226 RESULTS OF RECENT RESEARCHES ON THE 
Suppose we denote an element of the mass of a spheroid by m, and its distance from 
the axis of rotation by d; then the moment of inertia is 
IL = Sa 
If the spheroid be rotating with an angular velocity y, then Jy will be the moment 
of momentum of the body about its axis. For a second body whose moment of inertia is 
I’, and angular velocity z, the moment of momentum is /’z. 
Foilowing the analogy of Darwin’s procedure, we choose a system of units designed 
to simplify the resulting equations. Let us take as the unit of mass 
UM. 
MM” 
and as the unit of length a space 1 such that the moment of inertia of the spheroid about 
its axis of rotation shall be equal to the moment of inertia of the two spheroids treated as 
material points, about their common centre of inertia when distant apart [. Then 
we have 
wr)? Mr)? 
M | M+ a | + MW ea } = /Lor 
I (M+ W))3 
ae a | 
Let the unit of time be the interval in which one spheroid describes 57°.3 in its 
orbital motion about the other when distant I’. In this case, 1 is the orbital angular 
velocity of the body. The generalization of Kepler’s law gives 
G7 i= Gh-— i). and 
9 = {Pore my: 
ine pe (ILM)? 
Now suppose the two stars to revolve about their common centre of inertia in a cir- 
cular orbit, with an angular velocity ©, when the radius vector is p. Then the orbital 
moment of momentum is 
a eal OM, 7) 2 rs (rsa) pO 
In a circular orbit the law of Kepler gives 0’p? = uw (M + M); and Qo’ 
= w (M+ IM)? p; andon inserting for Op’ its value, we have u? MMW (M+ M’)~? ¢', 
