Page:Newton's Principia (1846).djvu/437

 pm an arc which it describes in the least moment of time; N and n the nodes joined by the line Nn; pK and mk perpendiculars upon the axis Qq, produced both ways till they meet the circle in P and M, and the line of the nodes in D and d. And if the moon, by a radius drawn to the earth, describes an area proportional to the time of description, the horary motion of the node in the ellipsis will be as the area pDdm and AZ² conjunctly.

For let PF touch the circle in P, and produced meet TN in F; and pf touch the ellipsis in p, and produced meet the same TN in f, and both tangents concur in the axis TQ at Y. And let ML represent the space which the moon, by the impulse of the above-mentioned force 3IT or 3PK, would describe with a transverse motion, in the meantime while revolving in the circle it describes the arc PM; and ml denote the space which the moon revolving in the ellipsis would describe in the same time by the impulse of the same force 3IT or 3PK; and let LP and lp be produced till they meet the plane of the ecliptic in G and g, and FG and fg be joined, of which FG produced may cut pf, pg, and TQ, in c, e, and R respectively; and fg produced may cut TQ in r. Because the force 3IT or 3PK in the circle is to the force 3IT or 3pK in the ellipsis as PK to pK, or as AT to aT, the space ML generated by the former force will be to the space ml generated by the latter as PK to pK; that is, because of the similar figures PYKp and FYRc, as FR to cR. But (because of the similar triangles PLM, PGF) ML is to FG as PL to PG, that is (on account of the parallels Lk, PK, GR), as pl to pe, that is (because of the similar triangles plm, cpe) as lm to ce; and inversely as LM is to lm, or as FR is to cR, so is FG to ce. And therefore if fg was to ce as fy to cY, that is, as fr to cR (that is, as fr to FR and FR to cR conjunctly, that is, as fT to FT, and FG to ce conjunctly), because the ratio of FG to ce, expunged on both sides, leaves the ratios fg to FG and fT to FT, fg would be to FG as fT to FT; and, therefore, the angles which FG and fg would subtend at the earth T would be equal to each other. But these angles (by what we have shewn in the preceding Proposition) are the motions of the nodes, while the moon describes in the circle the arc PM, in the ellipsis the arc pm; and therefore the motions of the nodes in the circle and in the ellipsis would be equal to each other. Thus, I say, it would be, if fg was to ce as fY to cY, that is, fg was equal to $$\scriptstyle \frac{ce\times fY}{cY}$$. But because of the similar triangles fgp, cep, fg is to ce as fp to cp; and therefore fg is equal to $$\scriptstyle \frac{ce\times fp}{cp}$$; and therefore the angle which fg subtends in fact is to the former angle which FG subtends, that is to say, the motion of the nodes in the ellipsis is to the motion of the same in the circle as this fg or $$\scriptstyle \frac{ce\times fp}{cp}$$ to the fromer fg or $$\scriptstyle \frac{ce\times fY}{cY}$$, that is, as fp $$\scriptstyle \times$$