A Treatise on the Precession of the Equinoxes, and in General on the Motion of the Nodes, and the Alteration of the Inclination of the Orbit of a Planet to the Ecliptic. Inscribed to the Gentlemen of the Royal Society, by M. De St. Jaques Silvabelle Translated from the French M. S. by J. Bevis, M. D.

LVIII. A treatise on the precession of the equinoxes, and in general on the motion of the nodes, and the alteration of the inclination of the orbit of a planet to the ecliptic. Inscribed to the gentlemen of the Royal Society, by M. De St. Jaques Silvabelle. Translated from the French M. S. by J. Bevis, M. D. INTRODUCTION. Read March 12, 1752. If the earth were perfectly spherical, the action of the sun on all the parts which compose it, would not produce any effect to make it turn round its centre; because the moment, which would be produced on one side, would be always counterbalanced by an equal moment on the opposite side of the centre. It would be the same, if the earth were a spheroid flatted at the poles, and the sun was always in the equator, or in the ninetieth degree of declination: But in every other degree of declination its action on the excess of matter about the equator has a tendency to make the equator approach towards the sun's place, or to diminish the angle of the sun's declination, by making the earth's axis to turn round its centre in the plane of the circle of the sun's declination. The earth has then, at every instant, two motions of rotation; one about the axe of the equator, called also the earth's axe; and this is the diurnal motion, which is uniform; the other motion of rotation is performed about the axe of the circle of the sun's declination, which is a diameter of the equator; and this motion is produced by the action of the sun on the redundant matter about the equator, and is continually accelerated, from the continual application of the solar action producing it. The point $E$, fig. 1. n° 2. which is the intersection of the circumference of the equator and the circumference of the circle of declination, has two motions, whose directions are perpendicular to each other. Let $Ee$ be the space, which it runs through in an instant $dt$, in the circumference of the equator, by the uniform diurnal motion, and let $E\epsilon$ be the space it runs through in the same instant, in the circumference of the circle of declination, by an accelerated motion, as has been explained. The point $E$, in virtue of these two motions $Ee$ and $E\epsilon$, will not circulate either in the circumference $EeQe'QE$ of the equator, or in the circumference $E\epsilon PE'PE$ of the circle of declination, but forming the rectangular parallelogram $Eee'\epsilon$, the diagonal $Ee'$ will be the elementary arc of the circumference $EeqE'$, in which the point $E$ will circulate, and the angle $ee'E$ will be equal to the angle $QCq$, and equal to the angle $P'Cp'$ which the pole $P'$ runs through in an instant in the circumference $P'p'QgPQ'P'$ whose plane is perpendicular to the plane of the circle of declination; and when the lines $Ee$, $E\epsilon$ are known at every instant, $P'p'$ will also be known, since the angle $ee'E = \text{to the angle } QCq = \text{to the angle } P'Cp'$. The instantaneous motion of the pole, which is $P'p'$, or $Pp$, fig. 2. n° 1. & n° 2*. may be resolved into two, $PR$ and $PM$, perpendicular to each other, and both to the earth's axe. The former causes the pole $P$ to move parallel to the ecliptic $\gamma \varphi \omega$, and alters the place of the solstice $\Xi$, and consequently also that of the equinoctial points $\gamma$ and $\omega$; the latter, which is according to $PM$, alters the inclination of the earth's axe to the ecliptic. To have the motion of the pole parallel to the ecliptic, or, which is the same, the motion of the node $\gamma$, or the precession, in the same time that the sun passes from the equinox $\gamma$ to the solstice $\Xi$, take the integral of the lines $PR$, supposing $PR$ generally to express the instantaneous precession for any declination of the sun $S$. And to have the alteration of the inclination in the same time that the sun is passing from $\gamma$ to $\Xi$, take the sum, or the integral of the lines $PM$, supposing $PM$ generally to express the instantaneous alteration of the inclination of the earth's axe to the ecliptic for any given declination of the sun. The sum of the lines $PR$ is always the same, and has the same sign, or the same direction, during every quarter of the sun's revolution, whether he moves from $\gamma$ to $\Xi$, or from $\Xi$ to $\omega$, or from $\omega$ to $\varphi$, or from $\varphi$ to $\gamma$; so that the precession answering to any one quarter of the sun's revolution about the earth, or to three months, being known, that multiplied by 4 will be the annual precession; by 8 will give it for two years; by 16 for 4 years, &c. Likewise * Fig. 2. is explained at the beginning of Prob. 5. Likewise the sum of the lines $PM$ is ever the same for every quarter of the solar revolution; but it has alternatively a contrary sign; that is, a contrary direction. During the quarter from $\gamma$ to $\delta$, the alteration of the inclination of the earth's axe to the ecliptic is positive, and the angle of the inclination increases; but during the succeeding quarter, or from $\delta$ to $\epsilon$, the alteration of the inclination is negative, and the angle of the inclination diminishes: And as the diminution from $\delta$ to $\epsilon$ is equal to the augmentation from $\gamma$ to $\delta$, it follows, that at the end of the semirevolution the inclination of the earth's axe to the plane of the ecliptic will become again the same, having undergone an oscillation, which is completed in a semirevolution. It is the same, when the sun passes from $\epsilon$ to $\gamma$. The angle of the inclination increases from $\epsilon$ to $\varphi$, and decreases from $\varphi$ to $\gamma$, where it becomes again the same it was at $\epsilon$. And hence the inclination of the earth's axe to the ecliptic may be considered as constant, tho' subject to this oscillation, and indeed to several others, which will be presently explained, they being all regular, and performed in regular periods. The earth's inclination to the ecliptic being constant, and the motion of the pole which produces the precession, being always parallel to the plane of the ecliptic, the earth's pole moves in a parallel to the ecliptic, about 23 degrees and a half distant from the pole of the ecliptic, and the terrestrial axe describes a conic surface. To this motion of the terrestrial axe or pole is to be ascribed the apparent motion of the stars about the pole of the ecliptic. But But hitherto we have not considered, that to the precession, thus caused by the sun, we are to add that likewise produced by the moon; and it remains, that we examine into the motion of the earth's pole, caused by the action of the moon on the redundant matter about the earth's equator. All, that has been said concerning the sun, is alike applicable to the moon, which we may put in the place of the sun; the moon's orbit in the place of the ecliptic; and the time of the moon's revolution round the earth in the place of the revolution of the sun round the earth: And we shall find the motion of the earth's pole parallel to the lunar orbit, which is always the same at every quarter of the time of the revolution of the moon round the earth, and the oscillation of the earth's axe to the plane of the lunar orbit, which is completed, in each semirevolution of the moon round the earth. But whereas the plane of the lunar orbit, which is always inclined to the plane of the ecliptic in an angle of about 5 degrees, never continues in a constant position, like the plane of the ecliptic, so that its pole describes a small circle parallel to the ecliptic, at the distance of about 5 degrees from the pole thereof; it follows, that the precession, with respect to the lunar orbit, is not the same as with respect to the ecliptic; and that the motion of the pole parallel to the lunar orbit should be referred to the plane of the ecliptic: Which is done by resolving the motion of the pole, parallel to the plane of the lunar orbit, into two motions, the one parallel to the plane of the ecliptic, and the other perpendicular thereto, and in the plane of the solsticial colure. The former of these two motions gives the precession with respect to the ecliptic, and has its direction always the same way. The latter motion has two opposite directions, in the two semirevolutions of the pole of the lunar orbit round the pole of the ecliptic, and causes an oscillation of the terrestrial axe on the plane of the ecliptic, which is compleated in a revolution of the pole of the lunar orbit round the pole of the ecliptic. From all that has been said, it follows, that there are five distinct motions of the pole of the earth; namely, two of precession, which are parallel to the plane of the ecliptic, and three of oscillation on the plane of the ecliptic. The two of precession are caused, the one by the sun, the other by the moon. That, which is caused by the sun, is constantly the same at every quarter of the time of the revolution of the sun round the earth, that is, every three months: That which is caused by the moon, is constantly the same at every quarter of the time of the revolution of the moon round the earth; that is, about every seven days. Of the three motions of oscillation, one is caused by the sun, and is completed in the time of the semirevolution of the sun round the earth, taken from one equinox to the following one; that is, in six months. The other is caused by the moon; and each oscillation is completed in the space of a semirevolution of the moon round the earth; that is, in about 14 days. The third is caused likewise by the moon, and arises from the plane of her orbit being different from the plane of the ecliptic, and from the pole of the lunar orbit making its revolution about the pole of the ecliptic. ecliptic in about 18 years and two thirds. And this oscillation is completed in the time of the revolution of the pole of the lunar orbit about the pole of the ecliptic; that is, in about 18 years and two thirds. It will appear in the memoir, that there is a relation purely geometrical between the quantity of the nutation, during the time of the semirevolution of the pole of the lunar orbit, and the quantity of the precession, caused likewise by the moon in the same time. This relation is quite independent of the force of the moon, of the quantity of the earth's flatness, of the quantity of the terrestrial matter, and, in a word, of every thing of a physical nature that can enter into the problem. We are content to examine the motions of the pole of the earth produced by the sun and the moon. The same method, and the same formulæ, will give likewise the motions of the terrestrial pole arising from any other planet, as Saturn, Jupiter, &c. but these motions are too minute to merit attention. Whatever has been said of the action of the sun on the redundant matter about the earth's equator, is also applicable to his action on a simple ring placed at the equator, without adhering to the terrestrial globe; and the motion of the pole of such ring may be determined by the same method, and consequently the motion of its nodes on the plane of the ecliptic, and the alteration of the inclination of its axe to the same plane. And since these motions are the same, whether the ring be supposed entire, or a small portion of it only be considered, or a mere point thereof, the motions of the nodes, and the alteration of the inclination of a moon, or a satellite of a planet, may thereby thereby be known. And the formulae differ in nothing from those of the motion of the nodes of the earth's equator, and of the alteration of the obliquity of the earth's axe to the plane of the ecliptic, but in this; that the action of the sun on the ring to make it turn, is exerted entirely thereon; whereas in the problem of the precession this force must necessarily be distributed throughout the whole mass of the earth, on account of the adherence of the ring to the globe of the earth. DIVISION of the WORK. This memoir is divided into four sections. The 1st section treats of the motion of the pole of the terrestrial equator caused by the sun. The 2d section treats of the motion of the pole of the terrestrial equator caused by the moon. The 3d section treats of the motion of the pole of a ring, or of the orbit of a moon, caused by the sun. The 4th section contains the application of the formulae found in the other sections. SECT. I. Of the motion of the pole of the terrestrial equator caused by the action of the sun. PROBLEM I. Article 1. To find the moment, which results from the attraction of each particle of the earth towards the sun, in the inverse ratio of the square of the distance, to make the earth's axis turn upon its centre C. Let E F E' F E, fig. 1. n° 1. be the section of the earth by the plane of the circle of the sun's declination; tion; $P E P' E' P$ the circumference whose diameter is the earth's equator, $E E'$ or $P P'$. Let $E E'$ be perpendicular to $P P'$, and in the equator. Let $I I'$ be perpendicular to $C S$, and assume this diameter $II'$ for the lever to which all the moments are to be referred. The motion of any point, as $g$, towards the sun, is $\frac{S}{S g^2}$. By resolving this motion into two, one according to $g C$, the other parallel to $C S$; the motion according to $g C$ has no tendency to make the point $g$ turn round the centre $C$. The motion of the point $g$ parallel to $C S$ will be $\frac{S}{S g^2} \times \frac{S C}{S g}$ or $\frac{S \times S C}{S g^3}$; but the motion of the centre $C$ according to $C S$ is $\frac{S}{S C^3}$; therefore the relative motion of the point $g$, in regard to the centre $C$, is $\frac{S \times S C}{S g^3} - \frac{S}{S C}$ or $S \times S C \times \left(\frac{1}{S g^3} - \frac{1}{S C^3}\right)$. The moment of the point $g$ to turn about the centre $C$, will therefore be $S \times S C \times \left(\frac{1}{S g^3} - \frac{1}{S C^3}\right) \times g \times CI$, the moment being the product of the motion by the mass of the body, and by the arm of the lever. And this moment causes the point $I$ to approach towards the sun, when it is positive, or when $S C$ is greater than $S g$; and causes the point $I$ to recede from the sun when it is negative, or when $S g$ is greater than $S C$. It will be found, in like manner, that the moment of the point $g'$ to turn about the centre $C$, is $S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{SC^3} \right) \times g' \times CI$; or with regard to the point $I$ which is on the opposite side of the centre to the point $I'$, this moment, by changing the signs, will be, $S \times SC \times \left( \frac{1}{SC^3} - \frac{1}{Sg'^3} \right) \times g' \times CI = S \times SC \times \left( \frac{1}{SC^3} - \frac{1}{Sg'^3} \right) \times g \times CI$. Therefore the moment of the two equal points $g$ and $g'$ to cause the point $I$ to turn about the centre $C$, is $S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{SC^3} \right) \times g \times CI + S \times SC \times \left( \frac{1}{SC^3} - \frac{1}{Sg'^3} \right) \times g \times CI = g \times CI \times S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{Sg'^3} \right)$. If the points $\gamma$ and $\gamma'$ be taken equally distant from the points $D$ and $D'$, as the points $g$ and $g'$, it will likewise be found, that the moment of the equal points $\gamma$ and $\gamma'$ to cause the point $I$ to turn about the centre $C$, is $\gamma \times CI \times S \times SC \times \left( \frac{1}{S\gamma^3} - \frac{1}{S\gamma'^3} \right) = \gamma \times CI \times S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{Sg'^3} \right)$. Therefore the moment of all the four points together, $g$, $g'$, $\gamma$, $\gamma'$, to cause the point $I$ to turn about the centre $C$, is $(g - \gamma) \times CI \times S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{Sg'^3} \right)$, which which becomes = 0 when the point g can be taken = γ, as may be done throughout the whole extent of the circle PEP'E'P; but if the zone EPE'P'EF'E'FE be taken away from this circle, to have the true figure of the earth, we shall have \( g = gf \times gr \), and \( γ = -γf \times γg \), taking the points g and γ for the elements of the zone, and \( g - γ = gf \times γg - gr \), because \( γf = gf \). Let the lines CS and CP, or CE, be called s and a, respectively; the radius r, or unity; the sine of the angle SCP, V; its cosine, or the sine of the angle SCE, u; CK, x; gK, y; gm, z; γM, z'; gf or γf', du; PF, aa; we shall have gn = az; γv = az' (by the property of the ellipse); also calling gr, r; and γg, g; and regarding the little triangle grn as a right-lined one, and similar to the triangle gcm, we shall have \( r = \frac{z}{a} \times a z \), and likewise \( g = \frac{z'}{a} \times a z' \). By the proportion of sines to the sides of triangles, we shall have \( γG = γV: CN = xu \). Finally, if we consider the point S, or the sun, as at an infinite distance, we shall have \( Sg = SK = SC - CK = s - x \), and \( Sg' = SK' = SC + CK' = s + x \). Therefore \( \frac{1}{Sg^3} = \frac{1}{s^3 - 3s^2x + 3sx^2 - x^3} \) and \( \frac{1}{Sg'^3} = \frac{1}{s^3 + 3s^2x + 3sx^2 + x^3} \). Therefore \( \frac{1}{Sg^3} - \frac{1}{Sg'^3} = \frac{1}{6s^2x} \). \[ \frac{6s^2x + 2x^3}{(s-x)^3 \times (s+x)^3} \text{; and, rejecting the infinitely small magnitudes, } = \frac{6x}{s^4}. \] Therefore the moment wherewith the four points \( g, g', y, y' \), cause the point \( I \) to turn, will, by substituting in the place of the letters their analytical values, be \((gdu - rdu) \times S \times y \times \frac{6x}{s^3}\); or putting, instead of \( g \) and \( r \), the values above found, we shall have for the moment \(\left( \frac{a}{a} z'z' - \frac{a}{a} zz \right) \times du \times S \times y \times \frac{6x}{s^3} = \frac{6aS}{s^3a} \times y \times du \times (z' + z) \times (z' - z)\). But \( z' + z = \gamma M + gm = 2 \gamma G = 2yV \); and \( z' - z = \gamma M - gm = 2GM = 2CN = 2xu \). Therefore \((z' + z) \times (z' - z) = 4Vu \times y \); and the moment is \(\frac{6aS}{s^3} \times 4Vu \times x \times y^2du = \frac{6aS}{s^3} \times 4Vu \times xdx\). But in the circle \( DgI \) we have \( ydu = adx \), and the moment is \(\frac{6aS}{s^3} \times 4Vu \times x \times ydx = \frac{6aS}{s^3} \times 4Vu \times dx \sqrt{aa - xx}\). But the integral of \( x^2dx \sqrt{aaxx} \) is \(\frac{1}{4} a \int dx \sqrt{aa - xx} - \frac{x}{4} (aa - xx)^{\frac{3}{2}} \), and when \( x = a \), this integral is \(\frac{1}{4} a \int dx \sqrt{aa - xx} \), or, calling the ratio of the circumference to the diameter \( \pi \), we shall have \(\int dx \sqrt{aa - xx} \), or the area of the quadrant whereof \( a \) is the radius, \( = \frac{1}{2} a \pi \times \frac{1}{2} a = \frac{1}{4} a^2 \pi \). Therefore the integral of \(\frac{6aS}{s^3} \times 4Vu \times x \times dx \sqrt{aa}\) \[ \sqrt{aa-xx}, \text{or the sum of the moments of all the points which compose the elliptic zone } E P' E' P E F E F E \] is \( \frac{6aS}{s^3} \times \frac{\nu u \pi a^4}{4}. \) If the terrestrial spheroid be cut by any plane parallel to the plane \( P E P' E' P \) of the circle of declination, fig. 1. n° 2. the section will be an ellipse similar to the ellipse \( E F' E' F E \); and if the greater semiaxe \( c' L \) of this ellipse be called \( X \), it will be found, as has been already in the ellipse \( E F' F' F E \), that the sum of the moments of all the points of this ellipse to turn about its centre \( C \), is \( \frac{6aS}{s^3} \times \frac{\nu u \pi X^4}{4}. \) It has not been taken into consideration, that here the centre \( S \) of the sun is a little below the plane of the section, the line \( c' S \) making an insensible angle with the line \( C S \). Calling \( C c' Y \); the moment of all the sections parallel to \( E F' E' F E \) to turn about the axe \( C c' \) will be the integral of \( \frac{3aS}{s^3} \times \frac{\nu u \pi X^4}{2} \times d Y \), or the integral of \( \frac{3aS}{s^3} \times \frac{\nu u \pi}{2} \times (aa YY)^2 \times d Y = \frac{3aS}{s^3} \times \frac{\nu u \pi}{2} \times (a^4 d Y - 2a^2 Y^2 d Y + Y^4 d Y) \), and this integral is \( \frac{3aS}{s^3} \times \frac{\nu u \pi}{2} \times (a^4 Y - \frac{2}{3} a^2 Y^3 + \frac{1}{5} Y^5) \); and when \( Y = a \), the integral is \( \frac{3aS}{s^3} \times \frac{\nu u \pi}{2} \times \frac{8}{15} a^5 \). This is the sum of the moments of all the points of the hemispheroid of the earth formed by the section \( E P' E' P E \) of the circle of declination, and the sum of the moments of all the points of the whole spheroid. roid to turn about the axe \( QQ' \) is \( \frac{3aS}{s^3} \times V_u \pi \times \frac{8}{15} a^5 \). Which was to be found. **Corollary.** 2. If it were required to find the sum of the moments of all the points of a single crown \( ELE'E'Q'QE \), fig. 1. no 2. placed at the equator, and detached from the earth, to turn about the axe \( QQ' \) of the circle of the sun's declination, the thickness of the crown being \( a \): The moment of the point \( L \) to turn about \( c' \) is the same as that of the point \( y \) to turn about \( C \); as is manifest from the resolution of the motion according to \( SL \) into two, \( Ly \) and \( yS* \); and from the great distance of the sun \( yS \) will not differ from \( LS \); the triangle \( LyS \) being right-angled at \( y \). By the preceding problem, if \( c'L \), or \( CY \), be called \( X \); and \( Cc' \), \( Y \); and all the other denominations of that problem be retained; we shall have \( yx = Xu \), and \( Cx = XV \). And the moment of the point \( y \) is \( y \times yx \times S \times SC \times \left( \frac{1}{S_y^3} - \frac{1}{SC^3} \right) = y \times Xu \times S \times \frac{3XV}{s^3} \). And putting instead of \( y \), the element of the crown, which is \( a \times du \), calling, in this case, \( du \) the element of the circumference \( ELE'E' \), the moment of an element of the crown will be \( aa \times \frac{3S}{s^3} \times V_u \times \) --- * The lines \( SL \) and \( yS \) are not drawn in the fig. to avoid confusion. \( XXdu \) But in the circle $EQQ'E'$ we have, $Xdu = adY$; therefore the moment of an element of the crown is $aa \times \frac{3S}{s^3} \times Vu \times a \times dY = \frac{3S}{s^3} \times aa^2Vu \times dY \sqrt{aa - YY}$; and the sum of the moments of the quadrant $ELQ$ of the crown is $\frac{3S}{s^3} \times aa^2Vu \times \frac{1}{4}a\pi \times \frac{1}{4}a = \frac{3S}{s^3} \times aVu \times \frac{1}{4}\pi a^4$. And every other quadrant of the crown having an equal motion, the sum of the moments of all the points of the crown, to turn, in virtue of the solar action, about the axe $QQ'$, is $\frac{3S}{s^3} \times aVu \times \pi a^4$. **Problem II.** 3. To find the sum of the moments of all the points of the terrestrial spheroid turning about the axe $QQ'$, the motion of the point $E$ being given, and $= \mu$. We must first seek the sum of the moments of all the points of the ellipse $EE'FE$, fig. 1. No. 2, turning about its centre $C$; and to have the sum of the moments of all the points of this ellipse, we shall first find the sum of the moments of all the points of the circle $EP'E'PE$, and then we shall subduct therefrom the sum of the moments of the whole elliptic zone $EP'E'PEF'E'FE$. The motion of any point of the circle $EP'E'PE$, placed at the distance $x$ from the centre $C$, will be $\frac{x}{a}\mu$; the motions being here in the same ratio as the distances stances from the centre, since all the points complete their revolution in the same time; and the moment of all the points which compose the circumference $2 \pi x$ will be $2 \pi x \times \frac{x}{a} \mu \times x$; the moment being produced from the mass by the motion, and by the arm of the lever. Therefore the motion of all the concentric circumferences which compose the circle, is $\int \frac{2 \pi \mu}{a} \times x^3 dx$ $= \frac{\pi \mu}{a} \times x^4 = \frac{\pi \mu}{a} \times a^4$, when $x = a$, as for the circle $E P' E' P E$. To have the sum of the moments of all the points of the elliptical zone $E P' E' P E F' E' F E$, all the points of the zone may be considered as placed at the distance $a$, from the centre $C$, and having, consequently, the same motion as the point $E$, because the greatest thickness $P F$ of the zone is very small. The sum of the moments of the zone will be equal to the quantity of matter multiplied by the motion $\mu$, and by the arm of the lever $a$, or $2 a \pi \times \frac{1}{2} a \mu \times \mu \times a = \pi \mu a a^3$. Therefore the sum of the moments of all the points of the ellipse $E F' E' F E$ turning about its centre $C$, the motion of the point $E$ being $\mu$, will be $\frac{\pi \mu}{a} \times a^4 — \pi \mu a a^3$. It will likewise be found generally, that the sum of the moments of a similar ellipse, of which $C L$, or $X$, is half the greater axe, and which turns about its centre $C'$, is $\frac{\pi \mu}{a} \times X^4 — \frac{\pi \mu a}{a} \times X^3$. Therefore Therefore the sum of all the similar ellipses which compose the terrestrial hemispheroid, is \[ \int \frac{1 - 2a}{2a} \times \pi \mu \times X^4 \, dY = \frac{1 - 2a}{2a} \times \pi \mu \times \int (aa - YY)^4 \times dY, \] which, as has been seen at the end of Prob. I. when, after the integration \( Y \) has been made \( = a \), will be \[ \frac{1 - 2a}{2a} \pi \mu \times \frac{8}{15} a^5, \] and multiplying by 2, the sum of all the points of the terrestrial spheroid will be \[ (1 - 2a) \times \pi \mu \times \frac{8}{15} a^4. \] Which was to be found. **Corollary.** 4. If you would find the sum of the moments of all the points of the ring \( E Q E' Q' B \), fig. 1. n° 2. turning about the axe \( Q Q' \): The moment of an element \( L \) of the crown placed at the distance \( X \) from the axe of rotation \( Q Q' \), is \[ a a d u \times \frac{\mu X}{a} \times X, \] which is the product of the quantity of matter by the motion, and by the arm of the leaver. But in the circle \( E Q E' E \) we have, \( du \frac{adY}{X} \); therefore the moment of an element of the crown is \[ a a \mu \times X \, dY, \] whose integral, which is \( a a \mu \times a a \pi \), gives the sum of the moments of all the points of the crown. Which was to be found. Problem III. 5. Having given, the moment wherewith the earth turns about the axe QQ' of the circle of the sun's declination, To find the motion of the pole P, or, which is the same, of the point E in the plane of the circle of the sun's declination. By Prob. I. the moment wherewith the earth turns about the axe CQ, by the action of the sun, is $\frac{3S}{s^3} \times a \pi v u \times \frac{8}{15} a^5$. By Prob. II. the moment wherewith the earth turns about the same axe CQ, the velocity of the point P, or of the point E, being $\mu$, is $(1-2a) \times \pi \mu \times \frac{8}{15} a^4$. We shall have then $\frac{3S}{s^3} a \pi v u \times \frac{8}{15} a^5 = (1-2a) \times \pi \mu \times \frac{8}{15} a^4$; whence we get $\mu = \frac{3Saa\pi u}{s^3 \times (1-2a)}$. And this is the motion of the pole P, or of the point E, about the centre C. Which was to be found. Corollary. 6. In like manner, the moment wherewith the crown EE'E'E' turns about the axe QQ' being given, the motion of the point E may be known. For, By the Coroll. of Prob. I. the moment of this crown, turning about the axe CQ by the action of the sun, is $\frac{3S}{s^3} \times a \pi v u \times a^4$. By By the Coroll. of Prob. II. the moment of this crown is also $a \mu \times a^2 \pi$. We have then $\frac{3S}{s^3} \times a \pi v u \times a^4 = a \mu \times a^2 \pi$; whence we get $\mu = \frac{3S}{s^3} \times a v u$. Which was to be found. Remark. It is proper to observe, that the motion of the point $E$ of the crown is the same, whether the crown be entire, or there be no more of it but the point $E$. For, by Prob. I. the motion of any point $g$, parallel to $CS$, fig. 1. n° 2. and respectively to the centre $C$, is $S \times SC \times \left( \frac{1}{Sg^3} - \frac{1}{SC^3} \right)$, which is $= S \times \frac{3x}{S^3}$; and when the point $g$ becomes the point $E$, we have $x = av$, $v$ being the cosine of the angle $DC E$. Therefore the respective motion of the point $E$ parallel to the motion $CS$ of the centre $C$, is $S \times \frac{3av}{S^3}$; and resolving this motion into two, the one according to $EC$, and the other perpendicular to $EC$, this latter will be $= S \times \frac{3av}{s^3} \times \frac{au}{a} = \frac{3S}{s^3} avu$, being the same as that which was found for the point $E$ when it is united to the crown. This shews that the motion of the other points of the crown produce no alteration in the motion of the point $E$, which is owing to this; that all the points of the crown have the same angular motion about the axe $CQ$. Problem IV. 7. To find at every instant the variation of the place of the pole of the earth. The earth has two motions of rotation, the one about the axe of the equator, which is likewise the earth's axis; and the other about the axe $CQ$ of the circle of the sun's declination; this motion is caused by the sun's action on the redundant matter about the equator. The point $E$, fig. 1. n° 2., which is the intersection of the circumference $EQE'Q'E$ of the equator and the circle $PEP'E'P$ of the sun's declination, will therefore have at every instant two motions, whose directions are perpendicular to each other. The former of these two motions, which is the diurnal motion, is uniform; and if we call it $m$, the space run through in an instant, $dt$, is $mdt$; the motion being always equal to the space divided by the time, or the space equal to the product of the time. Also let $Ee$ be that space. The second motion of the point $E$ which is performed in the circumference $EP'E'P'E$ of the circle of the sun's declination, and arises from the action of the sun, is continually accelerated, from the continual application of the sun's action; and if we call the initial motion $\mu$, at the first term of the instant, $dt$, or the increment of the motion, the motion at the end of the instant $dt$, is $\mu dt$; and the space gone thro' uniformly by the motion during the instant $dt$, is $\mu dt \times dt$, or $\mu dt^2$. Let $Ee$ be that space. The point $E$, in consequence of these two motions together, $Ee$ and $Ee'$, circulates neither in the circumference $EQE'Q'E$ of the equator, nor in the circumference $EP'EP'E$ of the sun's declination. But if we form the rectangle $Ee'e'\epsilon E$, the diagonal $Ee'$ will be the elementary arc of the circumference $Ee'qE'E$ wherein the point $E$ will circulate; and the angle $eEe'$, equal to the angle $Qcq$, equal to the angle $P'Cp'$, will be the angle whereby the pole $P'$ is elevated, and the pole $P$ depressed, below the circle of declination, by moving in the circumference $P'p'QqPP'P'$, whose plane is perpendicular to the plane of delineation; and the point $p'$ is the true place of the pole at the end of the instant $dt$, and the small arc $P'p'$ expresses the instantaneous variation of the place of the earth's pole. Which was to be found. **Corollary I.** 8. The similar sectors $eEe'$, $QCq$, or $P'Cp'$, whose three sides of the one are each parallel to the three sides of the other, give this proportion, $Ee$; $ee'$ or $Ee :: CP': P'p' = \frac{Ee}{Ee'} \times CP' =$ (by the preceding Prob.) $\frac{\mu dt^2}{mdt} \times CP' = \frac{\mu dt}{m} \times a$. But by Problem III. $\mu = \frac{3Saa\upsilon u}{s^3(1-2a)}$; therefore $P'p' = \frac{3Saa\upsilon u}{s^3 \times m \times (1-2a)} \times adt = A\upsilon uadt$, making $A = \frac{3Saa}{s^3 \times m \times (1-2a)}$. Corollary II. 9. If, instead of the terrestrial spheroid, we supposed only a single ring, it would be found in like manner, that the pole $P'$ of such ring would shift its place during the instant $dt$, by running through the arc $P'p' = \frac{E_i}{E_e} \times CP' = \frac{\mu d t}{m} \times a$. But then we should take the value of $\mu$ as in Coroll. of Prob. III. which is $\frac{3S}{s^3} \times a \nu u$. Therefore we shall have $P'p' = \frac{\mu d t}{m} \times a = \frac{3S}{s^3 \times m} \times \nu u adt$. To apply, then, the formula of the preced. Coroll., it will suffice to make $\frac{a}{1 - 2a}$ in that Coroll. = 1. for then the formula of Coroll. I. becomes the same with that just now found in this Coroll. This will be of use hereafter for applying the formulae of the motion of the nodes of the terrestrial equator to the motion of the nodes of a moon or satellite of a planet. Corollary III. 10. It should be well observed in the preceding Coroll. that the direction $P'p'$ of the pole $P'$ is ever parallel to the direction $Ee$ of the point $E$ of the equator, since they are both perpendicular to the plane of the circle of the sun's declination, and the direction $P'p'$ goes on the same side as the direction $Ee$ of the point $E$, which is placed on the same side of the the pole $P'$, with respect to the line $S C D'$, and the direction of the opposite pole $P$, goes on the same side as the direction of the point $E$ of the equator which is placed on the same side of the pole $P$ with respect to the line $S C D'$. This should be carefully attended to in regard to the following Problem. **Problem V.** 11. To determine at every instant the motion of the pole of the earth with respect to the ecliptic. $PP'$, fig. 2. n° 1. & n° 2*. is the earth's axe; $\gamma \approx \varphi$ represents the ecliptic; $S E p E S$ is the tangent plane to the pole $P$, and parallel to the equator; the line $D \approx D'$ is the intersection of this plane with that of the ecliptic, and is tangential at the common point $S$ to the circles $S E p E S$ and $\gamma \approx \varphi$. The angle $P C S$ is the angle of inclination of the earth's axe to the plane of the ecliptic, being likewise the solsticial angle. The earth, by its diurnal motion, turns about the axe $P' P$ according to the order of $S E p E S$; and the sun, by the annual motion in the ecliptic, moves according to the order of $\gamma \approx \varphi$. Let $S$ be any place of the sun in the ecliptic. By Art. 10. the motion of the pole $P$ making the acute angle $P C S$, will be according to $P p$ perpendicular to the plane $P C S D P$, and consequently to the line $P D$. *Note, That the fig. n° 2. is only to represent the circle $S E p S$, which could not be represented but by projection in fig. n° 1.* Forming, therefore, the rectangle $P M \rho R$, the motion of the pole according to $P \rho$ will be resolved into two, according to $PR$ and $PM$. The motion of the pole according to $PR$ is parallel to the plane of the ecliptic, and to the line of the nodes $\gamma C \omega$. This motion tends to move the pole $P$ about the axe of the ecliptic, and in a parallel to the ecliptic, and causes the equinoctial point $\gamma$ to recede, and the point $\omega$, and the node $\omega$, to advance towards the sun; which produces a precession of the equinox equal to the angular motion of the pole $P$ about the axe of the ecliptic; for it is indifferent whether the pole $P$ of the earth moves about the axe of the ecliptic, whilst the plane $\gamma \omega \omega$ remains fix’d; or the same be supposed to be fix’d, and the plane of the ecliptic $\gamma \omega \omega$ to turn about its centre $C$, and the axe of the ecliptic. If we take another place $S^*$ of the sun at the same distance from the solstice $\omega$, but on the side contrary to $S$, the direction of the pole $P$ will be, by Art. 10. according to $PP^*$, perpendicular to the plane $PCS^*D^*P$, and consequently to the line $PD^*$; and the sun’s declination being the same in $D^*$ as in $D$, the motion $PP^*$ will be equal to the motion $PP$. Resolving the motion according to $PP^*$ into two, according to $PK$ and $PM'$, the motion according to $PR$ produces the same precession, and with the same sign, or the same direction, as when the sun was on the other side of the solstice $\omega$, and in $D$; and the precession is ever the same at every quarter revolution of the sun. The motion according to $PM'$ causes the same variation in the inclination of the earth’s axe to the ecliptic, ecliptic, as when the sun was on the other side in $S$, but with a contrary sign, or direction; and the sum of the lines $PM'$, from the solstice $S$ to $\omega$, will be the same as the sum of the lines $PM$, from $r$ to $S$, but with a contrary sign: so that the angle of inclination of the earth's axe to the plane of the ecliptic is continually increasing from $r$ to $S$, and then decreasing from $S$ to $\omega$; and as the diminution is equal to the increase, the inclination becomes again the same in $\omega$ as it was in $r$, having undergone an oscillation whose period is six months. Whilst the sun passes from $\omega$ to $w$, there are the same motions of the pole, as whilst it was passing from $S$ to $\omega$. Thus is the motion of the pole determined at every instant, with respect to the ecliptic. Which was to be done. **Problem VI.** 12. To find the quantity of the precession, or, which is the same, the arc of the parallel to the ecliptic run through by the pole of the earth $P$ in the space of time the sun is passing from the equinox to the solstice, or from the solstice to the equinox; which may be taken for a quarter of the time of the sun's revolution about the earth; the angle comprehended between the equinox and the ensuing solstice differing but by an infinitely small quantity from 90 degrees. Calling, as in Prob. I. the semi-axe $CP$, $a$; fig. 2. $n° 1.$ & $n° 2.$ the sine of the angle $PCS$, or $PCD$, $V$; and its cosine, or the sine of the sun's declination, $u$: Let \( P \) be \( b \); \( C \), \( r \); we shall have \( CD = \frac{a}{u} \); \( PD = \frac{a}{u} V \); the difference \( Ss \) of the arc \( r S \), which is the sun's distance from the node \( r \), will be \( \frac{rr \times dCD}{CD \times SD} \), as is found by these two proportions, \( r : Ss :: CD : D \omega = \frac{CD \times Ss}{r} \), and \( D \omega = \frac{CD \times Ss}{r} : \delta \omega = \) \( dCD : Cs = r : SD \); whence we have \( CD \times Ss \) \( \times SD = rr \times dCD \), and \( Ss = \frac{rr \times dCD}{CD \times SD} = \) \( \frac{rd}{\sqrt{aa - rr}} \) (putting instead of \( CD \) and \( SD \), their values \( \frac{a}{u} \) and \( \sqrt{\frac{aa}{uu} - rr} \)) = to the space run through by the sun \( S \) in the ecliptic in an instant, \( = 2\pi r \times \frac{dt}{T} \), calling \( T \) the time the sun employs in running thro' the whole circumference \( 2\pi r \). Whence we have \( dt = \frac{T}{2\pi r} \times \frac{rd}{\sqrt{aa - uu}} = \frac{T}{2\pi} \) \( \times \frac{du}{\sqrt{aa - uu}} \). By By Art. 8, we have \( a P p = AV u \times ad t \); therefore, by Prob. V. the instantaneous precession will be \[ AV u \times ad t \times \frac{PR}{PP}, \text{ or } AV u \times ad t \times \frac{PD}{PD}, \] the triangles \( PP \), \( PD \), being similar, since the three sides of the one are perpendicular to the three sides of the other. Putting, therefore, instead of \( PD \), \( PD \), and \( dt \), their values above found, we shall have the instantaneous precession \[ AV u \times ab \times \frac{T}{2\pi} \times \frac{\sqrt{aa - uu}}{rr} \] \[ = Ab \times \frac{T}{2\pi} \times \frac{udu}{\sqrt{aa - uu}}. \] But \( \int \frac{udu}{\sqrt{aa - uu}} = - \) \[ u \sqrt{aa - uu} + \int du \sqrt{aa - uu}, \text{ or simply } \int \frac{udu}{\sqrt{aa - uu}} \] when \( u = \frac{a}{r} \), as it happens when the sun is in the solstice \( S \), and then \( \int du \sqrt{aa - uu} \), which is the area of a quadrant of a circle whose radius is \( \frac{a}{r} \), will be \[ \frac{a\pi}{2r} \times \frac{a}{2r} = \frac{aa\pi}{4rr}. \] Therefore the precession, whilst the sun is passing from \( r \) to \( S \), is \( Ab \times \frac{T}{2\pi} \times \frac{aa\pi}{4rr} = Ab \times \frac{T}{2} \times \frac{aa}{4rr}; \) and calling $CH$, which is = the radius of the circle parallel to the ecliptic which the pole $P$ runs thro', $g$; we shall have $g = \frac{aa}{r}$, and the precession $Ab \times \frac{T}{2} \times \frac{aa}{4rr}$ will be likewise $Ab \times \frac{T}{2} \times \frac{g}{4r}$. Which was to be found. And this also is the quantity of the precession during the time the sun passes from the solstice to the equinox, or generally, during the time of each quadrant of the sun's revolution; as was seen in Prob. V. Remark. 13. The difference of the sun's distance $Sr$ from the equinoctial point, is equal to the arc the sun goes through in the ecliptic, when the node $\gamma$ is fix'd, as was suppos'd in the foregoing Prob. but when the node $\gamma$ moves in the ecliptic the contrary way to the sun's motion, the difference of the sun's distance from the node is equal to the arc gone through by the sun, increased by the arc gone through by the node $\gamma$. Therefore always taking $Ss$, fig. 2. n° 1. for the difference of the sun's distance from the node, we shall have $Ss$ equal to the arc described by the sun, more by the arc described by the node. But, by the former Prob. we have seen, that the arc run through by the pole $P$ in an instant $dt$, and in a parallel to the ecliptic, whereof $g$ is the radius, is $$AVu \times adt \times \frac{PD}{PD} = AVu \times \frac{ab}{av} \times dt = Abuudt.$$ The arc run through by the node $\gamma$ will be then $Abu$. \[ A b u u d t \times \frac{r}{S}, \text{ since these arcs, which measure equal angles, are as their radii } g \text{ and } r. \] We shall have, therefore, in the above Problem, \[ Ss, \text{ or } \frac{r \, du}{\sqrt{aa-uu}} = 2 \pi r \times \frac{dt}{T} + Abuu \times \frac{r}{g} \, dt; \] whence we get \( dt = \frac{T \, du}{2 \pi \sqrt{aa-uu}} \times \frac{1}{rr} \times \frac{1}{1 + Abuu \times \frac{T}{2 \pi g}}. \) But \[ \frac{1}{1 + Abuu \times \frac{T}{2 \pi g}} = 1 - Abuu \times \frac{T}{2 \pi g} + \&c. \] where the other terms may be neglected which contain the powers of \( A, \) because \( A \) is a very small fraction. We shall have, therefore, \( dt = \frac{T \, du}{2 \pi \sqrt{aa-uu}} \times (1 - Abuu \times \frac{T}{2 \pi g}); \) and the instantaneous precession will be \( Ab \times \frac{T}{2 \pi} \times \frac{uu \, du}{\sqrt{aa-uu}} \times (1 - Abuu \times \frac{T}{2 \pi g}) \) whose integral for the first term, in the solsticial point, where \( u = \frac{a}{r}, \) is \( Ab \times \frac{T}{2 \pi} \times \frac{aa \pi}{4 rr}, \) as was seen in the preceding Problem. But the integral of \( uu \, du \) \[ \frac{u^2 d u}{\sqrt{aa - uu}} = -u^3 \sqrt{\frac{aa - uu}{rr}} - \frac{1}{4} u \times \left( \frac{aa}{rr} - uu \right) + \frac{3}{4} \frac{aa}{rr} \int d u \sqrt{\frac{aa - uu}{rr}}, \text{ or simply, } = \frac{3}{4} \frac{aa}{rr} \int d u \sqrt{\frac{aa - uu}{rr}}, \text{ when we suppose, after the integration, } \] \[ \frac{a}{r} = u; \text{ and } = \frac{3}{4} \frac{aa}{rr} \times \frac{a \pi}{2 r} = \frac{3}{4} \frac{aa}{rr} \times \frac{aa \pi}{4 rr}. \text{ Therefore the precession from the sun's departure from the equinox till his arrival at the solstice, is } Ab \times \frac{T}{2} \times \frac{aa}{4 rr} \times \left( 1 - Ab \times \frac{T}{2 \pi g} \times \frac{3}{4} \frac{aa}{rr} \right) = (g \text{ being } = \frac{aa}{r}) Ab \times \frac{T}{2 r} \times \frac{g}{4} \times \left( 1 - Ab \times \frac{T}{2 \pi r} \times \frac{3}{4} \right). \] When the motion of the node is not very sensible, it will suffice to make use of the formula of the preceding Problem, which is the first term of this: But when that motion is sensible, which is the case of the moon's node, when it is to be determined for a quarter of the revolution of the sun, it will be requisite to employ this formula, which contains two terms of the series. Nay, it would be very easy to take a greater number of terms, so as to neglect nothing; but this would be absolutely useless, because of the quick converging of the series. PRO- Problem VII. 14. To find the alteration of the inclination of the earth's axe to the ecliptic, during the sun's passage from the equinox to the solstice, or from the solstice to the next equinox; which is the same, only affected with a contrary sign, as has been seen in Prob. V. By Art. 8. we have \( aPp = AVu \times adt \); and by Prob. V. the instantaneous variation of the inclination of the earth's axe to the ecliptic is \( AVu \times adt \times \frac{PM}{PP} = AVu \times adt \times \frac{\text{§D}}{PD} \), because of the similar triangles \( PP, PD \), (taking the denominations of Prob. VI.) \( AVu \times ax \times \frac{T}{2\pi} \times \frac{r \times dCD}{CD \times \text{§D}} \times \frac{\text{§D}}{PD} = AVu \times \frac{dT}{2\pi} \times \frac{r \times dCD}{CD \times PD} = AVu \times \frac{dT}{2\pi} \times \frac{r \times du}{ar \times u} = AV \times \frac{T}{2\pi} \times udu \), whose integral is \( \frac{1}{2} Ar \times \frac{T}{2\pi} \times uu \); and when the sun is at the point \( \text{§} \), we have \( 1 : u :: r : a \), and \( a = ru \); therefore the preceding integral is then \( \frac{1}{2} Ar \times \frac{T}{2\pi r} \times aa \). Which was to be found. Remark. 15. When the motion of the node is not insensible, you are to take the value of \( adt \) of the Remark that follows. follows Art. 12. which gives \( dt = \frac{T}{2\pi} \times \frac{r \times dCD}{CD \times SD} \times (1 - Ab \times \frac{T}{2\pi g} \times uu) \), and the instantaneous variation of the inclination of the earth's axe to the ecliptic will be \( Ar \times \frac{T}{2\pi} \times u du \times (1 - Ab \times \frac{T}{2\pi g} \times uu) \), whose integral, when the sun arrives at the solstice, where we have \( ru = a \), becomes \( \frac{1}{2} Aaa \times \frac{T}{2\pi r} - \frac{1}{4} \frac{Aa^4}{r^2} \times \frac{T}{2\pi r} \times Ab \times \frac{T}{2\pi g} = (\text{because } g = \frac{aa}{r}) \) \( \frac{1}{4} Aaa \times \frac{T}{2\pi r} \times (1 - \frac{1}{2} Ab \times \frac{T}{2\pi r}) \); and this is the formula to be made use of when the motion of the nodes is sensible. **Sect II.** *Of the motion of the pole of the equator of the earth caused by the action of the moon.* 16. Whatever has been said in Sect. I. of the precession, and of the oscillation of the earth's axe on the plane of the ecliptic, through the action of the sun, is equally applicable to the action of the moon; it being requisite only to substitute every-where the moon instead of the sun; \( L \) instead of \( S \), and \( \lambda \) instead of \( s \); the time which the moon employs in completing her revolution about the earth, instead of the time the sun employs in performing his his revolution about the earth, or $M$ instead of $T$. Lastly, the lunar orbit, instead of the ecliptic. Likewise, instead of $A$, which (Art. 8.) was equal to $\frac{3S}{s^3m}$, we must put $B = \frac{3L}{\lambda^3m} \times \frac{aa}{1-2a}$. 17. We shall find then, by Art. 12. that the arc described by the pole $P$ of the earth parallel to the plane of the lunar orbit, whilst the moon makes one quarter of a revolution about the earth, is $Bb \times \frac{g}{2r} \times \frac{M}{4}$, which gives the arc described by the mean motion in an instant, $dt$, equal to $Bb \times \frac{g}{2r} \times dt = Bb \times \frac{aa}{2rr} \times dt$; putting, instead of $g$, its value $\frac{aa}{r}$. 18. We shall find also, by Art. 14. that the variation of the inclination of the earth's axe to the plane of the lunar orbit, when the moon is at its point of station 90 degrees distant from the node of the equator with the lunar orbit, is $\frac{1}{2} B \times \frac{M}{2\pi r} \times aa$, whereby the angle of inclination of the earth's axe to the plane of the lunar orbit is increased when the moon is at its point of station. 19. We therefore should have nothing to add, if the plane of the lunar orbit were the same as that of the ecliptic, or if it always retained the same position: But as this plane continually varies its position, and its pole describes a small circle about the pole of the ecliptic, from whence it is always about 5 degrees distant; it still remains that we examine what is the precession caused by the moon, with respect to the ecliptic; which may be done by resolving the motion of the earth's pole, parallel to the plane of the lunar orbit, into two motions, one whereof is parallel to the plane of the ecliptic, and the other perpendicular thereto, and which varies the inclination of the earth's axe to the ecliptic. This is what we shall determine in the following Problem. **Problem VIII.** 20. Having given, the motion of the pole P of the earth parallel to the plane of the lunar orbit, to find its motion parallel to the plane of the ecliptic. * C P, fig. 3. n° 1. & n° 2. is the semi-axe of the earth. The plane of the figure is the plane of the solsticial colure; C Z is the axe of the ecliptic, consequently perpendicular to C S; S the solsticial point, and P C S the solsticial angle: P Z is parallel to C S, and perpendicular to C Z; and it is equal to the line C H of fig. 2. The circumference l, l', l'', l''', l is the circumference described by the pole, l, of the lunar orbit; L, L', L'', L''' is the section of the cone C l, l' l'' l''' prolonged, by the plane that is tangential at the pole of the earth, P. Let l' be any place of the pole of the moon's orbit. If upon the point P be erected P R, perpendicular to the plane P C l', this perpendicular will be the direction of the motion of the earth's pole, parallel to --- * The lines which are in the plane which touches the earth's pole l, are separately represented in fig. 2. n° 3. the plane of the lunar orbit, the plane $PCl^\circ$ being perpendicular to that plane, $PCl^\circ$ being the axe thereof. Letting fall from the point $R$ the perpendicular $RM$ upon the line $\approx PZ$, the motion of the pole of the earth according to $PR$ may be resolved into two; the one according to $PM$, the other parallel and equal to $RM$; and this latter expresses the precession with regard to the ecliptic, and the former gives the alteration of the inclination of the earth's axe to the plane of the ecliptic. The motion of the pole represented by $MR$, which gives the precession with regard to the ecliptic, is always on the same side during the entire revolution of the pole of the lunar orbit, and is the same on either side the point $l$ at the like distance from it; the motion according to $PR$ being then the same, and always on the same side, as has been explained in Prob. V. * The motion according to $PM$ is the same on either side of the point $l$ at the same distance from it, but with a contrary sign, that is, in an opposite direction. For it is manifest, that if another point, as $l'$, be taken on the other side, and at an equal distance from the point $l$, the point $M$, where the perpendicular $RM$ falls, will be on the other side with regard to $P$, than the point $M$, since the angle $RPZ'$, which is here obtuse, will then be acute, and less than the right angle of an angle equal to $L^\circ PZ'$. * The point $l'$ placed on the other side of the point $l$, that is, between $l$ and $l'$, is not marked in the fig. nor the point $M'$ where the perpendicular $KM'$ falls; for the sake of not embarrassing the fig. with too many lines, it being easy to conceive. For this reason, the motion according to \( PM \) must produce an oscillation of the earth's axis, whose period is the time of an entire revolution of the pole of the lunar orbit about the pole of the ecliptic. **Problem IX.** 21. To find the quantity of the precession caused by the moon with regard to the ecliptic; that is, the arc parallel to the ecliptic, run through by the earth's pole \( P \), during the time of a semi-revolution \( l^1 l^2 \) of the pole \( l \) of the lunar orbit. Let there be drawn the lines \( NK \), \( l^o K \) perpendicular to the axe of the earth \( CP \), and parallel to the lines \( NP \), \( l^o P \), fig. 3. no 1. & no 2. The triangles \( PRM \), \( PN L^o \), \( KN l^o \) are right-angled and similar; for the three sides of the first are perpendicular to the three sides of the second, and the three sides of the second are parallel to the three sides of the third. Still retaining the foregoing denominations, let \( zl = x \); \( ZN = z \); \( PZ = g \), as in fig. 2. the line \( CH \); \( CZ = f \); \( Cl \) or \( Cl^o = C \). We shall have \( Nl^o = \sqrt{xx - zz} \); \( NP = g - z \); \( NK = \frac{f}{a} \times g - z \), or \( \frac{f}{a} \times g + z \), when the point \( l^o \) lies on the other side of \( l^o \); \( PR = \frac{g}{a} \times g - z \), or \( \frac{g}{a} \times g + z \) when the point \( l^o \) is on the other side of \( l^o \). The triangle \( Cl^o K \) is here similar to the triangle whose three sides are represented by the letters \( a, b, r \), since since $\frac{b}{r}$ expresses here the sine of the angle of inclination of the earth's axe to the plane of the lunar orbit, and $\frac{a}{r}$ is its cosine; we shall have then $l° K = \frac{ac}{r}$; $C K = \frac{bc}{r} = CP - PK = a - \frac{g}{a} \times g - z = \frac{aa - gg + gz}{a} = \frac{ff + gz}{a}$, or $\frac{ff - gz}{a}$, when the point $l°$ is on the other side with respect to $l°$; $\frac{RM}{PR}$ $= NK = \frac{f}{a} \times \frac{g - z}{ac}$, or $\frac{f}{a} \times \frac{g + z}{ac}$ for the other side of $l°$; $\frac{PM}{PR} = \frac{Nl°}{l° K} = \sqrt{\frac{xx - zz}{ac}}$: Lastly, the difference of the arc $l° l° = du = \frac{x dz}{\sqrt{xx - zz}} = \frac{2 \pi x}{nT}$ $x dt$, calling $nT$ the time which the pole of the lunar orbit employs in performing its revolution in the circumference $2 \pi x$, whence we get $dt = \frac{nT}{2 \pi} \times \frac{dz}{\sqrt{xx - zz}}$. The space run through in an instant $dt$ by the pole $P$ of the earth parallel to the plane of the lunar orbit being being by Art. 17. \(= B b \times \frac{aa}{2r} \times dt\); and by Prob. VIII. the space run through parallel to the plane of the ecliptic, will be \(B b \times \frac{aa}{2rr} \times dt \times \frac{x}{P} =\) (by the preceding denominations) \(Bb \times \frac{aa}{2rr} \times \frac{nT}{2\pi} \times \frac{dz}{\sqrt{xx - zz}} \times \frac{f \times g - z}{ac}\) and putting instead of \(\frac{b}{r}\), its value, which was found to be \(\frac{ff + gz}{ac}\), we shall have \(B \times \frac{nT}{4\pi} \times \frac{f}{acc} \times \frac{ff + gz}{ac} \times \frac{g - z}{\sqrt{xx - zz}} \times \frac{dz}{\sqrt{xx - zz}}\); and when the point \(l^o\) is on the other side with regard to \(l^i\), we shall have \(B \times \frac{nT}{4\pi} \times \frac{f}{acc} \times \frac{ff - gz}{ac} \times \frac{g + z}{\sqrt{xx - zz}} \times \frac{dz}{\sqrt{xx - zz}}\). Therefore always taking the precession for the two points together placed on either side of \(l^i\), and at an equal distance from it, we shall have, by joining the two values now found, after making reduction, \(B \times \frac{nT}{4\pi} \times \frac{f}{aa} \times \frac{2gf - 2gz}{\sqrt{xx - zz}} \times \frac{dz}{\sqrt{xx - zz}}\), or \(B \times \frac{nT}{2\pi} \times \frac{fg}{aa} \times \frac{ff - zz}{\sqrt{xx - zz}} \times \frac{dz}{\sqrt{xx - zz}}\). But But the integral of \( f f \times \frac{d z}{\sqrt{\kappa \kappa - zz}} = \frac{ff}{x} \times \frac{d z}{\sqrt{\kappa \kappa - zz}} \) is \( \frac{ff}{x} \times \frac{x \pi}{2} = \frac{ff \pi}{2} \), when we take all the points \( l^o \), from \( l^i \) to \( l \); and the integral of \( \frac{zz d z}{\sqrt{\kappa \kappa - zz}} \), which is \( -z \sqrt{\kappa \kappa - zz} + \int dz \sqrt{\kappa \kappa - zz} \); when, after the integration, we make \( z = x \), becomes \( \int dz \sqrt{\kappa \kappa - zz} \), or the area of the quadrant \( l^i \times l \), which is \( \frac{x \pi}{2} \times \frac{x}{2} = \frac{x \kappa \pi}{4} \). Therefore the integral of \( B \times \frac{nT}{2\pi} \times \frac{fg}{acc} \times \frac{ff - zz}{\sqrt{\kappa \kappa - zz}} \times \frac{d z}{\sqrt{\kappa \kappa - zz}} \) is \( B \times \frac{nT}{2\pi} \times \frac{fg}{acc} \times \left( \frac{ff \pi}{2} - \frac{x \kappa \pi}{4} \right) \) \( = B \times \frac{nT}{2} \times \frac{fg}{acc} \times \left( \frac{ff}{2} - \frac{x \kappa}{4} \right) = B \times \frac{nT}{2} \times \frac{gf^3}{2acc} \times \left( 1 - \frac{x \kappa}{2ff} \right) \). This is the sum of all the instantaneous precessions for all places of the lunar orbit from \( l \) to \( l^i \), because we have always taken two points together placed on either side of \( l^i \). And this is the quantity of the precession with regard to the ecliptic, during the time of a semi-revolution of the lunar orbit. Which was to be found. Problem X. 22. To find the quantity of the nutation, or of the variation of the inclination of the earth's axe to the ecliptic, caused by the moon during the semi-revolution 1 l° 1° of the pole of the lunar orbit. The space run through in an instant \( dt \) by the pole of the earth, parallel to the plane of the lunar orbit, being, by Art. 17. \( = B b \times \frac{aa}{2rr} \times dt \), the instantaneous nutation for any one place \( l° \) of the lunar orbit will be, by Prob. VIII. \( = B b \times \frac{aa}{2rr} \times dt \times \frac{PM}{PR} \) \( = \) (by the denomination of the preceding Problem) \( B b \times \frac{aa}{2rr} \times \frac{nT}{2\pi} \times \frac{dz}{\sqrt{\kappa\kappa - zz}} \times \sqrt{\kappa\kappa - zz} \times \frac{ac}{r} = B \times \frac{b}{r} \times \frac{aa}{2} \times \frac{nT}{2\pi} \times \frac{dz}{\sqrt{\kappa\kappa - zz}} \). Putting instead of \( \frac{b}{r} \) its value, which was found to be \( \frac{ff + gz}{ac} \), we shall have \( B \times \frac{nT}{2\pi \times 2cc} \times ff + gz \times dz; \) and when the point \( l° \) is on the other side with regard to \( l' \), we shall have \( B \times \frac{nT}{2\pi \times 2cc} \times ff - gz \times dz. \) Taking therefore the nutation for the two points together, \( l° \), placed on either side of \( l' \), we have \( B \times \) \[ B \times \frac{nT}{2\pi \times 2cc} \times 2ff dz, \text{ whose integral, when } z \text{ is } = x, \text{ is } B \times \frac{nT}{2\pi \times 2cc} \times 2ff x = B \times \frac{nT}{2\pi} \times \frac{ff}{cc} \times x. \] This is the quantity of the nutation caused by the moon during the semi-revolution \( l l' l'' \) of the pole of the lunar orbit. Which was to be found. **Corollary.** 23. By *Prob. IX.* the quantity of the precession caused by the moon, with regard to the ecliptic, is, during the space of a semi-revolution \( l l' l'' \) of the pole of the lunar orbit, \( B \times \frac{nT}{2} \times \frac{gf^3}{2acc} \times \left(1 - \frac{1}{2} \times \frac{xx}{ff}\right). \) By the preceding *Problem,* the quantity of the nutation during the same time of a semi-revolution of the pole of the lunar orbit \( l l' l'' \), caused also by the moon, is \( B \times \frac{nT}{2\pi} \times \frac{ff}{cc} \times x. \) Therefore the precession, during the time of a semi-revolution of the pole of the lunar orbit, is to the nutation, during the same time, \[ \text{As } \frac{gf}{2a} \times \left(1 - \frac{1}{2} \times \frac{xx}{ff}\right), \text{ is to } \frac{x}{\pi}; \text{ or as } gf \pi \times \left(1 - \frac{1}{2} \times \frac{xx}{ff}\right) \text{ is to } 2a \pi. \] But the precession is performed in a parallel to the ecliptic, whereof \( g \) is the radius; and the nutation is performed in the plane of the solsticial colure, and in a circle whereof \( a \) is the radius: Dividing, viding, therefore, each circular arc by its radius, we shall have the quantity of the angle: And The angle of the precession, during the time of a semi-revolution of $l' l'$ of the pole of the lunar orbit, will be to the angle of the nutation, in the same time, $$\text{As } \pi \times \left(1 - \frac{\chi \chi}{ff}\right) \text{ is to } 2 \chi, \text{ or as } \pi \left(1 - \frac{\chi \chi}{ff}\right) \text{ is to } \frac{2 \chi}{f}.$$ **Scholium.** 24. Whatever has been said as to the moon and the sun, may be equally applicable to all the other planets; but as the effect of the action of the rest of the planets upon the earth is not sensible, we shall not take it into consideration. **Sect. III.** *Of the motion of the pole of a ring, caused by the action of the sun: or of the motion of the pole of the moon's orbit.* 25. We have seen, in Art. 9, that if, instead of the terrestrial spheroid, we were to suppose a simple ring placed at the circumference of the equator, the motion of the pole of such ring would be found by the same formula as that for the motion of the pole of the earth; supposing only, in that formula, $\frac{a}{1-2a} = 1.$ By By this means we may apply all the formulae which have been found in Sect. I. to the motion of the nodes of this ring, and to the alteration of the inclination of its axe to the plane of the ecliptic. But by the Remark on Art. 6. the motion of the plane of this ring is the same, whether the ring be entire, or there be only a single point which circulates in the ring's circumference: Whence it follows, that These same formulae do likewise give the motion of the pole of the orbit of a moon or satellite; the motion of the node of the orbit of such moon in the ecliptic; and the variation of the inclination of its axe of the orbit to the plane of the ecliptic: Observing to put the time of the revolution of this moon about the sun, instead of the time of the earth's revolution about the sun, and the motion of the moon in its orbit instead of the motion of a point of the equator. Remark. It may be observed, that although the motion of the pole of a ring be the same as that of a moon, during the time of the revolution of the ring, or of the moon, which is the same; yet there are some particular motions which take place when there is only one moon revolving in the circumference of the ring, and which cease to exist when it is an entire ring that revolves. For example; in Art. 2. the force, according to \( L \gamma \), fig. i. n° 2. of the point \( L \), is destroyed by an equal and directly opposite force of another point \( A \) placed on the other side, and at the same distance from the point \( E \), as the point \( L \), when the ring is entire; but but when there is but one point, as $L$, or, which is the same, one moon, which revolves in the circumference of the ring; this motion subsists, and disturbs the motion of the moon in the plane of its orbit. In like manner the force of the point $L$ parallel to $\gamma C$, or according to $c' L$, is destroyed, in a ring, by the force of another equal point $\lambda'$ placed on the other side with regard to $Q$, and at the same distance as the point $L$. But this force subsists in the case where there is but one single point $L$, or one moon, and this motion disorders, besides, the motion of the moon in the plane of its orbit. Lastly, If the ring, which circulates about the centre $C$, be elliptical, the motion of every point of the ring in the orbit is proportional to its distance from the centre; because, on account of the continuity of the ring, all the points complete their revolution in the same time, and have equal angular motions; but when it is but one single moon that revolves, in an elliptic orbit, it describes areas proportional to the times, and has an angular motion so much the greater, as it approaches nearer the centre; which is the contrary of the motion of the points of the elliptic ring, the more distant whereof from the centre, have the most motion. And it arises from hence, that the motion of the pole of the lunar orbit is much the same as that of the ring, taking it during the entire revolution; but that there is a small variation, or inequality, in the instantaneous motion of the pole of a moon; whereas this instantaneous motion is constant through every instant of the revolution of the ring. I shall I shall spend no more time on the ring, with respect to those motions which have nothing to do with the subject in question; but they shall be thoroughly discussed, and applied, in a second memoir, which I shall publish, on the particular motions of the planets. S E C T. IV. Containing the application of the formulæ of the preceding Sections. 26. The motion of the sun, or, to speak more exactly, of the earth, in the ecliptic, is $\frac{2\pi s}{T}$, which is the space divided by the time. 27. The earth's central force will therefore be $\left(\frac{2\pi s}{T}\right)^2 \times \frac{1}{2s}$, the central force being equal to the square of the motion divided by the diameter of the circumference in which the body moves; and the earth's central force being the force $\frac{S}{ss}$ wherewith it is attracted towards the sun; we shall have $\frac{S}{ss} = \left(\frac{2\pi s}{T}\right)^2 \times \frac{1}{2s} = \frac{2\pi \pi s}{TT}$; whence we get $\frac{S}{s^3} = \frac{2\pi \pi}{TT}$. 28. The motion of a point of the equator whose radius is $a$, will be, calling the time of its revolution $t$, $\frac{2A\pi}{t}$, which is the space divided by the time. We have therefore $m = \frac{2A\pi}{t}$. 29. By Art. 8. \( A = \frac{3S}{s^3m} \times \frac{aa}{1-2a} \) = (by Articles 27. & 28.) \( 3 \times \frac{2\pi}{TT} \times \frac{aa}{2a\pi} \times \frac{aa}{1-2a} = \frac{3t}{TT} \times \pi \times \frac{a}{1-2a} \) = (by Art. 31. following) \( \frac{3t}{TT} \times \pi \times \frac{1}{176} \). 30. Putting \( q \) for the ratio of the lunar force \( \frac{L}{\lambda^3} \), to the solar force \( \frac{S}{s^3} \), we shall have \( \frac{L}{\lambda^3} = \frac{S}{s^3} \times q \); and \( B \), which (by Art. 16.) is \( \frac{3L}{\lambda^3m} \times \frac{aa}{1-2a} \) will also be \( = \frac{3S}{s^3m} \times q \times \frac{aa}{1-2a} \) = (by Art. 29.) \( \frac{3qt}{TT} \times \pi \times \frac{1}{176} \). 31. Supposing \( a = \frac{1}{178} \), which is conformable to the observations of the gentlemen of the Parisian academy, we shall have \( \frac{a}{1-2a} = \frac{1}{178 \times 1 - \frac{2}{178}} = \frac{1}{176} \). Problem XI. 32. To find the precession caused by the sun during the time between the equinox and the solstice; that is to say, in a quarter of a year, or three months. By Art. 12. the precession, during this time, is \( Ab \times \frac{T}{2} \times \frac{\xi}{4r} \). Putting, instead of \( A \), the value found found in Art. 29. we shall have \( \frac{3}{T} \times \pi \times \frac{1}{176} \times \frac{bT}{2} \times \frac{g}{4r} = \frac{3}{T} \times \frac{1}{176} \times \frac{b}{r} \times \frac{\pi g}{8} \). But \( \frac{b}{r} \) is the ratio of the sine of the inclination of the axe of the earth to the ecliptic, to the radius, or of the sine of 66 degrees 32 minutes to the radius, \( = \frac{9172919}{10000600} \); and \( \pi g \) is the semi-circumference of the parallel to the ecliptic wherein the pole of the earth moves, and, consequently, equal to 180 deg. \( t \), which is the time of the revolution of a point of the equator, is equal to one day; and \( T \) equal to 365\(\frac{1}{4}\) days; we shall have, then, by substituting all the values in the preceding formula, \( \frac{3}{365\frac{1}{4}} \times \frac{9172919}{10000000} \times \frac{1}{176} \times \frac{180}{8} \) degrees. This is the precession produced by the sun during the space of 3 months. 33. Multiplying by 4, we shall have the annual precession caused by the sun \( = \frac{3}{365\frac{1}{4}} \times \frac{9172919}{10000000} \times \frac{1}{176} \times \frac{180}{8} \) degrees, \( = \frac{3}{365\frac{1}{4}} \times \frac{9172919}{10000000} \times \frac{45}{88} \) degrees, \( = \text{(after due reduction)} 13'' \cdot 52'' \cdot 11'' \). Problem XII. 34. To find the precession caused by the moon, during the time of a semi-revolution of the pole of the lunar orbit about the pole of the ecliptic; that is, in about $9\frac{1}{3}$ years, or $9\frac{1}{3} \times T$, as $T = 1$ year. By Art. 21, we have found this precession $= B \times \frac{nT}{2} \times \frac{g f^3}{2acc} \times (1 - \frac{1}{2} \times \frac{x x}{ff})$. By Art. 30, $B = \frac{3qt}{TT} \times \pi \times \frac{1}{176}$: Substituting this value in the preceding formula, it becomes $\frac{3qt}{T} \times \frac{n}{2} \times \frac{f}{2a} \times \frac{ff}{cc} \times \frac{1}{176} \pi g \times (1 - \frac{1}{2} \times \frac{x x}{ff})$. But $\frac{f}{a}$ is the ratio of the fine of 66 deg. 32 min. to the radius, $= \frac{9172919}{10000000}$. $\frac{f}{c}$ is the ratio of the fine of 85 deg. to the radius, the angle $l c z$, whereby the pole of the lunar orbit is distant from the pole of the ecliptic, being, at all times, about 5 degrees. Therefore $\frac{f}{c} = \frac{9961947}{10000000}$. $\frac{x}{f}$ is the ratio of the fine of 5 deg. to the fine of 85 degrees, $= \frac{871557}{9961947}$. Therefore Therefore substituting these values, and those of \( t, T, \) and \( \pi g \) found in the preceding Problem, we shall find that the precession, during a semi-revolution of the lunar pole, which is \( \frac{nT}{2} \), by Art. 21, becomes \[ = q \times \frac{n}{2} \times \frac{3}{365^{\frac{1}{4}}} \times \frac{9172919}{10000000 \times 2} \times \left( \frac{9961947}{10000000} \right)^2 \times \frac{1}{176} \] \[ \times 180 \text{ deg.} \times (1 - \frac{1}{2} \times \left( \frac{871557}{9961947} \right)^2). \] This is the precession caused by the moon during the time of a semi-revolution of the pole of the lunar orbit, \( \frac{nT}{2} = 9^{\frac{1}{2}} \times T \). 35. Therefore the mean precession caused by the moon during the time \( T \), which is the mean annual precession caused by the moon, will be, by dividing the preceding formula by \( \frac{n}{2} \), \[ = q \times \frac{3}{365^{\frac{1}{4}}} \times \frac{9172919}{10000000 \times 2} \times \left( \frac{9961947}{10000000} \right)^2 \times \frac{180 \text{ deg.}}{176} \times (1 - \frac{1}{2} \times \left( \frac{871557}{9961947} \right)^2) \] where it only remains to determine the value of \( q \), which, by Art. 30, is the ratio of the lunar force to the solar force; and making this ratio \( = \frac{5}{2} \), conformable to the observations of the tides related by Mr. Daniel Bernoulli, we shall have for the mean annual precession caused by the moon, \[ = \frac{5}{2} \times \frac{3}{365^{\frac{1}{4}}} \times \frac{9172919}{10000000 \times 2} \times \left( \frac{9961947}{10000000} \right)^2 \times \frac{180 \text{ deg.}}{176} \times (1 - \frac{1}{2} \times \left( \frac{871557}{9961947} \right)^2) \] \( = \) (after due reduction) \( 34'' \cdot 16'' \cdot 51'' \). Problem XIII. 36. To find the nutation of the earth's axe caused by the moon, during the time of a semi-revolution of the pole of the moon's orbit, that is, in $9\frac{1}{3}$ years. By Art. 23. $\pi \times (1 - \frac{x}{f} \times \frac{xx}{ff})$ is to $\frac{2x}{f}$, as the angle of the precession, found in the preceding Problem for the time $\frac{nT}{2}$ of the semi-revolution of the pole of the lunar orbit, is to the nutation during the same time. By it, therefore, may be found the nutation, having the quantity of the precession of the preceding Prob. or, reciprocally, knowing the nutation during the time of a semi-revolution of the pole of the lunar orbit, the precession will be known to a geometrical certainty, for the same time. The ratio, in numbers, is $\frac{1095798}{61236}$. 37. But we may also know the nutation during the time of a semi-revolution of the pole of the lunar orbit, by the formula of Art. 22. which is $B \times \frac{nT}{2} \times \frac{ff}{cc} \times x$; which, without alteration, may be multiplied by $\frac{a\pi}{a\pi}$, or by $\frac{a\pi}{a \times \frac{22}{7}}$; since $\pi$, which is the ratio of the circumference to the diameter, is $\frac{22}{7}$, and we shall have have the formula \( B \times \frac{nT}{2} \times \frac{ff}{cc} \times \pi \times \frac{a\pi}{a \times 22} = B \times \frac{nT}{2} \) \[ \times \frac{f}{a} \times \frac{f}{c} \times \frac{\pi}{c} \times \frac{7}{22} \times a\pi; \] which, by substituting the values of the preceding Problem, and 180 degrees instead of \( a\pi \), which is the semi-circumference of the circle wherein the nutation is performed, becomes \[ \frac{39t}{T} \times \frac{n}{2} \times \frac{9172919}{10000000} \times \frac{9961947}{10000000} \times \frac{871557}{10000000} \] \[ \times \frac{7}{22} \times \frac{180}{176} \text{ deg.} = \frac{3}{365\frac{1}{4}} \times \frac{5}{9} \times \frac{9172919}{10000000} \times \frac{9961947}{10000000} \] \[ \times \frac{871557}{10000000} \times \frac{7}{22} \times \frac{180}{176} \text{ deg. equal, after reduction, to } 17". 51'". 14'". \] **Problem XIV.** 38. To find the difference of the inclination of the earth's axe, between the point of the equinox and that of the solstice, caused by the action of the sun. By Art. 14. the formula is \( \frac{1}{2} A \times \frac{T}{2\pi r} \times a^2 \). Putting, instead of \( A \), its value of Art. 29. instead of \( \frac{a}{r} \), which is the ratio of the sine of the obliquity of the ecliptic to the radius \( \frac{3982155}{10000000} \), we shall have \[ \frac{1}{2} \times \frac{3}{365\frac{1}{4}} \times \frac{1}{2} \times \frac{1}{176} \times \frac{3982155}{10000000} \times a; \] and putting, instead of \(a\), \(\frac{a\pi}{\pi} = \frac{a\pi}{22} = \frac{7}{22} a \pi = \frac{7}{22} \times 180\) deg. (since \(a \pi\) is the semi-circumference of the circle wherein this inclination is performed) we shall have \(\frac{3}{365\frac{1}{4}} \times \frac{1}{4 \times 176} \times \frac{3982155}{10000000} \times \frac{7}{22} \times 180\) deg. \(= (\text{after reduction}) 57''\). 25''. **Problem XV.** 39. To find the difference of the inclination of the earth's axe to the plane of the lunar orbit, when the moon arrives at its point of station, 90 degrees distant from the node of the equator with the lunar orbit, this difference of inclination being caused by the moon. By Art. 18. the variation of inclination will be \[\frac{1}{2} B \times \frac{M}{2\pi r} \times a^2.\] Putting, instead of \(B\), its value of Art. 30.; in- instead of \(M\), which expresses the time of the moon's revolution about the earth, 27 \(\frac{1}{3}\) days; in- stead of \(\frac{a}{r}\), the mean value \(\frac{3982155}{10000000}\); I say mean value, because the earth's axe is not always equally inclined to the plane of the lunar orbit, as has been shewn in Art. 19. &c. we shall have \(\frac{1}{2} \times \frac{3 \times \frac{5}{2}}{(365\frac{1}{4})^2} \times \frac{27\frac{1}{3}}{2 \times 176} \times \frac{3982155}{10000000} \times a = (\text{putting, as in the pre-}\) ceding ceding Problem, instead of \( a \), \( \frac{7}{22} \times 180 \) degrees) \( \frac{3}{2} \times \frac{35}{(365\frac{1}{4})^2} \times \frac{27\frac{1}{2}}{2 \times 176} \times \frac{3982155}{10000000} \times \frac{7}{22} \times 180 \) degrees, \( = \) (after reduction) \( 10''''. 19'''' \). **Problem XVI.** 40. To find the motion of the node of Jupiter's fourth satellite, caused by the sun, during the time of \( \frac{1}{4} \) of a revolution of that satellite about the sun. We need only make use of the formula of Problem XI. wherein, conformably to Art. 25, we must make \( \frac{a}{1-2a} = 1 \), \( r = 16\frac{3}{4} \) days, being the time of the revolution of the satellite about Jupiter; \( T = 365\frac{1}{4} \times 12 \) days, which is the time of the satellite's revolution about the sun, or about 12 years; \( \frac{b}{r} = 1 \), because the orbit of the satellite is but little inclined to the ecliptic, and, consequently, the axe thereof nearly at right angles thereto. We shall have then \( \frac{3t}{T} \times \frac{\pi g}{8} \), or \( \frac{3 \times 16\frac{3}{4}}{365\frac{1}{4} \times 12} \times \frac{180}{8} \) deg. This is the motion of the node of this satellite in 3 years, or during \( \frac{1}{4} \) of its revolution about the sun. **Remark.** 41. By this may be found the motion of the nodes of the rest of Jupiter's satellites. For it appears, from the formula, that the motion of the nodes of different different satellites of the same planet, is as the times of their revolutions about the planet. Thus the motion of the nodes of the 3d satellite, in the space of 3 years, is \(\frac{3 \times 7\frac{1}{2}}{365\frac{1}{4} \times 12} \times \frac{180}{8}\) deg. The motion of the nodes of the 2d is \(\frac{3 \times 3\frac{1}{2}}{365\frac{1}{4} \times 12} \times \frac{180}{8}\) deg. The motion of the nodes of the 1st is \(\frac{3 \times 1\frac{1}{2}}{365\frac{1}{4} \times 12} \times \frac{180}{8}\) deg. **Problem XVII.** 42. To find the motion of the nodes of the 5th satellite of Saturn, caused by the sun, during the time of \(\frac{1}{4}\) of the revolution of this satellite about the sun. Conformably to the preceding Problem we shall have, by putting instead of \(t\), 79\(\frac{1}{2}\) days; instead of \(T\) 365\(\frac{1}{4}\) \(\times\) 30 days, which is nearly the time of the revolution of this satellite about the sun, and we shall have \(\frac{3 \times 79\frac{1}{2}}{365\frac{1}{4} \times 30} \times \frac{180}{8}\) deg. 43. In the same manner may be found the motion of the nodes of all the rest of Saturn's satellites. Problem XVIII. 44. To find the motion of the moon's node during the time that the sun is passing from the node of the moon's orbit with the ecliptic, to its point of station; that is, in three months, wanting a few days. Here we are to make use of the formula of Art. 13, because the motion of the moon's nodes is very sensible in respect of the motion of the sun. We shall have \( Ab \times \frac{T}{2r} \times \frac{g}{4} \times (1 - Ab \times \frac{T}{2\pi r} \times \frac{1}{4}) \); and employing the values of Prob. XI. except that here \( t = 27\frac{1}{3} \) days, which is the time of the moon's revolution about the earth, and that \( \frac{b}{r} = \) to the sine of 85 deg. divided by the radius; the axe of the lunar orbit being inclined in an angle of about 85 degrees to the ecliptic. Therefore \( \frac{b}{r} = \frac{9961947}{10000000} \). Lastly, by Art. 25, we must take \( \frac{a}{1-2a} = 1 \), and the preceding formula will be changed into this, \( \frac{3 \times 27\frac{1}{3}}{365\frac{1}{4}} \times \frac{9661947}{10000000} \times \frac{180 \text{deg.}}{8} \times (1 - \frac{3 \times 27\frac{1}{3}}{365\frac{1}{4}} \times \frac{9961947}{10000000} \times \frac{3}{2 \times 8}) \) = (after reduction) 4°. 36′. 33″. This is the motion of the moon's node during the time that the sun is passing from the node of the moon's orbit with the ecliptic to its point of station; that is, during the time the sun is moving 90° - 4 degrees grees, 36 minutes, 33 seconds, or $85^\circ 23' 27''$. And if the motion of the node be required for one year, 'tis only multiplying the motion, just found, by $\frac{360^\circ}{85^\circ 23' 27''}$, and it will be found $19^\circ 25' 39''$. being a very small matter more than the truth, for reasons which will be explained elsewhere, and because we have taken $27\frac{1}{4}$ days for the time of the moon's revolution about the earth, which is really a little less than that quantity. **Problem XIX.** 45. To find the greatest difference of the inclination of the lunar orbit to the ecliptic; which happens when the sun arrives at his point of station with regard to the moon. Here the formula of Art. 15. must be made use of, which gives $\frac{1}{2} Aaa \times \frac{T}{2\pi r} \times \left(1 - \frac{1}{2} Ab \times \frac{T}{2\pi r}\right)$, or, which is the same, $\frac{1}{2} A \times \frac{aT}{2\pi r} \times \frac{a\pi}{22} \times \left(1 - \frac{1}{2} A \times \frac{bT}{2\pi r}\right)$, substituting the values as in the preceding Problem, and instead of $\frac{a}{r}$, putting $\frac{871557}{10000000}$, which is the ratio of the sine of $5^\circ$ to radius, we shall have $\frac{3 \times 27\frac{1}{4}}{365\frac{1}{4}} \times \frac{1}{2 \times 2} \times \frac{871557}{10000000} \times \frac{7}{22} \times 180 \text{ deg.} \times \left(1 - \frac{1}{4} \times \frac{3 \times 27\frac{1}{4}}{365\frac{1}{4}} \times \frac{9961947}{10000000}\right) = 15'.51''$. Remark. Remark. 46. It should be observed, that, to shorten the computations, I have contented myself with taking the times of the revolutions pretty near the truth; but if the utmost exactness be required, the accurate times of the revolutions must be employed. 47. This might be a proper place to add the method of determining the perturbation of the orbit of any planet, as derived from another planet; but since this depends upon no other than the very same principles that have been made use of in this memoir, and as their application will be shewn, in its full extent, in the memoir which I am going to print, and intend myself the honour of sending to the Royal Society, I shall desist, that I may not run this paper to a greater length. LIX. A Letter to the Right Honourable George Earl of Macclesfield, P. R. S. concerning the ages of Homer and Hesiod. By George Costard, M. A. My Lord, Read Dec. 13, 1753. It seems to be an opinion pretty generally received, that Homer and Hesiod lived much about the same time. If this be true, and they did so, whatever arguments prove the age of one, will equally serve for fixing that of the other. What that age was, is indeed not at all agreed on among writers; the only thing in which they con-