15 



.'Substituting this expression in the preceding equation, 



)i{Ma?IEm) (3 cos^ e- 1) =a^ul{a-^uy. (15) 



Since u is very small in comparison with a, the value of (a+w)^ is 

 always very close to that of a^. Equation (15) then becomes: 



u=}(2{Ma?IER^)a{^ cos^ d-l) (16) 



This is the equation of the lunar equilibrium tide. 



29. The volume of the tidal surface of equilibrium evidently is the 

 same as the volume of the undisturbed sphere if the total positive tidal 

 volume over the zones P1P2A and P^PiP^, in figure 6 is equal to the 

 negative tidal volume over the zone P1P3P4P2; or, what is the same 

 thing, if the positive and negative tidal volumes in the hemisphere 

 PJ^'P" are equal. 



In figure 7, P is any point on the tidal-equilibrium surface; CP its 

 radius vector, r; PiP the equilibrium tide, n, at that point; CPy the 



radius, a, of the undisturbed sphere; Q its angle with the line CO 

 directed toward the moon; and CP' iP' the position of CPiP when B 

 is increased by the differential angle dQ. The equilibrium surface is, 

 as has been seen, a surface of revolution whose axis is CO. A well 

 known theorem establishes the volume of a solid formed by rotating 

 a plane figure about an axis in the same plane as the product of the 

 area of the plane figure by the length of the circumference of the circle 

 described by its center of gravity. 



The area of the differential triangle PP'C is )i r- dQ and the radius of 

 the circle described by its center of gravity is 2/3 r sin Q. The differ- 

 ential volume resulting from the rotation of the triangle about CO 

 is therefore: 



2/3 Ttr^ sin QdQ=2\'6 Tria^nf sin QdQ 



The corresponding differential volume of the sphere is 



2/3 Tra^ sin B dd 



The difference between these volumes is the elementary tidal 

 volume, dq^, generated by the rotation of PPiPi'P' and is: 



c?g=2/3 Tr[{a+uy-a^] sin 6 dd 



= 2/3 7ra^(3w/a+3MVa'+'«^Va') sin 6 dd 



