Orbital Mechanics For Engineering Students - Solutions by Howard D. Curtis (z-lib.org)

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 + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ ( ) ) ( ) ( = Ax iˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ + Ay ˆj ⋅ Ax iˆ + Ay ˆj + Az kˆ + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ ) =  Ax 2 ( iˆ ⋅ iˆ ) + Ax Ay iˆ ⋅ ˆj + Ax Az ( iˆ ⋅ kˆ )  +  Ay Ax ˆj ⋅ iˆ + Ay 2 ˆj ⋅ ˆj + Ay Az ˆj ⋅ kˆ      +  Az Ax ( kˆ ⋅ iˆ ) + Az Ay kˆ ⋅ ˆj + Az 2 ( kˆ ⋅ kˆ )    2 =  Ax (1) + Ax Ay ( 0 ) + Ax Az ( 0 ) +  Ay Ax ( 0 ) + Ay 2 (1) + Ay Az ( 0 ) +  Az Ax ( 0 ) + Az Ay ( 0 ) + Az 2 (1)       ( ) ( ) ( ) ( ) ( ) = Ax 2 + Ay 2 + Az 2 But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of the vector A . Thus A ⋅ A = A2 . (b) iˆ A ⋅ ( B × C ) = A ⋅ Bx ˆj By kˆ Bz Cx Cy Cz ( ) ( = Ax iˆ + Ay ˆj + Az kˆ ⋅  iˆ By Cz − Bz C y − ˆj ( Bx Cz − Bz Cx ) + kˆ Bx C y − By Cx    = Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx ( ) ) ( ) ( ) or A ⋅ ( B × C) = AxBy C z + Ay Bz C x + Az BxC y − AxBz C y − Ay BxC z − Az By C x (1) Note that ( A × B) ⋅ C = C ⋅ ( A × B) , and according to (1) C ⋅ ( A × B) = C x Ay Bz + C y Az Bx + C z AxBy − C x Az By − C y AxBz − C z Ay Bx (2) The right hand sides of (1) and (2) are identical. Hence A ⋅ ( B × C) = ( A × B) ⋅ C . (c) iˆ A × ( B × C ) = Ax iˆ + Ay ˆj + Az kˆ × Bx ˆj By kˆ Bz = Cx Cy Cz ( ) iˆ Ax ˆj Ay kˆ Az By Cz − Bz C y Bz Cx − Bx C y Bx C y − By Cx =  Ay Bx C y − By Cx − Az ( Bz Cx − Bx Cz )  î +  Az By Cz − Bz C y − Ax Bx C y − By Cx  ˆj     +  Ax ( Bz Cx − Bx Cz ) − Ay By Cz − Bz C y  k̂   = Ay Bx C y + Az Bx Cz − Ay By Cx − Az Bz Cx iˆ + Ax By Cx + Az By Cz − Ax Bx C y − Az Bz C y ˆj ( ) ( ( ) ( ) ) ( ) ( ) + ( Ax Bz Cx + Ay Bz C y − Ax Bx Cz − Ay By Cz ) k̂ = Bx ( Ay C y + Az Cz ) − Cx ( Ay By + Az Bz )  î + By ( Ax Cx + Az Cz ) − C y ( Ax Bx + Az Bz )  ĵ     + Bz ( Ax Cx + Ay C y ) − Cz ( Ax Bx + Ay By )  k̂   Add and subtract the underlined terms to get 1 Solutions Manual Orbital Mechanics for Engineering Students ( ) ( ) A × ( B × C ) = Bx Ay C y + Az Cz + Ax Cx − Cx Ay By + Az Bz + Ax Bx  î   + By Ax Cx + Az Cz + Ay C y − C y Ax Bx + Az Bz + Ay By  ĵ   + Bz Ax Cx + Ay C y + Az Cz − Cz Ax Bx + Ay By + Az Bz  kˆ   ˆ ˆ ˆ ˆ ˆ = Bx i + By j + Bz k Ax Cx + Ay C y + Az Cz − Cx i + C y j + Cz kˆ Ax Bx + Ay By + Az Bz ( ) ( ) ( ) ( ) ( Chapter 1 )( ) ( )( ) 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 = v o + Ω × rrel + v rel . (1) From the given information we have ˆ v o = −10Iˆ + 30 Jˆ − 50K (2) ( ) ( ) ˆ − 300Iˆ + 200 Jˆ + 100K ˆ = −150Iˆ − 400 Jˆ + 200K ˆ rrel = r − ro = 150Iˆ − 200 Jˆ + 300K Iˆ Jˆ (3) ˆ K ˆ Ω × rrel = 0.6 −0.4 1.0 = 320Iˆ − 270 0 Jˆ − 300K −150 −400 200 2 (4) Solutions Manual Orbital Mechanics for Engineering Students Chapter 1 v rel = −20iˆ + 25 ˆj + 70kˆ ( ˆ = −20 0.57735Iˆ + 0.57735 Jˆ + 0.57735K ) ( ˆ + 25 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K ) ( ˆ + 70 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K ) 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 Jˆ − 300K ˆ + −53.826Iˆ − 28.115 Jˆ + 47.300K ˆ v = −10Iˆ + 30 Jˆ − 50K ) ˆ v = 256.17Iˆ − 268.12 Jˆ − 302.7K = 478.68 0.53516Iˆ − 0.56011Jˆ − 0.63236K ( m s ) ( ) Acceleration analysis From Equation 1.42, a = a O + Ω × rrel + Ω × ( Ω × rrel ) + 2Ω × v rel + a rel (6) Using the given data together with (4) and (5) we obtain ˆ ao = 25Iˆ + 40 Jˆ − 15K Iˆ (7) Jˆ ˆ K ˆ −1.0 = −340Iˆ + 230 Jˆ + 205K Ω × rrel = −0.4 0.3 −150 −400 Iˆ (8) 200 Jˆ ˆ K ˆ Ω × ( Ω × rrel ) = 0.6 −0.4 1.0 = 390IIˆ + 500 Jˆ − 34K 320 −270 −300 (9) ˆ Iˆ Jˆ K ˆ 2Ω × v rel = 2 0.6 −0.4 1.0 = 2 9.151Iˆ − 82.206 Jˆ − 38.399K −53.826 −28.115 47.300 ( ) (10) a rel = 7.5iˆ − 8.5 ˆj + 6.0kˆ ( ˆ = 7.5 0.57735Iˆ + 0.57735 Jˆ + 0.57735K ) ( ˆ − 8.5 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K ) ( + 6.0 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K̂ ) ˆ a rel = 8.6134Iˆ − 4.1649 Jˆ + 8.5418K (11) Substituting (7), (8), (9), (10) and (11) into (6) yields ( ) ( ) ( ˆ + −340Iˆ + 230 Jˆ + 205K ˆ + 390Iˆ + 500 Jˆ − 34K ˆ a = 25Iˆ + 40 Jˆ − 15K ( ) ( ) ˆ ˆ  + 8.6134Iˆ − 4.1649 Jˆ + 8.5418K +  2 9.151Iˆ − 82.206 Jˆ − 38.399K  3 ) Solutions Manual Orbital Mechanics for Engineering Students Chapter 1 ˆ a = 102Iˆ + 601.42 Jˆ + 87.743K )( ( ˆ m s2 = 616.29 0.16551Iˆ + 0.97588 Jˆ + 0.14327K ) Problem 1.4 From Example 2.8, we have F = ω × F + 2ω × ( ω × F ) + ω × ω × F + ω × ( ω × F )  Substituting the given values for the quantities on the right hand side, ω × F = 0 × 10iˆ = 0 2ω × ( ω × F ) = 2 ( −2kˆ ) × ( 3kˆ ) × (10iˆ )  = 2 ( −2kˆ ) × 30 ˆj = 120iˆ ( ) ω × ( ω × F ) = ( 3kˆ ) × ( −2kˆ ) × (10iˆ )  = ( 3kˆ ) × ( −20 ĵj) = 60iˆ ω × ω × ( ω × F )  = ( 3kˆ ) × {( 3kˆ ) × ( 3kˆ ) × (10iˆ )} = ( 3kˆ ) × ( 3kˆ ) × ( 30ˆj) = ( 3kˆ ) × ( −90iˆ ) = −270 ĵ ( ) Thus, F = 0 + 120iˆ + 60iˆ − 270 ˆj = 120iˆ − 270 ˆj N s3 . Problem 1.5 iˆ = sin θ Iˆ + cos θ Jˆ ˆj = − cos θ Iˆ + 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 = sin θ ˆ h ˆ h i= sin θ Iˆ + cos θ Jˆ = h I + hJˆ cos θ cos θ cos θ ( ) (4) The angular velocity of the moving frame is Ω = −θ K̂ (5) The velocity of the airplane relative to the moving frame is, making use of (1) ( ) v rel = vrel iˆ = vrel sin θ Iˆ + cos θ Jˆ = vrel sin θ Iˆ + vrel cos θ Jˆ 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 sin θ Iˆ + v cos θ Jˆ vIˆ = 0 + ( −θ K rel rel  cos θ  ( or 4 ) (6) Solutions Manual Orbital Mechanics for Engineering Students   sin θ  ˆ  vIˆ = 0 + ( hθ ) Iˆ −  hθ J + vrell sin θ Iˆ + vrel cos θ Jˆ  cos θ    ( Chapter 1 ) Collecting terms, sin θ   vIˆ = hθ + vrel sin θ Iˆ +  vrel cos θ − hθ  Ĵ  cos θ  ( ) Equate the Î and Ĵ components on each side to obtain hθ + vrel sin θ = v − hθ sin θ + vrel cos θ = 0 cos θ Solving these two equations for θ and vrel yields v cos2 θ h (7) vrel = v sin θ (8) θ= 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 ˆ Ω = −θ K ao = 0 (9) The acceleration of the airplane relative to the moving frame is, making use of (1), ( ) a rel = arel iˆ = arel sin θ Iˆ + cos θ Jˆ = arel sin θ Iˆ + arel cos θ Jˆ (10) Substituting (7) into (5), the angular velocity of the moving frame becomes ˆ = − v cos2 θ K ˆ Ω = −θ K h (11) Substituting (8) into (6) yields ( ) v rel = vrel iˆ = v sin θ sin θ Iˆ + cos θ Jˆ = v sin 2 θ Iˆ + v sin θ cos θ Jˆ (12) From (4) and (9) we find ˆ ) ×  h sin θ Iˆ + hJˆ  = hθ Iˆ − hθ sin θ Jˆ Ω × rrel = ( −θ K   cos θ cos θ (13) Using (5) and (7) we get sin θ ˆ v  v  sin θ ˆ Ω × rrel = hθ Iˆ − hθ J = h  cos2 θ  Iˆ − h  cos2 θ  J = v cos2 θ Iˆ − v sin θ cos θ Jˆ h  h  cos θ cos θ From (11) and (14) we have 5 (14) Solutions Manual Orbital Mechanics for Engineering Students Iˆ Jˆ 0 0 v cos2 θ − v sin θ cos θ Ω × ( Ω × rrel ) = Chapter 1 ˆ K − v2 v2 v cos2 θ = − sin θ cos3 θ Iˆ − cos4 θ Jˆ h h h (15) 0 From (11) and (12), 2Ω × v rel = 2 Iˆ Jˆ ˆ K 0 0 v sin 2 θ v sin θ cos θ − v v2 v2 cos2 θ = 2 sin θ cos3 θ Iˆ − 2 sin 2 θ cos2 θ Jˆ h h h (16) 0 According to Equation 1.42, a = a o + Ω × rrel + Ω × ( Ω × rrel ) + 2Ω × v rel + a rel . Substituting (9), (10), (13), (15) and (16) yields  sin θ ˆ   v 2 v2  0 = 0 +  hθ Iˆ − hθ J +  − sin θ cos3 θ Iˆ − cos4 θ Jˆ    cos θ   h h  v2  v2 +2 sin θ cos3 θ Iˆ − 2 sin 2 θ cos2 θ Jˆ  + arel sin θ Iˆ + arel cos θ Jˆ  h  h ( ) Collecting terms   v2 v2 0 =  hθ − sin θ cos3 θ + 2 sin θ cos3 θ + arel sin θ  Iˆ   h h   sin θ v 2 v2 +  − hθ − cos4 θ − 2 sin 2 θ cos2 θ + arel cos θ  Jˆ   cos θ h h or     v2 sin θ v 2 0 =  hθ + sin θ cos3 θ + arel sin θ  Iˆ +  − hθ − cos2 θ (1 + sin 2 θ ) + arel cos θ  Jˆ     h h cos θ Equate the Î and J√ components on each side to obtain hθ + arel sin θ = − − hθ v2 sin θ cos3 θ h v2 sin θ cos 2 θ (1 + sin 2 θ ) + arel cos θ = h cos θ Solving these two equations for θ««and arel yields θ = −2 v2 h2 cos3 θ sin θ arel = v2 cos3 θ h Problem 1.6 From Equation 2.58b with z = 0 we have  y2  a = −2Ω y sin φ iˆ + Ω 2 RE sin φ cos φ ˆj −  + Ω 2 RE cos2 φ  k̂  RE  where 6 (1) Solutions Manual Orbital Mechanics for Engineering Students Chapter 1 RE = 6378 × 103 m φ = 30° 1000 × 103 = 27.78 m s 3600 2π Ω= = 7.2921 × 10−5 rad s 23.934 × 3600 y= Substituting these numbers into (1), we find a = −0.0020256iˆ + 0.014686 ˆj − 0.025557kˆ ( m s ) From F = ma , with m = 1000 kg , we obtain the net force on the car, F=-2.0256 iˆ + 14.686 ˆj − 25.557kˆ ( N ) Flateral = Fx = −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 , z a = Ω 2 RE sin l cos lˆj − Ω 2 RE cos2 lkˆ ∑ Fy = ma y From we get T sin θ = mΩ 2 RE sin φ cos φ T= mΩ 2 RE sin φ cos φ sin θ ∑ Fz = ma z From L = 30 m g we obtain T cos θ − mg = − mΩ 2 RE cos2 φ y North 2 mΩ RE sin φ cos φ cos θ − mg = − mΩ 2 RE cos2 φ sin θ tan θ = Ω 2 RE sin φ cos φ g − Ω 2 RE cos2 φ Since d = L tan θ , we deduce d=L Ω 2 RE sin φ cos φ g − Ω 2 RE cos2 φ Setting 7 d Solutions Manual Orbital Mechanics for Engineering Students L = 30 m RE = 6378 × 1000 = 6.378 × 10 6 m φ = 29° yields g = 9.81 m s2 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 = 3t 4 Iˆ + 2t 3 Jˆ + 9t 2K ( 3t 4Iˆ + 2t3 Jˆ + 9t 2Kˆ ) ⋅ ( 3t 4Iˆ + 2t3 Jˆ + 9t 2Kˆ ) = r = r= 9t 8 + 4t 6 + 81t 4 d r 36t7 + 12t 5 + 162t 3 = 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 3Iˆ + 6t 2 Jˆ + 18tK r = (12t3Iˆ + 6t 2 Jˆ + 18tKˆ ) ⋅ (12t3Iˆ + 6t 2 Jˆ + 18tKˆ ) = 144t 6 + 36t 4 + 324t 2 At t = 2 sec , r« = 9216 + 576 + 1296 = 105.3 m s Problem 2.2 uˆ r ⋅ uˆ r = 1 ⇒ duˆ duˆ duˆ d (uˆ ⋅ uˆ ) = 0 ⇒ dtr ⋅ uˆ r + uˆ r ⋅ dtr = 0 ⇒ uˆ r ⋅ dtr = 0 dt r r Or, dr dr −r r duˆ r d  r dt dt = rr − rr =  = 2 dt dt  r  r r2 duˆ r rr − rr r ⋅ r rr uˆ r ⋅ r = ⋅ = − dt r r2 r2 r2 duˆ But according to Equation 2.25, r ⋅ r = rr . Hence uˆ r ⋅ r = 0 dt 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 û 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 uˆ m The absolute acceleration of m is normal to its circular path around c.m., a = ω2 d uˆ 2 From Newton’s second law, Fg = ma , so that G m2 d 2 uˆ = mω 2 9 d uˆ , or 2 m d Fg c.m. Solutions Manual ω = Orbital Mechanics for Engineering Students 2Gm 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= Chapter 2 3 m Fg 2 2 h = d o sin 60° 3 3 1 3 m x 2 h 3 Fg 1 2 2 ω d sin 60° 3 o o o c.m. 30° 30° Relative to an inertial frame with the center of mass as its origin, the acceleration of each particle is a = ω o2 r = y do 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 = Fg 1−2 cos 30° + Fg 1−3 cos 30° = G m ⋅m d o2 cos 30° + G m ⋅m d o2 cos 30° = 2 Gm2 d o2 cos 30° Setting Fnet = ma , we get 2 Gm2 d o2 2 cos 30° = m ω o2 d o sin 60° 3 3Gm cos 30° 3Gm = d o3 sin 60° d o3 3Gm ωo = d o3 ω o2 = Problem 2.5 (a) (b) µ 398 , 600 = = 7.697 km s r 6378 + 350 3 2π 32 2π (6378 + 350) 2 = 5492 sec = 91 min 32 s T= r = µ 398 , 600 v= Problem 2.6 The mass of the moon is 7.348 × 10 22 kg . Therefore, for a satellite orbiting the moon,  km 3  km 3 22 µ = Gm moon =  6.67259 × 10 −20 7 . 348 × 10 kg = 4903  kg − s2  s2  ( ) The radius of the moon as 1738 km. Hence µ 4903 = = 1.642 km s r 1738 + 80 3 2π 32 2π (1738 + 80) 2 = 6956 sec = 1 hr 56 min T= r = µ 4903 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 = ( REarth + z)3/2 Ω Earth REarth µ where d = 3000 km is the distance between ground tracks, z is the altitude of the orbit, REarth = 6378 km and Ω Earth = 2π (23.934 ⋅ 3600) = 7.2921 × 10 −5 rad s . Thus 3000 (7.2921 × 10 )(6378) −5 = 2π 398 600 (6378 + z)3/2 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) vGEO = 0.41421 ⋅ 3.0747 = 1.2736 km s . Problem 2.9 µsun = 1.3271 × 1011 km 3 s2 rearth = 149.6 × 10 6 km vearth = µsun 1.3271 × 1011 = = 29.784 km s rearth 149.6 × 10 6 vesc = 2 ⋅ 29.784 = 42.121 km s vrelative = 42.121 − 29.784 = 12.337 km s Problem 2.10 A πab = T 3 T πab A= = 1.0472 ab 3 Problem 2.11 µ e sin θ h h v⊥ = = 2 r h vr = h 1 µ 1 + e cos θ v = v r2 + v ⊥ 2 µ 2( 2 = e cos θ + sin 2 θ ) + 2e cos θ + 1 h µ 2 v= e + 2e cos θ + 1 h Problem 2.12 For the ellipse, according to Problem 2.11, 11 Solutions Manual vellipse2 = Orbital Mechanics for Engineering Students µ2 (e 2 + 2e cos θ + 1) h2 For the circle, at the point of intersection with the ellipse, vcircle2 = µ µ µ2 = 2 = 2 (1 + e cos θ ) r h 1 h µ 1 + e cos θ Setting vcircle2 = vellipse2 , µ2 h2 2 (1 + e cos θ ) = µ (e 2 + 2e cos θ + 1) 2 h yields e cos θ = −e 2 , or θ = cos −1 ( −e ) . Problem 2.13 From Equation 3.42 tan γ = e sin θ 1 + e cos θ From Problem 2.12 θ = cos −1 ( −e ) . Hence tan γ = [ e sin cos −1 ( −e ) 1 −e [ ] 2 ] But sin cos −1 ( −e ) = 1 − e 2 . Therefore, tan γ = e 1 − e2 1 −e 2 e = 1 − e2 Problem 2.14 (a) e= rapogee − rperigee rapogee + rperigee = 70 000 − 7000 = 0.81818 (ellipse) 70 000 + 7000 = 77 000 = 38 500 km 2 (b) a= rapogee + rperigee 2 (c) T= 2π µ a 3/2 = 2π 398 600 (38 500)3/2 = 75 180 s (d) ε=− 398 600 µ =− = −5.1766 km 2 s2 2a 2 ⋅ 38 500 (e) From Equation 3.62, 38 500(1 − 0.81818 2 ) 1 + 0.81818 cos θ cos θ = 0.88615 ⇒ θ = 27.607° 6378 + 1000 = 12 (20.88 h ) Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students (f) From Equation 3.40 h = µ(1 + e )rperigee = 398 600 ⋅ (1 + 0.81818) ⋅ 7000 = 71 226 km 2 s Then h 71 226 = = 9.6538 km s r 7378 398 600 µ v r = e sin θ = ⋅ 0.81818 ⋅ sin 27.607° = 2.1218 km s h 71 226 v⊥ = (g) h vperigee = rperigee h vapogee = rapogee = 71 226 = 10.175 km s 7000 = 71 226 = 1.0175 km s 70 000 Problem 2.15 rperigee = 6378 + 250 = 6628 km rapogee = 6378 + 300 = 6678 km a= T= rperigee + rapogee 2π µ 2 a 3/2 = = 2π 6628 + 6678 = 6653 km 2 398 600 tperigee to apogee = ⋅ 6653 3/2 = 5400.5 s (90.009 m ) T = 45.005 m 2 Problem 2.16 (a) e= rapogee − rperigee rapogee + rperigee = (6378 + 1600) − (6378 + 600) = 0.066863 (6378 + 1600) + (6378 + 600) (b) h = µ(1 + e )rperigee = 398 600 ⋅ (1 + 0.066863) ⋅ (6378 + 600) = 54 474 km 2 s vperigee = 54 474 h = = 7.8065 km s rperigee 6378 + 600 vapogee = 54 474 h = = 6.8280 km s rapogee 6378 + 1600 (c) 3 3 T= 54 474 2π  h  2π   = = 6435.6 s = 107.26 m 2   2  2 2 µ  1 −e  398 600  1 − 0.066863  Problem 2.17 h = rperigee vperigee = (6378 + 1270) ⋅ 9 = 68 832 km 2 s rperigee = h2 1 1 µ +e 2 13 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students 68 832 2 1 ⇒ e = 0.55416 1 + e 398 600 e sin θ 0.55416 ⋅ sin 100° tan γ = = = 0.60385 ⇒ γ = 31.13° 1 + e cos θ 1 + 0.55416 cos 100° 6378 + 1270 = z + Rearth = z + 6378 = 2 h2 1 µ 1 + e cos θ 68 832 2 1 ⇒ z = 6773.8 km 398 600 1 + 0.55416 cos 100° h = rperigee vperigee = (6378 + 1270) ⋅ 9 = 68 832 km 2 s h2 1 rperigee = 2 µ 1+e 68 832 2 1 ⇒ e = 0.55416 1 +e 398 600 e sin θ 0.55416 ⋅ sin 100° = = 0.60385 ⇒ γ = 31.13° tan γ = 1 + e cos θ 1 + 0.55416 cos 100° 6378 + 1270 = z + Rearth = z + 6378 = 2 h2 1 µ 1 + e cos θ 68 832 2 1 ⇒ z = 6773.8 km 398 600 1 + 0.55416 cos 100° Problem 2.18 v r = v sin γ = 9.2 ⋅ sin 10° = 1.5976 km s v ⊥ = v cos γ = 9.2 ⋅ cos 10° = 9.0602 km s ∴ h = rv ⊥ = (6378 + 640) ⋅ 9.0602 = 63 585 km 2 s r= h2 1 µ 1 + e cos θ 63 5852 1 398 600 1 + e cos θ e cos θ = 0.445 29 6378 + 640 = µ e sin θ h 398 600 e sin θ 1.5976 = 63 585 e sin θ = 0.254 84 sin θ 0.254 84 tan θ = = = 0.572 31 ⇒ θ = 29.783° cos θ 0.445 29 vr = e sin 29.783° = 0.254 84 ⇒ e = 0.51306 3 3 63 585 2π  h  2π   T= 2 = = 16 075 s = 4.4654 h  2  2 2 µ  1 −e  398 600  1 − 0.51306  14 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students Problem 2.19 rperigee + rapogee (6378 + 250) + (6378 + 42 000) = = 27 503 km 2 2 2π 3/2 2π T= a = 27 503 3/2 = 45 392 s = 12.61 h µ 398 600 a= e= rapogee − rperigee (6378 + 42 000) − (6378 + 250) = = 0.75901 rapogee − rperigee (6378 + 42 000) + (6378 + 250) h = µ(1 + e )rperigee = 398 600 ⋅ (1 + 0.75901) ⋅ (6378 + 250) = 68 170 km 2 s h vperigee = rperigee 68 170 = 10.285 km s 6378 + 250 = Problem 2.20 h = rperigee vperigee = (6378 + 640) ⋅ 8 = 56 144 km s rperigee = 7018 = h2 1 µ 1+e 56 144 2 1 ⇒ e = 0.126 82 398 600 1 + e 56 144 2 h2 1 1 = = 9056.6 km µ 1 − e 398 600 1 − 0.126 82 = 9056.6 − 6378 = 2678.6 km rapogee = zapogee a= T= rperigee + rapogee = 2 2π µ (7018) + (9056.6) 2 2π a 3/2 = 398 600 = 8037.3 km 8037.3 3/2 = 7171 s = 1.992 h Problem 2.21 T= 2π µ a 3/2 2π 2 ⋅ 3600 = 398 600 a 3/2 ⇒ a = 8059 km Using the energy equation vperigee2 2 − µ rperigee =− µ 2a 2 398 600 398 600 8 − =− ⇒ rperigee = 7026.2 km 2 rperigee 2 ⋅ 8059 zperigee = 7026.2 − 6378 = 648.25 km 15 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students Problem 2.22 T= 2π a 3/2 µ 2π 90 ⋅ 60 = µ a 3/2 ⇒ a = 6652.6 km rperigee + rapogee = 2 a (6378 + 150) + rapogee = 2 ⋅ 6652.6 ⇒ rapogee = 6777.1 km e= rapogee − rperigee rapogee + rperigee = 6777.1 − 6528 = 0.018 723 6777.1 + 6528 Problem 2.23 (a) vesc = 2 v∞ = 398 600 µ = 2 = 10.926 km s r 6378 + 300 vperigee2 − vesc 2 = 152 − 10.926 2 = 10.277 km s (b) h = rperigee vperigee = 6678 ⋅ 15 = 100 170 km 2 s rperigee = 6678 = r= h2 1 µ 1+e 100 170 2 1 ⇒ e = 2.7696 398 600 1 + e 100 170 2 h2 1 1 = = 48 497 km s µ 1 + e cos θ 398 600 1 + 2.7696 cos 100° (c) 398 600 µ e sin θ = ⋅ 2.7696 ⋅ sin 100° = 10.853 km s h 100 170 h 100 170 v⊥ = = = 2.0655 km s r 48 497 vr = Problem 2.24 (a) From Equation 3.47 ε= v 2 µ 2.23 2 398 600 − = − = 1.4949 km 2 s2 2 2 402 000 r From Equation 3.50 h2 = − 2 1 µ2 ( 1 398 600 ( 1 − e2 ) = − 1 − e2 ) 2 ε 2 1.4949 h2 = (5.3141e 2 − 1) × 1010 From the orbit equation 16 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students h2 = µr (1 + e cos θ ) = 398 600 ⋅ 402 000 ⋅ (1 + e cos 150°) h2 = (16.024 − 13.877 e ) × 1010 Equating the two expressions for h2 , (5.3141e 2 − 1) × 1010 = (16.024 − 13.877 e) × 1010 yields e 2 + 22.6113e − 4.0153 = 0 which has the positive root e = 1.086 (b) Using this value of the eccentricity we find h2 = (16.024 − 13.877 ⋅ 1.086) × 1010 = 9.5334 × 10 9 km 4 s2 so that h2 1 9.5334 × 10 9 1 = = 11 466 km µ 1+e 398 600 1 + 1.086 = 11 466 − 6378 = 5087.6 km rperigee = zperigee (c) vperigee = h rperigee = 9.5334 × 10 9 = 8.5158 km s 11 466 Problem 2.25 From the energy equation v ∞ 2 + vesc 2 = v 2 2µ 2 = (1.1v ∞ ) r µ r = 9.5238 2 v∞ v ∞2 + Substituting Equation 3.105 µ r = 9.5238 µ 2  e −1 h  2 = 9.5238 h2 1 µ e2 − 1 Using Equation 3.40 [ r = 9.5238 rperigee (1 + e ) 1 ] e 2 − 1 = 9.5238 rperigee e −1 Problem 2.26 v∞ = 398 600 2 µ 2 e −1 = 3 − 1 = 10.737 km s h 105 000 17 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students Problem 2.27 (a) r1 = 1 h2 µ 1 + e cos θ1 6378 + 1700 = h2 1 398 600 1 + e cos 130° h2 = (3.2199 − 2.0697 e ) × 10 9 r2 = 1 h2 µ 1 + e cos θ 2 6378 + 500 = h2 1 398 600 1 + e cos 50° h2 = (2.7416 + 1.7622e ) × 10 9 (3.2199 − 2.0697 e ) × 10 9 = (2.7416 + 1.7622e ) × 10 9 3.832e = 0.478 32 e = 0.124 82 (b) h2 = (2.7416 + 1.7622 ⋅ 0.124 82) × 10 9 = 2.9615 × 10 9 km 4 s2 2.9615 × 10 9 1 h2 1 = = 6605.4 km µ 1+e 398 600 1 + 0.124 82 zperigee = 6605.4 − 6378 = 227.35 km rperigee = (c) From Equation 3.63, a= rperigee 1 −e = 6605.4 = 7547.5 km 1 − 0.124 82 Problem 2.28 (a) vr = tan γ = tan 15° = 0.26795 ⇒ v r = 0.26795v ⊥ v⊥ v2 = v r2 + v ⊥ 2 2 7 2 = (0.26795v ⊥ ) + v ⊥ 2 ∴ v ⊥ = 6.7615 km s v r = 1.8117 km s h = rv ⊥ = 9000 ⋅ 6.7615 = 60 853 km 2 s µ e sin θ h 398 600 1.8117 = e sin θ 60 853 e sin θ = 0.27659 vr = 18 Chapter 2 Solutions Manual r= Orbital Mechanics for Engineering Students 1 h2 µ 1 + e cos θ 60 853 2 1 398 600 1 + e cos θ e cos θ = 0.032259 9000 = tan θ = 0.27659 e sin θ = = 8.574 ⇒ θ = 83.348° e cos θ 0.032259 (b) e cos 83.348° = 0.032259 ⇒ e = 0.27847 Problem 2.29 From Equation 3.50, 1 µ2 ( 1 − e2 ) 2 h2 1 398 600 2 ( −20 = − 1 − e 2 ) ⇒ e = 0.30605 2 60 000 2 ε=− 60 000 2 h2 1 1 = = 6915.2 km µ 1 + e 398 600 1 + 0.30605 = 6915.2 − 6378 = 537.21 km rperigee = zperigee 60 000 2 h2 1 1 = = 13 015 km µ 1 − e 398 600 1 − 0.30605 = 13 015 − 6378 = 6636.8 km rapogee = zapogee Problem 2.30 (a) v ⊥ = v cos γ = 8.85 ⋅ cos 6° = 8.8015 km s h = rv ⊥ = (6378 + 550) ⋅ 8.8015 = 60 977 km 2 s r= h2 1 µ 1 + e cos θ 6928 = 60 977 2 1 ⇒ e cos θ = 0.34644 398 600 1 + e cos θ v r = v sin γ = 8.85 ⋅ sin 6° = 0.92508 km s µ v r = e sin θ h 398 600 0.92508 = e sin θ ⇒ e sin θ = 0.14152 60 977 e sin θ = 0.40849 ⇒ θ = 22.22° e cos θ e sin 22.22° = 0.14152 ⇒ e = 0.37423 ∴ tan θ = (b) 3 3 T= 60 977 2π  h  2π   = = 11 243 s = 187.39 m 2   2  µ  1 − e2  398 600  1 − 0.37423 2  19 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students Problem 2.31 v ⊥ = v cos γ = 10 ⋅ cos 30° = 8.6603 km s h = rv ⊥ = (10 000) ⋅ 8.6603 = 86 603 km 2 s r= 10 000 = h2 1 µ 1 + e cos θ 86 603 2 1 ⇒ e cos θ = 0.88159 398 600 1 + e cos θ v r = v sin γ = 10 ⋅ sin 30° = 5 km s µ e sin θ h 398 600 5= e sin θ ⇒ e sin θ = 1.0863 86 603 vr = ∴ tan θ = e sin θ = 1.2323 ⇒ θ = 50.94° e cos θ Problem 2.32 v ⊥ = v cos γ = 10 ⋅ cos 20° = 9.3969 km s h = rv ⊥ = (15 000) ⋅ 9.3969 = 140 950 km 2 s r= 15 000 = h2 1 µ 1 + e cos θ 140 950 2 1 ⇒ e cos θ = 2.323 398 600 1 + e cos θ v r = v sin γ = 10 ⋅ sin 20° = 3.4202 km s µ v r = e sin θ h 398 600 3.4202 = e sin θ ⇒ e sin θ = 1.2095 140 950 ∴ tan θ = e sin θ = 0.52065 ⇒ θ = 27.504° e cos θ Problem 2.33 Using the orbit equation r = h2 1 we find µ 1 + e cos θ h2 = 10 000(1 + e cos 30°) = 10 000 + 8660.3e µ h2 = 30 000(1 + e cos 105°) = 30 000 − 7764.6e µ ∴ 10 000 + 8660.3e = 30 000 − 7764.6e 16 425e = 20 000 ⇒ e = 1.2177 Problem 2.34 v1 = 398 600 µ = = 7.6127 km s r 6378 + 500 20 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students v v2 = v1 + 1 = 11.419 km s 2 2 v∞ µ v 2 µ − = 2 − ∞ r 2 2 2 2 398 600 11.419 v∞ = − = 7.2441 ⇒ v ∞ = 3.8062 km s 2 2 6878 Problem 2.35 398 600 µ = = 7.7143 km s r 6378 + 320 = v1 + 0.5 = 8.2143 km s v1 = vperigee h = rperigee vperigee = 6698 ⋅ 8.2143 = 55 019 rperigee = 6698 = h2 1 µ 1+e 55 019 2 1 ⇒ e = 0.13383 398 600 1 + e 55 019 2 h2 1 1 = = 8767.8 km µ 1 − e 398 600 1 − 0.13383 = 8767.8 − 6378 = 2389.8 km rapogee = zapogee Problem 2.36 µ v= rperigee vperigee = v + αv = (1 + α ) µ rperigee h = rperigee vperigee = rperigee (1 + α ) µ = (1 + α ) µrperigee rperigee 2 rperigee (1 + α ) µrperigee 1 h2 1 = = µ 1+e µ 1+e ∴ 1 + e = (1 + α ) 2 1 + e = 1 + 2α + α 2 ⇒ e = α (α + 2) Problem 2.37 (a) v1 = v⊥2 398 600 µ = = 7.6686 km s r 6378 + 400 = v1 + 0.24 = 7.9086 km s h2 = rv ⊥ 2 = 6778 ⋅ 7.9086 = 53 605 km 2 s r= h2 2 1 µ 1+e 53 6052 1 ⇒ e = 0.063572 398 600 1 + e zperigee = 400 km 6778 = 2 21 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students 6378 + zapogee = 2 Chapter 2 53 6052 h2 2 1 1 = ⇒ zapogee = 1320.3 km 2 µ 1 − e 398 600 1 − 0.063572 (b) 398 600 µ = = 7.6686 km s 6378 + 400 r v⊥2 = h2 = rv ⊥ 2 = 6778 ⋅ 7.6686 = 51 978 km 2 s µ e sin θ h2 2 398 600 0.24 = e sin θ ⇒ e2 sin θ = 0.031296 51 978 2 v r2 = 1 h2 r= 2 µ 1 + e2 cos θ 6778 = 51 978 2 1 ⇒ e2 cos θ = 0 ⇒ θ = 90° (since e cannot be zero if v r ≠ 0) 398 600 1 + e2 cos θ e2 sin 90° = 0.031296 ⇒ e2 = 0.031296 51 978 2 h2 1 1 6378 + zperigee = 2 = ⇒ zperigee = 196.49 km 2 2 µ 1 + e 398 600 1 + 0.031296 51 978 2 h2 1 1 6378 + zapogee = 2 = ⇒ zapogee = 631.3 km 2 2 µ 1 − e 398 600 1 − 0.031296 Problem 2.38 In Figure 3.30 m1 = msun = 1.989 × 10 30 kg m2 = mearth = 5.974 × 10 24 kg r12 = 149.6 × 10 6 km From Equation 3.169 π2 = m2 = 3.0035 × 10 −6 m1 + m2 Substitute π 2 into Equation 3.195, f (ξ ) = 1 − π2 ξ + π2 3 (ξ + π 2 ) + π2 ξ + π2 − 1 3 (ξ + π 2 − 1) − ξ The graph of f (ξ ) is similar to Figure 3.33, with the two crossings on the right much more closely spaced. Zeroing in on the regions where f (ξ ) = 0 , with the aid of a computer, reveals ξ1 = 0.990 026 6 ξ2 = 1.010 034 ξ3 = −1.000 001 Then, x1 = ξ1r12 = 148.108 × 10 6 km x2 = ξ2 r12 = 151.101 × 10 6 km x3 = ξ3 r12 = −149.600 × 10 6 km 22 Solutions Manual Orbital Mechanics for Engineering Students These are the locations of L1 , L2 and L3 relative to the center of mass of the sun-earth system (essentially the center of the sun). 23 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students 24 Chapter 2 Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 Problem 3.1 Graph the function f = x2 − 5x + 4 − 10e sin x f(x) to get an idea where the roots lie. 5 f ′ = 2 x − 5 − 10 cos xe xi+1 = xi − sin x 0 f ( xi ) x -5 f ′( xi ) x 2 − 5xi + 4 − 10e sin xi xi+1 = xi − i 2 xi − 5 − 10 cos xie sin xi 5 6 7 First root: x0 = 4 (Estimate) x1 = 4.7733482 x2 = 4.9509264 x3 = 4.9577657 Second root: x0 = 7 (Estimate) x1 = 6.7673080 x2 = 6.7732223 x3 = 6.7732128 Third root: x0 = 8 (Estimate) x1 = 7.9259101 x2 = 7.9198260 x3 = 7.9197836 x4 = 4.9577768 x5 = 4.9577768 x5 = 6.7732128 x4 = 7.9197836 Problem 3.2 Graph the function f(x) 1 f = tan x − tanh x to get an idea where the roots lie. Clearly, the first root is x = 0 . 0 x 2 -1 4 6 8 f ′ = sec 2 x − sech 2 x f ( xi ) x i +1 = x i − f ′ ( xi ) tan xi − tanh xi x i +1 = x i − sec 2 xi − sech 2 xi Second root: x0 = 4 (Estimate) x1 = 3.9322455 x2 = 3.9266343 x3 = 3.9266023 x4 = 3.9266023 Third root: x0 = 7 (Estimate) x1 = 7.0730641 x2 = 7.0686029 x3 = 7.0685827 x5 = 7.0685827 Problem 3.3 1 1 rapogee + rperigee = (6978 + 6578) = 6778 km 2 2 2π 3/2 2π T= a = 6778 3/2 = 5553.5 s µ 398 600 rapogee − rperigee 6978 − 6578 e= = = 0.029 507 rapogee + rperigee 6978 + 6578 a= ( 8 ) 25 Fourth root: x0 = 10 (Estimate) x1 = 10.247568 x2 = 10.211608 x3 = 10.210178 x4 = 10.210176 x5 = 10.210176 10 Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 Let B denote the point where the satellite flies through 400 km altitude on the way to apogee. rB = 6378 + 400 = E tan B = 2 a (1 − e 2 ) 1 + e cos θ B ( 6778 1 − 0.029 507 2 ) 1 + 0.029 507 cos θ B ⇒ θ B = 91.691° 1 − 0.029 507 θ 1 −e 91.691° tan B = tan ⇒ EB = 1.5708 rad 1+e 2 1 + 0.029 507 2 MB = EB − e sin EB = 1.5708 − 0.029 507 sin 1.5708 = 1.5413 rad M T 1.5413 ⋅ 5553.5 tB = B = = 1362.3 s 2π 2π tB is the time after perigee at which the spacecraft goes above 400 km. Let C denote the point at which the satellite flies downward through 400 km altitude on its way to perigee. The time of flight tBC from B to C is tBC = T − 2tB = 5553.5 − 2 ⋅ 1362.3 = 2828.9 s = 47.148 m Problem 3.4 (a) 1 1 r +r = (10 000 + 7000) = 8500 km 2 apogee perigee 2 2π 3/2 2π T= a = 8500 3/2 = 7799 s µ 398 600 rapogee − rperigee 10 000 − 7000 e= = = 0.176 47 rapogee + rperigee 10 000 + 7000 a= ( ) t1 = 0.5 ⋅ 3600 = 1800 s 2πt1 2π ⋅ 1800 M1 = = = 1.4501 rad T 7799 E1 − e sin E1 = M1 E1 − 0.176 47 sin E1 = 1.4501 E1 = 1.6263 rad (Algorithm 3.1) ∴ tan 1 + 0.176 47 θ1 E 1+e 1.6263 = tan 1 = tan ⇒ θ1 = 103.28° 2 1 −e 2 1 − 0.176 47 2 t2 = 1.5 ⋅ 3600 = 5400 s 2πt2 2π ⋅ 5400 M2 = = = 4.3504 rad T 7799 E2 − e sin E2 = M2 E2 − 0.176 47 sin E2 = 4.3504 E2 = 4.1969 rad (Algorithm 3.1) 1 + 0.176 47 θ2 1+e E 4.1969 tan 2 = tan = ⇒ θ 2 = 231.99° 2 1 −e 2 1 − 0.176 47 2 ∆θ = θ 2 − θ1 = 128.7° tan (b) h = µ rperigee (1 + e ) = 398 600 ⋅ 7000 ⋅ (1 + 0.176 47 ) = 57 294 km 2 s ∆A h = ∆t 2 ∆t = 3600 s 26 Solutions Manual ∴ ∆A = Orbital Mechanics for Engineering Students 1 1 h∆t = 57 294 ⋅ 3600 = 103.13 × 106 km 2 2 2 Problem 3.5 (a) T= 15.743 ⋅ 3600 = 2π µ a 3/2 2π 398 600 a 3/2 ⇒ a = 31890 km 1 r +r 2 perigee apogee 1 31890 = 12756 + rapogee ⇒ rapogee = 51024 km 2 rapogee − rperigee 51024 − 12756 = = 0.6000 e= rapogee + rperigee 51024 + 12756 a= ( ) ( M = 2π ) 10 t = 2π = 3.9911 rad 15.743 T E − e sin E = M E − 0.6 sin E = 3.9911 E = 3.6823 rad (Algorithm 3.1) tan θ E 1+e 1 + 0.6 3.6823 = tan = tan ⇒ θ = 195.78° 2 1 −e 2 1 − 0.6 2 r= 31890(1 − 0.6 2 ) a (1 − e 2 ) = = 48 924 km 1 + e cos θ 1 + 0.6 cos 195.78° (b) v2 µ µ − =− 2 r 2a 398 600 v 2 398 600 − =− ⇒ v = 2.0019 km s 2 48 924 2 ⋅ 31890 (c) h = µa (1 − e 2 ) = 398 600 ⋅ 31 890 ⋅ (1 − 0.6 2 ) = 90 196 km 2 s vr = 398 600 µ e sin θ = ⋅ 0.6 ⋅ sin 195.78° = −0.720 97 km s h 90 196 Problem 3.6 (a) MPB = E − e sin E 2πtPB π π = − e sin T 2 2 π T   tPB = −e 2  2π 1 e tDPB = 2tPB = − T 2 π (b) T −t 2 PB T π e   T 1 tBA = − −e = + T    2 2 2π 4 2π  tBA = 27 Chapter 3 Solutions Manual Orbital Mechanics for Engineering Students tBAD = 2tBA = Chapter 3 1 e + T 2 π Problem 3.7 tan EB θ π 1 −e 1 − 0.3 tan B = tan = 0.733 80 ⇒ EB = 1.2661 rad = 2 1+e 2 1 + 0.3 4 MB = EB − e sin EB = 1.2661 − 0.6 sin 1.2661 = 0.97992 rad M 0.97992 tB = B T = T = 0.48996T 2π 2π Problem 3.8 a= e= T= rapogee + rperigee rapogee 2 − rperigee rapogee + rperigee 2π µ a 3/2 = 14 000 + 7000 = 10 500 km 2 14 000 − 7000 = = 0.33333 14 000 + 7000 = 2π 398 600 10 500 3/2 = 10708 s E 1 −e 1 − 0.33333  60°  tan θ = 60° = tan = tan (30°) = 0.40825 ⇒ Eθ = 60° = 0.77519 rad  2  2 1+e 1 + 0.33333 Mθ = 60° = 0.77519 − 0.33333 sin (0.77519) = 0.54191 rad M 0.54191 tθ = 60° = θ = 60° T = 10708 = 923.51 s 2π 2π tθ = tθ = 60° + 30 ⋅ 60 = 2723.5 s t 2723.5 Mθ = 2π θ = 2π = 1.5981 rad T 10708 Eθ − e sin Eθ = Mθ Eθ − 0.33333 sin Eθ = 1.5981 ⇒ Eθ = 1.9122 rad (Algorithm 3.1) tan θ 1+e E 1 + 0.33333 1.9122 = tan θ = tan ⇒ θ = 126.95° 2 1 −e 2 1 − 0.33333 2 Problem 3.9 a cos θ + b sin θ = c b c cos θ + sin θ = a a b sin φ ≡ tan φ = a cos φ sin φ c cos θ + sin θ = cos φ a cos θ cos φ + sin θ sin φ = c cos φ a c cos φ a c  θ − φ = ± cos −1 cos φ a  cos(θ − φ ) = θ = φ ± cos −1 c  cos φ a  28 Solutions Manual Orbital Mechanics for Engineering Students Problem 3.10 MB = EB − e sin EB π π t 2π B = − e sin T 2 2 π π − e sin 2 T = (0.25 − 0.15915e )Τ tB = 2 2π Problem 3.11  1 h2  rperigee (1 + e ) µ 1 + e cos θ   ⇒ r= 2 1 + e cos θ 1  h rperigee =  µ 1+e r= 2rperigee = rperigee (1 + 0.5) 1 + 0.5 cos θ B cos θ B = −0.5 ⇒ θ B = 120° θ 1 −e 1 − 0.5 120° E π tan B = tan B = tan ⇒ EB = rad 2 1+e 2 1 + .5 2 2 π π MB = EB − e sin EB = − 0.5 sin = 1.0708 rad 2 2 1.0708 MB tB = T= T = 0.17042T 2π 2π Problem 3.12 From Example 3.3 we have e = 0.24649 T = 8679.1 s θ c = 143.36 Thus tan 1 −e 1 − 0.24649 143.36° Ec θ ⇒ Ec = 2.3364 rad = tan c = tan 2 1+e 2 1 + 0.24649 2 Mc = Ec − e sin Ec = 2.3364 − 0.24649 ⋅ sin 2.3364 = 2.1587 rad 2.1587 M ⋅ 8679.1 = 2981.8 s tc = c T = 2π 2π Problem 3.13 rSOI = 925 000 km rperigee (1 + e ) r= 1 + e cos θ (6378 + 500)(1 + 1) 925 000 = ⇒ θ = 170.11° 1 + 1 ⋅ cos θ θ 1 θ 1 1 170.11° 1 170.11° M p = tan + tan 3 = tan + tan 3 = 262.82 2 2 6 2 2 2 6 2 h2 rperigee = ⇒ h = 2µrperigee = 2 ⋅ 398 600 ⋅ 6878 = 74 048 km 2 s 2µ 29 Chapter 3 Solutions Manual t= Orbital Mechanics for Engineering Students Chapter 3 74 048 3 h3 Mp = ⋅ 262.82 = 671 630 s = 7d 18h 34m 398 600 µ Problem 3.14 (a) h = µrperigee (1 + e ) = 398 600 ⋅ 7500 ⋅ (1 + 1) = 77 324 km 2 s 1 90° 1 90° Mp = tan + tan 3 = 0.66667 θ = 90° 2 2 6 2 77 324 3 h3 0.66667 = 1939.9 s tθ = 90° = 2 M p = θ = 90° 389 600 2 µ t −90° to +90° = 2 ⋅ 1939.9 = 3879.8 s = 1.0777 h ) ) (b) Mp = µ2 t = h3 398 600 3 ⋅ (24 ⋅ 3600) 77 324 3 = 29.692 1/3 −1/3 θ     2 2 = 3 M p + (3 M p ) + 1  − 3 M p + (3 M p ) + 1      2 1/3 θ 2 2 tan = 3 ⋅ 29.692 + (3 ⋅ 29.692) + 1 − 3 ⋅ 29.692 + (3 ⋅ 29.692) + 1 2 θ = 159.2° tan [ r= ] [ 77 324 2 1 1 h2 = = 230 200 km µ 1 + cos θ 398 600 1 + cos 159.2° Problem 3.15 (a) vperigee = 1.1 2 ⋅ 398 600 2µ = 1.1 = 11.341 km s rperigee 7500 h = rperigee vperigee = 7500 ⋅ 11.341 = 85056 km 2 s rperigee = 7500 = h2 1 µ 1+e 85 056 2 1 ⇒ e = 1.4200 398 600 1 + e 90° 1.42 − 1 90° F90° e −1 = tan = tan ⇒ F90° = 0.887 14 2 2 1.42 + 1 2 e +1 M h )90° = e sinh F90° − F90° = 1.42 ⋅ sinh 0.88714 − 0.88714 = 0.544 46 tanh M h )90° = µ2 0.544 46 = t -90° to 90° 3 (e 2 − 1)3/2 t90° h 398 600 2 3 (1.42 2 − 1)3/2 t90° 85056 = 2t90° = 4115.7 s = 1.1433 h ⇒ t90° = 2057.9 s (b) Mh = µ2 2 (e 2 − 1)3/2 t = 398 6003 (1.42 2 − 1)3/2 ⋅ 24 ⋅ 3600 = 22.859 3 h 85056 e sinh F − F = M h 1.42 sinh F − F = 22.859 ⇒ F = 3.6196 (Algorithm 3.2) 30 ] −1/3 = 5.4492 Solutions Manual Orbital Mechanics for Engineering Students tan e +1 F 1.42 + 1 3.6196 θ = ⇒ θ = 132.55° tanh = tanh 2 e −1 2 1.42 − 1 2 r= 85 056 2 h2 1 1 = = 455 660 km µ 1 + e cos θ 398 600 1 + 1.42 ⋅ cos 132.55° Chapter 3 Problem 3.16 h = rperigee vperigee = (6378 + 300) ⋅ 11.5 = 76 797 km 2 s rperigee = e= a= h2 1 µ 1+e h2 µrperigee −1 = 76 797 2 − 1 = 1.2157 ( hyperbola) 398 600 ⋅ 6878 76 797 2 h2 1 1 = = 30 964 km 2 µ e − 1 398 600 1.2157 2 − 1 At 6 AM: v2 µ µ − = 2 r 2a 2 398 600 398 600 10 − = ⇒ r = 9149.9 km 2 r 2 ⋅ 30 964 r= 9149.9 = 1 h2 µ 1 + e cos θ 76 797 2 1 ⇒ θ = −59.494° ( flying towards earth ) 398 600 1 + 1.2157 cos θ 1.2157 − 1 θ F e −1 −59.494° tan = tan = = −0.10384 ⇒ F = −0.360 45 2 2 1.2157 + 1 2 e +1 M h = e sinh F − F = 1.2157 sinh (−0.360 45) − (−0.360 45) = −0.087 287 tanh 3/2 µ2 M h = 3 (e 2 − 1) t h 398 600 2 (1.2157 2 − 1)3/2 t ⇒ t = −753.3 s −0.087 287 = 3 76 797 ( negative means time until perigee) At 11 AM: t = 5 ⋅ 3600 − −753.3 = 17 247 s Mh = µ2 3 ( time since perigee at 11 AM) 2 (e 2 − 1)3/2 t2 = 398 6003 (1.2157 2 − 1)3/2 ⋅ 17 247 = 1.9984 h e sinh F − F = M h 76 797 1.2157 sinh F − F = 1.9984 ⇒ F = 1.8760 (Algorithm 3.2) tanh θ = 2 e +1 F 1.8760 1.2157 + 1 tan = tan ⇒ θ = 133.96° e −1 1.2157 − 1 2 2 76 797 2 1 1 h2 = = 94 771 km µ 1 + e cos θ 398 600 1 + 1.2157 ⋅ cos 133.96° z = r − 6378 = 88 393 km r= 31 Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 Problem 3.17 v ⊥ = v cos γ = 8 ⋅ cos( −65°) = 3.3809 km s h = rv ⊥ = (37 000 + 6378) ⋅ 3.3809 = 146 660 km 2 s r= 43 378 = h2 1 µ 1 + e cos θ 146 660 2 1 ⇒ e cos θ = 0.24397 398 600 1 + e cos θ v r = v sin γ = 8 ⋅ sin ( −65°) = −7.2505 km s µ v r = e sin θ h 398 600 −7.2505 = e sin θ ⇒ e sin θ = −2.6677 146 660 e sin θ −2.6677 tan θ = = = −10.935 ⇒ θ = −84.775° e cos θ 0.24397 e sin ( −84.775°) = −2.6677 ⇒ e = 2.6788 ( hyperbola) rperigee = 146 660 2 1 h2 1 = = 14 668 km ( no impact ) µ 1 + e 398 600 1 + 2.6788 F e −1 θ 2.6788 − 1  −84.775°  = tan = tan ⇒ F = −1.4389   e +1 2 2 2.6788 + 1 2 M h = e sinh F − F = 2.6788 sinh ( −1.4389) − ( −1.4389) = −3.8906 tanh 3/2 µ2 M h = 3 (e 2 − 1) t h 398 600 2 (2.6788 2 − 1)3/2 t ⇒ t = −5032.5 s = −1.3979 h −3.8906 = 3 46 660 1.3979 hours until perigee passage. Problem 3.18 Write the following MATLAB script to use M-function kepler_U to implement Algorithm 3.3. clear global mu mu = 398600; ro vro a dt = = = = 7200; 1; 10000; 3600; x = kepler_U(dt, ro, vro, 1/a); fprintf('\n\n-----------------------------------------------------\n') fprintf('\n Initial radial coordinate = %g',ro) fprintf('\n Initial radial velocity = %g',vro) fprintf('\n Elapsed time = %g',dt) fprintf('\n Semimajor axis = %g\n',a) fprintf('\n Universal anomaly = %g',x) fprintf('\n\n-----------------------------------------------------\n\n') Running this program produces the following output in the MATLAB Command Window: 32 Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 ----------------------------------------------------Initial radial coordinate = 7200 Initial radial velocity = 1 Elapsed time = 3600 Semimajor axis = 10000 Universal anomaly = 229.341 ----------------------------------------------------That is, χ = 229.34 km1/2 . To check this with Equation 3.55, proceed as follows. h2 1 µ 1 + e cos θ r= 7200 = h2 1 398 600 1 + e cos θ ∴ cos θ = 3.844 × 10 −10 h2 − 1 e (1) µ e sin θ h 398 600 1= e sin θ h 1 h ∴ sin θ = 398 600 e vr = (2) sin 2 θ + cos2 θ = 1 Substitute (1) and (2): 1   398 600  2 2  h 3.844 × 10 −10 h2 − 1   =1 +    e e (3) h2 1 µ 1 − e2 h2 1 10 000 = 398 600 1 − e 2 a= ∴ h = 3.986 × 10 9 (1 − e 2 ) (4) Substitute (4) into (3):  1   398 600 2 2   3.844 × 10 −10 3.986 × 10 9 (1 − e 2 ) − 1  3.986 × 10 9 (1 − e 2 )  =1  + e e    [ Expanding and collecting terms yields 1 1993e 2 (3844.5e 4 − 4195.9e 2 + 351.41) = 0 or 3844.5e 4 − 4195.9e 2 + 351.41 = 0 33 ] Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 The positive roorts of this equation are e = 1.0000 and e = 0.30233 . Obviously, we choose the latter. e = 0.30233 (5) Substituting (5) into (4), we get h = 60180 km 2 s (6) Substituting (5) and (6) into (1) or (2) yields θ1 = 29.959° (7) Compute the time at this initial true anomaly as follows: tan E1 = 2 1 −e 1 − 0.30233 29.959° θ = 0.19584 tan 1 = tan 1+e 2 1 + 0.30233 2 ∴ E1 = 0.38678 rad (8) M1 = E1 − e sin E1 = 0.38678 − 0.30233 ⋅ sin 0.38678 = 0.27273 rad M 0.27273 t1 = 1 T = ⋅ 9952.0 = 431.99 s 2π 2π Obtain E one hour later. t2 = t1 + 3600 = 4032 s t 4032 M2 = 2 π 2 = 2 π = 2.5456 rad T 9952.0 E2 − e sin E2 = M2 E2 − 0.30233 sin E2 = 2.5456 Using Algorithm 3.1, E2 = 2.6802 rad (9) According to Equation 3.55, χ = a (E2 − E1 ) = 10 000 (2.6802 − 0.38678) = 229.34 km1/2 This is the same as the value obtained via Algorithm 3.3 Problem 3.19 Write the following MATLAB script to use the M-function rv_from_r0v0 to execute Algorithm 3.4. clear global mu mu = 398600; R0 V0 t = [20000 -105000 -19000]; = [ 0.9 -3.4 -1.5]; =2*3600; [R V] = rv_from_r0v0(R0, V0, t); fprintf('-----------------------------------------------------') fprintf('\n Initial position vector (km):') fprintf('\n r0 = (%g, %g, %g)\n', R0(1), R0(2), R0(3)) fprintf('\n Initial velocity vector (km/s):') 34 Solutions Manual Orbital Mechanics for Engineering Students Chapter 3 fprintf('\n v0 = (%g, %g, %g)', V0(1), V0(2), V0(3)) fprintf('\n\n Elapsed time = %g s\n',t) fprintf('\n Final position vector (km):') fprintf('\n r = (%g, %g, %g)\n', R(1), R(2), R(3)) fprintf('\n Final velocity vector (km/s):') fprintf('\n v = (%g, %g, %g)', V(1), V(2), V(3)) fprintf('\n-----------------------------------------------------\n') The output to the MATLAB Command Window is as follows: ----------------------------------------------------Initial position vector (km): r0 = (20000, -105000, -19000) Initial velocity vector (km/s): v0 = (0.9, -3.4, -1.5) Elapsed time = 7200 s Final position vector (km): r = (26337.8, -128752, -29655.9) Final velocity vector (km/s): v = (0.862796, -3.2116, -1.46129) ----------------------------------------------------- 35 Solutions Manual Orbital Mechanics for Engineering Students 36 Chapter 3 Solutions Manual Orbital Mechanics for Engineering Students Problem 4.1 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8): (1) r = r = 16 850 km ( 2 ) v = v = 5.7415 km s ( 3 ) vr = r ⋅ v = 0.001 885 6 km s ( > 0 ) r ( 4 ) h = r × v = 82 234Iˆ − 23 035Jˆ + 41 876Kˆ ( km 2 s) ( 5 ) h = h = 95 360 km 2 s ( 6 ) i = cos−1  hZ  = cos−1  41 876  = 63.952°   h  96 350  ( 7 ) N = Kˆ × h = 24 035Iˆ + 82 234 Jˆ ( km 2 s) ( NY > 0 ) 2 ( 8 ) N = N = 85 674 km s ( 9 ) Ω = cos−1  N X  = cos−1  24 035  = 73.707°    85 674  N (10 ) e = 1  5.7415 2 − 398 600 16 850 2615Iˆ + 15 881Jˆ + 3980K ˆ 398 600  ( )( ( ) ) ˆ  − (16 580 ) ( 0.001 855 6 ) −2.767Iˆ − 0.7905 Jˆ + 4.98K  ˆ ( e > 0) = 0.059 521Iˆ + 0.360 32 Jˆ + 0.08988K Z (11) e = e = 0.37602 (12) ω = cos −1 (13) θ = cos −1  N ⋅e = 15.43°  Ne   e⋅r = 0.067 42°  er  Problem 4.2 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8): (1) r = r = 12 670 km ( 2 ) v = v = 3.9538 km s ( 3 ) vr = r ⋅ v = −0.7905 km s ( < 0 ) r ( 4 ) h = r × v = 49 084Iˆ ( km 2 s) ( 5 ) h = h = 49 084 km 2 s ( 6 ) i = cos−1  hZ  = cos−1    0  = 90° 49  084  h ( 7 ) N = Kˆ × h = 49 084 Jˆ ( km 2 s) ( NY > 0 ) ( 8 ) N = N = 49 084 km 2 s ( 9 ) Ω = cos−1  N X  = cos−1    N (10 ) e = 0  = 90°  49 084  1  3.9538 2 − 398 600 12 670 12 670K ˆ 398 600  ( )( ( ) ) ˆ  − (12 670 ) ( −0.7905 ) −3.8744 Jˆ − 0.7905K  ˆ ˆ = −0.097342 J − 0.52296K ( eZ < 0 ) (11) e = e = 0.53194 37 Chapter 4 Solutions Manual Orbital Mechanics for Engineering Students (12) ω = 360° − cos −1 (13) θ = 360° − cos −1 Chapter 4  N ⋅e = 360° − 100.54° = 259.46°  Ne   e⋅r = 360° − 169.46° = 190.54°  er  Problem 4.3 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8): (1) r = 10189 km ( 2 ) v = 5.8805 km s ( 3 ) vr = r ⋅ v = 1.2874 km/s ( > 0 ) r ( 4 ) h = r × v = 31 509Iˆ + 11 468Jˆ + 47 888Kˆ ( km 2 s) ( 5 ) h = h = 58 461 km 2 s ( 6 ) i = cos−1  hZ  = cos−1  47 888  = 35°   h  58 461  ( 7 ) N = Kˆ × h = −11 468Iˆ + 31 509Jˆ ( km 2 s) ( NY > 0 ) ( 8 ) N = N = 33 532 km 2 s ( 9 ) Ω = cos−1  N X  = cos−1  −11 468  = 110°   N (10 ) e =  33 532  1  5.8805 2 − 398 600 10189 6472.7Iˆ − 7470.8 Jˆ − 2469.8K ˆ 398 600  ( )( ( ) ˆ  − (10 189 )(1.2874 ) 3.9914Iˆ + 2.7916 Jˆ − 3.2948K  ˆ ˆ ˆ = −0.2051I − 0.006 738 2 J + 0.13657K ( eZ > 0 ) (11) e = e = 0.2465 (12) ω = cos −1 (13) θ = cos −1  N ⋅e = 74.996°  Ne   e⋅r = 130°  er  Problem 4.4 v⋅r = −2.30454 km s (Flying towards perigee) r  e⋅r θ = 360° − cos −1 = 360° − 30° = 330°  er  vr = Problem 4.5 ˆ r×e −3353.1Iˆ + 6361.8 Jˆ + 2718.1K = 7687.9 r×e ˆ ˆ ˆ w = −0.43616I + 0.827 51Jˆ + 0.353 55K ˆ = w ˆ ) = cos−1 ( 0.353 55 ) = 69.295° ˆ ⋅K i = cos−1 ( w Problem 4.6 (a) → AB = ( 4 − 1) iˆ + ( 6 − 2 ) ˆj + ( 5 − 3 ) kˆ = 3iˆ + 4 ˆj + 2kˆ → AC = ( 3 − 1) iˆ + ( 9 − 2 ) ˆj + ( −2 − 3 ) kˆ = 2iˆ + 7 ˆj − 5kˆ 38 ) Solutions Manual Orbital Mechanics for Engineering Students → X ′ = AB = 3iˆ + 4 ˆj + 2kˆ → Z′ = X ′ × AC = −34 iˆ + 19 ˆj + 13kˆ Y ′ = Z′ × X ′ = −14 iˆ + 107 ˆj − 193kˆ (b) X′ iˆ′ = = 0.557 09iˆ + 0.742 78 ˆj + 0.371 39kˆ X′ ˆj′ = Y ′ = −0.063 314 iˆ + 0.483 90 ˆj − 0.872 83kˆ Y′ Z′ kˆ ′ = = −0.828 04 iˆ + 0.462 73 ˆj + 0.316 60kˆ Z′   iˆ′      0.5557 09 0.742 78 0.371 39  ˆ [Q] =   j′   =  −0.063 314 0.483 90 −0.872 83       kˆ ′    −0.828 04 0.462 73 0.316 60  0.557 09 −0.063 314 −0.828 04   2    0.483 90 0.462 73  −1   0.371 39 −0.872 83 0.316 60   3  {v} = [Q ]T {v ′} =  0.742 78 −1.3066  {v} =  2.3898   2.5654    ( v = −1.3066iˆ + 2.3898ˆj + 2.5654k̂ˆ ) Problem 4.7  0.267 26 0.534 52 0.801 78  −50  {V}u = [Q ]xu {V}x =  −0.443 76 0.806 84 −0.389 97  100   −0.855 36 −0.251 58 0.452 84   75  100.22  {V}u = 73.624  ( V = 100.22uˆ + 73.624 vˆ + 51.573wˆ ) 51.573    Problem 4.8 1 0 0   [R1 ] = 0 cos 40° sin 40°  0 − sin 40° cos 40° cos 25° 0 − sin 25°  0 [R2 ] =  0 1   sin 25° 0 cos 25°  cos 25° sin 40° sin 25° − cos 40° sin 25°   cos 40° sin 40° [Q ] = [R2 ][R1 ] =  0   sin 25° − sin 40° cos 25° cos 40° cos 25°  0.906 31 0.27165 −0.32374   0.766 04 0.64279 [Q ] =  0  0.422 62 −0.582 56 0.694 27  Problem 4.9 ro = −5102Iˆ − 8228Jˆ − 2106Kˆ ( km ) v o = −4.348Iˆ + 3.478Jˆ − 2.846Kˆ ( km s) Method 1 39 Chapter 4 Solutions Manual Orbital Mechanics for Engineering Students Chapter 4 Use Algorithm 3.4 (MATLAB M-function rv_from_r0v0 in Appendix D.7): (1a ) ro = 9907.6 km vo = 6.2531 km s (1b) vr o = −0.044 678 km s (1c ) α = 103.77 × 10−6 km-1 ( 2 ) χ = 195 km1/2 ( 3 ) f = −0.365 39 g = 1394.4 s-1 ( 4 ) r = ( −0.36 539 ) ( −5102Iˆ − 8228Jˆ − 2106Kˆ ) + 1394.4 ( −4.348Iˆ + 3.478Jˆ − 2.846Kˆ ) ˆ ( km ) = −4198.4Iˆ + 7856.1Jˆ − 3199.2K ( 5 ) f = −6.0467 × 10−4 s-1 g = −0.429 31 ( 6 ) v = ( −6.0467 × 10−4 ) ( −5102Iˆ − 8228Jˆ − 2106Kˆ ) + ( −0.429 31) ( −4.348Iˆ + 3.478Jˆ − 2.846Kˆ ) ˆ ( km s ) = 4.9517Iˆ + 3.4821Jˆ + 2.4946K Method 2 Compute the orbital elements using Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8): (1) r = r = 9907.6 km ( 2 ) v = v = 6.2531 km s ( 3 ) vr = r ⋅ v = −0.044 678 km s ( < 0 ) r ( 4 ) h = r × v = 30 738Iˆ − 5367.8Jˆ − 53 520Kˆ ( km 2 s) ( 5 ) h = h = 61 952 km 2 s ( 6 ) i = cos−1  hZ  = cos−1  −53 520  = 149.76°     h 61 952 ˆ ˆ ( 7 ) N = K × h = 5367.8I + 30 738Jˆ km 2 s ( ) ( NY > 0 ) ( 8 ) N = N = 31 203 km 2 s ( 9 ) Ω = cos−1  N X  = cos−1  5367.8  = 80.094°   N (10 ) e =  31 203  1  5.7415 2 − 398 600 9907.6 −5102Iˆ − 8228 Jˆ − 2106K ˆ 398 600  ( )( ( ) ˆ  − ( 9907.6 ) ( −0.044 678 ) −4.348Iˆ + 3.478 Jˆ − 2.846K  ˆ ˆ ˆ = 0.009 638 51I + 0.027193 J + 0.002 808 3K ( e > 0 ) Z (11) e = e = 0.028 987 (12) ω = cos −1  N ⋅e = 11.09°  Ne  (13) θ = 360° − cos −1  e⋅r = 360° − 166.14° = 193.86°  er  3 2π  h  T= 2 = 9414.9 s µ  1 − e 2  Determine the time since perigee passage at true anomaly θ o = 193.86° : 40 ) Solutions Manual E tan o = 2 Orbital Mechanics for Engineering Students Chapter 4 1 − 0.028 987 1 −e 193.86° θ ⇒ E o = −2.8926 rad tan o = tan 1+e 2 1 + 0.028 987 2 M o = E o − e sin E o = −2.8926 − 0.028 987 ⋅ sin ( −2.8926) = −2.8855 rad M −2.8855 9414.9 = −4323.7 s ( minus means time until perigee passage) to = o T = 2π 2π Update the true anomaly of the spacecraft. t = t o + 50 ⋅ 60 = −1323.7 s . M = 2π t −1323.7 = 2π = −0.883 41 rad 9414.9 T E − e sin E = M E − 0.028 987 sin E = −0.883 41 ⇒ E = −0.90623 rad (Algorithm 3.1) tan 1 + 0.028 987 1+e −0.90623 E θ = ⇒ θ = −53.242° tan = tan 2 1 −e 2 1 − 0.028 987 2 Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): (1) 2 1 {r } x = h µ 1 + e cos θ cos( −53.242°)  cos θ  2 1     61 952 sin ( −53.242°)  sin θ  = ( −53.242°)  398 600 1 + 0 . 028 987 cos   0  0      5663.9    =  −7582.8  ( km )  0     − sin θ  (2) {v} x = µ e + cos θ  = (3) [Q]xX   5.1548   − sin ( −53.242°) 398 600     ( ) 0.028 987 + cos −53.242°  = 4.0368  (km s) 61 952    0  0     h   0  cos Ω cos ω − sin Ω sin ω cos i  = sin Ω cos ω + cos Ω cos i sin ω  sin i sin ω − cos Ω sin ω − sin Ω cos i cos ω − sin Ω sin ω + cos Ω cos i cos ω sin i cos ω 0.33251 0.80204 0.49616    = 0.93811 −0.33532 −0.086644  0.09688 0.49426 −0.8639  −4198.4   ( 4 ) {r}X = [Q ]xX {r}x =  7856.1  ( km ) or  −3199.2    sin Ω sin i   − cos Ω sin i   cos i ˆ ( km ) r = −4198.4Iˆ + 7856.1Jˆ − 3199.2K  4.9517  {v}X = [Q ]xX {v}x =  3.4821  ( km s) or v = 4.9517Iˆ + 3.4821Jˆ + 2.4946Kˆ ( km s)  2.4946    Problem 4.10 e = 1.5 Ω = 130° i = 35° ω = 115° θ = 0° rperigee = 6678 km h = µ(1 + e )rperigee = 398 600 ⋅ (1 + 1.5) ⋅ 6678 = 81 576 km 2 s Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): 41 Solutions Manual 2 1 (1) {r}x = h µ 1 + e cos θ Orbital Mechanics for Engineering Students Chapter 4 cos ( 0 ) 6678  cos θ  2    1  81 576    sin ( 0 )  =  0  ( km )  sin θ  = ( ) 398 600 . cos + 1 1 5 0  0   0   0        r = 6678pˆ ( km )  − sin ( 0 )   0   − sin θ  398 600    µ    ( 2 ) {v}x = e + cos θ  = 1.5 + cos ( 0 ) = 12.216  ( km s ) h   0   81 576  0     0   v = 12.216qˆ ( km s ) (3) cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω [Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω  sin i sin ω sin i cos ω  −0.297 06 0.847 76 0.439 39    =  −0.800 95 −0.47175 0.368 69   0.519 84 −0.2424 0.81915  sin Ω sin i   − cos Ω sin i   cos i −1983.8  ( 4 ) {r}X = [Q ]xX {r}x = −5348.8  ( km ) or r = −1983.8Iˆ − 5348.8 Jˆ + 3471.5K̂ ( km )  3471.5     10.356  {v}X = [Q ]xX {v}x = −5.7627  ( km s) or v = 10.356Iˆ − 5.7627 Jˆ − 2.9611Kˆ ( km s)  −2.9611    Problem 4.11 h = 81 576 km 2 s e = 1.5 Ω = 130° i = 35° ω = 115° . t = 7200 s Mh = µ2 2 (e 2 − 1)3/2 t = 398 6003 (1.52 − 1)3/2 ⋅ 7200 = 2.945 3 h 81 576 e sinh F − F = M h 1.5 sinh F − F = 2.945 ⇒ F = 1.886 ( Algorithm 3.2) tan θ = 2 e +1 F 1.5 + 1 1.886 ⇒ θ = 117.47° tanh = tan e −1 2 1.5 − 1 2 Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): cos (117.47° ) −25 007  cos θ  81 576 2    1 1 h2    (1) {r}x =  siin (117.47° )  =  48 093  ( km )  sin θ  = ( ) µ 1 + e cos θ    0   398 600 1 + 1.5 cos 117.47°  0  0      r = −25 007pˆ + 48 093qˆ ( km )  − sin (117.47° )  −4.3352   − sin θ  398 600    µ    ( 2 ) {v}x = e + cos θ  = 1.5 + cos (117.47° ) =  5.0752  ( km s ) 8 h 1 576    0   0     0   ˆ ˆ v = −4.3352p + 5.0752q ( km s ) 42 Solutions Manual (3) [Q]xX Orbital Mechanics for Engineering Students  −0.297 06 0.847 76 0.439 39    =  −0.800 95 −0.47175 0.368 69   0.519 84 −0.2424 0.81915  Chapter 4 (Problem 4.10)  48 200  ˆ ( km ) ( 4 ) {r}X = [Q ]xX {r}x =  −2658  ( km ) or r = 48 200Iˆ − 2658 Jˆ − 24 658K −24 658     5.5903  {v}X = [Q ]xX {v}x =  1.0781  ( km s) or v = 5.5903Iˆ + 1.0781Jˆ − 3.4838Kˆ ( km s) −3..4838    Problem 4.12 ro = 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ( km ) vo = 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ ( km s2 ) Method 1 Use Algorithm 3.4 (MATLAB M-function rv_from_r0v0 in Appendix D.7): (1a ) ro = 10189 km vo = 5.805 km s (1b) vr o = 1.2874 km s (1c ) α = 109.54 × 10−6 km-1 ( 2 ) χ = 171.31 km1/2 ( 3 ) f = −0.093 379 g = 1870.6 s-1 ( 4 ) r = ( −0.093 379 ) ( 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ) + 1870.6 ( 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ ) ˆ ( km ) = 6861.9Iˆ + 5919.6 Jˆ − 5932.7K ( 5 ) f = −5.3316 × 10−4 s-1 g = −0.028 475 ( 6 ) v = ( −5.3316 × 10−4 ) ( 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ) + ( −0.028 475 ) ( 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ ) ˆ ( km s ) = −3.5647Iˆ + 3.9037 Jˆ + 1.4106K Method 2 From Problem 4.3 h = 58 461 km 2 s , e = 0.2465, i = 35°, Ω = 110°, ω = 74.996°, θ = 130° . 3 3 T= tan 58 461  2π  2π h   = = 8680.3 s 2  2   398 600  1 − 0.24652  µ  1 − e2  1 −e 1 − 0.2465 130° Eo θ = tan o = tan ⇒ E o = 2.0612 rad 2 1+e 2 1 + 0.2465 2 M o = E o − e sin E o = 2.0612 − 0.2465 ⋅ sin (2.0612) = 1.8437 rad M 1.8437 to = o T = 8680.3 = 2547.1 s ( minus means time until perigee passage) 2π 2π Update the true anomaly: t = t o + 50 ⋅ 60 = 5547.1 s 43 Solutions Manual Orbital Mechanics for Engineering Students M = 2π Chapter 4 t 5547.1 = 2π = 4.0153 rad T 8680.3 E − e sin E = M E − 0.2465 sin E = 4.0153 ⇒ E = 3.8541 rad (Algorithm 3.1) tan 3.8541 θ 1+e E 1 + 0.2465 tan = tan = ⇒ θ = −147.73° 2 1 −e 2 1 − 0.2465 2 Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): (1) 2 1 {r } x = h µ 1 + e cos θ cos( −147.73°)  cos θ  2 1     58 461 sin ( −147.73°)  sin θ  = ( −147.73°)  398 600 1 + 0 . 2465 cos   0  0      −9158.1    =  −5783.8  ( km )  0    (2) (3)   3.6408   − sin ( −147.73°)  − sin θ  398 600  µ     {v} x = e + cos θ  = 0.2465 + cos( −147.73°)  =  −4.0841 (km s) h 8 461    0   0  0      cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω sin Ω sin i  [Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω − cos Ω sin i    sin i sin ω sin i cos ω cos i  −0.8321 0.13078 0.538 99    =  −0.026 991 −0.9802 0.19617   0.553 97 0.148 69 0.81915   6864  ( 4 ) {r}X = [Q ]xX {r}x =  5916.5  ( km ) or −5933.3    ˆ ( km ) r = 6864Iˆ + 5916.5 Jˆ − 5933.3K −3.5636  {v}X = [Q ]xX {v}x =  3.905  ( km s) or v = −3.5636Iˆ + 3.905Jˆ + 1.4096Kˆ ( km s)  1.40996    Problem 4.13 e = 1.2 Ω = 75° i = 50° ω = 80° θ = 0° rperigee = 6578 km h = µ(1 + e )rperigee = 398 600 ⋅ (1 + 1.2) ⋅ 6578 = 79 950 km 2 s Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): cos ( 0 ) 6578  cos θ  79 950 2    1 1 h2    (1) {r}x =  sin ( 0 )  =  0  ( km )  sin θ  = ( ) 398 600 µ 1 + e cos θ  . cos 1 1 2 0 +  0   0        0  r = 6578pˆ ( km )  − sin ( 0 )   0   − sin θ  398 600    µ    ( 2 ) {v}x = e + cos θ  = 1.2 + cos ( 0 ) = 11.546  ( km s ) h 9 950 7   0    0     0   v = 11.546qˆ ( km s ) 44 Solutions Manual (3) Orbital Mechanics for Engineering Students cos Ω cos ω − sin Ω sin ω cos i [Q]xX = sin Ω cos ω + cos Ω cos i sin ω  sin i sin ω − cos Ω sin ω − sin Ω cos i cos ω − sin Ω sin ω + cos Ω cos i cos ω sin i cos ω Chapter 4 sin Ω sin i   − cos Ω sin i   cos i  −0.56 561 −0.3627 0.739 94    =  0.331 57 −0.92236 −0.198 27   0.754 41 0.133 02 0.64279  −3726.5  ( 4 ) {r}X = [Q ]xX {r}x =  2181.1  ( km )  4962.5    −4.1878  {v}X = [Q ]xX {v}x =  −10.65   1.5359    Problem 4.14 h = 75950 km 2 s or ˆ ( km s ) or v = −4.1878Iˆ + 10.65 Jˆ + 1.5359K ( km s) e = 1.2 ˆ ( km ) r = −3726.5Iˆ + 2181.1Jˆ + 49962.5K Ω = 130° i = 50° ω = 80° t = 7200 s Mh = 2 µ2 (e 2 − 1)3/2 t = 398 6003 (1.2 2 − 1)3/2 ⋅ 7200 = 0.762 09 3 h e sinh F − F = M h 75950 1.2 sinh F − F = 0.762 09 ⇒ F = 1.3174 ( Algorithm 3.2) tan θ = 2 0.762 09 e +1 F 1.2 + 1 tanh = tan 86° ⇒ θ = 124.86 e −1 2 1.2 − 1 2 Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9): cos (124.86° ) −26 336  cos θ     h2 1 1 75950 2    (1) {r}x =  siin (124.86° )  =  37 806  ( km )  sin θ  = µ 1 + e cos θ    0   398 600 1 + 1.2 cos (124.86° )  0      0  r = −26 336pˆ + 37 806qˆ ( km )  − sin (124.86° )  −4.3064   − sin θ  398 600    µ    ( 2 ) {v}x = e + cos θ  = 1.5 + cos (124.86° ) =  3.298  ( km s ) h 1 576 8   0    0    0    ˆ ˆ v = −4.3064p + 3.298q ( km s ) (3) [Q]xX  −0.56 561 −0.3627 0.739 94    =  0.331 57 −0.92236 −0.198 27   0.754 41 0.133 02 0.64279   1207.2  ( 4 ) {r}X = [Q ]xX {r}x =  −43 603  ( km ) −14 839     1.2434  {v}X = [Q ]xX {v}x = −4.4698  −2.8100    or ( km s) 45 (Problem 4.13) ˆ ( km ) r = 1207.2Iˆ − 43 603 Jˆ − 14 8399K ˆ ( km s ) or v = 1.2434Iˆ − 4.4698 Jˆ − 2.8100K Solutions Manual Orbital Mechanics for Engineering Students Problem 4.15 h = 75000 km 2 s e = 0.7 Chapter 4 θ = 25°  − sin (25°)   −2.2461 398 600     0.7 + cos(25°)  =  8.537  (km s) h   0   75000  0  0       −0.83204 0.02741 0.55403   −2.2461  2.1028       = [Q ]xX {v} x =  -0.13114 -0.98019 -0.14845  8.537  =  −8.0733  (km s)  0.53899 -0.19617 0.81915   0   −2.8853   − sin θ  {v} x = µ e + cos θ  = {v} X or ˆ ( km s ) v = 2.1028Iˆ − 8.0733 Jˆ − 2.8853K Problem 4.16 Ω = 60° ω = 0 B Z i = 90° perigee apogee  −3.208  {v} x =  −0.8288  (km s)  0    Y 60° X cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω [Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω  sin i sin ω sin i cos ω  0.5 0 0.86603    = 0.86603 0 −0.5   0 1 0   −1.604  {v}X = Q  {v}x = −2.7782  ( km s) xX −0.8288    or ˆ ( km s ) v = −1.604Iˆ − 2.7782 Jˆ − 0.8288K Problem 4.17 a = 7016 km ( 2) {r } x = a 1 − e 1 + e cos θ e = 0.05 i = 45° Ω =0 ω = 20° θ = 10° cos θ  2 cos 10°  6568.7    7016(1 − 05 )     sin θ  = sin 10°  = 1158.2  ( km ) + . cos ° 1 0 05 10  0   0   0        46 sin Ω sin i   − cos Ω sin i   cos i Solutions Manual Q  xX Orbital Mechanics for Engineering Students cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω  = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω  sin i sin ω sin i cos ω 0  0.93969 -0.34202    = 0.24184 0.66446 -0.70711  0.24184 0.66446 0.70711  Chapter 4 sin Ω sin i   − cos Ω sin i   cos i 5776.4  {r}X = Q  {r}x =  2358.2  ( km ) xX  2358.2    or ˆ ( km ) r = 5776.4Iˆ + 2358.2 Jˆ + 2358.2K Problem 4.18 Ω=− 2 3 µ J 2 Rearth cos i 2 2( 1 − e 2 ) a7/2 1 1 r +r = ( 6878 + 7378 ) = 7128 km 2 perigee apogee 2 rapogee − rperigee 7378 − 6878 = 0.035 073 e= = rapogee + rperigee 7378 + 68788 ( a= Ω= 2π = 1.991 × 10−7 raad s 365.26 ⋅ 24 ⋅ 3600 1.991 × 10−7 = − 1.991 × 10 −7 ) 3 2 398 600 ⋅ 0.001082 6 ⋅ 6378 2 (1 − 0.035 073 ) 2 = −1.367 × 10 −6 2 7128 cos i 7/2 cos i ⇒ i = 98.372° Problem 4.19 T= Ω =- ω earth 2π µ 3 2 r 3/2 = 2π 398 600 µ J 2 Rearth 2 ( 6378 + 180 )3/2 = 5285.3 s cos i = - 3 2 398 600 ⋅ 0.001082 6 ⋅ 6378 2 r 7/2 ( 6378 + 180 )7/2 2π 2π + .26 = 7.2921 × 10−5 rad s 365 = 24 ⋅ 3600 cos 30° = −1.5814 × 10−6 rad s Change in east longitude of the ascending node after 1 orbit of the satellite: ∆λ = ω earth − Ω T = 7.2921 × 10−5 − ( −1.5814 × 10−6 )  ⋅ 5285.3 = 0.393 77 rad ( ) Spacing s, s = Rearth ∆λ = 6378 ⋅ 0.393 77 = 2511.4 km Problem 4.20 The change in east longitude λ of the ascending node of a satellite after n s orbits is 47 Solutions Manual Orbital Mechanics for Engineering Students ( Chapter 4 ) ∆λ = ω earth − Ω nsT If ∆λ is an integral multiple n e of earth rotations ( 2π ) then the ground track will close on itself, ( ) 2π ne = ω earth − Ω nsT ( ) Let ν = n s n e . Then 2π = ω earth − Ω ν T , or T= 1 2π ν ω earth − Ω But T = 2πr 3/2 (1) µ , where r is the radius of the orbit. Thus 2π 3/2 1 2π r = ν ω earth − Ω µ or  µ r=  ν ω earth − Ω ( )    2/3 (2) For a circular orbit Ω= − 3 2 µ J 2 Rearth 2 r 7/2 cos i or cos i = − 2 3 r 7/2 µ J 2 Rearth 2 Ω Substituting (2) we get   µ   2  ν ω earth − Ω  cos i = − 3 µ J 2 Rearth 2 ( 7/3 ) (3) Ω Substituting ω earth = 7.2921 × 10−5 rad J 2 = 0.001082 6 Ω = 1.997 × 10−7 rad s Rearth = 6378 km µ =398 600 km 3 s2 into (1), (2) and (3) we get  73.948  i = cos −1  − 7/3   ν  T= 86 400 ν z= 42 241 ν 2/3 From these we obtain the following table of scenarios: 48 − 6378 Solutions Manual ν 17 16 15 14 13 12 11 10 9 8 Orbital Mechanics for Engineering Students z ( km ) i (deg) T (h) 11.042 95.711 1.4118 274.55 567.03 96.583 97.658 1.5000 1.6000 893.93 99.006 1.7143 1262.2 100.72 2.000 1681.0 102.96 105.95 2.1818 2162.6 2722.6 110.07 2.4000 2.6667 3384.8 116.03 3.0000 4182.3 125.29 3.4286 Problem 4.21 5  ω = − f  sin 2 i − 2 2  ω 7 f =− =− = 7.2384 deg day 5 5 2 2 sin i − 2 sin 40° − 2 2 2 Ω = − f cos i = −7.2384 ⋅ cos 40° = −5.545 deg day Problem 4.22 From ˆ ( km ) ro = −2429.1Iˆ + 4555.1Jˆ + 4577.0K ˆ ( km s ) v o = −4.7689Iˆ − 5.6113 Jˆ + 3.0535K we obtain the orbital elements by means of Algorithm 4.1: ( h = 55000 km 2 s ) e = 0.1 Ω o = 70° i = 50° ω o = 60° θo = 0 55 000 2 1 h2 1 = = 7665.8 km µ 1 − e 2 398 600 1 − 0.12 2π 3/2 2π 7665.8 3/2 = 6679.5 s T= a = µ 398 600 a= The satellite is at perigee ( θ o = 0 ) so t o = 0 . After 72 hours, t f = t o + 72 ⋅ 3600 = 259 000 s . t f T = 38.805 , so t f is in orbit 39. The time since perigee in orbit 39 is t39 = (38.805 − 38)T = 5378.3 s 2π 2π t = 5378.3 = 5.0952 rad T 39 6679.5 = M39 ∴ M39 = E39 − e sin E39 E39 − 0.1 sin E39 = 5.0952 ⇒ E39 = 4.9623 rad ( Algorithm 3.1) tan 1+e 1 + 0.1 4.9623 θ 39 E = tan 39 = tan ⇒ θ 39 = 278.68° 2 1 −e 2 1 − 0.1 2 The perifocal state vector is 49 Chapter 4 Solutions Manual Orbital Mechanics for Engineering Students Chapter 4 cos(278.68°)   1128.9  cos θ  55000 2 h2 1 1       {r } x = sin (278.68°)  =  −7390.5 ( km ) sin θ  = ( ) µ 1 + e cos θ    0   398 600 1 + 0.1 cos 278.68°  0  0       − sin (278.68°)  7.1642   − sin θ  398 600     {v} x = µ e + cos θ  = 0.1 + cos(278.68°)  = 1.8191  (km s) h   0   55000  0  0      Update Ω and ω : Ω=− 2 3 µ J 2 Rearth 3 cos i = − 2 2( 2 1 − e 2 ) a7/2 3998 600 ⋅ 0.0010826 ⋅ 6378 2 (1 − 0.12 ) 2 7665.87/2 cos 50° = −1.0789 × 10−6 cos 50° = −6.9352 × 10−7 rad s = −3.9736 × 10−5 deg s ∴ Ω39 = Ω1 + Ω∆t = 70° − 3.9736 × 10−5 ⋅ 259 200 = 59.701° ω=− 2 3 µ J 2 Rearth 2 2( 1 − e 2 ) a7/2 5   2 −6  5 2  sin i − 2 = −1.0789 × 10  sin 50° − 2 2 2 = 5.7599 × 10−7 rad s=3.2945 × 10-5 deg s ∴ ω 39 = ω1 + ω∆t = 60° + 3.2945 × 10-5 ⋅ 259 200 = 68.539° Update the transformation matrix between perifocal and geocentric equatorial coordinates: cos Ω39 cos ω 39 − sin Ω39 sin ω 39 cos i [Q]xX = sin Ω39 cos ω 39 + cos Ω39 cos i sin ω 39  sin i sin ω − cos Ω39 sin ω 39 − sin Ω39 cos i cos ω 39 − sin Ω39 sin ω 39 + cos Ω39 cos i cos ω 39 sin i cos ω 39 sin Ω39 sin i   − cos Ω39 sin i   cos i  −0.33192 −0.672 58 0.66141    =  0.6177 −0.684 89 −0.386 48   0.712 94 0.280 27 0.64278  Compute the geocentric equatorial state vector at t f :  4596  ˆ ( km ) {r}X = [Q ]xX {r}x =  5759  ( km ) or r = 4596Iˆ + 5759 Jˆ − 1266.5K −1266.5    −3.6014  {v}X = [Q ]xX {v}x =  3.1794  ( km s) or v = −3.6014Iˆ + 3.1794 Jˆ + 5.6174K̂ˆ ( km s)  5.6174    50 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Problem 5.1 The following MATLAB script uses the given data to compute v 2 by means of Algorithm 5.1, which is implemented as the M-function gibbs in Appendix D.10. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_01 % ~~~~~~~~~~~~ % % This program uses Algorithm 5.1 (Gibbs method) to obtain the state % vector from the three coplanar position vectors provided in % Problem 5.1. % % mu - gravitational parameter (km^3/s^2) % r1, r2, r3 - three coplanar geocentric position vectors (km) % ierr - 0 if r1, r2, r3 are found to be coplanar % 1 otherwise % v2 - the velocity corresponding to r2 (km/s) % % User M-function required: gibbs % -------------------------------------------------------------------clear global mu mu = 398600; r1 = [5887 -3520 -1204]; r2 = [5572 -3457 -2376]; r3 = [5088 -3289 -3480]; %...Echo the input data to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.1: Gibbs Method\n') fprintf('\n Input data:\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n r1 (km) = [%g %g %g]', r1(1), r1(2), r1(3)) fprintf('\n r2 (km) = [%g %g %g]', r2(1), r2(2), r2(3)) fprintf('\n r3 (km) = [%g %g %g]', r3(1), r3(2), r3(3)) fprintf('\n\n'); %...Algorithm 5.1: [v2, ierr] = gibbs(r1, r2, r3); %...If the vectors r1, r2, r3, are not coplanar, abort: if ierr == 1 fprintf('\n These vectors are not coplanar.\n\n') return end %...Output the results to the command window: fprintf(' Solution:') fprintf('\n'); fprintf('\n v2 (km/s) = [%g %g %g]', v2(1), v2(2), v2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.1: Gibbs Method Input data: Gravitational parameter (km^3/s^2) 51 = 398600 Solutions Manual r1 (km) = [5887 r2 (km) = [5572 r3 (km) = [5088 Orbital Mechanics for Engineering Students -3520 -3457 -3289 Chapter 5 -1204] -2376] -3480] Solution: v2 (km/s) = [-2.50254 0.723248 -7.13125] ----------------------------------------------------- ˆ v 2 = −2.5025Iˆ + 0.72325 Jˆ − 7.1312K ( km/s ) Problem 5.2 The following MATLAB script uses r2 and v 2 from Problem 5.1 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_02 ~~~~~~~~~~~~ This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.1. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.2: r = [ 5572 -3457 -2376]; v = [-2.50254 0.723248 -7.13125]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.2: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... 52 Solutions Manual Orbital Mechanics for Engineering Students r(1), r(2), r(3)) = [%g %g %g]', ... v(1), v(2), v(3)) fprintf('\n v (km/s) disp(' ') fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n Chapter 5 Angular momentum (km^2/s) Eccentricity Right ascension (deg) Inclination (deg) Argument of perigee (deg) True anomaly (deg) Semimajor axis (km): = = = = = = = %g', %g', %g', %g', %g', %g', %g', coe(1)) coe(2)) coe(3)/deg) coe(4)/deg) coe(5)/deg) coe(6)/deg) coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.2: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [5572 -3457 -2376] = [-2.50254 0.723248 -7.13125] Angular momentum (km^2/s) = 52948.9 Eccentricity = 0.0127382 Right ascension (deg) = 150.003 Inclination (deg) = 95.0071 Argument of perigee (deg) = 151.688 True anomaly (deg) = 48.3093 Semimajor axis (km): = 7034.71 Period: Seconds = 5871.93 Minutes = 97.8655 Hours = 1.63109 Days = 0.0679622 ----------------------------------------------------- 52 949 2 h2 1 1 = = 6945.1 km µ 1 + e 398 600 1 + 0.012 738 = 6945.1 − 6378 = 567.11 km rperigee = zperigee Problem 5.3 ( ) As in Example 5.3, we set r1 = r1 iˆ and r2 = r2 cos ∆θ iˆ + sin ∆θ ˆj , where r1 = 6978 km , r2 = 6678 km and ∆θ = 60° . The following MATLAB script uses this data to compute v1 and v 2 by means of Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem 5_03a 53 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % ~~~~~~~~~~~~~ % % This program uses Algorithm 5.2 to solve Lambert's problem for the % data provided in Problem 5.3. % % % % % % % % % % deg pi mu r1, r2 dt dtheta R1, R2 string - factor for converting between degrees and radians 3.1415926... gravitational parameter (km^3/s^2) initial and final radii (km) time between r1 and r2 (s) change in true anomaly during dt (degrees) initial and final position vectors (km) = 'pro' if the orbit is prograde = 'retro if the orbit is retrograde V1, V2 - initial and final velocity vectors (km/s) % User M-function required: lambert % ----------------------------------------------------------clear global mu mu = 398600; %km^3/s^2 deg = pi/180; r1 r2 dt dtheta = = = = 6378 + 600; 6378 + 300; 15*60; 60; %km %km %sec %degrees R1 = [r1 0 0]; R2 = [r2*cos(dtheta*deg) r2*sin(dtheta*deg) 0]; %...Algorithm 5.2: string = 'pro'; [V1 V2] = lambert(R1, R2, dt, string); %...Echo the input data and output results to the command window: fprintf('\n-----------------------------------------------------') fprintf('\n Problem 5.3: Lambert''s Problem\n') fprintf('\n Input data:\n'); fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n Radius 1 (km) = %g', r1) fprintf('\n Position vector R1 (km) = [%g %g %g]\n',... R1(1), R1(2), R1(3)) fprintf('\n Radius 2 (km) = %g', r2) fprintf('\n Position vector R2 (km) = [%g %g %g]\n',... R2(1), R2(2), R2(3)) fprintf('\n Elapsed time (s) = %g', dt) fprintf('\n Change in true anomaly (deg) = %g', dtheta) fprintf('\n\n Solution:\n') fprintf('\n Velocity vector V1 (km/s) = [%g %g %g]',... V1(1), V1(2), V1(3)) fprintf('\n Velocity vector V2 (km/s) = [%g %g %g]',... V2(1), V2(2), V2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.3: Lambert's Problem Input data: 54 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Gravitational parameter (km^3/s^2) = 398600 Radius 1 (km) Position vector R1 (km) = 6978 = [6978 0 Radius 2 (km) Position vector R2 (km) = 6678 = [3339 5783.32 0] 0] Elapsed time (s) = 900 Change in true anomaly (deg) = 60 Solution: Velocity vector V1 (km/s) = [-0.544135 7.68498 0] Velocity vector V2 (km/s) = [-6.98129 3.96849 0] ----------------------------------------------------To find the perigee altitude, we need the orbital elements. The following MATLAB script uses r = 6978iˆ ( km ) and v = −0.544135iˆ + 7.68498 ˆj ( km/s ) from above to compute the orbital elements 1 1 by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_03b ~~~~~~~~~~~~ This program employs Algorithm 4.1 to obtain the orbital elements from the state vector found from the solution of Lambert's problem using the data given in Problem 5.3. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.3: r = [6978 0 0]; v = [-0.544135 7.68498 0]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: 55 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 fprintf('-----------------------------------------------------') fprintf('\n Problem 5.3: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.3: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [6978 0 0] = [-0.544135 7.68498 0] Angular momentum (km^2/s) = 53625.8 Eccentricity = 0.0806743 Right ascension (deg) = 0 Inclination (deg) = 0 Argument of perigee (deg) = 0 True anomaly (deg) = 294.849 Semimajor axis (km): = 7261.83 Period: Seconds = 6158.57 Minutes = 102.643 Hours = 1.71071 Days = 0.0712798 ----------------------------------------------------- 53 626 2 h2 1 1 = = 6676 km µ 1 + e 398 600 1 + 0.0806743 = 6676 − 6378 = 298 km rperigee = zperigee Problem 5.4 The following MATLAB script uses r1 , r2 and ∆t to compute v1 and v 2 by means of Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the MATLAB Command Window is listed afterwards. 56 Solutions Manual % % % % % % % % % % % % % % % % % Orbital Mechanics for Engineering Students Chapter 5 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_04 ~~~~~~~~~~~~ This program uses Algorithm 5.2 to solve Lambert's problem for the data provided in Problem 5.4. mu r1, r2 dt string - gravitational parameter (km^3/s^2) initial and final position vectors (km) time between r1 and r2 (s) = 'pro' if the orbit is prograde = 'retro if the orbit is retrograde v1, v2 - initial and final velocity vectors (km/s) coe - orbital elements [h e RA incl w TA a] User M-function required: lambert ----------------------------------------------------------- clear global mu deg = pi/180; %...Data mu = r1 = r2 = dt = string = %... declaration for Problem 5.4: 398600; [-3600 3600 5100]; [-5500 -6240 -520]; 30*60; 'pro'; %...Algorithm 5.2: [v1, v2] = lambert(r1, r2, dt, string); %...Echo the input data and output the results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.4: Lambert''s Problem\n') fprintf('\n Input data:\n'); fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu); fprintf('\n r1 (km) = [%g %g %g]', ... r1(1), r1(2), r1(3)) fprintf('\n r2 (km) = [%g %g %g]', ... r2(1), r2(2), r2(3)) fprintf('\n Elapsed time (s) = %g', dt); fprintf('\n\n Solution:\n') fprintf('\n v1 (km/s) = [%g %g %g]', ... v1(1), v1(2), v1(3)) fprintf('\n v2 (km/s) = [%g %g %g]', ... v2(1), v2(2), v2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.4: Lambert's Problem Input data: Gravitational parameter (km^3/s^2) = 398600 r1 (km) = [-3600 r2 (km) = [-5500 Elapsed time (s) = 1800 3600 5100] -6240 -520] 57 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Solution: v1 (km/s) = [-5.68521 -5.19833 0.348733] v2 (km/s) = [3.42204 -3.24131 -4.71994] ----------------------------------------------------- ˆ ( km/s ) v = 3.4220Iˆ − 3.2413 Jˆ − 4.7199K ˆ ( km/s ) v1 = −5.6852Iˆ − 5.1983 Jˆ + 0.348 73K 2 Problem 5.5 The following MATLAB script uses r1 = −3600Iˆ + 3600Jˆ + 5100Kˆ ( km ) and ˆ ( km/s ) from Problem 5.4 to compute the orbital elements by v1 = −5.6852Iˆ − 5.1983 Jˆ + 0.348 73K means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_05 ~~~~~~~~~~~~ This program employs Algorithm 4.1 to obtain the orbital elements from the state vector found in Problem 5.4. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.5: r = [ -3600 3600 5100]; v = [-5.68521 -5.19833 0.348733]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.5: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... 58 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 v(1), v(2), v(3)) disp(' ') fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n Angular momentum (km^2/s) Eccentricity Right ascension (deg) Inclination (deg) Argument of perigee (deg) True anomaly (deg) Semimajor axis (km): = = = = = = = %g', %g', %g', %g', %g', %g', %g', coe(1)) coe(2)) coe(3)/deg) coe(4)/deg) coe(5)/deg) coe(6)/deg) coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.5: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [-3600 3600 5100] = [-5.68521 -5.19833 0.348733] Angular momentum (km^2/s) = 55458 Eccentricity = 0.0982445 Right ascension (deg) = 45.0287 Inclination (deg) = 45.0497 Argument of perigee (deg) = 46.034 True anomaly (deg) = 43.9458 Semimajor axis (km): = 7791.19 Period: Seconds = 6844.1 Minutes = 114.068 Hours = 1.90114 Days = 0.0792142 ----------------------------------------------------- 55 458 2 h2 1 1 = = 7025.7 km µ 1 + e 398 600 1 + 0.098 244 = 7025.7 − 6378 = 647.74 km rperigee = zperigee Problem 5.6 The following MATLAB script uses r1 , r2 and ∆t to compute v1 and v 2 by means of Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the MATLAB Command Window is listed afterwards. (a) % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_06 % ~~~~~~~~~~~~ % % This program uses Algorithm 5.2 to solve Lambert's problem for the % data provided in Problem 5.6. % 59 Solutions Manual % % % % % % % % % % mu r1, r2 dt string Orbital Mechanics for Engineering Students Chapter 5 - gravitational parameter (km^3/s^2) initial and final position vectors (km) time between r1 and r2 (s) = 'pro' if the orbit is prograde = 'retro if the orbit is retrograde v1, v2 - initial and final velocity vectors (km/s) coe - orbital elements [h e RA incl w TA a] User M-function required: lambert ----------------------------------------------------------- clear global mu deg = pi/180; %...Data mu = r1 = r2 = dt = string = %... declaration for Problem 5.6: 398600; [ 5644 -2830 -4170]; [-2240 7320 -4980]; 20*60; 'pro'; %...Algorithm 5.2: [v1, v2] = lambert(r1, r2, dt, string); %...Echo the input data and output the results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.6: Lambert''s Problem\n') fprintf('\n Input data:\n'); fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu); fprintf('\n r1 (km) = [%g %g %g]', ... r1(1), r1(2), r1(3)) fprintf('\n r2 (km) = [%g %g %g]', ... r2(1), r2(2), r2(3)) fprintf('\n Elapsed time (s) = %g', dt); fprintf('\n\n Solution:\n') fprintf('\n v1 (km/s) = [%g %g %g]', ... v1(1), v1(2), v1(3)) fprintf('\n v2 (km/s) = [%g %g %g]', ... v2(1), v2(2), v2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.6: Lambert's Problem Input data: Gravitational parameter (km^3/s^2) = 398600 r1 (km) = [5644 -2830 r2 (km) = [-2240 7320 Elapsed time (s) = 1200 -4170] -4980] Solution: v1 (km/s) = [-4.13223 9.01237 -4.3781] v2 (km/s) = [-7.28524 6.31978 2.5272] ----------------------------------------------------- 60 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 ˆ ( km/s ) v = −7.2852Iˆ + 6.3198 Jˆ + 2.5272K ˆ ( km/s ) v1 = −4.1322Iˆ + 9.0124 Jˆ − 4.3781K 2 Problem 5.7 The following MATLAB script uses r1 = 5644Iˆ − 2830Jˆ − 4170Kˆ ( km ) and ˆ ( km/s ) from Problem 5.4 to compute the orbital elements by means v1 = −4.1322Iˆ + 9.0124 Jˆ − 4.3781K of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_07 ~~~~~~~~~~~~ This program employs Algorithm 4.1 to obtain the orbital elements from the state vector found in Problem 5.6. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.7: r = [ 5644 -2830 -4170]; v = [-4.13223 9.01237 -4.3781]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.7: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) 61 Solutions Manual Orbital Mechanics for Engineering Students fprintf('\n Semimajor axis (km): Chapter 5 = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.7: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [5644 -2830 -4170] = [-4.13223 9.01237 -4.3781] Angular momentum (km^2/s) = 76096.4 Eccentricity = 1.20053 Right ascension (deg) = 130.007 Inclination (deg) = 59.0184 Argument of perigee (deg) = 259.98 True anomaly (deg) = 320.023 Semimajor axis (km): = -32922.3 ----------------------------------------------------- 76 096 2 h2 1 1 = = 6601.8 km µ 1 + e 398 600 1 + 1.2005 = 6601.8 − 6378 = 223.82 km rperigee = zperigee Problem 5.8 The following MATLAB script uses the M-function J0 in Appendix D.12 to compute the Julian day for the date given in part (a) of Problem 5.8. The output to the MATLAB Command Window is listed afterwards, as are the results for the dates (b) through (c). % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_08a ~~~~~~~~~~~~~ This program computes J0 and the Julian day number using the data in Problem 5.8. year month day hour minute second ut j0 jd - range: 1901 - 2099 range: 1 - 12 range: 1 - 31 range: 0 - 23 (Universal Time) rage: 0 - 60 range: 0 - 60 universal time (hr) Julian day number at 0 hr UT Julian day number at specified UT User M-function required: J0 -------------------------------------------------------------------- 62 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 clear %...Data year = month = day = declaration for Problem 5.8a: 1914; 8; 14; hour = 5; minute = 30; second = 00; %... ut = hour + minute/60 + second/3600; %...Equation 5.46: j0 = J0(year, month, day); %...Equation 5.47: jd = j0 + ut/24; %...Echo the input data and output the results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Example 5.8a: Julian day calculation\n') fprintf('\n Input data:\n'); fprintf('\n Year = %g', year) fprintf('\n Month = %g', month) fprintf('\n Day = %g', day) fprintf('\n Hour = %g', hour) fprintf('\n Minute = %g', minute) fprintf('\n Second = %g\n', second) fprintf('\n Julian day number = %11.3f', jd); fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Example 5.8a: Julian day calculation Input data: Year Month Day Hour Minute Second = = = = = = 1914 8 14 5 30 0 Julian day number = 2420358.729 ----------------------------------------------------Example 5.8b: Julian day calculation Input data: Year Month Day Hour Minute Second = = = = = = 1946 4 18 14 0 0 Julian day number = 2431929.083 ----------------------------------------------------Example 5.8c: Julian day calculation 63 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Input data: Year Month Day Hour Minute Second = = = = = = 2010 9 1 0 0 0 Julian day number = 2455440.500 ----------------------------------------------------Example 5.8d: Julian day calculation Input data: Year Month Day Hour Minute Second = = = = = = 2007 10 16 12 0 0 Julian day number = 2454390.000 ----------------------------------------------------- Problem 5.9 This is similar to Example 5.5. The MATLAB script listed in the solution to Problem 5.8 can be used to obtain the Julian day numbers. Problem 5.10 The following MATLAB script uses Algorithm 5.3, which is implemented in MATLAB by the M-function LST in Appendix D.13, to compute the local sidereal time for the data given in part (a) of Problem 5.10. The output to the MATLAB Command Window is listed afterwards, as are the results for the data in (b) through (e). % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_10a ~~~~~~~~~~~~~ This program uses Algorithm 5.3 to obtain the local sidereal time from the data provided in Problem 5.10. lst EL - local sidereal time (degrees) - east longitude of the site (west longitude is negative): degrees (0 - 360) minutes (0 - 60) seconds (0 - 60) WL - west longitude year - range: 1901 - 2099 month - range: 1 - 12 day - range: 1 - 31 ut - universal time hour (0 - 23) minute (0 - 60) second (0 - 60) User M-function required: LST -------------------------------------------------------------------- clear %...Data declaration for Problem 5.10a: 64 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % East longitude: degrees = 18; minutes = 3; seconds = 0; % Date: year = 2008; month = 1; day = 1; % Universal time: hour = 12; minute = 0; second = 0; %... %...Convert negative (west) longitude to east longitude: if degrees < 0 degrees = degrees + 360; end %...Express the longitudes as decimal numbers: EL = degrees + minutes/60 + seconds/3600; WL = 360 - EL; %...Express universal time as a decimal number: ut = hour + minute/60 + second/3600; %...Algorithm 5.3: lst = LST(year, month, day, ut, EL); %...Echo the input data and output the results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.10a: Local sidereal time calculation\n') fprintf('\n Input data:\n'); fprintf('\n Year = %g', year) fprintf('\n Month = %g', month) fprintf('\n Day = %g', day) fprintf('\n UT (hr) = %g', ut) fprintf('\n West Longitude (deg) = %g', WL) fprintf('\n East Longitude (deg) = %g', EL) fprintf('\n\n'); fprintf(' Solution:') fprintf('\n'); fprintf('\n Local Sidereal Time (deg) = %g', lst) fprintf('\n Local Sidereal Time (hr) = %g', lst/15) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.10a: Local sidereal time calculation Input data: Year Month Day UT (hr) West Longitude (deg) East Longitude (deg) = = = = = = 2008 1 1 12 341.95 18.05 65 Solutions Manual Orbital Mechanics for Engineering Students Solution: Local Sidereal Time (deg) = 298.572 Local Sidereal Time (hr) = 19.9048 ----------------------------------------------------Problem 5.10b: Local sidereal time calculation Input data: Year Month Day UT (hr) West Longitude (deg) East Longitude (deg) = = = = = = 2007 12 21 10 215.033 144.967 Solution: Local Sidereal Time (deg) = 24.5646 Local Sidereal Time (hr) = 1.63764 ----------------------------------------------------Problem 5.10c: Local sidereal time calculation Input data: Year Month Day UT (hr) West Longitude (deg) East Longitude (deg) = = = = = = 2005 7 4 20 118.25 241.75 Solution: Local Sidereal Time (deg) = 104.676 Local Sidereal Time (hr) = 6.9784 ----------------------------------------------------Problem 5.10d: Local sidereal time calculation Input data: Year Month Day UT (hr) West Longitude (deg) East Longitude (deg) = = = = = = 2006 2 15 3 43.1 316.9 Solution: Local Sidereal Time (deg) = 146.884 Local Sidereal Time (hr) = 9.79228 ----------------------------------------------------Problem 5.10e: Local sidereal time calculation Input data: Year Month Day UT (hr) West Longitude (deg) East Longitude (deg) = = = = = = 2006 3 21 8 228.067 131.933 66 Chapter 5 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Solution: Local Sidereal Time (deg) = 70.6348 Local Sidereal Time (hr) = 4.70899 ----------------------------------------------------- Problem 5.11. θ = 117° φ = 51° Α = 28° a = 68° . From Equation 5.83a, δ = sin −1 (cos φ cos A cos a + sin φ sin a )= sin −1 (cos 51° cos 28° cos 68° + sin 117° sin 68°) δ = 68.235° From Equation 5.83b, since A < 180° , h = 360° − cos −1 = 360° − cos −1  cos φ sin a − sin φ cos A cos a    cos δ  cos 51° sin 68° − sin 51° cos 28° cos 68°    cos 68.235° = 331.69° From Equation 5.83c α = θ − h = 117° − 331.69° = −214.69 Placing this within the range 0 ≤ θ ≤ 360° , α = 145.31° Problem 5.12 The following MATLAB script uses Algorithm 5.4, which is implemented in MATLAB by the M-function rv_from_observe in Appendix D.14, to compute the state vector of a space object from the data given in Problem 5.12. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_12 ~~~~~~~~~~~~ This program uses Algorithm 5.4 to obtain the state vector from the observational data provided in Problem 5.12. deg pi mu - conversion factor between degrees and radians - 3.1415926... - gravitational parameter (km^3/s^2) Re f wE omega - equatorial radius of the earth (km) earth's flattening factor angular velocity of the earth (rad/s) earth's angular velocity vector (rad/s) in the geocentric equatorial frame - slant range of object (km) range rate (km/s) azimuth (deg) of object relative to observation site time rate of change of azimuth (deg/s) elevation angle (deg) of object relative to observation % rho % rhodot % A % Adot % a site 67 Solutions Manual Orbital Mechanics for Engineering Students % adot - time rate of change of elevation angle (degrees/s) % theta % phi % H - local sidereal time (deg) of tracking site - geodetic latitude (deg) of site - elevation of site (km) % r % v - geocentric equatorial position vector of object (km) - geocentric equatorial velocity vector of object (km) % % % % % % % % % % % % % % coe Chapter 5 - orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = inclination of the orbit (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - perigee radius (km) - period of elliptical orbit (s) rp T User M-function required: rv_from_observe -------------------------------------------------------------------- clear global deg f Re wE mu f Re wE mu = = = = = %...Data rho = rhodot = A = Adot = a = adot = theta = phi = H = %... pi/180; 0.0033528; 6378; 7.2921e-5; 398600; declaration for Problem 5.12: 988; 4.86; 36; 0.59; 36.6; -0.263; 40; 35; 0; %...Algorithm 5.4: [r,v] = rv_from_observe(rho, rhodot, A, Adot, a, adot, theta, phi, H); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.12') fprintf('\n\n Input data:\n'); fprintf('\n Slant range (km) = %g', rho); fprintf('\n Slant range rate (km/s) = %g', rhodot); fprintf('\n Azimuth (deg) = %g', A); fprintf('\n Azimuth rate (deg/s) = %g', Adot); fprintf('\n Elevation (deg) = %g', a); fprintf('\n Elevation rate (deg/s) = %g', adot); fprintf('\n Local sidereal time (deg) = %g', theta); fprintf('\n Latitude (deg) = %g', phi); fprintf('\n Altitude above sea level (km) = %g', H); fprintf('\n\n'); 68 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 fprintf(' Solution:') fprintf('\n\n State vector:\n'); fprintf('\n r (km) = [%g, %g, %g]', ... r(1), r(2), r(3)); fprintf('\n v (km/s) = [%g, %g, %g]', ... v(1), v(2), v(3)); fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.12 Input data: Slant range (km) Slant range rate (km/s) Azimuth (deg) Azimuth rate (deg/s) Elevation (deg) Elevation rate (deg/s) Local sidereal time (deg) Latitude (deg) Altitude above sea level (km) = = = = = = = = = 988 4.86 36 0.59 36.6 -0.263 40 35 0 Solution: State vector: r (km) = [3794.66, 3792.71, 4501.31] v (km/s) = [-7.72483, 7.72134, 0.0186586] ----------------------------------------------------- ˆ ( km ) r = 3794.7Iˆ + 3792.7 Jˆ + 4501.3K ˆ ( km/s ) v = −7.7248Iˆ + 7.72134 Jˆ + 0.018 659K Problem 5.13 The following MATLAB script uses the state vector found in Problem 5.12 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_13 ~~~~~~~~~~~~ This program employs Algorithm 4.1 to obtain the orbital elements from the state vector found in Problem 5.12. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) 69 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % User M-function required: coe_from_sv % -------------------------------------------------------------------clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.13: r = [ 3794.66, 3792.71, 4501.31]; v = [-7.72483, 7.72134, 0.0186586]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.13: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.13: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [3794.66 3792.71 4501.31] = [-7.72483 7.72134 0.0186586] Angular momentum (km^2/s) = 76490.5 Eccentricity = 1.09593 Right ascension (deg) = 315.13 Inclination (deg) = 39.9968 Argument of perigee (deg) = 89.8097 True anomaly (deg) = 0.0797759 Semimajor axis (km): = -73003.5 ----------------------------------------------------- 70 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Problem 5.14 The local sidereal time θ , azimuth A, angular elevation a and slant range ρ are provided at three observation times. The rates are not provided, but we can still use Algorithm 5.4, implemented in MATLAB as rv_from_observe in Appendix D.14, to find just the position vectors at each of the times. The following MATLAB script carries out this procedure, passing zeros to rv_from_observe as values for the rates. The output to the MATLAB Command Window follows. % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_10a ~~~~~~~~~~~~~ This program uses Algorithm 5.4 to find the geocentric position vectors corresponding to the three sets of azimuth, elevation and slant range data given in Problem 5.14 % deg % pi - conversion factor between degrees and radians - 3.1415926... % Re % f % wE % - equatorial radius of the earth (km) - earth's flattening factor - angular velocity of the earth (rad/s) (not required in this problem) % % % % % % % - vector of three observation times (min) - vector of slant ranges (km) of the object at the three observation times - vector of azimuths (deg) of the object relative to the observation site at the three observation times - vector of elevation angles (deg) of the object relative to the observation site at the three observation times t rho az el % theta at % % phi % H - vector of local sidereal times (deg) of the tracking site % r % v % given) - geocentric equatorial position vector of object (km) - geocentric equatorial velocity vector of object (km) (not computed since the rates of rho, az and el are not the three observation times - geodetic latitude (deg) of site - elevation of site (km) % User M-function required: rv_from_observe % -------------------------------------------------------------------clear global f Re wE deg Re f wE = = = = pi/180; 6378; 0.0033528; 7.292115e-5; %...Data declaration for Problem 5.14: phi = -20; H = 0.5; t = [0 2 4]; theta = [60.0 60.5014 61.0027]; az = [165.931 145.967 2.40962]; el = [9.53549 45.7711 21.8825]; rho = [1214.89 421.441 732.079]; %... 71 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 %...Echo the input data to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.14') fprintf('\n\n Input data (angles in degrees):\n'); fprintf('\n Time') fprintf('\n (min) Azimuth Elevation Slant range\n') for i = 1:3 fprintf('\n %5.1f%15.5e%15.5e%15.5e',t(i), az(i), el(i), rho(i)) end %...Output the solution to the command window: fprintf('\n\n Solution:') fprintf('\n\n Time') fprintf('\n (min) Geocentric position vector (km)\n') for i = 1:3 %...Algorithm 5.4: [r,v] = rv_from_observe(rho(i), 0, az(i), 0, el(i), ... 0, theta(i), phi, H); fprintf('\n %5.1f [%g %g %g]',t(i), r(1), r(2), r(3)) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.14 Input data (angles in degrees): Time (min) 0.0 2.0 4.0 Azimuth 1.65931e+02 1.45967e+02 2.40962e+00 Elevation Slant range 9.53549e+00 4.57711e+01 2.18825e+01 1.21489e+03 4.21441e+02 7.32079e+02 Solution: Time (min) Geocentric position vector (km) 0.0 [2641.68 5158.02 -3328.73] 2.0 [2908.04 5474.36 -2500.03] 4.0 [3118.6 5685.65 -1623.34] ----------------------------------------------------- ˆ ( km ) r1 = 2641.7Iˆ + 5158.0 Jˆ − 3328.7K ˆ ( km ) r = 2908.0Iˆ + 5474.4 Jˆ − 2500.0K 2 ˆ ( km ) r3 = 3118.6Iˆ + 5685.6 Jˆ − 1623.3K Using these three position vectors we employ Gibbs’ method, Algrorithm 5.1, which is implemented in MATLAB as the M-function gibbs in Appendix D.10. The following MATLAB script calls upon gibbs to find the velocity vector v 2 corresponding to the position vector r2 . The output is listed afterward. % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_14b ~~~~~~~~~~~~~ This program uses Algorithm 5.1 (Gibbs method) to obtain the state 72 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % vector from the three coplanar position vectors found in the first % part of Problem 5.14. % % mu - gravitational parameter (km^3/s^2) % r1, r2, r3 - three coplanar geocentric position vectors (km) % ierr - 0 if r1, r2, r3 are found to be coplanar % 1 otherwise % v2 - the velocity corresponding to r2 (km/s) % % User M-function required: gibbs % -------------------------------------------------------------------clear global mu mu = 398600; r1 = [2641.68 5158.02 -3328.73]; r2 = [2908.04 5474.36 -2500.03]; r3 = [3118.6 5685.65 -1623.34]; %...Echo the input data to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.14: Gibbs Method\n') fprintf('\n Input data:\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n r1 (km) = [%g %g %g]', r1(1), r1(2), r1(3)) fprintf('\n r2 (km) = [%g %g %g]', r2(1), r2(2), r2(3)) fprintf('\n r3 (km) = [%g %g %g]', r3(1), r3(2), r3(3)) fprintf('\n\n'); %...Algorithm 5.1: [v2, ierr] = gibbs(r1, r2, r3); %...If the vectors r1, r2, r3, are not coplanar, abort: if ierr == 1 fprintf('\n These vectors are not coplanar.\n\n') return end %...Output the results to the command window: fprintf(' Solution:') fprintf('\n'); fprintf('\n v2 (km/s) = [%g %g %g]', v2(1), v2(2), v2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.14: Gibbs Method Input data: Gravitational parameter (km^3/s^2) = 398600 r1 (km) = [2641.68 5158.02 -3328.73] r2 (km) = [2908.04 5474.36 -2500.03] r3 (km) = [3118.6 5685.65 -1623.34] Solution: v2 (km/s) = [1.99357 2.20552 7.12881] ----------------------------------------------------- ˆ ( km/s ) v 2 = 1.9936Iˆ + 2.2055Jˆ + 7.1288K 73 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Problem 5.15 The following MATLAB script uses r2 and v 2 from Problem 5.14 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_15 ~~~~~~~~~~~~ This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.14. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.15: r = [2908.04 5474.36 -2500.03]; v = [1.99357 2.20552 7.12881]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.15: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 74 Solutions Manual Orbital Mechanics for Engineering Students T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds fprintf('\n Minutes fprintf('\n Hours fprintf('\n Days = = = = %g', %g', %g', %g', Chapter 5 T) T/60) T/3600) T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.15: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [2908.04 = [1.99357 5474.36 2.20552 -2500.03] 7.12881] Angular momentum (km^2/s) = 51626.3 Eccentricity = 0.00102595 Right ascension (deg) = 60.0001 Inclination (deg) = 95.0003 Argument of perigee (deg) = 270.34 True anomaly (deg) = 67.6075 Semimajor axis (km): = 6686.59 Period: Seconds = 5441.5 Minutes = 90.6916 Hours = 1.51153 Days = 0.0629803 ----------------------------------------------------- Problems 5.16 and 5.17 The following MATLAB script uses Equations 5.55 and 5.56 to convert the data given in Problem 5.16 into three tracking site position vectors ( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρ̂ ρ1 , ρ̂ρ2 , ρ̂ρ3 ). These vectors together with the three observation times are then handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss Algorithm 5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which iteratively improves it. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_16 ~~~~~~~~~~~~ This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute the state vector from the angles only data provided in Problem 5.16. deg pi mu Re f H phi t ra dec theta R - factor for converting between degrees and radians 3.1415926... gravitational parameter (km^3/s^2) earth's equatorial radius (km) earth's flattening factor elevation of observation site (km) latitude of site (deg) vector of observation times t1, t2, t3 (s) vector of topocentric equatorial right ascensions at t1, t2, t3 (deg) - vector of topocentric equatorial right declinations at t1, t2, t3 (deg) - vector of local sidereal times for t1, t2, t3 (deg) - matrix of site position vectors at t1, t2, t3 (km) 75 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % rho - matrix of direction cosine vectors at t1, t2, t3 % fac1, fac2 - common factors % r_old, v_old - the state vector without iterative improvement (km, km/s) % r, v - the state vector with iterative improvement (km, km/s) % % User M-function required: gauss % -------------------------------------------------------------------clear global mu deg mu Re f = = = = pi/180; 398600; 6378; 1/298.26; %...Data declaration for Problen 5.16: H = 0; phi = 29*deg; t = [ 0 60 120 ]; ra = [ 0 6.59279e+01 7.98500e+01]*deg; dec = [5.15110e+01 2.79911e+01 1.46609e+01]*deg; theta = [ 0 2.50684e-01 5.01369e-01]*deg; %... %...Equations 5.56, 5.57: fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2); fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi); for i = 1:3 R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i)); R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i)); R(i,3) = fac2; rho(i,1) = cos(dec(i))*cos(ra(i)); rho(i,2) = cos(dec(i))*sin(ra(i)); rho(i,3) = sin(dec(i)); end %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problems 5.16 and 5.17: Orbit determination') fprintf('\n by the Gauss method\n') fprintf('\n Radius of earth (km) = %g', Re) fprintf('\n Flattening factor = %g', f) fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu) fprintf('\n\n Input data:\n'); fprintf('\n Latitude (deg) of tracking site = %g', phi/deg); fprintf('\n Altitude (km) above sea level = %g', H); fprintf('\n\n Observations:') fprintf('\n Right') fprintf(' Local') fprintf('\n Time (s) Ascension (deg) Declination (deg)') fprintf(' Sidereal time (deg)') for i = 1:3 fprintf('\n %9.4g %11.4f %19.4f %20.4f', ... 76 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg) end fprintf('\n\n Solution:\n') fprintf('\n Without iterative improvement (Problem 5.16)...\n') fprintf('\n r (km) = [%g, %g, %g]', r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3)) fprintf('\n'); fprintf('\n\n With iterative improvement (Problem 5.17)...\n') fprintf('\n r (km) = [%g, %g, %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problems 5.16 and 5.17: Orbit determination by the Gauss method Radius of earth (km) = 6378 Flattening factor = 0.00335278 Gravitational parameter (km^3/s^2) = 398600 Input data: Latitude (deg) of tracking site = 29 Altitude (km) above sea level = 0 Observations: Time (s) (deg) 0 60 120 Right Ascension (deg) Declination (deg) 0.0000 65.9279 79.8500 Local Sidereal time 51.5110 27.9911 14.6609 0.0000 0.2507 0.5014 Solution: Without iterative improvement (Problem 5.16)... r (km) = [5788.09, 484.257, 3341.52] v (km/s) = [-0.460072, 8.05816, -0.265618] With iterative improvement (Problem 5.17)... r (km) = [5788.42, 485.007, 3341.96] v (km/s) = [-0.460926, 8.0706, -0.266112] ----------------------------------------------------- Problem 5.18 The following MATLAB script uses r2 and v 2 from Problem 5.17 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_18 ~~~~~~~~~~~~ This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.17. pi - 3.1415926... 77 Solutions Manual % % % % % % % % % % % % % % % % deg mu r v coe T Orbital Mechanics for Engineering Students Chapter 5 - factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.18: r = [ 5788.42, 485.007, 3341.96]; v = [-0.460926, 8.0706, -0.266112]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.18: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') Problem 5.18: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 78 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 State vector: r (km) v (km/s) = [5788.42 485.007 3341.96] = [-0.460926 8.0706 -0.266112] Angular momentum (km^2/s) = 54201.2 Eccentricity = 0.100054 Right ascension (deg) = 270 Inclination (deg) = 30.0001 Argument of perigee (deg) = 89.9993 True anomaly (deg) = 4.15098 Semimajor axis (km): = 7444.75 Period: Seconds = 6392.73 Minutes = 106.546 Hours = 1.77576 Days = 0.07399 ----------------------------------------------------- Problems 5.19 The following MATLAB script uses Equations 5.55 and 5.56 to convert the data given in Problem 5.19 into three tracking site position vectors ( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 ,ρˆ2 ,ρˆ3 ) . These vectors together with the three observation times are then handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss Algorithm 5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which iteratively improves it. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_19 % ~~~~~~~~~~~~ % % This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute % the state vector from the angles only data provided in Problem 5.16. % % deg - factor for converting between degrees and radians % pi - 3.1415926... % mu - gravitational parameter (km^3/s^2) % Re - earth's equatorial radius (km) % f - earth's flattening factor % H - elevation of observation site (km) % phi - latitude of site (deg) % t - vector of observation times t1, t2, t3 (s) % ra - vector of topocentric equatorial right ascensions % at t1, t2, t3 (deg) % dec - vector of topocentric equatorial right declinations % at t1, t2, t3 (deg) % theta - vector of local sidereal times for t1, t2, t3 (deg) % R - matrix of site position vectors at t1, t2, t3 (km) % rho - matrix of direction cosine vectors at t1, t2, t3 % fac1, fac2 - common factors % r_old, v_old - the state vector without iterative improvement (km, km/s) % r, v - the state vector with iterative improvement (km, km/s) % % User M-function required: gauss % ------------------------------------------------------------------- 79 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 clear global mu deg mu Re f = = = = pi/180; 398600; 6378; 1/298.26; %...Data declaration for Problem 5.19: H = 0; phi = 29*deg; t = [ 0 60 120]; ra = [15.0394 25.7539 48.6055]*deg; dec = [20.7487 30.1410 43.8910]*deg; theta = [ 90 90.2507 90.5014]*deg; %... %...Equations 5.56, 5.57: fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2); fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi); for i = 1:3 R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i)); R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i)); R(i,3) = fac2; rho(i,1) = cos(dec(i))*cos(ra(i)); rho(i,2) = cos(dec(i))*sin(ra(i)); rho(i,3) = sin(dec(i)); end %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problems 5.19 and 5.20: Orbit determination') fprintf('\n by the Gauss method\n') fprintf('\n Radius of earth (km) = %g', Re) fprintf('\n Flattening factor = %g', f) fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu) fprintf('\n\n Input data:\n'); fprintf('\n Latitude (deg) of tracking site = %g', phi/deg); fprintf('\n Altitude (km) above sea level = %g', H); fprintf('\n\n Observations:') fprintf('\n Right') fprintf(' Local') fprintf('\n Time (s) Ascension (deg) Declination (deg)') fprintf(' Sidereal time (deg)') for i = 1:3 fprintf('\n %9.4g %11.4f %19.4f %20.4f', ... t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg) end fprintf('\n\n Solution:\n') fprintf('\n Without iterative improvement (Problem 5.19)...\n') fprintf('\n r (km) = [%g, %g, %g]', r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3)) fprintf('\n'); 80 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 fprintf('\n\n With iterative improvement (Problem 5.20)...\n') fprintf('\n r (km) = [%g, %g, %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3)) fprintf('\n-----------------------------------------------------\n') Problems 5.19 and 5.20: Orbit determination by the Gauss method Radius of earth (km) = 6378 Flattening factor = 0.00335278 Gravitational parameter (km^3/s^2) = 398600 Input data: Latitude (deg) of tracking site = 29 Altitude (km) above sea level = 0 Observations: Time (s) (deg) 0 60 120 Right Ascension (deg) Declination (deg) 15.0394 25.7539 48.6055 Local Sidereal time 20.7487 30.1410 43.8910 90.0000 90.2507 90.5014 Solution: Without iterative improvement (Problem 5.19)... r (km) = [765.19, 5963.6, 3582.88] v (km/s) = [-7.50919, 0.705265, 0.423729] With iterative improvement (Problem 5.20)... r (km) = [766.265, 5964.12, 3583.57] v (km/s) = [-7.51882, 0.706233, 0.42431] ----------------------------------------------------Approximate state vector: ˆ ( km ) r = 765.19Iˆ + 5963.60Jˆ + 3582.88K ˆ ( km/s ) v = −7.50919Iˆ + 0.705265Jˆ + 0.423729K Problem 5.20 From the MATLAB output in the previous problem solution, the refined state vector is ˆ ( km ) r = 766.265Iˆ + 5964.12Jˆ + 3583.57K ˆ ( km/s ) v = −7.518 82Iˆ + 0.706 233Jˆ + 0.424 31K Problem 5.21 The following MATLAB script uses r and v from Problem 5.20 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_21 ~~~~~~~~~~~~ This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.20. pi deg - 3.1415926... - factor for converting between degrees and radians 81 Solutions Manual % % % % % % % % % % % % % % % mu r v coe T Orbital Mechanics for Engineering Students Chapter 5 - gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.21: r = [ 766.265, 5964.12, 3583.57]; v = [-7.51882, 0.706233, 0.42431]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.21: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.21: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 82 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 State vector: r (km) v (km/s) = [766.265 5964.12 3583.57] = [-7.51882 0.706233 0.42431] Angular momentum (km^2/s) = 52946.7 Eccentricity = 0.00474691 Right ascension (deg) = 360 Inclination (deg) = 30.9997 Argument of perigee (deg) = 90.3282 True anomaly (deg) = 353.388 Semimajor axis (km): = 7033.16 Period: Seconds = 5869.98 Minutes = 97.833 Hours = 1.63055 Days = 0.0679396 ----------------------------------------------------- Problem 5.22 The following MATLAB script uses the given three tracking site position vectors ( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) together with the three observation times to find the state vector ( r , v) by means of Algorithm 5.5 and then iteratively improve it using Algorithm 5.6. Both algorithms are implemented in the MATLAB M-function gauss in Appendix D.15. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_22 % ~~~~~~~~~~~~ % % This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute % the state vector from the angles only data provided in Problem 5.16. % % deg - factor for converting between degrees and radians % pi - 3.1415926... % mu - gravitational parameter (km^3/s^2) % t - vector of observation times t1, t2, t3 (s) % theta - vector of local sidereal times for t1, t2, t3 (deg) % R - matrix of site position vectors at t1, t2, t3 (km) % rho - matrix of direction cosine vectors at t1, t2, t3 % r_old, v_old - the state vector without iterative improvement (km, km/s) % r, v - the state vector with iterative improvement (km, km/s) % % User M-function required: gauss % -------------------------------------------------------------------clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.22: t = [ 0 60 120]; R = [ -1825.96 -1841.63 -1857.25 3583.66 3575.63 3567.54 rho = [-0.301687 -0.793090 0.200673 -0.210324 4933.54 4933.54 4933.54]; 0.932049 0.571640 83 Solutions Manual Orbital Mechanics for Engineering Students -0.873085 -0.362969 Chapter 5 0.325539]; %... %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problems 5.22 and 5.23: Orbit determination') fprintf('\n by the Gauss method\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu) fprintf('\n\n Input data:\n'); fprintf('\n Site position vector (R) and space object') fprintf('\n direction cosine vector (rho) at three times:\n') for i = 1:3 fprintf('\n t = %g s:\n',t(i)) fprintf('\n R = [%g %g %g]', R(i,1), R(i,2), R(i,3)) fprintf('\n rho = [%g %g %g]', rho(i,1), rho(i,2), rho(i,3)) disp(' ') end fprintf('\n\n Solution:\n') fprintf('\n Without iterative improvement (Problem 5.22)...\n') fprintf('\n r (km) = [%g, %g, %g]', r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3)) fprintf('\n'); fprintf('\n\n With iterative improvement (Problem 5.23)...\n') fprintf('\n r (km) = [%g, %g, %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problems 5.22 and 5.23: Orbit determination by the Gauss method Gravitational parameter (km^3/s^2) = 398600 Input data: Site position vector (R) and space object direction cosine vector (rho) at three times: t = 0 s: R = [-1825.96 3583.66 4933.54] rho = [-0.301687 0.200673 0.932049] t = 60 s: R = [-1841.63 rho = [-0.79309 3575.63 4933.54] -0.210324 0.57164] t = 120 s: R = [-1857.25 3567.54 4933.54] rho = [-0.873085 -0.362969 0.325539] Solution: 84 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Without iterative improvement (Problem 5.22)... r (km) = [-2350.74, 3440.62, 5300.49] v (km/s) = [-6.61345, -3.88226, -0.413321] With iterative improvement (Problem 5.23)... r (km) = [-2351.59, 3440.39, 5301.1] v (km/s) = [-6.62403, -3.8885, -0.414013] ----------------------------------------------------Approximate state vector: ˆ ( km ) r = −2350.74Iˆ + 3440.62Jˆ + 5300.49K ˆ ( km/s ) v = −6.613 45Iˆ − 3.882 26Jˆ − 0.413 321K Problem 5.23 From the MATLAB output listed in the previous problem solution, the iteratively improved state vector is ˆ ( km ) r = −2351.59Iˆ + 3440.39Jˆ + 5301.1K ˆ ( km/s ) v = −6.624 03Iˆ − 3.8885Jˆ − 0.414 013K Problem 5.24 The following MATLAB script uses r and v from Problem 5.23 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_24 % ~~~~~~~~~~~~ % % This program uses Algorithm 4.1 to obtain the orbital % elements from the state vector obtained in Problem 5.23. % % pi - 3.1415926... % deg - factor for converting between degrees and radians % mu - gravitational parameter (km^3/s^2) % r - position vector (km) in the geocentric equatorial frame % v - velocity vector (km/s) in the geocentric equatorial frame % coe - orbital elements [h e RA incl w TA a] % where h = angular momentum (km^2/s) % e = eccentricity % RA = right ascension of the ascending node (rad) % incl = orbit inclination (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) % T - Period of an elliptic orbit (s) % % User M-function required: coe_from_sv % ------------------------------------------------------------------clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.24: r = [-2351.59, 3440.39, 5301.1]; v = [-6.62403, -3.8885, -0.414013]; 85 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.24: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.24: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [-2351.59 = [-6.62403 3440.39 -3.8885 5301.1] -0.414013] Angular momentum (km^2/s) = 51868.3 Eccentricity = 0.000957299 Right ascension (deg) = 28.0006 Inclination (deg) = 51.9999 Argument of perigee (deg) = 88.9231 True anomaly (deg) = 4.99817 Semimajor axis (km): = 6749.43 Period: Seconds = 5518.38 Minutes = 91.973 Hours = 1.53288 Days = 0.0638701 ----------------------------------------------------- Problems 5.25 The following MATLAB script uses Equations 5.55 and 5.56 to convert the data given in Problem 5.25 into three tracking site position vectors ( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) . These vectors together with the three observation times are then 86 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss Algorithm 5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which iteratively improves it. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_25 % ~~~~~~~~~~~~ % % This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute % the state vector from the angles only data provided in Problem 5.25. % % deg - factor for converting between degrees and radians % pi - 3.1415926... % mu - gravitational parameter (km^3/s^2) % Re - earth's equatorial radius (km) % f - earth's flattening factor % H - elevation of observation site (km) % phi - latitude of site (deg) % t - vector of observation times t1, t2, t3 (s) % ra - vector of topocentric equatorial right ascensions % at t1, t2, t3 (deg) % dec - vector of topocentric equatorial right declinations % at t1, t2, t3 (deg) % theta - vector of local sidereal times for t1, t2, t3 (deg) % R - matrix of site position vectors at t1, t2, t3 (km) % rho - matrix of direction cosine vectors at t1, t2, t3 % fac1, fac2 - common factors % r_old, v_old - the state vector without iterative improvement (km, km/s) % r, v - the state vector with iterative improvement (km, km/s) % % User M-function required: gauss % -------------------------------------------------------------------clear global mu deg mu Re f = = = = pi/180; 398600; 6378; 1/298.26; %...Data declaration for Problem 5.25: H = 0.5; phi = 60*deg; t = [ 0 300 600]; ra = [157.783 159.221 160.526]*deg; dec = [24.2403 27.2993 29.8982]*deg; theta = [ 150 151.253 152.507]*deg; %... %...Equations 5.56, 5.57: fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2); fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi); for i = 1:3 R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i)); R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i)); R(i,3) = fac2; rho(i,1) = cos(dec(i))*cos(ra(i)); 87 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 rho(i,2) = cos(dec(i))*sin(ra(i)); rho(i,3) = sin(dec(i)); end %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problems 5.25 and 5.26: Orbit determination') fprintf('\n by the Gauss method\n') fprintf('\n Radius of earth (km) = %g', Re) fprintf('\n Flattening factor = %g', f) fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu) fprintf('\n\n Input data:\n'); fprintf('\n Latitude (deg) of tracking site = %g', phi/deg); fprintf('\n Altitude (km) above sea level = %g', H); fprintf('\n\n Observations:') fprintf('\n Right') fprintf(' Local') fprintf('\n Time (s) Ascension (deg) Declination (deg)') fprintf(' Sidereal time (deg)') for i = 1:3 fprintf('\n %9.4g %11.4f %19.4f %20.4f', ... t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg) end fprintf('\n\n Solution:\n') fprintf('\n Without iterative improvement (Problem 5.25)...\n') fprintf('\n r (km) = [%g, %g, %g]', r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3)) fprintf('\n'); fprintf('\n\n With iterative improvement (Problem 5.26)...\n') fprintf('\n r (km) = [%g, %g, %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3)) fprintf('\n-----------------------------------------------------\n' ----------------------------------------------------Problems 5.25 and 5.26: Orbit determination by the Gauss method Radius of earth (km) = 6378 Flattening factor = 0.00335278 Gravitational parameter (km^3/s^2) = 398600 Input data: Latitude (deg) of tracking site = 60 Altitude (km) above sea level = 0.5 Observations: Time (s) (deg) 0 300 600 Right Ascension (deg) Declination (deg) 157.7830 159.2210 160.5260 24.2403 27.2993 29.8982 88 Local Sidereal time 150.0000 151.2530 152.5070 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 Solution: Without iterative improvement (Problem 5.25)... r (km) = [-19050.2, 7702.56, 14469.6] v (km/s) = [-3.27477, -0.482844, 5.07464] With iterative improvement (Problem 5.26)... r (km) = [-19081, 7714.25, 14486.6] v (km/s) = [-3.27846, -0.484358, 5.08206] ----------------------------------------------------Approximate state vector: ˆ ( km ) r = -19050.2Iˆ + 7702.56Jˆ + 14469.6K ˆ ( km/s ) v = −3.27477Iˆ − 0.482844Jˆ + 5.07464K Problem 5.26 From the MATLAB output listed in the previous problem solution, the iteratively improved state vector is ˆ ( km ) r = −19081Iˆ + 7714.25Jˆ + 14486.6K ˆ ( km/s ) v = −3.27846Iˆ − 0.484 358Jˆ + 5.082 06K Problem 5.27 The following MATLAB script uses r and v from Problem 5.26 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_5_27 ~~~~~~~~~~~~ This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.26. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv -------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.27: r = [ -19081, 7714.25, 14486.6]; v = [-3.27846, -0.484358, 5.08206]; %... 89 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.27: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.27: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [-19081 7714.25 14486.6] = [-3.27846 -0.484358 5.08206] Angular momentum (km^2/s) Eccentricity Right ascension (deg) Inclination (deg) Argument of perigee (deg) True anomaly (deg) Semimajor axis (km): = = = = = = = 76005.8 1.08937 136.949 62.9772 287.335 112.915 -77612.3 Problem 5.28 The following MATLAB script uses the given three tracking site position vectors ( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) together with the three observation times to find the state vector ( r , v) by means of Algorithm 5.5 and then iteratively improve it using Algorithm 5.6. Both algorithms are implemented in the MATLAB M-function gauss in Appendix D.15. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_28 % ~~~~~~~~~~~~ % 90 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 % This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute % the state vector from the angles only data provided in Problem 5.28. % % deg - factor for converting between degrees and radians % pi - 3.1415926... % mu - gravitational parameter (km^3/s^2) % t - vector of observation times t1, t2, t3 (s) % theta - vector of local sidereal times for t1, t2, t3 (deg) % R - matrix of site position vectors at t1, t2, t3 (km) % rho - matrix of direction cosine vectors at t1, t2, t3 % r_old, v_old - the state vector without iterative improvement (km, km/s) % r, v - the state vector with iterative improvement (km, km/s) % % User M-function required: gauss % -------------------------------------------------------------------clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.28: t = [ 0 300 600]; R = [ 5582.84 5581.50 5577.50 0 122.122 244.186 3073.90 3073.90 3073.90]; rho = [ 0.846428, 0.749290, 0.529447, 0, 0.463023, 0.777163, 0.532504 0.473470 0.340152]; %... %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problems 5.28 and 5.29: Orbit determination') fprintf('\n by the Gauss method\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu) fprintf('\n\n Input data:\n'); fprintf('\n Site position vector (R) and space object') fprintf('\n direction cosine vector (rho) at three times:\n') for i = 1:3 fprintf('\n t = %g s:\n',t(i)) fprintf('\n R = [%g %g %g]', R(i,1), R(i,2), R(i,3)) fprintf('\n rho = [%g %g %g]', rho(i,1), rho(i,2), rho(i,3)) disp(' ') end fprintf('\n\n Solution:\n') fprintf('\n Without iterative improvement (Problem 5.28)...\n') fprintf('\n r (km) = [%g, %g, %g]', r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3)) fprintf('\n'); fprintf('\n\n With iterative improvement (Problem 5.29)...\n') fprintf('\n r (km) = [%g, %g, %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3)) 91 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problems 5.28 and 5.29: Orbit determination by the Gauss method Gravitational parameter (km^3/s^2) = 398600 Input data: Site position vector (R) and space object direction cosine vector (rho) at three times: t = 0 s: R = [5582.84 0 3073.9] rho = [0.846428 0 0.532504] t = 300 s: R = [5581.5 122.122 3073.9] rho = [0.74929 0.463023 0.47347] t = 600 s: R = [5577.5 244.186 3073.9] rho = [0.529447 0.777163 0.340152] Solution: Without iterative improvement (Problem 5.28)... r (km) = [8282.6, 1791.26, 4780.7] v (km/s) = [-1.07108, 5.89508, -0.618321] With iterative improvement (Problem 5.29)... r (km) = [8306.27, 1805.89, 4795.66] v (km/s) = [-1.07872, 5.94219, -0.622807] ----------------------------------------------------Approximate state vector ˆ ( km ) r = 8282.6Iˆ + 1791.26Jˆ + 4780.7K ˆ ( km/s ) v = -1.07108Iˆ + 5.895 08Jˆ − 0.618 321K Problem 5.29 From the MATLAB output listed in the previous problem solution, the iteratively improved state vector is ˆ ( km ) r = 8306.27Iˆ + 1805.89Jˆ + 4795.66K ˆ ( km/s ) v = −1.078 72Iˆ + 5.94219Jˆ − 0.622 807K Problem 5.30 The following MATLAB script uses r and v from Problem 5.29 to compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem_5_30 % ~~~~~~~~~~~~ 92 Solutions Manual % % % % % % % % % % % % % % % % % % % % % - Orbital Mechanics for Engineering Students Chapter 5 This program uses Algorithm 4.1 to obtain the orbital elements from the state vector obtained in Problem 5.29. pi deg mu r v coe T - 3.1415926... factor for converting between degrees and radians gravitational parameter (km^3/s^2) position vector (km) in the geocentric equatorial frame velocity vector (km/s) in the geocentric equatorial frame orbital elements [h e RA incl w TA a] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = orbit inclination (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) - Period of an elliptic orbit (s) User M-function required: coe_from_sv ------------------------------------------------------------------- clear global mu deg = pi/180; mu = 398600; %...Data declaration for Problem 5.30: r = [ 8306.27, 1805.89, 4795.66]; v = [-1.07872, 5.94219, -0.622807]; %... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 5.30: Orbital elements from state vector\n') fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu) fprintf('\n State vector:\n') fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ') fprintf('\n Angular momentum (km^2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg) fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg) fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period (Equation 2.73): if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)^1.5; fprintf('\n Period:') fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end 93 Solutions Manual Orbital Mechanics for Engineering Students Chapter 5 fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 5.30: Orbital elements from state vector Gravitational parameter (km^3/s^2) = 398600 State vector: r (km) v (km/s) = [8306.27 1805.89 4795.66] = [-1.07872 5.94219 -0.622807] Angular momentum (km^2/s) = 59242.6 Eccentricity = 0.0995646 Right ascension (deg) = 270 Inclination (deg) = 30.0002 Argument of perigee (deg) = 269.945 True anomaly (deg) = 190.718 Semimajor axis (km): = 8893.18 Period: Seconds = 8346.36 Minutes = 139.106 Hours = 2.31843 Days = 0.0966014 ----------------------------------------------------- 94 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 Problem 6.1 Orbit 1 (circle): 398 600 µ = = 7.726 km /s r 6378 + 300 vc = Orbit 2 (ellipse): e= rapogee = 6378 + 300 = vapogee = rapogee − rperigee rapogee + rperigee = (6378 + 300) − (6378 + 200) = 0.003758 (6378 + 300) − (6378 + 200) h2 1 µ 1 −e h2 1 ⇒ h = 51 500 km 2 s 398 600 1 − 0.003758 h rapogee = 51 500 = 7.711 km s 6678 ∆v = 7.726 − 7.711 = 0.01453 km s = 14.53 m s Thrust ⋅ ∆t = m∆v 53 400 ⋅ ∆t = 125000 ⋅ 14.53 ⇒ ∆t = 34.01 s (a) vcircle + ( vcircle + ∆v) ∆v 14.53 = vcircle + = 7.726 + = 7.733 m s 2 2 2 ∆s = v avg ∆t = 7.733 ⋅ 34.01 = 263 km v avg = (b) ∆s 263 = = 0.006 268 or 0.63% orbit circumference 2π ⋅ 6600 (c) Problem 6.2 vperigee = 8.2 km s 1 rperigee = 6378 + 480 = 6858 km 1 rapogee = rperigee = 6858 km 2 1 rperigee = 6378 + 160 = 6538 km 2 e2 = rapogee − rperigee 2 2 rapogee + rperigee 2 2 h2 2 1 rapogee = 2 µ 1 − e2 = 6858 − 6538 = 0.023 89 6858 + 6538 h2 2 1 ⇒ h2 = 51 660 km 2 s 398 600 1 − 0.023 89 51 660 h2 = = = 7.532 km s rapogee 6858 6858 = vapogee 2 2 ∆v = vapogee − vperigee = 7.532 − 8.2 = −0.6678 km s 2 1 95 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 Problem 6.3 (a) Orbit 1 (circle): v1 = 398 600 µ = = 7.726 km s r 6378 + 300 Orbit 2 (transfer ellipse): e2 = rapogee − rperigee 2 rapogee + rperigee 2 rperigee 2 2 2 = (6378 + 3000) − (6378 + 300) = 0.1682 (6378 + 3000) + (6378 + 300) h2 1 = 2 µ 1 + e2 h2 2 1 ⇒ h2 = 55760 km 2 s 398 600 1 + 0.1682 55760 h2 = = = 8.35 km s rperigee 6678 6678 = vperigee 2 vapogee = 2 2 h2 rapogee = 2 55760 = 5.946 km s 9378 Orbit 3 (circle): v3 = 398 600 µ = = 6.519 km s r 6378 + 3000 ∆v1 = vperigee − v1 = 8.350 − 7.726 = 0.6244 km s 2 ∆v2 = v3 − vapogee = 6.519 − 5.946 = 0.5734 km s 2 ∆vtotal = ∆v1 + ∆v2 = 1.198 km s (b) 1 1 r = (6678 + 9378) = 8028 km s +r 2 perigee2 apogee2 2 2π 3/2 2π T2 = a = 8028 3/2 = 7159 s ( period of tranfer ellipse) µ 398 600 a2 = tperigee to apogee = ( ) T2 = 3579 s = 59.65 m 2 Problem 6.4 To determine where the projectile B impacts the earth we need the orbital elements. rapogee = 7000 km B vapogee = 7.1 km s B hB = rapogee vapogee = 7000 ⋅ 7.1 = 49 700 km 2 s B rapogee = B 7000 = 2 B hB 1 µ 1 + eB 49 700 2 1 ⇒ eB = 0.1147 398 600 1 + eB hB 2π  TB = 2  µ  1 − e 2 B 3 3  49 700   2π  =  = 4952 s ( period of B's orbit ) 2  2 398 600  1 − 0.1147   96 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 At impact, rB = Rearth . Rearth = hB2 1 µ 1 + eB cos θ impact 6378 = (At impact, rB = Rearth ) 49 700 2 1 ⇒ θ impaact = 104.3° ( from perigee of B's elliptical orbit ) 398 600 1 + 0.1147 cos θ impact Determine the time of flight ( tof ) to impact by first finding t impact , the time from perigee to B’s impact point. E impact θ impact 1 − eB 1 − 0.1147 104.3° tan tan = ⇒ E impact = 1.708 rad 2 1 + eB 2 1 + 0.1147 2 M impact = E impact − eB sin E impact = 1.708 − 0.1147 sin 1.708 = 1.594 rad tan = t impact = TB M impact = 4952 2π 1.594 = 1257 s (from impact point to perigee) 2π Then T 4952 tof = B − t impact = − 1257 = 1220 s 2 2 Find the orbital elements of spacecraft S trajectory. rperigee = 7000 km S vperigee = 1.3 v esc = 1.3 S 2µ rperigee = 1.3 S 2 ⋅ 398 600 = 13.97 km s 7000 hS = rperigee vperigee = 7000 ⋅ 13.97 = 97 110 km 2 s S rperigee = S 7000 = 2 S hS 1 µ 1 + eS 97 110 2 1 ⇒ eS = 2.38 398 600 1 + eS Location of S on its hyperbolic trajectory when B impacts the earth: Mh = µ2 2 (e 2 − 1)3/2 tof = 398 6003 ⋅ (2.382 − 1)3/2 ⋅1220 = 2.131 rad 3 S 97 110 hS eS sinh F − F = M h 2.38 sinh F − F = 2.131 ⇒ F = 1.118 (Algorihm 3.2) θ e +1 F 2.38 + 1 1.118 ⇒ θ S = 76.87° tan S = S tanh = tanh 2 eS − 1 2 2.38 − 1 2 rS = 97 110 2 hS2 1 1 = = 15 360 km µ 1 + eS cos θ s 398 600 1 + 0.28 cos 76.87° distance = rS − 6378 = 8978 km Problem 6.5 (a) For the transfer ellipse 97 Solutions Manual Orbital Mechanics for Engineering Students 1 1 (r + r ) = 2 (227.9 + 149.6) × 10 6 = 188.8 × 10 6 km 2 Mars earth 2π 3/2 2π (188.8 × 10 6 )3/2 = 44.73 × 10 6 s = 517.7 days T= a = 9 µ 132.7 × 10 T tof = = 258.8 days ( time of flight from earth to Mars ) 2 a= (b) Period of Mars in its orbit, 2π 2π (227.9 × 10 6 )3/2 = 59.34 × 10 6 s = 686.8 days TMars = rMars3/2 = µ 132.7 × 10 9 180° − α 258.8 tof ∴ = = = 0.7537 T 180° 343.4 Mars 2 α = 44.33° Problem 6.6 rA = 7000 km rC = 32 000 km e1 = 0.3 r −r e1 = B A rB + rA r − 7000 0.3 = B ⇒ rB = 13 000 km rB + 7000 e2 = 0.5 r −r e2 = D C rD + rC r − 32 000 0.5 = D ⇒ rD = 96 000 km rD + 32 000 h1 = 2µ 7000 ⋅ 13 000 rArB = 2 ⋅ 398 600 = 60 230 km 2 s rA + r B 7000 + 13 000 h2 = 2µ 32 000 ⋅ 96 000 rC rD = 2 ⋅ 398 600 = 138 300 km 2 s rC + r D 32 000 + 96 000 h3 = 2µ 7000 ⋅ 96 000 rArD = 2 ⋅ 398 600 = 72120 km 2 s rA + r D 7000 + 96 000 h4 = 2µ 13 000 ⋅ 32 000 rBrC = 2 ⋅ 398 600 = 85850 km 2 s 13 000 + 32 000 rB + r C 60 230 h v A1 = 1 = = 8.604 km s rA 7000 60 230 h vB1 = 1 = = 4.633 km s rB 13 000 138 300 h vC 2 = 2 = = 4.323 km s rC 32 000 138 300 h vD2 = 2 = = 1.441 km s rD 96 000 72120 h v A3 = 3 = = 10.3 km s rA 7000 85850 h v B4 = 4 = = 6.604 km s rB 13 000 85850 h vC 4 = 4 = = 2.683 km s rC 32 000 72120 h vD3 = 3 = = 0.7512 km s rD 96 000 98 Chapter 6 Solutions Manual Orbital Mechanics for Engineering Students 2π  rA + rD  T3 = µ 2  2π  rB + rC  T4 = µ 2  3/2 3/2  7000 + 96 000    =  2 398 600  2π Chapter 6 3/2  13 000 + 32 000    =  2 398 600  2π = 116 300 s = 32.31 h 3/2 = 33 590 s = 9.33 h (a) Transfer orbit 3: ∆v A = v A3 − v A1 = 10.3 − 8.604 = 1.699 km s ∆vD = vD2 − vD3 = 1.441 − 0.7512 = 0.6896 km s ∆vtotal = ∆v A + ∆vD = 2.389 km s tof 3 = T3 32.32 = = 16.15 h 2 2 (b) Transfer orbit 4: ∆vB = vB4 − vB1 = 6.604 − 4.633 = 1.971 km s ∆vC = vC 2 − vC 4 = 4.323 − 2.683 = 1.64 km s ∆vtotal = ∆vB + ∆vC = 3.611 km s T 9.33 tof 4 = 4 = = 4.665 h 2 2 Problem 6.7 Orbit 1 is the original circular orbit and orbit 2 is the impact trajectory. v A1 = 398 600 µ = = 7.613 km s r 6378 + 500 h2 1 rapogee = 2 2 µ 1 − e2 h2 1 h2 6378 + 500 = 2 ⇒ 2 = 6878(1 − e2 ) µ 1 − e2 µ 1 h2 rB2 = 2 µ 1 + e2 cos θ B 1 h2 h2 6378 = 2 ⇒ 2 = 6378(1 + 0.5e2 ) µ 1 + e2 cos 60° µ h2 2 h2 2 = ⇒ 6378(1 + 0.5e2 ) = 6878(1 − e2 ) ⇒ e2 = 0.04967 µ µ ∴ h2 = 6878µ(1 − e2 ) = 6878 ⋅ 398 600 ⋅ (1 − 0.04967 ) = 51 040 km 2 s (a) 51 040 h v A2 = 2 = = 7.421 km s rA 6878 ∆v = v A2 − v A1 = 7.421 − 7.613 = −0.1915 km s (b) To fall through the point directly below, we must remove completely the transverse component of velocity: ∆v = 0 − v A1 = −7.613 km s 99 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 Problem 6.8 A is apogee of impact trajectory, I is the impact point, a is the semimajpr axis. rA = a (1 + e ) r ∴e = A −1 a (1) A spacecraft 6578 km From Equation 3.22, I rI = a (1 − e cos E) projectile  r   = a 1 − A − 1 cos E    a   = (1 + cos E) a − rA cos E r + r cos E ∴a = I A 1 + cos E (2) Substitute (2) into (1) to get e= rA rA − rI −1 = rI + rA cos E rA cos E + rI 1 + cos E (3) Mean anomaly of the impact point (measured ccw from perigee) is T  +t 2  µt t t M = 2π = π + 2π = π + 2π = π + 3/2 2π 3/2 T T a a µ Let f (E) = M − E + e sin E . Then Kepler’s equation is f (E) = 0 . f (E) = M − E + e sin E f (E) = π + µt a 3/2 µt − E + e sin E = π + 3/2  ri + ra cos E   1 + cos E  Setting rA = 6578 km , rI = 6378 km , t = 30 ⋅ 60 = 1800 s , f (E) = π + 1136 400  6378 + 6578 cos E    1 + cos E 3/2 −E + −E + ra − ri sin E ra cos E + ri 200 sin E 6578 cos E + 6378 Graphing f (E) reveals that f (E) = 0 at E = 5.319 rad . Substituting this into (1) and (2) yields e = 0.019 75 a = 6451 km True anomaly of the impact point: tan 1 + 0.019 75 θI 1+e Ε 5.319 = ⇒ θ = 303.8° tan I = tan 2 1 −e 2 1 − 0.019 75 2 ( ) h = µa (1 − e 2 ) = 398 600 ⋅ 6451 ⋅ 1 − 0.019752 = 50700 km 2 s 100 (123.8° cw from apogee) Solutions Manual vA = vc = Orbital Mechanics for Engineering Students Chapter 6 h 50700 = = 7.707 km s velocity of projectile at apogee. rA 6578 398 600 µ = = 7.784 km s r 6578 velocity of spacecraft in circular orbit. ∆v = v a − vc = 7.707 − 7.784 = −0.07725 km s Problem 6.9 rapogee = 6378 + 302 = 6680 km rperigee = 6378 + 296 = 6674 km rapogee = 6378 + 291 = 6669 km rperigee = 6378 + 259 = 6637 km r3 = 6378 + 259 = 6637 km rapogee = 6378 + 255 = 6633 km rperigee = 6378 + 194 = 6572 km 1 1 2 2 4 h1 = 2µ 4 rapogee rperigee 1 1 rapogee + rperigee 1 vapogee = 1 vperigee = 1 1 1 h1 = rperigee 1 51 590 = 7.730 km s 6674 rapogee rperigee 2 2 rapogee + rperigee 2 2 vperigee = 2 6680 ⋅ 6674 = 51 590 km 2 s 6680 + 6674 51 590 h1 = = 7.723 km s rapogee 6680 h2 = 2µ vapogee = = 2 ⋅ 398 600 h2 rapogee 2 h1 rperigee = 2 ⋅ 398 600 2 6669 ⋅ 6637 = 51 500 km 2 s 6669 + 6637 51 500 = = 7.722 km s 6669 = 2 51 500 = 7.759 km s 6637 398 600 µ = = 7.750 km s r3 6637 v3 = h4 = 2µ rapogee rperigee 4 rapogee + rperigee 4 vapogee = 4 vperigee = 4 4 = 2 ⋅ 398 600 4 6633 ⋅ 6572 = 51 300 km 2 s 6633 + 6572 51 300 h4 = = 7.734 km s rapogee 6633 4 h4 rperigee = 4 51 300 = 7.806 km s 6572 Apogee of orbit 1 to perigee of orbit 2: h12 = 2µ rapogee rperigee 1 rapogee + rperigee 1 ∆v12 = 2 = 2 ⋅ 398 600 2 6680 ⋅ 6637 = 51 520 km 2 s 6680 + 6637 h12 h12 − vapogee + − vperigee = 7.712 − 7.723 + 7.762 − 7.759 = 0.013 93 km s 1 2 rapogee rperigee 1 2 Perigee of orbit 2 to orbit 3 (tangent): 101 Solutions Manual Orbital Mechanics for Engineering Students ∆v23 = vperigee − v3 = 7.759 − 7.750 = 0.009 313 km s 2 Orbit 3 to perigee of orbit 4: h34 = 2µ ∆v34 = r3 rperigee 4 r3 + rperigee = 2 ⋅ 398 600 4 6572 ⋅ 6637 = 51310 km 2 s 6572 + 6637 h34 h34 − v3 + − vperigee = 7.731 − 7.75 + 7.807 − 7.806 = 0.020 26 km s 4 r3 rperigee 4 ∆vtotal = ∆v12 + ∆v23 + ∆v34 = 0.013 93 + 0.009 313 + 0.020 26 = 0.04351 km s Problem 6.10 rA = 6878 km v A1 = rB = 7378 km 398 600 µ = = 7.613 km s rA 6878 h2 = 2µ h v A2 = 2 rA h v B2 = 2 rB rArB 6878 ⋅ 7378 = 2 ⋅ 398 600 = 53 270 km 2 s rA + rB 6878 + 7378 53 270 = = 7.745 km s 6878 53 270 = = 7.220 km s 7378 Alternatively, the energy equation, v 2 2 − µ r = − µ (2 a ) , implies v=  2 1 − µ  r a so that, since a = (rA + rB ) 2 = 7128 km , 1   2 1  2 v A2 =  − µ = − ⋅ 398 600 = 7.745 km s  6878 7128   rA a  1   2 1  2 v B2 =  −  µ = − ⋅ 398 600 = 7.220 km s  7378 7128   rB a  v B3 = 398 600 µ = = 7.35 km s rB 7378 ∆v = v A2 − v A1 + vB3 − vB2 = 0.1323 + 0.1300 = 0.2624 km s Problem 6.11 v A1 = µ r r (12r ) = 1.359 µr r + 12r h µ = 2 = 1.359 r r h2 = 2µ v A2 102 Chapter 6 Solutions Manual v B2 = Orbital Mechanics for Engineering Students µ h2 = 0.1132 12r r Alternatively, using the energy equation, a2 = r + 12r = 6.5r 2 µ µ 2 1  v A2 =  −  µ = 1.846 = 1.359 r r  r a2  µ µ 1  2 v B2 =  −  µ = 0.01282 = 0.1132 r r  12r a2  v B3 = µ µ = 0.2887 r 12r ∆v = v A2 − v A1 + vB3 − vB2 = 0.3587 µ µ µ + 0.1754 = 0.5342 r r r Problem 6.12 µ 2µ µ v A2 = = 1.414 r r r 2µ µ µ µ v B3 = v B4 = = 0.4082 = 0.2887 r r 12r 12r µ µ µ ∆v = v A2 − v A1 + vB4 − vB3 = 0.4142 + 0.1196 = 0.5338 r r r v A1 = Problem 6.13 rA = r v A1 = v1 = h2 = 2µ rB = 3r µ µ = rA r rArB r (3r ) = 1.225 µr = 2µ rA + rB r + 3r 1.225 µr h µ = 1.225 v A2 = 2 = rA r r (Alternatively, use the energy equation.) 1.225 µr h µ = 0.4082 v B2 = 2 = rB 3r r v B3 = v 3 = µ µ µ = = 0.5774 3r rB r ∆v = v A2 − v A1 + vB3 − vB2 = 0.2247 Problem 6.14 rA = 6678 km (a) Orbit 1: v A1 = µ µ µ + 0.1691 = 0.3938 = 0.3938 v1 r r r rC = 9378 km 398 600 µ = = 7.726 km s rA 6678 Orbit 2: 103 Chapter 6 Solutions Manual Orbital Mechanics for Engineering Students rB − rA = e2 rB + rA rB − 6678 = 0.3 ⇒ rB = 12 402 km rB + 6678 h2 = 2µ 6678 ⋅ 12 402 rArB = 2 ⋅ 398 600 = 58 830 km 2 s rA + rB 6678 + 12 402 58 830 h v A2 = 2 = = 8.809 km s rA 6678 58 830 h v B2 = 2 = = 4.743 km s rB 12 402 Orbit 3: h3 = 2µ 12 402 ⋅ 9378 rBrC = 2 ⋅ 398 600 = 65 250 km 2 s rB + rC 12 402 + 9378 65 250 h v B3 = 3 = = 5.261 km s rB 12 402 65 250 h vC 3 = 3 = = 6.957 km s rC 9378 Orbit 4: 398 600 µ = = 6.519 km s vC 4 = 9 378 rC ∆vtotal = v A2 − v A1 + vB3 − vB2 + vC 4 − vC 3 = 1.083 + 0.5177 + 0.4379 = 2.039 km s (b) 2π  rA + rB  µ 2  3/2 T2 = 2π  rB + rC  µ 2  3/2 T3 = ttotal =  6678 + 12 402     2 398 600  3/2 =  12 402 + 9378     2 398 600  3/2 = 2π 2π = 9273 s = 11 310 s 1 (T + T2 ) = 10 290 s = 2.859 hr 2 1 Problem 6.15 rA = rC = rE = 15 000 km (a) Orbit 1: v A )1 = rB = 22 000 km rD = 6878 km 398 600 µ = = 5.155 km s rA 15000 γ A1 = Orbit 2: 22 000 − 6878 r −r = 0.5237 e2 = B D = rB + rD 22 000 + 6878 h2 = 2µ 22 000 ⋅ 6878 rBrD = 2 ⋅ 398 600 = 64 630 km 2 s 22 000 + 6878 rB + rD At the maneuvering point A: 104 Chapter 6 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 1 h2 rA = 2 µ 1 + e2 cos θ A 64 630 2 1 ⇒ θ A = 125.1° 398 600 1 + 0.5237 cos θ A 64 630 h v A2 ) = 2 = = 4.309 km s ⊥ rA 15000 398 600 µ v A2 ) = e sin θ A = 0.5237 sin 125.1° = 2.641 km s r h2 2 64 630 15 000 = v A2 = v A2 ) γ A2 = tan −1 2 ⊥ + v A2 ) v A2 ) v A2 ) r 2 r = 4.309 2 + 2.6412 = 5.054 km s = tan −1 ⊥ 2.641 = 0.5499 ⇒ γ A2 = 31.51° 4.309 ∆γ A = γ A2 − γ A1 = 31.51° − 0 = 31.51° ∆v A = v A12 + v A2 2 − 2 v A1 v A2 cos ∆γ A = 5.1552 + 5.054 2 − 25.1555.054 cos ∆γ A = 2.773 km s T (b) Try Hohmann transfer (orbit 3) from point E on orbit 1 to point B on orbit 2. h3 = 2µ 15 000 ⋅ 22 000 rErB = 2 ⋅ 398 600 = 84 320 km 2 s rE + rB 15 000 + 22 000 vE1 = v A1 = 5.155 km s h3 84 320 = = 5.621 km s rE 15000 84 320 h = 3 = = 3.833 km s rB 22 000 64 630 h = 2 = = 2.938 km s rB 22 000 v E3 = v B3 v B2 ∆vtotal = vE3 − vE1 + vB2 − vB3 = 0.4665 + 0.985 = 1.362 km s Try Hohmann transfer (orbit 4) from point C on orbit 1 to point D on orbit 2. h4 = 2µ 15 000 ⋅ 6878 rC rD = 2 ⋅ 398 600 = 61 310 km 2 s rC + rD 15 000 + 6878 vC1 = v A1 = 5.155 km s 61 310 h vC 4 = 4 = = 4.088 km s rC 15 000 61 310 h vD4 = 4 = = 8.914 km s rD 6878 64 630 h vD4 = 2 = = 9.397 km s rD 6878 ∆vtotal = vC 4 − vC1 + vD2 − vD4 = 1.067 + 0.4824 = 1.55 km s This is larger than the total computed above; thus for minimum Hohmann transfer ∆v = 1.362 km s 105 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 Problem 6.16 Orbit 1: rperigee = 6378 + 1270 = 7648 km 1 vperigee = 9 km s 1 h1 = rperigee vperigee = 7648 ⋅ 9 = 68 832 km 2 s 1 1 2 rperigee = 1 7648 = h1 1 µ 1 + e1 68 832 2 1 ⇒ e1 = 0.5542 398 600 1 + e1 At the maneuver point, θ = 100° . 68 832 2 h12 1 1 = = 13 150 km µ 1 + e1 cos θ 398 600 1 + 0.5542 cos 100° 68 832 h v1 ⊥ = 1 = = 5.234 km s r 13150 r= v1 r = 398 600 µ e sin θ = 0.5542 sin 100° = 3.16 km s h1 1 68 832 v1 = v1 ⊥ 2 + v1 r 2 = 5.234 2 + 3.16 2 = 6.114 km s v1 3.16 = 31.13 γ 1 = tan −1 r = tan −1 v1 ⊥ 5.234 Orbit 2: e2 = 0.4 h2 1 r= 2 µ 1 + e2 cos θ h2 2 1 ⇒ h2 = 69 840 km 2 s 398 600 1 + 0.4 cos 100° 69 840 h v2 ⊥ = 2 = = 5.311 km s 13150 r 13150 = v2 r = 398 600 µ 0.4 sin 100° = 2.248 km s e2 sin θ = 69 840 h2 v2 = v2 ⊥ 2 + v2 r 2 = 5.3112 + 2.248 2 = 5.767 km s v2 2.248 = 22.94° γ 2 = tan −1 r = tan −1 5.767 v2 ⊥ ∆γ = γ 2 − γ 1 = 22.94° − 31.13° = −8.181° ∆v = v12 + v2 2 − 2 v1 v2 cos ∆γ = 6.114 2 + 5.767 2 − 2 ⋅ 6.114 ⋅ 5.767 cos( −8.181) = 0.9155 km s Problem 6.17 rA = 12 756 km v A = 6.5992 km γ A = 20° Orbit 1: v A ⊥ = v A cos γ A = 6.5992 cos 20° = 6.20122 km s ∴ h1 = rA v A ⊥ = 12756 ⋅ 6.20122 = 79102.8 km 2 s 106 Solutions Manual Orbital Mechanics for Engineering Students v A r = v A sin γ A = 6.5992 ⋅ sin 20° = 2.25706 km s µ e sin θ A h1 1 398 600 2.25706 = e sin θ A ⇒ e1 sin θ A = 0.447 917 79 102.8 1 vAr = 2 B 79 102.8 2 1 ⇒ e1 cos θ A = 0.230 641 398 600 1 + e1 cos θ A ( ) ∴ e12 sin 2 θ A + cos2 θ A = 0.447 917 2 + 0.230 6412 e12 = 0.253 825 ⇒ e1 = 0.503 81 ∴ sin θ A = 0.447 917 = 0.889 058 ⇒ θ A = 62.7552° or θ A = 117.235° 0.503 81 Since cos θ A > 0 , θ A = 62.755 2° . 79 102.8 2 h2 1 1 rB = 1 = = 27 848.9 km µ 1 + e1 cos θ B1 398 600 1 + 0.503 81 cos 150° 79 102.8 h = 2.840 43 km s vB ⊥ ) = 1 = 1 rB 27 848.9 398 600 µ ⋅ 0.503 81 ⋅ sin 150° = 1.269 45 km s vB r ) = e sin θ B1 = 1 h1 1 79 102.8 Orbit 2: ∆vB ⊥ = 0.758 20 km s ∴ vB ⊥ ) = vB ⊥ ) + ∆vB ⊥ = 2.840 43 + 0.758 20 = 3.598 63 km s 2 1 h2 = rB vB ⊥ ) = 27 848.9 ⋅ 3.598 63 = 100 218 km 2 s 2 ∆vB r = 0 ∴ vB r ) = vB r ) + ∆vB ⊥ = 1.269 45 + 0 = 1.269 45 km s 2 1 µ e sin θ B2 h2 2 398 600 e sin θ B2 ⇒ e2 sin θ B2 = 0.319 146 1.269 45 = 100 218 2 vB r ) = 2 rB = 27 848.9 = A 150° 1 1 h2 rA = 1 µ 1 + e1 cos θ A 12 756 = Chapter 6 h2 2 1 µ 1 + e2 cos θ B2 100 218 2 1 ⇒ e2 cos θ B2 = −0.095216 6 398 600 1 + e2 cos θ B2 107 Solutions Manual ( Orbital Mechanics for Engineering Students ) 2 e2 2 sin 2 θ B2 + cos2 θ B2 = 0.319146 2 + (−0.095216 6) e2 2 = 0.110 921 e2 = 0.333 048 0.319146 = 0.958 261 ⇒ θ B2 = 73.387 7° or 106.612° 0.333 048 < 0 , θ B2 = 106.612° . ∴ sin θ B2 = Since cos θ B2 ∆θ = 150 − 106.612° = 43.3877° That is, the apse line is rotated 43.387 7° ccw from that of orbit 1. Problem 6.18 r1 = r2 h2 h2 1 1 = µ 1 + e cos θ µ 1 + e cos(η − θ ) cos θ = cos(η − θ ) ∴ θ = η − θ ⇒ 2θ = η ⇒ θ = η 2 Problem 6.19 Orbit 1: rP1 = h12 1 µ 1+e Orbit 2: rP1 = ∴ h2 2 1 h2 = 2 µ 1 + e cos 90° µ h2 2 h12 1 h1 = ⇒ h2 = µ µ 1+e 1+e Problem 6.20 At A: r= h2 µ µ µ e sin 90° = e h h µ µ v r 2 = e sin ( −90°) = − e h h h v ⊥1 = v ⊥ 2 = r ∴ ∆v ⊥ = 0 µ ∆v r = v r 2 − v r1 = −2 e h v r1 = 108 Chapter 6 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 2µe ∴ ∆v = ∆v r = h Problem 6.21 For the circular orbit of the space station, r = 6728 km vc = 398 600 µ = = 7.697 km s r 6728 Tc = 2π µ r 3/2 = 2π 398 600 6728 3/2 = 5492 s = 91.54 m (a) The time required for spacecraft A to reach the space station is the time it takes for the space station to fly around to the original position of spacecraft A. tSA = Tc 2π ⋅ 6728 − 600 2πr − 600 = 5492 = 5414 s = 90.2 min 2πr 2π ⋅ 6728 The time required for spacecraft B to reach the space station is the time it takes for the space station to fly around to the original position of spacecraft B. tBS = Tc 2π ⋅ 6728 + 600 2πr + 600 = 5492 = 5570 s = 92.8 min 2πr 2π ⋅ 6728 (b) The period of spacecraft A’s phasing orbit, is tSA , which determines the semimajor axis of that orbit: 5414 = 2π 398 600 a A3/2 ⇒ a A = 6664 km Spacecraft A is at the apogee of its phasing orbit. From the energy equation 1  1   2 2 v A = µ − − = 7.660 km s  = 398 600 6728 6664   r aA  The delta-v required to drop into the phasing orbit is ∆v A = v A − vc = 7.660 − 7.697 = −0.036 94 km s Spacecraft A must therefore slow down in order to speed up (i.e., catch the space station). After one circuit of its phasing orbit, this delta-v must be added in order to rejoin the circular orbit. Thus ∆v Atotal = 2 ∆v A = 0.073 88 km s orbit: The period of spacecraft B’s phasing orbit, is tBS , which determines the semimajor axis of that 5570 = 2π 398 600 aB3/2 ⇒ aB = 6791 km Spacecraft B is at the perigee of its phasing orbit. From the energy equation 1  2 1   2 vB = µ −  = 398 600 − = 7.733 km s  6728 6791   r aB  The prograde delta-v required to enter the phasing orbit is 109 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 ∆vB = vB − vc = 7.733 − 7.697 = +0.03576 km s Spacecraft B must therefore speed up in order to slow down (i.e, allow the space station to catch up). After one circuit of its phasing orbit, this delta-v must be subtracted in order to rejoin the circular orbit. Thus ∆vBtotal = 2 ∆vB = 0.071 53 km s Problem 6.22 T 2 2π 3/2 1 2π 3/2 a = r 2 µ µ 1 a 3/2 = r 3/2 2 Tphasing = a= Problem 6.23  1 3/2  r 2  2/3 ⇒ a = 0.63r rA = 13 000 km rP = 8000 km Orbit 1: rA + rP = 10 500 km 2 r −r e1 = A P = 0.2381 rA + rP a1 = h1 = µ(1 + e )rP = 398 600(1 + 0.2381) ⋅ 8000 = 62 830 km 2 s T1 = 2π µ a 3/2 = 2π 398 600 10 500 3/2 = 10 710 s Time of flight from P to C:  1 − e1  1 − 0.2381 θ  30°  EC = tan −1  tan C  = tan −1  tan  = 0.4144 rad  1 + 0.2381 2  2   1 + e1 MC = EC − e1 sin EC = 0.4144 − 0.2381 sin 0.4144 = 0.3185 rad 0.3185 M ⋅ 10 710 = 542.8 s tC = C T = 2π 2π Time of flight from P to D:  1 − e1  1 − 0.2381 θ  90°  ED = tan −1  tan D  = tan −1  tan  = 1.330 rad  2  1 + 0.2381 2   1 + e1 MD = ED − e1 sin Ed = 1.330 − 0.2381 sin 1.330 = 1.099 rad 1.099 M tD = D T = ⋅ 10 710 = 1873 s 2π 2π Time of flight from C to D: tCD = tD − tC = 1873 − 542.8 = 1330 s To determine the trajectory from P to D is Lambert’s problem. Note that rD = 62 830 2 1 1 h12 = = 9905 km µ 1 + e1 cos θ D 398 600 1 + 0.2381 cos 90° 110 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 so that in perifocal coordinates rP = 8000pˆ km rD = 9905qˆ km Note as well, that on orbit 1, v P1 = 398 600 µ  − sin θ P pˆ + ( e + cos θ P ) qˆ  =  − sin 0pˆ + ( 0.2381 + cos 0 ) qˆ  h1  62 830  = 7.854qˆ ( km s ) v D11 = 398 600 µ  − sin θ Dpˆ + ( e + cos θ D ) qˆ  =  − sin 90°pˆ + ( 0.2381 + cos 90° ) qˆ  h1  62 830  = −6.344pˆ + 1.510q̂ ( km s ) The following MATLAB script calls upon Algorithm 5.2, implemented as the M-function lambert in Appendix D.11, to solve Lamberts’ problem for the velocities on orbit 2 at P and D. The output to the MATLAB Command Window is listed afterwards. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Problem 6_23 % ~~~~~~~~~~~~ % % This program uses Algorithm 5.2 to solve Lambert's problem for the % data of in Problem 6.23. % % % % % % % % % % deg pi mu r1, r2 dt dtheta R1, R2 string - factor for converting between degrees and radians 3.1415926... gravitational parameter (km^3/s^2) initial and final radii (km) time between r1 and r2 (s) change in true anomaly during dt (degrees) initial and final position vectors (km) = 'pro' if the orbit is prograde = 'retro if the orbit is retrograde V1, V2 - initial and final velocity vectors (km/s) % User M-function required: lambert % ----------------------------------------------------------clear global mu mu = 398600; %km^3/s^2 deg = pi/180; r1 r2 dt dtheta = = = = 8000; 9905; 1330; 90; %km %km %sec %degrees R1 = [r1 0 0]; R2 = [r2*cos(dtheta*deg) r2*sin(dtheta*deg) 0]; %...Algorithm 5.2: string = 'pro'; [V1 V2] = lambert(R1, R2, dt, string); %...Echo the input data and output results to the command window: fprintf('\n-----------------------------------------------------') fprintf('\n Problem 6.23: Lambert''s Problem\n') fprintf('\n Input data:\n'); 111 Solutions Manual fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n Orbital Mechanics for Engineering Students Chapter 6 Gravitational parameter (km^3/s^2) = %g\n', mu) Radius 1 (km) = %g', r1) Position vector R1 (km) = [%g %g %g]\n',... R1(1), R1(2), R1(3)) Radius 2 (km) = %g', r2) Position vector R2 (km) = [%g %g %g]\n',... R2(1), R2(2), R2(3)) Elapsed time (s) = %g', dt) Change in true anomaly (deg) = %g', dtheta) fprintf('\n\n Solution:\n') fprintf('\n Velocity vector V1 (km/s) = [%g %g %g]',... V1(1), V1(2), V1(3)) fprintf('\n Velocity vector V2 (km/s) = [%g %g %g]',... V2(1), V2(2), V2(3)) fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 6.23: Lambert's Problem Input data: Gravitational parameter (km^3/s^2) = 398600 Radius 1 (km) Position vector R1 (km) = 8000 = [8000 Radius 2 (km) Position vector R2 (km) = 9905 = [-0 9905 0 0] 0] Elapsed time (s) = 1330 Change in true anomaly (deg) = 90 Solution: Velocity vector V1 (km/s) = [-2.53168 9.57638 0] Velocity vector V2 (km/s) = [-7.73458 4.37347 0] ----------------------------------------------------- v P 2 = −2.532pˆ + 9.576qˆ ( km s ) v D 2 = −7.734pˆ + 4.373qˆ ( km s ) ∆v P = v P 2 − v P1 = ( −2.532pˆ + 9.576qˆ ) − 7.854qˆ = −2.532pˆ + 1.722qˆ ( km s ) 2 ∆v P = ( −2.532 ) + 1.722 2 = 3.062 ( km s ) ∆v D = v D 2 − v D1 = ( −7.734pˆ + 4.373qˆ ) − ( −6.344pˆ + 1.510qˆ ) = −1.391pˆ + 2.863qˆ ( km s ) 2 ∆v D = ( −1.391) + 2.863 2 = 3.183 ( km s ) ∆vtotal = ∆v P + ∆v D = 3.062 + 3.183 = 6.245 km s Problem 6.24 h = µa (1 − e 2 ) = 398 600 ⋅ 15 000 ⋅ (1 − 0.52 ) = 66 960 km 2 s rascending node = 66 960 2 h2 1 1 = = 7851 km µ 1 + e cos( −ω ) 398 600 1 + 0.5 cos( −30°) 112 Solutions Manual Orbital Mechanics for Engineering Students rdescending node = Chapter 6 66 960 2 h2 1 1 = = 19 840 km µ 1 + e cos( −ω + π ) 398 600 1 + 0.5 cos( −30° + 180°) Rotate the orbital plane 10 degrees around the node line. That means hold v r fixed and rotate v ⊥ 10 degrees. For minimum delta-v, do this maneuver at the furthest distance from the focus (at the descending node, rather than the ascending node). v⊥ = h rdescending node ∆v = 2 v ⊥ sin = 66 960 = 3.375 km s 19 840 ∆i 10° = 2 ⋅ 3.375 ⋅ sin = 0.5883 km s 2 2 (Note: if the maneuver is done at the ascending node, v⊥ = h rascending node ∆v = 2 v ⊥ sin = 66 960 = 8.53 km s 7851 10° ∆i = 2 ⋅ 8.53 ⋅ sin = 1.487 km s 2 2 Over twice the delta-v requirement.) Problem 6.25 For the circular orbit v1 = 398 600 µ = = 7.668 km s 6778 r Assume the maneuver is done at apogee of the ellipse (orbit 2). h2 1 r= 2 µ 1 − e2 6778 = 1 h2 2 ⇒ h2 = 36 750 km 2 s 398 600 1 − 0.5 Then rperigee = 36 750 2 h2 2 1 1 = = 2259 km µ 1 + e2 398 600 1 + 0.5 which is inside the earth. So the maneuver cannot occur at apogee. Assume it occurs at perigee. h2 1 r= 2 µ 1 + e2 6778 = 1 h2 2 ⇒ h2 = 63 660 km 2 s 398 600 1 + 0.5 63 660 h = 9.392 km s v2 = 2 = 6778 r ∆v = v12 + v2 2 − 2 v1 v2 cos δ = 7.668 2 + 9.392 2 − 2 ⋅ 7.668 ⋅ 9.392 cos δ = 3.414 km s 113 Solutions Manual Orbital Mechanics for Engineering Students Chapter 6 Problem 6.26 60 000 2 1 h2 1 = = 12 900 km µ 1 − e 398 600 1 − 0.3 60 000 h vapogee = = = 4.65 km s rapogee 12 900 rapogee = ∆v = 2 vapogee sin δ 90° = 2 ⋅ 4.65 sin = 6.577 km s 2 2 Problem 6.27 µ ro vB1 = h2 h2 h2 1 1 ro = 2 = 2 = 2 ⇒ h2 = µro µ 1 + e cos θ µ 1 + 0.25 cos( −90°) µ vB ⊥ = µro µ h2 = = ro ro ro vB r = µ µ µ e sin θ = ⋅ 0.25 ⋅ sin ( −90°) = −0.25 h2 2 r µro o 2 2 ∆v = (vB r2 − vB r1 )2 + vB ⊥12 + vB ⊥ 22 − 2 vB ⊥1vB ⊥ 2 cos δ 2   µ µ ∆v =  −0.25 − 0 + r r   o o ∆v = 0.0625 µ µ µ + + −0 ro ro ro ∆v = 2.0625 µ ro ∆v = 1.436 2 µ + ro 2 −2 µ ro µ cos( −90°) ro µ ro Problem 6.28 The initial and target orbits are “1” and “2”, respectively, and “3” is the transfer orbit. r1 = 6678 km 398 600 µ = = 7.726 km s 6678 r1 r2 = 6978 km v1 = 398 600 µ = = 7.558 km s 6978 r2 6678 + 6978 r +r = 6828 km a3 = 1 2 = 2 2 1  2 1   2 − = 7.810 km s vperigee = µ −  = 398 600   3 6678 6828 r a  1 3 v2 = 1 1  2  2 vapogee = µ −  = 398 600 − = 7.474 km s  3 6978 6828   r2 a3  114 Solutions Manual Orbital Mechanics for Engineering Students (a) ( ) ( ∆v = vperigee − v1 + v2 − vapogee 3 ∆i 3 ) + 2 ⋅ v2 sin 2 = (7.810 − 7.726) + (7.558 − 7.474) + 2 ⋅ 7.558 sin 20° 2 = 0.0844 + 0.083 48 + 2.625 = 2.793 km s (b) ( ) ∆v = vperigee − v1 + vapogee 3 2 3 + v2 2 − 2 vapogee v2 cos ∆i 3 = (7.810 − 7.726) + 7.474 2 + 7.588 2 − 2 ⋅ 7.474 ⋅ 7.558 cos 20° = 0.0844 + 2.612 = 2.696 km s (c) ∆v = vperigee 2 3 + v12 − 2 vperigee v1 cos ∆i + v2 − vapogee 3 ( 3 ) = 7.812 + 7.726 2 − 2 ⋅ 7.81 ⋅ 7.726 cos 20° + (7.558 − 7.474) = 2.699 + 0.083 48 = 2.783 km s Problem 6.29 Design problem. Problem 6.30  cos i  A = sin −1    cos φ  (a) A = sin −1  cos 116.57°  = sin −1 ( −0.5088) = 329.4°  cos 28.5°  (b) A = sin −1  cos 116.57°  = sin −1 ( −0.5427 ) = 327.1°  cos 34.5°  (c) A = sin −1  cos 116.57°  = sin −1 ( −0.4493) = 333.3°  cos 34.5°  115 Chapter 6 Solutions Manual Orbital Mechanics for Engineering Students Chapter 7 Problem 7.1 ˆ iˆ = 0Iˆ + Jˆ + 0K ˆj = 0Iˆ + 0 Jˆ + K ˆ ˆ kˆ = Iˆ + 0 Jˆ + 0K 0 1 0    ∴ [Q ]Xx = 0 0 1  1 0 0  ˆ = 6628K ˆ ( km ) rB = ( 6378 + 250 ) K 398 600 ˆ µˆ J=− J = −7.754 92 Jˆ ( km s ) rB 6628 vB = − vB 2 ˆ 7.755 2 ˆ ˆ km 2 s K = −0.009 073 45K K=− rB 6628 ( aB = − ) ˆ = 6678 Jˆ ( km ) rA = ( 6378 + 300 ) K vA = 398 600 ˆ µ ˆ ˆ ( km s ) K=− K = 7.725 84K rA 6678 aA = − vA 2 ˆ 7.755 2 ˆ J = −0.008 938 08 Jˆ km 2 s J=− 6628 rA ( ) v 7.726 ˆ Ωxyz = A Iˆ = I = 0.001 156 91Iˆ ( rad s ) rA 6678 Ωxyz = 0 ˆ ( km ) rrel = rB − rA = −6678 Jˆ + 6628K {rrel }xyz = [Q]Xx {rrel }XYZ 0 1 0   0  −6678       = 0 0 1  −6678  =  6628  1 0 0   6628   0  rrel = −6678iˆ + 6628 ˆj ( km ) v B = v A + Ωxyz × rrel + v rel ˆ + 0.0001 156 91Iˆ × −6678 Jˆ + 6628K ˆ +v −7.754 92 Jˆ = 7.725 84K rel ˆ ˆ ˆ ˆ −7.754 92 J = 7.725 84K + −7.667 99 J − 7.725 84K + v ) ( ( ( ) ) rel v reel = −0.086 931 6 Ĵ ( km s ) 0  −0.086 931 6  0 1 0      0 {v rel }xyz = [Q]Xx {v rel }XYZ = 0 0 1  −0.086 931 6  =      1 0 0   0 0    ˆ v rel = −0.086 931 6i ( km s ) ( ) aB = a A + Ωxyz × rrel + Ωxyz × Ωxyz × rrel + 2Ωxyz × v rel + a rel ( ) ( ) ( ) ) ( ) ˆ = −0.008 983 08 Jˆ + 0 + 0.001 156 91Iˆ ×  0.001 156 91Iˆ × −6678 Jˆ + 6628K ˆ  −0.009 073 45K   ˆ   + 2  0.001 156 91I × −0.086 931 6 Ĵ  + a rel ( 116 Solutions Manual Orbital Mechanics for Engineering Students Chapter 7 ˆ = −8.938 08 (10−3 ) Jˆ + 0 + 1.156 91 (10−3 ) Iˆ  × −7.667 99 Jˆ − 7.725 84K ˆ −0.009 073 45K   ( ) ˆ +a + 2  −0.000 100 572K  reel ˆ = −8.938 08 (10−3 ) Jˆ + 0 + 0.008 938 08 Jˆ − 0.008 87116K ˆ −0.009 073 45K ( ( ) ) ˆ +a + 2 −0.000100 572K rel ˆ = −0.009 072 31K ˆ +a −0.009 073 45K rel ( ˆ km s2 a rel = −1.140 18 × 10−6 K ) 0  0   0 1 0       −6  0 {arel }xyz = [Q]Xx {arel }XYZ = 0 0 1    = −1.140 18 × 10    1 0 0  −1.140 18 × 10−6   0    a rel = −1.140 18 × 10−6 ˆj km s2 ( ) Problem 7.2 vA = 398 600 ˆ µˆ j=− j = 7.058 68 ˆj ( km s ) 8000 rA vB = 398 600 ˆ µˆ j=− j = 7.546 05 ˆj ( km s ) rB 7000 v 7.058 68 ˆ k = 0.000 882 335kˆ ( rad s ) Ωxyz = A kˆ = 8000 rA Ωxyz = 0 rrel = rB − rA = −1000iˆ ( km ) v B = v A + Ωxyz × rrel + v rel ( ) ( ) 7.546 05 ˆj = 7.058 68 ˆj + 0.000 882 335kˆ × −1000iˆ + v rel 7.546 05 ˆj = 7.058 68 ˆj − 0.882 335 ˆj + v rel v rel = 1.369 70 ˆj ( km s ) aA = − vA 2 ˆ 7.058 68 2 ˆ i=− i = −0.006 228 12iˆ km m s2 8000 rA aB = − vB 2 ˆ 7.546 05 2 ˆ i=− i = −0.008 134 69iˆ km s2 7000 rB ( ( ( ) ) ) aB = a A + Ωxyz × rrel + Ωxyz × Ωxyz × rrel + 2Ωxyz × v rel + a rel ( ) ( ) −0.008 134 69iˆ = −0.006 228 12iˆ + 0 + 0.000 882 335kˆ ×  0.000 882 335kˆ × ( −1000iˆ )  + 2 0.000 882 335kˆ × 1.369 70 ĵ + a ( ) ( ) 117 rel Solutions Manual Orbital Mechanics for Engineering Students ( ) ( −0.008 134 69iˆ = −0.006 228 12iˆ + 0.000 882 335kˆ × −0.882 335 ˆj ( ) ) + 2 −0.001 208 54 î + a rel 0.008 134 69iˆ = −0.006 228 12iˆ + 0.000 778 516kˆ − 0.002 417 07iˆ + a rell −0.008 134 69iˆ = −0.007 866 68iˆ + a rel ( a rel = −0.000 268 012iˆ km s2 ) Problem 7.3 r = ro + δr (a) 1/2 r = [( ro + δr ) ⋅ ( ro + δr ) ] 1/2 = ( ro ⋅ ro + 2 ro ⋅ δr + δr ⋅ δr ) )1/2 ( = ro2 + 2 ro ⋅ δr + δr 2 1/2  2 r ⋅ δr  δr  2  = ro 1 + o 2 +     ro   ro  Keep only terms of the first order in δr : 2 r ⋅ δr   r = ro  1 + o 2  ro   1/2 2 r ⋅ δr   ∴ r = ro  1 + o 2  ro   1/4 By means of the binomial theorem, 1 2 ro ⋅ δr  1 ro ⋅ δr  r = ro  1 + = ro +  2 4 ro  2 ro3/2  1 ro ⋅ δr ∴ O(δr ) = 2 ro3/2 (b) ro = 3iˆ + 4 ˆj + 5kˆ δ r = 0.01iˆ − 0.01ˆj + 0.03kˆ 1/2 ro = (3 2 + 4 2 + 52 ) (δ r = 0.033 166 2) = 7.071 07 ro = 2.659 15 O (δ r ) = ( )( ) 1 ro ⋅ δ r 1 3iˆ + 4 ˆj + 5kˆ ⋅ 0.01iˆ − 0.01ˆj + 0.03kˆ 1 0.140 000 = 0.003 722 81 = = 3/2 2 ro3/2 2 2 18.803 00 7.071 07 r = 3.01iˆ + 3.99 ˆj + 5.03kˆ 1/2 r = (3.012 + 3.99 2 + 5.03 2 ) = 7.090 92 r = 2.662 88 r − ro = 0.003 729 58 118 Chapter 7 Solutions Manual ( Orbital Mechanics for Engineering Students ) r − ro − O(δr ) r − ro ⋅ 100 = Thus, O(δr ) is within 0.2% of (c) δ r = iˆ − ˆj + 3kˆ O (δ r ) = 0.003 729 58 − 0.003 722 81 ⋅ 100 = 0.181 565% 0.003 729 58 r − ro , (δ r = 3.316 62, δ r ro = 4.69 × 10−1 ( )( ) ) k 1 ro ⋅ δ r 1 3iˆ + 4 ˆj + 5kˆ ⋅ iˆ − ˆj + 3k̂ 1 14 = = = 0.372 281 3 / 2 3 / 2 2 r 2 2 18.803 00 7.071 07 o r = ro + δ r = 4 iˆ + 3 ˆj + 8kˆ 1/2 r = (4 2 + 3 2 + 8 2 ) = 9.433 98 r = 3.071 48 r − ro = 0.412 331 ( ) r − ro − O(δr ) r − ro ⋅ 100 = 0.412 331 − 0.372 281 ⋅ 100 = 9.713 08% 0.412 331 In this case O(δr ) is a poor approximation, exceeding Problem 7.4 For e << 1 : r= a (1 − e 2 ) 1 + e cos θ ≈ a (1 − e 2 )(1 − e cos θ ) = a (1 − e cos θ ) − ae 2 + ae 3 cos θ ≈ a (1 − e cos θ ) Problem 7.5 x + 9x = 10 10 9 x = 3 A cos 3t − 3B sin 3t x = A sin 3t + B cos 3t + At t = 0 , x=5: 5 = A sin (0) + B cos(0) + 5 = B+ 10 9 10 9 35 9 At t = 0 , x = −3 : ∴B = −3 = 3 A cos(0) − 3 ⋅ 35 sin (0) 9 −3 = 3 A A = −1 119 r − ro by almost 10 percent. Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students Chapter 7 Thus 35 10 cos 3t + 9 9 35 x = −3 cos 3t − sin 3t 3 x = − sin 3t + At t = 1.2 : 35 10 cos ( 3 ⋅ 1.2 ) + = −1.934 9 9 35 x = −3 cos ( 3 ⋅ 1.2 ) − sin ( 3 ⋅ 1.2 ) = 7.853 3 x = − sin ( 3 ⋅ 1.2 ) + Problem 7.6 x + 10 x + 2 y = 0 y + 3x = 0 (1) (2) Initial conditions (at t = 0 ): x0 = 1 (3) x0 = −3 y0 = 2 y0 = 4 (4) (5) (6) From (2) d ( y + 3x ) = 0 dt y + 3 x = const y + 3 x = y0 + 3 x0 = 4 + 3 ⋅ 1 = 7 y = 7 − 3x (7) Substitute (7) into (1): x + 10 x + 2 ( 7 − 3 x ) = 0 x + 4 x = −14 General solution: 7 2 x = 2 A cos 2t − 2B sin 2t (8) x = A sin 2t + B cos 2t − (9) Evaluating (8) at t = 0 and using (3), 1 = A sin (0) + B cos(0) − 1 =B− B= 9 2 7 2 7 2 (10) Evaluating (9) at t = 0 and using (4), 120 Solutions Manual Orbital Mechanics for Engineering Students −3 = 2 A cos(0) − 2B sin (0) 3 A=− 2 Chapter 7 (11) With (10) and (11), (8) becomes 3 9 7 x = − sin 2t + cos 2t − 2 2 2 (12) Substituting (12) into (7) yields 9  3 y = 7 − 3  − sin 2t + cos 2t −  2 2 9 27 35 y = sin 2t − cos 2t + 2 2 2 7  2 (13) Then 9 27 35 y = − cos 2t − sin 2t + t +C 4 4 2 (14) Evaluating (14) at t = 0 and using (5), we get 9 27 35 (0) + C 2 = − cos(0) − sin (0) + 4 4 2 2=− C= 9 +C 4 17 4 (15) Substituting this into (14) yields 9 27 35 17 y = − cos 2t − sin 2t + t+ 4 4 2 4 (16) At t = 5 , (12) and (14) yield, respectively, 3 9 7 x = − sin (2 ⋅ 5) + cos(2 ⋅ 5) − = −6.460 2 2 2 9 27 35 17 y = − cos(2 ⋅ 5) − sin (2 ⋅ 5) + ⋅5 + = 97.31 4 4 2 4 Problem 7.7 2π = 0.001163 6 s -1 90 ⋅ 60 t = 15 ⋅ 60 = 900 s n= [Φrr (t)]  4 − 3 cos nt 0 0   2.5 0 0      = 6(sin nt − nt ) 1 0  =  −1.087 1 0   0 0 cos nt   0 0 0.5 121 Solutions Manual [Φrv (t)] Orbital Mechanics for Engineering Students  1  n sin nt 2 =  (cos nt − 1) n  0  1  {δr0 } = 0  ( km ) 0    2 (1 − cos nt ) n       744.29 859.44 0   =  −859.44 1 0  0   0 0 0 .5 1  sin nt   n 0 1 (4 sin nt − 3 nt ) n 0  0  {δv 0 } = 0.01 (km s)  0    {δr } = [Φrr (t ) ]{δr0 } + [Φrv (t ) ]{δv 0 } {δr } = [Φrr (t ) ]{δr0 } + [Φrv (t ) ]{δv 0 }  2.5 0 0  1   744.29 859.44 0   0        1 0  0.01 =  −1.087 1 0  0  +  −859.44  0 0 0.5 0   0 0 0.5  0   2.5  8.5944      =  −1.087  + 2.7718   0   0      11.094    = 1.6847   0    δr = 11.222 km Problem 7.8 398 600 µ = = 7.7714 km s r 6600 v 7.7714 n= = = 0.001177 5 s -1 r 6600 2π T= = 5446.1 s n T t = = 1778.7 s 3  4 − 3 cos nt 0 0   5.5 0 0     [Φrr (t)] = 6(sin nt − nt) 1 0  = −7.3702 1 0   0 0 cos nt   0 0 −0.5 v= [Φrv (t)] [Φvr (t)] [Φvv (t)]  1 2  0 (1 − cos nt )  n sin nt   n 2547.8 0  2   735.49 1   =  −2547.8 −2394.2 0  =  (cos nt − 1) 0 (4 sin nt − 3 nt ) n n   0 735.49  1  0  sin nt  0 0   n   3 n sin nt  0 0 0  0.003 059 2 0     = 6 n(cos nt − 1) 0 0 0 . 010 597 0 0 = −     0 0 − n sin nt   0 0 −0.0010197   cos nt 2 sin nt 0   −0.5 1.7321 0      =  −2 sin nt 4 cos nt − 3 0  =  −1.7321 0  −5  0 0 cos nt   0 0 −0.5 122 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students {δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + } 0 0  1  735.49 2547.8 0  0   5.5        0  1 +  −2547.8 −2394.2 0  δv 0 + 0  =  −7.3702 1 0   0 0 −0.5 1  0 0 735.49      735.49  5.5 2547.8 0  0 0  1  −5.5         + 0  δv 0 = −  −7.3702 1 0  1 = 6.3702   −2547.8 −2394.2  0  0 0 735.49  0 −0.5 1  0.5  −1  735.49 2547.8 0   −5.5   −0.000 647 34        0  6.3702  =  −0.001 971 8  (km s) δv 0 + =  −2547.8 −2394.2  0 0 735.49   0.5   0.000 679 82  { { { } } }  −0.000 647 34   0   −0.000 647 34  {∆v1} = δv 0 − δv 0 =  −0.0019718  −  0  =  −0.0019718  (km s)  0.000 679 82  0.005  −0.004 320 2        ∆v1 = ∆v1 = 0.004792 8 km s { + } { − } {δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + } 0.003 059 2 0 0 1.7321 0   −0.000 647 34   1  −0.5        δv f =  −0.010 597 0 0 1 + −1.7321 −5 0   −0.0019718      0 0 −0.0010197  1  0 0 −0.5  0.000 679 82    0.003 059 2   −0.003 0917      δv f − =  −0.010 597  +  0.010 98   −0.0010197   −0.000 339 91      −0.000 032 48    δv f − =  0.000 38312  (km s)  −0.001359 6    0    δv f + = 0  0    0   −0.000 032 48   0.000 032 48  + − {∆v 2 } = δv f − δv f = 0  −  0.000 38312  = −0.000 38312  (km s) 0   −0.001359 6   0.001359 6        ∆v2 = ∆v 2 = 0.001413 km s { − } { } { } { } { } { } vtotal = ∆v1 + ∆v2 = 0.004792 8 + 0.001413 = 0.006 2058 km s = 6.206 m s Problem 7.9 T0 2 2π nt = =π T0  4 − 3 cos nt 0 0   7  [Φrr (t)] = 6(sin nt − nt) 1 0  = −6π  0 0 cos nt   0 t= 0 0  1 0 0 −1 123 Chapter 7 Solutions Manual (a) Orbital Mechanics for Engineering Students  1   2 0 (1 − cos nt )  n sin nt   0 n 2   [Φrv (t)] =  n (cos nt − 1) 1n (4 sin nt − 3 nt) 0  = − 4n    1 0  0 0 sin nt      n  3 n sin nt   0 0 0 0 0     0 [Φvr (t)] = 6 n(cos nt − 1) 0  =  −12 n 0 0   0 0 − n sin nt   0 0 0   cos nt 2 sin nt 0   −1 0 0    [Φvv (t)] = −2 sin nt 4 cos nt − 3 0  =  0 −7 0   0 0 0 −1 cos nt   0 {δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + } 0   7    0  =  −6π 0   0      0  0 0  δρ   4   1 0  δy0  +  −  n 0 −1  0   0  4 n 3π − n 0  0   0  0  δv0    0  0    4δv0    7δρ 0     n       3πδv0  + − = − + 0 6 πδρ δ y    0  n  0     0       0  4δv0   7δρ +   n  0    −6πδρ − 3πδv0 + δ  = 0  y0     n   0  0   4δv0 7 7δρ + = 0 ⇒ δv0 = − nδρ n 4 3π  7  3 δy0 = 6πδρ + − nδρ = πδρ  4 n  4  δρ   0  3   7  + ∴ {δr0 } =  πδρ  δv 0 =  − vδρ  4 4    0   0    { } (b) {δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + } {δv f − } {δv f − }  0 0 0   δρ   −1 0 0  0      3   7 =  −12 n 0 0   πδρ  + 0 −7 0   − vδρ    4 4  0 0 0   0   0 0 −1  0       0   0   0   0     49   1 πδρ   1 =  −12 nδρ  +  nδρ  =  nδρ  =    0  4   4 0   2 T0  0      0    124 4 n 3π − n 0 Chapter 7  0  0  0  Solutions Manual Orbital Mechanics for Engineering Students x 3π δρ 4 t=0 0 y Problem 7.10 nt = 2π 2π T  4 − 3 cos nt  = 6(sin nt − nt )  0  1  n sin nt 2 =  (cos nt − 1) n  0   3 n sin nt  = 6 n(cos nt − 1)  0 n= [Φrr (t)] [Φrv (t)] [Φvr (t)] [Φvv (t)] 0 0   1   1 0  =  −12π 0 cos nt   0 2 (1 − cos nt ) n 1 (4 sin nt − 3 nt ) n 0 0  1 0 0 1    0 0   6π  = 0 − 0 n   1 0 0 0 sin nt    n  0 0 0  0 0    0 0  = 0 0 0  0 −n sin nt  0 0 0   cos nt 2 sin nt 0  1 0 0      0  = 0 1 0  =  −2 sin nt 4 cos nt − 3  0 0 cos nt  0 0 1  0 {δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + } 0 0  δu0 +  0 0   0  0    6π   1 0  δy0  + 0 − 0  δv0 +  n   0 1   0  0 0 0   0  0  0   0    6πδv0 +       0  = δy0  +  − n  0   0   0       0  0   nδy0 6πδv0 +     +  ⇒ δv0 = 0  = δy0 − n 6π  0   0     0   1    0  =  −12π 0   0    Thus, assuming a Hohmann transfer, δu0 + = 0 , {δv0 } +  0   nδy0  =   60π    125 0  0 0  Chapter 7 Solutions Manual {∆v1} = {δv 0 Orbital Mechanics for Engineering Students + ∆v1 = ∆v1 = } − {δv0 } −  0  0   0   nδy0     nδy0  =  − 0  =    60π  0   60π        nδy0 6π {δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + } 0 0 0   0  1 0 0   0        nδy0  δv f = 0 0 0  δy0  + 0 1 0    6π 0 0 0   0  0 0 1   0     0   nδy0  δv f − =    60π    0   0   0  nδy nδy + − {∆v 2 } = δv f − δv f = 0  −  6π 0  = − 6π 0  0         0   0  nδy0 ∆v2 = ∆v 2 = 6π { { − } } { } { } ∆vtotal = ∆v1 + ∆v2 = nδy0 nδy0 nδy0 2δy0 + = = 6π 6π 3π 3T x y0 y t=0 y0 Problem 7.11 2π = 0.000 872 66 2 ⋅ 3600 t = 30 ⋅ 60 = 1800 s π nt = 2  4 − 3 cos nt [Φrr (t)] = 6(sin nt − nt)  0  1  n sin nt  [Φrv (t)] =  n2 (cos nt − 1)   0   3 n sin nt  [Φvr (t)] = 6 n(cos nt − 1)  0 n= s -1 0 0   4 0 0    1 0  =  −3.425 1 0  0 cos nt   0 0 0  2  0 (1 − cos nt )   n 2292 0    1146 1   =  −2292 −816.3 0 0  (4 sin nt − 3 nt ) n   0 1146  1  0 sin nt  0  n   0.002 618 0  0 0 0    0 0 0  =  −0.005236 0  0 − n sin nt   0 0 −0.000 8727  126 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students  cos nt 2 sin nt 0  0 2 0    [Φvv (t)] = −2 sin nt 4 cos nt − 3 0  = −2 −3 0   cos nt   0 0 0 0 0  {δr f } = [Φrr ]{δr0 } + [Φrv ] δv0 + { }  4 0 0  0   1146 2292 0  0        0   −0.003  δr f =  −3.425 1 0  6  +  −2292 −816.3  0 0 0  0   0 0 1146   0  0   −6.875  −6.875       δr f = 6  +  2.449  =  8.449  ( km ) 0   0   0        δr f = 10.89 km { } { } {δv f } = [Φvr ]{δr0 } + [Φvv ]{δv0 + }  0.002 618 0  0   0 0 2 0  0        δv f =  −0.005236 0 0  6  +  −2 −3 0   −0.003   0 0 −0.000 8727  0   0 0 0   0  0   −0.006  0   −0.006          δv f = 0  +  0.009  = 0  +  0.009  (km s) 0   0  0   0          δv f = 0.010 82 km s { } { } x (km) y (km) t=0 8 4 6 2 0 -2 -4 -6 t = 30 min Problem 7.12 Use C-W frame attached to original location of satellite in GEO. 2π 365.26 = 7.292 × 10 −5 s -1 24 ⋅ 3600 2π + n= First determine the relative position and velocity after two hours. t = 2 ⋅ 3600 = 7200 s nt = 0.1671π 127 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students  4 − 3 cos nt 0   1.404 0 = =   6(sin nt − nt ) 1   −0.1427 1   1 2  (1 − cos nt )   6874 3694   n sin nt n [Φrv (t)] =  2  = 1  (cos nt − 1) (4 sin nt − 3 nt )   −3694 5895 n  n  3 n sin nt 0   10.97 × 10 −5 0  [Φvr (t)] = 6 n(cos nt − 1) 0  = −5.893 × 10 −5 0       cos nt 2 sin nt   0.8653 1.002  [Φvv (t)] = −2 sin nt 4 cos nt − 3  = −1.002 0.4612      [Φrr (t)] {δr2 } = [Φrr ]{δr0 } + [Φrv ]{δv 0 } 0  0   6874 3694   −10   1.404  =   +  {δv 0 }  10   −0.1427 1  0   −3694 5895  6874 3694   −10   −3694 5895{δv 0 } =  10       6874 {δv 0 } =  −3694  −1 3694    −10   −0.00177    =  (km s) 5895   10  0.000 587  {δv2 − } = [Φvr ]{δr0 } + [Φvv ]{δv0 }  97 × 10 −5 0  0   0.8653 {δv2 − } = −105..893   +  −1.002 × 10 −5 0 0 {δv2 − }      −0.000 943 4  =  (km s)  0.002 045   1.002   −0.00177    0.4612  0.000 587  Initiate rendezvous with the origin. t = 6 ⋅ 3600 = 21600 s nt = 0.5014π {δr6 } = [Φrr ]{δr2 } + [Φrv ]{δv 2 + } 0   4.013 0   −10   13710 27 540  +  = +   δv 2 0   −3.451 1   10   −27 540 −9947   13710 27 540   4.013 0   −10  +  −  δv 2 = −   − 27 540 9947    −3.451 1   10  { { } }  13710 27 540   40.13  +  −27 540 −9947  δv 2 =  −44.51     { {δv2 + } = −13710 27 540  } 27 540  −9947  −1  40.13   0.001329   =  (km s)  −44.51 0.0007954  {δv6 − } = [Φvr ]{δr0 } + [Φvv ]{δv2 + } 218 8 0   −10   −0.004 3 {δv6 − } = −00.000  + .000 439 4 0  10 −2 2   0.001329    −3.017  0.0007954      − 0 . 002188 0 . 001 585   −0.000 602 5    δv 6 − =  + =  (km s)  0.004 394   −0.005057   −0.000 663  { } 128 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students  0.001 329   −0.000 943 4   0.002 272  =  (km s)    0.002 045   −0.001 25 {∆v 2 } = {δv 2 + } − {δv 2 − } = 0.000 795 4  −  ∆v2 = ∆v 2 = 0.002 593 km s 0   −0.000 602 5 0.000 602 5 =  (km s) 663   0.000 663     {∆v 6 } = {δv 6 + } − {δv 6 − } = 0  −  −0.000 ∆v6 = ∆v 6 = 0.000 8958 km s ∆vtotal = ∆v2 + ∆v6 = 0.002 593 + 0.000 8958 = 0.003 489 km s = 3.489 m s x (km) t=0 0 y (km) 10 8 6 4 2 -2 Return -4 Outbound -6 -8 t = 2 hr -10 Problem 7.13 Design problem. Problem 7.14 398 600 6600 n= = 0.001 177 5 s -1 6600 3 3 δv = − nδx = − ⋅ 0.001 177 5 ⋅ 5 = 0.008 8311 km s = 8.8311 m s 2 2 Problem 7.15 nt = π 2 0  4 − 3 cos nt 0   4 = =   6(sin nt − nt ) 1   −3.425 1   1 2   1 (1 − cos nt )    n sin nt n n [Φrv (t)] =  2 = 2 1  (cos nt − 1) (4 sin nt − 3 nt )   − n   n n δr f = [Φrr ]{δr0 } + [Φrv ]{δv 0 } [Φrr (t)]  2  n 0.7124   − n  { }  1   nπδr  2  4 0   δr   n   16  n δr f =   + 2  0.7124   7 nδr   −3.425 1  πδr   −   − −  n 4  n   4δr   −3.304δr   0.6963δr  δr f =  + =   −0.2832δr   0.854δr  0.85708δr  { } { } 129 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students d = δr f = 0.9004δr Problem 7.16 nt = π nt = π 0  4 − 3 cos nt 0   7 [Φrr (t)] = 6(sin nt − nt) 1  = −18.85 1       1 2   (1 − cos nt )   0  n sin nt n [Φrv (t)] =  2 = 4 1  (cos nt − 1) (4 sin nt − 3 nt )   − n   n n 4  n  9.425   − n  {δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 }  4  0  δx0   0 0   7 n {δv }  =  0  +  4 9 . 425  0 0 18 85 1 − .   −     −  n n   4   0 n {δv } =  −7δx0     4 9.425  0 18.85δx0   − −  n n  or 4    0  {δv 0 } =  4 9n.425  −  − n   n  −0.589 nδx  {δv 0 } =  −1.75nδx 0  0   −1  −7δx0    18.85δx0  δ v 0 = −0.589nδ x0 iˆ − 1.75nδ x0 ˆj ( km s ) Problem 7.17 nt = π 4 [Φrr (t)] [Φrv (t)] 0  4 − 3 cos nt 0   1.879 = =   6(sin nt − nt ) 1   −0.4697 1  2   0.7071  1 (1 − cos nt )    n sin nt n n =  =  0.5858 2 1  (cos nt − 1) (4 sin nt − 3 nt )   −   n n n {δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 }  0.7071 0.5858  0   −d   n n  0    +  1  δy0   0.5858 0.4722  δv0  −   n n   0.5858δv0  0   −1.879 d    n  =  +  0.4722δv  0 0  0.4697 d + δy0   n   0.5858δv0   − 1.879 d   0  n  0.4722δv =  0 0  + 0.4697 d + δy0    n   0   1.879  = 0   −0.4697 130 0.5858  n  0.4722   n  Chapter 7 Solutions Manual  0  1  Orbital Mechanics for Engineering Students 0.5858  n  δy0  =  1.879 d  0.4722  δv0   −0.4697 d   n  −1 0.5858   δy0  0 n   1.879 d   =   −0.4697 d  . 0 4722 δv0  1    n   δy0   −0.8062 1   1.879 d   −1.984 d   = =   δv0   1.707 n 0   −0.4697 d  3.207 nd  δy0 = −1.984 d 131 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students 132 Chapter 7 Solutions Manual Orbital Mechanics for Engineering Students Problem 8.1 µsun = 132.7 × 10 9 km 3 s2 µearth = 398 600 km 3 s2 Rearth = 147.4 × 10 6 km rearth = 6378 km a= 1 1 + R2 ) = (147.4 × 10 6 + 120 × 10 6 ) = 138.7 × 10 6 km (R 2 earth 2 Heliocentric spacecraft velocity at earth’s sphere of influence: V ( v) 1 2 1  2   = µsun  −  = 132.7 × 10 9  −  = 28.43 km s 6 6  147.4 × 10  Rearth a  138.7 × 10  Heliocentric velocity of earth: Vearth = µsun 132.7 × 10 9 = = 30.06 km s Rearth 149.6 × 10 6 ∴ v ∞ = Vearth − V ( v) = 30.06 − 28.43 = 1.579 km s Geocentric spacecraft velocity of spacecraft at perigee of departure hyperbola: v p = v ∞2 + 2 ⋅ 398 600 2µearth = 1.579 2 + = 11.12 km s 6378 + 200 rp Geocentric spacecraft velocity in its circular parking orbit: vc = 398 600 µearth = = 7.784 km s r 6378 + 200 ∴ ∆v = v p − vc = 11.12 − 7.784 = 3.337 km s Problem 8.2 µsun = 132.7 × 10 9 km 3 s2 µearth = 398 600 km 3 s2 µMercury = 22 930 km 3 s2 Rearth = 149.6 × 10 6 km RMercury = 57.91 × 10 6 km rearth = 6378 km rMercury = 2440 km Vearth = µsun 132.7 × 10 9 = = 29.78 km s Rearth 149.6 × 10 6 VMercury = µsun RMercury = 132.7 × 10 9 57.91 × 10 6 = 47.87 km s 133 Chapter 8 Solutions Manual Orbital Mechanics for Engineering Students Semimajor axis of Hohmann transfer ellipse: a= 1 1 Rearth + RMercury = (149.6 × 10 6 + 57.91 × 10 6 ) = 103.8 × 10 6 km 2 2 ( ) Departure from earth: Spacecraft heliocentric velocity: V ( v) 1 2 1  2   = µsun  −  = 132.7 × 10 9  −  = 22.25 km s 6 6  149.6 × 10  Rearth a  103.8 × 10  ∴ v ∞ = Vearth − V ( v) = 29.78 − 22.25 = 7.532 km s Spacecraft geocentric velocity in circular parking orbit: vc = 398 600 µearth = = 7.814 km s rp 6378 + 150 Spacecraft geocentric velocity at perigee of departure hyperbola: v p = v ∞2 + 2 ⋅ 398 600 2µearth = 7.532 2 + = 13.37 km s rp 6378 + 150 ∴ ∆v1 = v p − vc = 13.37 − 7.814 = 9.611 km s Arrival at Mercury: Spacecraft heliocentric velocity: V ( v) 2 1 2 1    = µsun  −  = 132.7 × 10 9  − = 57.48 km s 6 6   R a 57.91 × 10 103.8 × 10  Mercury  ( ) ∴ v ∞ = V v − VMercury = 57.48 − 47.87 = 9.611 km s Spacecraft velocity relative to Mercury at periapse of approach hyperbola: v p = v ∞2 + 2µMercury rp = 9.6112 + 2 ⋅ 22 930 = 10.49 km s 2440 + 150 Spacecraft parking orbit speed relative to Mercury: vc = µMercury rp = 22 930 = 2.975 km s 2440 + 150 ∴ ∆v2 = v p − vc = 10.49 − 2.975 = 7.516 km s ∆vtotal = ∆v1 + ∆v2 = 9.611 + 7.516 = 15.03 km s Problem 8.3  mp  rSOI = R   msun  msun = 1.989 × 10 30 kg Mercury: 134 Chapter 8 Solutions Manual Orbital Mechanics for Engineering Students R = 57.91 × 10 6 km m p = 3.302 × 10 23 kg  3.302 × 10 23  5 rSOI = 57.91 × 10 6   = 1.124 × 10 km  1.989 × 10 30  Venus: R = 108.2 × 10 6 km m p = 4.869 × 10 24 kg  4.869 × 10 24  5 rSOI = 108.2 × 10 6   = 6.162 × 10 km  1.989 × 10 30  Mars: R = 227.9 × 10 6 km m p = 6.419 × 10 23 kg  6.419 × 10 23  5 rSOI = 227.9 × 10 6   = 5.771 × 10 km  1.989 × 10 30  Jupiter: R = 778.6 × 10 6 km m p = 1.899 × 10 27 kg  1.899 × 10 27  7 rSOI = 778.6 × 10 6   = 4.882 × 10 km  1.989 × 10 30  Problem 8.4  mp  rSOI = R   msun  msun = 1.989 × 10 30 kg Saturn: R = 1433 × 10 6 km m p = 5.685 × 10 26 kg  5.685 × 10 26  7 rSOI = 1433 × 10 6   = 5.479 × 10 km  1.989 × 10 30  Uranus: R = 2872 × 10 6 km m p = 8.683 × 10 25 kg  8.683 × 10 25  7 rSOI = 2872 × 10 6   = 5.178 × 10 km  1.989 × 10 30  Neptune: R = 4495 × 10 6 km m p = 1.024 × 10 26 kg  1.024 × 10 26  7 rSOI = 1.024 × 10 26   = 8.658 × 10 km  1.989 × 10 30  135 Chapter 8 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 Pluto: R = 5870 × 10 6 km m p = 1.25 × 10 22 kg  1.25 × 10 22  6 rSOI = 5870 × 10 6   = 3.076 × 10 km  1.989 × 10 30  Problem 8.5 µsun = 132.7 × 10 9 km 3 s2 µJupiter = 126.7 × 10 6 km 3 s2 Rearth = 149.6 × 10 6 km RJupiter = 778.6 × 10 6 km rJupiter = 71490 km Semimajor axis of Hohmann transfer ellipse: a1 = 1 1 Rearth + RJupiter = (149.6 × 10 6 + 778.6 × 10 6 ) = 464.1 × 10 6 km 2 2 ( ) VJupiter = µsun 132.7 × 10 9 = = 13.06 km s RJupiter 778.6 × 10 6 Use the energy equation to obtain the spacecraft’s velocity upon arrival at Jupiter’s sphere of influence: 2 1  ( ) V1 v = µsun  −  = 132.7 × 10 9 R a  Jupiter 1 2 1   −   = 7.412 km s  778.6 × 10 6 464.1 × 10 6  ( ) v ∞ = VJupiter − V1 v = 13.06 − 7.412 = 5.643 km s Sun û S VJupiter = 13.06 km/s V1 v 778.6(106) km 7.412 km s 149.6(106) km ûV Eccentricity of hyperbolic swing by trajectory: e =1 + rp v ∞ 2 µJupiter =1 + 271490 ⋅ 5.643 2 126.7 × 10 6 = 1.068 Turn angle:  1  1  = 2 sin −1 = 138.8°  e  1.068  Angle between VJupiter and v ∞ at inbound crossing: φ1 = 180° . At the outbound crossing, δ = 2 sin −1 136 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 φ2 = 180° + δ = 318.8° At the outbound crossing ( ) V2v = VJupiter + v ∞ 2 = VJupiter uˆ V + ( v∞ cos φ 2uˆ V + v∞ sin φ 2 ) uˆ S = 13.06uˆ V + ( 5.643 ⋅ cos 318.8°uˆ V + 5.643 sin 318.8° ) uˆ S p û S = 17.30uˆ V − 3.716uˆ S ( km s ) ( v) ∆V ( ) 318.8° v ( ) = V2v − V1 v 2 ûV = (17.30uˆ V − 3.716uˆ S ) − 7.412uˆ V = 9.890uˆ V − 3.716uˆ S ( km s ) ∴ ∆V ( v) = ∆V ( v) V = 10.57 km s v v v 1 v 1 2 138.8° δ ( ) Alternatively, ∆V v = 2 v ∞ sin = 2 ⋅ 5.643 ⋅ sin = 10.57 km s . 2 2 Angular momentum of the new orbit after flyby: ( ) h2 = RJupiterV⊥v = 778.6 × 10 6 ⋅ 17.30 = 13.47 × 10 9 km 2 s Obtain the semimajor axis a2 from the energy equation: ( )2 V2 v 2 − µsun µ = − sun RJupiter 2 a2 17.70 2 132.7 × 10 9 132.7 × 10 9 − = − ⇒ a2 = 4.791 × 10 9 km 2 2 a2 778.6 × 10 6 Observe that a2 a1 = 10.32 . The new orbit dwarfs the original one in size and, therefore, energy. Use Equation 3.61 to find the eccentricity: a2 = 1 h2 2 µsun 1 − e2 2 9 4.791 × 10 = 13.47 × 10 9 132.7 × 10 2 9 1 1 − e2 2 ⇒ e2 = 0.8453 Use the orbit equation to find the true anomaly in the new orbit: RJupiter = h2 2 1 µsun 1 + e2 cos θ 2 778.6 × 10 6 = 13.47 × 10 9 132.7 × 10 2 9 1 ⇒ θ 2 = 26.5° 1 + 0.8453 cos θ 2 The new and original heliocentric orbits are illustrated below. 137 Solutions Manual Orbital Mechanics for Engineering Students Periapse of orbit 2 Chapter 8 1 26.5° Sun Jupiter Earth at launch 2 Problem 8.6 Algorithm 8.1, which makes use of the data in Table 8.1, is implemented in MATAB as the M-function planet_elements_and_sv in Appendix D.17. The MATLAB script Example_8_07, which also appears in Appendix D.17, calls upon planet_elements_and_sv to calculate the orbital elements of earth on the date specified in Example 8.7. The output to the Command Window is also listed in Appendix D.17 for comparison with the results presented in Example 8.7. By changing planet_id to 4, the following Command Window output is obtained for Mars. ----------------------------------------------------Problem 8.6 Input data: Planet: Year : Month : Day : Hour : Minute: Second: Mars 2003 August 27 12 0 0 Julian day: 2452879.000 Orbital elements: Angular momentum (km^2/s) = 5.47595e+09 Eccentricity = 0.0934167 Right ascension of the ascending node (deg) = 49.5682 Inclination to the ecliptic (deg) = 1.85035 Argument of perihelion (deg) = 286.488 True anomaly (deg) = 358.131 Semimajor axis (km) = 2.27936e+08 Longitude of perihelion (deg) Mean longitude (deg) Mean anomaly (deg) Eccentric anomaly (deg) = = = = 138 336.057 334.513 358.457 358.298 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 State vector: Position vector (km) = [1.85954e+08 Magnitude = 2.06653e+08 Velocity (km/s) = [11.4744 Magnitude = 26.4984 ----------------------------------------------------- -8.99155e+07 23.8842 -6.45661e+06] 0.218255] Problem 8.7 The following MATLAB script calls upon Algorithm 8.1, implemented as the MATLAB M-function planet_elements_and_sv in Appendix D.17, to compute the distance of the earth from the sun on the first day of each month of the year 2005, at 12:00:00 UT. The output to the MATLAB command window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_8_07a ~~~~~~~~~~~~~ This program uses Algorithm 8.1 to compute the orbital elements and state vector of Mars at the date and time specified in Example 8.7. mu - gravitational parameter of the sun (km^3/s^2) coe - vector [h e where h e RA incl w TA a w_hat L M E r v of heliocentric orbital elements RA incl w TA a w_hat L M E], = = = = = = = = = = = angular momentum (km^2/s) eccentricity right ascension (deg) inclination (deg) argument of perihelion (deg) true anomaly (deg) semimajor axis (km) longitude of perihelion ( = RA + w) (deg) mean longitude ( = w_hat + M) (deg) mean anomaly (deg) eccentric anomaly (deg) - heliocentric position vector (km) - heliocentric velocity vector (km/s) planet_id - planet identifier: 1 = Mercury 2 = Venus 3 = Earth 4 = Mars 5 = Jupiter 6 = Saturn 7 = Uranus 8 = Neptune 9 = Pluto year month day hour minute second - range: range: range: range: range: range: 1901 - 2099 1 - 12 1 - 31 0 - 23 0 - 60 0 - 60 User M-functions required: planet_elements_and_sv, month_planet_names 139 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 % -------------------------------------------------------------------global mu mu = 1.327124e11; deg = pi/180; %...Data declaration for Problem 8.6 : planet_id = 3; year = 2005; day = 1; hour = 12; minute = 0; second = 0; %... fprintf('\n-----------------------------------------------------\n') fprintf(' Problem 8.7a: Determine the month\n') fprintf('\n Year = %g Time = 12:00:00 UT\n', year) for month = 1:12 %...Algorithm 8.1: [coe, r, v, jd] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second); %...Convert the month numbers into names for output: [month_name, planet_name] = month_planet_names(month, planet_id); fprintf('\n %10s 1st: Distance = %11.5e (km)', month_name, norm(r)) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 8.7a: Determine the month Year = 2005 Time = 12:00:00 UT January 1st: Distance = 1.47100e+08 (km) February 1st: Distance = 1.47417e+08 (km) March 1st: Distance = 1.48243e+08 (km) April 1st: Distance = 1.49505e+08 (km) May 1st: Distance = 1.50745e+08 (km) June 1st: Distance = 1.51707e+08 (km) July 1st: Distance = 1.52093e+08 (km) August 1st: Distance = 1.51830e+08 (km) September 1st: Distance = 1.50963e+08 (km) October 1st: Distance = 1.49758e+08 (km) November 1st: Distance = 1.48464e+08 (km) December 1st: Distance = 1.47505e+08 (km) ----------------------------------------------------This list reveals that the greatest distance occurs in the month of July. We can modify the above MATLAB script to loop through the days of July: % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_8_07b ~~~~~~~~~~~~~ This program uses Algorithm 8.1 to determine the distance of the earth from the sun, according to Problem 8.7 User M-functions required: planet_elements_and_sv, month_planet_names -------------------------------------------------------------------- 140 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 global mu mu = 1.327124e11; %...Data declaration for Problem 8.7 : planet_id = 3; year = 2005; month = 7; hour = 12; minute = 0; second = 0; %... fprintf('\n-----------------------------------------------------\n') fprintf(' Problem 8.7b: Determine the day\n') fprintf('\n Year = %g Time = 12:00:00 UT\n', year) %...Convert the planet_id and month numbers into names for output: [month_name, planet_name] = month_planet_names(month, planet_id); for day = 1:31 %...Algorithm 8.1: [coe, r, v, jd] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second); %...Convert the planet_id and month numbers into names for output: [month_name, planet_name] = month_planet_names(month, planet_id); fprintf('\n %5s %4g: Distance = %14.7e (km)',... month_name, day, norm(r)) end fprintf('\n-----------------------------------------------------\n') The output to the MATLAB command window is: ----------------------------------------------------Problem 8.7b: Determine the day Year = 2005 July July July July July July July July July July July July July July July July July July July July July July July July July July Time = 12:00:00 UT 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21: 22: 23: 24: 25: 26: Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance Distance = = = = = = = = = = = = = = = = = = = = = = = = = = 1.5209314e+08 1.5209524e+08 1.5209664e+08 1.5209732e+08 1.5209728e+08 1.5209653e+08 1.5209506e+08 1.5209287e+08 1.5208998e+08 1.5208637e+08 1.5208204e+08 1.5207701e+08 1.5207126e+08 1.5206481e+08 1.5205765e+08 1.5204978e+08 1.5204122e+08 1.5203195e+08 1.5202198e+08 1.5201132e+08 1.5199996e+08 1.5198792e+08 1.5197519e+08 1.5196178e+08 1.5194769e+08 1.5193292e+08 141 (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) (km) Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 July 27: Distance = 1.5191748e+08 (km) July 28: Distance = 1.5190138e+08 (km) July 29: Distance = 1.5188461e+08 (km) July 30: Distance = 1.5186718e+08 (km) July 31: Distance = 1.5184910e+08 (km) ----------------------------------------------------The furthest distance occurs on July 4. Problem 8.8 For the data given in this problem, the following MATLAB script invokes Algorithm 8.2, which is implemented as the MATLAB M-function interplanetary in Appendix D.18. interplanetary uses Algorithms 8.1 and 5.2 to compute v ∞ at the home and target planets. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_8_08 ~~~~~~~~~~~~ This program uses Algorithm 8.2 to obtain v-infinities in Problem 8.8. mu deg pi - gravitational parameter of the sun (km^3/s^2) - conversion factor between degrees and radians - 3.1415926... planet_id - planet identifier: 1 = Mercury 2 = Venus 3 = Earth 4 = Mars 5 = Jupiter 6 = Saturn 7 = Uranus 8 = Neptune 9 = Pluto year month day hour minute second - depart - [planet_id, year, month, day, hour, minute, second] at departure - [planet_id, year, month, day, hour, minute, second] at arrival arrive range: range: range: range: range: range: 1901 - 2099 1 - 12 1 - 31 0 - 23 0 - 60 0 - 60 planet1 planet2 trajectory - [Rp1, Vp1, jd1] - [Rp2, Vp2, jd2] - [V1, V2] coe - orbital elements [h e RA incl w TA] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = inclination of the orbit (rad) w = argument of perigee (rad) TA = true anomaly (rad) 142 Solutions Manual % % % % % % % % % % % % % % % % % % Orbital Mechanics for Engineering Students a Chapter 8 = semimajor axis (km) jd1, jd2 tof - Julian day numbers at departure and arrival - time of flight from planet 1 to planet 2 (days) Rp1, Vp1 Rp2, Vp2 R1, V1 - state vector of planet 1 at departure (km, km/s) - state vector of planet 2 at arrival (km, km/s) - heliocentric state vector of spacecraft at departure (km, km/s) - heliocentric state vector of spacecraft at arrival (km, km/s) R2, V2 vinf1, vinf2 - hyperbolic excess velocities at departure and arrival (km/s) User M-functions required: interplanetary, coe_from_sv, month_planet_names -------------------------------------------------------------------- clear global mu mu = 1.327124e11; deg = pi/180; %...Data declaration for Problem 8.8: %...Departure planet_id = 3; year = 2005; month = 12; day = 1; hour = 0; minute = 0; second = 0; depart = [planet_id year month day hour minute second]; %...Arrival planet_id = 2; year = 2006; month = 4; day = 1; hour = 0; minute = 0; second = 0; arrive = [planet_id year month day hour minute second]; %... %...Algorithm 8.2: [planet1, planet2, trajectory] = interplanetary(depart, arrive); R1 = planet1(1,1:3); Vp1 = planet1(1,4:6); jd1 = planet1(1,7); R2 = planet2(1,1:3); Vp2 = planet2(1,4:6); jd2 = planet2(1,7); V1 V2 = trajectory(1,1:3); = trajectory(1,4:6); tof = jd2 - jd1; 143 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 %...Use Algorithm 5.1 to find the orbital elements of the % spacecraft trajectory based on [Rp1, V1]... coe = coe_from_sv(R1, V1); % ... and [R2, V2] coe2 = coe_from_sv(R2, V2); %...Equations 8.102 and 8.103: vinf1 = V1 - Vp1; vinf2 = V2 - Vp2; %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 8.8: Earth to Venus') fprintf('\n\n Departure:\n'); fprintf('\n Planet: %s', planet_name(depart(1))) fprintf('\n Year : %g', depart(2)) fprintf('\n Month : %s', month_name(depart(3))) fprintf('\n Day : %g', depart(4)) fprintf('\n Hour : %g', depart(5)) fprintf('\n Minute: %g', depart(6)) fprintf('\n Second: %g', depart(7)) fprintf('\n\n Julian day: %11.3f\n', jd1) fprintf('\n Planet position vector (km) = [%g %g %g]', ... R1(1),R1(2), R1(3)) fprintf('\n Magnitude = %g\n', norm(R1)) fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp1(1), Vp1(2), Vp1(3)) fprintf('\n Magnitude = %g\n', norm(Vp1)) fprintf('\n Spacecraft velocity (km/s) = [%g %g %g]', ... V1(1), V1(2), V1(3)) fprintf('\n Magnitude = %g\n', norm(V1)) fprintf('\n v-infinity at departure (km/s) = [%g %g %g]', ... vinf1(1), vinf1(2), vinf1(3)) fprintf('\n Magnitude = %g\n', norm(vinf1)) fprintf('\n\n Time of flight = %g days\n', tof) fprintf('\n\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n\n fprintf('\n Arrival:\n'); Planet: %s', planet_name(arrive(1))) Year : %g', arrive(2)) Month : %s', month_name(arrive(3))) Day : %g', arrive(4)) Hour : %g', arrive(5)) Minute: %g', arrive(6)) Second: %g', arrive(7)) Julian day: %11.3f\n', jd2) Planet position vector (km) = [%g %g %g]', ... R2(1), R2(2), R2(3)) fprintf('\n Magnitude fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp2(1), Vp2(2), Vp2(3)) = %g\n', norm(R1)) 144 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 fprintf('\n Magnitude = %g\n', norm(Vp2)) fprintf('\n Spacecraft Velocity (km/s) = [%g %g %g]', ... V2(1), V2(2), V2(3)) fprintf('\n Magnitude = %g\n', norm(V2)) fprintf('\n v-infinity at arrival (km/s) = [%g %g %g]', ... vinf2(1), vinf2(2), vinf2(3)) fprintf('\n Magnitude = %g', norm(vinf2)) fprintf('\n\n\n Orbital elements of flight trajectory:\n') fprintf('\n Angular momentum (km^2/s) = %g',... coe(1)) fprintf('\n Eccentricity = %g',... coe(2)) fprintf('\n Right ascension of the ascending node (deg) = %g',... coe(3)/deg) fprintf('\n Inclination to the ecliptic (deg) = %g',... coe(4)/deg) fprintf('\n Argument of perihelion (deg) = %g',... coe(5)/deg) fprintf('\n True anomaly at departure (deg) = %g',... coe(6)/deg) fprintf('\n True anomaly at arrival (deg) = %g\n',... coe2(6)/deg) fprintf('\n Semimajor axis (km) = %g',... coe(7)) % If the orbit is an ellipse, output the period: if coe(2) < 1 fprintf('\n Period (days) = %g',... 2*pi/sqrt(mu)*coe(7)^1.5/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 8.8: Earth to Venus Departure: Planet: Year : Month : Day : Hour : Minute: Second: Earth 2005 December 1 0 0 0 Julian day: 2453705.500 Planet position vector (km) Magnitude = [5.33243e+07 = 1.47517e+08 Planet velocity (km/s) Magnitude = [-28.2595 = 30.202 10.6564 -6.01367e-05] Spacecraft velocity (km/s) Magnitude = [-27.0436 = 27.9729 6.58196 2.7931] v-infinity at departure (km/s) = [1.2159 145 1.37541e+08 -4.07444 2.79316] -1830.84] Solutions Manual Orbital Mechanics for Engineering Students Magnitude Chapter 8 = 5.08735 Time of flight = 121 days Arrival: Planet: Year : Month : Day : Hour : Minute: Second: Venus 2006 April 1 0 0 0 Julian day: 2453826.500 Planet position vector (km) Magnitude = [-5.74135e+07 = 1.47517e+08 -9.1938e+07 Planet velocity (km/s) Magnitude = [29.4613 = 34.955 -18.7099 -1.95644] Spacecraft Velocity (km/s) Magnitude = [30.3918 = 37.8351 -22.2324 -3.68154] v-infinity at arrival (km/s) Magnitude = [0.930541 = 4.0311 -3.52248 2.05581e+06] -1.7251] Orbital elements of flight trajectory: Angular momentum (km^2/s) = 4.0914e+09 Eccentricity = 0.183291 Right ascension of the ascending node (deg) = 68.8159 Inclination to the ecliptic (deg) = 5.77973 Argument of perihelion (deg) = 142.255 True anomaly at departure (deg) = 217.738 True anomaly at arrival (deg) = 26.8913 Semimajor axis (km) = 1.30519e+08 Period (days) = 297.66 ----------------------------------------------------The output shows that at departure from earth, v ∞ = 5.087 km s . Hence, the spacecraft velocity at perigee of the departure hyperbola is v p = v ∞2 + 2µearth rp2 = 5.087 2 + 2 ⋅ 398 600 (6378 + 180)2 = 12.14 km s The spacecraft velocity in its circular 180 km parking orbit is vc = 398 600 µearth = = 7.796 km s rp 6378 + 180 Hence, the delta-v requirement at earth is ∆v1 = v p − vc = 12.14 − 7.796 = 4.346 km s 146 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 At Venus ( µVenus = 324 900 km 3 s2 , rVenus = 6052 km ) the above output shows that v ∞ = 4.031 km s . The speed at the 300 km altitude periapse on the arrival hyperbola is therefore vp hyperbola = v ∞2 + 2µVenus rp2 = 4.0312 + 2 ⋅ 324 900 (6052 + 300)2 = 10.89 km s The semimajor axis of the elliptical capture orbit is a= 1 + 300) + (rVenus + 9000) ] = 10 702 [(r 2 Venus Therefore the velocity at periapse on the ellipse is, using the energy equation, vp ellipse 2 1   2 1  = µVenus  −  = 324 900 −  = 8.482 km s  6052 + 300 10702   rp a  It follows that the delta-v requirement at Venus is ∆v2 = v p hyperbola − vp ellipse = 10.89 − 8.482 = 2.406 km s The total delta-v requirement is ∆vtotal = ∆v1 + ∆v2 = 4.346 + 2.406 = 6.753 km s Problem 8.9 For the data given in this problem, the following MATLAB script invokes Algorithm 8.2, which is implemented as the MATLAB M-function interplanetary in Appendix D.18. interplanetary uses Algorithms 8.1 and 5.2 to compute v ∞ at the home and target planets. The output to the MATLAB Command Window is listed afterwards. % % % % % % % % % % % % % % % % % % % % % % % % % % % % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Problem_8_09 ~~~~~~~~~~~~ This program uses Algorithm 8.2 to obtain v-infinities in Problem 8.9. mu deg pi - gravitational parameter of the sun (km^3/s^2) - conversion factor between degrees and radians - 3.1415926... planet_id - planet identifier: 1 = Mercury 2 = Venus 3 = Earth 4 = Mars 5 = Jupiter 6 = Saturn 7 = Uranus 8 = Neptune 9 = Pluto year month day hour minute second - range: range: range: range: range: range: 1901 - 2099 1 - 12 1 - 31 0 - 23 0 - 60 0 - 60 147 Solutions Manual % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % depart arrive Orbital Mechanics for Engineering Students Chapter 8 - [planet_id, year, month, day, hour, minute, second] at departure - [planet_id, year, month, day, hour, minute, second] at arrival planet1 planet2 trajectory - [Rp1, Vp1, jd1] - [Rp2, Vp2, jd2] - [V1, V2] coe - orbital elements [h e RA incl w TA] where h = angular momentum (km^2/s) e = eccentricity RA = right ascension of the ascending node (rad) incl = inclination of the orbit (rad) w = argument of perigee (rad) TA = true anomaly (rad) a = semimajor axis (km) jd1, jd2 tof - Julian day numbers at departure and arrival - time of flight from planet 1 to planet 2 (days) Rp1, Vp1 Rp2, Vp2 R1, V1 - state vector of planet 1 at departure (km, km/s) - state vector of planet 2 at arrival (km, km/s) - heliocentric state vector of spacecraft at departure (km, km/s) - heliocentric state vector of spacecraft at arrival (km, km/s) R2, V2 vinf1, vinf2 - hyperbolic excess velocities at departure and arrival (km/s) User M-functions required: interplanetary, coe_from_sv, month_planet_names -------------------------------------------------------------------- clear global mu mu = 1.327124e11; deg = pi/180; %...Data declaration for Problem 8.9: %...Departure planet_id = 3; year = 2005; month = 8; day = 15; hour = 0; minute = 0; second = 0; depart = [planet_id %...Arrival planet_id = year = month = day = hour = minute = second = year month day 4; 2006; 3; 15; 0; 0; 0; 148 hour minute second]; Solutions Manual Orbital Mechanics for Engineering Students arrive = [planet_id year month day hour minute Chapter 8 second]; %... %...Algorithm 8.2: [planet1, planet2, trajectory] = interplanetary(depart, arrive); R1 = planet1(1,1:3); Vp1 = planet1(1,4:6); jd1 = planet1(1,7); R2 = planet2(1,1:3); Vp2 = planet2(1,4:6); jd2 = planet2(1,7); V1 V2 = trajectory(1,1:3); = trajectory(1,4:6); tof = jd2 - jd1; %...Use Algorithm 5.1 to find the orbital elements of the % spacecraft trajectory based on [Rp1, V1]... coe = coe_from_sv(R1, V1); % ... and [R2, V2] coe2 = coe_from_sv(R2, V2); %...Equations 8.102 and 8.103: vinf1 = V1 - Vp1; vinf2 = V2 - Vp2; %...Echo the input data and output the solution to % the command window: fprintf('-----------------------------------------------------') fprintf('\n Problem 8.9: Earth to Mars') fprintf('\n\n Departure:\n'); fprintf('\n Planet: %s', planet_name(depart(1))) fprintf('\n Year : %g', depart(2)) fprintf('\n Month : %s', month_name(depart(3))) fprintf('\n Day : %g', depart(4)) fprintf('\n Hour : %g', depart(5)) fprintf('\n Minute: %g', depart(6)) fprintf('\n Second: %g', depart(7)) fprintf('\n\n Julian day: %11.3f\n', jd1) fprintf('\n Planet position vector (km) = [%g %g %g]', ... R1(1),R1(2), R1(3)) fprintf('\n Magnitude = %g\n', norm(R1)) fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp1(1), Vp1(2), Vp1(3)) fprintf('\n Magnitude = %g\n', norm(Vp1)) fprintf('\n Spacecraft velocity (km/s) = [%g %g %g]', ... V1(1), V1(2), V1(3)) fprintf('\n Magnitude = %g\n', norm(V1)) fprintf('\n v-infinity at departure (km/s) = [%g %g %g]', ... vinf1(1), vinf1(2), vinf1(3)) fprintf('\n Magnitude = %g\n', norm(vinf1)) 149 Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 fprintf('\n\n Time of flight = %g days\n', tof) fprintf('\n\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n fprintf('\n\n fprintf('\n Arrival:\n'); Planet: %s', planet_name(arrive(1))) Year : %g', arrive(2)) Month : %s', month_name(arrive(3))) Day : %g', arrive(4)) Hour : %g', arrive(5)) Minute: %g', arrive(6)) Second: %g', arrive(7)) Julian day: %11.3f\n', jd2) Planet position vector (km) = [%g %g %g]', ... R2(1), R2(2), R2(3)) fprintf('\n Magnitude fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp2(1), Vp2(2), Vp2(3)) fprintf('\n Magnitude = %g\n', norm(Vp2)) fprintf('\n Spacecraft Velocity (km/s) = [%g %g %g]', ... V2(1), V2(2), V2(3)) fprintf('\n Magnitude = %g\n', norm(V2)) fprintf('\n v-infinity at arrival (km/s) = [%g %g %g]', ... vinf2(1), vinf2(2), vinf2(3)) fprintf('\n Magnitude = %g\n', norm(R1)) = %g', norm(vinf2)) fprintf('\n\n\n Orbital elements of flight trajectory:\n') fprintf('\n Angular momentum (km^2/s) = %g',... coe(1)) fprintf('\n Eccentricity = %g',... coe(2)) fprintf('\n Right ascension of the ascending node (deg) = %g',... coe(3)/deg) fprintf('\n Inclination to the ecliptic (deg) = %g',... coe(4)/deg) fprintf('\n Argument of perihelion (deg) = %g',... coe(5)/deg) fprintf('\n True anomaly at departure (deg) = %g',... coe(6)/deg) fprintf('\n True anomaly at arrival (deg) = %g\n',... coe2(6)/deg) fprintf('\n Semimajor axis (km) = %g',... coe(7)) % If the orbit is an ellipse, output the period: if coe(2) < 1 fprintf('\n Period (days) = %g',... 2*pi/sqrt(mu)*coe(7)^1.5/24/3600) end fprintf('\n-----------------------------------------------------\n') ----------------------------------------------------Problem 8.9: Earth to Mars Departure: Planet: Earth 150 Solutions Manual Year : Month : Day : Hour : Minute: Second: Orbital Mechanics for Engineering Students Chapter 8 2005 August 15 0 0 0 Julian day: 2453597.500 Planet position vector (km) Magnitude = [1.19728e+08 = 1.51517e+08 -9.28572e+07 798.533] Planet velocity (km/s) Magnitude = [17.7711 = 29.405 23.4274 -0.000315884] Spacecraft velocity (km/s) Magnitude = [20.6107 = 33.0431 25.7677 1.75181] v-infinity at departure (km/s) = [2.83967 Magnitude = 4.07557 2.34021 1.75212] Time of flight = 212 days Arrival: Planet: Year : Month : Day : Hour : Minute: Second: Mars 2006 March 15 0 0 0 Julian day: 2453809.500 Planet position vector (km) Magnitude = [-8.33472e+07 = 1.51517e+08 2.26736e+08 Planet velocity (km/s) Magnitude = [-21.8221 = 22.7173 -6.30169 0.40447] Spacecraft Velocity (km/s) Magnitude = [-20.4825 = 20.9374 -4.25753 -0.845206] v-infinity at arrival (km/s) Magnitude = [1.33959 = 2.74496 2.04416 -1.24968] Orbital elements of flight trajectory: Angular momentum (km^2/s) = 5.00602e+09 Eccentricity = 0.246978 Right ascension of the ascending node (deg) = 322.198 Inclination to the ecliptic (deg) = 3.03935 Argument of perihelion (deg) = 355.671 True anomaly at departure (deg) = 4.33479 True anomaly at arrival (deg) = 152.278 Semimajor axis (km) = 2.01098e+08 Period (days) = 569.273 ----------------------------------------------------- 151 6.79991e+06] Solutions Manual Orbital Mechanics for Engineering Students Chapter 8 The output shows that at departure from earth, v ∞ = 4.076 km s . Hence, the spacecraft velocity at perigee of the departure hyperbola is v p = v ∞2 + 2µearth rp2 = 4.076 2 + 2 ⋅ 398 600 (6378 + 190)2 = 11.75 km s The spacecraft velocity in its circular 190 km parking orbit is vc = 398 600 µearth = = 7.790 km s rp 6378 + 190 Hence, the delta-v requirement at earth is ∆v1 = v p − vc = 11.75 − 7.790 = 3.957 km s At Mars ( µMars = 42 830 km 3 s2 , rMars = 3396 km ) the above output shows that v ∞ = 2.745 km s . The speed at the 300 km altitude periapse on the arrival hyperbola is therefore vp hyperbola = v ∞2 + 2µMars = 2.745 2 + rp2 2 ⋅ 42 830 (3396 + 300)2 = 5.542 km s The semimajor axis of the capture ellipse is found from the required 35 hour period. T= 2π µMars 35 ⋅ 3600 = a 3/2 2π 42 830 a 3/2 ⇒ a = 25 830 km Therefore the velocity at periapse on the ellipse is, using the energy equation, vp ellipse 2 1    2 1 = µMars  −  = 42 830 −  = 4.639 km s 3396 300 25830 r a +    p  It follows that the delta-v requirement at Mars is ∆v2 = v p hyperbola − vp ellipse = 5.542 − 4.639 = 0.9030 km s The total delta-v requirement is ∆vtotal = ∆v1 + ∆v2 = 3.957 + 0.9030 = 4.860 km s Problem 8.10 Rearth = 149.6 × 10 6 km RSaturn = 1433 × 10 6 km Semimajor axis of Hohmann transfer ellipse: a= 1 1 Rearth + RSaturn ) = 149.6 × 10 6 + 1433 × 10 6 = 791.3 × 10 6 km ( 2 2 ( ) Period of Hohmann transfer ellipse: 152 Solutions Manual T= 2π µsun Orbital Mechanics for Engineering Students a 3/2 = 2π 132.7 × 10 9 Chapter 8 (791.3 × 10 6 )3/2 = 383.9 × 10 6 s = 12.17 y Therefore, the time of flight to Saturn’s orbit is T 2 = 6.083 y . Cassini departed on 15 October 1997 (Julian day 2 450 736.5) and arrived on 1 July 2004 (Julian day 2 453 187.5). The number of years for Cassinni’s flight was 2 453 187.5 - 2 450 736.5 = 6.71 y 365.25 Cassini, with its several flyby maneuvers, required a flight time only about 10 percent longer than the Hohmann transfer. The velocity of the spacecraft at the outbound crossing of the earth’s sphere of influence is V ( v) 1 2 1  2   = µsun  −  = 132.7 × 10 9  −  = 40.08 km s  149.6 × 10 6 791.3 × 10 6   Rearth a  The velocity of earth in its (assumed) circular orbit is Vearth = µsun 132.7 × 10 9 = = 29.78 km s Rearth 149.6 × 10 6 Thus v∞ = V ( v) − Vearth = 40.08 − 29.78 = 10.3 km s The spacecraft velocity at the 180 km altitude perigee of the departure hyperbola is v p = v ∞2 + 2 ⋅ 398 600 2µearth = 10.3 2 + = 15.09 km s rp 6378 + 180 The velocity in the circular parking orbit is vc = 398 600 µearth = = 7.796 km s rp 6378 + 180 Hence, ∆v = v p − vc = 15.09 − 7.796 = 7.289 km s From Equation 6.1   7.289  ∆v     ∆m = m o 1 − exp − = m o 1 − exp −  = 0.916m o   − 3  300 ⋅ 9.81 × 10    I sp go     But ∆m equals the mass m p of propellant expended, and the initial mass m o equals m p plus the mass of the spacecraft (2000 kg). Thus, ( m p = 0.916 2000 + m p ) ⇒ m p = 21810 kg 153 Solutions Manual Orbital Mechanics for Engineering Students 154 Chapter 8 Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Problem 9.1 ( ) ˆ as the basis. Method 1: Use Iˆ , Jˆ , K ˆ iˆ = sin θ sin φ Iˆ − sin θ cos φ Jˆ + cos θ K ˆj = cos φ Iˆ + sin φ Jˆ (1) ˆ kˆ = − cos θ sin φ Iˆ + cos θ cos φ Jˆ + sin θ K (3) ˆ + 3 ˆj + 4 iˆ ω = 2K (4) (2) α = ω = 3 ˆj + 4 iˆ ( ) ˆj = ( 2K ˆ ) × ˆj = ( 2K ˆ ) × cos φ Iˆ + sin φ Jˆ = −2 sin φ Iˆ + 2 cos φ Jˆ ( ) ( ) ( ˆ + 3 ˆj × iˆ = 2K ˆ + 3 cos φ Iˆ + 3 sin φ Jˆ × sin θ sin φIˆ − sin θ cos φ Jˆ + cos θ K ˆ iˆ = 2K ˆ = ( 3 cos θ sin φ + 2 sin θ cos φ ) Iˆ + ( 2 sin θ sin φ − 3 cos θ cos φ ) Jˆ − 3 sin θ K ( ) ) ˆ α = 3 −2 sin φ Iˆ + 2 cos φ Jˆ + 4 ( 3 cos θ sin φ + 2 sin θ cos φ ) Iˆ + ( 2 sin θ sin φ − 3 cos θ cos φ ) Jˆ − 3 sin θ K  ˆ ˆ ˆ = ( −6 sin φ + 12 cos θ sin φ + 8 sin θ cos φ ) I + ( 6 cos φ − 12 cos θ cos φ + 8 sin θ sin φ ) J − 12 sin θ K α = α ⋅α = ( −6 sin φ + 12 cos θ sin φ + 8 sin θ cos φ )2 + ( 6 cos φ − 12 cos θ cos φ + 8 sin θ sin φ )2 + 144 sin 2 θ = (sin 2 φ + cos2 φ ) ( 36 + 144 cos2 θ − 144 cosθ + 64 sin 2 θ ) + 144 sin 2 θ = 36 + 144 ( cos2 θ + sin 2 θ ) − 144 cos θ + 64 sin 2 θ = 36 + 144 − 144 cos θ + 64 sin 2 θ α = 180 − 144 cos θ + 64 sin 2 θ ( ) Method 2: Use iˆ , ˆj, kˆ as the basis. Multiply (1) through by cos θ and (2) by sin θ to obtain ˆ = cos θ iˆ sin θ cos θ sin φ Iˆ − sin θ cos θ cos φ Jˆ + cos2 θ K ˆ = sin θ kˆ − sin θ cos θ sin φ Iˆ + sin θ cos θ cos φ Jˆ + sin 2 θ K Adding these two equations yields ˆ = cos θ iˆ + sin θ kˆ K Then (4) can be written ω = 2 ( cos θ iˆ + sin θ kˆ ) + 3 ˆj + 4 iˆ = ( 4 + 2 cos θ ) iˆ + 3 ˆj + 2 sin θ kˆ α= dω   +Ω×ω dt  rel dω   = −2θ sin θ iˆ + 2θ cos θ kˆ = −6 sin θ iˆ + 6 cos θ kˆ dt  rel 155 (6) Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 ˆ + 3 ˆj = 2 ( cos θ iˆ + sin θ kˆ ) + 3 ˆj Ω = 2K iˆ Ω × ω = 2 cos θ 4 + 2 cos θ ˆj kˆ 3 2 sin θ = 8 sin θ ˆj − 12k̂ 3 sin θ ∴ α = ( −6 sin θ iˆ + 6 cos θ kˆ ) + 8 sin θ ˆj − 12kˆ = −6 sin θ iˆ + 8 sin θ ˆj + ( 6 cos θ − 12 ) kˆ α = α ⋅ α = 36 sin 2 θ + 64 sin 2 θ + ( 36 cos2 θ − 144 cos θ + 144 ) = 36 ( sin 2 θ + cos2 θ ) + 64 sin 2 θ − 144 cos θ + 144 = 36 + 64 sin 2 θ − 144 cos θ + 144 α = 180 + 64 sin 2 θ − 144 cos θ Problem 9.2 (a) { } ˆ + ψ nˆ ω plate = (θ kˆ ) + φ ˆj  + ν m (1) ˆ = sin φ iˆ + cos φ kˆ ˆ × ˆj = − cos φ iˆ + sin φ kˆ m pˆ = m nˆ = cos ν ˆj + sin ν pˆ = − cos φ sin ν iˆ + cos ν ˆj + sin φ sin ν kˆ Substituting m̂ and n̂ into (1) yields the result, ω plate = (ν sin φ − ψ cos φ sin ν ) iˆ + (φ + ψ cos ν ) ˆj + θ + ν cos φ + ψ sin φ sin ν k̂ ( ) (2) (b) α plate = dω plate dt = dω plate   + Ω × ω plate dt  rel (3) where Ω is the angular velocity of the xyz frame, which is θ k̂ That is, Ω = θ k̂ (4) From (2) dω plate   d  d  d θ + ν cos φ + ψ sin φ sin ν  kˆ  =  (ν sin φ − ψ cos φ sin ν )  iˆ +  (φ + ψ cos ν )  ˆj +  dt   dt   dt  rel  dt ( ) Taking the derivatives, bearing in mind that θ , φ and ψ are all constant, we get dω plate  os φ cos ν  iˆ  = φ (ν cos φ + ψ sin φ sin ν ) − ψν co dt  rel  − ψν sin ν ˆj + φ ( −ν sin φ + ψ cos φ sin ν ) + ψν cos ν sin φ  k̂ From (2) and (4), 156 (5) Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Ω × ω plate = θ kˆ × (ν sin φ − ψ cos φ sin ν ) iˆ  + (φ + ψ cos ν ) ˆj + θ + ν cos φ + ψ sin φ sin ν kˆ   ˆ = −θ (ψ cos ν + φ ) i + θ (ν sin φ − ψ cos φ sin ν ) ˆj ( ) (6) Substituting (5) and (6) into (3) and collecting terms yields the result, α plate = ν φ cos φ − ψ cos φ cos ν + ψφ sin φ sin ν − ψθ cos ν − φθ  iˆ   ˆ   + ν θ sin φ − ψ sin ν − ψθ cos φ sin ν j   + ψν cos ν sin φ + ψφ cos φ sin ν − φν sin φ  kˆ ( ( ) ) (7) (c) ( aC = a B + α BC × rC B + ω BC × ω BC × rC B ) (8) ˆ = l sin φ iˆ + cos γ kˆ rC B = lm ω = θ kˆ + φ ˆj (9) (10) BC α BC = dω BC dω BC  =  + Ω × ω BC = 0 + θ kˆ × θ kˆ + φ ˆj = −θφ iˆ dt dt  rel ( a B = a O + α AB × rB O ( + ω AB × ω AB × rB O ) (11) ) (12) = 0 + 0 × 1.25ljˆj + θ kˆ × θ kˆ × 1.25lˆj = −1.25θ 2 lˆj ( ) ( ) Substituting (9), (10), (11) and (12) into (8) yields ( ) ( aC = −1.25θ 2 lˆj + −θφ iˆ × l sin φ iˆ + cos γ kˆ + θ kˆ + φ ˆj ×  θ kˆ + φ ˆj × l sin φ iˆ + cos γ kˆ  ) ( ) ( ) ( Upon expanding and collecting terms, we get the result ( ) ( ) aC = − l φ 2 + θ 2 sin φ iˆ + 2lφθ cos φ − 1.25lθ 2 ĵj − lφ 2 cos φ kˆ Problem 9.3 dv   +ω×v dt  rel d 3ˆ = t i + 4 ˆj + ( 2t 2kˆ ) × t 3 iî + 4 ˆj dt = 3t 2 iˆ + −8t 2 iˆ + 2t 5 ˆj aG = ( ) ( ( ) ) = −5t 2 iˆ + 2t 5 ˆj At t = 2 s ( aG = −20iˆ + 64 ˆj m 2 s ) Problem 9.4 α= dω   +Ω×ω dt  rel α= d ω iˆ + ω y ˆj + ω z kˆ + ω x iˆ + ω y ˆj × ω x iˆ + ω y ˆj + ω z k̂ dt x ( ) ( ) ( 157 ) ) ( ) Solutions Manual Orbital Mechanics for Engineering Students iˆ α = 0 + ωx ˆj kˆ ωy 0 Chapter 9 ωx ω y ωz α = ω yω z iˆ − ω xω z ˆj Problem 9.5 About the origin O: y 2 + z 2 i  i [I i ] = mi  −xi y i  −x z i i   20 −10 [I1 ] =  −10 20  −10 −10  256 128 I = [ 3 ]  128 256  −128 128 216 108  [I 5 ] = 108 216 108 −108 − xi y i 2 xi + z i 2 − y i zi − xi z i   − y i zi  xi2 + y i2  −10  −10  kg - m 2  20  −128   128  kg - m 2 256  108  −108  kg - m 2  216  ( ) ( ) ( )  20 [I 2 ] =  −10  −10  64  I = [ 4 ]  32  −32 216  [I 6 ] = 108 108 792 356 36  [I O ] = ∑ [I i ] = 356 792 −76  kg - m 2 i =1  36 −76 792  6 ( −10 −10  20 −10  kg - m 2  −10 20  32 −32   64 32  kg - m 2 32 64  108 108  216 −108  kg - m 2 −108 216  ( ) ( ) ( ) ) 6 m= ∑ mi = 60 kg i =1 6 ∑ m i xi xG = i =1 m = 16 = 0.2667 m 60 = −16 = −0.2667 m 60 = 16 = 0.2667 m 60 6 ∑ mi y i y G = i =1 m 6 ∑ m i zi zG = [ i =1 m y 2 + z 2 G  G Ι mP = m  − xG yG  −x z G G  ] [I G ] = [I P ] − [ Ι mP ] − xG yG 2 xG + zG − yG zG 2 − xG zG   8.533  − yG zG  =  4.267  xG2 + yG2   −4.267 792 356 36   8.533 4.267 = 356 792 −76  −  4.267 8.533     36 −76 792   −4.267 4.267 158 4.267 8.533 4.267 −4.267  4.267   8.533  −4.267  4.267  kg - m 2  8.533  ( ) Solutions Manual 783.5 [I G ] = 351.7 40.27 Orbital Mechanics for Engineering Students 351.7 783.5 −80.27 40.27  −80.27  kg - m 2 783.5  ( ) Problem 9.6 From the previous problem 792 356 36   [I O ] = 356 792 −76  kg ⋅ m 2  36 −76 792  ( iˆ′ = iˆ + 2 ˆj + 2kˆ 12 + 2 2 + 2 2 ) = 0.3333iˆ + 0.6667 ˆj + 0.6667kˆ ˆ  = 0.3333 0.6667 0.6667   i′  792 356 36  T     I x′ =  î′   356 792 −76   iˆ′   36 −76 792   7922 356 36  0.3333     = 0.3333 0.6667 0.6667   356 792 −76  0.6667   36 −76 792  0.6667   535.3    = 0.3333 0.6667 0.6667   596   489.3    I x′ = 898.7 kg ⋅ m 2 Problem 9.7  b2 + c 2 m [I G ] = 12  −ab  − ac  − ab a + c2 − bc 2 − ac   − bc  a 2 + b2  a = l cos θ b = l sin θ c =0  ( l sin θ )2 + (0)2 −( l cos θ )( l sin θ )  −( l cos θ )(0)  m 2 2 −( l sin θ )(0) [I G ] = 12  −( l cos θ )( l sin θ ) ( l cos θ ) + (0)  2 2  −( l cos θ )(0) ( )( ) ( ) ( ) l l l sin θ cos θ sin θ 0 − +    2  sin θ ml2  sin 2θ [I G ] = 12  − 2   0 − sin 2θ 2 cos2 θ 0  0  0  1  159 Chapter 9 Solutions Manual Orbital Mechanics for Engineering Students Problem 9.8 (a)  b2 + c 2 0 0   m 2 2 a +c 0  [I G ] = 12  0  0 a 2 + b2  0  2 2 + 12 0 0   1000  2 2 = 3 +1 0   0 12  0 3 2 + 2 2   0 416.7 0 0   = 0 833.3 0  kg ⋅ m 2    0 0 1083  ( y 2 + z 2 G  G I mO = m  − xG yG  −x z G G  [ ] xG = a = 1.5 m 2 [ ∴ I mO ] − xG yG xG2 + zG2 − yG zG yG = ) − xG zG   − yG zG  xG2 + yG2  b =1 m 2 zG = c = 0.5 m 2  1250 −1500 −750    =  −1500 2500 −500  kg ⋅ m 2  −750 −500 3250  ( ) 0 0   1250 −1500 −750  416.7  0 833.3 0  +  −1500 2500 −500  [I O ] = [I G ] + I mO =   0 0 1083   −750 −500 3250   1667 −1500 −750  [I O ] =  −1500 3333 −500  kg ⋅ m 2  −750 −500 4333  [ ] ( ) (b) 1667 − λ −1500 −1500 −750 3333 − λ −500 −750 −500 = 0 4333 − λ λ3 − 9333λ2 + 24.16(10 6 )λ − 10.91(10 9 ) = 0 λ1 = 568.9 kg ⋅ m 2 λ2 = 4209 kg ⋅ m 2 λ3 = 4556 kg ⋅ m 2 Principal direction 1: ( ) 1667 − 568.9 −1500 −750   v x1  0    −1500  v (1)  0  3333 − 568.9 −500   y  =    −750 −500 4333 − 568.9   v (z1)  0     1098 −1500 −750   1  0   −1500 2764 −500   v (1)  = 0    y     −750 −500 3764   v (z1)  0    (1 )    2764 −500   v y  1500   −500 3764   (1)  =  750     v z    160 Chapter 9 Solutions Manual Orbital Mechanics for Engineering Students Chapter 9  v (1)   2764 −500  −1 1500  0.5929  y =   (1)  =     v z   −500 3764   750   0.278  iˆ + 0.5829 ˆj + 0.278kˆ ( ) vˆ 1 = = 0.8366iˆ + 0.496 ˆj + 0.2326kˆ 2 2 2 1 + 0.5829 + 0.278 Principal direction 2: ( ) 1667 − 4209 −1500 −750   v x2  0    −1500  v (2 )  0  3333 − 4209 − 500   y  =    −750 −500 4333 − 4209   v (z2 )  0     −2542 −1500 −750   1  0   −1500 −875.5 −500   v (2 )  = 0    y     −750 −500 124.5  v (z2 )  0     −875.5 −500   v (y2 )  1500   −500 124.5  (2 )  =  750     v z    − 1 ( ) 2  v   −875.5 −500  1500   −1.565  y  (2 )  =  =     v z   −500 124.5  750   −0.2601 iˆ − 1.565 ˆj − 0.2601kˆ ( ) = 0.5333iˆ − 0.8345 ˆj − 0.1387kˆ vˆ 2 = 2 2 2 ( 1 + −1.565 ) + ( −0.2601) Principal direction 3: ( ) 1667 − 4556 −1500 −750   v x3  0    −1500  (3 )  0  3333 − 4556 −500   v y  =    −750 −500 4333 − 4556   v (z3 )  0     −2889 −1500 −750   1  0   −1500 −1222 −500   v (3 )  = 0    y      ( )  −750 −500 −222.3   v z3  0   −1222 −500   v (y3 )  1500   −500 −222.3   (3 )  =  750     v z     v (3 )   −1222 −500  −1 1500   1.916  y  (3 )  =   =  −500 −222.3   750   −7.685 v  z   iˆ + 1.916 ˆj − 7.685kˆ ( ) vˆ 3 = = 0.1253iˆ + 0.2401ˆj − 0.9626kˆ 2 12 + 1.916 2 + ( −7.685 ) The following MATLAB script uses the built-in function eig to obtain these results, as shown in the Command Window output which follows. fprintf('\n---------------------------------------------------\n') Matrix = [ 1666.7 -1500 -750 -1500 3333.3 -500 -750 -500 4333.3] [eigvector, eigvalue] = eig(I); fprintf('\n Eigenvalue for i = 1:3 Eigenvector') 161 Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 fprintf('\n %g [%g %g %g]', eigvalue(i,i), eigvector(1,i), eigvector(2,i), eigvector(3,i)) end fprintf('\n---------------------------------------------------\n') --------------------------------------------------Matrix = 1666.7 -1500 -750 -1500 3333.3 -500 -750 -500 4333.3 Eigenvalue Eigenvector 4208.83 [-0.533304 0.834478 0.138683] 4555.59 [0.125278 0.240046 -0.962644] 568.881 [0.836596 0.496008 0.232559] --------------------------------------------------(c) iˆ′ = 3iˆ + 2 ˆj + kˆ 2 2 3 + 2 +1 2 = 0.8018iˆ + 0.5345 ˆj + 0.2673k̂  iˆ′  = 0.8018 0.5345 0.2673    Ix′  1667 −1500 −750  0.8018  T    ˆ ˆ       I =  i′   O   i′  = 0.8018 0.5345 0..2673   −1500 3333 −500   0.5345   −750 −500 4333   0.2673  334.1    I x′ = 0.8018 0.5345 0.2673  445.4  239.5    I x′ = 583.3 kg ⋅ m 2 Problem 9.9 1   ml2ω  0 0  ω   12 1  ml2  sin 2θ     1 2 2 {H C } = [I C ]{ω } = 12 0 sin θ − 2   0  = − 24 ml Ω sin 2θ     Ω   1 sin 2θ 2 2   Ω θ ml cos2 θ  cos 0 −  12    2 1 1 1 HC = ml 2ω iˆ − ml 2Ω sin 2θ ˆj + ml 2Ω cos2 θ kˆ 12 24 12 H P = H C + rC P × mv C rC P = d iˆ v C = Ω kˆ × d iˆ = Ω dˆj H P = HC + d iˆ × mΩ dˆj = HC + mΩ d 2kˆ 1 1   1 H P =  ml 2ω iˆ − ml 2Ω sin 2θ ˆj + ml 2Ω cos2 θ kˆ  + mΩ d 2kˆ  12  24 12 1 1  1  ml 2ω iˆ − ml 2Ω sin 2θ ˆj +  ml 2 cos2 θ + md 2  Ω kˆ HP =   12 24 12 162 Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Problem 9.10 1000 0 −300  ω x  ω x       = 1000 500  ω y  1000 ω y   0 ω   −300 500 1000  ω z   z 1000ω x − 300ω z   1000ω x      1000 ω + 500 ω  y z  = 1000ω y   −300ω x + 500ω y + 1000ω z  1000ω z      −300ω z   0      500ω z   = 0   −300ω + 500ω  0  x y    ωz =0 −300ω x + 500ω y = 0 ⇒ ω y = 3 ω 5  ωx  3  ∴ω =  ω x  5   0  ω = 1.166ω x = 20 ⇒ ω x = 17.15 17.15    ∴ ω = 10.29  or ω = 17.15iˆ + 10.29jˆ ( s-1 )  0    Problem 9.11 ω = 2t 2 iˆ + 4 ˆj + 3tkˆ dω  dω  +ω ×ω =  dt rel dt  rel α = 4tiˆ + 3kˆ t = 3: ω = 18iˆ + 4 ˆj + 9kˆ α= α = 12iˆ + 3kˆ {M} = [I G ]{α } + {ω } × [I G ]{ω } 10 0 0  12  18  10 0 0  18        20 0   0  +  4  ×  0 20 0   4       0 0 30   3   9   0 0 30   9  120  18  180  {M} =  0  +  4  ×  80   90   9  270        {M} =  0 ˆj iˆ kˆ M = (120iˆ + 90kˆ ) + 18 4 9 = (120iˆ + 90k̂ k ) + 360iˆ − 3240 ˆj + 720kˆ 180 80 270 ( M = 480iˆ − 3240 ˆj + 810kˆ 2 M = M = 480 2 + ( −3240) + 810 2 = 3374 N ⋅ m 163 ) Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Problem 9.12 {M G } = [I G ]{α } + {ω } × [I G ]{ω } {ω } = 0 ˆ  × 100Iˆ = 34.29 ˆj + 25.36kˆ ( N ⋅ m ) MG = r × F = ( 0 − 0.075 ) Iˆ + ( 0 − 0.2536 ) Jˆ + ( 0 − 0.05714 ) K  −0.03975 0.0120  α X   0   0.1522      34.29  =  −0.03975 0.07177 0.040 57  αY  25.36   0.0120 0.040 57 0.1569  α Z     −1 −0.039 75 0.0120   0  143.9  α X   0.1522         2 αY  =  −0.039 75 0.07177 0.040 57  34.29  =  553.1 m s α   0.0120 0.040 57 0.1569  25.36   7.61   Z  ( ) Problem 9.13 (a) ∑ Fx = maG x  v2  cos θ  mg sin θ = m  R  tan θ = v2 gR (b) iˆ M = ω × H = ωx Aω x ˆj ωy kˆ ω z = ( C − B ) ω yω z iˆ + ( A − C ) ω xω z ˆj + ( B − A ) ω xω y kˆ Bω y Cω z v v sin θ ωy =0 ω z = cos θ R R  v  v  M = ( A − C )  − sin θ   cos θ  ˆj  R R  ωx = − M y = (C − A) v2 R2 sin θ cos θ = (C − A) v2 2 R2 sin 2θ Problem 9.14 iˆ M = ω × H = ωx Aω x ωx =0 ˆj kˆ ωy ω z = ( C − B ) ω yω z iˆ + ( A − C ) ω xω z ˆj + ( B − A ) ω xω y kˆ Bω y Cω z ω y = ω Z sin α ω z = ω Z cos α M = ( C − B ) (ω Z sin α )(ω Z cos α ) iˆ Mx = 1 (C − B)ω Z2 sin 2α 2 My = 0 Mz = 0 G (–l[1 + cos ]/12, lsin /12, 0) Problem 9.15 l  l l xG1 = − − cos = − cos θ  2 3 6 l yG1 = sin θ 6 zG1 = 0 y 2l/ 2 3 1 C (0,0,0) B(l/3, 0, 0) x A(–2l/3, 0, 0) l/ 3 164 2l/3 l/3 Solutions Manual xG2 = − Orbital Mechanics for Engineering Students Chapter 9 l 6 yG2 = zG2 = 0 xG = yG = mxG1 + mxG2 l l 1 l − cos θ − = − (1 + cos θ ) 2 6 6 12 l 1 l  sin θ = sin θ =   2 6 12 = 2m myG1 + myG2 2m zG = 0 Free-body diagram (no bearing couples and no thrust components of bearing force): y A Az Ay z Bz x By l/3 2l/3 ∑ Fx = 2maG x : B C 0 =0 ∑ Fy = 2maG y : Ay + By = − ∑ Fz = 2maG z : Az + Bz = 0 mω 2 l sin θ 6 (1) (2) Moments of inertia about G1 (inferred from results of Exercise 9.7):  2  sin θ ml2  1 ( )  sin 2θ I G1 = 1 12  2  0 [ ] 1 sin 2θ 2 cos2 θ 0  0  0  1  From Equation 9.60: [I( ) ] 1 mC y 2 + z 2 G1  G1 = m  − xG1 yG1   − xG1 zG1 − xG1 yG1 xG12 + zG12 − yG1 zG1 2  l   sin θ 6     l  l = m  − − cos θ sin θ  6 6   0   − xG1 zG1   − yG1 zG1   xG12 + yG12   l  l − − cos θ sin θ  6 6 2  l  − cos θ  6  0 165     0   2 2  l  l   − cos θ + sin θ   6  6   0 Solutions Manual Orbital Mechanics for Engineering Students 1 2  3 sin θ ml2  1 ( )  sin 2θ I m1 C = 12  6  0  [ 1 sin 2θ 6 1 cos2 θ 3 ] [I (C1) ] = [I (G1) ] + [I (m1)C ] 1  0  0  1 3  0 [ ] 2 sin 2θ 3 4 cos2 θ 3 0 0 0 0  ml2  IG = 0 1 0   2 12 0 0 1  y 2 + z 2 G2  G2 ( ) I m2 C = m  − xG2 yG2   − xG2 zG2  1 2 0  3 sin θ  ml2  1  sin 2θ 0 +  12  6 1   0  1 sin 2θ 2  2  sin θ ml2  1  sin 2θ = 12  2  0 4 2  3 sin θ ml2  2 ( )  sin 2θ I C1 = 12  3  0  Chapter 9 cos2 θ 0 1 sin 2θ 6 1 cos2 θ 3 0  0  0  1 3   0  0  4 3  [ ] (2 ) [ ] [I ] (2 ) mC 0 ml2  = 0 12   0  0 1 3 0  ml 0 ( ) I C2 = 12  0  [ ] 2 xG2 2 + zG2 2 − yG2 zG2 0 0 − xG2 zG2    l 2  − yG2 zG2  = m 0 −   6   xG2 2 + yG2 2   0 0  0  0 1  3 0 0 0 0 0  2  ml  ml 0 ( )  I m2 C = 0 1 0  + 12  12  0 0 1  0  0 0 4  0 3 4 0  3 [I (C2 ) ] = [I (G2 ) ] + [ 2 − xG2 yG2 ] 4 2  3 sin θ ml2  2 [I C ] = I C(1) + I C(2 ) = 12  3 sin 2θ   0  [ ] [ ] 0 1 3 2 2 sin 2θ 3 4 cos2 θ 3 0 0 0  0 1  3  0 0 2  ml 0  0+  12  0 4   3 166 0 4 3 0 0  0 4  3 0    0  2  l   −  6   Solutions Manual Orbital Mechanics for Engineering Students  ml2 sin 2 θ  9  2 ml [I C ] =  18 sin 2θ   0  ml2 sin 2θ 18 2 ml ( 1 + cos2 θ ) 9  0    0   2ml2  9  0  ml2 sin 2 θ  9  2 ml { } {H C } = [I C ] ω =  18 sin 2θ   0     ml2ω sin 2 θ  0    ω   92       ml ω sin 2θ  0  0  =  18     0 2ml2   0    9  ml2 sin 2θ 18 ml2 ( 1 + cos2 θ ) 9 0 iˆ ω ˆj 0 ml 2ω sin 2 θ 9 ml 2ω sin 2θ 18 MC = ω × H C = kˆ ml 2ω 2 sin 2θ k̂ 0 = 18 0 MC x = 0 MC y = 0 MC z = Chapter 9 (3) (4) ml2ω 2 sin 2θ 18 (5) Calculate the moments of the bearing reactions in the above free body diagram: 1 1   2   2  l  2 MC =  − liˆ  × Ay ˆj + Az kˆ +  iˆ  × By ˆj + Bz kˆ =  Az l − Bz l  ˆj +  − Ay l + By l  ĵ   3   3  3  3 3 3 ( ) ( ) MC x = 0 2 1 MC y = Az l − Bz l 3 3 2 1 MC z = − Ay l + By l 3 3 (6) (7) (8) From (4) and (7) 2 Az − Bz = 0 (9) From (5) and (8) 2 1 ml2ω 2 − Ay l + By l = sin 2θ 3 3 18 (10) From (1) we have By = − mω 2 l sin θ − Ay 6 (11) Substituting this into (10): −  2 1  mω 2 l ml2ω 2 ml2ω 2 Ay l +  − sin θ − Ay  l = sin 2θ ⇒ Ay = − sin θ (1 + 2 cos θ )  3 3 6 18 18 167 (12) Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Therefore, from (11), By = −  ml2ω 2  mω 2 l mω 2 l sin θ −  − sin θ (1 + 2 cos θ )  = − sin θ (1 − cos θ )   6 18 9 (13) From (2) we have Bz = − Az Substituting this into (9) 2 Az − ( − Az ) = 0 ⇒ Az = 0 Therefore, Bz = 0 . The only reactions at each bearing are in the plane of the rod and shaft, normal to the shaft, as given by Equations (12) and (13). H = m l2sin /9 y A B l/3 2l/3 mw2lsin (1 + 2cos )/18 x m l2sin (1 – cos )/9 Problem 9.16 MG = HG )rel + ω × H iˆ MG = Aω x iˆ + Bω y ˆj + Cω z kˆ + ω x Aω x ˆj ωy kˆ ωz Bω y Cω z ˆj iˆ k̂ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 600i + MG y j + MGz k = 5ω x i + 5 0 j + 10 0 k + 0 0.5 100 5 ( 0 ) 5 ⋅ 0.5 10 ⋅ 100 600iˆ + MGx ˆj + MG y kˆ = ( 5ω x + 250 ) iˆ 5ω x + 250 = 600 ⇒ ω x = 70 rad s2 Problem 9.17 MO = IOxω x iˆ + IO yω y ˆj + IOzω z kˆ + iˆ ωx ˆj ωy kˆ ωz IO xω x IO yω y IOzω z 2 1 mL2  L mL2 + m = IOy = 0  6 12 9 ω«x = ω«y = ω«z = 0 ωx =0 ω y = −ω cos θ ω z = ω sin θ IOx = IOz = 168 Solutions Manual Orbital Mechanics for Engineering Students ˆj iˆ MO = 0 −ω cos θ 0 kˆ ω sin θ =− mL2 ω sin θ 9 0 Chapter 9 1 mω 2 L2 sin θ cos θ iˆ 9 1 ∴ MOx = − mω 2 L2 sin θ cos θ 9 Moment of the weight vector about O: L MOx = −mg sin θ 6 1 L 3 g ∴ − mω 2 L2 sin θ cos θ = −mg sin θ ⇒ ω = 9 6 2 L sin θ Problem 9.18 )rel + Ω × H MG = HG MG = IGxα x iˆ + IG yα y ˆj + IGzα z kˆ + iˆ 10 ⋅ 9.81 ⋅ 0.25iˆ = 0 + 0 iˆ Ωx ˆj Ωy kˆ Ωz IGxω x IG yω y IGzω z ˆj ωp 0 0.014 06ω p kˆ 630 0.02812 ⋅ 630 24.52iˆ = 17.72ω p iˆ ⇒ ω p = 1.384 rad s Or, using Equation 9.96, ωp = mgd 10 ⋅ 9.81 ⋅ 0.25 = = ω p = 1.384 rad s I zω s 0.02812 ⋅ 630 Problem 9.19 ω wheel = v = r î 1000 3600 = 120.4 rad s 0.3 130 k̂ MB I wheel = 25 ⋅ 0.2 2 = 1 kg ⋅ m 2 120.4 rad/s M = ω p × H s = 0.8 ˆj × [1 ⋅ 120.4kˆ ] = 96.3iˆ ( N ⋅ m ) 0.8 rad/s ĵ 169 Solutions Manual Orbital Mechanics for Engineering Students Chapter 9 Problem 9.20 I rotor = 4 ⋅ 0.07 2 = 0.0196 kg ⋅ m 2 2π ˆ   k  = 41.05 ˆj ( N ⋅ m ) M = ω p × H s = 2iˆ ×  0.0196 ⋅ 10 000 ⋅  60  By = F Ay = − F 0.04 F = 41.05 F = 1026 N Problem 9.21 i 0.07222 rad/s 1570.8 rad/s G C = 12.5 kg - m 2 j k MG I rotor = 200 ⋅ 0.25 2 = 12.5 kg ⋅ m 2 2π = 1571 rad s 60 650 000 v ω p = = 3600 = 0.072 22 rad s r 2500 M = ω p × H s = 0.072 22iˆ × 12.5 ( −1571kˆ )  = 1418iˆ ( N ⋅ m ) ω s = 15 000 ⋅ The moment reaction on the airframe is clockwise, pitching the nose down. Problem 9.22 MG = ω p × H s ω p = 20î ( rad s ) 1 ⋅ 10 ⋅ 0.05 2 ⋅ 200kˆ = 2.5kˆ kg ⋅ m s2 2 MG = 20iˆ × 2.5kˆ = −50 ˆj ( N ⋅ m ) ( Hs = 0.6RB ˆj = −50 ˆj RB = −83.33 N ) RA = − RB = 83.33 N Problem 9.23 lx = cosψ cos φ − sin φ sin ψ cos θ = cos 70° cos 50° − sin 50° sin 70° cos 25° = 0.6428 ⋅ 0.3420 − 0.7660 ⋅ 0.9397 ⋅ 0.9063 = −0.4326 α xX = cos −1 ( −0.4326) = 115.6° 170 forward Solutions Manual Orbital Mechanics for Engineering Students Problem 9.24  20 {H } = [I ]{ω } =  −10   0 −10 0  10   0      30 0  20  =  500  0 40  30  1500  10  1 1   T = H ⋅ ω = 0 500 1500  20  = 23 000 J 2 2 30    171 ( J ⋅ s) Chapter 9 Solutions Manual Orbital Mechanics for Engineering Students 172 Chapter 9 Solutions Manual Orbital Mechanics for Engineering Students Problem 10.1 ωp = ωs C 1200 6 = = 5.171 rad s A − C cos θ 2600 − 1200 cos 6° H = Aω p = 2600 ⋅ 5.171 = 13 450 kg ⋅ m 2 s Problem 10.2 C 500 6 ωs = = −15.06 rad s A − C cos θ 300 − 500 cos 10° 2π 2π T= = = 0.4173 s ω p 15.06 ωp = Problem 10.3 C mr 2 ωs ωs ωs = = −2 A − C cos θ 1 2 θ θ cos cos mr − mr 2 2 θ2 cos θ = 1 − 2 ωp = −1  θ2  θ2 1 = 1 −  = 1 + 2  2 cos θ  2  θ   ∴ ω p = −2ω s  1 +  2  (θ << 0) Problem 10.4 H = Aω p = 1000 ⋅ 2 = 2000 kg ⋅ m 2 s Problem 10.5 HG )rel + ω × HG = 0 [I G ]{α } + ω × H G = 0 385.4   0  0 385.4   0  0 385.4   0  0 385.4   0  0 0 416.7 0 0 416.7 0 0 416.7 0 0 416.7 0 0  0 0   0.01  0   0.01  385.4       {α }   +  −0.03  ×  0 0  416.7 0   −0.03  = 0   0.02   0 52.08  0 52.08   0.02  0     0   0.01   3.854  0        {α }  +  −0.03  ×  −12.50  = 0  0   0.02   1.042  0  52.08        0   0.2188  0       0  {α } +  0.066 6  = 0   −0.009 39  0  52.08      0   0.2188  0       0  {α } +  0.066 6  = 0   −0.009 39  0  52.08      173 Chapter 10 Solutions Manual Orbital Mechanics for Engineering Students 385.4 0 0  {α } = − 0 416 . 7 0    0 0 52.08  2 α = 0.000 6167 m s −1  0.2188   −0.000 567 6       0.066 6  =  −0.000 16   −0.009 39   0.000 180 3      Chapter 10 ( m s2 ) Problem 10.6 C 0.72 30 ωs = = −62.12 rad s C − A cos θ 0.72 − 0.36 cos 15° ωn = 0 ωp = ω x = ω p sin θ sin ψ + ω n cosψ = −62.12 sin 15° sin ψ + 0 ⋅ cosψ = −16.08 sin ψ ω y = ω p sin θ cosψ − ω n cosψ = −62.12 sin 15° cosψ − 0 ⋅ cosψ = −16.08 cosψ ω z = ω s + ω p cos θ = 30 + ( −62.12) cos15° = −30 1 Aω x2 + Bω y 2 + Cω z 2 2 1 2 2 2 = 0.36(−16.08 sin ψ ) + 0.36(−16.08 cosψ ) + 0.72( −30) 2 1 = 93.05 sin 2 ψ + cos2 ψ + 648 2 1 = (93.05 + 648) 2 = 370.5 J TR = ( [ [ ) ( ) ] ] Or, H = Aω p = 0.36( −62.12) = −22.36 kg ⋅ m 2 / s TR = 1 H2  C−A  1+ sin 2 θ   2 C A 2 1 ( −22.36)  0.72 − 0.36  1+ sin 2 15°  2 0.72  0.36 1 = ⋅ 694.5 ⋅ (1 + 0.066 99) 2 = 370.5 J = Problem 10.7 1  12 m  [I G ] =     [l2 + (2 l)2 ] 0 0 1 2 m l2 + (3 l) 12 0 0 5 2  12 ml  {H 0 } = [I G ]{ω } =  0   0  2 H 0 = 1.121ml ω 0 [ 0 5 2 ml 6 0  5 2   12 ml   = 0 0   1 2 2 m (2 l) + (3 l)   0   12 0 ] [ ]   2    1.5ω 0   0.625ml ω 0  0  0.8ω 0  = 0.6667 ml2ω 0    2   13 2  0.6ω 0   0.65ml ω 0  ml  12 0 174 0 5 2 ml 6 0    0   13 2  ml  12 0 Solutions Manual T0 = Orbital Mechanics for Engineering Students 1 1 ω ⋅ H 0 = 1.5ω 0 2 2 0.8ω 0 (a) Chapter 10  0.625ml2ω 0    0.6ω 0  0.6667 ml2ω 0  = 0.9305ml2ω 0 2  0.65ml2ω  0   1 1 13 2 2 Cω 2 = ml ω = 0.5417 ml2ω 2 2 2 12 T = T0 T= 0.5417 ml2ω 2 = 0.9305ml2ω 0 2 ω = 1.718ω 0 = 1.311ω 0 (b) H = Cω = 13 2 ml ω 12 H = H0 13 2 ml ω = 1.121ml2ω 0 12 ω = 1.035ω 0 Problem 10.8 H initial = 1000 ⋅ 6 = 6000 kg ⋅ m 2 s 1 Tinitial = ⋅ 1000 ⋅ 6 2 = 18 000 kg ⋅ m 2 s2 2 H final = 5000ω final H final = H initial 5000ω final = 6000 ω final = 1.2 rad s 1 1 ⋅ 5000 ⋅ ω final2 = ⋅ 5000 ⋅ 1.2 2 = 3600 kg ⋅ m 2 s2 2 2 ∆T = T final − Tinitial = 3600 − 18 000 = −14 400 J T final = Problem 10.9  0.1522 −0.03975 0.012  10   1.522   {H G0 } = −0.03975 0.07177 0.04057  0  = −0.3975 kg ⋅ m 2 s  0.012 0.04057 0.1569   0   0.1200  2 H G0 = H G0 = 1.5776 kg ⋅ m s ( T0 = 10  1 1   H G0 ⋅ ω 0 = 1.522 −0.3975 0.1200   0  = 7.610 J 2 2 0    I max = 0.1747 kg ⋅ m 2 H G f = I maxω f ( from Example 10.11) = 0.1747ω f H G f = H G0 0.1747ω f = 1.578 ω f = 9.03 rad s 175 ) Solutions Manual Orbital Mechanics for Engineering Students Chapter 10 1 1 ⋅I ω 2 = ⋅ 0.1747 ⋅ 9.03 2 = 7.123 J 2 max f 2 ∆T = T f − T0 = 7.123 − 7.610 = −0.4867 J Tf = Problem 10.10 ( ( ) ( ) ( p) ) ( ) ( ( p) ( ) C rω 2r + C p ω 2r + ω rel 2 = C rω 1 r + C p ω 1 r + ω rel 1 (r) (r) ω 2 = ω1 + ( ) ω 2r = 3 + ( ( p) ( p) C p ω rel 1 − ω rel 2 ) ) Cp + Cr 500(1 − 0.5) = 3.167 rad s 500 + 1000 Problem 10.11 ωs C 1000 0.1 = = 0.0266 rad s A − C cos θ 5000 − 1000 cos 20° π π t= = = 118.1 s ω p 0.0266 ωp = Problem 10.12 A cos γ = A2 + C 2 tan 2 φ 2 m ( 2 2 ) 500 ( 1 1 3r + l = 3 ⋅ 0.52 + 2 2 ) = 197.9 kg ⋅ m 2 C = mr 2 = 500 ⋅ 0.52 = 62.5 kg ⋅ m 2 12 12 2 2   2   2   A  − A2    197.9  − 197.9 2    cos γ    −1   cos 5°  φ = 2 tan −1   = 2 tan   = 30.97° 2 C 62.52     A= Problem 10.13 Npd = I (ω 2 − ω 1 ) Npd ω 2 = ω1 + I 30 ⋅ 15 ⋅ 1.5 ω 2 = 0.01 ⋅ 2π + = 0.4003 rad s = 0.0637 rev s 2000 Problem 10.14 HG 0 = Aω 0 x iˆ + Bω 0 y ˆj + Cω 0 z kˆ = 2000 ⋅ 0.1iˆ + 4000 ⋅ 0.3 ˆj + 6000 ⋅ 0.5kˆ = 200iˆ + 1200 ˆj + 3000kˆ H G = H G 0 + ∆H G ( = 200iˆ + 1200 ˆj + 3000kˆ + 50iˆ − 100 ˆj + 300kˆ = 250iˆ + 1100 ˆj + 3300kˆ ) 176 Solutions Manual Orbital Mechanics for Engineering Students Aω x iˆ + Bω y ˆj + Cω z kˆ = 250iˆ + 1100 ˆj + 3300kˆ 250 250 = = 0.125 rad s A 2000 1100 1100 = = = 0.275 rad s B 4000 3300 3300 = = = 0.55 rad s C 6000 ωx = ωy ωz ω = ω x2 + ω y 2 + ω z 2 = 0.6275 rad s Problem 10.15 1 2 1 1 1 mr + ml2 = 300 ⋅ 1.52 + 300 ⋅ 1.52 = 225 kg ⋅ m 2 4 12 4 12 1 1 C = mr 2 = 300 ⋅ 1.52 = 337.5 kg ⋅ m 2 2 2 A=  2π ˆ  HG = Cω1kˆ = 337.5 ⋅  1 ⋅ k = 35.34kˆ kg ⋅ m 2 1  60  ( HG = HG + I 2 1 M ( HG = 35.34kˆ + −I M ˆj 2 ) H G2 = 35.34 2 + I M 2 = 1249 + I M 2 H G2 = Aω p 1249 + I M 2 = 225 ⋅ (0.1 ⋅ 2π ) = 141.4 I M 2 = 18740 I M = 136.9 N ⋅ m ⋅ s Problem 10.16 K =1 + (a) C 2mR lf = R K t= 2 =1 + ω0 −ω f ω0 + ω f 300 2 ⋅ 3 ⋅ 1.52 = 23.22 = 1.5 23.22 ⋅ 5 −1 = 5.902 m 5 +1 K ω0 −ω f 23.22 5 − 1 = = 0.7869 s 2 ω +ω 52 5 + 1 ω0 f 0 (b) lf = R K t= ω0 −ω f ω0 + ω f = 1.5 23.22 ⋅ 5 −0 = 7.228 m 5+0 K ω0 −ω f 23.22 5 − 0 = = 0.9636 s 2 ω +ω 52 5 + 0 ω0 0 f 177 ) Chapter 10 Solutions Manual Orbital Mechanics for Engineering Students Chapter 10 Problem 10.17 T0 = 1 1 1 1 A ω x2 + ω y 2 + Cω z 2 = AΩ 2 + Cω 0 2 2 2 2 2 ( ) C 60 2 ωs = = −4.141 rad s A − C cos θ 30 − 60 cos 15° 30 A ( −4.141) cos 15° = −2 rad s ω 0 = ω p cos θ = C 60 Ω = ω p sin θ = −4.141 sin 15° = −1.072 rad s ωp = ∴ T0 = (a) 1 1 2 2 ⋅ 30( −1.072) + ⋅ 60( −2) = 137.2 J 2 2 H 0 = Aω p = 30 ⋅ 4.141 = 124.2 kg ⋅ m 2 s H f = H0 60ω f = 124.2 ⇒ ω f = 2.071 rad s (b) 1 1 Cω f 2 = ⋅ 60 ⋅ 2.0712 = 128.6 J 2 2 ∆T = T f − T0 = 128.6 − 137.2 = −8.616 J Tf = (c) lf = R 1+ C 2 2mR =1⋅ 1 + 60 = 2.299 m 2 ⋅ 7 ⋅ 12 Problem 10.18 A 0 0  0   0  {H G } = [I G ]{ω } =  0 A 0   n  =  nA   0 0 C   −ω s   −ω sC  ˆj iˆ kˆ M =Ω×H = 0 n 0 = − Cω niˆ G G s 0 nA −ω sC 2π = 0.1047 rad /s 60 2π 2π + 365.26 = 7.292 × 10 −5 rad /s n= 24 ⋅ 3600 C = 550 kg ⋅ m 2 ωs =1⋅ ∴ MG x = −550 ⋅ 0.1047 ⋅ 7.292 × 10 −5 = −0.004 2 N ⋅ m Problem 10.19 ω 0 and ω f are the initial and final angular velocities of the spacecraft. ω is the angular velocity of the flywheel relative to the vehicle. [H (Gv) + H (Gv) ]0 = [H (Gv) + H ( w) ] f ( Aω iˆ + Aω ˆj + Cω kˆ ) + (I ω iˆ + I ω ˆj + I ω kˆ ) = ( Aω iˆ + Aω ˆj + Cω kˆ ) +  I (ω + ω ) iˆ + I ω ˆj + I ω kˆ    0x 0y x 0z y x 0x z x y 0y x z 0z y 178 y z z Solutions Manual Orbital Mechanics for Engineering Students ( A + I z )ω z = ( A + I z )ω 0 z ( ⇒ ω z = ω0z = 0 A + I y ω y = A + I y ω 0 y ⇒ ω y = ω 0 y = 0.05 rad s ) ( ) ( A + I x )ω x + I xω = ( A + I x )ω 0 x A  ⇒ ω =  1 +  (ω 0 x − ω x ) Ix   1000   (0.1 − 0.003) = 4.947 rad s ∴ω = 1 +  20  Problem 10.20 Given: I 3 > I 2 > I1 . Figure 10.29, Stable region I: I roll > I yaw > I pitch : I1 axis in pitch direction (normal to orbital plane) I 2 axis in yaw direction (radial) I 3 axis in roll direction (local horizon) Figure 10.29, Stable region II (preferred): I pitch > I roll > I yaw : I1 axis in yaw direction (radial) I 2 axis in roll direction (local horizon) I 3 axis in pitch direction (normal to orbital plane) 179 Chapter 10 Solutions Manual Orbital Mechanics for Engineering Students 180 Chapter 10 Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 Problem 11.1 mp = mp out + mp in = mp out + mp out 4 = 5 m 4 p out Outbound leg: m e + m p + mPL   ∆v = I sp go ln    m e + m p − m p out + mPL  5   m e + m p + 3500   out 4 4220 = 430 ⋅ 9.81 ⋅ ln   5  m e + m p − m p + 3500    out out 4 5   + 3500 m + m  e 4 p out  = 430 ⋅ 9.81 ⋅ ln   1  m e + m p + 3500    out 4 5 m e + m p + 3500 out 4 = 2.719 1 m e + m p + 3500 out 4 0.5702m p − 1.719m e = 6018 out (1) Return from GEO tp LEO: 1   m + m + 3500  e 4 p out  ∆v = I sp go ln   me   1   m + m  e 4 p out  4220 = 430 ⋅ 9.81 ⋅ ln   me   1 me + mp out 4 = 2.719 me m p = 6.876m e (2) out Substitute (2) into(1): 0.5702(6.876m e ) − 1.719m e = 6018 m e = 2733 kg Problem 11.2 First stage: c = I sp go = 235 ⋅ 9.81 = 2943 m s ( )  m  m0 − m f go  249.5  (249.5 − 170.1) ⋅ 9.81 = 1127 − 73.38 = 1054 m s v bo = c ln  0  − = 2943 ln −  170.1  « me 10.61 mf   mf   1  m0 − m f  2  ln   m f + m0 − m f  −   go   m0   2  m«e   1  249.5 − 170.1  2 2943   170.1  = ln ⋅ 170 . 1 + 249 . 5 − 170 . 1 9.81 −2  10.61   249.5  10.61  c h bo = « me 181 Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 = 3947 − 274.4 = 3673 m After 3 second staging delay: v = v bo − g∆t s = 1054 − 9.81 ⋅ 3 = 1024 m s 1 1 h = hbo + v bo ∆t s − g∆t s2 = 3673 + 1054 ⋅ 3 − ⋅ 9.81 ⋅ 3 2 = 3673 + 3117 = 6790 m 2 2 Second stage: v0 = 1024 m s h0 = 6790 m c = I sp go = 235 ⋅ 9.81 = 2305 m s ( )  m  m0 − m f go v bo = v0 + c ln  0  − m«e mf   113.4  (113.4 − 58.97 ) ⋅ 9.81 −  58.97  4.0573 = 1024 + 1508 − 131.7 = 1024 + 2305 ln = 2400 m s  1  m0 − m f  2  m0 − m f  c  mf  h bo = h0 + v0   ln   go +  m f + m0 − m f  −   m«e  m«e   m0   2  m«e   1  113.4 − 58.97  2  113.4 − 59.97  2305   59.97  + ln ⋅ 58.97 + 113.4 − 58.97  − ⋅ 9.81    4.053   113.4   4.053 4.063  2 = 6790 + 13 760 + 9028 − 884.7 = 28 690 m = 6790 + 1024 Coast to apogee: v0 = 2400 m s h0 = 28 690 m v 2400 0 = v0 − gt max ⇒ t max = 0 = = 244.7 s g 9.81 1 1 hmax = h0 + v0 t max − gt max 2 = 28 690 + 2400 ⋅ 244.7 − ⋅ 9.81 ⋅ 244.7 2 = 322 300 m 2 2 Problem 11.3 v0 = ω earth Rearth cos φ = 7.292(10 −5 ) ⋅ 6378 ⋅ cos 28° = 0.4107 km s 398 600 + 2 − 0.4107 = 9.315 km s 6678 = ∆v = 9.315 km s = v bo1 + v bo2 ∆v = v bo v bo  m0   m0  v bo = I sp go ln  1  + I sp go ln  2  1 2  m f1   m f2  182 Solutions Manual Orbital Mechanics for Engineering Students 2 ⋅ 525 000 + 30 000 + 600 000 + mPL   9315 = 290 ⋅ 9.81 ⋅ ln    2 ⋅ (525000 - 450000) + 3000 + 600000 + mPL   30 000 + 600 000 + mPL  + 450 ⋅ 9.81 ⋅ ln   30 000 + mPL    1 680 000 + mPL   630 000 + mPL  9315 = 2845 ⋅ ln   + 4414 ⋅ ln    753 000 + mPL   30 000 + mPL  To find the value of mPL satisfying this equation, graph the function  1 680 000 + mPL   630 000 + mPL  f = 9315 − 2845 ⋅ ln   − 4414 ⋅ ln    753 000 + mPL   30 000 + mPL  f 500 0 -500 100 000 120 000 mPL (kg) 110 800 f = 0 when mPL = 110 800 kg Problem 11.4 π PL = π PL1/3 λ= ε= 10 000 mPL = = 0.06667 150 000 m0 1 − π PL1/3 = 0.682 20 000 mE = = 0.1429 m0 − mPL 150 000 − 10 000 (a) 1+λ 1 + 0.682 = = 2.039 ε + λ 0.1429 + 0.682 ∆v = I sp go ln n 3 = 310 ⋅ 0.009 81 ⋅ ln 2.039 3 = 6.5 km s n= (b) mp = (1 − πPL1/3 )(1 − ε) m mp = (1 − πPL1/3 )(1 − ε) m = mp = (1 − πPL1/3 )(1 − ε) m = 1 2 3 π PL π PL2/3 π PL1/3 PL = PL PL (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 2/3 (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 1/3 183 ⋅ 10 000 = 76 440 kg ⋅ 10 000 = 30 990 kg ⋅ 10 000 = 12 570 kg Chapter 11 Solutions Manual Orbital Mechanics for Engineering Students (c) mE1 = (1 − πPL1/3 )ε m mE2 = (1 − πPL1/3 )ε m = mE3 = (1 − πPL1/3 )ε m = PL π PL π PL2/3 π PL1/3 PL PL = (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 2/3 (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 1/3 10 000 = 12740 kg 10 000 = 5166 kg 10 000 = 2095 kg (d) m0 3 = mE3 + m p + mPL = 2095 + 12 570 + 10 000 = 24 660 kg 3 m0 2 = mE2 + m p + m0 3 = 5166 + 30 990 + 24 660 = 60 820 kg 2 m01 = mE1 + m p + m0 2 = 12740 + 76 440 + 60 820 = 150 000 kg 1 Problem 11.5 c1 = I sp go = 300 ⋅ 0.009 81 = 2.943 km s 1 c2 = I sp go = 235 ⋅ 0.009 81 = 2.305 km s 2 ε 1 = 0.2 ε2 = 0.3 v bo = 6.2 km s 2  c iη − 1   c1η − 1   c 2η − 1    = c1 ln   + c2 ln  c iε iη   c1ε1η   c2 ε2η  i =1  2.943η − 1   2.305η − 1  6.2 = 2.943 ln   + 2.305 ln    2.943 ⋅ 0.2η   2.305 ⋅ 0.3η  v bo = ∑ c i ln   2.943η − 1   2.305η − 1  6.2 = 2.943 ln   + 2.305 ln    0.5886η   0.6915η  To find η , graph the function  2.943η − 1   2.305η − 1  f = 2.943 ln   + 2.305 ln   − 6.2  0.5886η   0.6915η  As shown below, f = 0 when η = 1.726 . c η − 1 2.943 ⋅ 1.726 − 1 = = 4.016 n1 = 1 2.943 ⋅ 0.2 ⋅ 1.726 c1ε1η c η − 1 2.305 ⋅ 1.726 − 1 = = 2.496 n2 = 2 c2 ε2η 2.305 ⋅ 0.3 ⋅ 1.726 2.496 − 1 n2 − 1 ⋅ 10 = 59.53 kg mPL = 1 − ε2 n2 1 − 0.3 ⋅ 2.496 4.016 − 1 n −1 m1 = 1 (m + mPL ) = 1 − 0.2 ⋅ 4.016 ( 59.53 + 10) = 1065 kg 1 − ε1 n1 2 m2 = M = m1 + m2 = 1065 + 59.53 = 1124 kg 184 Chapter 11 Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 f 0 -1 1.5 2 2.5 1.726 Problem 11.6 z = x2 + y 2 + 2 xy g = x2 − 2 x + y 2 ( h = z + λg = x2 + y 2 + 2 xy + λ x2 − 2 x + y 2 ) ∂h = 2 x + 2 y + λ (2 x − 2) = 0 ⇒ (λ + 1) x + y = λ ∂x ∂h = 2 y + 2 x + λ (2 y) = 0 ⇒ x + (λ + 1) y = 0 ∂y λ + 1 1   x  λ  ∴  =  λ + 1  y   0   1 1   x  λ + 1  = λ + 1 y  1 −1  λ +1  λ + 1 −1  λ   λ + 2  λ  1  = (     =   0  λ λ + 2)  −1 λ + 1  0   − 1   λ +2 x2 − 2 x + y 2 = 0 (λ + 1)2 (λ + 2)2 −2 1 λ +1 + =0 λ + 2 (λ + 2)2 2 Multiply through by (λ + 2) (this is okay since λ + 2 = 0 clearly does not correspond to a local extremum). Then (λ + 1)2 − 2(λ + 1)(λ + 2) + 1 = 0 or λ2 + 4λ + 2 = 0 The two roots are −0.5858 and − 3.414 . λ = −0.5858: λ + 1 −0.5858 + 1 = = 0.2929 λ + 2 −0.5858 + 2 1 1 y=− =− = −0.7071 λ +2 −0.585 + 2 x= 185 Solutions Manual Orbital Mechanics for Engineering Students 2 z1 = x2 + y 2 + 2 xy = 0.2929 2 + ( −0.7071) + 2 ⋅ 0.2929( −0.7071) = 0.1716 λ = −3.414: λ + 1 −3.414 + 1 = = 1.707 λ + 2 −3.414 + 2 1 1 y=− =− = 0.7071 −3.414 + 2 λ +2 z2 = x2 + y 2 + 2 xy = 1.707 2 + 0.70712 + 2 ⋅ 1.707 ⋅ 0.7071 = 5.828 x= Note that  ∂2z  ∂2z  ∂2z ∂ 2 g ∂2 g  ∂ 2 g 2 d 2 h =  2 + λ 2  dx2 + 2 +λ  dxdy +  2 + λ 2  dy  ∂x ∂x∂y   ∂x∂y ∂x  ∂y   ∂y d 2 h = (2 + λ ⋅ 2) dx2 + 2(2 + λ ⋅ 0) dxdy + (2 + λ ⋅ 2) dy 2 ( ) ( ) d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy For λ = −0.5858 , ( ) d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy = 2( −0.5858 + 1) dx2 + dy 2 + 4 dxdy ( ) = 0.8284 dx2 + dy 2 + 4 dxdy Since d 2 h > 0 , z1 = zmin . For λ = −3.414 , ( ) ( ) d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy = 2( −3.414 + 1) dx2 + dy 2 + 4 dxdy ( ) = −4.828 dx2 + dy 2 + 4 dxdy Since d 2 h < 0 , z2 = zmax . 186 Chapter 11 Solutions Manual Orbital Mechanics for Engineering Students 187 Chapter 11 Solutions Manual Orbital Mechanics for Engineering Students 188 Chapter 11

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