398 
PROFESSOR G. H. DARWIN ON FIGURES OF 
And, by symmetry, 
i 
D 
9 2^2 
4ir 
a s u + i(4Y 
i = 2 l 
0 
+*e^4i) 3 +i(f) s :i;™ ±1) y-'x; 
i — 1 
(70) 
Add (70) to (69), note that d + D = c, and y<ye = 3 w 2 /327t, and solve for or, and 
we have 
3 or 
4tt 
tM 3 
1 + f - 
- A (f J (f ‘ if < 7 ri ' - 1 a,- + ry - ■ x,) 
(71) 
Now let 1 -f- K denote the factor by which {A/cf + (a/c) 3 is multiplied in (71). 
Then, if the two masses were particles, K would be zero, and (71) would simply be 
the usual formula connecting masses, mean motion, and mean distance in a circular 
orbit. Hence 1 -f- K is an augmenting factor by which the value of the square of 
the angular velocity must be multiplied if it be derived from the law of the periodic 
time of two particles revolving about one another. K, in fact, gives the correction to 
Kepler’s law for the non-sphericity of the masses. 
This completes the solution of the problem, for we have determined the angular 
velocity in such a way as to justify the neglect of the harmonic terms of the first 
degree in §§ 2 and 4. 
§ 7. Solution of the Problem. 
We may now collect from the preceding paragraphs the complete solution of the 
problem. 
In (38) and (53) we have found that there are terms in rja as follows :— 
w„ , Pw, 
Now 
w. 2 = z 2 — fp — \\f, and 8 3 wq = 3 (r 2 — y~); 
hence 
w 2 — -l S'H = P — 2x 2 + f — P — 3r 2 , 
and these terms are therefore equal to e (^ — x 2 /r~). 
We note that e = 15 or/ 1677 - = \ or fir, and that or'l^ir is the ratio generally 
written m in works on the figure of the Earth. Then, from (17), (38), (53), the 
equations to the two surfaces are 
