Page 1 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
,Page 2 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
SOLUTIONS MANUAL
to accompany
ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
,Page 3 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
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
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
, Page 4 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
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
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
,Page 2 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
SOLUTIONS MANUAL
to accompany
ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
,Page 3 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
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
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
, Page 4 of 343 orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
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
orbital mechanics for engineering
orbital
students
mechanics
4th edition
for engineering
by howard
orbital
students
curtis
mechanics
solution
4th edition
formanual.pdf
engineering
by howardstudents
curtis solution
4th edition
manual.pdf
by howard curtis solution manual.pdf
