Page:Newton's Principia (1846).djvu/238

 place P, the ordinate DN was found to be as $$\scriptstyle \frac{DE^{2}\times PS}{PE\times V}$$. Therefore if IE be drawn, that ordinate for any other place of the corpuscle, as I, will become (mutatis mutandis) as $$\scriptstyle \frac{DE^{2}\times IS}{IE\times V}$$. Suppose the centripetal forces flowing from any point of the sphere, as E, to be to each other at the distances IE and PE as PEn to IEn (where the number n denotes the index of the powers of PE and IE), and those ordinates will become as $$\scriptstyle \frac{DE^{2}\times PS}{PE\times PE^{n}}$$ and $$\scriptstyle \frac{DE^{2}\times IS}{IE\times IE^{n}}$$ whose ratio to each other is as PS $$\scriptstyle \times$$ IE $$\scriptstyle \times$$ IEn to IS $$\scriptstyle \times$$ PE $$\scriptstyle \times$$ PEn. Because SI, SE, SP are in continued proportion, the triangles SPE, SEI are alike; and thence IE is to PE as IS to SE or SA. For the ratio of IE to PE write the ratio of IS to SA; and the ratio of the ordinates becomes that of PS $$\scriptstyle \times$$ IEn to SA $$\scriptstyle \times$$ PEn. But the ratio of PS to SA is subduplicate of that of the distances PS, SI; and the ratio of IEn to PEn (because IE is to PE as IS to SA) is subduplicate of that of the forces at the distances PS, IS. Therefore the ordinates, and consequently the areas which the ordinates describe, and the attractions proportional to them, are in a ratio compounded of those subduplicate ratios. Q.E.D.


 * To find the force with which a corpuscle placed in the centre of a sphere is attracted towards any segment of that sphere whatsoever.

Let P be a body in the centre of that sphere and RBSD a segment thereof contained under the plane RDS, and the sphærical superficies RBS. Let DB be cut in F by a sphærical superficies EFG described from the centre P, and let the segment be divided into the parts BREFGS, FEDG. Let us suppose that segment to be not a purely mathematical but a physical superficies, having some, but a perfectly inconsiderable thickness. Let that thickness be called O, and (by what Archimedes has demonstrated) that superficies will be as PF $$\scriptstyle \times$$ DF $$\scriptstyle \times$$ O. Let us suppose besides the attractive forces of the particles of the sphere to be reciprocally as that power of the distances, of which n is index; and the force with which the superficies EFG attracts the body P will be (by Prop. LXXIX) as $$\scriptstyle \frac{DE^{2}\times O}{PF^{n}}$$, that is, as $$\scriptstyle \frac{2DF\times O}{PF^{n-1}}-\frac{DF^{2}\times O}{PF^{n}}$$. Let the perpendicular FN drawn into