Page:Newton's Principia (1846).djvu/180

 pC, that will be to the distance which the other body P acquires from the line PC as the transverse motion of the body p to the transverse motion of the other body P. Therefore since kr is equal to the distance which the body P acquires from the line PC, and mr is to kr as the angle VCp to the angle VCP, that is, as the transverse motion of the body p to the transverse motion of the body P, it is manifest that the body p, at the expiration of that time, will be found in the place m. These things will be so, if the bodies p and P are equally moved in the directions of the lines pC and PC, and are therefore urged with equal forces in those directions, but if we take an angle pCn that is to the angle pCk as the angle VCp to the angle VCP, and nC be equal to kC, in that case the body p at the expiration of the time will really be in n; and is therefore urged with a greater force than the body P, if the angle nCp is greater than the angle kCp, that is, if the orbit upk, move either in consequentia or in antecedentia, with a celerity greater than the double of that with which the line CP moves in consequentia; and with a less force if the orbit moves slower in antecedentia. And the difference of the forces will be as the interval mn of the places through which the body would be carried by the action of that difference in that given space of time. About the centre C with the interval Cn or Ck suppose a circle described cutting the lines mr, mn produced in s and t, and the rectangle mn $$\scriptstyle \times$$ mt will be equal to the rectangle mk $$\scriptstyle \times$$ ms, and therefore mn will be equal to $$\scriptstyle \frac{mk\times ms}{mt}$$. But since the triangles pCk, pCn, in a given time, are of a given magnitude, kr and mr, and their difference mk, and their sum ms, are reciprocally as the altitude pC, and therefore the rectangle mk $$\scriptstyle \times$$ ms is reciprocally as the square of the altitude pC. But, moreover, mt is directly as ½mt, that is, as the altitude pC. These are the first ratios of the nascent lines: and hence $$\scriptstyle \frac{mk\times ms}{mt}$$, that is, the nascent lineola mn, and the difference of the forces proportional thereto, are reciprocally as the cube of the altitude pC.  Q.E.D.

. 1. Hence the difference of the forces in the places P and p, or K and k, is to the force with which a body may revolve with a circular motion from R to K, in the same time that the body P in an immovable orb describes the arc PK, as the nascent line mn to the versed sine of the nascent arc RK, that is, as $$\scriptstyle \frac{mk\times ms}{mt}$$ to $$\scriptstyle \frac{rk^{2}}{2kC}$$, or as mk $$\scriptstyle \times$$ ms to the square of rk; that is, if we take given quantities F and G in the same ratio to one another as the angle VCP bears to the angle VCp, as GG - FF to FF. And, therefore, if from the centre C, with any distance CP or Cp, there be described a circular sector equal to the whole area VPC, which the body