Solution Manual for Orbital Mechanics for Engineering Students 4th Edition by Howard D. Curtis All Chapters Fully Covered

, 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 + Ay ˆj + Azkˆ  Axiˆ + Ayˆj + Azkˆ )
( ) ( ) (
= Axiˆ  Axiˆ + Ayˆj + Azkˆ + Ayˆj  Axiˆ + Ayˆj + Azkˆ + Azkˆ  Axiˆ + Ayˆj + Azkˆ )
( ) ( ) ( ) (
= Ax2 (iˆ  iˆ) + Ax Ay iˆ  ˆj + Ax Az (iˆ  kˆ ) + Ay Ax ˆj  iˆ + Ay2 ˆj  ˆj + Ay Az ˆj  kˆ  )
   
( )
+ AzAx (kˆ  iˆ) + AzAy kˆ  ˆj + Az2 (kˆ  kˆ )
 
= Ax2 (1) + Ax Ay (0) + Ax Az (0) + Ay Ax (0) + Ay2 (1) + Ay Az (0) + Az Ax (0) + Az Ay (0) + Az2 (1)
     
= Ax2 + Ay2 + Az2


But, according to the Pythagorean Theorem, A 2x + A y2 + A z2 = 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 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 ByCz − BzCy BzCx − BxCy BxCy − ByCx
Cy Cz
( ) (
= Ay BxCy − ByCx − Az (BzCx − BxCz ) iˆ + Az ByCz − BzCy − Ax BxCy − ByCx  ˆj
   
) ( )
+ A (B C − B C ) − A B C − B C  kˆ

x z x x z y y z 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)
+ 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 y y)
+ B A C + A C + A C − C A B + A B + A B  kˆ
z x x z z

( )
(
= B x iˆ + B y ˆj + B z k ˆ ) (A C + A C + A C ) − ( C i ˆ + C ˆ j + C k ˆ ) (A B + A B
x x y y z z x y z x x y y + Az Bz )
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 0K̂ (2)


( ) ( )
rrel = r − ro = 150Iˆ − 200Jˆ + 300K̂ − 300Iˆ + 200Jˆ + 100K̂ = −150Iˆ − 400Jˆ + 200K̂ (3)


Iˆ Jˆ K̂
  rrel = 0.6 −0.4 1.0 = 320Iˆ − 270Jˆ − 300K̂ (4)
−150 −400 200




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