Hence for equal decrements of heat, or by the cooling in equal times, the hotter nucleus contracts more than does its envelope of solid matter.
The result is now, as at all periods since the signs changed of the tangential forces thus brought into play—i.e., since they became tangential pressures—that the nucleus tends to shrink away as it were from beneath the crust, and to leave the latter, unsupported or but partially supported, as a spheroidal dome above it.
Now what happens? If the hollow spheroidal shell were strong enough to sustain, as a spheric dome, the tangential thrust of its own weight and the attraction of the nucleus, the shell would be left behind altogether by the nucleus, and the latter might be conceived as an independent globe revolving, centrally or excentrically, within a shell outside of it. This, however, is not what happens.
The question then arises, Can the solid shell support the tangential thrust to which it would be thus exposed? By the application to this problem of an elegant theorem of Lagrange, I have proved that it cannot possibly do so, no matter what may be its thickness nor what its material, even were we to assume the latter not merely of the hardest and most resistant rocks we know anything of, but even were it of tempered cast-steel, the most resistant substance (unless possibly iridio-osmium exceed it) that we know anything about. Lagrange has shown that if P be the normal pressure upon any flexible plate curved in both directions, the radii of these principal curvatures being ρ' and ρ'', and T the tangential thrust at the point of application and due to the force P, then:
P = T (1/ρ' + 1/ρ'')
When the surface is spherical, or may be viewed as such, ρ' = ρ'' and
P = 2T / ρ or, T = P × ρ/2
In the present case P is for a unit square (taken relatively small and so assumed as plane) of the shell, suppose a square mile, equal to the effect of gravity upon that unit, ρ being the earth's radius, and if we assume the unit square be also a unit in thickness, P is then the weight of a cubic mile of its material; and if we take (roughly) the earth's radius as 4,000 miles, the tangential pressure, T, is, on each face of the cubic mile, equal to
(4000/2) P,