7
$\begingroup$

I am trying to understand a basic formula in a Celestial Mechanics reference. The formula is for the position of the Moon in a geocentric frame. The reference claims in line (981) that this should be given by $$ \vec{r}_M \approx a_M ( \cos \lambda_M, \sin \lambda_M, \iota_M \sin(\lambda_M - \bar{\omega}_M) ),$$ where $\lambda_M$ is the Moon's ecliptic longitude, $\bar{\omega}_M$ is the lunar ecliptic longitude of the ascending node, $\iota_M$ is the small inclination of the Moon's orbit with respect to the ecliptic, and $a_M$ is the distance from the Moon to the Earth.

Note that in the frame in question, the Earth's angular velocity is given by $\vec{\omega} = \omega (-\sin(\theta) \sin(\phi), \sin(\theta) \cos(\phi), \cos(\theta))$. $\phi$ is the angle of diurnal rotation, $\theta$ is the angle between the Earth's diurnal rotation axis and the ecliptic. Can someone show me the derivation of $\vec{r}_M$ in terms of explicit rotation matrices?

I can quickly reproduce the formula for the Earth's angular velocity about its axis $\vec{\omega}$. I assume that the Earth's rotation axis is about $(0,0,1)$ to begin with, and then apply a clockwise rotation of $\theta$ about the $x$-axis, followed by a counterclockwise rotation about the $z$-axis to obtain $$ \vec{\omega} = \left( \begin{matrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{matrix} \right) \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} - \sin \phi \sin \theta \\ \cos \phi \sin \theta \\ \cos \theta \end{matrix} \right), $$ precisely as claimed. How can we find the result for the lunar position $\vec{r}_M$?

$\endgroup$
  • $\begingroup$ I'm not sure you can do this since the formula is "too" approximate. Notice that the x and y terms are the sin and cos of the same number, so the vector length in the xy plane is already a(m), the distance from the Earth to the Moon. The non-zero z term puts the vector length at greater than a(m) which we know is wrong. I think you can regard this as a 2 dimensional transform with the third dimension tacked on as an afterthought. $\endgroup$ – barrycarter Dec 23 '16 at 4:36
  • $\begingroup$ However as you can see in the reference, the z-component term in $r_M$ is critical to the calculation of the period of Earth's precession. $\endgroup$ – viktor79 Dec 23 '16 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.