In the article Astronomy (vol. vii, p. 767), 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 show 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 of 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 produce 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 lunar 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, while 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 \( \omega \), receive an impulse which alone would cause it to revolve about an axis Bb, also passing through its centre of gravity, with a velocity \( \phi \), 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 \( \sin AOC : \sin BOC :: \phi : \omega \).

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 \( \sin BOC = \cos AOC \), and we have \( \frac{\sin AOC}{\cos AOC} = \frac{\phi}{\omega} \); that is, \( \tan AOC = \frac{\phi}{\omega} \).

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 \( \omega \) 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

Precession is the projection of a point in the interior of the spheroid on the plane PQ. 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 \( f' \) the force of its attraction on a particle at D, then the attraction being inversely as the square of the distance, we have \( f' = \frac{f}{SD^2} \). Now the force \( f' \), 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 PQ.

The resolved part of the force \( f' \), in the direction FD is \( f' \cdot \frac{SC}{SD^2} \cdot \frac{SC}{SD} \), 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 \( f' \cdot \left( \frac{SC}{SD} - 1 \right) = f' \cdot \frac{SC - SD}{SD} = f' \cdot \frac{(SC - SD)(SC + SC + SD)}{SD} \).

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 SC² which are so small as to be altogether insensible, the factor \( \frac{SC^2 + SC \cdot SD + SD^2}{SD^3} \) becomes \( \frac{3}{SD} \), and consequently the above expression is reduced to \( 3f' \cdot \frac{CE}{SC} \), or to \( 3f' \cdot \frac{CE}{SC} \), since the difference between 1 - SD and 1 - SC is a quantity divided by SC², 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 \( 3f' \cdot \frac{CE \cdot CF}{SC} \).

Assume \( r = SC \), and let S denote the absolute force of the sun, then \( f = S + r^2 \); and if \( k = \text{density at D} \), then \( kdm \) 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 \( \frac{3S}{r^2} \cdot \frac{CE \cdot CF}{SC} \cdot kdm \), 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 cos t + x sin t, and CF = MN - MO = y sin t - x cos t, whence the moment of the force impressed on D becomes

\[ \frac{3}{r^2} \cdot \frac{kdm}{SC} \left[ (x^2 - y^2) \sin t \cos t + xy (\cos^2 t - \sin^2 t) \right]. \]

In order to find the integrals \( \int kx^2 dm \), \( \int ky^2 dm \), \( \int kxy dm \) for every particle in the spheroid, let \( z \) be the co-ordinate Precession 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 \( hx^2 dm \) in respect of every particle is expressed by the triple integral \( \int \int \int hx^2 dx dy dz \).

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

\[ 2kax^2 dx dy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}. \]

This expresses the sum of \( hx^2 dm \) 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 \( hx^2 dm \) for the slice, the distance of which from the plane \( yz \) is \( x \). Make \( 1 - \frac{x^2}{a^2} = \frac{u^2}{b^2} \) or \( u^2 = b^2 - \frac{b^2 x^2}{a^2} \), and the expression becomes \( \frac{2kax^2 dx}{b} \sqrt{u^2 - y^2} dy \). 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 \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This equation gives \( y^2 = b^2 - \frac{b^2 x^2}{a^2} \); whence, at the limits \( y^2 = u^2 \), and therefore \( y = \pm u \). Now, by a known formula, the integral \( \int \sqrt{u^2 - y^2} dy \) from \( y = -u \) to \( y = +u \) is \( \frac{1}{2} \pi u^2 \) (\( \pi \) being the ratio of the circumference to the diameter); therefore the integral of \( x^2 dm \) for the slice corresponding to a given value of \( x \) is \( \frac{\pi ka^2 x^2 dx}{b} \); which, on substituting for \( u^2 \), its value \( b^2 - \frac{b^2 x^2}{a^2} \), becomes

\[ \frac{\pi ka^2 x^2 dx}{b}. \]

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

\[ \int \int \int hx^2 dx dy dz = \frac{4}{3} \pi ka^2 b^3. \]

On going through the same process for \( y^2 dm \), or \( \int ky^2 dm \), and observing that the integral \( \int \sqrt{u^2 - y^2} dy \) from \( y = -u \) to \( y = +u \) is \( \frac{1}{2} \pi u^2 \), there results for the whole spheroid

\[ \int \int \int ky^2 dx dy dz = \frac{4}{3} \pi ka^2 b^3. \]

With respect to the remaining integral \( \int kxy dm \), it is easy to see that its value in respect of the whole spheroid must Precession be 0, for on integrating \( \int x y d x d y d z \) with respect to \( x \), there results \( \frac{1}{2} k x^2 y d y d z \), which vanishes on giving \( x \) all values between \(-t\) and \(+t\), \( t \) being any definite quantity.

From these values of \( \int x^2 d m \), \( \int y^2 d m \), \( \int z^2 d m \), we obtain the following expression for the moment of all the forces impressed on the spheroid,

\[ \frac{3S}{r^3} \cdot \frac{4\pi}{15} ka^2 b (a^2 - b^2) \sin \theta \cos \delta. \]

**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 \( b \) 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 \( \beta \), for their moment on the elementary layer containing \( dm \), and then integrate this expression from \( \beta = 0 \) to \( \beta = b \). 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

\[ \frac{3S}{r^3} \cdot \frac{4\pi}{15} ka^2 b (a^2 - b^2) \sin \theta \cos \delta. \]

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

\[ \frac{3S}{r^3} \cdot \frac{8\pi}{15} F(\beta) \sin \theta \cos \delta. \]

**Prop. 3.** To determine the angular velocity \( \phi \) 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 \( \int (x^2 + y^2) dm \), supposing the spheroid homogeneous, and the density \( k = 1 \). But it has been shewn that in this case \( \int x^2 dm = \frac{4}{3} \pi a^2 b \), and \( \int y^2 dm = \frac{4}{3} \pi a^2 b \); therefore the moment of inertia of the spheroid is \( \frac{4}{3} \pi a^2 b (a^2 + b^2) \).

And by prop. 1, the moment of the impressed forces is

\[ \frac{3S}{r^3} \cdot \frac{4\pi}{15} ka^2 b (a^2 - b^2) \sin \theta \cos \delta, \]

therefore

\[ \phi = \frac{3S}{r^3} \cdot \frac{a^2 - b^2}{a^2 + b^2} \sin \theta \cos \delta. \]

In the case of the heterogeneous spheroid the moment Precession of inertia is thus found. As before, let \( a \) and \( b \) 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 shewn, is \( \frac{4}{3} \pi k a^2 b (a^2 + b^2) \). But \( a^2 + b^2 = 2a^2 + 2b^2e \), and \( a^2b^2 = b^2 + 2b^2e \); therefore \( a^2b^2(a^2 + b^2) = 2b^2 + 6b^2e \); 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 \( ka^2b(a^2 + b^2) = 2k b^3 \). For the spheroid whose semiaxis is \( b + db \), this quantity becomes \( 2k b^3 + \frac{d(k b^3)}{d\beta} db \), and therefore in respect of the elementary spheroidal layer the semiaxes of whose interior and exterior surfaces are \( b \) and \( b + db \), it is \( \frac{d(k b^3)}{d\beta} db \). Let \( 2F(\beta) \) denote the integral of this quantity from \( \beta = 0 \) to \( \beta = b \), that is, let \( \int \frac{d(k b^3)}{d\beta} db = 2F(\beta) \), and the moment of inertia of the heterogeneous spheroid becomes \( \frac{8\pi}{15} F(\beta) \). But the moment of the impressed forces is by prop. 2,

\[ \frac{3S}{r^3} \cdot \frac{8\pi}{15} F(\beta) \sin \theta \cos \delta, \]

therefore in the case of the heterogeneous spheroid,

\[ \phi = \frac{3S}{r^3} \cdot \frac{F(\beta)}{F(\beta)} \sin \theta \cos \delta. \]

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

\[ \phi = \frac{3S}{r^3} \cdot \frac{K}{F(\beta)} \sin \theta \cos \delta. \]

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 \( PQ \), or perpendicular to the plane of the paper, so as to bring the point \( Q \) nearer to the line \( SC \), with a velocity \( = \phi \). 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 \( = \frac{\phi}{v} \), 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 \( \frac{\phi}{v} \). 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.

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 properties of spherical triangles we have

\[ \sin SAC : \sin ACA' :: \sin A'C : \sin AA'. \]

But because ACA' is a very small angle, and AA' a very small arc, the arcs may be taken instead of the sines; whence, since

\[ A'C = AC, \]

the proportion gives

\[ AA' = ACA' \frac{\sin AC}{\sin SAC}. \]

But by the last proposition tan ACA' \(= \frac{\phi}{v} = \frac{3S.K}{vr^2} \sin \theta \cos \theta,\)

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

\[ AA' = \frac{3S.K}{vr^2} \frac{\sin AC}{\sin SAC} \sin \theta \cos \theta. \]

It will now be convenient to express this value of AA' 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) \(= 90^\circ - \delta.\) Let I = AS, (the sun's longitude), and I = SAC, (the obliquity of the ecliptic), then, in the right angled spherical triangle SAC, we have

\[ \cos SC \cdot \cos AC = \cos AS, \text{ or } \cos \delta \cos AC = \cos I; \]

and

\[ \sin SC = \sin SAC \sin AS, \text{ or } \cos \delta = \sin I \sin l, \]

therefore

\[ \sin \delta \cos \delta = \frac{\sin I \sin l \cos l}{\cos AC}. \]

Again, in the same triangle we have tan AC = cos SAC tan AS, whence sin AC = cos AC cos I tan l, and (dividing by sin SAC = sin I)

\[ \frac{\sin AC}{\sin SAC} = \frac{\cos AC \cos I \sin l}{\sin I \cos l}. \]

Substituting these values of sin \(\delta\) \cos \(\delta\) and \(\frac{\sin AC}{\cos AC}\) in the above value of AA', we get

\[ AA' = \frac{3S.K}{vr^2} \cos I \sin^2 l. \]

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

\[ \text{solar precession} = \frac{3S.K}{vr^2} \int \frac{\sin^2 l}{r^2} dt, \]

the integral being taken from \(t = 0\) to \(t = T.\)

To prepare this expression for integration, \(r^2\) 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 \(dt,\) \(r^2 dt\) 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

\[ dt : T :: \frac{1}{2} r^2 dl : \text{area of orbit}. \]

Now the area of the orbit is \(\pi a^2 \sqrt{1-e^2},\) therefore this portion gives

\[ dt = \frac{T r^2 dl}{2 \pi a^2 \sqrt{1-e^2}}. \]

But it is shewn in the article Astronomy, part iii. art. 10, that

\[ T = 2\pi \sqrt{\frac{a}{F}}, \]

where \(F\) 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 squares of the distances, \(F = S/a^2,\) whence

\[ T = \frac{2\pi a^3}{\sqrt{S}}. \]

From this formula we get

\[ S = \frac{4\pi^2 a^3}{T^2}, \]

and therefore

\[ \frac{S dt}{r^2} = \frac{2\pi adl}{Tr \sqrt{1-e^2}}. \]

Again, assuming \(l =\) longitude of sun's perigee, the polar equation of the ellipse gives

\[ r = \frac{a(1-e^2)}{1+e \cos(l-\lambda)}, \]

whence

\[ \frac{S dt}{r^2} = \frac{2\pi \{1+e \cos(l-\lambda)\} dl}{T(1-e^2)^{3/2}}, \]

and the above expression becomes

\[ \text{solar precession} = \frac{6\pi K \cos I}{Tr(1-e^2)^{3/2}} \int \sin^2 l \{1+e \cos(l-\lambda)\} dl. \]

The expression under the sign of integration consists of two parts, of which the first \(= \frac{1}{2} \sin^2 l dl = \frac{1}{2} (C+l \cos l \sin l).\) (See Fluxions, art. 153.) The second part, namely

\[ \int e \sin^2 l \cos(l-\lambda) dl, \]

when integrated becomes \(e \sin(l-\lambda) - \frac{e}{2} \sin(1+\lambda) + \frac{e}{6} \sin(3l-\lambda), \)

but by reason of the smallness of \(e\) these terms are insensible, and are therefore neglected. Rejecting also the terms in the divisor of the coefficient which are multiplied by \(e^2,\) and observing that \(\cos l \sin l = \frac{1}{2} \sin 2l,\) we obtain, finally,

\[ \text{solar precession} = \frac{3\pi K \cos I}{Tr} (C+l \frac{1}{2} \sin 2l). \]

The first term of this expression, which depends on \(C+l,\) or on the sun's longitude, is the constant or uniform precession. Its amount in one year is found by supposing \(l\) to be increased by \(2\pi,\) and is consequently

\[ \frac{6\pi K \cos I}{Tr}. \]

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°, and let ED be perpendicular to AB the ecliptic, and meet A'CB' in E', then EE' is the small change in the inclination in the time \(dt.\) In the triangle ECE', we have

\[ \sin EE' = \sin CE \sin ECE'. \]

But EE' being very small, the arc may be taken for the sine, and therefore

\[ EE' = \sin CE \sin ECE'. \]

Now \(\sin CE = \cos AC = \frac{\cos l}{\sin \delta};\) and it has been already seen that \(\sin ECE' = \sin ACA' = \frac{\phi}{v} = \frac{3S.K}{vr^2} \sin \delta \cos \delta;\) therefore

\[ EE' = \frac{3S.K}{vr^2} \cos \delta \cos l, \]

or since \(\cos \delta = \sin I \sin l,\) \(EE' = \frac{3S.K}{vr^2} \sin I \sin l \cos l.\)

Multiplying by \(dt,\) and making the same substitutions as in the last proposition, we have

\[ \int EE'dt = \frac{6\pi K}{Tr(1-e^2)^{3/2}} \sin I \int \sin l \cos l \{1+e \cos(l-\lambda)\} dl. \] Precession whence neglecting as before terms multiplied by \( \epsilon \), and integrating, there results

\[ \int EE' dt = -\frac{3\pi K}{2T} \sin I \cos 2t, \]

for the solar nutation in obliquity. This expression, depending 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 altogether 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,

\( r' = \) radius vector of the moon's orbit,

\( a' = \) semitransverse axis of moon's orbit,

\( e' = \) eccentricity of the lunar orbit,

\( T' = \) sidereal time of revolution,

then, by following exactly the same reasoning as was pursued 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 precession in prop. 4.), the expression

\[ \frac{3M.K \cos I'}{v} \int \frac{\sin \theta'}{r'^2} dt. \]

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

therefore \( \frac{dt}{r'^2} = \frac{T' \{1+e'\cos(\lambda'-\ell)\} d\theta'}{2\pi a'^2 (1-e'^2)^{3/2}} \). But in the present case \( T' = \frac{2\pi a'^2}{\sqrt{(E+M)}} \) (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 \( \frac{1}{a'^2} = \frac{4\pi^2}{T'^2(E+M)} \). The above expression therefore becomes

\[ \frac{6\pi M.K \cos I'}{T'(E+M)} \int \sin \theta' \{1+e'\cos(x-\ell')d\theta'\}, \]

the integral of which (rejecting, as before, terms multiplied by \( \epsilon' \)) gives

\[ \frac{3\pi M.K \cos I'}{T'(E+M)} (C + \ell' - \frac{1}{2} \sin 2\ell'). \]

for the regression of the equator on the plane of the lunar orbit. Suppose \( \ell' \) to be increased by \( 2\pi \), or a whole circumference, the regression caused by the moon's action in a sidereal revolution becomes \( \frac{6\pi^2 M.K \cos I'}{T'(E+M)} \).

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

Let \( AB \) be the ecliptic, \( ACB \) the equator, \( A'C'B' \) the new position of the equator, and \( FH \) the plane of the lunar orbit intersecting the new equator in \( F' \) and \( H' \), 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 momentary axis of rotation lies in the plane passing through that line and the pole of the equator, and the equator is consequently twisted about the equatorial diameter which is perpendicular to the intersection of the equator and lunar orbit. Hence, \( FC = CH \) and \( F'C = CH' \); and since \( FC + CH = 180^\circ \), therefore \( FC \) and \( F'C \) are quadrantial arcs, and the two triangles \( CFF' \) and \( CHH' \) are in all respects equal. Now

\[ \sin FC : CFF' :: \sin FF' : ACA', \]

that is,

\[ 1 : \sin I' :: FF' : ACA', \]

and

\[ \sin CA'A : \sin AC :: \sin ACA' : \sin AA', \]

that is,

\[ \sin I : \cos AF :: ACA' : AA', \]

therefore,

\[ \sin I : \sin I' \cos AF :: FF' : AA', \]

and consequently,

\[ AA' = \frac{FF' \sin I' \cos AF}{\sin I}. \]

But \( AA' \) represents the velocity along the ecliptic, and \( FF' \) the velocity along the plane of the moon's orbit, and we have seen that the motion along this plane is \( 6\pi^2 K.M \cos I' \), in the time \( T' \), or a sidereal revolution. Dividing this by \( T' \), we get the mean velocity in the plane of the orbit in the unit of time; whence \( FF' = \frac{6\pi^2 K.M \cos I'}{T'(E+M)} \).

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

\[ AA' = \frac{Q \cos I' \sin I' \cos AF}{\sin I}. \]

We must now express \( \cos I' \sin I' \cos AF \), in terms of the obliquity and inclination of the lunar orbit to the ecliptic. 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,

\[ \cos ANF = \cos NAF \cos NFA + \sin NAF \sin NFA \cos AF, \]

that is, since \( NAF = I, NFA = I' \),

\[ \cos I' = \cos I \cos I' + \sin I \sin I' \cos AF. \]

In like manner, in the same triangle,

\[ \cos I' = \cos I \cos I' + \sin I \sin I' \cos n. \]

From these two equations we obtain this other,

\[ \cos I' \sin I' \cos AF = \cos I \sin I' \cos n, \]

which, on substituting in it \( \cos 2I \) for \( \cos I \sin I' \cos n \), becomes

\[ \cos I' \sin I' \cos AF = \cos I \sin I' \cos n, \]

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

\[ n = \frac{2\pi t}{\tau}. \]

On making this substitution in the last equation, we obtain by means of it

\[ AA' = Q \left\{ \cos I (\cos i - \frac{1}{2} \sin i) \right\} \]

\[ - \frac{1}{2} \cos 2I \sin 2i \cos \frac{2\pi t}{\tau} \]

\[ - \frac{1}{2} \cos I \sin^2 i \cos \frac{4\pi t}{\tau} \]

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 \( AA' = \frac{dy}{dt} \), and the amount of this regression in a given time \( = \int \frac{dy}{dt} dt \). Multiplying therefore the right hand side of the last equation by \( dt \), and integrating, (observing that \( \int \cos \frac{2\pi t}{\tau} dt = \frac{\tau}{2\pi} \sin \frac{2\pi t}{\tau} \), and \( \int \cos \frac{4\pi t}{\tau} dt = \frac{\tau}{4\pi} \sin \frac{4\pi t}{\tau} \)), we obtain, finally, for the regression of the equinoctial points, or the precessional motion of the equinoxes, produced by the moon in the time \( t \),

\[ Q \left\{ \cos I \left( \cos^2 i - \frac{1}{2} \sin^2 i \right) t \right. \\ - \frac{\tau}{4\pi} \cos 2i \sin 2i \sin \frac{2\pi t}{\tau} \\ - \frac{\tau}{8\pi} \cos I \sin^2 i \sin \frac{4\pi t}{\tau} \right\} + \text{const.} \]

The first term of this expression increases uniformly with the time, and is called the lunar precession. The second term, being multiplied by \( \sin \frac{2\pi t}{\tau} \), 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,

lunar annual precession \( = Q \cos I \left( \cos^2 i - \frac{1}{2} \sin^2 i \right) T \).

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 perpendicular to \( AB \), and meeting \( A'B' \) in \( E' \); then the alteration of obliquity produced by the moon in the time \( dt \) is represented by \( EE' \). Now in the triangle \( CFF' \) we have

\[ \sin CF' : \sin CFF' :: \sin FF' : \sin FCF', \]

that is,

\[ 1 : \sin I' :: FCF' : FCF'. \]

and in the triangle \( CEE' \),

\[ \sin CE'E : \sin CE : \sin ECE' : \sin EE' \]

that is, by reason of \( CE = AF \) (since \( FC = AE = 90^\circ \)) and \( ECE' = FCF' \),

\[ 1 : \sin AF :: FCF' : EE'; \]

whence \( EE' = FF' \sin I' \sin AF \). Now it was shown in the last proposition that \( FF' = Q \cos I' \), therefore

\[ EE' = Q \cos I' \sin I' \sin AF. \]

To refer this to the ecliptic we have in the triangle \( NAF \),

\[ \sin NFA : \sin ANF :: \sin AN : \sin AF \]

which gives the equation

\[ \sin I' \sin AF = \sin i \sin n; \]

and, as before we have

\[ \cos I' = \cos I \cos i + \sin I \sin i \cos n; \]

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

\[ \cos I' \sin I' \sin AF = \frac{1}{2} \cos I \sin 2i \sin n + \frac{1}{2} \sin I \sin^2 i \sin 2n, \]

and, consequently, writing for \( n \) its value \( \frac{2\pi t}{\tau} \),

\[ EE' = Q \left\{ \frac{1}{2} \cos I \sin 2i \sin \frac{2\pi t}{\tau} + \frac{1}{2} \sin I \sin^2 i \sin \frac{4\pi t}{\tau} \right\}. \]

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

\[ -Q \left( \frac{\tau}{4\pi} \cos I \sin 2i \cos \frac{2\pi t}{\tau} + \frac{\tau}{8\pi} \sin I \sin^2 i \cos \frac{4\pi t}{\tau} \right). \]

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 \( = -Q \cdot \frac{\tau}{4\pi} \cos I \sin 2i \cos \frac{2\pi t}{\tau} \).

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

By prop. 4, the solar precession in one year \( = \frac{6\pi^2 K \cos I}{T'} \).

Now \( T' \) 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 \( = 2\pi \times 366.26 \). The solar precession therefore becomes \( \frac{3\pi K \cos (23^\circ 28')} {366.26} \) expressed in parts of the radius. To reduce it to seconds, we have, assuming radius \( = 1 \), \( = 180^\circ = 180 \times 60 \times 60 = 648000 \) seconds. Substituting this for \( \pi \), and computing the above expression by the logarithmic tables we get

solar annual precession \( = K \times 4869'' \).

By prop. 6, the lunar precession in a sidereal year (on substituting for \( Q \) its value) is \( \frac{6\pi^2 K \cdot M}{T'^2 (E+M)} \cos I (\cos^2 i - \frac{1}{2} \sin^2 i) T' \). But \( T' \) is the angle described by the diurnal rotation of the earth in one sidereal revolution of the moon, or 27.32 days (Astronomy, vol. iii. p. 792), and therefore \( = 2\pi \times 27.32 \). We have also (p. 791), \( i = 5^\circ 8' 47.9'' \), and, as before, \( T = 356.26 \) days. Now assuming the moon's mass \( = 1/70 \)th of the earth's mass, \( M = (E+M) = \frac{1}{71} \). By the substitution of these numbers, the lunar annual precession becomes

\[ \frac{3\pi \cdot K \times 366.26 \times \cos (23^\circ 28') \times \{1 - \frac{3}{5} \sin^2 (5^\circ 8' 50'')\}} {27.32 \times 27.32 \times 71}, \]

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

lunar annual precession \( = K \times 12176'' \).

Adding this to the solar annual precession, we obtain the effect produced by the joint action of the sun and moon, or luni-solar annual precession \( = K \times 17045'' \).

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 \( K = \frac{a^2 - b^2}{a^2 + b^2} = \frac{e}{1-e} \), \( e \) being the ellipticity \( = \frac{1}{301} \) (Figure of the Earth, vol. ix. p. 563), whence \( K = \frac{1}{300} \). This value of \( K \) gives

luni-solar annual precession \( = \frac{17045''}{300} = 56''82 \).

The observed quantity is only 50''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 \( \frac{3\pi K}{T_e} \cos I \times \frac{1}{2} \sin 2t \). Substituting \( 2\pi \times 365.26 \) for \( T_e \), and multiplying by \( \frac{180 \times 60 \times 60}{\pi} \) to reduce to seconds, the computation gives

1st part of solar nutation \( = K \times 387'' \times 5 \times \sin 2t \).

The second part is the nutation in obliquity, found by prop. 5, the value of which is \( \frac{3\pi K}{T_e} \sin I \times \frac{1}{2} \cos 2t \). This being computed in the same manner as the last gives

2d part of solar nutation \( = K \times 168'' \times 3 \times \cos 2t \).

Assuming the earth to be homogeneous, and consequently \( K = \frac{1}{300} \), we have for the sum of the two parts

solar nutation \( = -1''29 \sin 2t + 0''56 \cos 2t \),

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

\[ \frac{3\pi K \cdot M \cdot r}{2T^2v(E+M)} \cos 21 \sin 2i \cdot \frac{2\pi t}{\tau} \sin \frac{2\pi t}{\tau}, \]

which, since \( \tau = 18.6 \times 365.26 \) days, becomes, on substituting for the different quantities their numerical values, and reducing to seconds,

\[ K \times \frac{3 \times 64800 \times 18.6 \times 365.26 \times \cos (46^\circ 56') \sin (10^\circ 17' 36'')} {2\pi \times (27.32)^2 \times 271 \sin (29^\circ 28')} \sin \frac{2\pi t}{\tau}, \]

whence there is found from computation,

lunar nutation in longitude \( = K \times 6093'' \times \sin \frac{2\pi t}{\tau} \).

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

\[ \frac{3\pi K \cdot M \cdot r}{2T^2v(E+M)} \cos I \sin 2i \cos \frac{2\pi t}{\tau}. \]

Comparing the coefficient of \( \cos \frac{2\pi t}{\tau} \) in this expression with that of \( \sin \frac{2\pi t}{\tau} \) in the above, it is obvious that the latter is found by multiplying the former by \( \cos I \sin I \cdot \sin 2i \)

\[ = \frac{1}{2} \tan 2i = \frac{1}{2} \tan (46^\circ 56'). \]

The multiplication gives

lunar nutation in obliquity \( = K \times 3260'' \cos \frac{2\pi t}{\tau} \).

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

lunar nutation \( = 20''31 \sin \frac{2\pi t}{\tau} + 10''88 \cos \frac{2\pi t}{\tau} \).

The observed values of the coefficients are \( 18''36 \) and \( 9''239 \), 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 \( = \sin (23^\circ 28') \). The mean velocity corresponding to the regression of the equinoctial points, is \( 50''4 \) 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 \( \sin I : 1 \). Hence the motion of the pole in the plane of the small circle, is obtained by multiplying the expression in prop. 6, by \( \sin I \); and therefore the correction to be applied to the uniform motion of the pole, is the second term of that expression, multiplied by \( \sin I \), or \( \sin I \times \) lunar equation of the equinoxes in longitude. But by prop. 9, this term \( = K \times 6093'' \sin \frac{2\pi t}{\tau} \), therefore the corresponding motion of the pole \( = K \times 6093'' \sin (23^\circ 28') \sin \frac{2\pi t}{\tau} = K \times 2427'' \sin \frac{2\pi t}{\tau} \).

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 \( K \times 3260'' \cos \frac{2\pi t}{\tau} \). Now let \( K \times 2427'' = a \), and \( K \times 3260'' = b \), and let the motion of the pole in the two directions be respectively denoted by \( x \) and \( y \), we have then \( x = a \sin \frac{2\pi t}{\tau}, y = b \cos \frac{2\pi t}{\tau} \); and consequently the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1; \]

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 \( p \) (\( = 50''4 \)), and the coefficient of the lunar nutation in obliquity \( q \) (\( = 9''239 \)). Let \( s \) = the solar annual precession, and \( m \) = the lunar annual precession, then \( s = p - m \). Now, by prop. 7, \( q = \frac{Q \cdot r}{T^2} \cos I \sin 2i \), and, by prop. 6, \( m = Q \cos I (\cos 2i - \frac{1}{2} \sin 2i) T \), whence by eliminating \( Q \) cos I, we get

\[ m = \frac{q \cdot 4 \pi \cdot T (\cos 2i - \frac{1}{2} \sin 2i)}{\tau \sin 2i}, \]

and on computing \( m \) from the values of \( T, r \) and \( i \) above given, we find \( m = q \times 3''735 = 34''51 \); therefore, also, \( s = p - m = 15''89 \).

Again, from prop. 4, we have \( s = \frac{6\pi^2 \cdot K \cdot \cos I}{T^2v} \), and from prop. 6, (on substituting for \( Q \) its value), \( m = \frac{6\pi^2 \cdot K \cdot M \cos I}{T^2v(E+M)} (\cos 2i - \frac{1}{2} \sin 2i) T \); whence

\[ \frac{s}{m} = \frac{T^2(\cos 2i - \frac{1}{2} \sin 2i)}{E+M} \cdot \frac{M}{M}. \]

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

\[ \frac{E+M}{M} = 81.755, \]

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 Precordia of the tides, are the only astronomical facts which enable us to determine the moon's mass. From a long series of observations on the tides at the harbour of Brest, Laplace found the ratio of the masses of the moon and earth to be $1 : 75.77$.

The regression of the equinoctial points amongst the fixed stars, and consequent precession of the equinoxes, being a motion which, though extremely slow amounting only to a degree in about seventy-two years, increases constantly with the time, was detected at an early period in the history of astronomy, and its rate was determined with considerable accuracy by Hipparchus. Its physical cause was of course not suspected until after the discovery of gravitation; but Newton himself, by a process of reasoning, which, although not quite accurate, affords one of the most remarkable instances of his extraordinary sagacity, (Principia, lib.iii.prop. 39), shewed it to be a necessary consequence of the flattened form of the earth. D'Alembert was the first who gave a general and accurate solution of the problem, in his Recherches sur la Precession des Equinoxes (1749); and Euler also treated the subject in the Berlin Memoirs for the same year. Various solutions of the problem have since been given, amongst which may be mentioned those of Sylvabella and Walmesley in the Philosophical Transactions (vols. xlviii. and xlix); that of Simson (Miscellaneous Tracts, 1757); that of Frisi in his Theoria Geometrica Diurni Motus (Opera, tom. iii.); that of Lagrange in his Memoir on the Libration of the Moon, which obtained the prize of the Academy of Sciences of Paris for 1769; that of Landen (Mathematical Memoirs, 1780); and that of Vince (Philosophical Transactions, 1787.) Of these solutions, that of Frisi deserves to be noticed as perhaps the most perspicuous and elegant. An excellent elementary demonstration is given by Mr. Airy, the present astronomer royal, in his Mathematical Tracts, (1826 and 1831) of which we have freely availed ourselves in the present article; but for a complete investigation of the question in all its generality, we must refer the reader to the Mécanique Céleste, and still more particularly to a Memoir of Poisson, Sur le Mouvement de la Terre autour de son Centre de Gravité, in the Mémoires de l'Académie Royale des Sciences, tome vii. 1829.

The nutation, as has already been remarked, (see also Astronomy, vol. iii. p.768), was detected by Bradley, from a comparison of observations which were undertaken with a view to determine the parallax of the fixed stars. Bradley assigned to the coefficient or constant of nutation the value $9''$, which till a late period was adopted by most astronomers. Laplace computed its value from theory to be $9''63$; but as this result could only be obtained by having recourse to hypotheses respecting the ellipticity and density of the earth, and also the mass of the moon, which may possibly differ considerably from the truth, it cannot be regarded as of much weight. There are, however, three other determinations of the constant (besides that of Bradley), from observation, which may be supposed to give its value with all the precision that is capable of being attained. The first is that of Von Lindenau, from about 800 observations of Polaris made between the years 1750 and 1815, and consequently including three revolutions of the moon's node; as well as from those made by Bradley, Maskelyne, Bessel, Carlini, Piazzi, and Von Lindenau himself. From these observations he found the value of the constant of nutation to be $8''97718$. The second determination is that of Dr. Brinkley (Philosophical Transactions, 1821) deduced from his own observations with the Dublin Mural Circle; and the value which he found was $9''25$. The third determination, and that which has the greatest probability in its favour, is a very recent one by Dr. Robinson of Armagh, undertaken at the instance of the British Association. It is deduced from 11,000 observations made at Greenwich with the mural circle, between the years 1812 and 1834, and embraces more than a complete revolution of the node. The result gives the constant of nutation = $9''28913$. (See the Monthly Notices of the Royal Astronomical Society for May 1838).