# The Mathematical Principles of Natural Philosophy (1846)/BookIII-Prop3

PROPOSITION XXV. PROBLEM VI.
To find the forces with which the sun disturbs the motions of the moon.
Let S represent the sun, T the earth, P the moon, CADB the moon's orbit. In SP take SK equal to ST; and let SL be to SK in the duplicate proportion of SK to SP: draw LM parallel to PT; and if ST or SK is supposed to represent the accelerated force of gravity of the earth towards the sun, SL will represent the accelerative force of gravity of the moon towards the sun. But that force is compounded of the parts SM and LM, of which the force LM, and that part of SM which is represented by TM, disturb the motion of the moon, as we have shewn in Prop. LXVI, Book I, and its Corollaries. Forasmuch as the earth and moon are revolved about their common centre of gravity, the motion of the earth about that centre will be also disturbed by the like forces; but we may consider the sums both of the forces and of the motions as in the moon, and represent the sum of the forces by the lines TM and ML, which are analogous to thorn both. The force ML (in its mean quantity) is to the centripetal force by which the moon may be retained in its orbit revolving about the earth at rest, at the distance PT, in the duplicate proportion of the periodic time of the moon about the earth to the periodic time of the earth about the sun (by Cor. 17, Prop. LXVI, Book I); that is, in the duplicate proportion of 27$^d$.7$^h$.43′ to 365$^d$.6$^h$.9′; or as 1000 to 178725; or as 1 to 178$^{29}$⁄$_{40}$. But in the ​4th Prop. of this Book we found, that, if both earth and moon were revolved about their common centre of gravity, the mean distance of the one from the other would be nearly 60½ mean semi-diameters of the earth; and the force by which the moon may be kept revolving in its orbit about the earth in rest at the distance PT of 60½ semi-diameters of the earth, is to the force by which it may be revolved in the same time, at the distance of 60 semi-diameters, as 60½ to 60: and this force is to the force of gravity with us very nearly as 1 to 60 $\scriptstyle \times$ 60. Therefore the mean force ML is to the force of gravity on the surface of our earth as 1 $\scriptstyle \times$ 60½ to 60 $\scriptstyle \times$ 60 $\scriptstyle \times$ 60 $\scriptstyle \times$ 178$^{29}$⁄$_{40}$, or as 1 to 638092,6; whence by the proportion of the lines TM, ML, the force TM is also given; and these are the forces with which the sun disturbs the motions of the moon. Q.E.I.
PROPOSITION XXVI. PROBLEM VII.
To find the horary increment of the area which the moon, by a radius drawn to the earth, describes in a circular orbit.
We have above shown that 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; and 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 $\scriptstyle {\frac {3PK\times TK}{TP}}$. 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 ​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 27$^d$.7$^h$.43'. But since the moon's synodical period is really 29$^d$.12$^h$.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 T$^{100}$⁄$_{11023}$ 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.
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 $\scriptstyle {\frac {AT+CT}{2}}$, are nearly as $\scriptstyle {\frac {178725}{AT^{2}}}-{\frac {2000}{CT\times N}}$ and $\scriptstyle {\frac {178725}{CT^{2}}}+{\frac {1000}{AT\times N}}$, or as 178725N $\scriptstyle \times$ CT² - 2000AT² $\scriptstyle \times$ CT, and 178725N $\scriptstyle \times$ AT² + 1000CT² $\scriptstyle \times$ AT. 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 $\scriptstyle \times$ 178725AT² $\scriptstyle \times$ CT² $\scriptstyle \times$ N - 120406729 $\scriptstyle \times$ 2000AT$^4$ $\scriptstyle \times$ CT to 122611329 $\scriptstyle \times$ 178725AT² $\scriptstyle \times$ CT² $\scriptstyle \times$ N + 122611329 $\scriptstyle \times$ 1000CT$^4$ $\scriptstyle \times$ AT, that is, as 2151969AT $\scriptstyle \times$ CT $\scriptstyle \times$ N - 24081AT³ to 2191371AT $\scriptstyle \times$ CT $\scriptstyle \times$ 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 29$^d$.12$^h$.44′ to 27$^d$.7$^h$.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 - $^{16824}$⁄$_{100000}$CT² AT to CT³ + $^{16824}$⁄$_{100000}$AT² $\scriptstyle \times$ 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 27$^d$.7$^h$.43′, and 29$^d$.12$^j$.44′, applied to the square of the time 27$^d$.7$^h$.43′, and 27$^d$.7$^h$.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 $\scriptstyle \times$ CT, become 2062,79CT$^4$ - 2151969N $\scriptstyle \times$ CT³ + 368676N $\scriptstyle \times$ AT $\scriptstyle \times$ CT² + 36342AT² $\scriptstyle \times$ CT² - 362047N $\scriptstyle \times$ AT² $\scriptstyle \times$ CT + 2191371N $\scriptstyle \times$ AT³ + 4051,4AT$^4$ = 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 70$^1$⁄$_{24}$, and 69$^1$⁄$_{24}$. 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 69$^1$⁄$_{24}$ to 70$^1$⁄$_{24}$; or, in round numbers, as 69 to 70.
PROPOSITION XXIX. PROBLEM X.
To find the variation of the moon.
This inequality is owing partly to the elliptic figure of the moon's orbit, partly to the inequality of the moments of the area which the moon by a radius drawn to the earth describes. If the moon P revolved in the ellipsis DBCA about the earth quiescent in the centre of the ellipsis, and by the radius TP, drawn to the earth, described the area CTP, proportional to the time of description; and the greatest semi-diameter CT of the ellipsis was to the least TA as 70 to 69; the tangent of the angle CTP would be to the tangent of the angle of the mean motion, computed from the quadrature C, as the semi-diameter TA of the ellipsis to its semi-diameter TC, or as 69 to 70. But the description of the area CTP, as the moon advances from the quadrature to the syzygy, ought to be in such manner accelerated, that the moment of the area in the moon's syzygy may be to the moment thereof in its quadrature as 11073 to 10973; and that the excess of the moment in any intermediate place P above the moment in the quadrature may be as the square of the sine of the angle CTP; which we may effect with accuracy enough, if we diminish the tangent of the angle CTP in the subduplicate proportion of the number 10973 to the number 11073, that is, in proportion of the number 68,6877 to the number 69. Upon which account the tangent of the angle CTP will now be to the tangent of the mean motion as 68,6877 to 70; and the angle CTP in the octants, where the mean motion is 45°, will be found 44° 27′ 28″, which subtracted from 45°, the angle of the mean motion, leaves the greatest variation 32′ 32″. Thus it would be, if the moon, in passing from the quadrature to the syzygy, described an angle CTA of 90 degrees only. But because of the motion of the earth, by which the sun is apparently transferred in consequentia, the moon, before it overtakes the sun, describes an angle CT, greater than a right angle, in the proportion of the time of the synodic revolution of the moon to the time of its periodic revolution, that ​is, in the proportion of 29$^d$.12$^h$.44′ to 27$^d$.7$^h$.43′. Whence it comes to pass that all the angles about the centre T are dilated in the same proportion; and the greatest variation, which otherwise would be but 32′ 32″, now augmented in the said proportion, becomes 35′ 10″.
And this is its magnitude in the mean distance of the sun from the earth, neglecting the differences which may arise from the curvature of the orbis magnus, and the stronger action of the sun upon the moon when horned and new, than when gibbous and full. In other distances of the sun from the earth, the greatest variation is in a proportion compounded of the duplicate proportion of the time of the synodic revolution of the moon (the time of the year being given) directly, and the triplicate proportion of the distance of the sun from the earth inversely. And, therefore, in the apogee of the sun, the greatest variation is 33′ 14″, and in its perigee 37′ 11″, if the eccentricity of the sun is to the transverse semi-diameter of the orbis magnus as 16$^{15}$⁄$_{16}$ to 1000.
Hitherto we have investigated the variation in an orb not eccentric, in which, to wit, the moon in its octants is always in its mean distance from the earth. If the moon, on account of its eccentricity, is more or less removed from the earth than if placed in this orb, the variation may be something greater, or something less, than according to this rule. But I leave the excess or defect to the determination of astronomers from the phænomena.
PROPOSITION XXX. PROBLEM XI.
To 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;
Now 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. And 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. And 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 rectan​gle ML $\scriptstyle \times$ mP, that is, as the rectangle IT $\scriptstyle \times$ mP. And if Tml is a right angle, the angle mTl will be as $\scriptstyle {\frac {ml}{Tm}}$ and therefore as $\scriptstyle {\frac {IT\times Pm}{Tm}}$, that is (because Tm and mP, TP and PH are proportional), as $\scriptstyle {\frac {IT\times PH}{TP}}$; and, therefore, because TP is given, as IT $\scriptstyle \times$ 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 $\scriptstyle \times$ PH $\scriptstyle \times$ 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. And 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‴ 12½$^{iv}$, the horary motion of the node in this case will be 33″ 10‴ 33$^{iv}$.12$^v$. But in other cases the horary motion will be to 33″ 10‴ 33$^{iv}$.12$^v$. 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. And 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.
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 $\scriptstyle \times$ PH $\scriptstyle \times$ 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 $\scriptstyle \times$ AZ; and, by compounding those proportions, PK $\scriptstyle \times$ PH is as the solid content Kk $\scriptstyle \times$ PD $\scriptstyle \times$ AZ, and PK $\scriptstyle \times$ PH $\scriptstyle \times$ AZ as Kk $\scriptstyle \times$ PD $\scriptstyle \times$ AZ²; that is, as the area PDdM and AZ² conjunctly. Q.E.D.
Cor. 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‴ 16$^{iv}$.36$^v$. 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‴ 33$^{iv}$.12$^v$, their mean horary motion in this case will be 16″ 35‴ 16$^{iv}$.36$^v$. 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‴ 16$^{iv}$.36$^v$. as AZ² to AT². Q.E.D.
PROPOSITION XXXI. PROBLEM XII.
To find the horary motion of the nodes of the moon in an elliptic orbit.
Let Qpmaq represent an ellipsis described with the greater axis Qq, am the lesser axis ab; QAqB a circle circumscribed; T the earth in the common centre of both; S the sun; p the moon moving in this ellipsis; and
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 $\scriptstyle \times$ ​cY to fY $\scriptstyle \times$ cp, or as fp to fY, and cY to cp; that is, if ph parallel to TN meet FP in h, as Fh to FY and FY to FP; that is, as Fh to FP or Dp to DP, and therefore as the area Dpmd to the area DPMd. And, therefore, seeing (by Corol. 1, Prop. XXX) the latter area and AZ² conjunctly are proportional to the horary motion of the nodes in the circle, the former area and AZ² conjunctly will be proportional to the horary motion of the nodes in the ellipsis. Q.E.D.
Cor. Since, therefore, in any given position of the nodes, the sum of all the areas pDdm, in the time while the moon is carried from the quadrature to any place m, is the area mpQEd terminated at the tangent of the ellipsis QE; and the sum of all those areas, in one entire revolution, is the area of the whole ellipsis; the mean motion of the nodes in the ellipsis will be to the mean motion of the nodes in the circle as the ellipsis to the circle; that is, as Ta to TA, or 69 to 70. And, therefore, since (by Corol 2, Prop. XXX) the mean horary motion of the nodes in the circle is to 16″ 35‴ 16$^{iv}$.36$^v$. as AZ² to AT², if we take the angle 16″ 21‴ 3$^{iv}$.30$^v$. to the angle 16″ 35‴ 16$^{iv}$.36$^v$. as 69 to 70, the mean horary motion of the nodes in the ellipsis will be to 16″ 21‴ 3$^{iv}$.30$^v$. as AZ² to AT²; that is, as the square of the sine of the distance of the node from the sun to the square of the radius.
But the moon, by a radius drawn to the earth, describes the area in the syzygies with a greater velocity than it does that in the quadratures, and upon that account the time is contracted in the syzygies, and prolonged in the quadratures; and together with the time the motion of the nodes is likewise augmented or diminished. But the moment of the area in the quadrature of the moon was to the moment thereof in the syzygies as 10973 to 11073; and therefore the mean moment in the octants is to the excess in the syzygies, and to the defect in the quadratures, as 11023, the half sum of those numbers, to their half difference 50. Wherefore since the time of the moon in the several little equal parts of its orbit is reciprocally as its velocity, the mean time in the octants will be to the excess of the time in the quadratures, and to the defect of the time in the syzygies arising from this cause, nearly as 11023 to 50. But, reckoning from the quadratures to the syzygies, I find that the excess of the moments of the area, in the several places above the least moment in the quadratures, is nearly as the square of the sine of the moon's distance from the quadratures; and therefore the difference betwixt the moment in any place, and the mean moment in the octants, is as the difference betwixt the square of the sine of the moon's distance from the quadratures, and the square of the sine of 45 degrees, or half the square of the radius; and the increment of the time in the several places between the octants and quadratures, and the decrement thereof between the octants and syzygies, is in the same proportion. But the motion of the nodes, while the moon describes the several little equal parts of its orbit, is accelerated or retarded ​in the duplicate proportion of the time; for that motion, while the moon describes PM, is (cæteris paribus) as ML, and ML is in the duplicate proportion of the time. Wherefore the motion of the nodes in the syzygies, in the time while the moon describes given little parts of its orbit, is diminished in the duplicate proportion of the number 11073 to the number 11023; and the decrement is to the remaining motion as 100 to 10973; but to the whole motion as 100 to 11073 nearly. But the decrement in the places between the octants and syzygies, and the increment in the places between the octants and quadratures, is to this decrement nearly as the whole motion in these places to the whole motion in the syzygies, and the difference betwixt the square of the sine of the moon's distance from the quadrature, and the half square of the radius, to the half square of the radius conjunctly. Wherefore, if the nodes are in the quadratures, and we take two places, one on one side, one on the other, equally distant from the octant and other two distant by the same interval, one from the syzygy, the other from the quadrature, and from the decrements of the motions in the two places between the syzygy and octant we subtract the increments of the motions in the two other places between the octant and the quadrature, the remaining decrement will be equal to the decrement in the syzygy, as will easily appear by computation; and therefore the mean decrement, which ought to be subducted from the mean motion of the nodes, is the fourth part of the decrement in the syzygy. The whole horary motion of the nodes in the syzygies (when the moon by a radius drawn to the earth was supposed to describe an area proportional to the time) was 32″ 42‴ 7$^{iv}$. And we have shewn that the decrement of the motion of the nodes, in the time while the moon, now moving with greater velocity, describes the same space, was to this motion as 100 to 11073; and therefore this decrement is 17‴ 43$^{iv}$.11$^v$. The fourth part of which 4‴ 25$^{iv}$.48$^v$. subtracted from the mean horary motion above found, 16″ 21‴ 3$^{iv}$.30$^v$. leaves 16″ 16‴ 37$^{iv}$.42$^v$. their correct mean horary motion.
If the nodes are without the quadratures, and two places are considered, one on one side, one on the other, equally distant from the syzygies, the sum of the motions of the nodes, when the moon is in those places, will be to the sum of their motions, when the moon is in the same places and the nodes in the quadratures, as AZ² to AT². And the decrements of the motions arising from the causes but now explained will be mutually as the motions themselves, and therefore the remaining motions will be mutually betwixt themselves as AZ² to AT²; and the mean motions will be as the remaining motions. And, therefore, in any given position of the nodes, their correct mean horary motion is to 16″ 16‴ 37$^{iv}$.42$^v$. as AZ² to AT²; that is, as the square of the sine of the distance of the nodes from the syzygies to the square of the radius.
PROPOSITION XXXII. PROBLEM XIII.
To find the mean motion of the nodes of the moon.
The yearly mean motion is the sum of all the mean horary motions throughout the course of the year. Suppose that the node is in N, and that, after every hour is elapsed, it is drawn back again to its former place; so that, notwithstanding its proper motion, it may constantly remain in the same situation with respect to the fixed stars; while in the mean time the sun S, by the motion of the earth, is seen to leave the node, and to proceed till it completes its apparent annual course by an uniform motion. Let Aa represent a given least arc, which the right line TS always drawn to the sun, by its intersection with the circle NAn, describes in the least given moment of time; and the mean horary motion (from what we have above shewn) will be as AZ², that is (because AZ and ZY are proportional), as the rectangle of AZ into ZY, that is, as the area AZYa; and the sum of all the mean horary motions from the beginning will be as the sum of all the areas aYZA, that is, as the area NAZ. But the greatest AZYa is equal to the rectangle of the arc Aa into the radius of the circle; and therefore the sum of all these rectangles in the whole circle will be to the like sum of all the greatest rectangles as the area of the whole circle to the rectangle of the whole circumference into the radius, that is, as 1 to 2. But the horary motion corresponding to that greatest rectangle was 16″ 16‴ 37$^{iv}$.42$^v$. and this motion in the complete course of the sidereal year, 365$^d$.6$^h$.9′, amounts to 39° 38′ 7″ 50‴, and therefore the half thereof, 19° 49′ 3″ 55‴, is the mean motion of the nodes corresponding to the whole circle. And the motion of the nodes, in the time while the sun is carried from N to A, is to 19° 49′ 3″ 55‴ as the area NAZ to the whole circle.
Thus it would be if the node was after every hour drawn back again to its former place, that so, after a complete revolution, the sun at the year's end would be found again in the same node which it had left when the year begun. But, because of the motion of the node in the mean time, the sun must needs meet the node sooner; and now it remains that we compute the abbreviation of the time. Since, then, the sun, in the course of the year, travels 360 degrees, and the node in the same time by its greatest motion would be carried 39° 38′ 7″ 50‴, or 39,6355 degrees; and the mean motion of the node in any place N is to its mean motion in its quadrature as AZ² to AT²; the motion of the sun will be to the motion of the node ​in N as 360AT² to 39,6355 AZ²; that is, as 9,0827646AT² to AZ². Wherefore if we suppose the circumference NAn of the whole circle to be divided into little equal parts, such as Aa, the time in which the sun would describe the little arc Aa, if the circle was quiescent, will be to the time of which it would describe the same arc, supposing the circle together with the nodes to be revolved about the centre T, reciprocally as 9,0827646AT² to 9,0827646AT² + AZ²; for the time is reciprocally as the velocity with which the little arc is described, and this velocity is the sum of the velocities of both sun and node. If, therefore, the sector NTA represent the time in which the sun by itself, without the motion of the node, would describe the arc NA, and the indefinitely small part ATa of the sector represent the little moment of the time in which it would describe the least arc Aa; and (letting fall aY perpendicular upon Nn) if in AZ we take dZ of such length that the rectangle of dZ into ZY may be to the least part ATa of the sector as AZ² to 9,0827646AT² + AZ², that is to say, that dZ may be to ½AZ as AT² to 9,0827646AT² + AZ²; the rectangle of dZ into ZY will represent the decrement of the time arising from the motion of the node, while the arc Aa is described; and if the curve NdGn is the locus where the point d is always found, the curvilinear area NdZ will be as the whole decrement of time while the whole arc NA is described; and, therefore, the excess of the sector NAT above the area NdZ will be as the whole time. But because the motion of the node in a less time is less in proportion of the time, the area AaYZ must also be diminished in the same proportion; which may be done by taking in AZ the line eZ of such length, that it may be to the length of AZ as AZ² to 9,0827646AT² + AZ²; for so the rectangle of eZ into ZY will be to the area AZYa as the decrement of the time in which the arc Aa is described to the whole time in which it would have been described, if the node had been quiescent; and, therefore, that rectangle will be as the decrement of the motion of the node. And if the curve NeFn is the locus of the point e, the whole area NeZ, which is the sum of all the decrements of that motion, will be as the whole decrement thereof during the time in which the arc AN is described; and the remaining area NAe will be as the remaining motion, which is the true motion of the node, during the time in which the whole arc NA is described by the joint motions of both sun and node. Now the area of the semi-circle is to the area of the figure NeFn found by the method of infinite series nearly as 793 to 60. But the motion corresponding or proportional to the whole circle was 19° 49′ 3″ 55‴; and therefore the motion corresponding to double the figure NeFn is 1° 29′ 58″ 2‴, which taken from the former motion leaves 18° 19′ 5″ 53‴, the whole motion of the node with respect to the fixed stars in the interval between two of its conjunctions with the sun; and this motion subducted from the annual motion of the sun 360°, leaves 341° 40′ 54″ 7‴, ​the motion of the sun in the interval between the same conjunctions. But as this motion is to the annual motion 360°, so is the motion of the node but just now found 18° 19′ 5″ 53‴ to its annual motion, which will therefore be 19° 18′ 1″ 23‴; and this is the mean motion of the nodes in the sidereal year. By astronomical tables, it is 19° 21′ 21″ 50‴ . The difference is less than $^1$⁄$_{300}$ part of the whole motion, and seems to arise from the eccentricity of the moon's orbit, and its inclination to the plane of the ecliptic. By the eccentricity of this orbit the motion of the nodes is too much accelerated; and, on the other hand, by the inclination of the orbit, the motion of the nodes is something retarded, and reduced to its just velocity.
PROPOSITION XXXIII. PROBLEM XIV.
To find the true motion of the nodes of the moon.
In the time which is as the area NTA - NdZ (in the preceding Fig.) that motion is as the area NAe, and is thence given; but because the calculus is too difficult, it will be better to use the following construction of the Problem. About the centre C, with any interval CD, describe the circle BEFD; produce DC to A so as AB may be to AC as the mean motion to half the mean true motion when the nodes are in their quadratures (that is, as 19° 18′ 1″ 23‴ to 19° 49′ 3″ 55‴; and therefore BC to AC as the difference of those motions 0° 31′ 2″ 32‴ to the latter motion 19° 49′ 3″ 55‴, that is, as 1 to 38$^3$⁄$_{10}$). Then through the point D draw the indefinite line Gg, touching the circle in D; and if we take the angle BCE, or BCF, equal to the double distance of the sun from the place of the node, as found by the mean motion, and drawing AE or AF cutting the perpendicular DG in G, we take another angle which shall be to the whole motion of the node in the interval between its syzygies (that is, to 9° 11' 3") as the tangent DG to the whole circumference of the circle BED, and add this last angle (for which the angle DAG may be used) to the mean motion of the nodes, while they are passing from the quadratures to the syzygies, and subtract it from their mean motion while they are passing from the syzygies to the quadratures, we shall have their true motion; for the true motion so found will nearly agree with the true motion which comes out from assuming the times as the area NTA - NdZ, and the motion of the node as the area NAe; as whoever will please to examine and make the computations will find: and this is the semi-menstrual equation of the motion of the nodes. But there is also a menstrual equation, but which is by no means necessary for find​ing of the moon's latitude; for since the variation of the inclination of the moon's orbit to the plane of the ecliptic is liable to a twofold inequality, the one semi-menstrual, the other menstrual, the menstrual inequality of this variation, and the menstrual equation of the nodes, so moderate and correct each other, that in computing the latitude of the moon both may be neglected.
Cor. From this and the preceding Prop, it appears that the nodes are quiescent in their syzygies, but regressive in their quadratures, by an hourly motion of 16″ 19‴ 26$^{iv}$.; and that the equation of the motion of the nodes in the octants is 1° 30′; all which exactly agree with the phænomena of the heavens.
SCHOLIUM.
Mr. Machin, Astron., Prof. Gresh., and Dr. Henry Pemberton, separately found out the motion of the nodes by a different method. Mention has been made of this method in another place. Their several papers, both of which I have seen, contained two Propositions, and exactly agreed with each other in both of them. Mr. Machin's paper coming first to my hands, I shall here insert it.
OF THE MOTION OF THE MOON'S NODES.
“PROPOSITION I.
    “The mean motion of the sun from the node is defined by a geometric mean proportional between the mean motion of the sun and that mean motion with which the sun recedes with the greatest swiftness from the node in the quadratures.
“Let T be the earth's place, Nn the line of the moon's nodes at any given time, KTM a perpendicular thereto, TA a right line revolving about the centre with the same angular velocity with which the sun and the node recede from one another, in such sort that the angle between the quiescent right line Nn and the revolving line TA may be always equal to the distance of the places of the sun and node. Now if any right line TK be divided into parts TS and SK, and those parts be taken as the mean horary motion of the sun to the mean horary motion of the node in the quadratures, and there be taken the right line TH, a mean proportional between the part TS and the whole TK, this right line will be proportional to the sun's mean motion from the node.
and TN, let there be described an ellipsis NHnL; and in the time in which the sun recedes from the node through the arc Na, if there be drawn the right line Tba, the area of the sector NTa will be the exponent of the sum of the motions of the sun and node in the same time. Let, therefore, the extremely small arc aA be that which the right line Tba, revolving according to the aforesaid law, will uniformly describe in a given particle of time, and the extremely small sector TAa will be as the sum of the velocities with which the sun and node are carried two different ways in that time. Now the sun's velocity is almost uniform, its inequality being so small as scarcely to produce the least inequality in the mean motion of the nodes. The other part of this sum, namely, the mean quantity of the velocity of the node, is increased in the recess from the syzygies in a duplicate ratio of the sine of its distance from the sun (by Cor. Prop. XXXI, of this Book), and, being greatest in its quadratures with the sun in K, is in the same ratio to the sun's velocity as SK to TS, that is, as (the difference of the squares of TK and TH, or) the rectangle KHM to TH². But the ellipsis NBH divides the sector ATa, the exponent of the sum of these two velocities, into two parts ABba and BTb, proportional to the velocities. For produce BT to the circle in β, and from the point B let fall upon the greater axis the perpendicular BG, which being produced both ways may meet the circle in the points F and f; and because the space ABba is to the sector TBb as the rectangle ABβ to BT² (that rectangle being equal to the difference of the squares of TA and TB, because the right line Aβ is equally cut in T, and unequally in B), therefore when the space ABba is the greatest of all in K, this ratio will be the same as the ratio of the rectangle KHM to HT². But the greatest mean velocity of the node was shewn above to be in that very ​ratio to the velocity of the sun; and therefore in the quadratures the sector ATa is divided into parts proportional to the velocities. And because the rectangle KHM is to HT² as FBf to BG², and the rectangle ABβ is equal to the rectangle FBf, therefore the little area ABba, where it is greatest, is to the remaining sector TBb as the rectangle ABβ to BG². But the ratio of these little areas always was as the rectangle ABβ to BT²; and therefore the little area ABba in the place A is less than its correspondent little area in the quadratures in the duplicate ratio of BG to BT, that is, in the duplicate ratio of the sine of the sun's distance from the node. And therefore the sum of all the little areas ABba, to wit, the space ABN, will be as the motion of the node in the time in which the sun hath been going over the arc NA since he left the node; and the remaining space, namely, the elliptic sector NTB, will be as the sun's mean motion in the same time. And because the mean annual motion of the node is that motion which it performs in the time that the sun completes one period of its course, the mean motion of the node from the sun will be to the mean motion of the sun itself as the area of the circle to the area of the ellipsis; that is, as the right line TK to the right line TH, which is a mean proportional between TK and TS; or, which comes to the same as the mean proportional TH to the right line TS.
“PROPOSITION II.
    “The mean motion of the moon's nodes being given, to find their true motion.
“Let the angle A be the distance of the sun from the mean place of the node, or the sun's mean motion from the node. Then if we take the angle B, whose tangent is to the tangent of the angle A as TH to TK, that is,
“Cor. Hence the angle FTA is the equation of the moon's nodes; and the sine of this angle, where it is greatest in the octants, is to the radius as KH to TK + TH. But the sine of this equation in any other place A is to the greatest sine as the sine of the sums of the angles FTN + ATN to the radius; that is, nearly as the sine of double the distance of the sun from the mean place of the node (namely, 2FTN) to the radius.
“SCHOLIUM.
“If the mean horary motion of the nodes in the quadratures be 16″ 16‴ 37$^{iv}$.42$^v$. that is, in a whole sidereal year, 39° 38′ 7″ 50‴, TH will be to TK in the subduplicate ratio of the number 9,0827646 to the number 10,0827646, that is, as 18,6524761 to 19,6524761. And, therefore, TH is to HK as 18,6524761 to 1; that is, as the motion of the sun in a sidereal year to the mean motion of the node 19° 18′ 1″ 23⅔‴.
"But if the mean motion of the moon's nodes in 20 Julian years is 386° 50′ 15″, as is collected from the observations made use of in the theory of the moon, the mean motion of the nodes in one sidereal year will be 19° 20′ 31″ 58‴. and TH will be to HK as 360° to 19° 20′ 31″ 58‴; that is, as 18,61214 to 1: and from hence the mean horary motion of the nodes in the quadratures will come out 16″ 18‴ 48$^{iv}$. And the greatest equation of the nodes in the octants will be 1° 29′ 57″.“
PROPOSITION XXXIV. PROBLEM XV.
    To find the horary variation of the inclination, of the moon's orbit to the plane of the ecliptic.
an hour, since the angle GTg (by Prop. XXX) is to the angle 33″ 10‴ 33$^{iv}$. as IT $\scriptstyle \times$ PG $\scriptstyle \times$ AZ to AT³, the angle GPg (or the horary variation of the inclination) will be to the angle 33″ 10‴ 33$^{iv}$. as IT $\scriptstyle \times$ AZ $\scriptstyle \times$ TG $\scriptstyle \times$ $\scriptstyle {\frac {Pp}{PG}}$ to AT³. Q.E.I.
And thus it would be if the moon was uniformly revolved in a circular orbit. But if the orbit is elliptical, the mean motion of the nodes will be diminished in proportion of the lesser axis to the greater, as we have shewn above; and the variation of the inclination will be also diminished in the same proportion.
Cor. 1. Upon Nn erect the perpendicular TF, and let pM be the horary motion of the moon in the plane of the ecliptic; upon QT let fall the perpendiculars pK, Mk, and produce them till they meet TF in H and h; then IT will be to AT as Kk to Mp; and TG to Hp as TZ to AT; and, therefore, IT $\scriptstyle \times$ TG will be equal to $\scriptstyle {\frac {Kk\times Hp\times TZ}{Mp}}$, that is, equal to the area HpMh multiplied into the ratio $\scriptstyle {\frac {TZ}{Mp}}$: and therefore the horary variation of the inclination will be to 33″ 10‴ 33$^{iv}$. as the area HpMh multiplied into $\scriptstyle AZ\times {\frac {TZ}{Mp}}\times {\frac {Pp}{PG}}$ to AT³.
Cor. 2. And, therefore, if the earth and nodes were after every hour drawn back from their new and instantly restored to their old places, so as their situation might continue given for a whole periodic month together, the whole variation of the inclination during that month would be to 33″ ​10‴ 33$^{iv}$. as the aggregate of all the areas HpMh, generated in the time of one revolution of the point p (with due regard in summing to their proper signs + -), multiplied into AZ $\scriptstyle \times$ TZ $\scriptstyle \times {\frac {Pp}{PG}}$ to Mp $\scriptstyle \times$ AT³; that is, as the whole circle QAqa multiplied into AZ $\scriptstyle \times$ TZ $\scriptstyle \times {\frac {Pp}{PG}}$ to Mp $\scriptstyle \times$ AT³, that is, as the circumference QAqa multiplied into AZ $\scriptstyle \times$ TZ $\scriptstyle \times {\frac {Pp}{PG}}$ to 2Mp $\scriptstyle \times$ AT².
Cor. 3. And, therefore, in a given position of the nodes, the mean horary variation, from which, if uniformly continued through the whole month, that menstrual variation might be generated, is to 33″ 10‴ 33$^{iv}$. as AZ $\scriptstyle \times$ TZ $\scriptstyle \times {\frac {Pp}{PG}}$ to 2AT², or as Pp $\scriptstyle \times {\frac {AZ\times TZ}{{\frac {1}{2}}AT}}$ to PG $\scriptstyle \times$ 4AT; that is (because Pp is to PG as the sine of the aforesaid inclination to the radius, and $\scriptstyle {\frac {AZ\times TZ}{{\frac {1}{2}}AT}}$ to 4AT as the sine of double the angle ATn to four times the radius), as the sine of the same inclination multiplied into the sine of double the distance of the nodes from the sun to four times the square of the radius.
Cor. 4. Seeing the horary variation of the inclination, when the nodes are in the quadratures, is (by this Prop.) to the angle 33″ 10‴ 33$^{iv}$. as IT $\scriptstyle \times$ AZ $\scriptstyle \times$ TG $\scriptstyle \times {\frac {Pp}{PG}}$ to AT³, that is, as $\scriptstyle {\frac {IT\times TG}{{\frac {1}{2}}AT}}\times {\frac {Pp}{PG}}$ to 2AT, that is, as the sine of double the distance of the moon from the quadratures multiplied into $\scriptstyle {\frac {Pp}{PG}}$ to twice the radius, the sum of all the horary variations during the time that the moon, in this situation of the nodes, passes from the quadrature to the syzygy (that is, in the space of 177$^1$⁄$_6$ hours) will be to the sum of as many angles 33″ 10‴ 33$^{iv}$. or 5878″, as the sum of all the sines of double the distance of the moon from the quadratures multiplied into $\scriptstyle {\frac {Pp}{PG}}$ to the sum of as many diameters; that is, as the diameter multiplied into $\scriptstyle {\frac {Pp}{PG}}$ to the circumference; that is, if the inclination be 5° 1′, as 7 $\scriptstyle \times$ $^{874}$⁄$_{10000}$ to 22, or as 278 to 10000. And, therefore, the whole variation, composed out of the sum of all the horary variations in the aforesaid time, is 163″, or 2′ 43″.
PROPOSITION XXXV. PROBLEM XVI.
To a given time to find the inclination of the moon's orbit to the plant of the ecliptic.
Let AD be the sine of the greatest inclination, and AB the sine of the least. Bisect BD in C; and round the centre C, with the interval BC, describe the circle BGD. In AC take CE in the same proportion to EB
as EB to twice BA. And if to the time given we set off the angle AEG equal to double the distance of the nodes from the quadratures, and upon AD let fall the perpendicular GH, AH will be the sine of the inclination required.
For GE² is equal to GH² + HE² = BHD + HE² = HBD + HE² - BH² = HBD + BE² - 2BH $\scriptstyle \times$ BE = BE² + 2EC $\scriptstyle \times$ BH = 2EC $\scriptstyle \times$ AB + 2EC $\scriptstyle \times$ BH = 2EC $\scriptstyle \times$ AH; wherefore since 2EC is given, GE² will be as AH. Now let AEg represent double the distance of the nodes from the quadratures, in a given moment of time after, and the arc Gg, on account of the given angle GEg, will be as the distance GE. But Hh is to Gg as GH to GC, and, therefore, Hh is as the rectangle GH $\scriptstyle \times$ Gg, or GH $\scriptstyle \times$ GE, that is, as $\scriptstyle {\frac {GH}{GE}}\times$ GE², or $\scriptstyle {\frac {GH}{GE}}\times$ AH; that is, as AH and the sine of the angle AEG conjunctly. If, therefore, in any one case, AH be the sine of inclination, it will increase by the same increments as the sine of inclination doth, by Cor. 3 of the preceding Prop. and therefore will always continue equal to that sine. But when the point G falls upon either point B or D, AH is equal to this sine, and therefore remains always equal thereto. Q.E.D.
In this demonstration I have supposed that the angle BEG, representing double the distance of the nodes from the quadratures, increaseth uniformly; for I cannot descend to every minute circumstance of inequality. Now suppose that BEG is a right angle, and that Gg is in this case the horary increment of double the distance of the nodes from the sun; then, by Cor. 3 of the last Prop. the horary variation of the inclination in the same case will be to 33″ 10‴ 33$^{iv}$. as the rectangle of AH, the sine of the inclination, into the sine of the right angle BEG, double the distance of the nodes from the sun, to four times the square of the radius; that is, as AH, ​the sine of the mean inclination, to four times the radius; that is, seeing the mean inclination is about 5° 8½, as its sine 896 to 40000, the quadruple of the radius, or as 224 to 10000. But the whole variation corresponding to BD, the difference of the sines, is to this horary variation as the diameter BD to the arc Gg, that is, conjunctly as the diameter BD to the semi-circumference BGD, and as the time of 2079$^7$⁄$_{10}$ hours, in which the node proceeds from the quadratures to the syzygies, to one hour, that is as 7 to 11, and 2079$^7$⁄$_{10}$ to 1. Wherefore, compounding all these proportions, we shall have the whole variation BD to 33″ 10‴ 33$^{iv}$. as 224 $\scriptstyle \times$ 7 $\scriptstyle \times$ 2079$^7$⁄$_{10}$ to 110000, that is, as 29645 to 1000; and from thence that variation BD will come out 16′ 23½″.
And this is the greatest variation of the inclination, abstracting from the situation of the moon in its orbit; for if the nodes are in the syzygies, the inclination suffers no change from the various positions of the moon. But if the nodes are in the quadratures, the inclination is less when the moon is in the syzygies than when it is in the quadratures by a difference of 2′ 43″, as we shewed in Cor. 4 of the preceding Prop.; and the whole mean variation BD, diminished by 1′ 21½″, the half of this excess, becomes 15′ 2", when the moon is in the quadratures; and increased by the same, becomes 17′ 45″ when the moon is in the syzygies. If, therefore, the moon be in the syzygies, the whole variation in the passage of the nodes from the quadratures to the syzygies will be 17′ 45″; and, therefore, if the inclination be 5° 17′ 20″, when the nodes are in the syzygies, it will be 4° 59′ 35″ when the nodes are in the quadratures and the moon in the syzygies. The truth of all which is confirmed by observations.
Now if the inclination of the orbit should be required when the moon is in the syzygies, and the nodes any where between them and the quadratures, let AB be to AD as the sine of 4° 59′ 35″ to the sine of 5° 17′ 20″, and take the angle AEG equal to double the distance of the nodes from the quadratures; and AH will be the sine of the inclination desired. To this inclination of the orbit the inclination of the same is equal, when the moon is 90° distant from the nodes. In other situations of the moon, this menstrual inequality, to which the variation of the inclination is obnoxious in the calculus of the moon's latitude, is balanced, and in a manner took off, by the menstrual inequality of the motion of the nodes (as we said before), and therefore may be neglected in the computation of the said latitude.
SCHOLIUM.
By these computations of the lunar motions I was willing to shew that by the theory of gravity the motions of the moon could be calculated from their physical causes. By the same theory I moreover found that the annual equation of the mean motion of the moon arises from the various ​dilatation which the orbit of the moon suffers from the action of the sun according to Cor. 6, Prop. LXVI, Book 1. The force of this action is greater in the perigeon sun, and dilates the moon's orbit; in the apogeon sun it is less, and permits the orbit to be again contracted. The moon moves slower in the dilated and faster in the contracted orbit; and the annual equation, by which this inequality is regulated, vanishes in the apogee and perigee of the sun. In the mean distance of the sun from the earth it arises to about 11′ 50"; in other distances of the sun it is proportional to the equation of the sun's centre, and is added to the mean motion of the moon, while the earth is passing from its aphelion to its perihelion, and subducted while the earth is in the opposite semi-circle. Taking for the radius of the orbis magnus 1000, and 16$^7$⁄$_8$ for the earth's eccentricity, this equation, when of the greatest magnitude, by the theory of gravity comes out 11′ 49". But the eccentricity of the earth seems to be something greater, and with the eccentricity this equation will be augmented in the same proportion. Suppose the eccentricity 16$^{11}$⁄$_{12}$, and the greatest equation will be 11′ 51″.
Farther; I found that the apogee and nodes of the moon move faster in the perihelion of the earth, where the force of the sun's action is greater, than in the aphelion thereof, and that in the reciprocal triplicate proportion of the earth's distance from the sun; and hence arise annual equations of those motions proportional to the equation of the sun's centre. Now the motion of the sun is in the reciprocal duplicate proportion of the earth's distance from the sun; and the greatest equation of the centre which this inequality generates is 1° 56′ 20″, corresponding to the abovementioned eccentricity of the sun, 16$^{11}$⁄$_{12}$. But if the motion of the sun had been in the reciprocal triplicate proportion of the distance, this inequality would have generated the greatest equation 2° 54′ 30″; and therefore the greatest equations which the inequalities of the motions of the moon's apogee and nodes do generate are to 2° 54′ 30″ as the mean diurnal motion of the moon's apogee and the mean diurnal motion of its nodes are to the mean diurnal motion of the sun. Whence the greatest equation of the mean motion of the apogee comes out 19′ 43", and the greatest equation of the mean motion of the nodes 9′ 24″. The former equation is added, and the latter subducted, while the earth is passing from its perihelion to its aphelion, and contrariwise when the earth is in the opposite semi-circle.
By the theory of gravity I likewise found that the action of the sun upon the moon is something greater when the transverse diameter of the moon's orbit passeth through the sun than when the same is perpendicular upon the line which joins the earth and the sun; and therefore the moon's orbit is something larger in the former than in the latter case. And hence arises another equation of the moon's mean motion, depending ​upon the situation of the moon's apogee in respect of the sun, which is in its greatest quantity when the moon's apogee is in the octants of the sun, and vanishes when the apogee arrives at the quadratures or syzygies; and it is added to the mean motion while the moon's apogee is passing from the quadrature of the sun to the syzygy, and subducted while the apogee is passing from the syzygy to the quadrature. This equation, which I shall call the semi-annual, when greatest in the octants of the apogee, arises to about 3′ 45″, so far as I could collect from the phænomena: and this is its quantity in the mean distance of the sun from the earth. But it is increased and diminished in the reciprocal triplicate proportion of the sun's distance, and therefore is nearly 3′ 34″ when that distance is greatest, and 3′ 56″ when least. But when the moon's apogee is without the octants, it becomes less, and is to its greatest quantity as the sine of double the distance of the moon's apogee from the nearest syzygy or quadrature to the radius.
By the same theory of gravity, the action of the sun upon the moon is something greater when the line of the moon's nodes passes through the sun than when it is at right angles with the line which joins the sun and the earth; and hence arises another equation of the moon's mean motion, which I shall call the second semi-annual; and this is greatest when the nodes are in the octants of the sun, and vanishes when they are in the syzygies or quadratures; and in other positions of the nodes is proportional to the sine of double the distance of either node from the nearest syzygy or quadrature. And it is added to the mean motion of the moon, if the sun is in antecedentia, to the node which is nearest to him, and subducted if in consequentia; and in the octants, where it is of the greatest magnitude, it arises to 47″ in the mean distance of the sun from the earth, as I find from the theory of gravity. In other distances of the sun, this equation, greatest in the octants of the nodes, is reciprocally as the cube of the sun's distance from the earth; and therefore in the sun's perigee it comes to about 49″, and in its apogee to about 45″.
By the same theory of gravity, the moon's apogee goes forward at the greatest rate when it is either in conjunction with or in opposition to the sun, but in its quadratures with the sun it goes backward; and the eccentricity comes, in the former case, to its greatest quantity; in the latter to its least, by Cor. 7, 8, and 9, Prop. LXVI, Book 1. And those inequalities, by the Corollaries we have named, are very great, and generate the principal which I call the semi-annual equation of the apogee; and this semi- annual equation in its greatest quantity comes to about 12° 18′, as nearly as I could collect from the phænomena. Our countryman, Horrox, was the first who advanced the theory of the moon's moving in an ellipsis about the earth placed in its lower focus. Dr. Halley improved the notion, by putting the centre of the ellipsis in an epicycle whose centre ​is uniformly revolved about the earth; and from the motion in this epicycle the mentioned inequalities in the progress and regress of the apogee, and in the quantity of eccentricity, do arise. Suppose the mean distance of the moon from the earth to be divided into 100000 parts, and let T represent the earth, and TC the moon's mean eccentricity of 5505 such parts. Produce TC to B, so as CB may be the sine of the greatest semi-annual equation 12° 18′ to the radius TC; and the circle BDA described about the centre C, with the interval CB, will be the epicycle spoken of, in which the centre of the moon's orbit is placed, and revolved according to the order of the letters BDA. Set off the angle BCD equal to twice the annual argument, or twice the distance of the sun's true place from the place of the moon's apogee once equated, and CTD will be the semi-annual equation of the moon's apogee, and TD the eccentricity of its orbit, tending to the place of the apogee now twice equated. But, having the moon's mean motion, the place of its apogee, and its eccentricity, as well as the longer axis of its orbit 200000, from these data the true place of the moon in its orbit, together with its distance from the earth, may be determined by the methods commonly known.
In the perihelion of the earth, where the force of the sun is greatest, the centre of the moon's orbit moves faster about the centre C than in the aphelion, and that in the reciprocal triplicate proportion of the sun's distance from the earth. But, because the equation of the sun's centre is included in the annual argument, the centre of the moon's orbit moves faster in its epicycle BDA, in the reciprocal duplicate proportion of the sun's distance from the earth. Therefore, that it may move yet faster in the reciprocal simple proportion of the distance, suppose that from D, the centre of the orbit, a right line DE is drawn, tending towards the moon's apogee once equated, that is, parallel to TC; and set off the angle EDF equal to the excess of the aforesaid annual argument above the distance of the moon's apogee from the sun's perigee in consequentia; or, which comes to the same thing, take the angle CDF equal to the complement of the sun's true anomaly to 360°; and let DF be to DC as twice the eccentricity of the orbis magnus to the sun's mean distance from the earth, and the sun's mean diurnal motion from the moon's apogee to the sun's mean diurnal motion from its own apogee conjunctly, that is, as 33$^7$⁄$_8$ to 1000, and 52′ 27″ 16‴ to 59′ 8″ 10‴ conjunctly, or as 3 to 100; and imagine the centre of the moon's orbit placed in the point F to be revolved in an epicycle whose centre is D; and radius DF, while the point D moves in the circumference of the circle DABD: for by this means the centre of ​the moon's orbit comes to describe a certain curve line about the centre C, with a velocity which will be almost reciprocally as the cube of the sun's distance from the earth, as it ought to be.
The calculus of this motion is difficult, but may be rendered more easy by the following approximation. Assuming, as above, the moon's mean distance from the earth of 100000 parts, and the eccentricity TC of 5505 such parts, the line CB or CD will be found 1172¾, and DF 35$^1$⁄$_5$ of those parts; and this line DF at the distance TC subtends the angle at the earth, which the removal of the centre of the orbit from the place D to the place F generates in the motion of this centre; and double this line DF in a parallel position, at the distance of the upper focus of the moon's orbit from the earth, subtends at the earth the same angle as DF did before, which that removal generates in the motion of this upper focus; but at the distance of the moon from the earth this double line 2DF at the upper focus, in a parallel position to the first line DF, subtends an angle at the moon, which the said removal generates in the motion of the moon, which angle may be therefore called the second equation of the moon's centre; and this equation, in the mean distance of the moon from the earth, is nearly as the sine of the angle which that line DF contains with the line drawn from the point F to the moon, and when in its greatest quantity amounts to 2′ 25″. But the angle which the line DF contains with the line drawn from the point F to the moon is found either by subtracting the angle EDF from the mean anomaly of the moon, or by adding the distance of the moon from the sun to the distance of the moon's apogee from the apogee of the sun; and as the radius to the sine of the angle thus found, so is 2′ 25″ to the second equation of the centre: to be added, if the forementioned sum be less than a semi-circle; to be subducted, if greater. And from the moon's place in its orbit thus corrected, its longitude may be found in the syzygies of the luminaries.
The atmosphere of the earth to the height of 35 or 40 miles refracts the sun's light. This refraction scatters and spreads the light over the earth's shadow; and the dissipated light near the limits of the shadow dilates the shadow. Upon which account, to the diameter of the shadow, as it comes out by the parallax, I add 1 or 1⅓ minute in lunar eclipses.
But the theory of the moon ought to be examined and proved from the phenomena, first in the syzygies, then in the quadratures, and last of all in the octants; and whoever pleases to undertake the work will find it not amiss to assume the following mean motions of the sun and moon at the Royal Observatory of Greenwich, to the last day of December at noon, anno 1700, O.S. viz. The mean motion of the sun ♑ 20° 43′ 40″, and of its apogee ♋ 7° 44′ 30″; the mean motion of the moon ♒ 15° 21′ 00″; of its apogee, ♊ 8° 20′ 00″; and of its ascending node ♌ 27° 24′ 20″; and the difference of meridians betwixt the Observatory at Greenwich and ​the Royal Observatory at Paris, 0$^h$.9′ 20″: but the mean motion of the moon and of its apogee are not yet obtained with sufficient accuracy.