50G 
MR. G. H. DARWIN ON A PLANET 
In proceeding to consider the graphical method of solution by means of the contour 
lines of the energy surface, I shall choose the total momentum of the system to be 
greater than this third critical value, and the surface will have a maximum point. 
From the nature of the surface in this case we shall be able to see how it would differ 
if the total momentum bore any other position with reference to the three critical 
values. It will be sufficient if we only consider the case where the masses of the two 
satellites are small compared with that of the planet. 
By (17) we have, with an easily intelligible alternative notation, 
Now is an angular velocity, and if we choose 1/zq as the unit of time, we have 
also 
K l = 1 , K.,= 
m 2 
TOj 
X -S*L 
A, — - k , 
1 1 
Xo = 
2 
5 
M 
m 2 
O 
Now if we choose the mass of the first satellite as unit of mass, then m x — 1, and 
we have 
/q= 1 , K, = rn,, Aj = ?d/, A 2 =%Mm 2 
The unit of length has been already chosen as equal to the mean radius of the 
planet. 
Then substituting in (21) we have as the equation to the energy surface 
2z= 
And since we suppose ??? 1 and m. 2 to be small compared with M, we have 
On account of the abruptness of the curvatures, this surface is extremely difficult to 
illustrate unless the figure be of very large size, and it is therefore difficult to choose 
appropriate values of h, M, m 2 , so as to bring the figure within a moderate compass. 
In order to exhibit the influence of unequal masses in the satellites, I choose m 2 = 2, 
the mass of the first satellite being unity. I take M=50, so that ^M=20. 
Then with these values for M and m, the first critical value for h is 3'711, the 
second is 6‘241, and the third is 8'459. 
I accordingly take // = 9, which is greater than the third critical value. The surface 
to be illustrated then has the equation 
2 *= (9 - a* - 2yf-2o(\+^}j 
