A × B ( × D ) = [ A × B ) C D
) ⋅C ( ×
But

[ (A × B ) C D = [ × ( A × B ) D (1)
× − ]⋅
C
Using the bac̍ – c̍ab 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
Veloc ̍ity analysis

From Equation 1.38,

v = v o + Ω × r + v rel. (1)
rel
From the given information we have

v o= +30 J− 50 Kˆ (2)
−10 Iˆ ˆ
r rel= r − r =( 15− 20 J + 300 ) ( 30 + 20 J + 1 00 ) = − 40 J + 200 Kˆ (3)
− 0
o 0 Iˆ 0 ˆ Kˆ 0 ˆ Kˆ −150 0 ˆ
Iˆ Jˆ Kˆ Iˆ
Ω× r = 0 6 −0 4 1 0 = 320 −27 J− 300 (4)
rel −150 −400 200
Iˆ 0 ˆ Kˆ




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