where λ = 3⁄5(4h² − f²) / r4,   λ′ = 3⁄5(2h² − 3f²) / r4,   and r² = f² + h².

Now e being considered a function of c, we can at once express the attraction of a shell (density ρ) contained between the surface defined by c + dc, e + de and that defined by c, e upon an external point; the differentials with respect to c, viz. dX″ dZ″, must then be integrated with ρ under the integral sign as being a function of c. The integration will extend from c = 0 to c = c1. Thus the components of the attraction of the heterogeneous spheroid upon a particle within its mass, whose co-ordinates are f, 0, h, are

X = −4⁄3πk²f [ 1∫c10 ρ d{c³(1 + 2e)} − λ∫c10 ρ d(ec5) + 2⁄5∫c0c1 ρ de],
r3 r3
Z = −4⁄3πk²h [ 1∫c10 ρ d{c³(1 + 2e)} − λ′∫c10 ρ d(ec5) + 4⁄5∫c0c1 ρ de].
r3 r3

We take into account the rotation of the earth by adding the centrifugal force fω² = F to X. Now, the surface of constant density upon which the point f, 0, h is situated gives (1 − 2e) fdf + hdh = 0; and the condition of equilibrium is that (X + F)df + Zdh = 0. Therefore,

(X + F) h = Zf (1 − 2e),

which, neglecting small quantities of the order e² and putting ω²t² = 4π²k², gives

2e ∫c10 ρd{c³(1 + 2e)} −6 ∫c10 ρd(ec5) −6 ∫c10 ρde =3π .
r³ 5r55 t²

Here we must now put c for c1, c for r; and 1 + 2e under the first integral sign may be replaced by unity, since small quantities of the second order are neglected. Two differentiations lead us to the following very important differential equation (Clairault):

d²e +2ρc² ·de + (2ρc −6 ) e = 0.
dc² ∫ρc² dcdc ∫ρc² dcc²