344 Controversy between Archdeacon Pratt and Prof. Haughton. 
Let M be the entire mass of the earth. Then since 
a 
M=4mr( pa”, 
20 
if we introduce this value into the last term of equation (12), 
pais by a?, and substitute for e’ and p! their values, we shall 
ave 
ete) “yay —3 (ye) O_O Hee) 
mMa? _ 
Sia? 
This equation applies to the surface of the fluid nucleus. For 
the next surface de niveau within the fluid the equation will be 
(o—apege—m) (arya) —5 | way PO 
Baa * wanana) +4 Beye) b* 
a-h 
mM. 
— Sra? (a—h)°=0. 
Expanding and retaining terms of the first order only, we find 
; re 
~ (ab) +E go) (ee) eG He) + pa SOO 
0 
+0 (FWY) Ze Woge) + gas 05 
or striking out the terms which cancel each other, and dividing 
* If I rightly understand Archdeacon Pratt’s reasoning, he supposes 
that, under the conditions stated in the text, the differential coefficient of an 
expression such as 
a a 
m\ p'+n} pp! 
0 a 
my(a)—nk (a). 
This, however, is not so. It is easily seen by the mode of reasoning 
adopted in the text, that this coefficient will be either 
(m—n)y(a) or (m—n)I*(a), 
according to the region to which we suppose the differentiation to apply. 
The result, 
my(a)—nF(a), 
could only be obtained by the substitution of a—h for a in the first inte- 
gral, and of a+h for a in the second, a process which would be of course 
illogical. 
with regard to a would be 
