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航天动力学与控制 Lecture 2 2007年2月 Basic Concept and Dynamical Equation General Rigid Body Motion – The concept of Rigid Body • A rigid body can be defined as a system of particles whose relative distances are fixed with time. The internal potential energy of a rigid body is constant – Orbital Mechanics: translational motion of spacecraft under the influence of gravitational and other forces becomes orbital mechanics – Attitude Mechanics: Rotation about the center of mass under the influence of applied torques becomes attitude mechanics. Two-body and central force motion n-body problem – n-body problem has only 10 known integrals of motion: 3 velocity components, 3 position components, 3 angular momentum components, and kinetic energy. – only the two-body problem has an unrestricted solution. Special cases of the three-body problem have been treated in closed form •Assumption •Masses could be two bodies whose minimum distance apart is large compared to their largest dimensions or could have spherically symmetric mass distributions and never touch each other. This restriction allows these masses to be treated as particles in the following analysis. Two-body and central force motion (basic differential equation) Center of mass: c m1 (r1 rc ) m2 (r2 rc ) 0 Geometry dictates that r2 r1 r Which permits expressions to become r1 rc m2 r m1 m2 r2 rc m1 r m1 m2 then F1 m1r1 m1rc m1m2 r m1 m2 F2 m2r2 m2rc m1m2 r m1 m2 Two-body and central force motion (basic differential equation) Mutual attraction requires that F1 F2 m1rc m2rc then r 0 c Applying this results leads immediately to F1 F2 m1m2 r m1 m2 In terms of the gravitational attraction yields the basic differential equation of motion for the two-body system. d 2r 3r0 2 dt r G(m1 m2 ) Two-body and central force motion (Solution of the differential equation) Using a straightforward vector approach. d 2r r 2 3 .r r 0 dt r (1) Defining angular momentum per unit mass as h r r and dr dt d dr dr dr d 2r (r ) r 2 0 dt dt dt dt dt Leads to the conclusion that dh / dt 0 Thus, angular momentum is conserved and three integrals of motion are h constantand this plane must be inertially fixed Two-body and central force motion (Solution of the differential equation) Cross equation (1) with h d 2r dr h r h r ( r ) 2 3 3 dt r r dt Applying the standard identity for triple vector products and noting that r Leads to dr dr r dt dt d 2r d r h ( ) 2 dt dt r (2) Since h is constant, equation (2.) may be integrated directly dr dt h r (r re) (3) e is a constant of integration and is called the eccentricity vector. The orientation of e in this plane is taken as a reference direction Two-body and central force motion (Solution of the differential equation) Dot expression (3) with r and using the triple scalar product identity to obtain h2 / r 1 e cos cos r e re (4) Vector e is parallel to the direction of minimum r, and angular is measured from this point to the position in the orbit. This angle is known as the true anomaly Kinetic Energy 1 1 m1r1 r1 m2r2 r2 2 2 m1m22 2m1m2 1 m1rc rc r r r r c 2 (m1 m2 ) 2 (m1 m2 ) T m2 m12 2m1m2 1 m2rc rc r r r r c 2 (m1 m2 ) 2 (m1 m2 ) ( m2 m1 1 mm )rc rc ( 2 1 )r r 2 2 m2 m1 The first term on the right side is just translational kinetic energy of two-body system. The remaining term is rotational kinetic energy about the center of mass. Central Force Motion If m1 m2 then the motion of m2 about m1 is essentially the motion of a particle in an inertially fixed field. This type of motion is known as central force motion. rc r1 r2 r1 r T 1 m2r r 2 Notice that the two-body problem may be treated as central force motion of a particle of mass m2 m1 /(m2 m1 ) The force which is a function of r can be written as the gradient of gravitational potential F U Kinetic energy per unit mass is dr dr dt dt dr d ri r r i r dt dt Central Force Motion Which permits writing kinetic energy in a more convenient form 2 ( r r 2 2 ) The energy of a unit mass in a central fields is given 2 2 2 (r r ) U (r ) r d 1 dr 2 r 2 r d r d h introducing a new variable as thus 2 d 2U (r ) 2 h 2 d h2 2 Central Force Motion d d Consider the inverse-square force law characterized by 0 0 2 d 1 h sin 2 2 2 2 U (r ) 2 2 h2 h h 0 2 Define a parameter p as 0 U ( r ) 2 2 U ( r ) h2 Is replaced by cos( 2 1 1 r p 1 2 p 2 p p h sin( 0 1 1 r p 1 2 2 p p Rearranging gives r p 1 e cos( (5) e 1 2p Eccentricity (4) is identical to form (5) 0 Plus sign in (5) minus sign in (5) 0 / 2 0 p e=0, Circle 0<e<1 ellipse e=1 parabola e>1 Hyperbola Kepler’s Time Equation The relationship between time and position in orbit is considered h rv sin(r, Noting that v sin(r, p dr ) dt 3 dt dr d )r dt dt d (1 e cos A new variable is introduced ae r cos a p a(1 e2 ) cos Since cos ar ae h r2 d dt yields Kepler’s Time Equation Taking the time derivative of this gives radial speed r ae cos Evaluation of energy at periapsis Rearranging this leaves ar 2 r 2 Then ae2 (a r )2 r a 0 r 2 (r )2 2 r 2a a t e sin 3 p when Define n a3 nt p e sin This is known as Kepler’s equation for tp 0 relating time position in orbit Kepler’s Time Equation When eccentricity is less than 1.0, the orbit is closed and periodic. The associated period can be determined by considering the rate of area swept out by the radius vector which is the areal velocity. dA 1 2 1 r h dt 2 2 Kepler’s second law. Integrating over the entire orbit gives the period as A ab For an ellipse h2 / rp 1 e From geometry, 2A h h rp (1 e) rp a (1 e) 2 a3 2 n b a 1 e2 Kepler’s third law. 第一章 基本的物理定律 •相对运动(加速度) r d dr d (ω r) dt dt b dt d dr d dr ( xi yj zk ) xi yj zk ω dt dt b dt dt b The second term of the expression becomes d (ω r ) ω r ω r ω r ω rb ω (ω r ) dt R R 0 rb 2ω rb ω r ω (ω r ) R0 rb 2ω rb ωr ω (ω r ) is the acceleration of the origin of the moving frame is the apparent acceleration of particle in the moving frame is the Coriolis acceleration due to the motion of p in x, y, z is the acceleration of p due to the changes of the angular velocity is the centrifugal acceleration 第一章 基本的物理定律 •Motion of the Earth’s Surface RB F mgR m R R B Ω (Ω R E ) ab 2Ω vb Ω (Ω r ) RB F mgR m R ab g R Ω (Ω R E ) 2Ω v b Ω (Ω r ) R Ω 0.728 104 rad/s ab g R 2Ω v b R the two centrifugal acceleration terms are negligible compared to the Coriolis term for low altitude orbits. 第一章 基本的物理定律 •哥氏力的影响 To illustrate the effect of this Coriolis acceleration, consider a satellite traveling due east over the ground station, vb vj, F mgk Ω ( cos )i ( sin )k ab ( g 2v cos )k (2v sin )i This indicates that an observer would notice the satellite turing to the south or to the right of its initial flight path.