PRECESSION OF THE EQUINOXES.

In the article Astronomy (vol. vi. 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 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 processional 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 ecl iptic; 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, 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 ¾ 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 z, 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 A¢z, which passes through its centre of gravity O, with an angular velocity =v, receive an impulse which alone would cause it to revolve about an axis Bó, also passing through its centre of gravity, with a velocity =ψ, 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 B⅞, and so situated that the sine of its inclination to the axis Aα will be to the sine of its inclination to the axis BZ>, as the velocity about BZ> to the velocity about Aα; that is, the new axis will divide the angle AOB, so that sin AOC: sin BOC:: φ : 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 Aα, 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 B6 are at right sin jAO(√ angles, then sin BOC=cos AOC, and we have rτ=vz, % φ cosAOC

= —, that is, tan AOC= —.

v 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 byFrisi.

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

Let P∕> be the axis of rotation, Ctq the projection of the equator, S the sun, C the centre of the spheroid, and D [7:18:508]the projection of a point in the interior of the spheroid on the ])lane YQpq. 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

SC 2 square of the distance, we have f ’ =f. g^^∙ 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 VQpq. The resolved part of the force f ’ , in the direction FD is so 2 SC

/∙ gjj2 ∙ gß’ 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 p ∕SC2 φ SC ∖ p SC3—SD3 ∕ (SC—SD) (SC 2 +∙SC ∙ SD-∣-SD2) j ' ∖SD2 ' SD - 1 ∕~ l, ’ SD3 ~d ' ~ SD3

Nowbyreasonof the great distance of thesun in comparisonof 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

, , 1 . ... . r , t SC 2 + SC∙SD + SD 2 as to be altogether insensible, the factor ≈—srvr^ j

SD 3

3

becomes - -∙, and consequently the above expression is

CE CE

reduced to 3f., or to 3 A ~ z √-, since the difference j SD j SC

between l÷SD and l-÷-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 Qç; 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

CE<F

∙ ʹ ∙ SC ∙

'Assume r=SC, and let S denote the absolute force of the sun, then ∕=S÷r 2 ; and if k— the density at D, then hdm 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

3S

the forces on all the particles) is -— ∙ CE.CF. k.dm i the integral of which must be taken for the whole spheroid.

Draw DM parallel to P∕>, and MN perpendicular to DF, meeting SC in 0, and make

CM=X, DM=y, SCP=0,

(¢ being the complement of the sun’s declination) then CE=DN + NF='τ∕ cos i+^ , sin 0, and CF=MN—MO =ysintf—A7cos0, whence the moment of the force impressed on D becomes kdm (x 2 —# 2 ) sin ♯ cos ⅛∙ ∖ ∙xy (cos e 0—∙sin 2 0) j-.

In order to find the integrals fkx 2 dm,fky 2 dm,J'kxydm 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 ofτ∕z, the thickness of each slice being dχ ; 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 oflengths, each —dz. The element of the volume, dm, then becomes dxdydz ; and consequently the sum of kχ 2 din in respect of every particle is expressed by the triple integral fffkx 2 dxdydz.

Assuming x 2 , dx and dy to be constant, and integrating with respect to z, we obtain kχ 2 dxdyz A-const. Let ¢z= the semi-diameter of the equator, b~ the polar semi-axis, and the equation of the spheroid i≡~ + y^4 2^~B fr° m which the limits of z must be found. This equation gives for a point at the surface, z~z±za J ; '∑1 ^i,andbe-

a 2 b 2 tween those values of z the definite integral becomes

τ 2 77≡

1— ⅜-,

α 2 Z> 2

This expresses the sum of kχ i dm for the parallelepiped corresponding to a given value of x.

If we next suppose χ 2 and dx to be constant, and integrate this expression with respect to y, we shall have the sum of kχ-din for the slice, the distance of which from the plane ,, , . xU 2 ‘2 /2 ^ ~ X ÌÌ 1 1

yz is χ. Make 1 , or u—b —, and the j α “b 2 ' a υ - z ——

expression becomes ——— f J u v y2. dy. In integrat-

ing this expression, the limits of y, in respect, of any given value of a∙, are obtained from the equation of the section of • χ 2 ι∕ 2 the ellipsoid in the plane χy, namely ^y+ — 1∙ This

Z» 2 χ 2 equation gives# 2 -b 2 —; whence, at thelimitsy 2 =w 2 , and therefore #= z±zu. Now, by a known formula, the intcgrai y^∕u 2 — y-∙dy from y = — u to y = + u is ½ τr u 3 (π being the ratio of the circumference to the diameter); therefore the integral of x 2 dm for the slice corresponding to a 1 c . Tikau 2 χ i dX , . , ... A

given value of x is; winch, on substituting for b o - u 2 , its value 7> 2 — —- x q , becomes a- πk- (a 2 x 2 — x i )dx. α

We have lastly to integrate this expression from x= — a . m 1 1 1 i ,. . . c 2.∙nkb to x— + a∙ The integral between those limits is (d ei'* ∖ y — —}; whence we have ultimately, in respect of the whole spheroid, fff kx 2 dxdydz=f s τrka i b.

On going through the same process fory 2 dm, or y 2 dxdydz, and observing that the integral J'y 2 f Ju 2 ∙ — y-∙dy from y= — uloy—A-u is ½ mi 4 , there results for the whole spheroid fffky 2 dxdydz≈A s τrka s b s .

With respect to the remaining integral fxydm, it is easy to see that its value in respect of the whole spheroid must [7:18:509]be 0, for on integrating kxydxdydx with respect to x, there results ∖ kx 2 ydydz, which vanishes on giving x all values between — t and - ∣ -t, t being any definite quantity.

From these values of 'Jx 2 dm,fy 2 dm,fxydm,∙we, obtain the following expression for the moment of all the forces impressed on the spheroid,

—. Zζ l i a o -b(a 2 -b 2 } sin Ô cos ð. r 3 lo 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 α and β 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 β, for their moment on the elementary layer containing dm, and then integrate this expression from /3=0 to β≈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

3S 4ττ , , n . — . — ka 2 β(a z — ∕3 2 ) sm ô eos ô. r 3 15

The variable part of this expression isΛα 2 ∕3(α 2 -∕3 2 ). Now, since <ι=∕3(1 +β), we have, on neglecting terms multiplied by t 2 , which, by reason of the smallness of the earth’s ellipticity are altogether insensible, λ 2 =∕3 2 + 2∕3 2 e and α 2 — β 2 =z2β 2 e, whence Aα 2 ∕3(α 2 —∕3 2 )=2Λ∕3 5 e. Suppose now the semi-axis β to receive an infinitely small increment, and to become β- ∖ -dβ ; then since k and e are both functions of β, the expression Tkβ c, e will become 2kβ 5 e + ∙jqq s ; s t ; le va .

1 dβ riable part of the moment of the forces on the spheroid whose semiaxis is βA∙dβ ; consequently, in respect of the spheroidal layer which remains on subtracting the spheroid i ∙ ∙ „ ∙ i 2d(kβ s e) . whose semiaxis is β, it becomes ——-—⅛∕3. Let the mte- dβ gral of this expression between the limits /3=0 and ∕3=δbe denoted by 2 F (∕3); then the moment of all the forces impressed on the spheroid supposed heterogeneous is

3S 8π

—. — F (β) sm ð eos ð. r∙ y 15′ Prop. 3. To determine the angular velocity <p 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 f(x 2 -j-y 2 )dm, supposing the spheroid homogeneous, and the density Z, - =1. But it has been shewn that in thiseasej⅛ dmz=yÇ¶ra 4 b, and Jy 2 dm=f s πa 2 b 3 ; therefore the moment of inertia of the spheroid is τ ⅛τiz 2 0(n 2 +^ 2 ). And by prop. 1, the moment of the impressed forces is — . f s ττa 2 b^a 2 — 6 2 ) sin ō cos ô, therefore

__3S a 2 ~b o - φ ~^∙^+T≡ sin ^ cosa

In the ease of the heterogeneous spheroid the moment of inertia is thus found. As before, let α and β 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 ‰kα 2 β(α 2 + ∕3 2 ). But α 2 +∕3 2 =2 y δ 2 + 2∕3e, and a 2 β=β 3 + 2∕3∙¼ therefore a 2 β( a 2 +∕3 2 )=2∕3 5 + 6∕3 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 ∕3(α 2 +∕3 2 )=2λ∙∕3 5 . For the spheroid whose sermaxis is β-}-dβ, this quantity becomes 2A∙∕3 5 + ^dβ, dβ and therefore in respect of the elementary spheroidal layer the semiaxes of whose interior and exterior surfaces are β 2d(kβ 5y ) and β + dβ, it is ——— dβ. Let 2F ʹ (∕3) denote the integral of this quantity from /3=0 to β=b, that is, let Γ^^-dβ J dβ

=2F ʹ (β), and the moment of inertia of the heterogeneous Sττ

spheroid becomes — F ʹ (∕3). But the moment of the im-io 1 c . . „ 3S 8τr pressed forces is by prop. 2, =-. — F(i β) sin θ cos θ, therefore in the case of the heterogeneous spheroid, λ 3S F(∕3) . λ λ φ=z — . — ) < sin θ eos θ. r 3 F ʹ (∕3) α 2 — b 2 Let us assume K= ——— in the case of the homo<re- a 2 A-b~ o

F(∕3)

neous spheroid, and K= =τ~( in the ease of the heterose- F z (∕3) ö neous spheroid, and we have for both cases, 3S.K .

<⅛= —— sin ß eos ß. 'i r 5

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 PQpy, or perpendicular to the plane of the paper, so as to bring the point Q nearer to the line SC, with a velocity =<∕>. But the earth is already revolving about the axis Ρp, 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 = —, the pole P rising above the plane of the paper; and the new equator will intersect the former in the line Q,q, and make with it an angle whose tangent is also —. The v cffcet of the compound motion is thus to twist as it were the equator about the line Q,q 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 [7:18:510]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 ʹ CB ʹ be the new position of the equator, after the infinitely small time dt, intersecting the former in C, then the angle ACA ʹ is the measure of the momentary deviation of the axis, and AA ʹ, 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 ʹ:: sin A ʹ C: sin AA ʹ. But because ACA ʹ is a very small angle, and AA ʹ a very small arc, the arcs may be taken instead of the sines; whence, since ci tì A C 1

A ʹ C = AC, the proportion gives AA ʹ = ACA ʹ.

r b sin SAC

ó 3S K

But by the lastproposition tan AC A ʹ = — = —'~ sin ¢ cos i, therefore, as the small arc may be substituted for its tangent, 3S.K sin AC . .

AA = — ;———- sin ō cos ô. vr 5 sin SAC

It will now be convenient to express this value of AA ʹ in terms of the sun’s longitude and the obliquity of the ecliptic. Since the equator, as was shewn 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° — ê. Let Z=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 . cos AC=cos AS, or sin ô cos AC==cos l ; and sin SC= sin SAC sin AS, or cos ê— sin I sin l, therefore • λ λ ≡i n I sin cos sm θ cos θ — cos ac ’ Again, i n the same triangle we have tan AC=cos SAC tan AS, whence sin AC=cos AC cos I tan Z, and (dividing by sin SAC= sin I) — n A 9- — j sin SAC ~ cos AC cos I sin l .

- . . Substituting these values of sin ô cos ô olll X LUS V

, sin AC , and ~s^Xc ιn the above value of AA ʹ, we get

3S K

AA ʹ = —— cos I sin 2 Z.

This is the solar precession in the element of time dt ; consequently for a given time T we have solar precession = cos pg the integral being taken from i=0 to i=T.

To prepare this expression for integration, r ≡ and dt must be expressed in terms of l and known quantities Let α=semiaxis major of the earth’s orbit, ε=its eccentricity T≈a sidereal year, then r being the radius vector and dl the angle described in the time dt, l r *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 :: ½r 2 cZZ: area of orbit.

Now the area of the orbit is πα 2 x ∕l- f 2 , therefore this pro-

Tr 2 dl portion gives dt=z - — But it is shewn in the article Astronomy, part iii. art. 10, that T=2π√α÷ χ ∕p∖ 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÷α 2 , whence T= - π ~ a _

√S '

From this formula we get S= - 7- α , and therefore ~ di

1 2 r 5

2nadl

— —- 2 ∙ Again, assuming λ= longitude of sun’s perigee, the polar equation of the ellipse gives r =

α(1—ε 2 ) , Sxlt 2τrjΙ+εcos(Z—λ)i<ZZ

1+ecos(Z-λ) r 3 ~ T(1-ε 2 )2 ’ d the above expression becomes

. 6τrK cos I r∙

solar precession = ∕ s ; n 2Zn + ecos(λ-Z] i Z∕.

Tr(1- e 2)V i

The expression under the sign of integration consists of two parts, of which the first f sin 2 ZrfZ=½(C + Z— cos Z sin Z). (See Fluxions, art. 153.) The second part, namely fe sin 2 Z cos(Z—λ)cZZ, when integrated becomes e sin(Z—λ)

sin(Z+λ)— — sin(3Z—λ), but by reason of the smallness Zi 0

of ε these terms are insensible, and are therefore neglected. Rejecting also the terms in the divisor of the coefficient which are multiplied by ε 2 , and observing that cos Z sin l— ½ sin 2Z, we obtain, finally, 1 . 3χK cos I . r , , , .

solar precession= — (C+Z—½ sin 2Z).

The first term of this expression, which depends on C + Z, or on the sun’s longitude, is the constant or uniform precession. Its amount in one year is found by supposing , i , . 1 1 r> J ∙ 1 67r 2 K cos I

Z to be increased by 2π∙, and is consequently y .

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 ʹ CB ʹ in E ʹ, then EE ʹ is the small change in the inclination in the time dt. In the triangle ECE ʹ, we have sin EE ʹ = sin CE sin ECE ʹ. But EE ʹ being very small, the arc may be taken for the sine, and therefore EE ʹ == sin CE sin ECE'.

cos Z

Now sin CE= cos AC= —7; and it has been already sin ∂ 1 φ 3S.K

seen that sin ECE ʹ = AC A ʹ =- = — sinocos£; therefore v r 3 υ

3S.K

EE ʹ = —-— cos ô cos Z, or since cos ô = sin I sin Z, EE ʹ = r 0 ν -

3S K

--~- sin I sin Z cos Z. Multiplying by dt i and making the same substitutions as in the last proposition, we have

_ 6πK ∕ r *

∕eE z <Z∕ = ——- sin I /sinZcos Zp +ε>cos(Z-λ)WZ, J Tf(l-e≡p t ≠
[7:18:511]

whence neglecting as before terms multiplied by e, and integrating, there results íVM/dt— — sin I cos 2Z, J 2 lv 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

Μ — moon’s mass,

E — earth’s mass,

I' = inclination of moon’s orbit to the equator, l ’ = 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, d = 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

3M.K cos I ʹ /sin i l ’ , ∕ —— dt. v ,; τ 3 Tr r *dl' , α ʹ (1-√ 2 )

Now, we have αZ= . · and r ~ - — - —

2 7 rα' 2 √1-√^ 1 + e'cos(λ'-Z ʹ)

of T ’ fl +e ’ cos(λ ʹ-Z ʹ)7 dl ’ n . , therefore — = -—- -A — ∙ But in the pre-

√3 3 r 2πa r3 1 —√ 2 )⅛

2τra l sent case T ’ = - 77τ ,—√,χ (for the mass of the moon cannot √(E + M) v be neglected in comparison of that of the earth, as the mass of the earth is neglected in comparison of that of the sun),

1 4√-

hence ~- = _. _ , .,r∙ The above expression therefore α 3 I 2 (E+M)

becomes

6πM.KeosI' F . ., f , ,

∕ sm 2 Z{1 + ∈ cos (a·—Z ʹ)dZ ʹ, ], TT∙(E+M)(1- e ' 2 )⅜' 7 the integral of which (rejecting, as before, terms multiplied by e ʹ) gives

3a∙M.K cos I ʹ.. . .

7 l . 7- z -√,—r7τ- (C + Z ʹ —½ sin 2f).

1′ v(E+M) v 2 7 for the regression of the equator on the plane of the lunar orbit. Suppose Z ’ to be increased by 2π, or a whole circumference, the regression caused by the moon’s action in a

. 1 , , . . 6π 2 M .K cos I ʹ sidereal revolution becomes , r ,,∙ > , λ — ττ -—.

T ʹ r(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'CB ʹ the new position of the equator, and FH the plane of the lunar orbit intersecting the new equator in F ʹ and H ʹ, 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 hinar 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=CHandF ʹ C= CH'; and since FC + CH==180°, therefore FC and F ʹ C are quadrantal arcs, and the two triangles CFF and CHH ʹ are in all respects equal. Now sin FC: CF ʹ F:: sin FF':: ACA ʹ, that is, 1: sin I':: FF':: ACA ʹ, and sin CA ʹ A: sin AC:: sin ACA':: sin AA ʹ, that is, sin I: cos AF:: ACA ’:: AA ʹ, therefor e, sin I: sin I' cos AF:: FF ʹ: A A', and consequently,

. F'F sin I ʹ cos AF AA ʹ = : — .

sin I

But AA' represents the velocity along the ecliptic, and FF ʹ the velocity along the plane of the moon’s orbit, and we have seen that the motion along this plane is 6π 2 K.M cos I ʹ. v . r , v . 1 1 ⅛ ∣ ∕ /t? . ⅝< ∖ > ιn the time 1 , or a sidereal revolution. Di- 1′ t∙(L+M)

viding this by T ʹ, we get the mean velocity in the plane of the orbit in the unit of time; whence FF'= θΣ,...^^ c θ s r <.

T ʹ ⅛(E + M)

+ 2 K Μ

For the sake of brevity let Q= — f ——-,thenFF'=Q 1′ W(E + M)

cos I', and we have

... Q cos I ʹ sin I ’ cos AF AA ʹ ≈= :— .

sin 1

We must now express cos I ʹ sin I ʹ cos AF, in terms of the obliquity and inclination of the lunar orbit to the ecliptic. Let Z=ANF= inclination of moon’s orbit to the ecliptic, ri-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 NFAcos AF, that is, since NAF=I, NFA=I ʹ, cos i=cos I cos I'+sin I sin I' cos AF,

In like manner, in the same triangle, cos I ʹ = cos I cos z'+ sin I sin i cos n.

From these two equations we obtain this other, cos I ʹ sin I ʹ cos AF=cos I sin I cos 2 i

—(cos 2 1—sin 2 1) cos ∙ sin i cos n —cos I sin I sin 2 i cos 2 n, which, on substituting in it cos 21 for cos 2 I—sin 2 1, ½ sin 2i for cos i sin i, and ½+½ cos 2n for cos 9 n, becomes cos l’ sin I ʹ cos AF =cos I sin I (cos 2 i —½ sin 2 Z)

—½ cos 21 sin 2i cos η

—⅛ cos I sin I sin 2 i cos 2n.

Assuming z(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 w, (the longitude of the node), which, on the supposition of i constant, is proportional to the time. Let τ denote the time of a revolution of the node, then for any time t, Ave have 2ττt n— — . On making this substitution in the last equation, Ave obtain by means of it

A A'=Q cos I (cos 2 i — I sin i ι)

½ cos 21 . . 2⅛r<

— —;—sm 2í cos — sin I τ

1 T ∙ o∙ 4îT< 1 —⅛ cos 1 sm 2 ι cos >

Now, if we assume y=the regression of the equinoctial points on the ecliptic caused by the lunar action, and suppose[7:18:512] y a function of t, then AA'= ~. and the amount of dt this regression in a given time = f -'L dt. Multiplying therefore the right hand side of the last equation by dt, and integrating, ^observing that J cos dt— sin —ζ

, P 4π< _ τ . 4τri∖ n , „

and J cos dt ~ — sin ì, we obtain, finally, for the regression of the equinoctial points, or the processional motion of the equinoxes, produced by the moon in the time t,

Q cos I (cos 2 i — ½ sin 2 ι)t τ cos 21 sin 2? . 2τrt

— -7 :—ψ sin

4w sin I τ

r . . . 4πt )

— — cos 1 sin i i sin — > + const.

C7Γ T J

The first term of this expression increases uniformly with the time, and is called the lunar precession. The second

2τri term, being multiplied by sin —, 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 (cos 2 i — ½ sin 2 z)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 arc 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':: FF': FCF'; and in the triangle CEE',

Sin CE'E: sin CE:: sin ECE': sin EE'

that is, by reason ofCE=AF (since FC=AE=90°) and ECE'=FCF'

1: sin AF:: FCF': EE'; whence EE'=FF' sin I' sin AF. Now it was shewn in the last proposition that FF'≈QeosI', therefore

EE'= Q. cos I' sin I' sin AF.

To refer this to the ecliptic we have in the triangle NAF, sin NF A: sin ANF:: sin AN: sin A F

which gives the equation sin I' sin AF=sin i sin n ; and, as before we have cos I'= cos I cos z'+sin I sin i cos η ; whence multiplying the two equations together, and substituting ½ sin 2i for cos i sin i, and ½ sin 2n for cos η sin n, we get cos l’ sin I' sin AF=½ cos I sin 2i sin τz + ½ sin I sin 2 i sin 2n, 1 1 ∙ ∙ p ∙ 1 2/TÍ

and, consequently, writing for n its value —,

EE'=Q -∣ ½ cos I sin 2z sin —-+½ sin I sin 2 z* sin —— j- .

Multiplying this by t⅛and integrating, we obtain for the diminution of the inclination, or lunar nutation in obliquity „ ∕ τ τ . o . 2ττt τ . τ . 2 ∙ 4πt ∖

—.Q, I — cos I sin 2ι cos 1- — sm I sin z cos ).

∖4w τ 8τr τ ∕

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

τ . 2πt lunar nutation in obliquity =—Q. . — cos I sin 2i cos —-.

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

, . . 6τr 2 KeosI

By prop. 4. the solar precession in one year = — ∙

Now Tv 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 = 2τr × 36G ’ 26. The solar precession 1 ρ , 3τrK cos (23° 28′) therelore becomes ππ7⅛77 ' expressed in parts 0.

3ob∙2b l the radius. To reduce it to seconds, we have, assuming radius = 1, %∙=180° = 180 × GO ×60=G48000 seconds. Substituting this for τr, and computing the above expression by the logarithmic tables we get solar annual precession = K×4869".

By prop. 6, the lunar precession in a sidereal year (on sub- . . _ ^ . 1 λ . 6τr 2 K. Μ τ z ,.

stitting for Q its value) is c0s * (cos ‘ —

½sin 2 z)T. But T'v 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;r×27*32. We have also (p. 791), z=5° 8' 47 ’ 9", and, as before, T=366∙26 days. Now assuming the moon’s mass = 1-70th of the earth’s mass, M÷(E-j-M)= τ l 1 . By the substitution of these numbers, the lunar annual precession becomes

3π. K × 366∙26 × cos (23° 28') × {1—g sin 2 (5° 8' 50") ; 27∙32×27∙32×71

the calculation of which, reduced to seconds as before, gives lunar annual precession = K× 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 × 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

, , 1 1 τr ft2 e , .

the earth homogeneous, we have K= 9 —τ⅛=⅛ ,eeing a l A∙b 1— e the elipticity =—ιp (Figure of the Earth, vol. ix. p.

563), whence K=^ð. This value of K gives

17045" luni-solar annual precession =—- θ^- =5β"∙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, [7:18:513]its value is 4w~ cos I × ½ sin 2Z. Substituting 2τr × 366*26 1 v r oe ι 1 ∙ ι ∙ ι 180×60×60 ÷ 1 ÷ _

for 1 v, and multiplying by — to reduce to se-

7Γ

con ds, the computation gives

1st part of solar nutation ==K × 387 zz, 5 × sin 2Z.

The second part is the nutation in obliquity, found by prop.

3∙τriv

5, the value of which is ——sin I×½cos2Z. This being 1 v computed in the same manner as the last gives

2d part of solar nutation =K × 168 ″ ∙3 × cos 2Z. Assuming the earth to be homogeneous, and consequently

K = —we have for the sum of the two parts solar nutation =1 ″ ∙29 sin 2Z+0 ″ ∙56 cos 2Z, 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

3πK . Μ . τ cos 21 sin 2i . 2πt ——— — β ’ ⅞1Γ) -∙

2T ʹ ⅛(E + M) sin I . τ

which, since τ=18∙6 × 366∙26 days, becomes, on substituting for the different quantities their numerical values, and reducing to seconds, κ 3x64800×18∙6×366∙26×cos(46°56) sin (10° 17'36) . 2τi 2(r×(27∙32) 2 ×2×71 sin (23° 28) Sm V’

whence there is found from computation, lunar nutation in longitude = K× 6093" × sin —.

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

3%K. M. τ . . 2%√

2T'⅛(E+M) cos 1 × sm 2 ' × cos V

Comparing the coefficient of cos ^→n this expression with 2ττt that of sin — in the above, it is obvious that the latter r

. z. , , 1 . . . 1 „ , cos I sin I sin 21

is found by multiplying the former by — —

r cos 21 2 cos 21

=½ tan 2l=½tan (46°56′). The multiplication gives

2π∕

lunar nutation in obliquity =K× 3260" cos .

Assuming that the earth is homogeneous, and consequently K= 3^ q∙> these two parts added together give lunar nutation =20 ″ ∙31 sin —— + 10 ″ ∙88 cos

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° 2 8′). 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 1:1. 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 × lunar equation of the equinoxes in longitude. But by prop. 9, this term = K × 6093 ″ sin therefore the corresponding motion of the pole =

K × 6093 ″ sin (23° 28′)!sin — = K × 2427′ ' sin —.

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 2πt direction is K× 3260" cos —. Now let K × 2427 ″ =α, and K×326θ ″ =δ, and let the motion of the pole in the two directions be respectively denoted by x and y, we have 1 . 2π∕, 2πt then a≈αsm -— ,y=b cos ; and consequently the equa tion

— -+-—=1; a 2÷ b* ’ 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 — in. Now, by prop. 7, q~Q½- cos I sin 2i, 4χ

and, by prop. 6, wι=Qeos I (cos 2 z—½sin 2 z)T, whence by eliminating Q, cos I, we get __ q. 4π . T(cos 2 ? — ½sin *i) τ sin 2i ’ and on computing m from the values of T, r and i above given, we find ∞ = y×3 ″ ∙735 = 34 ″ ∙51; therefore, also, s=p — m— 15 zz, 89∙

• a ∙ c 4L 6w 2 ∙ K cos I 1 r

Again, from prop. 4, we have <?=— , and from

T⅛

prop. 6, (on substituting for Qits value), m— J~ (cos 2 z—½ sin 2 ∕)T; whence

_£ __ ___ 1+A 1 m T 2 (cos 2 z—½ sin 2 z) * Μ

By computing the coefficient of (E+M)-÷-M, and by means of the values <y and m now found, this equation gives

E + M

~⅛- = 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 [7:18:514]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 Re-cherches 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 andxlix); thatofSimson (Miscellaneous Tracts, 1757); thatof Frisiin 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 Celeste, and still more particularly to a Memoir of Poisson, Sur le Mouvement de la Terre autour de son Centre de Gravite, in the Mémoires de l’ Academie 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 theconstantof 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; andthevalue 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"∙23913. (See the Monthly Notices oíÜ\e Royal Astronomical Society for May 1838). s.