Orbital Mechanics Practice Questions with Solutions and Detailed Explanations for Exams

, SOLUTIONS MANUAL

to accompany


ORBITAL MECHANICS FOR ENGINEERING STUDENTS




Howard D. Curtis
Embry-Riddle Aeronautical
University
Daytona Beach, Florida

,Solutions Manual Orbital Mechanics ғor Chapter 1
Engineering Students

Problem 1.1
(a)
A=( A i + A y ˆ + A z k )⋅ ( A x i + A y ˆ + A z k)
x
ˆ j ˆ ˆ j ˆ
= A i ⋅ ( A iˆ + A y ˆ + A z k )+ Ay ˆ ⋅ ( A x i + A y ˆ + A k)+ A k ⋅ ( A x iˆ +Ay ˆ +Ak )
x ˆ x j ˆ j ˆ j z ˆ z ˆ j ˆ
=  A 2 ( i iˆ ) + A A x y ( i jˆ ) A A x ( i kˆ )   A A ( ˆ ) A y 2 ( ˆ ˆj ) A ( ˆ ) 
ˆ ˆ + z ˆ + ˆj + j + A y ˆj
x yx

+ AAz ( k ) + AAzy ( k ) Az 2 ˆ ( k ) 
x iˆ ˆ jˆ ˆ + kˆ
=  A 2 1 ( )+ A A y ( )+ A A x ( ) +  A A y ( )+ Ay 2 ( ) A A y z ( + A ( ) Ay ( A2 1
+ ) + ) (
x2 2 x 2 z x Az
=A + Ay + Az
x
But, according to the Pythagorean Theorem, A x 2+ A 2 + A2 = A 2 , where A = A , the
y magnitude oғ
the vector A. Thus A = A2.
A
(b)
iˆ ˆj kˆ
A⋅ ( B×C ) A ⋅ B x B y B z
=
C x C y Cz

=(A x ˆi + A
y
ˆ + A k )  i
( B C y− B y )− ˆj ( B z − B C z ) kˆ ( B y − B ) 
j zˆ⋅ ˆ +
z Cz Cx x C
= A x (B C y z − B y) A y ( B z − B C )+ A z ( B y − B C ) x C

or Cz − C z x Cx y x


A ⋅ ( B × CABC xy z +ABC y x + A y −A y − A B C z − A B C x (1
) z B B y x zy )
Note that A × B C =C ⋅ ( A × B ) , and according to (1)
)
C ⋅ ( A × BCAB x +
+ C A − C A − C A B y −C A B z
CAB y (2
) y z z x Bz Bx xz y x )
The right hand sides oғ (1) and (2) are identical. Hence A ⋅ ( B × C ( A × BC.
) )
(c)
iˆ ˆj kˆ ˆi ˆj kˆ
A×( B×C) ( A x ˆ + A y ˆ + A k )× B x By B z = A x Ay A z
= i j ˆ
z − B Cy
C x Cy C z BC y B − B C y B y −B C y x
z z Cz x Cx
= A y ( B C x y − B C ) A z (B C z − B C ) +ˆi A z ( B − B C y )− A ( B y−B )  ˆj
 y x − x x  C z x C x C 
+  A x ( B C z − B z ) A y ( B C y − B y )  ˆk
−
x Cx z Cz
=( A B C y y + A B C z − A B C y x − A z B C )+ iˆ ( A B C + A B C z − A B y− A y ) ˆ
x xz y z xyx y z C xx B j
+( A xz B C x + A B C y− A B C x z− ABC y ) ˆk
yz x yz
= B x ( A C y y + A C z ) C x ( A B y + A B ) +ˆi  By ( A x + A C z )− Cy ( A + A )  ˆj
 z − y zz  C z B x B
+ B z ( A C x + A y )− C z ( A B x + A B y )  ˆk
 x Cy x y
Add and subtract the underlined terms to get




1

, Solutions Orbital Mechanics ғor Engineering Students C
Manual h
a
p
A× ( B× C )  B ( A C y + A C z z + A C x )− C ( A B y + A B z z + A B x x )  ˆi t
=
x y x x y e
+  B y ( A C x + A C z z + A C y )−C y ( A B x + A B z + A yB y )  ˆj
x y x z
+  B (A Cx + AC yy + AC z )− C z ( A B + AB yy +A B zz )  kˆ
z x z xx
i +B ˆ + B k) A C
=( B x y z ( x x+A Cy + AC )− ( C x i + Cy ˆ + C z
k ) A+ A + A
( y )
ˆ j ˆ y z ˆ j ˆ
or

A × ( B × C) = BAC) − CAB)

Problem 1.2 Using the interchange oғ Dot and Cross we get
( A × B) ⋅ ( C × D) = [ ( A × B) × CD

But

[ ( A × B) × CD = − [ C × ( A × B ) ]⋅ D (1)

Using the bac – cab rule on the right, yields

[ ( A × B) × CD = −[ ACB) BCA) D
− ]⋅
or

[ ( A × B) × CD = −( ADCB ( BDCA (2)
) + )
Substituting (2) into (1) we get

[ ( A × B) × CD = ( ACBD ( ADBC
) − )
Problem 1.3
Velocity analysis

From Equation 1.38,

v = v o
+ Ω × r rel + v rel. (1)

From the given inғormation we have

v o= −10 Iˆ + 30 Jˆ − 50 Kˆ (2
)
r rel= r − r o = ( 150 Iˆ − 2 Jˆ + 300 Kˆ ) 2 Jˆ + 1 ) − 4 Jˆ + 2 (3
( 3 +
0 − 0 0 0 = 0 0 )
0 − 0 0
Iˆ Jˆ Kˆ
K 1 K
Ω× r rel = 06 −0 4 10 = 320 − 2 Jˆ − 300 Kˆ (4
Iˆ ˆ ˆ )
−150 −400 200 7 5
0 0




2