Newton's Easy Quadratures ``Omitted for the Sake of Brevity'

In the 1687 Principia, Newton gave a solution to the direct problem (given the orbit and center of force, find the central force) for a conic-section with a focal center of force (answer: a reciprocal square force) and for a spiral orbit with a polar center of force (answer: a reciprocal cube force). He did not, however, give solutions for the two corresponding inverse problems (given the force and center of force, find the orbit). He gave a cryptic solution to the inverse problem of a reciprocal cube force, but offered no solution for the reciprocal square force. Some take this omission as an indication that Newton could not solve the reciprocal square, for, they ask, why else would he not select this important problem? Others claim that ``it is child's play' for him, as evidenced by his 1671 catalogue of quadratures (tables of integrals). The answer to that question is obscured for all who attempt to work through Newton's published solution of the reciprocal cube force because it is done in the synthetic geometric style of the 1687 Principia rather than in the analytic algebraic style that Newton employed until 1671. In response to a request from David Gregory in 1694, however, Newton produced an analytic version of the body of the proof, but one which still had a geometric conclusion. Newton's charge is to find both ``the orbit' and ``the time in orbit.' In the determination of the dependence of the time on orbital position, t(r), Newton evaluated an integral of the form ∫dx/x <sup>n</sup> to calculate a finite algebraic equation for the area swept out as a function of the radius, but he did not write out the analytic expression for time t = t(r), even though he knew that the time t is proportional to that area. In the determination of the orbit, θ (r), Newton obtained an integral of the form ∫dx/√(1−x<sup>2</sup>) for the area that is proportional to the angle θ, an integral he had shown in his 1669 On Analysis by Infinite Equations to be equal to the arcsin(x). Since the solution must therefore contain a transcendental function, he knew that a finite algebraic solution for θ=θ(r) did not exist for ``the orbit' as it had for ``the time in orbit.' In contrast to these two solutions for the inverse cube force, however, it is not possible in the inverse square solution to generate a finite algebraic expression for either ``the orbit' or ``the time in orbit.' In fact, in Lemma 28, Newton offers a demonstration that the area of an ellipse cannot be given by a finite equation. I claim that the limitation of Lemma 28 forces Newton to reject the inverse square force as an example and to choose instead the reciprocal cube force as his example in Proposition 41. (Received August 14, 2002) Published online March 26, 2003 Communicated by G. Smith     

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