precession of the equinoxes

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Let us suppose an extended body body of mass $m\,$ is in a circular orbit around a distant body of mass $M\,$ , and let us choose, at an instant in time, a body-fixed basis $\mathbf{\hat{x}}_b,\mathbf{\hat{y}}_b,\mathbf{\hat{z}}_b\,$ in which the moment of inertia tensor $\mathbf{I}\,$ is diagonal.

Furthermore, choose spherical coordinates such that the distant mass $M\,$ be located in the direction

$\hat\mathbf{r} = \sin\xi \cos\zeta\mathbf{\hat{x}}_b + \sin\xi \sin\zeta\mathbf{\hat{y}}_b + \cos\xi \mathbf{\hat{z}}_b\,$ ,

at a distance $r\,$ , where $\xi\,$ is the angle subtended between $\hat{\mathbf{z}}_b\,$ and $\hat\mathbf{r}\,$ . The moment of inertia about an axis $\hat\mathbf{r}\,$ is $I_r = \hat\mathbf{r}^T \mathbf{I} \hat\mathbf{r}\,$ , so

$I_{r} = (\sin\xi\cos\zeta)^2 I^{b}_{xx} + (\sin\xi \sin\zeta)^2 I^b_{yy} + (\cos\xi)^2 I^b_{zz}\,$ .

Let us further suppose that $I_{xx} = I_{yy}\,$ so that

$I_{r} = I^{b}_{xx} \sin^2\xi + \cos^2\xi I^b_{zz}\,$ ,

and (dropping the superscript $b\,$ ), the gravitational potential energy is

$V\,$	$= -\frac{GMm}{r} + \frac{GM}{2 r^3} \left(I_{xx} - I_{zz} \right) \left(1 - 3 \cos^2\xi\right)\,$ ,
	$= -\frac{GMm}{r} - \frac{GM}{r^3} \left(I_{xx} - I_{zz} \right) P_2(\cos\xi)\,$ .

Now, in space axes aligned with the ecliptic, $\hat\mathbf{z}_b = \sin\theta \sin\psi\hat\mathbf{x} +\sin\theta\cos\psi\hat\mathbf{y} + \cos\theta \hat\mathbf{z}\,$ , $\hat\mathbf{r} = \cos\eta \hat\mathbf{x} - \sin\eta \hat\mathbf{y}\,$ , so that $\cos\xi = -\sin\theta \sin(\eta-\psi)\,$ . Normally precession happens over a time scale much slower than the orbital motion. If we average over one period,

$\frac{1}{2\pi}\int\!d\eta\, P_2(\cos\xi) = \frac{1}{2}\left( \tfrac{3}{2}\sin^2\theta -1\right)\,$ ,

so that

$\left\langle V \right\rangle = -\frac{GMm}{2r} - \frac{GM}{2r^3} \left(I_{xx} - I_{zz} \right) \left( \tfrac{3}{2}\sin^2\theta -1\right)\,$ .

The angular velocity in a body-fixed basis in terms of Euler angles is

$\boldsymbol{\omega} = \begin{pmatrix} \omega_{x_b}\\ \omega_{y_b}\\ \omega_{z_b} \end{pmatrix}= \begin{pmatrix} \sin\psi \sin\theta \dot{\phi} + \cos\psi \dot\theta\\ \cos\psi \sin\theta \dot{\phi} - \sin\psi \dot\theta\\ \cos\theta \dot{\phi} + \dot\psi \end{pmatrix}\,$ ,

so the kinetic energy $T = \tfrac{1}{2} \boldsymbol{\omega}^T I \boldsymbol{\omega}\,$ is

$T = \frac{1}{2}I_{xx} ( \sin^2\theta \dot{\phi}^2 + \dot{\theta}^2 ) + \frac{1}{2}I_{zz} (\cos\theta \dot\phi + \dot\psi)^2\,$ ,

and the effective Lagrangian governing precession is

$L = \frac{1}{2}I_{xx} ( \sin^2\theta \dot{\phi}^2 + \dot{\theta}^2 ) + \frac{1}{2}I_{zz} (\cos\theta \dot\phi + \dot\psi)^2 - \frac{GM}{ 2r^3} \left(I_{xx}- I_{zz} \right) \left(1 -\tfrac{3}{2} \sin^2\theta \right)\,$ .

Since $\frac{\partial L}{\partial \psi} = 0\,$ , we see that $\Omega = \cos\theta \dot\phi + \dot\psi\,$ is an integral of motion, and

$L = \frac{1}{2}I_{xx} ( \sin^2\theta \dot{\phi}^2 + \dot{\theta}^2 ) +\frac{1}{2} I_{zz} \Omega^2 - \frac{GM}{ 2r^3} \left(I_{xx}- I_{zz} \right) \left(1 -\tfrac{3}{2} \sin^2\theta \right)\,$ .

Then

$\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = I_{xx} \ddot{\theta} - \frac{\partial L}{\partial \theta} = 0\,$ .

For steady precession, $\dot{\phi} = 0\,$ , so

$\frac{\partial L}{\partial \theta} = I_{xx} \sin\theta\cos\theta \dot{\phi}^2 - I_{zz} \Omega \sin\theta \dot{\phi} + \frac{GM}{2r^3} (I_{xx} - I_{zz})(3\sin\theta\cos\theta) = 0\,$ ,

$I_{xx} \cos\theta \dot{\phi}^2 - I_{zz} \Omega \dot{\phi} + \frac{GM}{2r^3} (I_{xx} - I_{zz})(3\cos\theta) = 0\,$ .

Assuming $\dot\phi \ll \Omega\,$ ,

$\dot{\phi} \approx - \frac{GM}{2\Omega r^3}\frac{ (I_{zz} - I_{xx}) }{I_{zz}} (3\cos\theta) \,$ .

back to gravitational potential of an extended body

back to Gravitational potential theory

precession of the equinoxes

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