OF ROTATING COMPRESSIBLE MASSES. 
199 
In any arrangement of matter whatever, y A will have some value for each layer of 
constant density, being virtually defined by equation (172). There is no reason to 
expect that y A will be constant, or even approximately constant throughout the mass. 
But in the present complex problem we must be content to discover tendencies rather 
than exact results, and so may think of y A as a constant. An increase of atomic 
weight on passing to layers of greater density will give a positive value to y A , while 
similarly a diminution of atomic weight would give a negative value of y A . 
The whole of the foregoing analysis will now apply to the present case if we put 
- = — + —.(173) 
7 7m 7a 
It has been seen that for values of y less than a critical value, which may be taken 
(although only as a rough approximation) to be 2’2, it will be impossible for the 
rotating mass to assume the ellipsoidal form, and so impossible for it to separate into 
two detached masses. The condition for assuming the ellipsoidal form now becomes 
A + ±< A 
7m 7a 2 2 
Thus, if y A is positive (atomic weight increasing towards centre), the critical value 
of y M will be greater than 2'2 ; with y A negative, the critical value of y M will be less 
than 2'2. It is possible for a heterogeneous but perfectly incompressible mass 
(7m = oo) to fail to attain the ellipsoidal shape if y A is small enough— i.e., if the 
layers increase sufficiently rapidly in density from increase of atomic weight alone. 
44. In the foregoing' discussion we have considered only the effects of continuous 
changes of atomic weight. Entirely confirmatory results may be obtained from a 
consideration of excessively abrupt changes. 
Let us consider only the simplest case in which the mass is formed of completely 
separated layers of two substances only, one very light and the other very heavy— 
to make the picture definite, let us think of an inner mass of iron vapour surrounded 
by an atmosphere of hydrogen, and let us suppose the density of the hydrogen may 
be neglected in comparison with that of the iron, so that the gravitational field will 
be the same as if the iron alone were present. 
For any given value of w we can draw the equipotentials Q = cons., and the 
boundary of the mass of iron must be one of these, say Q = C,. Proceeding outwards 
and drawing the exterior equipotentials we must in time come to one having a double 
point, say Q = (3 X , and at the double point on this equipotential centrifugal force will 
be exactly equal to gravity (dQ/dn = 0). The space between the equipotentials Q — C x 
and Q = C x can be filled with hydrogen without matter being thrown off by rotation, 
but clearly no greater volume of hydrogen than this can be retained. 
As the rotation w increases from zero upwards, the equipotential = C z starts 
from infinity and moves inwards, so that the maximum volume of hydrogen which 
2 D 2 
