FIGURE AND STABILITY OF A LIQUID SATELLITE. 
165 
In both problems the rotational momentum of the first sphere is 
t(*irpa»)a’qA_) «, 
In the first problem the rotational momentum of the second sphere is 
2/4 3\ 2 1 
5 (s^a )a + 3 <u, 
and in the second problem it is nil. 
If, then, we write A for the total angular momentum of the two spheres, and A 
for that of the sphere and particle, we have 
A = f 77 -pa 5 
(O 
I+A 5 ' 3 
+ 
\r‘ 
L 2 = f 7rpa°w 
L 5 (l + X) 5/3 (1 + A) 2 a 2 
A 5 ' 3 X/- 2 
( 1 ) 
„ 5 (1+X) 5/3 (1+X) 2 a 2 _ ‘ 
On substituting for oj its value in terms of r, these expressions become 
A = ii^pY 
3/2 
a 
A = A Tip) 
3/2 
(i+x) 2 L 
n 5 
(iTx) 2 L 
f (1 + X 5/3 ) (! + X) 1/3 (^) + A(M 
3/2 
ix 5 ' 3 (i+x) 
,/aJ 
1/3 
3/2 
+ A - 
,\l/2‘ 
v a / 
To determine the minima of these functions, we differentiate with respect to r. and 
equate to zero. 
Then, if r u r 2 denote the two solutions, we find 
r, 
a 
- 2 = fX 2/3 (l+X) 1/3 . 
Whence 
Minimum A = (f 7rp) 3/2 a 5 . 4 (t! 5 ) 1/4 X ^^ 3 /2 > 
Minimum A = (f 7rp) 3/2 a 5 .4 (tAt) 1 ' 4 
X 7/6 
(1+X) 
23/12 " 
/ 1 \ 1/4 
The ratio of the minimum of A to that of A is f^+lj . Thus as X rises from 
0 to co this ratio falls from infinity to unity. 
All the possible cases of the first problem are comprised between X = 0 and A = 1. 
When X = 0, i\ = co; and when X = 1, 
r, 
a 
1= /(ii.21/3)^ 1738. 
