The attraction field of a spherical mass

consider the field of a homogeneous sphere with radius a. Taking into account the spherical symmetry of the mass distribution, it is natural to introduce a spherical system of coordinates with its origin at the center of the sphere, Fig. 1.5c. Then, the vector g p) is in general characterized by three components  

By definition, any plane 0 — constant is a plane of symmetry. In other words, there are always two elementary masses, which are equal to each other, and located at opposite sides of this plane but at the same distance. As is seen from Fig. 1.5d, the sum of 0-components, caused by both masses is equal to zero. Representing the total mass as a sum of such pairs we conclude that the 0-component, gg, due to the spherical mass is absent at every point outside and inside the body. In the same manner we can prove that — 0. Of course, volume integration, Equation (1.6), can prove this fact, but this procedure is much more complicated. Thus, the attraction field has only a radial component, g, and the field is directed toward the origin, 0. In order to determine this component we will proceed from Equation (1.26) and consider a spherical surface with radius R, as is shown in Fig. 1.5c. Inasmuch as dS — dSiR and the scalar component g is constant at points of the spherical surface, we have for the flux  

Therefore, the attraction field, g%, outside the sphere is equal to 

Here m is the mass enclosed by the surface of integration, S. As follows from Equation (1.126), the field outside coincides with that caused by a point source with the same mass, M, located at the sphere center. This is a well-known result, which is hard to predict. In fact, this behavior occurs regardless of how close the observation point is to the sphere, and it results from the superposition of fields, caused by elementary masses. This is rather an exception, since in general the field differs from that generated by an elementary mass. 

Next consider the field inside the sphere. Since 

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