26 
MR. J. H. JEANS ON THE INSTABILITY OF THE PEAR-SHAPED 
On substituting this value for r into equation (105) and equating 
of e and e 2 , we obtain 
2 f+- 
a 
x 
a — + /3“- a + y—+/c 
a b c 
0 
x 
f 2 + 2 g + f - 
a 
x 
:; w? +/3 b +y ? )+ ' t 
, 2 \ 
+ ¥ 
Lx 4 My 
a 4 + 1/ 
+ 
+ s 
the coefficients 
. . . (106) 
= 0. . (107) 
If dQ is an element of solid angle, equation (103) may be written in the form 
3 | j 
20t r JJ 
(a 2 x 2 + b 2 y 2 ) r 4 dr dQ, 
(a 2 x 2 + b 2 y 2 ) (l + 5ef+ 5e 2 g + 10e 2 f 2 )dQ. 
Hence we find that Id may be written in the form Jc 2 + A k 2 , where, as far as e 2 , 
t 
The integral is here taken over the sphere of unit radius, and so can be easily 
evaluated. Carrying out the necessary computations, I find 
h 2 = 1 (a 2 + b' 2 ) = 0-844105. 
Aid = — 0'079156e 2 . 
Thus the moment of inertia is given by 
M/c 2 = 0'844105M (l —0'09378e 2 ) 
20. This again differs from Sir G. Darwin’s result, in which it will be remembered 
it was found that the moment of inertia of the pear increased as e 2 increased. But 
we shall now see that the difference between the two results agrees exactly with 
what was to be expected from the different values of n" used in our two solutions. 
Sir G. Darwin chose his parameter e (which I shall denote by e D ) in such a way 
that the longest radius vector of the pear-shaped figure was a (l + 0’1482e D ). In 
my solution, the longest radius vector is found to be a (l + 0T309ej), where e 3 denotes 
my parameter e. Hence I find as the relation between our parameters 
e D = 0'8833ej. 
Darwin’s rotation was given by 
df- = 014200(1 —0T443066e D 2 )= 014200 (l-011259cj 2 ). 
27 rp 
