558 
MR. G. H. DARWIN ON PROBLEMS CONNECTED 
(Section 16) that if n be the earth’s rotation, r the moon’s distance at any time, v the 
ratio of the earth’s mass to the moon’s, then the whole energy both potential and 
kinetic of the moon-earth system is 
Av r 
Now c being the moon’s distance initially, since the lunar orbit is supposed to be 
circular, 
„ Q cd + V 
fl Q ~c 6 =ga~' -. 
Also 
Therefore 
n 
r \S2, 
1 
T 2 ' 
|2=»{@V (1 +, ) }V=*n°-*. 
according to the notation of the paper on “ Precession.” 
In that paper I also put -=sn 0 f. 2 0 k 
d 
Therefore % 
And the whole energy of the system is bc(iv — 
Therefore tbe rate of loss of energy is — 
dn 
d% n 
a 
But and as shown in the first part (19), fin Q — = 77 , also — ;"= 12 . 
dt C u dt C 
Therefore the rate of loss of energy is (n — Sl) or JToj, which expression agrees 
with that obtained above: The two methods therefore lead to the same result. 
I will'now return to the investigation in hand. 
The average throughout the earth of the rate of loss of energy is -f--§ 7 ra 3 , which 
quantity will be called H. Then 
H = = fMoA-g— sin 4:e.bj=\wc (?sin 4e. w. 
1 T* 
a 
Now 
1 10T . \~ T~ . 3 „ 1 0 T . H 
- — sm 2e a*= — cot 2e. — snr 2e.iva~ = Ta--kwa~. — sm 4e.a> = —. 
u\19 / 5a 19 ij a 10 
5n 
19 
Hence (26) may be written 
dE_H 
dt 19 
V 
— f - sin 2 0« 32 —(26+sin 2 0)(-- 
(28) 
