CMchap5 - Chapter 5 The Kepler Problem This problem is...

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Unformatted text preview: Chapter 5 The Kepler Problem This problem is defined by the equation of motion dp oar a? : "-71:3— 5 where p is the linear momentum of a particle of mass m, 7' is its position vector, and a is a positive constant. Eq. (5.1) represents a set of three second order differential equations, whose solutions give the trajectory r(t) of the particle. We will not solve the equation explicitly in this chapter. Instead, the geometrical nature of the trajectory Will be determined by the constants of motion of this problem. The force given by (5.1) is a special example of a central force (one which is along the radial direction). Consequently, the torque (with respect to the origin of the coordinate system) vanishes, and the angular momentum is conserved: dL_ :1? _ 0, (5.2) that is, L is a constant of motion. This fact can also be demonstrated explicitly as follows. From L = 7' X p we have dL_clr d _p a: _ E_Exp+rxa_mxp+( ——)r><r—. (5.3) Define the Runge-Lenz vector JVI as follows: (5.4) 26 CHAPTER 5. THE KEPLER PROBLEM We will show that M is also a constant of motion. Indeed, dM 1dp p dL d r _:___ L __ __ _ _ dt mdt X +m X dt adt (r) a cl 7' WWW—dab) oz r'r 1dr =—-7;3-r><(r><p)—a *T—2+;E (5.5) ___0‘_ . _ 2 Di _O‘_p — W301?" p) TP)+T2r W ——93 5i: 7' +afl——--a—Trr+ furl—O — r3 clt rg— r3 ar2_ ’ where the fourth equality follows from the “bac - cab” identity (1.51) and the next to last equality from (2.39), which implies that v - r 2 rr. The constants of motion L and M satisfy the following conditions: L-rzL-MzO, (5.6) which imply that L is perpendicular to the plane of the orbit and M lies on the plane of the orbit. Eq.(5.6) can be demonstrated easily: L-r:r-(rxp)=p'(rxr):O, (5.7) L 1 L-MzL-(px —9£>=—p-(LXL)—gr~L=O. (5.8) m r m r To find the equation of the orbit we calculate M -r: - L L2 M-rzr—(P—Q—arzfl—ar. (5.9) m m Choosing the :c—axis to be along the direction of M , we can write the above equation in terms of polar coordinates (r, 0) as (Fig. 5.1) L2 Mr cosé’ : — — or, (5.10) m or L2 r 2 ———~—]\—/—[— (5.11) am (1 + 3 cos 0) This is the equation of a conic section, with the eccentricity equal to M / a (Fig. 5.2). Fig. 5.1, Fig. 5.2 ...
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