Page:Newton's Principia (1846).djvu/234

 distance PE or PF, conjunctly. But (by the last Lemma) Dd is to Ff as PE to PS, and therefore Ff is equal to $$\scriptstyle \frac{PS\times Dd}{PE}$$; and DE² $$\scriptstyle \times$$ Ff is equal to Dd $$\scriptstyle \times \frac{DE^{2}\times PS}{PE}$$; and therefore the force of the lamina EFfe is as Dd $$\scriptstyle \times \frac{DE^{2}\times PS}{PE}$$ and the force of a particle exerted at the distance PF conjunctly; that is, by the supposition, as DN $$\scriptstyle \times$$ Dd, or as the evanescent area DNnd. Therefore the forces of all the laminae exerted upon the corpuscle P are as all the areas DNnd, that is, the whole force of the sphere will be as the whole area ANB. Q.E.D.

. 1. Hence if the centripetal force tending to the several particles remain always the same at all distances, and DN be made as $$\scriptstyle \frac{DE^{2}\times PS}{PE}$$ the whole force with which the corpuscle is attracted by the sphere is as the area ANB.

. 2. If the centripetal force of the particles be reciprocally as the distance of the corpuscle attracted by it, and DN be made as $$\scriptstyle \frac{DE^{2}\times PS}{PE^{2}}$$, the force with which the corpuscle P is attracted by the whole sphere will be as the area ANB.

. 3. If the centripetal force of the particles be reciprocally as the cube of the distance of the corpuscle attracted by it, and DN be made as $$\scriptstyle \frac{DE^{2}\times PS}{PE^{4}}$$, the force with which the corpuscle is attracted by the whole sphere will be as the area ANB.

. 4. And universally if the centripetal force tending to the several particles of the sphere be supposed to be reciprocally as the quantity V; and DN be made as $$\scriptstyle \frac{DE^{2}\times PS}{PE\times V}$$; the force with which a corpuscle is attracted by the whole sphere will be as the area ANB.


 * The things remaining as above, it is required to measure the area ANB.

From the point P let there be drawn the right line PH touching the sphere in H; and to the axis PAB, letting fall the perpendicular HI, bisect PI in L; and (by Prop. XII, Book II, Elem.) PE² is equal to