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By Herbert A. Pohl

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L(O+) - ql(o-) = -fq1(0) = -f, qz(o+) - &(O-) = -fq2(0) = 0. In the interval 0 < t < T we take q l ( t ) = cos(w0t) - f w o ' sin(w0t) , q z ( t ) = w0' sin(w0t) . 41) The matrix A, defined by xl(T-) = Ax1 (0-) and xz(T-) = Axz(O-), is A= ( c o s ( w o ~) fw;' sin(woT) w ~ sin(woT) ' -wo sin(woT) - f cos(w0T) cos(w0T) > . 42) We have det(A) = 1 and S = trace(A) = 2cos(w0T) - f w i ' sin(w0T). If 1 5'1 < 2, we find A1 = exp(iy), A2 = XI, COST = S/2 = cos(w0T) - fsin(woT)/2wo.

7 (i) Show that a particle with energy E (z= 1,y = O), (z = -1/2,y = 4 / 2 ) , (z = -1/2,y = 4 / 2 ) . (ii) These vertices are saddle points. Show this for the (1,O) vertex. 8. 5: Quasi-circular orbits Solutions to ch. 1 The force ( 7 , tension of the string) is constant if the motion of the puck is circular with center at the hole (r = M g ) , the period being T = 2 7 r J w . However, if m executes small oscillations about the circular orbit, one has mi: = -r I2/m~', Mi: = r - M g , ( m M)i: = - M g l'/rn~', + (the tension T ) is + given by T = M(i: + + g ) , and is clearly not constant.

15) CHAPTER 2. 16) For h > 0 (attractive cubic force), the precession is in the direction of motion. 1 (left 1 = 0, right 1 > 0). It is convenient to introduce the variabie s = l/r2. 17) The minimum of U, is for r = R, where R is the radius of the stable circular orbit R = Since Ue(R)= wd, for given E must be E > wl. 8) gives G. 4. THE KEPLER PROBLEM 27 we find the ellipse. 21) . 14). 5. 23) But 2nl = S,2”ld8 = li*ped8, with pe as defined in ( 2 . 24) + + Since d r = m l ( 5 dz 6 dy) = m $ ( + dr r28 do), we have Jr Je = J, Jv.

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