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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 = Ax iˆ + Ay ˆj + Azkˆ ⋅ Ax iˆ + Ay ˆj + Azkˆ ˆ ⋅ A ˆi + A ˆj + A kˆ = Ax iˆ ⋅ Axiˆ + A yˆj + Azkˆ + A yˆj ⋅ Axiˆ + Ayˆj + Azkˆ + Azk x y z

(

)

(

)

( ) ( )

(

( )

( )

)

( )

= A x2 ( ˆi ⋅ ˆi ) + A xA y ˆi ⋅ ˆj + A xA z ( ˆi ⋅ kˆ )  + A yA x ˆj ⋅ iˆ + A y 2 ˆj ⋅ jˆ + A yA z jˆ ⋅ kˆ      2   ˆ ˆ ˆ ˆ ˆ ˆ + Az Ax ( k ⋅ i ) + Az Ay k ⋅ j + Az ( k ⋅ k )   2 = A x (1 ) + A xA y (0 ) + A xA z (0 ) + A yAx ( 0) + Ay 2 (1) + Ay Az ( 0)  +  Az Ax (0 ) + Az Ay ( 0) + Az 2 ( 1)       = Ax 2 + A y 2 + Az 2 But, according to the Pythagorean Theorem, A x2 + A y2 + A z2 = A2 , where A = A , the magnitude of the vector A. Thus A ⋅ A = A2 . (b)

iˆ

ˆj

A ⋅ (B × C ) = A ⋅ B x

kˆ

By Bz

Cx C y Cz

(

) (

= Axiˆ + Ayˆj + Azkˆ ⋅  ˆi By Cz − Bz C y − ˆj (Bx Cz − Bz Cx ) + kˆ Bx Cy − By Cx    = A x B yC z − BzC y − A y ( BxCz − BzCx ) + Az BxC y − B yC x

(

)

)

( )

(

)

or

A ⋅ ( B × C) = A xB yC z + A yB zC x + A zB xC y − A xB zC y − A yB xC z − A zB yC x

(1)

Note that ( A × B) ⋅ C = C ⋅ (A × B) , and according to (1)

C ⋅ ( A × B) = Cx Ay Bz + Cy Az Bx + Cz Ax By − Cx Az By − Cy Ax Bz − C z Ay Bx

(2)

The right hand sides of (1) and (2) are identical. Hence A ⋅ ( B × C) = ( A × B) ⋅ C . (c)

(

iˆ

)

ˆj

kˆ

ˆi

ˆ ×B B A × ( B × C ) = A xˆi + A yˆj + A z k Ax x y Bz = C x C y C z By Cz − Bz Cy

ˆj

kˆ

Ay

Az

Bz Cx − BxC y

BxC y − By C x

=  A y B xC y − B yC x − A z (B zC x − B xC z ) ˆi +  A z B yC z − B zC y − Ax BxC y − By C x  ˆj     + A x (B zC x − B xC z ) − A y By Cz − BzC y  kˆ   = Ay Bx Cy + Az Bx Cz − Ay By Cx − Az BzC x iˆ + AxBy Cx + Az By Cz − Ax BxC y − Az BzC y ˆj + A B C + A B C − A B C − A B C kˆ

(

)

(

(

) ) ( z)

(

(

x z x

( (

y z y

)

x x z

(

y y

)

)

(

)

)

=  Bx Ay Cy + Az Cz − Cx Ay By + Az Bz  ˆi + By ( Ax Cx + Az Cz ) − Cy ( Ax Bx + Az Bz )  ˆj    +  Bz AxCx + Ay C y − Cz Ax Bx + Ay By  kˆ  

)

(

)

Add and subtract the underlined terms to get

1

Solutions Manual

Orbital Mechanics for Engineering Students

(

)

(

(

)

(

)

(

)

(

)

)

A × ( B × C ) =  Bx Ay Cy + Az Cz + Ax Cx − Cx Ay By + Az Bz + Ax Bx  iˆ    + By Ax Cx + Az Cz + Ay Cy − Cy Ax Bx + Az Bz + Ay By  ˆj   + Bz Ax Cx + Ay Cy + Az Cz − Cz Ax Bx + Ay By + Az Bz  kˆ   ˆ ˆ ˆ ˆ ˆ ˆ B B B A C A C A C C C C − + + = x i + y j + zk x x + y y + z z xi yj z k Ax Bx + Ay By + Az Bz or

(

Chapter 1

)(

) (

)(

)

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 = v o + Ω × rrel + v rel .

(1)

From the given information we have

ˆ vo = −10 Iˆ + 30 ˆJ − 50 K

(2)

(

) (

)

ˆ − 300 Iˆ + 200 ˆJ + 1 00K ˆ = − 150ˆI − 400ˆJ + 200K ˆ rrel = r − ro = 150 Iˆ − 200 Jˆ + 300 K

Iˆ

ˆJ

(3)

ˆ K

ˆ Ω × rrel = 0 .6 −0 .4 1 .0 = 320Iˆ − 270 0ˆJ − 300K − 150 − 400 200

2

(4)

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 1

v rel = −20ˆi + 25ˆj + 70kˆ

(

)

ˆ = −20 0 .57735 ˆI + 0.57735 ˆJ + 0.57735 K + 25 − 0 .74296Iˆ + 0.66475Jˆ + 0. 078206Kˆ

(

)

(

+ 70 −0 .33864 Iˆ − 0 .47410 Jˆ + 0.81274Kˆ

)

so that

ˆ ( m s) v rel = −53.826Iˆ − 28.115 Jˆ + 47.300K

(5)

Substituting (2), (3), (4) and (5) into (1) yields

(

) (

) (

)

ˆ + 320Iˆ − 270 ˆJ − 300K ˆ + −53. 826ˆI − 28. 115ˆJ + 47. 300K ˆ v = −10Iˆ + 30 Jˆ − 50K

ˆ v = 256 .17 Iˆ − 268 .12 Jˆ − 302 .7 K = 478 .68 0 .53516 ˆI − 0.56011Jˆ − 0.63236K (m s )

(

)

Acceleration analysis From Equation 1.42,  ×r a =a +Ω O

rel

+ Ω × ( Ω × rrel ) + 2Ω × v rel + arel

(6)

Using the given data together with (4) and (5) we obtain

ˆ ao = 25 ˆI + 40 ˆJ − 15 K Iˆ  Ω × rrel = −0.4

(7)

ˆJ

ˆ K

0.3 −1.0 = −340Iˆ + 230Jˆ + 205Kˆ − 150 − 400 200 Iˆ

Ω × ( Ω × rrel ) = 0. 6

ˆJ

(8)

ˆ K

ˆ 1. 0 = 390IˆI + 500ˆJ − 34K 320 − 270 − 300

2Ω × v rel = 2

(9)

− 0. 4

Iˆ

ˆJ

ˆ K

0 .6

−0 .4

1 .0

(

)

ˆ = 2 9.151ˆI − 82.206 ˆJ − 38.399 K

(10)

− 53. 826 − 28. 115 47. 300

arel = 7.5iˆ − 8.5ˆj + 6.0kˆ

(

ˆ = 7.5 0.57735 ˆI + 0.57735Jˆ + 0. 57735K

(

) )

− 8. 5 − 0. 74296Iˆ + 0. 66475Jˆ + 0. 078206Kˆ ˆ + 6. 0 −0. 33864ˆI − 0. 47410ˆJ + 0. 81274K

(

)

ˆ a rel = 8.6134Iˆ − 4. 1649Jˆ + 8.5418K

(11)

Substituting (7), (8), (9), (10) and (11) into (6) yields

(

) (

) (

)

ˆ + −340ˆI + 230ˆJ + 205K ˆ + 390ˆI + 500ˆJ − 34K ˆ a = 25ˆI + 40ˆJ − 15K

(

) (

ˆ ˆ  + 8.6134 ˆI − 4.1649 ˆJ + 8.5418 K +  2 9.151Iˆ − 82 .206 Jˆ − 38.399K 

3

)

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 1

ˆ a = 102Iˆ + 601.42Jˆ + 87. 743K

)(

(

ˆ m s2 = 616.29 0.16551Iˆ + 0.97588Jˆ + 0.14327K

)

Problem 1.4 From Example 2.8, we have   × F + 2 ω × (ω × F ) +ω × ω × F +ω × (ω × F ) F=ω Substituting the given values for the quantities on the right hand side,

 × F = 0 × 10 ˆi = 0 ω

2ω × ( ω × F) = 2 (−2 kˆ ) × ( 3 kˆ ) × (10 iˆ)  = 2 ( −2 kˆ ) × 30 ˆj = 120 iˆ

( ) ω × ( ω × F) = (3 kˆ ) × ( −2kˆ) × (10iˆ) = (3kˆ ) × ( −20 ˆj) = 60iˆ

{

}

ˆ ) × ( 3k ˆ ) × 30ˆj  = ( 3kˆ ) × ( − 90ˆi ) = −270 ˆj ω × ω × (ω × F)  = (3 kˆ ) × ( 3kˆ ) × ( 3kˆ ) × (10ˆi ) = ( 3k  

(

( )

)

F = 0 + 120iˆ + 60iˆ − 270ˆj = 120iˆ − 270ˆj N s3 . Thus, 

Problem 1.5 iˆ = sin θIˆ + cos θ Jˆ

ˆj = − cos θIˆ + sin θ Jˆ

ˆ kˆ = K

(1)

Velocity analysis The absolute velocity of the airplane is

v = vIˆ

(2)

The absolute velocity of the origin of the moving frame is

vo = 0

(3)

The position of the airplane relative to the moving frame is

rrel =

h ˆ h sinθ ˆ I + hˆJ i= sinθIˆ + cosθ Jˆ = h cosθ cosθ cosθ

(

)

(4)

The angular velocity of the moving frame is

ˆ Ω = −θK

(5)

The velocity of the airplane relative to the moving frame is, making use of (1)

(

)

v rel = vrel iˆ = vrel sin θIˆ + cos θ Jˆ = vrel sin θ Iˆ + v rel cos θ ˆJ

(6)

According to Equation 1.38, v = v o + Ω × rrel + v rel . Substituting (2), (3), (4), (5) and (6) yields

ˆ ) ×  h sin θ Iˆ + hJˆ + v sinθ Iˆ + v cos θ ˆJ vˆI = 0 + ( −θK  rel rel  cosθ

(

or

4

)

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 1

  sin θ  ˆ  vIˆ = 0 + ( hθ ) ˆI −  hθ  J + vrell sinθIˆ + vrel cosθ Jˆ  cos θ   

(

)

Collecting terms,

 sin θ  ˆ vIˆ = hθ + v rel sin θ Iˆ +  v rel cos θ − hθ J  cos θ 

(

)

Equate the Iˆ and ˆJ components on each side to obtain

hθ + v rel sinθ = v sinθ − hθ + v rel cosθ = 0 cosθ Solving these two equations for θ and vrel yields

v 2 θ = cos θ h

(7)

v rel = v sin θ

(8)

Acceleration analysis The absolute acceleration of the airplane, the absolute acceleration of the origin of the moving frame, and the angular acceleration of the moving frame are, respectively,

a=0

 = −θK ˆ Ω

ao = 0

(9)

The acceleration of the airplane relative to the moving frame is, making use of (1),

(

)

a rel = arelˆi = arel sin θIˆ + cos θ ˆJ = arel sin θ ˆI + a rel cosθ Jˆ

(10)

Substituting (7) into (5), the angular velocity of the moving frame becomes

v 2 Ω = −θKˆ = − cos θ Kˆ h

(11)

Substituting (8) into (6) yields

(

)

2 v rel = vrel iˆ = v sin θ sin θ Iˆ + cosθ Jˆ = v sin θ Iˆ + v sin θ cos θ ˆJ

(12)

From (4) and (9) we find

 sin θ ˆ  sin θ ˆ J I + hJˆ = hθIˆ − hθ Ω × rrel = ( −θKˆ ) ×  h  cosθ  cosθ

(13)

Using (5) and (7) we get

Ω × rrel = hθˆI − hθ

v sin θ ˆ  sin θ ˆ  v 2 J = h  cos 2 θ  ˆI − h  cos 2 θ  J v cos θIˆ − v sin θ cos θ ˆJ h  cos θ = h  cos θ

From (11) and (14) we have

5

(14)

Solutions Manual

Orbital Mechanics for Engineering Students

Ω × ( Ω × rrel ) =

Iˆ

ˆJ

0

0

Chapter 1

ˆ K v v2 v2 sinθ cos3 θ Iˆ − cos 4 θ Jˆ − cos 2 θ = − h h h

2 v cos θ − v sinθ cosθ

(15)

0

From (11) and (12),

2Ω × v rel = 2

ˆI

ˆJ

0

0

ˆ K −

v v2 v2 2 sin2 θ cos2 θJˆ cos θ = 2 sin θ cos3 θIˆ − 2 h h h

v sin2 θ v sin θ cos θ

(16)

0

 × r + Ω × ( Ω × r ) + 2 Ω × v + a . Substituting (9), (10), According to Equation 1.42, a = a o + Ω rel rel rel rel (13), (15) and (16) yields

sin θ  0 = 0 +  hθIˆ − hθ  cosθ

2 2  v   v sinθ cos3 θIˆ − Jˆ +  − cos4 θ Jˆ   h  h

  v2 v2 sin2 θ cos2 θ Jˆ + arel sinθ Iˆ + arel cosθ Jˆ +  2 sinθ cos3 θ Iˆ − 2  h h

(

)

Collecting terms 2 2   v v 0 =  hθ − sinθ cos 3θ + 2 sinθ cos 3θ + a rel sinθ  ˆI   h h 2 2   v sin θ v 2 +  −h − θ cos4 θ − 2 sin 2 θ cos θ + arel cosθ  ˆJ   h h cosθ

or 2     sin θ v2 v 0 =  hθ + sin θ cos 3θ + arel sinθ  Iˆ +  − hθ cos2 θ (1 + sin2 θ ) + a rel cosθ  Jˆ −     cosθ h h

Equate the Iˆ and J√ components on each side to obtain

hθ + arel sinθ = −

v2 sinθ cos3 θ h

v2 sinθ − hθ + a rel cos θ = cos 2 θ (1 + sin 2 θ ) h cosθ Solving these two equations for θ«« and arel yields 2

2

v 3 θ = −2 2 cos θ sin θ h

arel =

v 3 cos θ h

Problem 1.6 From Equation 2.58b with z = 0 we have   y 2 a = −2Ω y sin φ ˆi + Ω 2 RE sin φ cos φ ˆj −  + Ω 2 RE cos 2 φ kˆ   RE where

6

(1)

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 1

R E = 6378 × 103 m

φ = 30° 3

1000 × 10 = 27. 78 m s 3600 2π = 7 .2921 × 10 −5 rad s Ω= 23 .934 × 3600 y =

Substituting these numbers into (1), we find

a = −0 .0020256ˆi + 0.014686ˆj − 0. 025557kˆ ( m s ) From F = ma , with m = 1000 kg , we obtain the net force on the car,

ˆ ( N) F=-2.0256ˆi + 14. 686ˆj − 25. 557k Flateral = F x = −2.0256 N = − 0.4554 lb , that is

Flateral = 0. 4554 lb to the west The normal force N of the road on the car is given by N = Fz + mg , so that

N = −25.557 + 1000 × 9.81 = 9784 N

Problem 1.7 From Equation 1.61b, with z = 0 , a = Ω R E sin l cos l ˆj − Ω R E 2

From

2

∑ Fy = ma y

cos 2

z

l kˆ

we get

T sinθ = mΩ 2 RE sinφ cosφ 2

From

T=

mΩ RE sin φ cosφ sin θ

∑Fz

= ma z we obtain

L = 30 m g 2

2

T cosθ − mg = −mΩ RE cos φ

y North

2

mΩ RE sinφ cosφ cosθ − mg = −m Ω 2R E cos2 φ sinθ 2

tanθ =

Ω RE sinφ cosφ g − Ω 2RE cos 2 φ

Since d = L tan θ , we deduce 2

d=L

Ω R E sin φ cos φ g − Ω 2RE cos 2 φ

Setting

7

d

Solutions Manual

Orbital Mechanics for Engineering Students

L = 30 m 6 R E = 6378 × 1000 = 6 .378 × 10 m φ = 29° 2

yields

g = 9.81 m s 2π = 7.2921 × 10− 5 rad s Ω= 23.9344 × 3600 d = 44.1 mm (to the south )

8

Chapter 1

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 2

Problem 2.1 ˆ r = 3t4 ˆI + 2t3 ˆJ + 9t2 K

(3t 4 Iˆ + 2t 3 ˆJ + 9t2 Kˆ )⋅ ( 3t4 Iˆ + 2t3 ˆJ + 9t 2 Kˆ ) =

r = r =

9t 8 + 4t 6 + 81t 4

36t 7 + 12t 5 + 162t 3 d r = dt 9t 8 + 4t 6 + 81t 4

At t = 2 sec ,

r« =

4608 + 384 + 1296 2304 + 256 + 1296

= 101.3 m s

ˆ r =12t 3ˆI + 6t 2 ˆJ +18t K r =

(12t 3Iˆ + 6 t2 Jˆ + 18 tKˆ ) ⋅( 12t 3Iˆ + 6t2 Jˆ + 18tKˆ ) =

144 t6 + 36t4 + 324t2

At t = 2 sec ,

r« = 9216 + 576 + 1296 = 105.3 m s

Problem 2.2 uˆ r ⋅ uˆ r = 1 ⇒

ˆ ˆ u uˆ d (uˆ r ⋅ uˆ r ) = 0 ⇒ ddtur ⋅ uˆr + uˆr ⋅ ddt r = 0 ⇒ uˆ r ⋅ ddtr = 0 dt

Or,

dr dr duˆ r d  r  r dt − r dt r r − rr = =  = 2 2 dt dt  r  r r ˆ du r rr − r r r ⋅ r rr = − uˆ r ⋅ r = ⋅ 2 2 2 dt r r r r ˆr⋅ But according to Equation 2.25, r ⋅ r = rr . Hence u

ˆr du dt

=0

Problem 2.3 Both particles rotate with a constant angular velocity around the center of mass c.m., which lies midway along the line joining the two masses. Let uˆ be the unit vector drawn from one of the masses to c.m., which is the origin of an inertial frame. The only force on m is that of mutual gravitational attraction, Fg = G

m2 d2

c.m. uˆ

m

The absolute acceleration of m is normal to its circular path around c.m.,

a = ω2

m

d

d uˆ 2

m2 d ˆ = mω 2 u ˆ , or From Newton’s second law, F g = ma , so that G u 2 2 d

9

Fg

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 2

2 Gm

ω=

d3

Problem 2.4 The center of mass of the three equal masses lies at the centroid of the equilateral triangle, whose altitude h is given by h = d o sin 60° . The distance r of each mass from the center of mass is, therefore

r =

3 m y

do Fg 1

2 2 h = d o sin 60 ° 3 3

o

3

c.m.

Relative to an inertial frame with the center of mass as its origin, the acceleration of each particle is

m 1

2 a = ω o2 r = ω o2 do sin 60 ° 3

x

2 h 3

30° 30°

Fg 1

2

m 2

do

and this acceleration is directed toward the center of mass. The net force on each particle is the vector sum of the gravitational force of attraction of its two neighbors. This net force is directed towards the center of mass, so that its magnitude, focusing on particle 1 in the figure, is

Fnet = F g

1− 2

cos 30° + F g

1− 3

cos 30° = G

m ⋅m Gm2 G ° + ° = cos 30 cos 30 2 cos 30° do 2 do 2 do 2

m⋅m

Setting Fnet = ma , we get 2

2

Gm do

2

2 cos 30 ° = m ω o2 d o sin60 ° 3 2

ωo =

3Gm cos 30 ° 3Gm = do 3 sin 60° d o3

ωo =

3 Gm do 3

Problem 2.5 (a) (b)

398, 600 µ = = 7. 697 km s 6378 + 350 r 3 2π 32 2π (6378 +350 ) 2 = 5492 sec = 91 min 32 s T= r = 398 ,600 µ v=

Problem 2.6 The mass of the moon is 7. 348 × 1022 kg . Therefore, for a satellite orbiting the moon,  km 3 km 3  22 kg 7 . 348 10 4903 µ = Gmmoon =  6.67259 × 10 −20 × =  kg − s 2  s2 

(

)

The radius of the moon as 1738 km. Hence

4903 µ = = 1.642 km s 1738 + 80 r 3 2π 3 2π (1738 + 80) 2 = 6956 sec = 1 hr 56 min T= r2 = 4903 µ v=

10

Solutions Manual

Orbital Mechanics for Engineering Students

Chapter 2

Problem 2.7 The time between successive crossings of the equator equals the period of the orbit. That is

d 2π (R Earth + z )3/2 = Ω Earth R Earth µ where d = 3000 km is the distance between ground tracks, z is the altitude of the orbit, − R Earth = 6378 km and Ω Earth = 2π ( 23.934 ⋅ 3600) =7 .2921 ×10 5 rad s . Thus

3000

(7.2921 × 10 )(6378) −5

=

2π ( 6378 + z)3 /2 398 600

so that

z = 1440.7 km

Problem 2.8 From Example 2.3 we know that vGEO = 3.0747 km s . From Equation 2.82 we know that

vesc = 2 vcircular . Hence ∆v = ( 2 − 1)v GEO = 0.41421 ⋅ 3.0747 = 1. 2736 km s .

Problem 2.9 µsun = 1 .3271 × 1011 km 3 s 2 6 rearth = 149. 6 × 10 km

v earth =

µ sun 1.3271 × 1011 = = 29 .784 km s rearth 149 .6 × 10 6

v esc = 2 ⋅ 29 .784 = 42 .121 km s v relative = 42. 121 − 29. 784 = 12. 337 km s

Probl...