STABILITY OF THE PEAR-SHAPED FIGURE OF EQUILIBRIUM. 
19 
The Numerator divided by the Denominator is — Sa> 2 /47 rpe 2 , whence 
8a> 2 
log, = (-) 8-01054. 
1 4:77 pC 
With ~~ = -141990, from § 7 of the “ Pear-shaped Figure,” we have 
oj 2 + Scj 2 = a) 2 .[l-T443066e 2 ]. 
Thence we find 
f 2 = -195979c 2 , f 2 = -603177c 2 . 
These values differ sensibly from the old ones. 
The moment of inertia with log a = 9’8559759 is given by 
A,—A r = 
3Ma 
t , 
ph 
- [l + -157786e 2 ]. 
The moment of momentum is 
3APa 
O) 
2 7 T plvQ 
[l + -085633e 2 ]. 
As before, we find the pear-shaped figure to be stable, because the moment of 
momentum is greater than that of the critical Jacobian, provided that the infinite 
series does not amount to too great a sum. 
If e be the uncomputed residue of 2 
moment of momentum is 
I then find, as before, that the 
+ •085633e 2 -499-586ee 2 ]. 
^77 
The coefficient of e 2 will be positive and the pear stable, provided that 
499"586e < -085633, 
or 
e < -0001714. 
The eighth zonal harmonic gave a contribution of '0000196, and the tenth of 
"0000063. These are respectively a ninth and a twenty-seventh of the critical total. 
The pear is then stable unless the residue of the apparently highly convergent series 
shall amount to more than 27 times the value of the last term computed. 
M. Liapounoff claims in effect to prove that this is the case, but to me it seems 
incredible. I look for the discrepancy between our conclusions in some other 
direction. 
