Solutions Manual : Orbital Mechanics for Engineering Students – 4th Edition with MATLAB Examples | Howard D. Curtis 2026 |isbn 9780323853453

, SOLUTIONS MANUAL q




to accompany
q




ORBITAL MECHANICS FOR ENGINEERING STUDENTS
q q




Howard D. Curtis
q q



Embry-
Riddle Aeronautical University Daytona Beac
q q q q


h, Florida
q

, Orbital Mechanics for Engineering Students


Problem 1.1 q


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




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


q q
q q q q q q q
q q q q q q q q qq q q qq q q qq
q q q q q q q q

   
  2  ˆ ˆ
 
q q


 ˆ
q
ˆ ˆ ˆ
 AzAx k i  AzAy k  j  Az k k q
q
q
q
q
q q q
q
q
q
q
q q q q

 
 Ax2 1 AxAy 0 AxAz 0  AyAx 0 Ay2 1 AyAz 0  AzAx 0 AzAy 0 Az2 1
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q



     
q q q q q q q q q q q q

q q q q




 Ax  Ay2
q
2 q
q
q
 Az2 q





But, according to the Pythagorean Theorem, A 2 x A 2 y A 2 
q
2
z A , where A  A , the magnitude of
q q q q q q
q
q q
q
q q
q
q
q
q q q q q q q




the vector A . Thus A  A  A2 .
q q qq q q q q q
q




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

 q
q

q

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


or

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




(1)

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




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




(2)

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




(c)
iˆ ˆj kˆ iˆ ˆj kˆ
A B C Axiˆ  Ayˆj  Azkˆ  Bx
q q q q q q
 q
q
q
q
q q

q
q By Bz  q
Ax ByC q
Ay BzC q
Az
x  BxCy BxCy  ByCx Cx Cy Cz z  BzCyq
q
q
q
q
q





q


 

 Ay BxCy  ByCx  Az BzCx  BxCz  i  Az ByCz  BzCy  Ax BxCy  ByCx  ˆj
ˆ
q


q
q
q  q
q
q q
q
q
q
q
q

q q
q
q  q
q

q q
q
q  q




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




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




 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 q q q



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

q
q q q q q q q q q q q


 
q q q q q q q q q q q q q

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


 
q q q q q q q




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




1

, Orbital Mechanics for Engineering Students


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



q q q q



q  q
q
q
q

q q
q
q  q
q
q
q
q  q




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




 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

q  q
q
q
q
q  q
q
q q
q
q
q

q
q




q

 Bxiˆ  Byˆj  Bzkˆ q
q
q
q
q q
q
A C  A C  A C  C iˆ  C ˆj  C kˆ A B  A B  A B 
x x q
q
y y q
q
z z q q
q
x q
q
q
y q
q q
z
q

x x q
q
y y q
q
z z q


or

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



Problem 1.2 Using the interchange of Dot and Cross we get
q q q q q q q q q q




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




But

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

Using the bac – cab rule on the right, yields
q q q q q q q q q




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




or

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

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




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




Problem 1.3 Veloc q q




ity analysis From Eq
q q q




uation 1.38, q




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

From the given information we have
q q q q q




vo  10Iˆ  30Jˆ  50K̂
q
q
q
q
q
q (2)


q
q q q
q
q 
rrel  r  ro  150Iˆ  200Jˆ  300K̂  300Iˆ  200Jˆ  100K̂  150Iˆ  400Jˆ  200K̂ q
q
q
q
q
 q q
q
q
q
q
q
 q q
q
q
q
q (3)


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




2