TECHNISCHE MECHANIK, Band 15, Hefl4, (1995), 315-324 Manusktipteingang: 14. April 1995 Analogies Between the Chaplygin Problem and the Kepler Problem F. P. J. Rimrott; W. M. Szczygielski Chaplygin’s problem ofa powered airplane in a steady cross wind, and Kepler’s problem of a satellite subject to a central attraction force, both lead to elliptic orbits. The resulting analogies are discussed, established, and tabulated. Then Kepler's problem is reformulated using Chaplygin's approach, and finally Chaplygin’s problem is reformulated as a centralforce problem. 1 Introduction Chaplygin‘s is a classical problem of the calculus of variations, involving the kinematics of an aircraft in a steady cross wind, and formulated as an isoperimetric problem in parametric form. Kepler's is a classical problem of dynamics, involving the kinetics of a satellite in an inverse square central force field. It is typically posed as a vectorial problem involving force and acceleration. Both have as result an elliptical orbit, inviting the obvious question as to how far an analogy can be pursued. Specific questions that might be posed include: ls Chaplygin's problem, by any chance, also a central force problem? Can Kepler's problem be posed as a variational problem similar to Chaplygin's? Since both are problems in Newtonian mechanics, both can obviously be treated via Hamilton's principle. What then is the potential energy for Chaplygin's problem? What is the variational statement for Keplers' problem using an isoperimetric approach? 2 Chaplygin's Problem Chaplygin's problem involves an airplane flying in a horizontal plane, with an air velocity va (relative to air) of constant magnitude (Grigorian, 1965). The airplane is required to follow a closed path over a given period 'c . What path should it follow if the area boundary by the path is to be maximum, given that a constant wind velocity vW < v,z is prevailing? Let y be the direction of the wind velocity and suppose that 0c is the angle between the direction of the axis of the airplane and x. Let the closed path be described parameuically by x = x(t) y = y(t) If we associate t with time then the components of the ground velocity v (with respect to earth) of the airplane become (Figure 1) II x va cosoc (1a) y va sinoc + VW (1b) The area bounded by the airplane path is given by 1T . .A = §£(xy—yx)dt (2) 315 We require to maximize the area A subject to the nonholonomic constraint equations (1). Evidently to locate the airplane we need to know x, y and (l . Hence we allow variations of (x as well as x and y. Thus we construct a modified area functional (Pars, 1962) and express its variation as follows: 5J[%(xy‘—yic) + 7t}()'c—vacosoc) + Ä2(y'—v„ sinoc—vw)] dt 2 SJth z O (3) in which X10) and Mt) are Lagrange multipliers. The Euler-Lagrange equations of the functional JF dz emerges as follows: For 6x: 3—: — ä = äy— -g—t[—%y+klj = 0 (4a) 5y: gg— — ä = —%X — %(%x+7»2) = O (4b) 50c: 3—: = (Xlsinoc — K2 cosoc)va = 0 (4c) 5%: ä:— = x — Va cosoc = O (4d) 5kg: 8871: = y — va sinoc — vw = 0 (4e) From the first two equations we have, on integration, XI = y + q ÄZ = —(x+c2) (5) where (21 and (:2 are constants. Substituting into equation (4c) (x+c2)cosoc + (y+c1)sinoc = 0 (6) T0 satisfy this equation we let x + c2 = rsinoc y + c1 = —rcosoc (7) and then from the constraint equation (1) r'sinoc + rdcosoz = va cosoc (8) —r‘cosoc + rdcsinoc = Va sinoc + vw Multiplying the first by sinOL and the second by cosoc, and subtracting, we obtain r‘ : —vw cosoc (9) But from the first of equations (1) cosa = (10) 316 (11) hence (12) Integrating v r = ——Wx + p Va where p is an arbitrary constant. The x-component of the airplane's position, using polar coordinates emanating from a focus O (Figure l), is x = r c030, such that equation (12) can be rearranged to read (12)p cosO 1+—W Va which is the polar equation of conic (an ellipse if vw < Va ) of eccentricity 8 = (13) Va of semi-parameter (semi-latus-rectum) P (14) (15) of semi—major axis (16) and of semi-minor axis l P b : 2 1/2 v2 The area A of the ellipse covered is obtained from p2 A = nab Z TC 2 3/2 1_v_w l (18) The time period to circumnavigate the area A can be shown to be P1:21: 2 3/2 „_g)a For an eccentricity of e = O (i.e. for vanishing wind velocity VW) the ellipse degenerates into a circle. We conclude that the greater the wind speed vw, the flatter the ellipse (Figure l). 317 From equations (1), the Cartesian acceleration components are 5€ = — v o'tsintx .“ (19) = vaoccosoc which when combined, represent the airplane's acceleration v2p x2 + jy'z = vaöc = a (20),2 Since vw= constant in direction and magnitude, and v, = constant in magnitude only, it can be concluded readily that the airplane's acceleration is the result of the change of direction of the v,z vector. This change, which is perpendicular to the va vector, always points towards the origin O of the coordinate system. As a consequence of which we find = 9 + 90° (21) and conclude that the airplane's acceleration is pointing towards the origin 0 of the coordinate system (Figure 1). Consequently the airplane's motion is a "central acceleration" type motion, analogous the "central force" type motion of e.g. planets. The choice of polar coordinates emanating from the focus 0 of the ellipse, also means that the integration constants in equations (6) vanish, i.e. C1 = C2 = 0 Dynamics It is important to note that Chaplygin's problem is purely kinematic. All equations, and the solutions obtained, involve only kinematic statements. The Euler-Lagrange equations (4) are all in terms of velocities. In view of our intended comparison with the Kepler problem, it is of interest to have a look at the dynamics of the Chaplygin's problem. In particular we are interested in the forces which are involved in producing the kinematics of Chaplygin‘s airplane. Using the polar coordinate angle 9 , equation (21), rather than the air speed angle 06, and using equations (4d, e), the derivatives of equations (6) become x = r'cos9 — rÖ = —va sine (23a) )3 = r'sin9 + récose 2 Va cosG + vw (23b) Differentiating agam Jr' 2 (f—r92)cos6 — (2fÖ+rÖ)sinG = —va9c059 (24a) j} = (f—rtflsine + (2f6+ré)cos9 z —v„9sin6 (24b) Multiplying one equation with cost) and the other with sine , and then doing the reverse, adding, and multiplying with the airplane mass m the equations of the motion in polar coordinates are obtained m('r'~r92) = —mv„9 (253) m(rÖ+2fÖ) = o (25b) 318 It becomes immediately obvious that the airplane is acted upon by a resultant force which is directed towards the centre O of the coordinate system F, = —mv,,é (26) or, in other words Chaplygin's problem can be looked upon as a central force problem. In order to gain further information, we rewrite equation (25b) into 21' Z _T (27a) which, upon integration, gives r29 = h = constant (27b) where the integration constant h represents the airplane's specific angular momentum about the centre 0 of attraction, corroborating the well—known fact that central force motion is characterized by angular momentum constancy. Now , with the help of equations (27), the central force (26) can be written Fr:— 3 (28) Since m, va, and h are all constants it can be seen that the airplane moves indeed in an inverse square force field. It is interesting to note that, in principle, the airplane's pilot needs only follow a prescribed elliptic path at a constant height and a constant air speed. The obviously very complicated force system (thrust, drag, weight, lift, rudder and aileron effects etc.) acting on the plane, will then adjust itself such, that the resultant is a pure central force, as given by equation (28). With the help of equations (27), equations (25) can be written f _ r92 2 _ Vazh (29a) r ré + 2r‘é = 0 (29b) providing us with a relationship involving accelerations. 3 Kepler's Problem Variational Formulation Kepler's problem of the motion of a satellite in an inverse square central force field, is usually solved without recourse to variational methods (Rimrott, 1989). For purposes of the present study let us state Kepler's problem in variational form, for which Hamilton's principle (Goldstein, 1980) ($de: = 0 (30) is the obvious choice, with the Lagrangian L = T — V (31) and a kinetic energy of T = ä—m(r’2 „29) (32> 319 and a potential energy of V = —”—r'” (33) km3 where u is the gravitational attraction constant, with n = 398601 —2 for the earth as attractor. With 3 equations (31), (32) and (33) we can now express the satellite system’s variation (30) as follows: 1 . ($de: = 6Jm[—2—(r‘2+r292) A}: = 0 (34)r When one compares equations (34) and (3), then one notices, apart from polar coordinates in the one and Cartesian coordinates in the other, that the functional for Kepler’s problem is in action units, i.e. Js, while the functional (3) for Chaplygin’s problem was in area units, i.e. m2. This is confirmed when the Euler-Lagrange equations for variation (34) are formed, which emerge as follows: aL d 8L -2 .. u) or 5r ar dt 8f m(r6 r r2 0 ( a) 8L d 8L d 7* F 56: — — —— —. = —m— ‘6 : 35b or 66 dt 39 dr