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.
[Illustration: Fig. 8]
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.
[Illustration: Fig. 9]
{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