Projectivle

PROPOSITION XXVII. PROBLEM VIII: From the horary motion of the moon to find its distance from the earth.

The area which the moon, by a radius drawn to the earth, describes in every, moment of time, is as the horary motion of the moon and the square of the distance of the moon from the earth conjunctly. And therefore the distance of the moon from the earth is in a proportion compounded of the subduplicate proportion of the area directly, and the subduplicate proportion of the horary motion inversely. Q.E.I.

Cor. 1. Hence the apparent diameter of the moon is given; for it is reciprocally as the distance of the moon from the earth. Let astronomers try how accurately this rule agrees with the phænomena.

Cor. 2. Hence also the orbit of the moon may be more exactly defined from the phænomena than hitherto could be done.

PROPOSITION XXVIII. PROBLEM IX.

To find the diameters of the orbit, in which, without eccentricity, the moon would move.

The curvature of the orbit which a body describes, if attracted in lines perpendicular to the orbit, is as the force of attraction directly, and the square of the velocity inversely. I estimate the curvatures of lines compared one with another according to the evanescent proportion of the sines or tangents of their angles of contact to equal radii, supposing those radii to be infinitely diminished. But the attraction of the moon towards the earth in the syzygies is the excess of its gravity towards the earth above the force of the sun 2PK (see Fig. Prop. XXV), by which force the accelerative gravity of the moon towards the sun exceeds the accelerative gravity of the earth towards the sun, or is exceeded by it. But in the quadratures that attraction is the sum of the gravity of the moon towards the earth, and the sun’s force KT, by which the moon is attracted towards the earth. And these attractions, putting N for

are nearly as

For if the accelerative gravity of the moon towards the earth be represented by the number 178725, the mean force ML, which in the quadratures is PT or TK, and draws the moon towards the earth, will be 1000, and the mean force TM in the syzygies will be 3000; from which, if we subtract the mean force ML, there will remain 2000, the force by which the moon in the syzygies is drawn from the earth, and which we above called 2PK. But the velocity of the moon in the syzygies A and B is to its velocity in the quadratures C and D as CT to AT, and the moment of the area, which the moon by a radius drawn to the earth describes in the syzygies, to the moment of that area described in the quadratures conjunctly; that is, as 11073CT to 10973AT. Take this ratio twice inversely, and the former ratio once directly, and the curvature of the orb of the moon in the syzygies will be to the curvature thereof in the quadratures as 120406729 × {\displaystyle \scriptstyle \times } 178725AT² × {\displaystyle \scriptstyle \times } CT² × {\displaystyle \scriptstyle \times } N - 120406729 × {\displaystyle \scriptstyle \times } 2000AT4 × {\displaystyle \scriptstyle \times } CT to 122611329 × {\displaystyle \scriptstyle \times } 178725AT² × {\displaystyle \scriptstyle \times } CT² × {\displaystyle \scriptstyle \times } N + 122611329 × {\displaystyle \scriptstyle \times } 1000CT4 × {\displaystyle \scriptstyle \times } AT, that is, as 2151969AT × {\displaystyle \scriptstyle \times } CT × {\displaystyle \scriptstyle \times } N - 24081AT³ to 2191371AT × {\displaystyle \scriptstyle \times } CT × {\displaystyle \scriptstyle \times } N + 12261CT³.

Because the figure of the moon’s orbit is unknown, let us, in its stead, assume the ellipsis DBCA, in the centre of which we suppose the earth to be situated, and the greater axis DC to lie between the quadratures as the lesser AB between the syzygies. But since the plane of this ellipsis is revolved about the earth by an angular motion, and the orbit, whose curvature we now examine, should be described in a plane void of such motion we are to consider the figure which the moon, while it is revolved in that ellipsis, describes in this plane, that is to say, the figure Cpa, the several points p of which are found by assuming any point P in the ellipsis, which may represent the place of the moon, and drawing Tp equal to TP in such manner that the angle PTp may be equal to the apparent motion of the sun from the time of the last quadrature in C; or (which comes to the same thing) that the angle CTp may be to the angle CTP as the time of the synodic revolution of the moon to the time of the periodic revolution thereof, or as 29d.12h.44′ to 27d.7h.43′. If, therefore, in this proportion we take the angle CTa to the right angle CTA, and make Ta of equal length with TA, we shall have a the lower and C the upper apsis of this orbit Cpa. But, by computation, I find that the difference betwixt the curvature of this orbit Cpa at the vertex a, and the curvature of a circle described about the centre T with the interval TA, is to the difference between the curvature of the ellipsis at the vertex A, and the curvature of the same circle, in the duplicate proportion of the angle CTP to the angle CTp; and that the curvature of the ellipsis in A is to the curvature of that circle in the duplicate proportion of TA to TC; and the curvature of that circle to the curvature of a circle described about the centre T with the interval TC as TC to TA; but that the curvature of this last arch is to the curvature of the ellipsis in C in the duplicate proportion of TA to TC; and that the difference betwixt the curvature of the ellipsis in the vertex C, and the curvature of this last circle, is to the difference betwixt the curvature of the figure Cpa, at the vertex C, and the curvature of this same last circle, in the duplicate proportion of the angle CTp to the angle CTP; all which proportions are easily drawn from the sines of the angles of contact, and of the differences of those angles. But, by comparing those proportions together, we find the curvature of the figure Cpa at a to be to its curvature at C as AT³ - 16824⁄100000CT² AT to CT³ + 16824⁄100000AT² × {\displaystyle \scriptstyle \times } CT; where the number 16824⁄100000 represents the difference of the squares of the angles CTP and CTp, applied to the square of the lesser angle CTP; or (which is all one) the difference of the squares of the times 27d.7h.43′, and 29d.12j.44′, applied to the square of the time 27d.7h.43′, and 27d.7h.43′.

Since, therefore, a represents the syzygy of the moon, and C its quadrature, the proportion now found must be the same with that proportion of the curvature of the moon’s orb in the syzygies to the curvature thereof in the quadratures, which we found above. Therefore, in order to find the proportion of CT to AT, let us multiply the extremes and the means, and the terms which come out, applied to AT × {\displaystyle \scriptstyle \times } CT, become 2062,79CT4 - 2151969N × {\displaystyle \scriptstyle \times } CT³ + 368676N × {\displaystyle \scriptstyle \times } AT × {\displaystyle \scriptstyle \times } CT² + 36342AT² × {\displaystyle \scriptstyle \times } CT² - 362047N × {\displaystyle \scriptstyle \times } AT² × {\displaystyle \scriptstyle \times } CT + 2191371N × {\displaystyle \scriptstyle \times } AT³ + 4051,4AT4 = 0. Now if for the half sum N of the terms AT and CT we put 1, and x for their half difference, then CT will be = 1 + x, and AT = 1 - x. And substituting those values in the equation, after resolving thereof, we shall find x = 0,00719; and from thence the semi-diameter CT = 1,00719, and the semi-diameter AT = 0,99281, which numbers are nearly as 701⁄24, and 691⁄24. Therefore the moon’s distance from the earth in the syzygies is to its distance in the quadratures (setting aside the consideration of eccentricity) as 691⁄24 to 701⁄24; or, in round numbers, as 69 to 70.