Cassini states of a rigid body with a liquid core

Abstract

The purpose of this work is to determine the location and stability of the Cassini states of a celestial body with an inviscid fluid core surrounded by a perfectly rigid mantle. Both situations where the rotation speed is either non-resonant or trapped in a $p\!:\!1$ spin–orbit resonance where p is a half integer are addressed. The rotation dynamics is described by the Poincaré–Hough model which assumes a simple motion of the core. The problem is written in a non-canonical Hamiltonian formalism. The secular evolution is obtained without any truncation in obliquity, eccentricity or inclination. The condition for the body to be in a Cassini state is written as a set of two equations whose unknowns are the mantle obliquity and the tilt angle of the core spin axis. Solving the system with Mercury’s physical and orbital parameters leads to a maximum of 16 different equilibrium configurations, half of them being spectrally stable. In most of these solutions, the core is highly tilted with respect to the mantle. The model is also applied to Io and the Moon.

Appendices

Flattening coefficients

Let $A\le B \le C$ be the principal moments of inertia of a given body. The polar and equatorial flattening coefficients $\alpha$ and $\beta$ are, respectively, defined as (Van Hoolst and Dehant 2002)

$$\begin{aligned} \alpha = \frac{C-(A+B)/2}{C},\qquad \beta = \frac{B-A}{C}. \end{aligned}$$

(50)

We denote by $I = (A+B+C)/3$ the mean moment of inertia. The following relations holds

$$\begin{aligned} A = I\left( 1-\frac{1}{3}\alpha -\frac{1}{2}\beta \right) ,\qquad B = I\left( 1-\frac{1}{3}\alpha +\frac{1}{2}\beta \right) ,\qquad C = I\left( 1+\frac{2}{3}\alpha \right) . \end{aligned}$$

(51)

Spin operator

Let $\varvec{M}$ and $\varvec{N}$ be two matrices and $\varvec{M}' = \varvec{R} \varvec{M} {\varvec{R}}^{\mathrm T}$ where $\varvec{R}$ is a rotation matrix. Under an infinitesimal rotation increment, $\delta \varvec{R} = \delta \varvec{\Theta }\varvec{R}$, with $\delta \varvec{\Theta }\in \mathrm{skew}(3)$, the matrix $\varvec{M}'$ is transformed according to

$$\begin{aligned} \varvec{M}' + \delta \varvec{M}' = (\varvec{R} + \delta \varvec{R}) \varvec{M} {(\varvec{R} + \delta \varvec{R})}^{\mathrm T} = \varvec{M}' + [\delta \varvec{\Theta }, \varvec{M}'] + O(|\delta \varvec{\Theta }|^2). \end{aligned}$$

(52)

Let $f=\langle \varvec{M}', \varvec{N}\rangle$. Under the same infinitesimal rotation, the variation of f is given by

$$\begin{aligned} \begin{aligned} \delta f&= \langle [\delta \varvec{\Theta }, \varvec{M}'], \varvec{N}\rangle \\&= \langle \delta \varvec{\Theta }\varvec{M}', \varvec{N}\rangle - \langle \varvec{M}'\delta \varvec{\Theta }, \varvec{N}\rangle \\&= \langle \delta \varvec{\Theta }, \varvec{N} \varvec{M}^{\prime \mathrm {T}} \rangle - \langle \delta \varvec{\Theta }, \varvec{M}^{\prime \mathrm {T}}\varvec{N} \rangle \\&= \langle \delta \varvec{\Theta }, [\varvec{N} , \varvec{M}^{\prime \mathrm {T}}] \rangle . \end{aligned} \end{aligned}$$

(53)

But by definition, this variation $\delta f$ can also be written $\delta f = \langle \delta \varvec{\Theta }, \hat{\mathcal J}(f) \rangle .$ We thus deduce that $\hat{\mathcal J}(f) = -[\varvec{M}^{\prime \mathrm {T}}, \varvec{N}]$. In particular, if $\varvec{M}=\varvec{S}$ is symmetric ($\varvec{S} = {\varvec{S}}^{\mathrm T}$), then

$$\begin{aligned} \hat{\mathcal J}(\langle \varvec{S}', \varvec{N} \rangle ) = -[\varvec{S}', \varvec{N}], \end{aligned}$$

(54)

while if $\varvec{M} = \varvec{A}$ is skew-symmetric ($\varvec{A} = -{\varvec{A}}^{\mathrm T}$),

$$\begin{aligned} \hat{\mathcal J}(\langle \varvec{A}', \varvec{N}\rangle ) = [\varvec{A}', \varvec{N}]. \end{aligned}$$

(55)

Stability of Colombo’s top

Colombo’s top is an axisymmetric body whose orientation is determined by a single vector representing the direction of the figure axis. Both the kinetic and the potential energies can be expressed in terms of this vector. By consequence, there is no need to use the matrix formalism described in the main text. In this appendix, we take the standard notation up again and write vectors of ${\mathbb {R}}^3$ with bold font.

1.1 One-degree-of-freedom model

Let $\varvec{s}$ be the figure axis and $\alpha$ the precession constant (not the flattening coefficient). The orbit plane of normal $\varvec{k}$ is precessing at constant inclination i around the normal $\varvec{k}_L$ to the Laplace plane. In the frame rotating at the precession frequency $g<0$, where $\varvec{k}$ and $\varvec{k}_L = \varvec{R}_1(-i)\,\varvec{k}$ are both constant, the Hamiltonian describing the evolution of Colombo’s top is (e.g. Ward 1975)

$$\begin{aligned} H = -\frac{\alpha }{2}(\varvec{k}\cdot \varvec{s})^2 - g(\varvec{k}_L\cdot \varvec{s}). \end{aligned}$$

(56)

The related equations of motion are (Colombo 1966)

$$\begin{aligned} \varvec{\dot{s}} = -\hat{J}(H) = \frac{\partial H}{\partial \varvec{s}}\times \varvec{s} = -\alpha (\varvec{k} \cdot \varvec{s})(\varvec{k} \times \varvec{s}) - g (\varvec{k}_L \times \varvec{s}). \end{aligned}$$

(57)

The phase space of this problem is ${\mathscr {M}}_1 = \{\varvec{s} \in {\mathbb {R}}^3, |\varvec{s}| = 1\}$. It is of dimension 2; therefore, this problem has a single degree of freedom. The problem can be parametrised by two angles $(\phi , \theta )$ such that $\varvec{s} = \varvec{R}_3(\phi )\,\varvec{R}_1(-\theta )\,\varvec{k}$.

Cassini states are solution of $\varvec{\dot{s}} = \varvec{0}$. The left-hand side of Eq. (57) can vanish only if $\varvec{s}$ is coplanar with $\varvec{k}$ and $\varvec{k}_L$ (Cassini third law) which implies $\phi =0$. Setting $\phi =0$ in (57), one retrieve the well-known fact that Cassini states’ obliquities $\theta$ are solution of $\frac{\alpha }{g}\cos \theta \sin \theta + \sin (\theta -i) = 0$ (e.g. Ward and Hamilton 2004).

To ascertain the Lyapunov stability of a Cassini state, we evaluate the second derivative of the Hamiltonian in the vicinity of that given Cassini state. On this purpose, we set

$$\begin{aligned} \delta \varvec{s} = \delta \varvec{\theta } \times \varvec{s} ,\qquad \delta ^2\varvec{s} = \delta \varvec{\theta }\times (\delta \varvec{\theta }\times \varvec{s}) + \delta ^2\varvec{\theta } \times \varvec{s}, \end{aligned}$$

(58)

with

$$\begin{aligned} \delta \varvec{\theta } = \delta \phi \,\varvec{k} - \delta \theta \,\varvec{i},\qquad \delta ^2\varvec{\theta } = -\,\delta \phi \,\delta \theta \,\varvec{j}. \end{aligned}$$

(59)

As a result, we get

$$\begin{aligned} \delta ^2H = -\alpha \left( \left( \varvec{k}\cdot \delta \varvec{s}\right) ^2 + (\varvec{k}\cdot \varvec{s})(\varvec{k}\cdot \delta ^2\varvec{s})\right) -g(\varvec{k}_L\cdot \delta ^2\varvec{s}) = h_{\theta \theta }\,\delta \theta ^2 + h_{\phi \phi }\,\delta \phi ^2, \end{aligned}$$

(60)

where

$$\begin{aligned} h_{\theta \theta } = \alpha \cos 2\theta + g\cos (\theta -i),\qquad h_{\phi \phi } = g \sin \theta \sin i . \end{aligned}$$

(61)

The Hamiltonian is locally positive definite if both $h_{\theta \theta }$ and $h_{\phi \phi }$ are positive or negative. Besides, in this set of coordinates $\varvec{y} = (\theta , \phi )$, the Poisson matrix reads

$$\begin{aligned} B(\varvec{y}) = \frac{1}{\sin \theta } \begin{bmatrix} 0 &{} \quad 1 \\ -1 &{} \quad 0 \end{bmatrix}. \end{aligned}$$

(62)

Therefore, the eigenvalues $\lambda$ of the linearised equations of motion (37) are the such that $\lambda ^2 = - h_{\theta \theta }h_{\phi \phi }/\sin ^2\theta$. It follows that the system is spectrally stable if and only if $h_{\theta \theta }h_{\phi \phi } > 0$, i.e. if and only if the system is Lyapunov stable. The two criteria are equivalent. By virtue of the expression of $h_{\phi \phi }$ (61), if $-\pi<\theta <0$ (as is the case for Mercury and Io), then stable equilibrium states correspond to a minimum of H ($h_{\phi \phi }$ is positive because g is negative), but if $0<\theta <\pi$ (as is the case for the Moon), then the Cassini state is located on a maximum of H ($h_{\phi \phi }$ is negative). In the former case, we shall expect that the addition of a (positive definite) kinetic energy in the Hamiltonian will make the system Lyapunov unstable. This question is addressed in the following section.

1.2 Two-degree-of-freedom model

Hamiltonian (56) is only valid in the gyroscopic approximation. Here, we add a simple term accounting for the kinetic energy such that the Lagrangian of the problem reads

$$\begin{aligned} {\hat{L}} = \frac{1}{2}C{\hat{\omega }}^2 + \frac{3}{4}(C-A)n^2(\varvec{k}(t)\cdot \hat{\varvec{s}})^2, \end{aligned}$$

(63)

where $\hat{\varvec{\omega }}$ is the rotation speed, n the mean motion, A the equatorial moment of inertia and C the polar moment of inertia. The Lagrangian is defined up to a constant factor. Let us divide ${\hat{L}}$ by C and only then take the Legendre transform to get the Hamiltonian. Moreover, we choose units of time such that $n=1$. In that case, the moment is $\hat{\varvec{\pi }}= \hat{\varvec{\omega }}$ and the Hamiltonian ${\hat{H}}$ in the inertial frame reads

$$\begin{aligned} {\hat{H}} = \frac{{\hat{\pi }}^2}{2} - \frac{3}{4}\frac{C-A}{C}(\varvec{k}(t)\cdot \hat{\varvec{s}})^2, \end{aligned}$$

(64)

with equations of motion (e.g. Boué and Laskar 2006)

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}{\hat{\varvec{\pi }}} = \frac{\partial {\hat{H}}}{\partial \hat{\varvec{\pi }}} \times \hat{\varvec{\pi }}- {\hat{J}}({\hat{H}}) = \frac{\partial {\hat{H}}}{\partial \hat{\varvec{\pi }}} \times \hat{\varvec{\pi }}+ \frac{\partial {\hat{H}}}{\partial \hat{\varvec{s}}} \times \hat{\varvec{s}} ,\qquad \frac{\mathrm {d}}{\mathrm {d}t}\hat{\varvec{s}} = \frac{\partial {\hat{H}}}{\partial \hat{\varvec{\pi }}} \times \hat{\varvec{s}} . \end{aligned}$$

(65)

As in the main text, we apply a change of coordinates $(\hat{\varvec{\pi }}, \hat{\varvec{s}}) \rightarrow (\varvec{\pi }, \varvec{s})$ to study the problem in the frame rotating at the precession frequency g, i.e. we set $(\hat{\varvec{\pi }}, \hat{\varvec{s}}) = \varvec{R}_3(gt)\,(\varvec{\pi }, \varvec{s})$. To conserve the form of the equations of motion (65), the new Hamiltonian shall read $H(\varvec{\pi },\varvec{s}) = {\hat{H}}(\hat{\varvec{\pi }},\hat{\varvec{s}}) - g (\varvec{k}_L\cdot \varvec{\pi })$. Let $\gamma = \frac{3}{2}\frac{C-A}{C}$, we get

$$\begin{aligned} H = \frac{\pi ^2}{2} - \frac{1}{2}\gamma (\varvec{k}\cdot \varvec{s})^2 - g(\varvec{k}_L\cdot \varvec{\pi }). \end{aligned}$$

(66)

This expression is equivalent to Eq. (1) of Ward (1975).

The phase space of the problem ${\mathscr {M}}_2 = \{(\varvec{\pi }, \varvec{s})\in {\mathbb {R}}^3\times {\mathbb {R}}^3,\,|\varvec{s}| = 1 \text { and }\, (\varvec{s}\cdot \varvec{\pi }) = c\}$ where c is a constant. This is a manifold of dimension 4; hence, the problem has two degrees of freedom. The second condition in the definition of ${\mathscr {M}}_2$ makes it hard to define a ‘natural’ set of four coordinates to parametrise the phase space. Instead, we use the redundant state vector $\varvec{y} = (\varvec{\pi }, \varvec{s})$ where $\varvec{s}$ is parametrised by $(\phi , \theta )$ as in Sect. C.1, i.e. such that $\varvec{s} = \varvec{R}_3(\phi )\varvec{R}_1(-\theta )\varvec{k}$. For $\varvec{\pi }$, we use the rectangular coordinates $(\pi _x, \pi _y, \pi _z)$. Because the state vector is redundant, we have to add a Lagrange multiplier $\mu \in {\mathbb {R}}$ and we introduce the function F defined as

$$\begin{aligned} F = H + \mu \,(\varvec{s}\cdot \varvec{\pi }). \end{aligned}$$

(67)

The fixed points of the system are given by $\delta F = 0$ with

$$\begin{aligned} \delta F = (\varvec{\pi }- g \varvec{k}_L + \mu \varvec{s})\cdot \delta \varvec{\pi } + \left( -\gamma (\varvec{k}\cdot \varvec{s})\varvec{k} + \mu \varvec{\pi }\right) \cdot \delta \varvec{s}. \end{aligned}$$

(68)

Hence, $\varvec{\pi }$, $\varvec{s}$ and $\mu$ are solution of

$$\begin{aligned}&\varvec{\pi }- g \varvec{k}_L + \mu \varvec{s} = \varvec{0}, \end{aligned}$$

(69a)

$$\begin{aligned}&-\gamma (\varvec{k}\cdot \varvec{s})\varvec{k} + \mu \varvec{\pi } = \varvec{0}, \end{aligned}$$

(69b)

$$\begin{aligned}&\varvec{s}\cdot \varvec{\pi }= c. \end{aligned}$$

(69c)

From (69a) and (69c), one gets $\mu = g(\varvec{k}_L\cdot \varvec{s}) - c$. Substituting this result in Eq. (69b) leads to

$$\begin{aligned} -\gamma (\varvec{k}\cdot \varvec{s}) (\varvec{k}\times \varvec{s}) + \left( g(\varvec{k}_L\cdot \varvec{s})-c\right) g(\varvec{k}_L\times \varvec{s}) = \varvec{0}. \end{aligned}$$

(70)

Let $\omega _0 = c - g(\varvec{k}_L\cdot \varvec{s})$. We define the precession constant $\alpha$ as $\alpha = \gamma / \omega _0$. (This is a misuse of language since by construction $\alpha$ depends on the orientation $\varvec{s}$.) With this definition, the condition (70) becomes identical to (57). We thus retrieve the usual Cassini states.

To analyse the stability, we compute the second variation of F, viz.

$$\begin{aligned} \delta ^2 F = |\delta \varvec{\pi }|^2 -\gamma \left( (\varvec{k}\cdot \delta \varvec{s})^2 + (\varvec{k}\cdot \varvec{s})(\varvec{k}\cdot \delta ^2\varvec{s})\right) + \mu \left( \varvec{\pi }\cdot \delta ^2\varvec{s} + 2\delta \varvec{s}\cdot \delta \varvec{\pi }\right) . \end{aligned}$$

(71)

Substituting the expression of the Lagrange multiplier $\mu$ in this formula, one gets

$$\begin{aligned} \frac{\delta ^2 F}{\omega _0} = \frac{|\delta \varvec{\pi }|^2}{\omega _0} - \alpha \left( (\varvec{k}\cdot \delta \varvec{s})^2 + (\varvec{k}\cdot \varvec{s})(\varvec{k}\cdot \delta ^2\varvec{s})\right) - g\,(\varvec{k}_L\cdot \delta ^2\varvec{s}) - \omega _0\,(\varvec{s}\cdot \delta ^2\varvec{s}) - 2\delta \varvec{s}\cdot \delta \varvec{\pi }.\nonumber \\ \end{aligned}$$

(72)

Equivalently, the Hessian of F with respect to $\delta \varvec{y} = (\delta \pi _x, \delta \phi , \delta \pi _y, \delta \pi _z, \delta \theta )\in {\mathbb {R}}^5$ is

$$\begin{aligned} \nabla ^2 F = \omega _0\begin{bmatrix} 1/\omega _0 &{} \sin \theta &{} 0 \quad &{} 0 \quad &{} 0 \quad \\ \sin \theta &{} \quad g\sin \theta \sin i + \omega _0\sin ^2\theta &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 1/\omega _0 &{} \quad 0 &{} \quad -\cos \theta \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 1/\omega _0 &{} \quad \sin \theta \\ 0 &{} \quad 0 &{} \quad -\cos \theta &{} \quad \sin \theta &{} \quad \alpha \cos 2\theta + g\cos (\theta -i)+\omega _0 \end{bmatrix}. \qquad \end{aligned}$$

(73)

The seemingly odd order of the components of $\delta \varvec{y}$ has been chosen to highlight the block matrix structure of $\nabla ^2 F$. The Lyapunov stability of the system is guaranteed if and only if the matrix $\varvec{Q} \nabla ^2F \varvec{Q}$ is definite positive or definite negative where $\varvec{Q}$ is the projection matrix onto the tangent space, i.e. $\varvec{Q} = \mathbb {I}- |\varvec{q}|^{-2}\varvec{q}{\varvec{q}}^{\mathrm T}$ where $\varvec{q}$ is the gradient of the Casimir $C=\varvec{s}\cdot \varvec{\pi }$ of the problem (e.g. Boué et al. 2017). We have

$$\begin{aligned} \delta C = \varvec{s} \cdot \delta \varvec{\pi }+ (\varvec{s} \times \varvec{\pi })\cdot \delta \varvec{\theta }\end{aligned}$$

(74)

with $\varvec{s}\times \varvec{\pi }= g\,(\varvec{s} \times \varvec{k}_L)$ at equilibrium by virtue of (69a). From the expressions of $\delta \varvec{\pi }$ and $\delta \varvec{\theta }$, one gets

$$\begin{aligned} \varvec{q} = (0, 0, \sin \theta , \cos \theta , -g\sin (\theta -i)). \end{aligned}$$

(75)

At this stage, an important conclusion can be drawn without performing additional calculation. Let us decompose the tangent space of the phase space into two linear subspaces $V_1$ and $V_2$ defined as $V_1 = \{\delta \varvec{y}\in {\mathbb {R}}^5, \delta \pi _y = \delta \pi _z = \delta \theta = 0\}$ and $V_2 = \{\delta \varvec{y}\in {\mathbb {R}}^5, \delta \pi _x = \delta \phi = 0\}$. The vector $\varvec{q}$ belongs to $V_2$; therefore, the projection matrix $\varvec{Q}$ only acts on $V_2$. By consequence, the submatrix $\varvec{F}_1$ of $\nabla ^2F$ corresponding to the subspace $V_1$ is left unchanged by $\varvec{Q}$. The product of the eigenvalues of $\varvec{F}_1$ is equal to $\det \varvec{F}_1 = g\omega _0\sin \theta \sin i$. With $\omega _0>0$ (i.e. $\varvec{s}$ is chosen to point in the same direction as $\varvec{\pi }$), $\det \varvec{F}_1<0$ when $0<\theta <\pi$. As a result, the system cannot be Lyapunov stable as long as $0<\theta <\pi$. This is, in particular, the situation of the Moon. Nevertheless, the orientation of the Moon does not show any sign of instability. We thus conclude that the Lyapunov stability criterion is too stringent for this problem.