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Solar orbit

Let us set up a conventional Cartesian coordinate system which is such that the (apparent) orbit of the Sun about the Earth-Moon barycenter lies in the $ x$ -$ y$ plane. This implies that the $ x$ -$ y$ plane corresponds to the so-called ecliptic plane. Approximating the solar orbit as a low-eccentricity Keplerian orbit, we can write

    $\displaystyle \frac{x'}{a'}$ $\displaystyle \simeq \cos(n'\,t)-\frac{3}{2}\,e'+\frac{1}{2}\,e'\,\cos(2\,n'\,t),$ (11.41)
    $\displaystyle \frac{y'}{a'}$ $\displaystyle \simeq \sin(n'\,t)+\frac{1}{2}\,e'\,\sin(2\,n'\,t),$ (11.42)
and   $\displaystyle \frac{z'}{a'}$ $\displaystyle \simeq 0,$     (11.43)

where $ {\bf r}'\equiv (x',\,y',\,z')$ , and $ e'=0.016711$ is the orbital eccentricity (Yoder 1995). (See Section 4.13.) Here, we have neglected terms that are $ {\cal }O(e'^{\,2})$ , and smaller.

Richard Fitzpatrick 2016-03-31