EQUILIBRIUM OF ROTATING MASSES OF FLUID. 
415 
body. Now, if for equal masses equilibrium still subsists when the two masses are in 
contact, whilst for infinitesimal mass of one equilibrium is impossible with the masses 
in contact, it follows that for some ratio of masses equilibrium can just subsist when 
they are in contact. 
The question, therefore, remains to determine this limiting ratio of masses. 
I find, then, that when a = 1, A = 3'4, we have 
<3 = 4-4 + [2-25684] c~ 3 + [1 -94945] c“ 4 + [2-07156] c" 5 + [2-37619]<U G + [2-80737] c“ 7 , 
C= 1 — [2-15205] c“ 3 — [2-13850] c~ 4 - [2*24765] c~ 5 — [2'33480] c“ 6 - [2-40735] c ~ 7 , 
the numbers in [ ] being the logarithms of the coefficients. 
The solution of this is c = 5'57, which makes C — — "006. 
Again, when a = 1, A — 3 "3, we have 
c = 4-3 + [2-21556] c“ 3 + [1 -91353] c" 4 + [2-03266] c~ 5 + [2-32731] c" 6 + [274467] c~\ 
C= 1 — [2-11347] c —3 — [2-09961] c~ 4 — [3*20876] c~ 5 — [2-29591] c" 6 —[2-36846] c~ 7 , 
the solution of which is c — 5"45, which makes C = + *010. Since (3"4) 3 = 39"3, and 
(3"3) 3 = 35*9, it follows that the ratio of the masses in the first case is 1 : 39"3, and in 
the second L : 35"9. 
From this it appears that when the ratio of the masses is about 1 to 38 equilibrium 
is still just possible when the two masses touch. 
It must be borne in mind, however, that the nature of the approximations adopted 
in this investigation is such that the results in this limiting case are only given very 
roughly, and it is certain that actually the limiting size of the smaller of the two 
masses must be greater than as thus computed. 
We can only conclude that the limiting case occurs when the ratio of the masses is 
about 1 to 30, or the ratio of the radii about 1 to 3. 
There is one other case which it is interesting to consider, namely, to find the 
limiting proximity of the Moon to the Earth, both bodies being treated as homogeneous 
fluids of the same density, revolving as a rigid body. 
The case of Moon and Earth is well represented by a — 1, A = 4*333 ; for this 
gives 1 to 81 "35 as the ratio of the masses. With these values I find 
(7=1- [2-46626] c~ 3 — [2-45443] c~ 4 — [2*56358] c~ 5 — [2*65073] c“ 6 - [2*72328] c~ 7 , 
and 
r + R = 5-333 + [2-59898] c“ s + [2*24358] c" 4 + [2-38748] c~ 5 + [2-77563] c“ 6 
+ [3-32042] c“ 7 . 
Now c = 7*0 will be found to make (7 vanish, and, with this value of c, c — (y + R) 
= -414. 
If + be 4000 miles, c = 6500 miles, and c — (r + R ) = 380 miles. 
