Exam (elaborations) TEST BANK FOR Fundamentals of Dynamics and Control of Space Systems By Krishna Dev Kumar (Solution Manual)

Exam (elaborations) TEST BANK FOR Fundamentals of Dynamics and Control of Space Systems By Krishna Dev Kumar (Solution Manual) Fundamentals of Dynamics and Control of Space Systems Solution Manual Krishna Dev Kumar Professor and Canada Research Chair in Space Systems Department of Aerospace Engineering Ryerson University Toronto, Canada Contents Preface v 2 Kinematics, Momentum and Energy 1 3 Forces and Torques 33 4 Dynamics I 39 5 Dynamics II 45 6 Mathematical and Numerical Simulation 65 7 Control System 85 8 Formation Flying 115 Index 121 vii Chapter 2 Kinematics, Momentum and Energy Problem Set 2 2.1 The coordinate frames used in studying the dynamics of a spacecraft are as follows: a) Inertial reference frame, b) Orbital reference frame, c) Perifocal reference frame, c) Satellite body-fixed reference frame. 2.2 The inertial frames are those coordinate frames that are nonrotating and nonaccelerating frames. The inertial frames are relevant because in applying the Newton’s second law of motion ~F = m d~V dt (2.1) to derive the equation of motion of a system, the velocity ~V and the corresponding acceleration d~V /dt in the right-hand side of the above equation are to measured with respect to an inertial frame of reference. An Earth-fixed frame is not an inertial frame as it is spinning about its axis with a period of 24 hour. When viewed from space, the point on the surface of the earth moves in a circle as the earth spins on its axis. Thus, it is accelerating with an centripetal acceleration of r!2, 2 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY where r is the position of the point of the Earth center of mass and ! is the rate of spin of the Earth. With the earth a point on its surface also orbits the Sun. With the solar system, it orbits the center of the galaxy. Thus, the Earth-fixed frame is an accelerating frame and not an inertial frame. We consider just the effect of the spinning motion of the Earth and therefore the inertial acceleration can be written as d~V dt  inertial = d~V dt  body + ~! × ~Vbody (2.2) The corresponding error in considering an Earth-fixed frame as an inertial frame is Error = d~V dt  inertial − d~V dt  body = ~! × ~Vbody (2.3) The Earth’s spin rate ! is ~! = !kˆk = − 2 T ˆk = − 2 24 × 3600 ˆk = 7.275 × 10−5ˆk (2.4) where ˆk is a unit vector along the z-direction as taken for the aircraft body-fixed frame. The order of magnitude error would be 10−4×Vbody. As this magnitude is usually very small when compared to the magnitude of other relevant accelerations like the gravitational acceleration, which is 9.81 m/s2, and we often treat the Earth-fixed frame as an inertial frame. when solving problems. 2.3 The inertial position vectors for spacecraft m1 and m2 are ~ R1 = ~R − ~L (2.5) ~ R2 = ~R + (1 − )~L (2.6) where = m2/(m1 + m2). The corresponding magnitudes are R1 = [R2 + 2L2 − 2 ~R · ~ L]1/2 (2.7) R2 = [R2 + (1 − )2L2 + 2(1 − )~R · ~ L]1/2 (2.8) where L = L0 + vt. The nomenclature Lo defines the initial length of the cable while v is the speed by which the length of the cable varies. 3 Z Orbit E q b Y X S L m2 m1 R Figure 2.1: A dumbbell satellite system undergoing in-plane libration. Expressing ~R and ~L in terms of unit vectors of the respective coordinate frames as ~R = Rˆio, ~L = Lˆi (2.9) Applying the transformation between coordinate frames S−xoyozo and S − xyz, we get ˆi o ·ˆi = cos (2.10) and using this relation in Eqs. (2.7) and (2.8), we have the inertial positions of the spacecraft m1 and m2 as R1 = [R2 + 2L2 − 2 RLcos]1/2 (2.11) R2 = [R2 + (1 − )2L2 + 2(1 − )RLcos]1/2 (2.12) The inertial velocity vectors for spacecraft m1 and m2 are ~V1 = ˙~ R1 = ˙~R − ˙~L (2.13) ~V2 = ˙~ R2 = ˙~R + (1 − ) ˙~L (2.14) The corresponding magnitudes are V1 = [ ˙~R 2 + 2 ˙~L2 − 2 ˙~R · ˙~L ]1/2 (2.15) V2 = [ ˙~R 2 + (1 − )2 ˙~L2 + 2(1 − ) ˙~R · ˙~L ]1/2 (2.16) 4 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY ˙~R and ˙~L can be written as ˙~R =  ˙~R  xoyozo + ~!o × ~R (2.17) ˙~L =  ˙~L  xyz + ~! × ~L (2.18) Knowing the system is orbiting in a circular orbit (i.e., ˙R = 0), and the cable connecting the two spacecraft is moving with a constant speed of v, we get  ˙~R  xoyozo = 0,  ˙~L  xyz = ~v (2.19) where ~v is in the direction of ~ L. Substituting the above relations in Eqs. (2.17)-(2.18), we obtain ˙~R = ~!o × ~R (2.20) ˙~L = ~v + ~! × ~L (2.21) The terms ˙~R 2 and ˙~L 2 can be written as ˙~R 2 = (~!o × ~R )2 (2.22) ˙~L 2 = v2 + (~! × ~ L)2 + 2~v · (~! × ~ L) (2.23) Writing ~!o, ~R, ~!, ~ L, and ~v in terms of the unit vectors of the respective coordinate frames, we have !o = ˙ ˆko, ~R = Rˆio, ~! = ( ˙  + ˙)ˆk, ~L = Lˆi, ~v = vˆi (2.24) Inserting these expressions into Eqs. (2.22)-(2.23) and solving, we have ˙~R 2 = ˙2R2 (2.25) ˙~L 2 = v2 + ( ˙  + ˙)2L2 (2.26) Note that as ~v ? (~! × ~L), ~v · (~! × ~L) = 0. Next we derive ˙~R · ˙~L . Using Eqs. (2.20)-(2.21), we can write ˙~R · ˙~L = (~!o × ~R) · (~v + ~! × ~L) = (~!o × ~R) · ~v + (~!o × ~R) · (~! × ~L) = ˙ Rv(ˆjo · ˆi) + ˙ ( ˙  + ˙)RL(ˆjo · ˆj ) (2.27) 5 From the coordinate transformation between the coordinate frame S − ˆioˆjoˆko and S −ˆi ˆj ˆk, we have ˆj o ·ˆi = sin (2.28) ˆj o ·ˆj = cos (2.29) Thus, we can express ˙~R · ˙~L as ˙~R · ˙~L = ˙ Rvsin + ˙ ( ˙  + ˙)RLcos (2.30) Substituting the expressions for ˙~R 2 and ˙~L 2 from Eqs. (2.25)-(2.26) and the expression for ˙~R · ˙~L from Eq. (2.30) into Eqs. (2.15)-(2.16), we finally obtain the magnitudes of the inertial velocity vectors for spacecraft m1 and m2 as V1 ={ ˙2R2 + 2(˙L2 + ( ˙  + ˙ )2L2) − 2 ˙R[vsin + ( ˙  + ˙)Lcos]}1/2 (2.31) V2 ={ ˙2R2 + (1 − )2(˙L2 + ( ˙  + ˙ )2L2) + 2(1 − ) ˙ R[vsin + ( ˙  + ˙)Lcos]}1/2 (2.32) The inertial acceleration vectors for the spacecraft m1 and m2 are written using Eqs. (2.13)-(2.14) for their velocity vectors, as ~a1 = ˙~ V1 = ¨~R − ¨~L (2.33) ~a2 = ˙~ V2 = ¨~R + (1 − ) ¨~L (2.34) Here ¨~R and ¨~L can be expressed as ¨~R = ¨~Rxoyozo + 2(~!o × ˙~R xoyozo ) + ~!o × (~!o × ~R) + ˙~!o × ~R (2.35) ¨~L = ¨~Lxyz + 2(~! × ˙~L xyz) + ~! × (~! × ~ L) + ˙~! × ~L (2.36) The centripetal acceleration components ~!o×(~!o× ~ R) and ~!×(~!×~ L) can be simplified using the triple vector product relation ~a × (~b ×~c) = (~a · ~c)~b − (~a ·~b)~c, as ~!o × (~!o × ~ R) = (~!o · ~R)~!o − !2 o ~R = −!2 o ~R (2.37) ~!o × (~!o × ~ L) = (~!o · ~L)~!o − !2 o ~L = −!2~L (2.38) Note the terms ~!o · ~R = 0 and (~!o · ~ L) = 0 since !o ? ~R and ! ? ~L. Knowing that the system is in a circular orbit and the cable connecting the two spacecraft is deployed with constant velocity, i.e., ˙~R xoyozo = ¨~Rxoyozo = 0, ˙~!o = 0, ¨~Lxyz = v˙ = 0, ˙~! = ¨ ˆk (2.39) 6 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY and writing all vectors in Eqs. (2.35)-(2.36) in terms of the unit vectors along the respective coordinate frames, we obtain ¨~R = −˙2Rˆio (2.40) ¨~L = 2( ˙  + ˙)v(ˆk ׈i) − ( ˙  + ˙)2Lˆi + ¨L(ˆk ×~i) = 2( ˙  + ˙)vˆj − ( ˙  + ˙)2Lˆi + ¨ Lˆj = −( ˙  + ˙ )2Lˆi + [2( ˙  + ˙ )v + ¨ L]ˆj (2.41) Substituting the above relations in Eqs. (2.33)-(2.34), we have the inertial accelerations as ~a1 = −˙2Rˆio + ( ˙  + ˙)2Lˆi − [2( ˙  + ˙ )v + ¨ L]ˆj } (2.42) ~a2 = −˙2Rˆio − (1 − )( ˙  + ˙)2Lˆi − (1 − )[2( ˙  + ˙ )v − ¨ L]ˆj } (2.43) The corresponding magnitude for spacecraft m1 can be expressed using the algebraic relations ((a+b+c)2 = a2 +b2+c2+2ab+2ac+2bc)) as a1 = n (˙2R)2 + 2( ˙  + ˙ )4L2 + 2[2(˙ + ˙ )v + ¨ L]2 − 2 ˙2( ˙  + ˙ )2RL(ˆio · ˆi) + 2 ˙ 2 [2( ˙  + ˙ )v + ¨ L]R(ˆio · ˆj) o1/2 (2.44) From the coordinate transformation between the coordinate frames S− ˆioˆjoˆko and S −ˆi ˆj ˆk, we get ˆi o · ˆi = cos, ˆio · ˆj = cos(90 + ) = −sin (2.45) Using these relations, the corresponding magnitude for spacecraft m1 can be expressed as a1 = n (˙2R)2 + 2( ˙  + ˙)4L2 + 2[2( ˙  + ˙)v + ¨ L]2 − 2 ˙2( ˙  + ˙ )2RLcos − 2˙2 [2( ˙  + ˙)v + ¨ L]Rsin o1/2 (2.46) Similarly, the acceleration of spacecraft m2 can be obtained as a2 = n (˙2R)2 + (1 − )2( ˙  + ˙ )4L2 + (1 − )2[2( ˙  + ˙ )v − ¨ L]2 + 2(1 − )˙2( ˙  + ˙ )2RLcos − 2˙2(1 − )[2( ˙  + ˙ )v − ¨ L]Rsin o1/2 (2.47) Note the preceding expression can be derived simply by replacing replacing by (1 − ) and L by −L in Eq. (2.46). 7 2.4 The inertial position vectors for spacecraft m1 and m2 are ~ R1 = ~R − ~L (2.48) ~ R2 = ~R + (1 − )~L (2.49) where = m2/(m1 + m2). The corresponding magnitudes are R1 = [R2 + 2L2 − 2 ~R · ~L]1/2 (2.50) R2 = [R2 + (1 − )2L2 + 2(1 − )~R · ~ L]1/2 (2.51) Writing ~R and ~L in terms of the unit vectors of the respective coordinate frames, we have ~R = Rˆio; ~L = Lˆi (2.52) To expressˆi in terms of unit vectors in the frame S−xoyozo, we consider the transformation as 8>>>< >>>: ˆi ˆj ˆk 9>>>= >>>; = Rzy(, ) 8>>>< >>>: ˆi o ˆj o ˆko 9>>>= >>>; (2.53) where Rzy(, ) is Rzy(, ) = Ry()Rz() = 2 6664 cos 0 −sin 0 1 0 sin 0 cos 3 7775 2 6664 cos sin 0 −sin cos 0 0 0 1 3 7775 = 2 6664 coscos sincos −sin −sin cos 0 cossin sinsin cos 3 7775 (2.54) Using Eq. (2.52), we can write ˆi as ˆi= cos cos ˆi o + sincosˆjo − sinˆko (2.55) Applying the above relation and using Eqs. (2.52), we obtain the magnitudes of the position vectors as R1 = [R2 + 2L2 − 2 RLcoscos]1/2 (2.56) R2 = [R2 + (1 − )2L2 + 2(1 − )RLcoscos]1/2 (2.57) 8 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY