FIGURE AND STABILITY OF A LIQUID SATELLITE. 
227 
In order to obtain the expression for the angular momentum, which has to be a 
minimum for limiting stability, we must evaluate £. Now, from (24) and (15), we 
have 
l = (2 c 2 —a?—b 2 + same in A, B, C) 
+ Tg—j [3 (a 4 + 6 4 ) + 8 c 4 — 8c 2 (a 2 + b 2 ) + 2a 2 b 2 + same in A, B, C], 
+ [2 (^ 2 « 2 + B 2 b 2 + C 2 c 2 ) + (A 2 + B 2 +C 2 ) (a 2 + b 2 + c 2 ) - 5 C 2 (a 2 +& 2 ) 
ZU'T 
— 5c 2 (^4 2 +Z> 2 ) + 5c 2 C 2 ]. 
By means of the above series for a 2 , b 2 , c 2 , and of their analogues for A 2 , B 2 , C 2 , 
I find 
„5 8 10 
v 7 ct ct , ct 
C — l — 5 + n —>> 
t .8 *.10 
where 
1 _ 3 X 2/3 (7 + X) + 7X+1 
4 ' (1 + Xf 3 
m = 
_ 5 _ X 2;3 (82 + 38X + X 2 ) + 82X 2 +38X+l 
14 
(i+\) 
8,3 
2 2 5 X 4;3 (33 + 10X + X 2 ) + 33X 2 + 10X+ 1 , x 5 X 2/3 (15 + 102 X+ 15X 2 ) 
n 2 24 (l + X ) 10 ' 3 16 (1+X ) 10/3 
Now we have to evaluate the moment of momentum given above in § 18, viz. 
X /? o . o\ 1 1 / X 
3,2 
* = Q h TTx rb < S1+ (TTXF 
When b 2 , c 2 , B 2 , (7 2 , have their values attributed to them, we find 
X 
r 2 
p 2 = 
wher 
(1 +X ) 2 \r/ 
a \ 3/2 
-) (1 + 0 
1/2 
a 3 a 6 a 9 
R + S^+T- 6 +U% ... +r 2 
r r r 
:e 
2 
(1 + X 5,3 ) (1 + X) 1/3 , 
oX 
£ = 
r = 
u = 
1 X 2/3 (5 + 2X)+ i(5X + 2) 
A 
6(l + Xf 3 L 
5 
2.3 2 .7 ( 1 +Xf 3 L 
25 
2 2 .3 4 .7 2 (1 + X ) 8 ' 3 _ 
X 2,3 (131+ 34X+ 11X 2 )+ i(131X 2 +34X+ll) 
A 
X 2 ' 3 (40367 + 25548X + 2463X 2 -f 488X 3 ) 
+ ^(40367X 3 +25548X 2 + 2463X + 488) 
A 
In § 1 , where the same problem is treated for two spheres, we had l, m, n, S, T, U, 
all zero. 
2 G 2 
