FIGURE AND STABILITY OF A LIQUID SATELLITE. 
187 
For Roche’s problem, when the second body is a particle, V reduces to ( eE) 1 + (vv ), 
but in the modified form of the problem which I am going to solve the whole 
expression is required. 
The evaluation of (eE) is so complicated that I devote a special section to it. 
* 
§ 7. The Mutual Energy of Two Ellipsoids/ 1 ' 
The semi-axes of the ellipsoid of mass e are to be denoted a, h, c, and the 
corresponding notation for the other ellipsoid is E, A, B , C. The distance between 
the centres of e and E is v ; the axes c of e and O of E are in the same straight line, 
while a, A and h, B are respectively parallel. For brevity I imagine the densities ol 
the ellipsoids to be unity. 
If the external potential of e be U, and if dll be an element of volume or of mass 
of E, the lost energy to be evaluated is 
(eE)=\udn, 
integrated throughout the ellipsoid E. 
Let us suppose provisionally that the co-ordinates of the centres e and E are x, y, z 
and X, Y, Z, and let £ y, £ be the co-ordinates of the element dti ; the axes being 
respectively parallel to a, h, c or A, B, C with arbitrary origin. 
If R 2 = (£-x) 2 + {y-y) 2 +{l—z) 2 , ^ is well known that the potential U of e at the 
point y, £ is given by 
a 2 v i 
U=et 
(2n + l)(2n+3)2«! \ 
a 2 — + h 2 — + c 2 
^° + b dr, 2 0 V) R 
02 
Since satisfies Laplace’s equation, we may eliminate , and observing that 
a 2 a 2 a 2 a 2 
^, are the same as 5 , ^ respectively, we have 
dx Oy- or. 
2 a 2 /2 a 2 2 a 2 , 2 2X a 2 , 2 m 3 2 
“ gp +6 ay +c a?" {c hr ( J h^ 
dr, 2 
It follows therefore that 
_k 2 dx 2 dy 2 \ ’ 
U=et 
(~)*3 
F-|lA. + saVl 
0 ( 2 w + l)( 2 n + 3)2w! W dx 2 dy 2 j R 
Since the operator is independent of r,, f we have 
(eE) — e S 
(-) n 3 pn /1 _3i , 31Y. frfo 
0 (2n+l)(2n+3)2n\ W dx 2 dy 2 ) J R 
v The results of this section were arrived at originally by a longer method, I have to thank one of the 
referees for showing me the following procedure .-—February 26, 1906, 
2 B 2 
