The Orbit of the Moon

PROPOSITION XXVI. PROBLEM 7: Find the horary increment of the area which the moon, by a radius drawn to the earth, describes in a circular orbit

The area which the moon describes by a radius drawn to the earth is proportional to the time of description, excepting in so far as the moon’s motion is disturbed by the action of the sun.

Here, we propose to investigate the inequality of the moment, or horary increment of that area or motion so disturbed.

To render the calculus more easy, we shall suppose the orbit of the moon to be circular, and neglect all inequalities but that only which is now under consideration; and, because of the immense distance of the sun, we shall farther suppose that the lines SP and ST are parallel. By this means, the force LM will be always reduced to its mean quantity TP, as well as the force TM to its mean quantity 3PK. These forces (by Cor. 2 of the Laws of Motion) compose the force TL; and this force, by letting fall the perpendicular LE upon the radius TP, is resolved into the forces TE, EL; of which the force TE, acting constantly in the direction of the radius TP, neither accelerates nor retards the description of the area TPC made by that radius TP; but EL, acting on the radius TP in a perpendicular direction, accelerates or retards the description of the area in proportion as it accelerates or retards the moon. 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

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 × {\displaystyle \scriptstyle \times } Kk, or the area FKkf, will be as

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 27d.7h.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 × {\displaystyle \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 × {\displaystyle \scriptstyle \times } Kk, or by the rectangle ½TP × {\displaystyle \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 × {\displaystyle \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 × {\displaystyle \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 difference 11915 - 50, or 11865, the least moment thereof in the quadratures. Therefore the areas which in equal times are described in the syzygies and quadratures are one to the other as 11965 to 11865. And if to the least moment 11865 we add a moment which shall be to 100, the difference of the two former moments, as the trapezium FKCG to the triangle TCG, or, which comes to the same thing, as the square of the sine PK to the square of the radius TP (that is, as Pd to TP), the sum will represent the moment of the area when the moon is in any intermediate place P.

But these things take place only in the hypothesis that the sun and the earth are at rest, and that the synodical revolution of the moon is finished in 27d.7h.43’. But since the moon’s synodical period is really 29d.12h.41’, the increments of the moments must be enlarged in the same proportion as the time is, that is, in the proportion of 1080853 to 1000000. Upon which account, the whole increment, which was 100⁄11915 parts of the mean moment, will now become T100⁄11023 parts thereof; and therefore the moment of the area in the quadrature of the moon will be to the moment thereof in the syzygy as 11023 - 50 to 11023 + 50; or as 10973 to 11073: and to the moment thereof, when the moon is in any intermediate place P, as 10973 to 10973 + Pd; that is, supposing TP = 100.

The area, therefore, which the moon, by a radius drawn to the earth, describes in the several little equal parts of time, is nearly as the sum of the number 219,46, and the versed sine of the double distance of the moon from the nearest quadrature, considered in a circle which hath unity for its radius. Thus it is when the variation in the octants is in its mean quantity. But if the variation there is greater or less, that versed sine must be augmented or diminished in the same proportion.