604 APPENDIX A 2a1a0+a1a2J"+ J"a 0= 81 (A-13c) 2 a3=- 1「aa2J”+ J"a 24 (A-13d) It should be remembered that we already have shown,on the basis of slight deviations from circularity,that for closed orbits the force law must have the form of Eq.(A-8)or that J(u)is given by mk J=+P-职 (A-14) Keeping in mind the circularity condition,Eq.(A-3),the various derivatives at uo can be evaluated as J”= B2(1-B2) uo (A-15a) and J”=-B21-B21+B2) (A-15b) uo Equations(A-13a,b)thus say that ao/a and az/a are of the order of a/uo, which is by supposition a small number.Further,Eq.(A-13d)shows that a3/a,is of the order of(a/uo)2,whichjustifies theearlier statement that a is oflower order of magnitude than ao or a2. Equation (A-13c)is a condition on B only,the condition that in fact is the principal conclusion of Bertrand's theorem.Substituting Eqs.(A-13a,b)and (A-15)into Eq.(A-13c)yields the condition B2(1-B2)(4-B2)=0. (A-16) For deviations from a circular orbit,that is,B0,the only solutions are =10=合 (A-17a and B2=4,f)=-kr. (A-17b) Thus the only two possible force laws consistent with the solution are either the gravitational inverse-square law or Hooke's law! We started out with orbits that were circular.These are possible for all attractive force laws over a wide range of l and E,whose values in turn fix the orbital radius.The requirement that the circular orbit be stable for all radii already restricts the form of the force law through the inequality condition of B2>0(Eq.(3-48)).If we further seek force laws such that orbits that deviate only slightly from a circular orbit are still closed,no matter what the radius of the604 APPENDIX A (A- 13d) It should be remembered that we already have shown, on the basis of slight deviations from circularity, that for closed orbits the force law must have the form of Eq. (A-8) or that J(u) is given by (A- 14) Keeping in mind the circularity condition, Eq. (A-3), the various derivatives at uo can be evaluated as and J"' = -P2(1 - P2)(1 + P2) u; (A- 15a) Equations (A-13a, b) thus say that ao/a, and a2/a, are of the order of al/uo, which is by supposition a small number. Further, Eq. (A- 13d) shows that a3/al is of the order of (a,/~,)~, which justifies the earlier statement that a, is oflower order of magnitude than a, or a,. Equation (A-13c) is a condition on P only, the condition that in fact is the principal conclusion of Bertrand's theorem. Substituting Eqs. (A-13a, b) and (A-15) into Eq. (A-13c) yields the condition P2(1 - P2)(4 - P2) = 0. (A- 16) For deviations from a circular orbit, that is, p # 0, the only solutions are (A- 1 7a) and P2 = 4, f (r) = - kr. (A-17b) Thus the only two possible force laws consistent with the solution are either the gravitational inverse-square law or Hooke's law ! We started out with orbits that were circular. These are possible for all attractive force laws over a wide range of 1 and E, whose values in turn fix the orbital radius. The requirement that the circular orbit be stable for all radii already restricts the form of the force law through the inequality condition of P2 > 0 (Eq. (3-48)). If we further seek force laws such that orbits that deviate only slightly from a circular orbit are still closed, no matter what the radius of the