FIGURE AND STABILITY OF A LIQUID SATELLITE. 
175 
cannot create it. Hence, when two spheres revolve about one another, the opening 
of a channel of communication between them may destroy stability, but cannot create 
it. When two equal spheres revolve about one another at such a distance that they 
could be connected by a pipe and yet remain in equilibrium, their distance is 1‘666 ; 
but they are then unstable, because 1'666 is less than 1738. The opening of a pipe 
between them, being the removal of a constraint, cannot make the motion stable. 
A fortiori the like is true when the two spheres are unequal in mass. 
Hence the system of equilibrium of two spheres joined by a pipe is unstable in all 
cases. 
I will now consider the meaning of the vanishing of A. 
Having evaluated the angular momentum of the system corresponding to the 
several solutions tabulated above, I found it had a minimum when X 1/3 = 0'254. 
Such a solution is a critical one and is the starting point of two solutions of which 
one must have one fewer degrees of instability than the other. The vanishing of A 
must have the same meaning, but it remains to be proved that minimum angular 
momentum is secured by the vanishing of A. 
The angular momentum is Ioj, and is therefore proportional to fx, where 
.\!/2 
v a 
= F +4(2 - 
i /a V 
r 
3/2 
On equating ^ to zero so as to find its minimum, we have 
Now 
since 
we have 
2r(F’?- a + iG') + (F±-%Q)%-= 0 . 
a 
\ dr 
a 
/ dk 
dr 
dX 
zf + i + 5iL'r = 0 
a 3 +3 + 2 Q , a 
■^(1-X^) 2 \-^(1+X)- 4 
f G' - + F' 
a 
On substituting this value in the equation d/i/dX — 0 , I find that the result may 
be written 
(1-X 1 / 3 ) 2 r 2 
X 1/3 (l a 2 
*°- F v 
+ 
a 
a 
= 0. 
Hie first term of tbis is the same as the first term inside the bracket in the 
expression for A in (5). On comparing the two second terms together we see that 
A = 0 is the condition for minimum angular momentum, if 
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