PRECESSION

In the article ASTRONOMY (vol. iv. 14), the phenomenon of the annual precession of the equinoxes has been described, and its physical cause stated to be the attraction of the sun and moon upon the protuberant mass of matter accumulated about the earth's equator, combined with the diurnal rotation. We shall here discuss the subject more particularly, and shew in what manner the different forces which tend to displace the plane of the earth's equator give rise to the phenomenon in question, and how their effects are computed from the fundamental principles of dynamics. The general problem, which is that of determining the perturbations of the earth's axis of rotation, embraces the Nutation of the axis, as well as the precessional motion of the equinoctial points, and is one of the most important and interesting in physical astronomy.

If the earth were a perfect sphere, the attraction of the sun or moon would have no tendency to communicate to it any motion about its centre of gravity. In this case, all the particles being symmetrically disposed with reference to every plane passing through its centre, the forces acting on opposite sides of any plane passing through its centre and the centre of the attracting body, would exactly balance each other, and consequently would have no tendency to produce a rotatory motion. But by reason of the spheroidal form of the earth, and the intensity of the force of attraction varying with the distance, the action of a distant body which is not situated either in the plane of the equator, or in the prolongation of the axis of rotation, produces an unequal effect on the opposite sides of every plane passing through the earth's centre (excepting the meridian in which the body is situated), and tends to generate a rotatory motion about that diameter of the equator which is perpendicular to the line which joins the centre of the earth with the centre of the attracting body. Hence the sun exerts a force, which at every instant has a tendency to bring the plane of the earth's equator towards the plane of the ecliptic; and if the earth had no motion of rotation about its axis, the two planes would at length be brought to coincide. In consequence, however, of the rotatory motion, the inclination of the two planes, as we shall shew, undergoes no permanent alteration; but a motion is given to the earth's axis, such that the pole of the equator constantly revolves about the pole of the ecliptic in the direction opposite to that of the diurnal rotation, and the intersection of the equator and ecliptic, following the motion of the pole, is carried backwards along the ecliptic. The moon produces a similar effect in reference to the plane of the lunar orbit; and the motion produced by the combined action of the sun and moon, which is the phenomenon observed, is the luni-solar precession of the equinoxes.

As the efficacy of the disturbing force to turn the earth about an axis varies with the distance of the attracting body from the plane of the equator, the precessional motion of the equinoxes is not uniform. The efficacy of the sun's force continues to increase, whilst the sun passes from either equinox to the solstice, and to diminish while it passes from the solstice to the equinox. The period of the inequality is consequently half a year. The period in which the action of the moon passes through all its degrees of intensity is about nine years, being that in which the nodes of the lunar orbit accomplish half a revolution on the ecliptic. The apparent effect of this irregular action, is an alternate increase and diminution of the declinations of the fixed stars, most sensible for those nearest the pole, which is characteristically called the nutation of the earth's axis. The solar

nutation, however, is so small as to be insensible to observation; the lunar nutation is sufficiently sensible, amounting to about 18" between the extreme positions of the pole.

Before proceeding with the investigation of the problem, it will be convenient to premise the two following elementary theorems respecting the composition of rotatory motion, referring the reader for their demonstration to the article ROTATION.

Theorem 1. If a rigid body revolving about an axis $Aa$, which passes through its centre of gravity $O$, with an angular velocity $\backslash equiv\; v$, receive an impulse which alone would cause it to revolve about an axis $Bb$, also passing through its centre of gravity, with a velocity $\backslash equiv\; \backslash phi$, the body will now revolve about a third axis $Cc$, passing through its centre of gravity, and lying in the plane of the two axes $Aa$ and $Bb$, and so situated that the sine of its inclination to the axis $Aa$ will be to the sine of its inclination to the axis $Bb$, as the velocity about $Bb$ to the velocity about $Aa$; that is, the new axis will divide the angle $AOB$, so that $\backslash sin\; AOC\; :\; \backslash sin\; BOC\; ::\; \backslash phi\; :\; v$.

In order to determine whether the pole $C$ of the new axis lies between $A$ and $B$, or between $A$ and $b$, it is only necessary to consider that the new axis must evidently be that line in the body in which every point is at rest in respect of both motions. If, therefore, we suppose the original motion about $Aa$, to be in the direction which would raise the point $B$ above the plane of the paper, and to depress $b$ below it, and the new impulse to be given in the direction which would depress the point $A$ below the plane of the paper, and raise $a$ above it, then $C$ will lie between $A$ and $B$; but if the new impulse tends to raise $A$ above the plane of the paper, then $C$ will lie between $A$ and $b$.

Corollary 1. If the two axes $Aa$ and $Bb$ are at right angles, then $\backslash sin\; BOC\; =\; \backslash cos\; AOC$, and we have $\backslash frac\{\backslash sin\; AOC\}\{\backslash cos\; AOC\}\; =\; \backslash frac\{\backslash phi\}\{v\}$, that is, $\backslash tan\; AOC\; =\; \backslash frac\{\backslash phi\}\{v\}$.

Cor. 2. If the impulse is renewed at every instant of time, the axis about which the body actually revolves must have a uniform motion in space from $OA$ towards $OB$.

Theorem 2. If the force which tends to give the body a motion of rotation about an axis which is always perpendicular to the axis about which it is already revolving, and situated in the plane $AOB$, be uniform, the angular velocity of rotation remains unaltered, or $v$ is a constant quantity.

These theorems, which are true of bodies in general, whatever be their figure, were first demonstrated by Frisi.

Proposition 1. To determine the efficacy of the sun's attraction to turn the spheroid about its centre, the earth being supposed homogeneous.

Let $Pp$ be the axis of rotation, $Qq$ the projection of the equator, $S$ the sun, $C$ the centre of the spheroid, and $D$

Precession of the Equinoxes. the projection of a point in the interior of the spheroid on the plane PQpy. Join SC, SD, DC; and draw CG perpendicular to SC, and DE, DF respectively perpendicular to SC, CG.

Let $f$ represent the force of the sun's attraction on a particle at the centre C, and $f\text{'}$ the force of its attraction on a particle at D, then the attraction being inversely as the square of the distance, we have $f\text{'}\; =\; f\; \backslash cdot\; \backslash frac\{SC^2\}\{SD^2\}$. Now the force $f\text{'}$, which acts on D in the direction SD, may be resolved into two; one in the direction CD, which has no tendency to turn the spheroid about its centre, and the other in the direction FD parallel to SC, which tends to turn the spheroid in the direction PQ about an axis passing through C, and perpendicular to the plane PQpy. The resolved part of the force $f\text{'}$, in the direction FD is $f\text{'}\; \backslash cdot\; \backslash frac\{SC^2\}\{SD^2\}\; \backslash cdot\; \backslash frac\{SC\}\{SD\}$, since SD is the diagonal of a parallelogram

of which SC and CD are the sides. Now, if the resolved force in the direction FD parallel to SC were the same on every particle, and equal to $f$, it would have no tendency to produce rotation in the spheroid; we may therefore conceive the force on any particle which tends to produce rotation to be the difference between the part of the force acting on that particle in the direction parallel to SC, and the force $f$ acting on the particle at C. Hence the force on D tending to impress a rotatory motion on the spheroid is $f\; \backslash cdot\; \backslash left(\; \backslash frac\{SC^2\}\{SD^2\}\; \backslash cdot\; \backslash frac\{SC\}\{SD\}\; -\; 1\; \backslash right)\; =\; f\; \backslash cdot\; \backslash frac\{SC^2\; -\; SD^2\}\{SD^3\}\; =\; f\; \backslash cdot\; \backslash frac\{(SC-SD)(SC+SD)\}\{SD^3\}$.

Now, by reason of the great distance of the sun in comparison of the radius of the earth, SD is very nearly parallel to SC, and equal to SC-CE. Substituting therefore SC-CE for SD, and neglecting terms divided by $SC^2$ which are so small as to be altogether insensible, the factor $\backslash frac\{SC^2\; +\; SC\; \backslash cdot\; SD\; +\; SD^2\}\{SD^3\}$ becomes $\backslash frac\{3\}\{SD\}$, and consequently the above expression is reduced to $3f\; \backslash cdot\; \backslash frac\{CE\}\{SD\}$, or to $3f\; \backslash cdot\; \backslash frac\{CE\}\{SC\}$, since the difference between $1\; \backslash div\; SD$ and $1\; \backslash div\; SC$ is a quantity divided by $SC^2$, and therefore insensible. This is the part of the force on D which tends to turn the spheroid about the diameter of the equator which is perpendicular to Qq; and by the principles of mechanics, its efficacy in communicating a rotatory motion to the spheroid is proportional to the distance of its line of direction from the axis, that is, proportional to CF; whence the moment of the force at the point D becomes $3f\; \backslash cdot\; \backslash frac\{CE\; \backslash cdot\; CF\}\{SC\}$.

Assume $r\; =\; SC$, and let S denote the absolute force of the sun, then $f\; =\; S\; +\; r^2$; and if $k$ be the density at D, then $k\; dm$ is the quantity of matter in the particle $dm$ at D, and the moment of the force on that particle tending to produce in the spheroid a motion about its centre in the direction QP (which is the general direction of the motion produced by the forces on all the particles) is $-\backslash frac\{3S\}\{r^3\}\; \backslash cdot\; CE\; \backslash cdot\; CF\; \backslash cdot\; k\; dm$, the integral of which must be taken for the whole spheroid. Draw DM parallel to Pp, and MN perpendicular to DF, meeting SC in O, and make $CM\; =\; X$, $DM\; =\; y$, $SCP\; =\; t$, ($t$ being the complement of the sun's declination) then $CE\; =\; DN\; +\; NF\; =\; y\; \backslash cos\; t\; +\; x\; \backslash sin\; t$, and $CF\; =\; MN\; -\; MO\; =\; y\; \backslash sin\; t\; -\; x\; \backslash cos\; t$, whence the moment of the force impressed on D becomes $\backslash frac\{3\}\{r^3\}\; k\; dm\; \backslash left\backslash \{\; (x^2\; -\; y^2)\; \backslash sin\; t\; \backslash cos\; t\; +\; xy\; (\backslash cos^2\; t\; -\; \backslash sin^2\; t)\; \backslash right\backslash \}$.

In order to find the integrals $\backslash int\; kx^2\; dm$, $\backslash int\; ky^2\; dm$, $\backslash int\; kxy\; dm$

for every particle in the spheroid, let $z$ be the co-ordinate perpendicular to the plane of the figure, and conceive the whole spheroid to be divided into an infinite number of thin slices parallel to the plane of $yz$, the thickness of each slice being $dx$; suppose, again, each slice to be divided into an infinity of parallelepipeds, parallel to the axis $z$, and terminated by the surface of the spheroid, the breadth of each being $dy$; and, lastly, let each parallelepiped be divided into an infinite number of lengths, each $=\; dz$. The element of the volume, $dm$, then becomes $dx\; dy\; dz$; and consequently the sum of $kx^2\; dm$ in respect of every particle is expressed by the triple integral $\backslash iiint\; kx^2\; dx\; dy\; dz$.

Assuming $x^2$, $dx$ and $dy$ to be constant, and integrating with respect to $z$, we obtain $kx^2\; dx\; dy\; z\; +\; \backslash text\{const\}$. Let $a$ be the semi-diameter of the equator, $b$ be the polar semi-axis, and the equation of the spheroid is $\backslash frac\{x^2\}\{a^2\}\; +\; \backslash frac\{y^2\}\{b^2\}\; +\; \backslash frac\{z^2\}\{a^2\}\; =\; 1$, from which the limits of $z$ must be found. This equation gives for a point at the surface, $z\; =\; \backslash pm\; a\; \backslash sqrt\{1\; -\; \backslash frac\{x^2\}\{a^2\}\; -\; \backslash frac\{y^2\}\{b^2\}\}$, and between those values of $z$ the definite integral becomes

This expresses the sum of $kx^2\; dm$ for the parallelepiped corresponding to a given value of $x$.

If we next suppose $x^2$ and $dx$ to be constant, and integrate this expression with respect to $y$, we shall have the sum of $kx^2\; dm$ for the slice, the distance of which from the plane $yz$ is $x$. Make $1\; -\; \backslash frac\{x^2\}\{a^2\}\; =\; \backslash frac\{u^2\}\{b^2\}$, or $u^2\; =\; b^2\; -\; \backslash frac\{b^2\; x^2\}\{a^2\}$, and the expression becomes $\backslash frac\{2\; kax^2\; dx\}\{b\}\; \backslash sqrt\{u^2\; -\; y^2\}\; dy$. In integrating this expression, the limits of $y$, in respect of any given value of $x$, are obtained from the equation of the section of the ellipsoid in the plane $xy$, namely $\backslash frac\{x^2\}\{a^2\}\; +\; \backslash frac\{y^2\}\{b^2\}\; =\; 1$. This equation gives $y^2\; =\; b^2\; -\; \backslash frac\{b^2\; x^2\}\{a^2\}$; whence, at the limits $y^2\; =\; u^2$, and therefore $y\; =\; \backslash pm\; u$. Now, by a known formula, the integral $\backslash int\; \backslash sqrt\{u^2\; -\; y^2\}\; dy$ from $y\; =\; -u$ to $y\; =\; +u$ is $\backslash frac\{1\}\{2\}\; \backslash pi\; u^2$ ($\backslash pi$ being the ratio of the circumference to the diameter); therefore the integral of $kx^2\; dm$ for the slice corresponding to a given value of $x$ is $\backslash frac\{\backslash pi\; k\; a\; u^2\; x^2\; dx\}\{b\}$; which, on substituting for $u^2$, its value $b^2\; -\; \backslash frac\{b^2\; x^2\}\{a^2\}$, becomes $\backslash frac\{\backslash pi\; k\; b\}\{a\}\; (a^2\; x^2\; -\; x^4)\; dx$.

We have, lastly, to integrate this expression from $x\; =\; -a$ to $x\; =\; +a$. The integral between those limits is $\backslash frac\{2\; \backslash pi\; k\; b\}\{a\}\; \backslash left(\; \backslash frac\{a^5\}\{3\}\; -\; \backslash frac\{a^5\}\{5\}\; \backslash right)$; whence we have ultimately, in respect of the whole spheroid, $\backslash iiint\; kx^2\; dx\; dy\; dz\; =\; \backslash frac\{4\}\{3\}\; \backslash pi\; k\; a^4\; b$.

On going through the same process for $y^2\; dm$, or $y^2\; dx\; dy\; dz$, and observing that the integral $\backslash int\; y^2\; \backslash sqrt\{u^2\; -\; y^2\}\; dy$ from $y\; =\; -u$ to $y\; =\; +u$ is $\backslash frac\{1\}\{3\}\; \backslash pi\; u^4$, there results for the whole spheroid $\backslash iiint\; ky^2\; dx\; dy\; dz\; =\; \backslash frac\{4\}\{3\}\; \backslash pi\; k\; a^4\; b^2$.

With respect to the remaining integral $\backslash int\; kxy\; dm$, it is easy to see that its value in respect of the whole spheroid must

Precession be 0, for on integrating $kxydx\; dy\; dz$ with respect to $x$, there are results $\backslash frac\{1\}\{2\}kx^2y\; dy\; dz$, which vanishes on giving $x$ all values between $-t$ and $+t$, $t$ being any definite quantity.

From these values of $\backslash int\; x^2\; dm$, $\backslash int\; y^2\; dm$, $\backslash int\; xy\; dm$, we obtain the following expression for the moment of all the forces impressed on the spheroid,

Prop. 2. To determine the efficacy of the sun's attraction to turn the spheroid about its centre, the earth being supposed heterogeneous.

Conceive the spheroid to be composed of infinitely thin concentric layers, bounded by spheroidal surfaces, and suppose the ellipticity and density to be different for each layer, but both to be functions of the distance from the centre of the spheroid. Let $a$ and $\backslash beta$ be respectively the equatorial and polar semi-diameters of the layer which contains a particle $dm$, $e$ its ellipticity, and $k$ its density. In order to find the moment of all the forces impressed on the spheroid, we must first find an expression in terms of $\backslash beta$, for their moment on the elementary layer containing $dm$, and then integrate this expression from $\backslash beta=0$ to $\backslash beta=b$. Now, suppose all the matter of the spheroid exterior to the layer in question to be removed, and suppose also for a moment the matter in the interior of the spheroidal surface passing through $dm$ to be all of the same density, $k$, then the moment of the impressed forces on the spheroid whose surface passes through $dm$, is by the last proposition

The variable part of this expression is $ka^2\; \backslash beta\; (a^2\; -\; \backslash beta^2)$. Now, since $a\; =\; \backslash beta(1+e)$, we have, on neglecting terms multiplied by $e^2$, which, by reason of the smallness of the earth's ellipticity are altogether insensible, $a^2\; =\; \backslash beta^2\; +\; 2\backslash beta^2\; e$ and $a^2\; -\; \backslash beta^2\; =\; 2\backslash beta^2\; e$, whence $ka^2\; \backslash beta\; (a^2\; -\; \backslash beta^2)\; =\; 2k\backslash beta^3\; e$. Suppose now the semi-axis $\backslash beta$ to receive an infinitely small increment, and to become $\backslash beta\; +\; d\backslash beta$; then since $k$ and $e$ are both functions of $\backslash beta$, the expression $2k\backslash beta^3\; e$ will become $2k\backslash beta^3\; e\; +\; \backslash frac\{2d(k\backslash beta^3\; e)\}\{d\backslash beta\}\; d\backslash beta$. This is the variable part of the moment of the forces on the spheroid whose semiaxis is $\backslash beta\; +\; d\backslash beta$; consequently, in respect of the spheroidal layer which remains on subtracting the spheroid whose semiaxis is $\backslash beta$, it becomes $\backslash frac\{2d(k\backslash beta^3\; e)\}\{d\backslash beta\}\; d\backslash beta$. Let the integral of this expression between the limits $\backslash beta=0$ and $\backslash beta=b$ be denoted by $2F(\backslash beta)$; then the moment of all the forces impressed on the spheroid supposed heterogeneous is

Prop. 3. To determine the angular velocity $\backslash phi$ generated by the sun's force, and the position of the equator after the infinitely small time $dt$.

By dynamics, the angular velocity of rotation is equal to the moment of the impressed forces divided by the moment of inertia of the mass to be moved. Now, the moment of inertia of a body, with respect to a given axis of rotation, is the sum of the products obtained by multiplying each particle of the body into the square of its distance from the axis; that is, in respect of the axis $z$, about which the impressed forces tend to turn the spheroid, the moment of inertia is $\backslash int\; (x^2\; +\; y^2)\; dm$, supposing the spheroid homogeneous, and the density $k=1$. But it has been shown that in this case $\backslash int\; x^2\; dm\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^2\; b$, and $\backslash int\; y^2\; dm\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^2\; b^2$; therefore the moment of inertia of the spheroid is $\backslash frac\{1\}\{2\}\; \backslash pi\; a^2\; b\; (a^2\; +\; b^2)$. And by prop. 1, the moment of the impressed forces is $\backslash frac\{3S\}\{r^3\}\; \backslash cdot\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^2\; b\; (a^2\; -\; b^2)\; \backslash sin\; \backslash delta\; \backslash cos\; \backslash delta$, therefore

In the case of the heterogeneous spheroid the moment of inertia is thus found. As before, let $a$ and $\backslash beta$ be the equatorial and polar semiaxes of the spheroidal surface passing through the particle $dm$, then if the matter within this surface be supposed of uniform density $=k$, the moment of inertia of this spheroid, by what is already shown, is $\backslash frac\{1\}\{2\}\; \backslash pi\; ka^2\; \backslash beta\; (a^2\; +\; \backslash beta^2)$. But $a^2\; +\; \backslash beta^2\; =\; 2\backslash beta^2\; +\; 2\backslash beta^2\; e$, and $a^2\; \backslash beta\; =\; \backslash beta^3\; +\; 2\backslash beta^3\; e$, therefore $a^2\; \backslash beta\; (a^2\; +\; \backslash beta^2)\; =\; 2\backslash beta^5\; +\; 6\backslash beta^5\; e$; and since $e$ is a very small quantity, the second term of this expression is very small, and may be neglected in comparison of the first; therefore $ka^2\; \backslash beta\; (a^2\; +\; \backslash beta^2)\; =\; 2k\backslash beta^5$. For the spheroid whose semiaxis is $\backslash beta\; +\; d\backslash beta$, this quantity becomes $2k\backslash beta^5\; +\; \backslash frac\{2d(k\backslash beta^5)\}\{d\backslash beta\}\; d\backslash beta$,

and therefore in respect of the elementary spheroidal layer the semiaxes of whose interior and exterior surfaces are $\backslash beta$ and $\backslash beta\; +\; d\backslash beta$, it is $\backslash frac\{2d(k\backslash beta^5)\}\{d\backslash beta\}\; d\backslash beta$. Let $2F\text{'}(\backslash beta)$ denote the integral

of this quantity from $\backslash beta=0$ to $\backslash beta=b$, that is, let $\backslash int\; \backslash frac\{d(k\backslash beta^5)\}\{d\backslash beta\}\; d\backslash beta\; =\; 2F\text{'}(\backslash beta)$, and the moment of inertia of the heterogeneous spheroid becomes $\backslash frac\{8\backslash pi\}\{15\}\; F\text{'}(\backslash beta)$. But the moment of the im-

pressed forces is by prop. 2, $\backslash frac\{3S\}\{r^3\}\; \backslash cdot\; \backslash frac\{8\backslash pi\}\{15\}\; F(\backslash beta)\; \backslash sin\; \backslash delta\; \backslash cos\; \backslash delta$, therefore in the case of the heterogeneous spheroid,

Let us assume $K\; =\; \backslash frac\{a^2\; -\; b^2\}\{a^2\; +\; b^2\}$ in the case of the homogeneous spheroid, and $K\; =\; \backslash frac\{F(\backslash beta)\}\{F\text{'}(\backslash beta)\}$ in the case of the heterogeneous spheroid, and we have for both cases

Now to find the place of the pole, and the position of the equator after the small interval of time $dt$, we must apply the first of the two theorems above premised. If the earth had no diurnal rotation, the sun's force would cause it to revolve about an axis passing through $C$ perpendicular to the meridian $PQpq$, or perpendicular to the plane of the paper, so as to bring the point $Q$ nearer to the line $SC$, with a velocity $=\backslash phi$. But the earth is already revolving about the axis $Pp$, with a velocity $=v$, and in the direction which raises the point $q$ above the plane of the paper. Hence, by theorem 1, the new axis of rotation will be in the plane passing through $Pp$, perpendicular to the plane of the paper, and after the time $dt$ will make with $Pp$ an angle whose tangent

$=\backslash frac\{\backslash phi\}\{v\}$, the pole $P$ rising above the plane of the paper; and the new equator will intersect the former in the line $Qq$, and make with it an angle whose tangent is also $\backslash frac\{\backslash phi\}\{v\}$. The

effect of the compound motion is thus to twist as it were the equator about the line $Qq$ as an axis, or about that diameter of the equator which lies in the same meridian with the sun, instead of twisting it about the diameter perpendicular to that meridian, as would be the case if the earth had no diurnal motion.

Prop. 4. To find the amount of the solar precession after any given time.

Let $AB$ be the intersection of the plane of the ecliptic, with the surface of sphere whose centre is at the centre of

Precession of the Equinoxes. Precession the earth; let ACB be the equator, and S the projection of the sun's place in the ecliptic. Now, the plane of the

equator being always perpendicular to the axis of rotation, when the axis changes its position the equator will also change its position, and the new equator will intersect the former in an angle equal to the deviation of the axis. Let A'CB' be the new position of the equator, after the infinitely small time $dt$, intersecting the former in C, then the angle ACA' is the measure of the momentary deviation of the axis, and AA', which is the amount of variation in the place of the node, is the precession in the time $dt$.

From the known properties of spherical triangles we have $\backslash sin\; SAC\; :\; \backslash sin\; ACA\text{'}\; ::\; \backslash sin\; A\text{'}C\; :\; \backslash sin\; AA\text{'}$. But because $ACA\text{'}$ is a very small angle, and $AA\text{'}$ a very small arc, the arcs may be taken instead of the sines; whence, since

therefore, as the small arc may be substituted for its tangent,

It will now be convenient to express this value of $AA\text{'}$ in terms of the sun's longitude and the obliquity of the ecliptic. Since the equator, as was shown in the last proposition, is twisted about the diameter which is in the same meridian with the sun, it follows that the line joining S and C is a part of the meridian; whence ACS is a right angle, and SC (the sun's declination) = $90^\backslash circ\; -\; \backslash theta$. Let $l\; =\; AS$, (the sun's longitude), and $I\; =\; SAC$, (the obliquity of the ecliptic), then, in the right angled spherical triangle SAC, we have $\backslash cos\; SC\; \backslash cdot\; \backslash cos\; AC\; =\; \backslash cos\; AS$, or $\backslash sin\; \backslash theta\; \backslash cos\; AC\; =\; \backslash cos\; l$; and $\backslash sin\; SC\; =\; \backslash sin\; SAC\; \backslash sin\; AS$, or $\backslash cos\; \backslash theta\; =\; \backslash sin\; I\; \backslash sin\; l$, therefore

Again, in the same triangle we have $\backslash tan\; AC\; =\; \backslash cos\; SAC\; \backslash tan\; AS$, whence $\backslash sin\; AC\; =\; \backslash cos\; AC$

Substituting these values of $\backslash sin\; \backslash theta\; \backslash cos\; \backslash theta$ and $\backslash frac\{\backslash sin\; AC\}\{\backslash cos\; AC\}$ in the above value of $AA\text{'}$, we get

This is the solar precession in the element of time $dt$; consequently for a given time $T$ we have

the integral being taken from $t=0$ to $t=T$.

To prepare this expression for integration, $r^3$ and $dt$ must be expressed in terms of $l$ and known quantities. Let $a$ = semiaxis major of the earth's orbit, $e$ = its eccentricity, $T$ = a sidereal year, then $r$ being the radius vector, and $dl$ the angle described in the time $dt$, $\backslash frac\{1\}\{2\}\; r^2\; dl$ is the space passed over by the radius vector in the element of the time, and by Kepler's law of the equable description of areas, we have $dt\; :\; T\; ::\; \backslash frac\{1\}\{2\}\; r^2\; dl\; :\; \backslash text\{area\; of\; orbit\}$.

Now the area of the orbit is $\backslash pi\; a^2\; \backslash sqrt\{1-e^2\}$, therefore this pro-

But it is shown in the article ASTRONOMY, part iii. art. 10, that $T\; =\; 2\backslash pi\; \backslash sqrt\{a^3\; /\; \backslash mu\}$, where $\backslash mu$ is the attracting force at the mean distance $a$. But we have assumed $S$ to denote the sun's force at the unit of distance; therefore, the forces being inversely as the

Again, assuming $\backslash lambda$ = longitude of sun's perigee, the polar equation of the ellipse gives $r\; =$

The expression under the sign of integration consists of two parts, of which the first $\backslash int\; \backslash sin^2\; l\; dl\; =\; \backslash frac\{1\}\{2\}(C+l\; -\; \backslash cos\; l\; \backslash sin\; l)$. (See FLUXIONS, art. 153.) The second part, namely $\backslash int\; e\; \backslash sin^2\; l\; \backslash cos(l-\backslash lambda)\; dl$, when integrated becomes $e\; \backslash sin(l-\backslash lambda)\; -$

The first term of this expression, which depends upon $C+l$, or on the sun's longitude, is the constant or uniform precession. Its amount in one year is found by supposing

The second is periodic, and being proportional to twice the sine of the sun's longitude, it runs through its changes in half a year. It is usually regarded as a part of solar nutation, and called the solar equation of the equinoxes in longitude.

Prop. 5. To find the diminution of the obliquity of the ecliptic produced by the sun's attraction.

Referring to the last diagram, make $AE\; =\; EB\; =\; 90^\backslash circ$, and let $ED$ be perpendicular to $AB$ the ecliptic, and meet $A\text{'}CB\text{'}$ in $E\text{'}$, then $EE\text{'}$ is the small change in the inclination in the time $dt$. In the triangle $ECE\text{'}$, we have $\backslash sin\; EE\text{'}\; =\; \backslash sin\; CE\; \backslash sin\; ECE\text{'}$. But $EE\text{'}$ being very small, the arc may be taken for the sine, and therefore $EE\text{'}\; =\; \backslash sin\; CE\; \backslash sin\; ECE\text{'}$.

Multiplying by $dt$, and making the same substitutions as in the last proposition, we have

Precession whence neglecting as before terms multiplied by $\backslash epsilon$, and in-
of the tegrating, there results
Equinoxes.

for the solar nutation in obliquity. This expression, de-
pendent on twice the cosine of the sun's longitude, runs
through all its changes in half a year; but its greatest value
amounts to scarcely half a second, and is consequently al-
together insensible to observation.

Prop. 6. To investigate the precessional motion of the
equinoxes produced by the moon. Let

$M$ = moon's mass,
$E$ = earth's mass,
$I$ = inclination of moon's orbit to the equator,
$r$ = moon's distance from the intersection of her orbit with
the equator,
$r\text{'}$ = radius vector of the moon's orbit,
$a\text{'}$ = semitransverse axis of moon's orbit,
$e\text{'}$ = eccentricity of the lunar orbit,
$T\text{'}$ = sidereal time of revolution.

Then, by following exactly the same reasoning as was pur-
sued in prop. 4, there results for the retrograde motion of
the points in which the plane of the lunar orbit intersects
the plane of the equator (corresponding to the solar pre-
cession in prop. 4.), the expression

Now, we have $dt\; =\; \backslash frac\{T\text{'}\; r\text{'}^2\; d\backslash theta\}\{2\backslash pi\; a\text{'}\; \backslash sqrt\{1-e\text{'}^2\}\}$, and $r\text{'}\; =\; \backslash frac\{a\text{'}(1-e\text{'}^2)\}\{1+e\text{'}\; \backslash cos(\backslash theta\text{'}-\backslash theta)\}$

therefore $\backslash frac\{dt\}\{r\text{'}^2\}\; =\; \backslash frac\{T\text{'}\; \backslash \{1+e\text{'}\; \backslash cos(\backslash theta\text{'}-\backslash theta)\backslash \}\; d\backslash theta\}\{2\backslash pi\; a\text{'}\; \backslash sqrt\{1-e\text{'}^2\}\}$. But in the pre-

sent case $T\text{'}\; =\; \backslash frac\{2\backslash pi\; a\text{'}^\{3/2\}\}\{\backslash sqrt\{(E+M)\}\}$ (for the mass of the moon cannot

be neglected in comparison of that of the earth, as the mass
of the earth is neglected in comparison of that of the sun),

hence $\backslash frac\{1\}\{a\text{'}^2\}\; =\; \backslash frac\{4\backslash pi^2\}\{T\text{'}^2(E+M)\}$. The above expression therefore
becomes

the integral of which (rejecting, as before, terms multiplied
by $e\text{'}$) gives

for the regression of the equator on the plane of the lunar
orbit. Suppose $\backslash theta\text{'}$ to be increased by $2\backslash pi$, or a whole circum-
ference, the regression caused by the moon's action in a

sidereal revolution becomes $\backslash frac\{6\backslash pi^2\; M\; \backslash cdot\; K\; \backslash cos\; I\}\{T\text{'}(E+M)\}$.

It is now necessary to reduce this retrograde motion to
the plane of the ecliptic.

Let $AB$ be the ecliptic, $ACB$ the equator, $A\text{'}CB\text{'}$ the

new position of the equator, and $FH$ the plane of the lunar
orbit intersecting the new equator in $F\text{'}$ and $H\text{'}$, and the
ecliptic in $N$. The mean effect of the moon's action in the
course of a month, if the earth had no motion of rotation,
would be to bring the equator nearer the plane of the lunar

orbit, causing it to revolve about the line of its intersection
with the lunar orbit; therefore by Theorem I., the momen-
tary axis of rotation lies in the plane passing through that
line and the pole of the equator, and the equator is conse-
quently twisted about the equatorial diameter which is per-
pendicular to the intersection of the equator and lunar orbit.
Hence, $FC\; =\; CH$ and $F\text{'}C\; =\; CH\text{'}$; and since $FC\; +\; CH\; =\; 180^\backslash circ$,
therefore $FC$ and $F\text{'}C$ are quadrantal arcs, and the two tri-
angles $CFF\text{'}$ and $CHH\text{'}$ are in all respects equal. Now

that is, $\backslash sin\; FC\; :\; CF\text{'}F\; ::\; \backslash sin\; FF\text{'}\; :\; ACA\text{'}$,
and $\backslash sin\; CA\text{'}A\; :\; \backslash sin\; AC\; ::\; \backslash sin\; ACA\text{'}\; :\; \backslash sin\; AA\text{'}$,
that is, $\backslash sin\; I\; :\; \backslash cos\; AF\; ::\; ACA\text{'}\; :\; AA\text{'}$,
therefore, $\backslash sin\; I\; :\; \backslash sin\; I\; \backslash cos\; AF\; ::\; FF\text{'}\; :\; AA\text{'}$,
and consequently,

But $AA\text{'}$ represents the velocity along the ecliptic, and
$FF\text{'}$ the velocity along the plane of the moon's orbit,
and we have seen that the motion along this plane is
$6\backslash pi^2\; K\; \backslash cdot\; M\; \backslash cos\; I$, in the time $T\text{'}$, or a sidereal revolution. Di-

viding this by $T\text{'}$, we get the mean velocity in the plane of
the orbit in the unit of time; whence $FF\text{'}\; =\; \backslash frac\{6\backslash pi^2\; K\; \backslash cdot\; M\; \backslash cos\; I\}\{T\text{'}(E+M)\}$.

For the sake of brevity let $Q\; =\; \backslash frac\{6\backslash pi^2\; K\; \backslash cdot\; M\}\{T\text{'}(E+M)\}$, then $FF\text{'}\; =\; Q$
$\backslash cos\; I$, and we have

We must now express $\backslash cos\; I\; \backslash sin\; I\; \backslash cos\; AF$, in terms of
the obliquity and inclination of the lunar orbit to the eclip-
tic. Let $i\; =\; ANF\; =$ inclination of moon's orbit to the
ecliptic, $n\; =\; NA$, the longitude of the node; then in the
triangle $ANF$, we have by spherical trigonometry,
$\backslash cos\; ANF\; =\; \backslash cos\; NAF\; \backslash cos\; NFA\; +\; \backslash sin\; NAF\; \backslash sin\; NFA\; \backslash cos\; AF$,
that is, since $NAF\; =\; I$, $NFA\; =\; i$,

$\backslash cos\; i\; =\; \backslash cos\; I\; \backslash cos\; i\; +\; \backslash sin\; I\; \backslash sin\; i\; \backslash cos\; AF$,
In like manner, in the same triangle,
$\backslash cos\; I\; =\; \backslash cos\; I\; \backslash cos\; i\; +\; \backslash sin\; I\; \backslash sin\; i\; \backslash cos\; n$.
From these two equations we obtain this other,
$\backslash cos\; I\; \backslash sin\; I\; \backslash cos\; AF\; =\; \backslash cos\; I\; \backslash sin\; I\; \backslash cos\; i$

$-(\backslash cos^2\; I\; -\; \backslash sin^2\; I)\; \backslash cos\; i\; \backslash sin\; i\; \backslash cos\; n$
$-\backslash cos\; I\; \backslash sin\; I\; \backslash sin^2\; i\; \backslash cos^2\; n$,
which, on substituting in it $\backslash cos\; 2I$ for $\backslash cos^2\; I\; -\; \backslash sin^2\; I$,
$\backslash frac\{1\}\{2\}\; \backslash sin\; 2i$ for $\backslash cos\; i\; \backslash sin\; i$, and $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{2\}\; \backslash cos\; 2n$ for $\backslash cos^2\; n$, becomes
$\backslash cos\; I\; \backslash sin\; I\; \backslash cos\; AF\; =\; \backslash cos\; I\; \backslash sin\; I\; (\backslash cos^2\; i\; -\; \backslash frac\{1\}\{2\}\; \backslash sin^2\; i)$
$-\backslash frac\{1\}\{2\}\; \backslash cos\; 2I\; \backslash sin\; 2i\; \backslash cos\; n$
$-\backslash frac\{1\}\{2\}\; \backslash cos\; I\; \backslash sin\; I\; \backslash sin^2\; i\; \backslash cos\; 2n$.

Assuming $i$ (the inclination of the moon's orbit) to be con-
stant, which may be done in the present case without sen-
sible error, the only variable in this expression is $n$, (the
longitude of the node), which, on the supposition of $i$ con-
stant, is proportional to the time. Let $\backslash tau$ denote the time
of a revolution of the node, then for any time $t$, we have

$n\; =\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$. On making this substitution in the last equation,
we obtain by means of it

Now, if we assume $y$ = the regression of the equinoctial
points on the ecliptic caused by the lunar action, and sup-

Precession of the Equinoxes. pose $y$ a function of $t$, then $AA\text{'}\; =\; \backslash frac\{dy\}\{dt\}$, and the amount of

this regression in a given time $=\; \backslash int\; \backslash frac\{dy\}\{dt\}\; dt$. Multiplying

therefore the right hand side of the last equation by $dt$, and integrating, (observing that $\backslash int\; \backslash cos\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}\; dt\; =\; \backslash frac\{\backslash tau\}\{2\backslash pi\}\; \backslash sin\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$,

and $\backslash int\; \backslash cos\; \backslash frac\{4\backslash pi\; t\}\{\backslash tau\}\; dt\; =\; \backslash frac\{\backslash tau\}\{4\backslash pi\}\; \backslash sin\; \backslash frac\{4\backslash pi\; t\}\{\backslash tau\}$), we obtain, finally, for the regression of the equinoctial points, or the precessional motion of the equinoxes, produced by the moon in the time $t$,

The first term of this expression increases uniformly with the time, and is called the lunar precession. The second term, being multiplied by $\backslash sin\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$, is periodic, and depends on the mean longitude of the moon's ascending node. It is called the lunar equation of the equinoxes in longitude. The third term is also periodic, but its numerical value is so small as to be insensible, and it is therefore omitted in the calculation. The lunar precession in a sidereal year is found by substituting $T$ for $t$ in the first term, and we have, therefore,

Prop. 7. To find the diminution in the inclination of the equator to the ecliptic produced by the moon's action.

Let $AB$ be bisected in $D$, and let $DE$ be an arc perpendicular to $AB$, and meeting $A\text{'}B\text{'}$ in $E\text{'}$; then the alteration of obliquity produced by the moon in the time $dt$ is represented by $EE\text{'}$. Now in the triangle $CFE\text{'}$ we have

and in the triangle $CEE\text{'}$, $\backslash sin\; CE\text{'}E\text{'}\; :\; \backslash sin\; CE\text{'}\; ::\; \backslash sin\; ECE\text{'}\; :\; \backslash sin\; EE\text{'}$ that is, by reason of $CE\; =\; AF$ (since $FC\; =\; AE\; =\; 90^\backslash circ$) and $ECE\text{'}\; =\; FCF\text{'}$

whence $EE\text{'}\; =\; FF\text{'}\; \backslash sin\; I\text{'}\; \backslash sin\; AF$. Now it was shewn in the last proposition that $FF\text{'}\; =\; Q\; \backslash cos\; I$, therefore

To refer this to the ecliptic we have in the triangle $NAF$, $\backslash sin\; NFA\; :\; \backslash sin\; ANF\; ::\; \backslash sin\; AN\; :\; \backslash sin\; AF$

and, as before we have $\backslash cos\; I\text{'}\; =\; \backslash cos\; I\; \backslash cos\; i\; +\; \backslash sin\; I\; \backslash sin\; i\; \backslash cos\; n$;

whence multiplying the two equations together, and substituting $\backslash frac\{1\}\{2\}\; \backslash sin\; 2i$ for $\backslash cos\; i\; \backslash sin\; i$, and $\backslash frac\{1\}\{2\}\; \backslash sin\; 2n$ for $\backslash cos\; n\; \backslash sin\; n$, we get

and, consequently, writing for $n$ its value $\backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$,

Multiplying this by $dt$ and integrating, we obtain for the diminution of the inclination, or lunar nutation in obliquity

Both terms of this expression are periodic; but the second is omitted in the calculation, as being too small to be sensible. Hence we have

lunar nutation in obliquity $=\; -Q\; \backslash cdot\; \backslash frac\{\backslash tau\}\{4\backslash pi\}\; \backslash cos\; I\; \backslash sin\; 2i\; \backslash cos\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$. Precession of the Equinoxes.

Prop. 8. To compute the numerical value of the annual precession.

By prop. 4. the solar precession in one year $=\; \backslash frac\{6\backslash pi^2\; K\; \backslash cos\; I\}\{T\text{'}e\}$.

Now $T\text{'}e$ is the angle which any point of the earth describes about its axis of rotation in a sidereal year, or 366.26 days, and consequently $=\; 2\backslash pi\; \backslash times\; 366.26$. The solar precession

therefore becomes $\backslash frac\{3\backslash pi\; K\; \backslash cos\; (23^\backslash circ\; 28\text{'})\}\{366.26\}$ expressed in parts of the radius. To reduce it to seconds, we have, assuming radius $=\; 1$, $\backslash pi\; =\; 180^\backslash circ\; =\; 180\; \backslash times\; 60\; \backslash times\; 60\; =\; 648000$ seconds. Substituting this for $\backslash pi$, and computing the above expression by the logarithmic tables we get

solar annual precession $=\; K\; \backslash times\; 4869\text{'}\text{'}$.

By prop. 6, the lunar precession in a sidereal year (on substituting for $Q$ its value) is $\backslash frac\{6\backslash pi^2\; K\; \backslash cdot\; M\}\{T\text{'}^2\; v\; (E+M)\}\; \backslash cos\; I\; (\backslash cos^2\; i\; -\; \backslash frac\{1\}\{2\}\; \backslash sin^2\; i)\; T$. But $T\text{'}e$ is the angle described by the diurnal rotation of the earth in one sidereal revolution of the moon, or 27.32 days (ASTRONOMY, vol. iv.), and therefore $=\; 2\backslash pi\; \backslash times\; 27.32$. We have also $i\; =\; 5^\backslash circ\; 8\text{'}\; 47\text{'}\text{'}.9$, and, as before, $T\; =\; 366.26$ days. Now assuming the moon's mass $=\; 1/70$th of the earth's mass, $M\; +\; (E+M)\; =\; \backslash frac\{1\}\{71\}$. By the substitution of these numbers, the lunar annual precession becomes

the calculation of which, reduced to seconds as before, gives

Adding this to the solar annual precession, we obtain the effect produced by the joint action of the sun and moon, or

It is now necessary to assign a value to the quantity $K$ which depends on the law of the density of the earth. Supposing the earth homogeneous, we have $K\; =\; \backslash frac\{a^2\; -\; b^2\}\{a^2\; +\; b^2\}\; =\; \backslash frac\{e\}\{1-e\}$, $e$ being

the ellipticity $=\; \backslash frac\{1\}\{301\}$, (FIGURE OF THE EARTH,)

whence $K\; =\; \backslash frac\{1\}\{300\}$. This value of $K$ gives

The observed quantity is only $50\text{'}\text{'}.4$; the difference being occasioned chiefly by the erroneous assumption of the homogeneity of the earth. If the earth be denser towards the centre, (and it is known to be so from other phenomena), the momentum of the protuberant parts will not be so great as if it were equally dense with the interior parts, and the precession will be less. From Cavendish's experiment, and experiments on the attraction of mountains, it has been ascertained that the mean density of the whole earth is about five times greater than that of water, and twice as great as that of the solid substances composing its exterior crust. But we are entirely ignorant of the law according to which the density varies from the surface towards the centre; and an infinity of hypotheses may be made which would give the observed precession, and at the same time satisfy the condition of a superficial density equal to half the mean density.

Prop. 9. To compute the numerical value of the solar and lunar nutation.

The solar nutation consists of two parts. The first is the solar equation of the equinoxes in longitude, or the second term of the expression for the solar precession in prop. 4,

Precession
of the
Equinoxes.

its value is $\backslash frac\{3\backslash pi\; K\}\{T\_e\}\; \backslash cos\; I\; \backslash times\; \backslash frac\{1\}\{2\}\; \backslash sin\; 2I$. Substituting $2\backslash pi\; \backslash times\; 365.25$ for $T\_e$, and multiplying by $\backslash frac\{180\; \backslash times\; 60\; \backslash times\; 60\}\{\backslash pi\}$ to reduce to seconds, the computation gives

1st part of solar nutation $=\; K\; \backslash times\; 387\text{'}\text{'}.5\; \backslash times\; \backslash sin\; 2I$.
The second part is the nutation in obliquity, found by prop. 5, the value of which is $\backslash frac\{3\backslash pi\; K\}\{T\_e\}\; \backslash sin\; I\; \backslash times\; \backslash frac\{1\}\{2\}\; \backslash cos\; 2I$. This being computed in the same manner as the last gives

2d part of solar nutation $=\; K\; \backslash times\; 168\text{'}\text{'}.3\; \backslash times\; \backslash cos\; 2I$.
Assuming the earth to be homogeneous, and consequently $K\; =\; \backslash frac\{1\}\{300\}$, we have for the sum of the two parts

solar nutation $=\; 1\text{'}\text{'}.29\; \backslash sin\; 2I\; +\; 0\text{'}\text{'}.56\; \backslash cos\; 2I$,
both terms being so small as to be insensible to observation.

The lunar nutation is also composed of two parts; the first being the lunar equation of the equinoxes in longitude, or the second term of the expression in prop. 6; and the second the nutation in obliquity found in prop. 7. By prop. 6, the first of these parts is

which, since $\backslash tau\; =\; 18.6\; \backslash times\; 365.25$ days, becomes, on substituting for the different quantities their numerical values, and reducing to seconds,

whence there is found from computation,

By prop. 7, the lunar nutation in obliquity becomes, on substituting for $Q$ its value, and neglecting the sign,

Comparing the coefficient of $\backslash cos\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$ in this expression with

that of $\backslash sin\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$ in the above, it is obvious that the latter is found by multiplying the former by $\backslash frac\{\backslash cos\; I\; \backslash sin\; I\}\{\backslash cos\; 2I\}\; =\; \backslash frac\{\backslash sin\; 2I\}\{2\; \backslash cos\; 2I\}\; =\; \backslash frac\{1\}\{2\}\; \backslash tan\; 2I\; =\; \backslash frac\{1\}\{2\}\; \backslash tan\; (46^\backslash circ\; 56\text{'})$. The multiplication gives

Assuming that the earth is homogeneous, and consequently $K\; =\; \backslash frac\{1\}\{300\}$, these two parts added together give

The observed values of the coefficients are $18\text{'}\text{'}.36$ and $9\text{'}\text{'}.239$, the differences between the observed and computed values, as in the case of the precession, arising from the assumption of the uniform density of the earth.

Prop. 10. To determine the motion of the pole of the earth's axis of rotation.

As the inclination of the equator to the ecliptic undergoes no permanent alteration in consequence of the action of the sun and moon, and as the precessional motion of the equinoxes is proportional to the time, it follows that, abstracting the effects of lunar and solar nutation, the pole of the equator must describe a circle about the pole of the ecliptic, the plane of which is parallel to the ecliptic, and of which the radius is equal to the sine of the obliquity, or $=\; \backslash sin$

($23^\backslash circ\; 28\text{'}$). The mean velocity corresponding to the regression of the equinoctial points, is $50\text{'}\text{'}.4$ in a year, and consequently the period of a revolution is about 25900 years. Now in order to take account of the lunar nutation (the solar, as has already been remarked, is scarcely sensible), it is only necessary to remark that the absolute velocity of the pole in its small circle, is to the velocity with which the equinoctial points regress in the ecliptic, as the radius of the small circle to the radius of the ecliptic, or as $\backslash sin\; I\; :\; 1$. Hence the motion of the pole in the plane of the small circle, is obtained by multiplying the expression in prop. 6, by $\backslash sin\; I$; and therefore the correction to be applied to the uniform motion of the pole, is the second term of that expression, multiplied by $\backslash sin\; I$, or $\backslash sin\; I\; \backslash times$ lunar equation of the equinoxes in longitude. But by prop. 9, this term $=\; K\; \backslash times\; 6093\text{'}\text{'}\; \backslash sin\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$, therefore the corresponding motion of the pole $=$

With respect to the second part of the nutation, it is obvious that any change of obliquity produces an equal change in the place of the pole on the meridian, and therefore, by the last proposition, the motion of the pole in this direction is $K\; \backslash times\; 3260\text{'}\text{'}\; \backslash cos\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$. Now let $K\; \backslash times\; 2427\text{'}\text{'}\; =\; a$, and $K\; \backslash times\; 3260\text{'}\text{'}\; =\; b$, and let the motion of the pole in the two directions be respectively denoted by $x$ and $y$, we have then $x\; =\; a\; \backslash sin\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$, $y\; =\; b\; \backslash cos\; \backslash frac\{2\backslash pi\; t\}\{\backslash tau\}$; and consequently the equation

which is the equation to an ellipse, and shews that the pole describes a small ellipse about its mean place in the course of a revolution of the nodes, as was discovered by Bradley from observation.

Prop. 11. From the observed values of the precession and nutation, to determine the ratio of the moon's mass to the mass of the earth.

The quantities most accurately determined from observation, from which the moon's mass can be found, are the whole annual precession $p\; (=\; 50\text{'}\text{'}.4)$, and the coefficient of the lunar nutation in obliquity $q\; (=\; 9\text{'}\text{'}.239)$. Let $s$ be the solar annual precession, and $m$ be the lunar annual precession, then $s\; =\; p\; -\; m$. Now, by prop. 7, $q\; =\; Q\; \backslash frac\{\backslash tau\}\{4\backslash pi\}\; \backslash cos\; I\; \backslash sin\; 2i$, and, by prop. 6, $m\; =\; Q\; \backslash cos\; I\; (\backslash cos^2\; i\; -\; \backslash frac\{1\}\{2\}\; \backslash sin^2\; i)\; T$, whence by eliminating $Q\; \backslash cos\; I$, we get

and on computing $m$ from the values of $T$, $\backslash tau$ and $i$ above given, we find $m\; =\; q\; \backslash times\; 3\text{'}\text{'}.735\; =\; 34\text{'}\text{'}.51$; therefore, also, $s\; =\; p\; -\; m\; =\; 15\text{'}\text{'}.89$.

Again, from prop. 4, we have $s\; =\; \backslash frac\{5\backslash pi^2\; \backslash cdot\; K\; \backslash cos\; I\}\{T\_e\}$, and from

prop. 6, (on substituting for $Q$ its value), $m\; =\; \backslash frac\{6\backslash pi^2\; \backslash cdot\; K\; \backslash cdot\; M\; \backslash cos\; I\}\{T^2\; e(E+M)\}$

By computing the coefficient of $(E+M)\; \backslash div\; M$, and by means of the values $s$ and $m$ now found, this equation gives

whence it follows that the mass of the moon is to the mass of the earth in the ratio of 1 to 80.755.

The phenomena of the precession and nutation, and those