No. 8.] REMARKS ON THE EARTH'S CRUST. 65 
that the density, 9, in the firm earth’s crust beneath, changes on an average 
in one and the same manner downwards. Wemay assume that the case is similar 
with regard to the density, @,, beneath the portions of continent considered. 
o and g, are functions of the distance from the earth’s surface, and according to 
the above, the average value along a vertical must be greater for o than for 
o, in the outer spherical shell, of which we will call the thickness h,, when 
we have h, = R, — R,. If we now compare a portion of the earth’s crust 
belonging to the continental lowlands with a piece belonging to the ocean, 
the result of the equation for the flux of force is that 
hy —he hy 
he [re dw (0, —0) dh + 4x [x dw (0, —1)dh=0, 
0 hy —ha 
which means that the difference between the mass of the two pieces above 
equal elements of the inner spherical surface equals zero. 
Here R= Rf, +h=R, —h, +h; substituting this and expanding, we 
obtain, after division by 4” Rj do, 
hy —he hy hy 
he ly 
[te —ea+ le, —1)dh+ ite —el—hin+- [ie —1)(h—h,)ah|+ 
hy—he 0 hy — he 
ly —ha hy 
# pall — (b=) d+ [es — 1) (hy)? dh] =O. 
0 ly — Ia 
The radius of the earth must here be regarded as great, not only in 
proportion to the depth of the sea, h,, which we may take on an average 
as 3.5 km., but also in proportion to the thicknes of the shell, h,. The fourth 
term in the above equation must therefore be exceedingly small, and even 
the third term can only have a small value, as the two integrals in it have 
opposite signs. The first two terms can consequently only have a value that 
searcely differs from 0; moreover it is easy to show that their sum must be 
less than 0. If we inquire into the pressure that the separate parts of the 
earth’s crust each exert upon the interior nucleus on account of the attraction 
of the latter, we shall find that, taking M, as the mass of the interior nucleus, 
it will amount for every surface element, R{ dw, beneath the continents, to 
9 
