152-154] Roche's Model 151 



the ejection of streams of matter from the equator. It is obviously very 

 desirable to bridge over, if possible, the wide gap between these two extreme 

 cases, and this is to some extent effected by the consideration of a third 

 model, which combines some of the properties of both of the two models so 

 far discussed. 



154. Roche's model consisted of a nucleus of finite mass but infinitesimal 

 volume, surrounded by an atmosphere of zero mass but finite volume. The 

 density of the nucleus was accordingly infinite while that of the atmosphere 

 was zero. 



In the new model we take the nucleus to be of finite extent, and there- 

 fore of finite density, while supposing the atmosphere to remain of finite 

 extent but of infinitesimal density. Thus the potential of the mass V M may 

 no longer be put equal to Mjr but becomes equal to the potential of the 

 nucleus. The nucleus will be supposed to be incompressible and of uniform 

 density p 0) and the atmosphere will be supposed to exert no appreciable 

 pressure on the nucleus. 



Let V N denote the volume of the nucleus and V A that of the atmosphere. 

 The mass M is equal to p v N) so that the mean density p is given by 



V N 



P= 7W" 



Under a rotation co each particle of the nucleus will be subjected to 

 exactly the same forces as though the nucleus alone were rotating with 

 angular velocity co, the atmosphere being entirely non-existent. This deter- 

 mines the configurations of the nucleus ; they consist of Maclaurin spheroids, 

 Jacobian ellipsoids, etc. 



The boundary of the atmosphere must be one of the equipotentials 

 H = cons.; it must moreover be an equipotential of total volume V A + V N . 

 Thus to get a complete figure of equilibrium corresponding to a given rota- 

 tion co, we must first draw a figure of equilibrium appropriate to this 

 rotation for an incompressible mass of density /o and volume V N . The boun- 

 dary of this will be an equipotential Q = cons, of volume % . We must then 

 draw successive outer equipotentials until a further volume V A has been 

 enclosed. The equipotential which just includes a further volume V A will 

 be the required boundary. 



It may be that, in drawing these equipotentials, we shall find that closed 

 equipotentials give place to open ones before a volume V A has been enclosed. 

 If so, there can be no figure of equilibrium corresponding to the given rota- 

 tion. If V A ' is the volume enclosed by the last closed equipotential, the 

 greatest atmosphere which can be retained at the given rotation will be 

 one of volume V A , and of the atmosphere of the original model, a volume 

 U A VA must already have been thrown off at the equator. 



