Earth’s figure and gravitation - Earth Dynamics: Deformations and Oscillations of the Rotating Earth - page 327

potential GM / d ,andρ is expressed in units of the mean density 3 M /4π d 3 , while
the co-ordinates are given in units of d . The dimensionless version of (5.5) is then

2 d 3
GM .

2 Ω

2 U

∇ = 3ρ − (5.6)

The problem is now expressed solely in terms of the dimensionless parameter,

2 d 3
GM ≈

= Ω

1
290 ,

10 − 3 m 3.45 × = (5.7)

the ratio of centrifugal force to gravitational force on the surface of a hypothetical,
spherically symmetric mass distribution with the same mean radius and mean dens-
ity as the Earth. We shall see later that it is very close to the parameter
introduced by Newton in the first physical discussion of the Earth's figure. It would
now be called a Froude number in hydrodynamics. The utility of expressing the
problem in dimensionless form lies in the fact that, for the Earth, m is only about
1 290, so expansion of the dimensionless variables in powers of m permits uniform

approximation by rapidly converging series.
We expect the reference level surface to be a surface of revolution and to be
symmetrical under reflection in the equatorial plane. At least, there is no known
theoretical reason, at the relatively low rotation rate of the Earth, to expect a
departure of the overall figure from these symmetries. Thus, the dimensionless
radius of the reference level surface, R , is an even function of the vertical distance
to the equatorial plane, proportional to cosθ, and has the expansion

/

m 2 R 2 (θ) R (θ) = R 0 (θ) + mR 1 (θ) + +··· , (5.8)

where the functions of θ are expressible in series of Legendre polynomials of even
degree.
The function R 0 (θ) describes the figure free of the influence of rotation. A proof
that a sphere is the unique figure of equilibrium for a self-gravitating fluid at rest
was given by Liapounov in 1918 (Liapounov, 1930). Thus, in dimensionless form,

R 0 = 1 (5.9)

describes our zeroth-order approximation to the Earth's figure. With λ as the east
longitude, the volume enclosed by the true reference level surface is

2π π R (θ)

r 2 dr sinθ d θ d λ V=

0 0 0

π

2 3

R 3 (θ)sinθ d θ. = (5.10)

0

Now,

m 2 R 2 (θ), R (θ) = 1 + mR 1 (θ) + (5.11)