Page:Newton's Principia (1846).djvu/427

That acceleration of the moon, in its passage from the quadrature C to the conjunction A, is in every moment of time as the generating accelerative force EL, that is, as $\scriptstyle {\frac {3PK\times TK}{TP}}$. Let the time be represented by the mean motion of the moon, or (which comes to the same thing) by the angle CTP, or even by the arc CP. At right angles upon CT erect CG equal to CT; and, supposing the quadrantal arc AC to be divided into an infinite number of equal parts Pp, &c., these parts may represent the like infinite number of the equal parts of time. Let fall pk perpendicular on CT, and draw TG meeting with KP, kp produced in F and f; then will FK be equal to TK, and Kk be to PK as Pp to Tp, that is, in a given proportion; and therefore FK $\scriptstyle \times$ Kk, or the area FKkf, will be as $\scriptstyle {\frac {3PK\times TK}{TP}}$, that is, as EL; and compounding, the whole area GCKF will be as the sum of all the forces EL impressed upon the moon in the whole time CP; and therefore also as the velocity generated by that sum, that is, as the acceleration of the description of the area CTP, or as the increment of the moment thereof. The force by which the moon may in its periodic time CADB of 27^d.7^h.43′ be retained revolving about the earth in rest at the distance TP, would cause a body falling in the time CT to describe the length ½CT, and at the same time to acquire a velocity equal to that with which the moon is moved in its orbit. This appears from Cor. 9, Prop, IV., Book I. But since Kd, drawn perpendicular on TP, is but a third part of EL, and equal to the half of TP, or ML, in the octants, the force EL in the octants, where it is greatest, will exceed the force ML in the proportion of 3 to 2; and therefore will be to that force by which the moon in its periodic time may be retained revolving about the earth at rest as 100 to ⅔ $\scriptstyle \times$ 178721½, or 11915; and in the time CT will generate a velocity equal to ¹⁰⁰⁄₁₁₉₁₅ parts of the velocity of the moon; but in the time CPA will generate a greater velocity in the proportion of CA to CT or TP. Let the greatest force EL in the octants be represented by the area FK $\scriptstyle \times$ Kk, or by the rectangle ½TP $\scriptstyle \times$ Pp, which is equal thereto; and the velocity which that greatest force can generate in any time CP will be to the velocity which any other lesser force EL can generate in the same time as the rectangle ½TP $\scriptstyle \times$ CP to the area KCGF; but the velocities generated in the whole time CPA will be one to the other as the rectangle ½TP $\scriptstyle \times$ CA to the triangle TCG, or as the quadrantal arc CA to the radius TP; and therefore the latter velocity generated in the whole time will be ¹⁰⁰⁄₁₁₉₁₅ parts of the velocity of the moon. To this velocity of the moon, which is proportional to the mean moment of the area (supposing this mean moment to be represented by the number 11915), we add and subtract the half of the other velocity; the sum 11915 + 50, or 11965, will represent the greatest moment of the area in the syzygy A; and the