Find the variation of the moon

PROPOSITION 30 PROBLEM 11: Find the horary motion of the nodes of the moon, in a circular orbit.

Let S represent the sun, T the earth, P the moon, NPn the orbit of the moon, Npn the orthographic projection of the orbit upon the plane of the ecliptic; N, n the nodes, nTNm the line of the nodes produced indefinitely;

PI, PK perpendiculars upon the lines ST, Qq; Pp a perpendicular upon the plane of the ecliptic; A, B the moon’s syzygies in the plane of the ecliptic; AZ a perpendicular let fall upon Nn, the line of the nodes; Q, g the quadratures of the moon in the plane of the ecliptic, and pK a perpendicular on the line Qq lying between the quadratures.

The force of the sun to disturb the motion of the moon (by Prop. XXV) is twofold, one proportional to the line LM, the other to the line MT, in the scheme of that Proposition; and the moon by the former force is drawn towards the earth, by the latter towards the sun, in a direction parallel to the right line ST joining the earth and the sun. The former force LM acts in the direction of the plane of the moon’s orbit, and therefore makes no change upon the situation thereof, and is upon that account to be neglected; the latter force MT, by which the plane of the moon’s orbit is disturbed, is the same with the force 3PK or 3IT. And this force (by Prop. XXV) is to the force by which the moon may, in its periodic time, be uniformly revolved in a circle about the earth at rest, as 3IT to the radius of the circle multiplied by the number 178,725, or as IT to the radius there of multiplied by 59,575. But in this calculus, and all that follows, I consider all the lines drawn from the moon to the sun as parallel to the line which joins the earth and the sun; because what inclination there is almost as much diminishes all effects in some cases as it augments them in others; and we are now inquiring after the mean motions of the nodes, neglecting such niceties as are of no moment, and would only serve to render the calculus more perplexed.

Suppose PM to represent an arc which the moon describes in the least moment of time, and ML a little line, the half of which the moon, by the impulse of the said force 3IT, would describe in the same time; and joining PL, MP, let them be produced to m and l, where they cut the plane of the ecliptic, and upon Tm let fall the perpendicular PH. Now, since the right line ML is parallel to the plane of the ecliptic, and therefore can never meet with the right line ml which lies in that plane, and yet both those right lines lie in one common plane LMPml, they will be parallel, and upon that account the triangles LMP, lmP will be similar.

Seeing MPm lies in the plane of the orbit, in which the moon did move while in the place P, the point m will fall upon the line Nn, which passes through the nodes N, n, of that orbit.

Because the force by which the half of the little line LM is generated, if the whole had been together, and at once impressed in the point P, would have generated that whole line, and caused the moon to move in the arc whose chord is LP; that is to say, would have transferred the moon from the plane MPmT into the plane LPlT; therefore the angular motion of the nodes generated by that force will be equal to the angle mTl. But ml is to mP as ML to MP; and since MP, because of the time given, is also given, ml will be as the rectangle ML

…

that is, as the rectangle IT

..

if Tml is a right angle, the angle mTl will be as

and therefore as

..

that is (because Tm and mP, TP and PH are proportional), as …

and, therefore, because TP is given, as IT × PH.

But if the angle Tml or STN is oblique, the angle mTl will be yet less, in proportion of the sine of the angle STN to the radius, or AZ to AT. And therefore the velocity of the nodes is as IT × PH × AZ, or as the solid content of the sines of the three angles TPI, PTN, and STN.

If these are right angles, as happens when the nodes are in the quadratures, and the moon in the syzygy, the little line ml will be removed to an infinite distance, and the angle mTl will become equal to the angle mPl. But in this case the angle mPl is to the angle PTM, which the moon in the same time by its apparent motion describes about the earth, as 1 to 59,575.

For the angle mPl is equal to the angle LPM, that is, to the angle of the moon’s deflexion from a rectilinear path; which angle, if the gravity of the moon should have then ceased, the said force of the sun 3IT would by itself have generated in that given time; and the angle PTM is equal to the angle of the moon’s deflexion from a rectilinear path; which angle, if the force of the sun 3IT should have then ceased, the force alone by which the moon is retained in its orbit would have generated in the same time.

These forces (as we have above shewn) are the one to the other as 1 to 59,575. Since, therefore, the mean horary motion of the moon (in respect of the fixed stars) is 32′ 56″ 27‴ 12½iv, the horary motion of the node in this case will be 33″ 10‴ 33iv.12v. But in other cases the horary motion will be to 33″ 10‴ 33iv.12v. as the solid content of the sines of the three angles TPI, PTN, and STN (or of the distances of the moon from the quadrature, of the moon from the node, and of the node from the sun) to the cube of the radius.

As often as the sine of any angle is changed from positive to negative, and from negative to positive, so often must the regressive be changed into a progressive, and the progressive into a regressive motion. Whence it comes to pass that the nodes are progressive as often as the moon happens to be placed between either quadrature, and the node nearest to that quadrature. In other cases they are regressive, and by the excess of the regress above the progress, they are monthly transferred in antecedentia.

Corollary 1

Hence, if from P and M, the extreme points of a least arc PM, on the line Qq joining the quadratures we let fall the perpendiculars PK, Mk, and produce the same till they cut the line of the nodes Nn in D and d, the horary motion of the nodes will be as the area MPDd, and the square of the line AZ conjunctly.

For let PK, PH, and AZ, be the three said sines, viz., PK the sine of the distance of the moon from the quadrature,

PH the sine of the distance of the moon from the node, and AZ the sine of the distance of the node from the sun; and the velocity of the node will be as the solid content of PK × PH × AZ.

But PT is to PK as PM to Kk; and, therefore, because PT and PM are given, Kk will be as PK. Likewise AT is to PD as AZ to PH, and therefore PH is as the rectangle PD ×} AZ.

By compounding those proportions, PK × PH is as the solid content Kk × PD × AZ, and PK × PH × AZ as Kk × PD × AZ²; that is, as the area PDdM and AZ² conjunctly. Q.E.D.

Corollary 2

In any given position of the nodes their mean horary motion is half their horary motion in the moon’s syzygies; and therefore is to 16″ 35‴ 16iv.36v. as the square of the sine of the distance of the nodes from the syzygies to the square of the radius, or as AZ² to AT².

For if the moon, by an uniform motion, describes the semi-circle QAq, the sum of all the areas PDdM, during the time of the moon’s passage from Q to M, will make up the area QMdE, terminating at the tangent QE of the circle; and by the time that the moon has arrived at the point n, that sum will make up the whole area EQAn described by the line PD: but when the moon proceeds from n to q, the line PD will fall without the circle, and describe the area nqe, terminating at the tangent qe of the circle, which area, because the nodes were before regressive, but are now progressive, must be subducted from the former area, and, being itself equal to the area QEN, will leave the semi-circle NQAn.

While, therefore, the moon describes a semi-circle, the sum of all the areas PDdM will be the area of that semi-circle; and while the moon describes a complete circle, the sum of those areas will be the area of the whole circle. But the area PDdM, when the moon is in the syzygies, is the rectangle of the arc PM into the radius PT; and the sum of all the areas, every one equal to this area, in the time that the moon describes a complete circle, is the rectangle of the whole circumference into the radius of the circle; and this rectangle, being double the area of the circle, will be double the quantity of the former sum.

If, therefore, the nodes went on with that velocity uniformly continued which they acquire in the moon’s syzygies, they would describe a space double of that which they describe in fact; and, therefore, the mean motion, by which, if uniformly continued, they would describe the same space with that which they do in fact describe by an unequal motion, is but one-half of that motion which they are possessed of in the moon’s syzygies. Wherefore since their greatest horary motion, if the nodes are in the quadratures, is 33″ 10‴ 33iv.12v, their mean horary motion in this case will be 16″ 35‴ 16iv.36v. And seeing the horary motion of the nodes is every where as AZ² and the area PDdM conjunctly, and, therefore, in the moon’s syzygies, the horary motion of the nodes is as AZ² and the area PDdM conjunctly, that is (because the area PDdM described in the syzygies is given), as AZ², therefore the mean motion also will be as AZ²; and, therefore, when the nodes are without the quadratures, this motion will be to 16″ 35‴ 16iv.36v. as AZ² to AT². Q.E.D.