32 U. S. COAST AND GEODETIC SURVEY 
4 
_ pM’ 
7” 2GR 
Average V,= a aS iMG cos’ z—3 cos 2) sin z dz 
| 5/4 cos‘ 2+3/2 cos? z[=0 (113) 
96. Let V, represent the potential due to gravity at any point on 
the earth’s surface. Since the force of gravity at any point on or 
above the earth’s surface equals »E/r’, the corresponding potential 
becomes 
V,=Lil. = =— (114) 
If the earth is assumed to be a sphere with radius a, the gravitational 
potential at each point will equal n/a, which may be taken as the 
average gravitational potential over the surface of the earth. 
97. For a surface of equilibrium under the combined action of 
gravity and that part of the tide-producing force involving the cube 
of the moon’s distance the sum of the corresponding potentials must 
be a constant, and since the average tide-producing potential for the 
entire surface of the earth is zero (par. 95), the constant will be the 
average gravitational potential or pwH/a. Then from (110) and 
(114) we have 
3uM EF wk 
Vit Vi= ogre (cos 2 18)r (115) 
Transposing and omitting common factor », we may obtain 
as 2 
@— OT _ 3/2 (M/E) (ald)*(cos* 2-1/3) (116) 
Let 
r=ath (117) 
so that A represents the height of the equilibrium surface as referred 
to the undisturbed spherical surface of an equivalent sphere. Then 
(—a)a ha? 
r (ath)? 
As fraction h/a is very small, its greatest value being less than 
0.000001, the powers above the first may be neglected. Substituting 
in (116) and writing A with subscript 3 to identify it with the prin- 
cipal tide-producing force, we have 
hg [a=3/2 (M/E) (a/d)3 (cos? z—1/3) (119) 
98. Similarly, for a surface of equilibrium under the combined 
action of gravity and the part of the tide-producing force involving 
the 4th power of the moon’s distance, we have from (111) and (114) 
=h/a—3(h/a)?+6(h/a)?— ete. (118) 
Vit V,= 2a (5 cos? z—3 cos zp He ve (120) 
