Mathematical Introduction to Celestial Mechanics by Harry Pollard

By Harry Pollard

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Ll - P l. 1 ~J. (1 - p) . ] = O. ln, cases (a) and (e) this can be seen as follows. =,E=z_. P + I - (p + 1)-2 = - c ' whcre c/O. rt is easily verified that F'((1) > 0, so that Fis strictly inc'reusing. Moreover, nOi,) '/', nl) ,0. Therefore F assumes the valu~ -c at precisely one value of P between 0 and 1. The solutions are denoted by L t in case (a) and by L, in case (e), as indicated in Fig. 12. The case (b) is similar. > • 1, P- P . Jl. Fj(p) = _. L5 Figure 12 because fl ~ }. The function FI(p) is increasing in the interval t :::s:: P < I.

1. Ttere:·e1re. ' = O. ,1 triangles b,u",J elnlhclinej"ll1i:l~ 1 - ,.. ('1':,,': \; - ... l)l. lll·d in Fi~, I~. On the other hand, if '7 = 0 the Eqs. L = O. Pi P, P2 P2 *- O. Then 57 EQUILIBRIUM SOLUTIONS SEC, 12 CHAP. 2 INTRODUCTION TO THE c·BODY PROBLEM o. This means that the terms containing ~ in the first of Eqs. 3) drop out. ) cancels and we are left with -Jl. I· There are three. , pz = ~ + Jl. - I, = I - PI; pz = PI - 1. 4) in each of the cases as follows. (a) Let PI = p, P2 = 1 + p. P ...

L h(c, t). For example, suppose the equation of the opening sentence is modified k = I, ... ,no It is understood that each Xh' .. 10). If the determinant laxdoCl1 does not vanish, the Eqs. 13) k = I, ... ,n. As an example, consider the systm x = x + t + ax2, (16) where a is a constant. Since the original equation (with a = 0) has the solution x = - I - t + cet, it follows that ax/oc = e l • Becauseg(x, t)=ax', Eq.

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