Orbital Mechanics for Engineering Students 4th Edition 2026/2027 | Solution Manual, Worked Problems & Answers | Engineering Exam Prep

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l mechanics
orbital mechanics
orbital
for mechanics
engineering
for engineering
for
students
engineering
students
4th edition
students
4th edition
by howard
4th edition
by howard
curtis
by solution
howard
curtis solution
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
curtis solution
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*

,l mechanics
orbital mechanics
orbital
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for engineering
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engineering
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4th by
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SOLUTIONS MANUAL

to accompany


ORBITAL MECHANICS FOR ENGINEERING STUDENTS




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




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orbital mechanics
orbital
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mechanics
for engineering
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curtis solution
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,l mechanics
orbital mechanics
orbital
for engineering
mechanics
for engineering
for
students
engineering
students
4th edition
students
4th by
edition
howard
4th by
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howard
curtis
bysolution
howard
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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


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orbital mechanics
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for engineering
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,l mechanics
orbital mechanics
orbital
for engineering
mechanics
for engineering
for
students
engineering
students
4th edition
students
4th by
edition
howard
4th by
edition
howard
curtis
bysolution
howard
curtis solution
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
curtis solution
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
manual.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*



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




2


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orbital mechanics
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