172 
SIR G. H. DARWIN ON THE 
Fig. 1. Solutions for two spheres of liquid joined by a weightless pipe, for successive values of A. 1 ' 3 . 
In the case where the two bodies are no longer spheres, the equation corresponding 
to the cubic (3) becomes very complicated. It is therefore desirable to discover 
whether in any given solution of the figure of equilibrium the two detached masses 
are too far apart to admit of their being joined by a weightless pipe, or whether they 
are too near. This may be discovered in the following way :— 
Let r 0 be the solution of f (a/r 0 ) = 0, where 
l + \l/3 + X 2/3 
/( = ) = =- 
a 
-4-3. 
(1 4- X) 1/3 (14-X 1/3 ) r ’ .. 
There is only one solution of f = 0 between r equal to infinity and the case when 
the two spheres touch. Hence we can determine on which side of r 0 any given value 
of r lies by merely considering whether f changes from positive to negative or from 
negative to positive as r increases through the value r 0 . 
Now if — 4- hi — 
n 
r 
be any neighbouring value of —, we have approximately 
3 1 4- X 1/3 4- A 2/3 1 da\ 
2 (14-X) 1/3 (1 + X 1/3 )J W 
If we express a 2 /r 0 2 in terms of r 0 /a by means of the equation for r 0) this may be 
written 
4?J = ( i + xr 
1 4- X 1/3 — — — 
X 1/3 
ax 1 + X 1/3 
a 
where as before 
a 
3 _ 
a 
14-X 
Now the fourth column of the table shows that 1 4-X 1 —r/six is negative, and the 
