564 
MR. G. H. DARWIN ON PROBLEMS CONNECTED 
to which it should be equal. 
Then by substitution in (29) we have as the complete solution of the problem 
satisfying all the conditions 
3 = V+ 
V 
k 
120-907 cos — 
2 a 
, * „ L, 37 TX 
D107 cos —- 
2 a 
•048 cos 
The only quantity, which it is of interest to determine, is the temperature gradient 
dS 
at the surface, which is equal to —— when x=^a. 
Now when x—-^a. 
d!) _ \yd 7 t 
~dx = ~k 2 
120-907 
— 3-321 
•240 1—e 
( n? \ o 
— ) t is a small fraction, we have approximately 
and since 
k 7 
dS_ 
dx 
\yd° 
: T 
77- / 7T \ 3 
K ' 2a) 
o 
t\ 120-907 —9X3-321-25 X-240 
d$- 
dx 
This formula will give the temperature gradient at the surface when a proper value 
is assigned to f), and if t be not taken too large. 
With respect to the value of t, Sir W. Thomson took /c=400 in British units, the 
year being the unit of time ; and cc = 21 X 10° feet. 
Hence 
= 4X10- 
1-5 
2-1 xlO 7 
= ^ nearly, 
/ 07T \ O 
and k(— j = qqqii 1 ^ therefore t be 10 9 years, this fraction is Therefore the 
solution given above will hold provided the time t does not exceed 1,000 million years. 
We next have to consider what is the proper value to assign to i). 
By (27) and (28) it appears that f)a 4 is sirro of the average heat generated 
