SPHEROIDS, AND ON THE OCEAN TIDES UPON A YIELDING NUCLEUS. 13 
vt 
or if we put tan e = —, 
Ag sec e cos (r^+B+e) = «S cos (vt-fy). 
Hence A=-S cos e, and B = 17 — e. 
S ' 
Therefore the solution of (14) is, 
<r=^S cos e cos (vt -\-y — e) 
(15) 
Where tan e=—= 7 p—- 
g Zgaw 
But if the globe were a perfect fluid, and if the equilibrium theory of tides were 
true, we should have by (13), 
CT=— -aS COS (vt-\-y) = — COS (vt-fy). 
Thus we see that the tides of the viscous sphere are to the equilibrium tides of a 
6 
fluid sphere as cos e : 1 , and that there is a retardation in time of 
r v 
A parallel investigation will be applicable to the general case where the disturbing 
potential is wr { S* cos (vt-\-y) ; and the same solution will be found to hold save that 
2 (i-l)g 
(2i+l)a 
we now have tan £ _ 2 (^ + 1 ) + 1 _ that in place of (r v/e ha ve 
% gem x 
6 . Diminution of ocean tides on equilibrium theory. 
Suppose now that there is a shallow ocean on the viscous nucleus, and let us find 
the effects on the ocean tides of the motion of the nucleus according to the equilibrium 
theory, neglecting the gravitation of the water. 
The potential at a point outside the nucleus is 
3 
cr+r 2 S cos (vtf-rj), 
and if this be put equal to a constant, we get the form which the ocean must assume. 
Let r—a-\-u be the equation to the surface of the ocean. Then substituting for r in 
the potential, and neglecting u in the small terms, and equating the whole to a 
constant, we find, 
o 
~g uJ r'-qer-\-a 2 S COS (vt-\-rj) — 0 
D 
+ yS cos (vt-\-y)- 
or 
