SOLUTION MANUAL for Orbital Mechanics for Engineering Students 4th Edition By Howard Curtis

,SOLUTIONS MANUAL

to AccompAny


ORBITAL MECHANICS FOR ENGINEERING STUDENTS




HowArd D. Curtis
Embry-Riddle AeronAuticAl University
DAytonA BeAch, FloridA

, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1


Problem 1 .1
(A)


A  A  Ax iˆ  Ay ˆj  Azkˆ Ax iˆ  Ay ˆj  Azkˆ 
 Ax iˆ Ax iˆ  Ay ˆj  Azk ˆ  Ay ˆj Ax iˆ  Ay ˆj  Azkˆ  Azk ˆ Ax iˆ  Ay ˆj  Azkˆ 
2
    2
   
 Ax i ˆ  i ˆ   Ax Ay i ˆ  ˆ j  Ax Az i ˆ  k ˆ   Ay Ax ˆ j  i ˆ  Ay ˆ j  ˆ j  Ay Az ˆ j  kˆ 
   
 
 AzAx kˆ  iˆ  AzAy kˆ  ˆj  Az2 kˆ  kˆ 
 
 Ax 2 1   Ax Ay 0   Ax Az 0   Ay Ax 0   Ay 2 1   Ay Az 0   Az Ax 0   Az Ay 0   Az2 1 
     
 Ax 2  Ay 2  Az2

But, According to the PythAgoreAn Theorem, Ax 2  Ay 2  A
z
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  Azkˆ  iˆ ByCz  BzCy  ˆj BxCz  BzCx   kˆ BxCy  ByCx 
 

 
 Ax By Cz  BzCy  Ay Bx Cz  BzCx   Az Bx Cy  By Cx 
or

A  B  C  Ax By Cz  Ay BzCx  AzBx Cy  Ax BzCy  Ay Bx Cz  AzBy Cx (1)

Note thAt A  B   C  C  A  B  , And According to (1)

C  A  B   Cx Ay Bz  Cy AzBx  Cz Ax By  Cx AzBy  Cy Ax Bz  Cz 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  Azkˆ  Bx By Bz  Ax Ay Az
Cx B y C z  Bz C y Bz C x  Bx C y Bx C y  B y C x
Cy Cz
   
 Ay Bx Cy  By Cx  Az BzCx  Bx Cz  iˆ  Az By Cz  BzCy  Ax Bx Cy  By Cx  ˆj
   
 

x z y y z 
 A  B C  B C   A B C  B C k ˆ
x z x z y


 y x y z x z y y x z z x  x y x z y z x x y z z y
 A B C  A B C  A B C  A B C iˆ  A B C  A B C  A B C  A B C ˆj

 x z x y z y x x z y y z ˆ
 A B C  A B C  A B C  A B C kˆ
 Bx Ay Cy  AzCz   Cx Ay By  AzBz  i  By Ax Cx  AzCz   Cy AxBx  AzBz  ˆj

   
z  x x y y  z  x x y y 
 B A C  A C  C A B  A B  k ˆ
 

Add And subtrAct the underlined terms to get

, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1



  
A  B  C  Bx AyC y  AzC z  AxC x  Cx AyB y  AzB z  AxB x  i ˆ
 

  
 By Ax Cx  AzCz  Ay Cy  Cy Ax Bx  AzBz  Ay By  ˆj
 


z 
 B A C  A C  A C
x x y y z z  C z A B
x x  A 
B
y y  A B
z z  kˆ



 Bx i ˆ  By ˆ j  Bzkˆ AxCx  AyCy  AzCz   Cx iˆ  Cy ˆj  Czkˆ AxBx  AyBy  AzBz 
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  vo    rrel  vrel . (1)

From the given informAtion we hAve

vo  10Iˆ  30Jˆ  5 0 K̂ (2)

  
rrel  r  ro  150Iˆ  200Jˆ  3 0 0 K̂  300Iˆ  200Jˆ  1 0 0 K̂   150Iˆ  400Jˆ  2 0 0 K̂ (3)


Iˆ Jˆ K̂
  rrel  0.6 0.4 1.0  320Iˆ  270Jˆ  3 0 0K̂ (4)
150 400 200