Maclaurin spheroid

Example

Consider a homogeneous, incompressible fluid of density $\rho$ that attracts itself by Newtonian gravity and rotates rigidly with angular velocity $\Omega$ about a fixed axis. A Maclaurin spheroid is an oblate spheroid, symmetric about the axis of rotation, that is a figure of equilibrium of this fluid. This means that in the rotating frame the fluid is in hydrostatic equilibrium and the pressure vanishes on the boundary. With eccentricity $e$, where $0 < e < 1$, and $G$ the gravitational constant, equilibrium holds exactly when $$\frac{\Omega^2}{\pi G \rho} = \frac{2 \sqrt{1 - e^2}}{e^3} \left( 3 - 2 e^2 \right) \arcsin e - \frac{6 (1 - e^2)}{e^2}.$$ Check: inside a homogeneous spheroid the gravitational potential is a quadratic function of the coordinates, and so is the centrifugal potential $-\frac{1}{2} \Omega^2 (x^2 + y^2)$. So the total potential is quadratic. Let $a$ and $c$ be the equatorial and polar semi-axes. The total potential is constant on the surface exactly when its coefficients of $x^2 + y^2$ and of $z^2$ are proportional to $1/a^2$ and $1/c^2$. This condition is the displayed relation. As $e \to 0$ the right side tends to $0$, and one recovers the sphere at rest. The right side increases with $e$ up to a maximum of about $0.4494$ at $e \approx 0.9299$ and then decreases to $0$ as $e \to 1$.