As a starting point of our derivation of the equations of motion for a rigid body, we employ d Alembert s principle:

Transcription

1 MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 017) Lecture 10: Equations of Motion for a Rigid Body 10.1 D Alembert s Principle for A Free Rigid Body As a starting point of our derivation of the equations of motion for a rigid body, we employ d Alembert s principle: δz (df ρ z)d = 0 (10.1) in which ρ is the density of the rigid material, df is the external force per unit volume acting on the body, d is the infinitesimal volume that surrounds the particle point. The acceleration vector, z, has been already derived in (8.37). Hence, the only additional quantity we do not have, save for df, is the variation of the displacement or virtual displacement, δz, which can be expressed as δz = δz 0 + δr (10.) Note that δr must consist of the virtual translation of r with respect to the b-basis: δr = δx T b + δα r, δ α = δrr T (10.3) However, since we are dealing with a rigid body, we must have ẋ = ẍ = δx = 0. Therefore, the virtual displacement, δz, is expressed for as δz = δz 0 + δα r (10.4) Substitution of (8.37) and (10.4) into (10.1) and collecting the 1

2 resulting terms associated with δz and δα yield δz 0 (df ρ( z 0 + r))d + (δα r) (df ρ( z 0 + r))d dt = 0 (10.5) For a straightforward evaluation of the terms involving the relative acceleration ( r), viz., ρ(δα r) rd = ρ(δα r) ( ω r+ω (ω r) )d (10.6) we work with its angular momentum (p rot ), instead: δα ρ(δα r) ṙd = ρ(δα r) ṙd = δα ρ(δα r) (ω r)d ρr (ω r)d ρr (ω r)d = δα [ ρ(r ri rr)d ] ω = δα p rot p rot = J ω, J = ρ(r ri rr)d (10.7) Hence, we have ρ(δα r) rd = δα ṗ rot = δα (ω J ω + J ω) (10.8) The derivation of the equations of motion for a rigid body via d Alembert s principle is now reduced to the volume integrals in (10.5). 10. Mass Distributions The translational mass of a rigid body is given by m = ρd (10.9)

3 If the center of mass is not located at the origin of the body-fixed coordinates, we have ρrd = mr c (10.10) in which r c is the vector from the origin of the b-basis coordinates to the mass center of the rigid body Euler s Equations for a Free Rigid Body Substituting (10.8)-(10.10) into (10.5) and carrying out the necessary volume integrals, we obtain δz 0 (f m( z 0 + ω r c + ω (ω r c )) + δα (m mr c z 0 J ω ω J ω) = 0 f = v df d, m = r dfd (10.11) For unconstrained rigid bodies, we can perform the variations of δz and δα independently (why?). In addition, when the two variations are arbitrary, the quantities within the parentheses must vanish: Translation equations: m( z 0 + ω r c + ω (ω r c )) = f Rotational equations: J ω + ω J ω + mr c z 0 = m (10.1) The two equations in (10.1) become the classical Euler equations if the mass center coincides with the origin of the bodyfixed coordinates, i.e., r c = 0. We have derived the above equations of motion for cases when the mass center is deliberately away from the origin of b-basis coordinates as those cases are far more practical. Remark 10.1: If z 0 is dependent each other or on α, or α themselves are interdependent through various constraint con- 3

4 ditions, one must first express them in terms of his or her choice of independent variables. Once the linear dependency is taken care of, the variations within the time-definite integrals should be carried out. For example, the dynamics of rigid chains with various joints require such careful treatments of constraints. On the other hand, by incorporating constraints via Lagrange multipliers, one can treat both δz 0 and δα as independent variables. It is this very ability of d Alembert s principle (10.1) or its specialized forms (10.5) that makes their use practically superior to other classical formulations in dynamics. Remark 10.: The equations of motion for a single rigid body can be most conveniently derived by invoking Newton s Law II in conjunction with the linear and angular momentum expressions. However, as we would like to view that the derivation of the equations of motion for a single body as one of the simplest engineering systems, we prefer d Alembert s principle as it can handle not only conservative systems but also non-conservative systems. One compelling reason for this adoption is that d Alembert s principle allows us to incorporate nonholonomic constraints in a variational frame. The question that comes to our mind is: Why not Hamilton s principle which provides both the boundary and initial conditions as part of the variational process? An answer for not adopting Hamilton s principle is that our experience with it for solving complex dynamical systems is still maturing. In fact, only during the past two decades a serious attempt to apply Hamilton s principle to largescale dynamical problems have been made. Hence, applications of Hamilton s principle to practical engineering problems remain as a challenge and largely a research activity. Remark 10.3: In passing it is worth mentioning the so-called Kane s adaptation of the Gibbs formulation, which can be considered as a sophisticated version of the vector formulation as an alternative to the variational formulation. Undoubtedly, Kane s formulation can lead to a streamlined formulation for relatively straightforward rigid-body systems. However, its extensions to systems with flexible bodies with a full discrete companion of the flexible part has met only with a limited applicability. This is because the most widely used methodology to obtain the equations of motion for flexible bodies has been the finite element method whose foundation is based on the variation principles. 4

5 10.4 Derivation of Equations of Motion for a Free Rigid Body via Hamilton s Principle Let us derive the equations of motion for a rigid body via Hamilton s Principle: t t 1 {δ(t ) + δw }dt = 0 (10.13) Since the velocity of a free rigid body can be expressed as (see equation(8.36)): ż = ż 0 + ω r (10.14) its kinetic energy is obtained by T = 1 ρż ż d ρ{ż 0 ż 0 + ż 0 (ω r) = 1 + (ω r) (ω r)} d T = 1 ρ{ż T 0 ż 0 + ż0 T R T ωr + ω T r r T ω} d T = 1 {m(żt 0 ż 0 + ż T 0 R T r T c ω) + ω T J ω} (10.15) Using equation(7.44),viz., δω = δ θ + ωδθ (10.16) variation of the kinetic energy T can be expressed as δt = δż T 0 m(ż 0 + R T r T c ω) + ż T 0 δr T r T c ω + δω T (m r c Rż 0 + J ω) (10.17) δt = δż T 0 m(ż 0 + R T r T c ω) + δθ T r T c ωrż 0 + (δ θ T + δθ T ω T )(m r c Rż 0 + J ω) Time definite integral of δt can be expressed as t t 1 δt dt = t = {δż0 T m(ż 0 + R T r c T ω) t 1 + δθ T [m r c T ωrż 0 ω(m r c Rż 0 + J ω)] + δ θ T (m r c Rż 0 + J ω)} dt = 0 (10.18) 5

6 After carrying out variations and setting δθ(t 1 ) = δθ(t ) = δz 0 (t 1 ) = δz 0 (t ) = 0, we obtain t t 1 δt dt = = t t 1 { δz T 0 m( z 0 + R T RṘT r T c ω + R T r T c ω) + δθ T [m r c T ωrż 0 ω(m r c Rż 0 + J ω)] δθ T (m r c ṘR T Rż 0 + m r c R z 0 + J ω)} dt = 0 (10.19) where R T R = I is inserted for algebraic manipulation. Knowing that ω = ṘRT, the preceding equation leads to: t t 1 δt dt = = t t 1 { δz T 0 m( z 0 + R T ω T r T c ω + R T r T c ω) (10.0) + mδθ T ( r T c ω ω r c r c ω T )Rż 0 δθ T ( ωj ω + m r c R z 0 + J ω)} dt = 0 The second term in the above equation can be shown to be zero by the following identities: ã b = ba T a T bi, ãb = ba T ab T (10.1) Hence, finally the time definite integral of the variational kinetic energy (δt ) leads to the following equation: t t 1 δt dt = = t t 1 { δz T 0 m( z 0 + R T ω T r T c ω + R T r T c ω) δθ T ( ωj ω + m r c R z 0 + J ω)} dt = 0 (10.) With the applied force and moment given by (10.11), we have the following variationally derived equations of motion for a rigid body: m( z 0 + R T ω T r T c ω + R T r T c ω) = f J ω + ωj ω + m r c R z 0 = m (10.3) which are the same as the vectorially represented equations of motion derived by D Alembert s principle (10.1). 6

7 10.5 Stability Analysis of A Free-Floating Rigid Body An important use of the Euler equations of a rigid body for torque-free motions is to determine the stability of a free-floating rigid body. For example, an airplane hovering in the air, a satellite orbiting around the earth, and a sea surface vessel floating on the ocean should be designed to prevent the loss of their orientations, tumbling, disorienting and rolling flip-over. Complete assessment of the stability of floating rigid bodies requires not only incipient tendency to lose stability, but also subsequent nonlinear analysis subjected to applied torques. For the preliminary design evaluation, it is often adequate to perform linearized stability analysis. This is addressed in this section. Euler s rotational equations of motion for the torque-free case (10.38) are given by J 1 ω 1 + (J 3 J )ω ω 3 = 0 J ω 1 + (J 1 J 3 )ω 1 ω 3 = 0 J 3 ω 1 + (J J 1 )ω 1 ω = 0 (10.4) In classical perturbation techniques for the stability analysis of the torque-free motion about, say, the third axis, we introduce ω 1 = ω = 0, ω 3 = Ω = const (10.5) so that the perturbed angular velocity components become ω 1 = θ 1, ω = θ, ω 3 = Ω + θ 3 (10.6) Substitution of the above relations into Euler s equations and introducing an inertia-proportional damping to account for a realistic situation, we obtain the following linearized equation: J 1 θ1 + b 1 θ1 + (J 3 J )Ω θ = 0 J θ + b θ + (J 1 J 3 )Ω θ 1 = 0 J 1 θ3 + b 3 θ3 = 0, where b i > 0 (10.7) The characteristic equation of the above equation can be ob- 7

8 tained by assuming a nontrivial solution in the form so that we obtain θ = θ 0 e st (10.8) sj 1 + b 1 (J 3 J )Ω 0 s (J 1 J 3 )Ω sj + b 0 θ 0 = 0 (10.9) 0 0 sj 1 + b 3 whose characteristic equation is give by s(s J 1 J +s(b 1 J +b J 1 ) Ω (J 3 J )(J 1 J 3 )+b 1 b ) = 0 (10.30) Note that, in order for the above equation to have its roots to lie in the left-hand s-plane (otherwise, the solution will grow exponentially, thus making the system unstable!), we must have as b i 0 J 3 > J 1 J 3 > J or J 1 > J 3 J > J 3 (10.31) This result indicates that the rigid body will remain stable if it spins about the axis of either the smallest or the largest moment of inertia. This is inconclusive as we shall see below by reexamining the same problem by a consistent linearization. From a physical consideration, the above result is due to the neglect of the perturbation of the new body-fixed frame (angular orientations due to perturbed rotations) that deviates slightly from a frame rotating at constant angular velocity. A possible resolution to account for such angular perturbations (in addition to angular velocity and angular accelerations) is to adopt the following new orientation: R = I θ (10.3) so that the angular velocity on the new frame is given by ω = θ + R ω 0 R T ω θ + (I θ)ω 0 = ω 0 + ω 0 θ + θ ω0 T = 0 0 Ω (10.33) 8

9 Linearizing the Euler equations(10.4) utilizing (10.33), we obtain J 1 0 b 1 (J 1 + J J 3 )Ω 0 0 J 0 θ + (J 1 + J J 3 )Ω b 0 θ 0 0 J b 3 (J 3 J )Ω (J 3 J 1 )Ω 0 θ By introducing the change of variables (10.34) J i θ i θ i, ξ = (J 1 + J J 3 )Ω/ J 1 J, k 1 = (J 3 J )Ω /J 1, k = (J 3 J 1 )Ω /J The characteristic equation for the above linearized equation can be shown to be s 4 +s 3 (b 1 +b )+s (b 1 b +k 1 +k +ξ )+s(b 1 k +b k 1 )+k 1 k ) = 0 (10.35) The stability conditions for the above characteristic equation is thus given by: b 1 + b > 0 (b 1 b + ξ )(b 1 + b ) + (b 1 k + b k 1 ) > 0 k 1 k > 0 (b 1 k + b k 1 )(b 1 b + ξ )(b 1 + b ) + b 1 b (k 1 k ) > 0 (10.36) If k 1 > 0 and k > 0 ( which corresponds to J 3 > J 1 and J 3 > J, a major axis rotation criterion), then the system is stable for all value of the gyroscopic coupling term ξ. If k 1 k < 0 (a intermediate axis rotation criterion), there will be instability for any value of ξ. If we assume k 1 < 0 and k < 0, in order for b 1 + b > 0 and (b 1 k + b k 1 ) > 0 to hold, we must introduce a negative damping, viz., b < b 1 < b k 1 k (10.37) which can be realized only by a rate feedback control strategies. 9

10 Finally, we note that if the first three conditions are satisfied in the above stability conditions, by increasing ξ (usually by increasing Ω or (J 1 +J J 3 )/ J 1 J ). This, again, is a stabilization by a gyroscopic coupling Torque Free Motions of A Free Rigid Body A simplest model of a spacecraft is either a single rigid body or a rigid body with symmetric rotors. Hence, the study of the behavior of a torque-free rigid body has a fundamental importance in spacecraft dynamics as well as floating ocean vehicle. In the absence of external torques, we have from (10.1) with (m = 0, r c = 0): J ω + ω J ω = 0 (10.38) First, let us perform scalar multiplications by ω and J ω to obtain ω (J ω + ω J ω) = ω J ω = 1 d (ω J ω) = 0 dt J ω (J ω + ω J ω) = J ω J ω = 1 d dt (J ω) = 0 (10.39) The preceding equation set shows that the rate of change of the kinetic energy and the angular momentum are zero: ω J ω = T = const (J ω) = L = const (10.40) In particular, when the moment of inertia dyadic is expressed in terms of the principal axes, we have T = J 1 ω 1 + J ω + J 3 ω 3 = const L = J 1 ω 1 + J ω + J 3 ω 3 = const = DT (10.41) where D is a parameter which has the dimension of the moment of inertia. The preceding derivations manifest that, for a torque-free rigid- 10

11 body, the kinetic energy and the angular momentum remain constant. The first and the second of (10.41) are called the momentum ellipsoid and the energy ellipsoid, respectively. To render a geometrical interpretation for the torque-free angular velocity trajectory, we visualize that we construct the two ellipsoids relative to the body-fixed axes. Then ω moves along the intersection curve of the two ellipsoidal surfaces. These intersection curves are called polhodes. Let us examine the relations between the momentum and the kinetic energy. To this end, we rearrange the system kinetic energy in (10.41) as T = 1 L L L , J 1 J J 3 T = L J (J 1 J ) J 1 J L + 1 L i = J i ω i (J 1 J 3 ) J 1 J 3 L 3 (10.4) Without the loss of generality, we may assume J 1 would be the major axis: J 3 J J 1 (10.43) Before we discuss the general case for the intersection curves of the energy and the momentum, it is useful to examine the following special cases: Minimum energy motion: If H = H 3 = 0 (i.e., ω = ω 3 = 0), then we have T 1 = H = J 1 ω1 = J 1ω1 J 1 J 1 (10.44) For this case, the body spins about the largest inertia axis, b 1. Intermediate energy motion: H 1 = H 3 = 0 T = L = J ω J (10.45) so that the body spins about b 11

12 Maximum energy motion: H 1 = H = 0 T 3 = L = J 3ω3 J 3 (10.46) so that the body spins about b 3 In order to illustrate the general case, we multiply the first of (10.41) by J 1, J 3, andj and subtract from them the second of (10.41) to give J (J 1 J )ω + J 3 (J 1 J 3 )ω 3 = T (J 1 D) J 1 (J 1 J 3 )ω 1 + J (J J 3 )ω = T (D J 3 ) J 1 (J 1 J )ω 1 J 3 (J J 3 )ω 3 = T (D J ) (10.47) Since the left-hand sides of the first and the second of (10.47) are non-negative, we conclude that J 3 D J 1 (10.48) Hence, thefirst of (10.47) represents a family of ellipses on the ω -ω 3 plane, the second of (10.47) those on the ω 1 -ω plane. On the other hand, the third of (10.47) represents a family of hyperbolas on the ω 1 -ω 3 plane. These polhodes on the energy ellipsoid of an unsymmetric rigid body are extensively discussed in most of classical dynamics texts(e.g., p.47 of J. Wittenburg, Dynamics of Systems of Rigid Bodies). So far we have used the conservation principles of the angular momentum and the kinetic energy for a torque-free motion to characterize the locus of the angular velocity vector on the b- basis frame. These two principles can be used to asses the motion of a torque-free rigid body relative to the inertial frame due to Poinsot. Since the magnitude and the direction of L = J ω is constant, L ω = 0 (or 3 i=1 J i ω i dω i = 0) defines an invariable plane perpendicular to L. We also recall that ω i lie the surface of the energy ellipsoid and the point of contact between the energy ellipsoid and the invariable plane is located on the instantaneous axis of rotation. Hence, we can visualize that the body moves as 1

13 if the energy ellipsoid were rolling without slip on the invariable plane with its instantaneous radius of T/L. Finally, the study of a torque-free rigid body provides us a heuristic insight into the attitude stability of a satellite. Suppose that there exists an internal energy dissipation mechanism (damping, heat sink, magnetic converter, etc.) in the satellite. However, as long as there is no external torque, L remains constant. Thus, the kinetic energy T will be reduced without affecting L. Since the global minimum energy state is pure spin about b 3, during such energy reduction due to internal dissipation, it is the only stable state. This is a fundamental consideration for the deployment of satellites. In fact, the Earth, the Moon, the Sun and other planets whose moments of inertia all spin about their axes of largest inertia! 10.7 Further Applications There are numerous applications of Euler s equations (10.1), ranging from orbital determinations to gyroscopes. Since most applications deal with symmetric bodies, we will concentrate on the motions of symmetric bodies Precession of Symmetric Spinning Body Euler s equations for a free rigid body (10.1) with J 1 = J reduce to J 1 ω 1 = (J 1 J 3 )ω ω 3 J ω = (J 1 J 3 )ω 1 ω 3 J 3 ω 3 = 0 (10.49) If we take the z-axis to be the symmetry axis. Since ω 3 = 0, we have ω 3 = const (10.50) so that one can rearrange the first two equations in the following 13

14 form: ω 1 = Ωω, Ω = J 3 J 1 J 1 ω 3 ω = Ωω 1 (10.51) which can be expressed as ω = Ω ω ω = ω 1 i + ω j, Ω = Ωk (10.5) Equation (10.5) is analogous to the well-known expression for the velocity of a rigid link ṙ with its angular velocity ω, viz., ṙ = ω r (10.53) which states that the mass located at the tip of r rotates with the angular velocity of ω. Similarly, ω = Ω ω states that the angular velocity precesses about the z-axis with its frequency Ω. Observe that the closer J 1 is to J 3, the slower will be the precession frequency of Ω compared to the rotational frequency of ω 3. In physical terms, the precession motion is relative to the body axes which in turn rotate with ω 3. As an example, the earth is symmetrical about the polar axis (to be taken as the z-axis here) and semi-spheroidal with J 1 = J slightly smaller than J 3. The ratio of the two moments is known to be J 3 J 1 J Ω = ω (10.54) Since ω ω 3 k, the precession period Ω is 306 days. Field measurements have confirmed that the precession axis moves with an amplitude about 10m but its movements consist of different irregular wobbles and the dominant wobble frequency is about 40 days instead of 306. It is speculated that the elasticity of the earth s core, tidal friction and the dissipative effects of tectonic movements are responsible for the irregular wobbles and the discrepancies between the theory and the measurements. However, for spacecrafts designed for deep space missions, this phenomena 14

15 Fig. 1. Euler sequential rotations must be accounted for in making necessary orientation corrections. There exists still another slow precession of the earth about the normal to the ecliptic or known as the ecliptic precession of the equinoxes. This phenomenon is due to the gravitational torques of the sun and the moon, whose period is about 81,000 years. We will first study the dynamics of a heavy symmetric top and then discuss the ecliptic precession due to gravitation afterwards Heavy Symmetrical Top The second application is the motion of a symmetric heavy top as shown in Fig For physical reasons, the three quantities, ( φ, θ, ψ), are called the spin, the precession and the nutation rates. The angular velocity in terms of these quantities are given by ω = ( θ cos ψ + φ sin θ sin ψ)i + ( φ sin θ cos ψ θ sin ψ)j + ( ψ + φ cos θ)k (10.55) 15

16 so that the system kinetic and the potential energy due to the gravitation can be obtained as T = J 1 ( θ + φ sin θ) + J 3 ( ψ + φ cos θ), = mgl cos θ (10.56) Introducing the Lagrangian as L = T we find that φ and ψ are ignorable variables since the Lagrangian does not depend these two angles. Thus, the Routhian can be constructed as 1 R = L p φ ψ p φ φ = J θ 1 J 1 (b a cos θ) sin mgl cos θ = θ T (10.57) where the momenta, p φ and p ψ, are given by p ψ = L ψ = J 3ω 3 = J 3 ( ψ + φ cos θ) = J 1 a p φ = L φ = (J 1 sin θ + J 3 cos θ) φ + J 3 ψ cos θ = J 1 b (10.58) where φ and ψ have been eliminated in terms of a and b. The net result is that the system is reduced to an equivalent onedimensional problem whose kinetic energy and potential energy are given by T = 1 J 1 θ, = mgl cos θ + J 1 a cos θ (b ) (10.59) sin θ Furthermore, since the system is conservative, its total energy E is constant, viz., E = T + = J 1 θ + J 1 (b a cos θ) sin θ + mgl cos θ = const (10.60) To gain physical insight into the general nature of its motion, we simplify E through u = cos θ, α = E J 1, β = mgl J 1 (10.61) 16

17 Substituting these expressions into (10.60), we obtain u = (1 u )(α βu) (b au) = f(u) (10.6) The solution of (10.6) can be expressed as t = u(t) u(o) du (1 u )(α βu) (b au) (10.63) Most of the practical applications (except the slowly spinning gyroscope!) satisfy 1 J 3ω 3 >> mgl (10.64) which means that the system kinetic energy is much larger than the potential energy. Extent of Nutation: Let us compute the extent of nutation, (cos θ o cos θ), the nutation frequency ( θ ave ) and the precession ( φ ave ) for this fast top case. Given θ o and θ o = ψ o = 0, we have E = mglu o if b a cos θ 0 = 0 so that we have from (10.6) f(u) = (u o u)[β(1 u ) a (u o u)] (10.65) The case of u = 0 yields (1 u ) a β (u o u) = 0 (10.66) If we define the extend of nutation n as n = cos θ o cos θ = u o u (10.67) 17

18 we can rewrite (10.65) as n + pn q = 0 p = a β cos θ o, q = sin θ o (10.68) Since (cos θ o, sin θ o ) 1, and a β = (J 3 ω 3 ) J 1 J 1 mgl = 1 (J 3ω3 mgl ) J3 J 1 Observe that, unless J 3 /J 1 << 1 which correspond to a long rod (hence not a top!), we conclude p >> q (10.69) so that we can neglect cos θ 0 compared with a /β. Hence, the approximate extent of nutation is given by n = cos θ 0 cos θ q p β sin θ o = [ J 1 mgl sin 1 θ a 0 ] J 3 J 3 ω3 (10.70) which indicates that the extent of nutation is inversely proportional to ω3. Frequency of Nutation: Since n << 1, we have from (10.68) Hence, we have u = cos θ = cos θ o n cos θ o (10.71) 1 u 1 cos θ o = sin θ o (10.7) Under this condition, f(u) becomes u = ṅ = a n(n n), n(0) = 0 (10.73) A change of variable, y = n n produces ẏ = a ( n 4 y ) ÿ = a y (10.74) so that its solution is given by n = n (1 cos at), a = J 3 J 1 ω 3 (10.75) 18

19 which states that the nutation frequency is proportional to ω 3. Precession: From (10.58), (10.70) and (10.75), we have φ = (b au) sin θ (au 0 au) sin θ 0 β a (1 cos J 3 J 1 ω 3 t) (10.76) The rate of precession thus varies harmonically with time and the average precession frequency is given by φ = β a = mgl I 3 ω 3 (10.77) Gravitational Precession Due To Oblateness We have seen from the preceding section that the difference between the moment of inertia about the spin axis and the other two give rise to precession of the heavy top. An extension of the preceding analysis can be used to analyze the effect of the oblateness of the earth on its precession. To this end, we need to replace the potential energy of the top with the gravitational potential: dm = Gm 1 (10.78) m r In the above equation, m 1 is the mass of an interacting body with the earth (say, the sun) and m is the mass of the earth. r is the distance between the two bodies measured from the body to body 1 given by r = R 1 R c R + ( c R ) (10.79) in which c is the distance measure from the mass center of the earth to an arbitrary point in the earth. A power series expansion of r 1 gives r 1 = R 1 +R 3 R c+ 1 R 5 [3(R c) ( c R ) ] (10.80) A more convenient representation of the above series has been 19

20 worked out in terms of the Legendre polynomials as r 1 = R 1 n=0 where the Legendre polynomials are given by (R c) n P n (cos β) (10.81) P 0 (x) = 1, P 1 (x) = x, P (x) = 1 (3x 1),... etc. Since R c = Rc cos β we have (c cos β) = R c R c R = 1 R R [c I cc] R + c (10.8) Note that the first term in the above expression, when integrated over, yields the inertia dyadic operator and the second term gives one-half of the trace of the inertia matrix. Thus we have m (c cos β) dm = R J R R (10.83) Substituting this expression into (4.63), while noting that the integration of the second expansion term vanishes, we obtain = Gm 1m R + 1 Gm 1 R [3J 3 R 1(J 1 + J + J 3 )] (10.84) where J R is the moment of inertia of the earth expressed in the direction of R. If we adopt the spherical coordinates for the orientation of the earth as shown in Fig. 7.3, we have J R = J 1 (sin λ cos φ) +J 1 (sin λ sin φ) +J 3 (cos λ) = J 1 +(J 3 J 1 ) cos λ (10.85) in which we have used J 1 = J due to the earth s symmetry. Substituting the above expression and after some algebraic manipulations we finally obtain = Gm 1m R + Gm 1(J 3 J 1 ) R 3 P (cos λ) (10.86) To relate the angle λ to the Euler angles, we observe from Fig. 7.3 cos λ = sin θ cos λ (10.87) Furthermore, since the orbital rate ( λ ) is much larger than the precession rate ( θ), we integrate the gravitational potential with 0

21 respect to λ, yielding the following averaged potential: = Gm 1(J 3 J 1 ) R 3 P (cos θ) (10.88) Physically, the averaged gravitational potential implies that the mass of the earth or the rotating body is approximated as a ring around the other mass m 1 with the radius R. Hence, by replacing the potential energy of the symmetric heavy top (10.60), the Lagrangian due to the earth s oblateness for the system becomes L = J 1 ( θ + φ sin θ)+ J 3 ( ψ+ φ cos θ) Gm 1(J 3 J 1 ) R 3 P (cos θ) (10.89) If we neglect the nutation variables, θ and θ, then θ becomes an ignorable variable. Hence, we obtain the following equation of motion: L θ = 0 J 3ω 3 φ J1 φ cos θ = 3Gm 1(J 3 J 1 ) cos θ R 3 (10.90) In view of φ << ω 3, that is to say, the spin rate is much larger than the precession rate, we obtain an approximate expression for the precession rate φ = 3Gm 1 R 3 ω 3 J 3 J 1 J 3 cos θ (10.91) Note that if the body didn t have any oblateness, viz., J 1 = J = J 3, we have φ = 0 from the above equation. We will now utilize (10.91) to determine the gravitational precession of the earth and of earth-orbiting satellites Precession of the Earth due to the Sun s Gravitation To compute the precession of the earth, we note that Kepler s third law gives for the orbital rate of the earth by ω O = ( Gm 1 R 3 )1/ ( π ) rad/day (10.9)

22 and the inclination axis of the earth is approximately θ = 3.4 o and the measured oblateness is known to be J 3 J 1 J (10.93) Also ω 3 in (10.91) denotes the sidereal rotational rate of the earth or ω E = ω 3 π/3.93 rad/hr. Substituting these numerical values into (10.91) we find φ ω O 81, 143 (10.94) In other words, it would take about 81,000 years for the earth to complete one precessional rotation due to the sun s gravitational effect Precession of Satellites due to the Earth s Oblateness For this case, equation (10.91) needs to be modified as φ = 3Gm 1 r 3 j 3 ω 3 (J 3 J 1 ) cos θ (10.95) where j 3 is the moment of inertia of the orbiting satellite with respect to the center of the earth. The reason for this is due to the fact that the averaged gravitational potential (10.88) approximates the moment of inertia about the earth s axis of rotation as a ring around the earth. Hence we have for this case ω 3 = ( Gm E ) 1/ π r 3 90 rad/min. (10.96) and j 3 is approximately j 3 = m 1 r (10.97) Substituting these into (10.95) and noting that the observed principal moment of inertia of the earth J m E R E 1 3 m ER E, (10.98)

23 we finally obtain the precessional rate of the earth-orbiting satellites as φ J 3 J 1 ( R E ω O J 3 r ) cos θ 1 (10.99) 700 which indicates that the low-orbit satellites need 700 orbits to make one complete precession due to earth s oblateness, or about 44 days. 3