Hydrostatic Equilibrium

Abstract

The most important characteristic of any celestial body is, indeed, its mass; and in the case of the Moon this is known (cf. Chapter 1) to be 7.35 × 10²⁵ g or 1.2 per cent of that of the Earth. This mass is, moreover, contained inside a globe of r _c = 1738 km mean radius, rendering its mean density ρ _m = 3.34 g/cm³. What is the corresponding pressure inside such a configuration? If the latter were in hydrostatic equilibrium — an assumption which will have to be tested on its merits — the pressure P(r) and the density ρ(r) at a distance r from the Moon’s center should, to a high degree of approximation* be related by the well-known equation

$$ \frac{{dP}}{{dr}} = - g\rho $$

((7–1))

, of hydrostatic equilibrium where the gravitational acceleration g inside a spherically-symmetrical configuration will be given by

$$ g = G\frac{{m(r)}}{{{r^2}}} $$

((7–2))

, where G is the constant of gravitation and m(r), the mass interior to a volume of radius r, follows from the equation

$$ \frac{{dP}}{{dr}} = 4\pi \rho {r^2} $$

((7–3))

.