560 
ME. G. H. DARWIN ON PROBLEMS CONNECTED 
The heat lost by] 44 
earth per annum J =' earfch ’ 8 surface in S( l uare centimeters x — 3 x n —— 3 x 31557 x 10 ? 
= earth’s surface x 45’9 (centimeter-gram-second heat units). 
Now it J be Joule’s equivalent 
Earth’s kinetic energy-i „ 2 / 0 \ 
of rotation in heat L =f-^- =—(f) 2 ( where C = 4M « 3 
units J J ° V 9 J 
= earth’s surface x ^^ (■§-)%, where e 0 =f n ^- ( — -- 1 — 
3J g 
= earth’s surface x { * 3 *3^ * 1Q16 * C 4 ) 2 , * 1Q8 
3 x 4-34 x 10 4 x232 
= earth’s surface x 1‘2 x 10 10 nearly. 
centimeters. 
J = 4'34 x 10 4 gramcentim. 
and w = o\. 
Therefore at the present rate of loss the earth is losing energy by cooling equivalent 
1-2 x 10 10 
to its kinetic energy of rotation in ———— —262 million years. 
If we had taken the earth as heterogeneous and C=^Ma a we should have found 218 
million years. 
We will next find how much energy is lost to the moon-earth system in the series 
of changes investigated in the paper on “ Precession.” 
In that paper (Section 16) it was shown that the whole energy of the system is 
-g-Mcrfrr— ~V where v is earth 4- moon, r moon’s distance, n earth’s diurnal rotation. 
2 vn Q * \r r, 
Hence the loss of energy=TMa 2 n n 2 | (~\ —1— ’ ' 
to n 0> and r from r to r 0 . 
25 / 4/7 \ 100 x 232 
0/j 
Now 
2 vn 0 z 8v \5 n^a 
a— 
32x82 
—a=8'84cq taking n=82, and yy-=23i 
0 ?2'q Ct 
while n passes from n 
4 g 
n l) n 
If D be the length of the day, ; and if n be the moon’s distance in earth’s 
n 0 D 
radii, then 
loss of energy= 
D °V — 1 1 
D 
n ip 
0 /j 
X earth’s present he. of rotation. 
But in the paper on “Precession” we showed the system passing from a day of 5 hours 
40 minutes,* and a lunar distance of 2'547 earth’s radii, to a day of 24 hours, and a 
lunar distance of 60 - 4 earth’s radii. 
* A recalculation in the paper on “ Precession ” gave 5 hours 36 minutes, hut I have not thought it worth 
while to alter this calculation. 
