Part 7

Let the following figure represent a globe of wood of uniform density throughout. Let this globe be rotated round the axis. It is evident that no change of position of the axis would be produced by the rotation. If we add two equal masses of lead at m and m', on opposite sides of the axis, the globe is still in equilibrium, as far as gravity is concerned, and if perfectly spherical and homogeneous it might be suspended from its centre in any position, or assume indifferently any position in a vessel of water. If, however, the globe is now put into a state of rapid rotation round the axis, and then allowed to float freely in the water, we perceive that it is no longer in a state of equilibrium. The mass m being more dense than its antagonist particle at n, and having equal velocity, its momentum is greater, and it now tends continually to pull the pole from its perpendicular, without affecting the position of the centre. The same effect is produced by m', and consequently the axis describes the surface of a double cone, whose vertices are at the centre of the globe. If these masses of lead had been placed at opposite sides of the axis on the equator of the globe, no such motion would be produced; for we are supposing the globe formed of a hard and unyielding material. In the case of the ethereal vortex of the earth, we must remember there are two different kinds of matter,--one ponderable, the other not ponderable; yet both subject to the same dynamical laws. If we consider the axis of the terral vortex to coincide with the axis of the lunar orbit, the moon and earth are placed in the equatorial plane of the vortex, and consequently there can be no derangement of the equilibrium of the vortex by its own rotation. But even in this case, seeing that the moon's orbit is inclined to the ecliptic, the gravitating power of the sun is exerted on the moon, and of necessity she must quit the equatorial plane of the vortex; for the sun can exert no influence on the matter of the vortex by his attracting power. The moment, however, the moon has left the equatorial plane of the vortex, the principle of momentum comes into play, and a conical motion of the axis of the vortex is produced, by its seeking to follow the moon in her monthly revolution. This case is, however, very different to the illustration we gave. The vortex is a fluid, through which the moon freely wends her way, passing through the equatorial plane of the vortex twice in each revolution. These points constitute the moon's nodes on the plane of the vortex, and, from the principles laid down, the force of the moon to disturb the equilibrium of the axis of the vortex, vanishes at these points, and attains a maximum 90d from them. And the effect produced, in passing from her ascending to her descending node, is equal and contrary to the effect produced in passing from her descending to her ascending node,--reckoning these points on the plane of the vortex.

INCLINATION OF THE AXIS.

By whatever means the two planes first became permanently inclined, we see that it is a necessary consequence of the admission of these principles, not only that the axis of the vortex should be drawn aside by the momentum of the earth and moon, ever striving, as it were, to maintain a dynamical balance in the system, in accordance with the simple laws of motion, and ever disturbed by the action of gravitation exerted on the grosser matter of the system; but also, that this axis should follow, the axis of the lunar orbit, at the same mean inclination, during the complete revolution of the node. The mean inclination of the two axes, determined by observation, is 2d 45', and the monthly equation, at a maximum, is about 15', being a plus correction in the northern hemisphere, where the moon is between her descending and ascending node, reckoned on the plane of the vortex, and a minus correction, when between her ascending and descending node. And the mean longitude of the node will be the same as the true longitude of the moon's orbit node,--the maximum correction for the true longitude being only about 5d +/-.

In the following figure, P is the pole of the earth; E the pole of the ecliptic; L the pole of the lunar orbit; V the mean position of the pole of the vortex at the time; the angle [ARIES]EL the true longitude of the pole of the lunar orbit, equal to the true longitude of the ascending node +/- 90d. VL is therefore the mean inclination +/- 2d 45'; and the little circle, the orbit described by the pole of the vortex twice in each sidereal revolution of the moon. The distance of the pole of the vortex from the mean position V, may be approximately estimated, by multiplying the maximum value 15' by the sine of twice the moon's distance from the node of the vortex, or from its mean position, viz.: the true longitude of the ascending node of the moon on the ecliptic. From this we may calculate the true place of the node, the true obliquity, and the true inclination to the lunar orbit. Having indicated the necessity for this correction, and its numerical coefficient, we shall no longer embarrass the computation by such minutiae, but consider the mean inclination as the true inclination, and the mean place of the node as the true place of the node, and coincident with the ascending node of the moon's orbit on the ecliptic.

POSITION OF THE AXIS OF THE VORTEX.

It is now necessary to prove that the axis of the vortex will still pass through the centre of gravity of the earth and moon.

Let XX now represent the axis of the lunar orbit, and C the centre of gravity of the earth and moon, X'X' the axis of the vortex, and KCR the inclination of this axis. Then from

similarity Ct : Tt :: Cm : Mm but Tt : Mm :: Moon's mass : Earth's mass. That is Tt : Mm :: TC : MC.

Therefore the system is still balanced; and in no other point but the point C, can the intersection of the axes be made without destroying this balance.

It will be observed by inspecting the figure, that the arc R'K' is greater than the arc RK. That the first increases the arc AR, and the second diminishes that arc. The arc R'K' is a plus correction therefore, and the smaller arc RK a minus correction. If the moon is between her descending and ascending node, (taking now the node on the ecliptic,) the correction is negative, and we take the smaller arc. If the moon is between her ascending and descending node, the correction is positive, and we take the larger arc. If the moon is 90d from the node, the correction is a maximum. If the moon is at the node, the correction is null. In all other positions it is as the sine of the moon's distance from the nodes. We must now find the maximum value of these arcs of correction corresponding to the mean inclination of 2d 45'.

To do this we may reduce TC to Tt in the ratio of radius to cosine of the inclination, and taking TS for radius.

{TC x Cos &c. (inclination 2d 45')}/R is equal the cosine of the arc SK' and SK' + AS = AK' and AK' + AR' = R'K'. But from the nature of the circle, arc RK + arc R'K' = angle RCK + angle R'CK', or equal to double the inclination; and therefore, by subtracting either arc from double the inclination, we may get the other arc.

The maximum value of these arcs can, however, be found by a simple proportion, by saying; as the arc AR, plus the inclination, is to the inclination, so is the inclination to the difference between them; and therefore, the inclination, plus half the difference, is equal the greater arc, and the inclination, minus half the difference, is equal the lesser; the greater being positive, and the lesser negative.

Having found the arc AR, and knowing the moon's distance from either node, we must reduce these values of the arcs RK and R'K' just found, in the ratio of radius to the sine of that distance, and apply it to the arc AR or A'R', and we shall get the first correction equal to the arc AK or AK'.

Call the arc AR = a " inclination = n " distance from the node = d " arc AK = k

and supposing the value of AK be wanted for the northern hemisphere when the moon is between her descending and ascending node, we have

n^2 ------- a + n (n - ------- ) sin d. 2 k = a - ---------------------- R

If the moon is between her ascending and descending node, then

n^2 ------- a + n (n - ------- ) sin d. 2 k = a + ---------------------- R

The computation will be shorter, however, if we merely reduce the inclination to the sine of the distance from the node for the first correction of the arc AR, if we neglect the semi-monthly motion of the axis; for this last correction diminishes the plus corrections, and the first one increases it. If, therefore, one is neglected, it is better to neglect the other also; especially as it might be deemed affectation to notice trifling inequalities in the present state of the elements of the question.

There is one inequality, however, which it will not do to neglect. This arises from the displacement of the axis of the vortex.

DISPLACEMENT OF THE AXIS.

We have represented the axis of the terral vortex as continually passing through the centre of gravity of the earth and moon. Now, by following out the principles of the theory, we shall see that this cannot be the case, except when the moon is in quadrature with the sun. To explain this:

Let the curve passing through C represent a portion of the orbit of the earth, and S the sun. From the principles laid down, the density of the ethereal medium increases outward as the square roots of the distances from the sun. Now, if we consider the circle whose centre is C to represent the whole terral vortex, it must be that the medium composing it varies also in density at different distances from the sun, and at the same time is rotating round the centre. That half of the vortex which is exterior to the orbit of the earth, being most dense, has consequently most inertia, and if we conceive the centre of gravity of the earth and moon to be in the orbit (as it must be) at C, there will not be dynamical balance in the terral system, if the centre of the vortex is also found at C. To preserve the equilibrium the centre of the vortex will necessarily come nearer the sun, and thus be found between T and C, T representing the earth, and the moon, and C the centre of gravity of the two bodies. If the moon is in opposition, the centre of the vortex will fall between the centre of gravity and the centre of the earth, and have the apparent effect of diminishing the mass of the moon. If, on the other hand, the moon is in conjunction, the centre of the vortex will fall between the centre of gravity and the moon, and have the apparent effect of increasing the mass of the moon. If the moon is in quadrature, the effect will be null. The coefficient of this inequality is 90', and depends on the sun's distance from the moon. When the moon is more than 90d from the sun, this correction is positive, and when less than 90d from the sun, it is negative. If we call this second correction C, and the moon's distance from her quadratures Q, we have the value of C = +/-(90' x sin Q)/R.

This correction, however, does not affect the inclination of the axis of the vortex, as will be understood by the subjoined figure. If the moon is in opposition, the axis of the vortex will not pass through C, but through C', and QQ' will be parallel to KK'. If the moon is in conjunction, the axis will be still parallel to KK', as represented by the dotted line qq'. The correction, therefore, for displacement, is equal to the arc KQ or Kq, and the correct position of the vortex on the surface of the earth at a given time will be at the points Q or q and Q' or q', considering the earth as a sphere.

In the spherical triangle APV, P is the pole of the earth, V the pole of the vortex, A the point of the earth's surface pierced by the radius vector of the moon, AQ is the corrected arc, and PV is the obliquity of the vortex. Now, as the axis of the vortex is parallel to the pole V, and the earth's centre, and the line MA also passes through the earth's centre, consequently AQV will all lie in the same great circle, and as PV is known, and PA is equal to the complement of the moon's declination at the time, and the right, ascensions of A and V give the angle P, we have two sides and the included angle to find the rest, PQ being the complement of the latitude sought.

We will now give an example of the application of these principles.

Example. Required the latitude of the central vortex at the time of its meridian passage in longitude 88d 50' west, July 2d, 1853.