SOLUTION MANUAL FOR Orbital Mechanics for Engineering Students 4th Edition by Howard D. Curtis ISBN: 978-0128240250 COMPLETE GUIDE ALL CHAPTERS COVERED 100% VERIFIED A+ GRADE ASSURED!!!!!NEW LATEST UPDATE!!!!!

z l

, Solutions Manual z l Orbital Mechanics for Engineering Students zl zl zl zl Chapter 1
zl




Problem 1.1 zl


(a)
zl zl zl zl

A A  Axiˆ  Ayˆj  Azkˆ  Axiˆ  Ayˆj  Azkˆ zl
zl
zl
zl
zl zl
zl
 zl
zl
zl
zl
zl zl

zl




zl
zl

 Axiˆ Axiˆ  Ayˆj  Azkˆ  Ayˆj Axiˆ  Ayˆj  Azkˆ  Azkˆ  Axiˆ  Ayˆj  Azkˆ
zl
zl
zl zl zl

zl
zl zl zl
 zl
zl zl zl

zl
zl
zl

zl
zl
zl zl zl
zl

 Ax2 iˆ iˆ AxAy iˆ  ˆj  AxAz iˆ kˆ   AyAx ˆjiˆ  Ay2 ˆj ˆj  AyAz ˆjkˆ         
zl zl zl zl zl zl zl zl


zl zl
zl zl zl zl zl zl zl
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
zl zl zl zl zl zl zl zl

   
 AzAx kˆ iˆ AzAy kˆ ˆj  Az2 kˆ kˆ   
z l z l

zl
zl zl zl zl zl
zl zl zl zl zl zl zl
zl zl zl zl

 
 Ax2 1 AxAy 0 AxAz 0  AyAx 0 Ay2 1 Ay Az 0  AzAx 0 AzAy 0 Az2 1
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl



   
zl zl zl zl zl zl zl zl zl zl zl zl

z l z l


 z l z l



 Ax2  Ay2  Az2
zl
zl
zl
zl
zl





But, according to the Pythagorean Theorem, Ax 2  Ay 2  zA 2  A2 , where A  A , the magnitude of
zl zl zl zl zl z l z l
z l
zl z l
z l
zl z l
z l
zl
zl
zl z l zl zl zl zl zl



2
the vector A . Thus A  A  A .
zl z l zl zl z l zl zl zl zl
zl




(b)
iˆ ˆj kˆ
A B  C  A  Bx
zl zl zl zl zl zl z l By Bz
Cx Cy Cz

 Axiˆ  Ayˆj  Azkˆ iˆ ByCz  BzCy  ˆjBxCz  BzCx  kˆ BxCy  ByCx 
zl



zl
zl
zl



zl
zl zl
 
zl
zl
zl
zl

zl zl
zl zl
zl
zl
zl
zl zl

zl

zl
zl
zl 
zl

 Ax ByCz  BzCy  Ay BxCz  BzCx  Az BxCy  ByCx
zl zl
zl

zl zl
zl
zl zl
zl
zl
zl zl
zl  zl
zl

zl


or

A  B  C  AxByCz  AyBzCx  AzBxCy  AxBzCy  AyBxCz  AzByCx
zl zl zl zl
zl
zl
z l
zl
z l
zl
z l
zl
z l
zl
z l
zl




(1)

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




C  A  B  CxAyBz  Cy AzBx  Cz AxBy  CxAzBy  Cy AxBz  Cz AyBx
zl zl zl zl
zl
zl
z l
zl
zl z l
zl
zl z l
zl
z l
zl
zl z l
zl
zl




(2)

The right hand sides of (1) and (2) are identical. Hence A   B  C  A  B C .
zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl zl zl zl zl zl zl zl




(c)
iˆ ˆj kˆ iˆ ˆj kˆ
A B C Axiˆ  Ayˆj  Azkˆ  Bx
zl zl zl zl zl zl
 zl
zl
zl
zl
zl zl
zl
 z l By Bz z l
Ax By z l
Ay Bz z l
Az
Cx Cy  Cz  BzCy zl
zl Cx  BxCy zl
zl BxCy  ByCx zl
zl



Cz


zl


 
zl 
 Ay BxCy  ByCx  Az BzCx  BxCz iˆ  Az ByCz  BzCy  Ax BxCy  ByCx ˆj

zl
zl

zl zl
zl
zl zl
zl
zl
zl
zl
zl
zl  zl
zl
zl  zl
zl
zl  zl
zl
zl  zl




 A B C  B C  A B C  B C kˆ

x z x x z y y z z y

zl zl
zl
zl
zl zl
zl  zl
zl
zl  zl





zl
 
 AyBxCy  AzBxCz  AyByCx  AzBzCx iˆ  AxByCx  AzByCz  AxBxCy  AzBzCy ˆj zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl
zl




 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 zl zl zl



 Bx AyCy  AzCz  Cx AyBy  AzBz iˆ  By AxCx  AzCz  Cy AxBx  AzBz ˆj
zl zl zl zl

zl
zl zl zl zl zl zl zl zl zl zl zl
zl zl zl zl zl zl zl zl zl zl zl zl zl

1

,Solutions Manual
z l Orbital Mechanics for Engineering Students
zl zl zl zl Chapter 1
zl




   


x x
zl 
 B A C  A C  C A B  A B kˆ
z y y z x x
zl y y
zl


zl zl
zl
zl  zl
zl

zl
zl




Add and subtract the underlined terms to get
zl zl zl zl zl zl zl




2

, Solutions Manual z l Orbital Mechanics for Engineering Students zl zl zl zl Chapter 1
zl




A  B  C  Bx AyCy  AzCz  AxCx  Cx AyBy  AzBz  AxBx  iˆ
zl zl



zl zl zl



zl
zl  z l
zl
z l
zl
zl  zl
zl
zl  z l
zl
z l
zl
zl  zl




 By AxCx  AzCz  AyCy  Cy AxBx  AzBz  AyBy ˆj
 

zl zl
zl
zl
zl
zl   zl
zl zl
zl
zl
zl

zl
zl




 B A C  A C  A C  C A B  A B  A B kˆ

z x x y y z z z x x y y z z

zl  z l
zl
z l
zl
zl  zl
zl
zl  z l
zl
z l
zl
zl  zl





 Bxiˆ  Byˆj  Bzkˆ
zl
zl
zl
zl
zl
zl zl
zl
  A C  A C  A C   C i ˆ  C ˆ j  C k ˆ   A B  A B
x x zl
zl
y y zl
zl
z z zl zl
zl
x zl
zl
zl
y zl
zl zl
z
zl

x x zl
zl
y y zl
 AzBz
zl

zl


or

A  B C  BA C CA B
zl zl zl zl zl zl zl zl zl zl zl zl



Problem 1.2 Using the interchange of Dot and Cross we getzl zl zl zl zl zl zl zl zl zl




A  BC D 
zl zl zl zl zl zl z l A  B CD zl zl zl zl zl




But

A  B CD   C A  BD
zl zl zl zl zl zl zl zl zl zl zl zl zl (1)

Using the bac – cab rule on the right, yields
zl zl z l zl zl zl zl zl zl




A  B CD  ACB  BCAD
zl zl zl zl zl zl zl zl zl zl zl zl zl zl




or

A  B CD  A DCB  BDCA
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl (2)

Substituting (2) into (1) we get zl zl zl zl zl





A  B CD  A CBD A DBC
zl zl
zl
zl zl zl zl zl zl zl zl
zl
zl zl zl zl zl




Problem 1.3 Veloc zl zl




ity analysis From Eq
zl zl zl




uation 1.38, zl




v  vo    rrel  vrel .
zl zl
zl
zl zl zl
zl
zl
zl
(1)

From the given information we have
zl zl zl zl zl




vo  10Iˆ  30Jˆ  5 0K̂
zl
zl
zl
zl
zl
zl (2)


rrel  r  ro  150Iˆ  200Jˆ  300K̂  300Iˆ  200Jˆ 1 00K̂  150Iˆ  400Jˆ  200K̂
zl
zl zl zl
z l
zl  zl
zl
zl
zl
zl
 zl zl
zl
zl
zl
zl
zl
 zl zl
zl
zl
zl
zl (3)


Iˆ Jˆ K̂
  rrel  0.6
zl zl
zl
zl 0.4 1.0  320Iˆ  270Jˆ  300K̂
z l zl
zl
zl
zl
zl (4)
150 400 z l z l 200




3