 Solution

On a sphere of radius $r$, the radial gravitational field has constant magnitude. Thus Gauss's law for gravity becomeswhere $M(r)$ is the enclosed mass. In the hollow core $M(r)=0$. In the material shell, put $u=r/r_0$ and use the spherical coordinates volume element:Outside the shell the enclosed mass is the totalConsequently,At $r=0$ the displayed value refers to the radial derivative of the constant core value. The expressions match at both shell boundaries. There is no surface sheet of mass, so a jump in the volume mass density does not produce a jump in the gravitational field. The core has zero gravitational field, and the exterior gravitational field equals that of a central point mass $M_{\ast }$, as in the shell theorem.