Orbital Mechanics for Engineering Students 4th Edition Complete Solutions Manual: Chapter-by-Chapter Problem Solutions with MATLAB Code & Detailed Explanations – Author: Howard D. Curtis

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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  Axiˆ Ayˆj  Azkˆ  Axiˆ Ayˆj  Azkˆ 
   
 Axiˆ Axiˆ Ayˆj  Azkˆ  Ayˆj  Axiˆ Ayˆj  Azkˆ  Azkˆ  Axiˆ Ayˆj  Azkˆ  
     
 A 2 iˆ  iˆ  A A iˆ  ˆj  A A iˆ  kˆ   A A ˆj  iˆ  A 2 ˆj  ˆj  A A ˆj  kˆ   
x x y x z y x y y z
 AzAx kˆ iˆ AzAy kˆ ˆj A 2 kˆ kˆ
    
z
 
 A 2 1  A A 0  A A 0  A A 0  A 2 1  A A 0  A A 0  A A 0  A 2 1
x x y x z y x y z z x z y z
   y   
 Ax2  Ay2  Az2


But, according to the Pythagorean Theorem, A 2x  A 2y  A 2z  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
 
 
 Axiˆ A yˆj  A zkˆ  iˆ ByCz  B zC y  ˆjBxCz  BzCx   kˆ BxCy  ByCx 
 
 
 
 Ax ByCz  BzCy  Ay BxCz  BzCx   Az BxCy  ByCx  
or

A  B  C  AxByCz  AyBzCx  AzBxCy  AxBzCy  AyBxCz  AzByCx (1)

Note that A  B  C  C  A  B , and according to (1)

C  A  B  CxAyBz  Cy AzBx  Cz AxBy  CxAzBy  Cy AxBz  Cz AyBx (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 Axiˆ Ayˆj  Azkˆ  Bx  By Bz  Ax Ay Az
Cx Cy Cz ByCz  BzCy BzCx  BxCy BxCy  ByCx

  
 Ay BxCy  ByCx  Az BzCx  BxCz  iˆ  Az ByCz  BzCy  Ax BxCy  ByCx  ˆj
   
  

x z y y z z y 
 A B C  B C  A B C  B C  kˆ
x z x


  
 AyBxCy  AzBxCz  AyByCx  AzBzCx iˆ  AxByCx  AzByCz  AxBxCy  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 AyCy  AzCz  Cx AyBy  AzBz  iˆ  By AxCx  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  kˆ
 

Add and subtract the underlined terms to get




1

, Solutions Manual Orbital Mechanics for Engineering Students Chapter 1



  
A  B  C  Bx AyCy  AzCz  AxCx  Cx AyBy  AzBz  AxBx  iˆ
 

  
 By AxCx  AzCz  AyCy  Cy AxBx  AzBz  AyBy  ˆj
  
 y y z z z x x 
 B A C  A C  A C  C A B  A B  A B  kˆ
 z x x y y 
z z
 

 Bx iˆ  By ˆj  Bzkˆ A C  A C  A C   C iˆ  C ˆj  C kˆ A B  A B
x x y y z z x y z x x y y  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  50 Kˆ (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  300K̂ (4)
150 400 200




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