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

, SOLUTIONS
MANUALTO
ACCOMPANYORBI
TAL MECHANICS
FOR
ENGINEERING
STUDENTS

,SOLUTIONS MANUALTO ACCOMPANYORBITAL MECHANICS FOR
ENGINEERING STUDENTS

Problem 1.1
(A)


A  A  Ax Iˆ  Ay ˆj  Azkˆ Ax Iˆ  Ay ˆj  Azkˆ 
 Ax Iˆ  Ax Iˆ  Ay ˆj  Azkˆ  Ay ˆj  Ax Iˆ  Ay ˆj  Azkˆ  Azk ˆ Ax Iˆ  Ay ˆj  Azkˆ 
      
 Ax2 Iˆ  Iˆ   Ax Ay Iˆ  ˆ j  Ax Az Iˆ  Kˆ   Ay Ax ˆ j  Iˆ  Ay2 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  A
y
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 Iˆ ˆj Kˆ
Kˆ


A  B  C  Ax Iˆ  Ay ˆj  Azkˆ  Bx  By Bz  Ax Ay Az
Cx CyBy cz  Bzcy
Cz Bzcx  Bx cy Bx cy  By cx

  
 Ay Bx cy  By cx  Az Bzcx  Bx cz  Iˆ  Az By cz  Bzcy  Ax Bx cy  By cx  ˆj
  


 
 A B c  B c   A B c  B c  Kˆ

x z x x z y y z 
z y


   
 Ay bx cy  Azbx cz  Ay by cx  Azbzcx Iˆ  Ax by cx  Azby cz  Ax bx cy  Azbzcy ˆj

 x z x y z y x x z y y z
 A b c  A b c  A b c  A b c Kˆ
 Bx Ay cy  Azcz   Cx Ay by  Azbz  Iˆ  By Ax cx  Azcz   Cy Ax bx  Azbz  ˆj
   
z  x x y y   Cz Axb x  Ayby  K
 B A c  A c ˆ
 

Add And Subtract The Underlined Terms To Get

, SOLUTIONS MANUALTO ACCOMPANYORBITAL MECHANICS FOR
ENGINEERING STUDENTS

  
A  B  C  Bx Ayc y  Azc z  Axcx  Cx Ayby  Azb z  Axbx  Iˆ
 

  
 By Ax cx  Azcz  Ay cy  Cy Ax bx  Azbz  Ay by  ˆj
 


z x x
 B A c  A c  A c
yy zz  C z A b
x x 
 A b
y y A b
z z  Kˆ



 Bx Iˆ  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  (3)
2 0 0 K̂

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