SOLUTIONS MANUAL - Fourth Edition Howard D. Curtis NOTE: For Compl...

SOLUTIONS MANUAL

for

ORBITAL MECHANICS FOR ENGINEERING STUDENTS

Fourth Edition Howard D. Curtis NOTE: For Complete File, Download link at the end of this File 1 / 4

Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1

Howard D. Curtis 1–1

Problem 1.1 Given the three vectors

A=A x ˆ i+A y ˆ j+A z ˆ k,

B=B x ˆ i+B y ˆ j+B z ˆ k and

C=C x ˆ i+C y ˆ j+C z ˆ k, show analytically that

(a) A⋅A=A 2

(b)

A⋅B×C

( )=A×B( )⋅C

(c)

A×B×C

( )=B A⋅C( )−C A⋅B( )

Solution

A⋅A=A

x ˆ i+A y ˆ j+A z ˆ k

( ) ⋅A

x ˆ i+A y ˆ j+A z ˆ k

• )

=A x ˆ i⋅A x ˆ i+A y ˆ j+A z ˆ k

( ) +A

y ˆ j⋅A x ˆ i+A y ˆ j+A z ˆ k

( ) +A

z ˆ k⋅A x ˆ i+A y ˆ j+A z ˆ k

• )

=A x 2 ˆ i⋅ ˆ i( )+A x A y ˆ i⋅ ˆ j( )+A x A z ˆ i⋅ ˆ k( )⎡ ⎣ ⎤ ⎦ +A y A x ˆ j⋅ ˆ i( )+A y 2 ˆ j⋅ ˆ j( )+A y A z ˆ j⋅ ˆ k( ) ⎡ ⎣ ⎤ ⎦ +A z A x ˆ k⋅ ˆ i( )+A z A y ˆ k⋅ ˆ j( )+A z 2 ˆ k⋅ ˆ k( )⎡ ⎣ ⎤ ⎦

=A x 2

1( )+A

x A y

0( )+A

x A z

0( )⎡

⎣ ⎤ ⎦ +A y A x

0( )+A

y 2

1( )+A

y A z

0( )⎡

⎣ ⎤ ⎦ +A z A x

0( )+A

z A y

0( )+A

z 2

1( )⎡

⎣ ⎤ ⎦

=A x 2 +A y 2 +A z 2

But, according to the Pythagorean Theorem,

A x 2+A y 2+A z 2=A 2 , where

A=A, the magnitude of the vector

• Thus

A⋅A=A

2 .

(b)

A⋅B×C

( )=A⋅

ˆ i ˆ j ˆ k B x B y B z C x C y C z

=A x ˆ i+A y ˆ j+A z ˆ k( ) ⋅ ˆ iB y C z −B z C y( ) − ˆ jB x C z −B z C x

A( ) +

ˆ kB x C y −B y C x( ) ⎡ ⎣ ⎤ ⎦

=A x B y C z −B z C y( ) −A y B x C z −B z C x( ) +A z B x C y −B y C x( )

or

A⋅B×C

• )=A

x B y C z +A y B z C x +A z B x C y −A x B z C y −A y B x C z −A z B y C x (1)

Note that

A×B

( )⋅C=C⋅A×B ( )

, and according to (1)

C⋅A×B

• )=C

x A y B z +C y A z B x +C z A x B y −C x A z B y −C y A x B z −C z A y B x (2)

The right hand sides of (1) and (2) are identical. Hence

A⋅B×C

( )=A×B( )⋅C

.

(c) 2 / 4

Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1

Howard D. Curtis 1–2

A×B×C

• )=A

x ˆ i+A y ˆ j+A z ˆ k

( ) ×

ˆ i ˆ j ˆ k B x B y B z C x C y C z= ˆ i ˆ j ˆ k A x A y A z B y C z −B z C y B z C x −B x C y B x C y −B y C x

=A y B x C y −B y C x( ) −A z B z C x −B x C z( )⎡ ⎣ ⎤ ⎦ ˆ i+A z B y C z −B z C y( ) −A x B x C y −B y C x( ) ⎡ ⎣ ⎤ ⎦ ˆ j +A x B z C x −B x C z( ) −A y B y C z −B z C y( ) ⎡ ⎣ ⎤ ⎦ ˆ k

=A y B x C y +A z B x C z −A y B y C x −A z B z C x( ) ˆ i+A x B y C x +A z B y C z −A x B x C y −A z B z C y( ) ˆ j +A x B z C x +A y B z C y −A x B x C z −A y B y C z( ) ˆ k

=B x A y C y +A z C z( ) −C x A y B y +A z B z( ) ⎡ ⎣ ⎤ ⎦ ˆ i+B y A x C x +A z C z( ) −C y A x B x +A z B z( )⎡ ⎣ ⎤ ⎦ ˆ j +B z A x C x +A y C y( ) −C z A x B x +A y B y( ) ⎡ ⎣ ⎤ ⎦ ˆ k

Add and subtract the underlined terms to get

A×B×C

• )=B

x A y C y +A z C z +A x C x

• )

−C x A y B y +A z B z +A x B x( ) ⎡ ⎣ ⎤ ⎦ ˆ i +B y A x C x +A z C z +A y C y

• )

−C y A x B x +A z B z +A y B y( ) ⎡ ⎣ ⎤ ⎦ ˆ j +B z A x C x +A y C y +A z C z

• )

−C z A x B x +A y B y +A z B z( ) ⎡ ⎣ ⎤ ⎦ ˆ k

=B x ˆ i+B y ˆ j+B z ˆ k

• ) A

x C x +A y C y +A z C z( ) −C x ˆ i+C y ˆ j+C z ˆ k

• ) A

x B x +A y B y +A z B z( )

=B x ˆ i+B y ˆ j+B z ˆ k

( ) A⋅C( )−C

x ˆ i+C y ˆ j+C z ˆ k

( ) A⋅B( )

Or,

A×B×C

( )=B A⋅C( )−C A⋅B( )

• / 4

Solutions Manual Orbital Mechanics for Engineering Students Fourth Edition Chapter 1

Howard D. Curtis 1–3

Problem 1.2 Use just the vector identities in Problem 1.1 to show that

A×B

( )⋅C×D( )=A⋅C( )B⋅D( )−A⋅D( )B⋅C( )

Solution

From Problem 1.1(b)

A×B

( )⋅C×D( ) = A×B( )×C[ ] ⋅D

(1)

But

A×B

( )×C[ ] ⋅D=− C×A×B ( )[ ] ⋅D

Using Problem 1.1(c) 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( )

• / 4