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3 Orbital Dynamics Chapter 3 Orbital Dynamics In 1608, Johannes Kepler 1571{1630 published twolaws of planetary motion that he deduced from an analysis of the accurate observational data he inherited from his employer, the Danish astronomer Tycho Brahe 1546{1601. Ten years later, Kepler published a third law called the Harmonic Law. These three laws, which fty years later led Isaac Newton 1642{1727 to the discovery of the law of gravitation, can be expressed as follows: 1. The planets move in el lipses, with the Sun at one focus; 2. The radius vector from the Sun to a planet sweeps out equal areas in equal times; 3. The period of revolution squared is proportional to the semi-major axis cubed. The second law is the most general b ecause it is true for any two-body central force problem. The rst law is a consequence of the fact that the gravitational force of one b o dy on another dep ends on the inverse square of their separation, and the third law re ects the fact that the gravitational force is prop ortional to the masses of the b o dies. We will derive Kepler's Laws from Newton's Law of Gravitation, and then will discuss what happ ens when there are more than two b o dies to consider. 3.1 Two-Bo dy Central Force 3{1 As a prelude to understanding the motions of solar system ob jects, we treat the problem of two b o dies that move under the in uence of a mutual central force. Consider two b o dies with masses m and m whose p osition vectors relative to the center of mass 1 2 are r t and r t, resp ectively. A central force is one that dep ends only on the di erence 1 2 2 2 1=2 vector, r r r where jrj = jr + r j and its time derivatives 2 1 x y _ F = Fr; r; :::: 3:1 A conservative central force, like gravity,is one that can be expressed as the gradient of a p otential that dep ends only on the magnitude of the di erence vector, jrj. We use 2 2 2 2 1=2 p olar co ordinates in whichx; y = r cos; r sin and so jrj =[r cos + r sin ] = r . In this system 1 @ @ ^ ^ : 3:2 + e Fr = rr = e r @r r @ Along with the di erence vector there is a second natural p osition vector, the center of mass of the system, Rt: P m r t m r + m r i i 1 1 2 2 i P Rt = : 3:3 m m + m i 1 2 i where i represents the summation over the particles in the system. By expressing r and 1 r in terms of R and r : 2 m 2 r = R r; 3:4 1 m + m 1 2 m 1 r = R + r; 3:5 2 m + m 1 2 the kinetic energy K maybe written: X 1 1 1 2 2 2 _ _ _ m jr j = m jr j + m jr j ; K = i i 1 1 2 2 2 2 2 i 2 2 1 1 m m 2 1 _ _ _ _ = + ; m R m R + r r 1 2 2 m + m 2 m + m 1 2 1 2 " 2 2 1 m 1 m 1 2 1 2 2 _ _ = m + m jRj + m + m jrj ; 1 2 1 2 2 2 m + m 2 m + m 1 2 1 2 1 m m 1 1 2 2 2 _ _ = m + m jRj + jrj : 3:6 1 2 2 2 m + m 1 2 In Equation 3:6, the rst term on the RHS corresp onds to the motion of the center of mass and the second term represents the motion about the center of mass. The quantity m m 1 2 3:7 m + m 1 2 3{2 is called the reduced mass of the system, which can be rewritten 1 1 1 + : = m m 1 2 Because there are no external forces in this problem, no equations of the motion for r contain terms with R. The center of mass is either at rest or is moving uniformly, meaning that this term is constant and can be dropp ed. Thus, for conservative central forces, it is p ossible to express a two-body problem as an equivalent one-b o dy problem, with a center of force at the origin and with a single body of mass lo cated at a distance rt from the origin. 3.1.1 Constants of Motion For a body under the in uence of a central conservative force, the p otential is only a function of the radial distance from the center of mass. The problem thus has spherical symmetry in whichany rotation ab out a xed axis will have no e ect on the solution. This yields imp ortant simpli cations. In a spherically symmetric system of the form r Fr = Fr ; jrj angular momentum, L, is conserved. In such a system r must be normal to the direction of L, and the time rate of change of angular momentum, the torque, N,is zero: dL 1 N = r F =rr Fr= 0: 3:8 dt jrj Consequently, the b o dy will never b e torqued out of its plane of motion. It is convenientto use p olar co ordinates to describ e this system. Let r and denote the radial and azimuthal ^ co ordinates, with corresp onding unit vectorsr ^ and . Asketch of the geometry reveals that ^ the cartesian comp onents of the unit vectors are r^ = cos ; sin and =sin ; cos , which implies the identities: ^ dr^ d ^ = ; = r:^ 3:9 d d Since the p osition vector is a pro duct of two functions, r = r r^, the velo cityvector b ecomes a sum of two terms: d dr^ d _ r = r r^= r_r^+r v ; dt dt d _ ^ = r_r^+ r; 3:10 where dots indicate time derivatives, and we have used 3.9 and the chain rule for di er- entiation. A second time derivative yields the acceleration vector: ^ dr^ d d _ _ ^ ^ _ _ v =rr^+r_ + r_ + r + r ; a dt d d 2 _ ^ _ ^ ^ _ =rr^+r_+r_+rr r;^ 2 _ _ ^ =rr ^r +r +2_r: 3:11 3{3 Thus, the radial and azimuthal comp onents of the vector equation of motion, a = F, are d 2 _ r r = Fr= ; 3:12 dr _ r +2r_ =0: 3:13 This is a system of two second-order, ordinary di erential equations. The constants of motion that lead to the solutions for r t and t are straightforward to nd. First, multiply 3:13 by r , and then rearrange the equation into the form d=dt: d 2 2 _ _ r +2r r_ = r =0: 3:14 dt The constant of motion in 3.14 is the magnitude of the angular momentum, L: 2 _ jLj L = r = constant: 3:15 _ Equation 3.15 will b e used rep eatedly in what follows to eliminate the term whenever 2 _ it app ears. Our rst opp ortunity is to eliminate from 3:12, after which we multiply by the radial velo city r_ : 2 2 L d d 1 L 2 r_ r r_ + r_ + =0: 3:16 = r_ + 3 2 r dr dt 2 2r The new constant of motion revealed in 3.16 is the total energy, E : tot 2 1 L 2 E = +: 3:17 r_ + tot 2 2 2r The total energy is the sum of the kinetic and p otential energies: 2 1 L 2 K r_ ; +; 3:18 e 2 2 2r 2 2 where is the e ective p otential energy. The term L =2r arises from the cen- e 2 _ trip etal acceleration term, r , in 3:12. 3.1.2 Kepler's 2nd Law Kepler's 2nd Law is equivalent to the conservation of angular momentum. To see this, note that the di erential area dA swept out by the radius vector when the body moves through a di erential angle d is a triangle whose area is given by one-half its base times its height as 1 dA = rdr: 3:19 2 3{4 Consider the areal velo city, which is the area swept out by the radius vector per unit time. Conservation of angular momentum is equivalent to saying that areal velo city is constant. We divide the di erentials in 3:19 by dt and use 3:15 to obtain dA L 1 1 2 _ _ = A = r = = constant; 3:20 dt 2 2 which is Kepler's 2nd Law. If the motion is p erio dic with p erio d , then the total area enclosed by the orbit is given by: L 1 A = : 3:21 2 3.1.3 Formal solution For problems with three or more b o dies, chaotic tra jectories are part of the solution space and it is imp ossible to write down a complete analytical solution. But, for the two- body problem, there are a sucient number of constants of motion, namely L and E , tot to solve for r t and t. First, we can use 3:17 to solve for r_ as 1=2 2 dr 2 L r_ = = : 3:22 E r tot 2 dt 2r Next wemay obtain an implicit formula for r t in the form tr byintegrating 3:22 with resp ect to r : Z 1=2 r 2 2 L tr = dr ; 3:23 E r tot 2 2r r 0 where r is the radius at t = 0.
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