Earth–Moon–Sun system

Consider the Earth–Moon–Sun system. See Figure 11.1. Let the $m_i$, for $i=1, 3$, denote the masses of the Earth, the Moon, and the Sun, respectively. Furthermore, let the ${\bf r}_i$, for $i=1, 3$, denote the position vectors of the Earth, the Moon, and the Sun, respectively, in a non-rotating reference frame in which the Earth–Moon–Sun barycenter, $C$, is at rest at the origin. The equations of motion of the three bodies that make up the Earth–Moon–Sun system can be written

$$m_i\,\ddot{\bf r}_i=-\frac{\partial U}{\partial {\bf r}_i},$$ (11.1)

for $i=1, 3$, where

$$U({\bf r}_1,{\bf r}_2,{\bf r}_3) = -G\left(\frac{m_1\,m_2}{r_{12}}+\frac{m_1\,m_3}{r_{13}}+ \frac{m_2\,m_3}{r_{23}}\right).$$ (11.2)

(See Chapter 4.) Here,

$$r_{ij} = \vert{\bf r}_i-{\bf r}_j\vert.$$ (11.3)

Furthermore,

$$\sum_{i=1,3} m_i\,{\bf r}_i={\bf0},$$ (11.4)

because the barycenter of the Earth–Moon–Sun system lies at the origin.

Figure: 11.1 The Earth–Moon–Sun system. Here, $1$ denotes the Earth, $2$ the Moon, and $3$ the Sun. Furthermore, $C$ is the barycenter of the Earth–Moon–Sun system, whereas $B$ is the barycenter of the Earth–Moon system.

Let

$${\bf r}= {\bf r}_2-{\bf r}_1$$ (11.5)

be the position vector of the Moon relative to the Earth, and let

$${\bf r}' = {\bf r}_3-\left(\frac{m_1\,{\bf r}_1+m_2\,{\bf r}_2}{m_1+m_2}\right)$$ (11.6)

be the position vector of the Sun relative to the Earth–Moon barycenter, $B$. It follows that

$$\ddot{\bf r} =-\frac{1}{m_2}\,\frac{\partial U}{\partial {\bf r}_2}+\frac{1}{m_1}\,\frac{\partial U}{\partial{\bf r}_1},$$ (11.7)

$$\ddot{\bf r}' = -\frac{1}{m_3}\,\frac{\partial U}{\partial {\bf r}_3}+\frac{1}{... ...{\partial U}{\partial {\bf r}_1}+ \frac{\partial U}{\partial {\bf r}_2}\right).$$ (11.8)

Now, assuming that $U({\bf r}_1,{\bf r}_2,{\bf r}_3)\equiv U({\bf r},{\bf r}')$, we can write

$$\frac{\partial U}{\partial {\bf r}_i}= \frac{\partial U}{\partial... ...{\partial U}{\partial {\bf r} '}\,\frac{\partial {\bf r}'}{\partial {\bf r}_i}.$$ (11.9)

We conclude that

$$\frac{\partial U}{\partial {\bf r}_1} =-\frac{\partial U}{\partial {\bf r}}-\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial U}{\partial{\bf r}'},$$ (11.10)

$$\frac{\partial U}{\partial {\bf r}_2} = \frac{\partial U}{\partial {\bf r}}-\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial U}{\partial{\bf r}'},$$ (11.11)

$$\frac{\partial U}{\partial {\bf r}_3} =\frac{\partial U}{\partial {\bf r}'},$$ (11.12)

so that

$$\ddot{\bf r} = -\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\frac{\partial U}{\par... ...r}}= -\left(\frac{m_1+m_2}{m_1\,m_2}\right)\frac{\partial U}{\partial {\bf r}},$$ (11.13)

$$\ddot{\bf r}' =-\left(\frac{1}{m_1+m_2}+\frac{1}{m_3}\right)\frac{\partial U}{\... ...[\frac{m_1+m_2+m_3}{(m_1+m_2)\,m_3}\right]\frac{\partial U}{\partial {\bf r}'}.$$ (11.14)

Equations (11.4)–(11.6) yield

$${\bf r}_1 =-\left(\frac{m_2}{m_1+m_2}\right){\bf r} -\left(\frac{m_3}{m_1+m_2+m_3}\right){\bf r}',$$ (11.15)

$${\bf r}_2 = \left(\frac{m_1}{m_1+m_2}\right){\bf r} -\left(\frac{m_3}{m_1+m_2+m_3}\right){\bf r}',$$ (11.16)

$${\bf r}_3 = \left(\frac{m_1+m_2}{m_1+m_2+m_3}\right){\bf r}',$$ (11.17)

which implies that

$${\bf r}_2-{\bf r}_1 ={\bf r},$$ (11.18)

$${\bf r}_3-{\bf r}_1 = {\bf r}'+\left(\frac{m_2}{m_1+m_2}\right){\bf r},$$ (11.19)

$${\bf r}_3-{\bf r}_2 = {\bf r}' -\left(\frac{m_1}{m_1+m_2}\right){\bf r}.$$ (11.20)

Let

$${\bf r}\cdot{\bf r}' = r\,r'\,\cos\theta.$$ (11.21)

It follows from Equations (11.18)–(11.20) that

$$r_{12} = \vert{\bf r}_2-{\bf r}_1\vert = r,$$ (11.22)

$$r_{13} = \vert{\bf r}_3-{\bf r}_1\vert =r'\left[1+2\,\cos\theta\,\left(\... ...\frac{m_2}{m_1+m_2}\right)^{\,2}\left(\frac{r}{r'}\right)^{\,2}\right]^{\,1/2},$$ (11.23)

$$r_{23} = \vert{\bf r}_3-{\bf r}_2\vert =r'\left[1-2\,\cos\theta\left(\fr... ...\frac{m_1}{m_1+m_2}\right)^{\,2}\left(\frac{r}{r'}\right)^{\,2}\right]^{\,1/2}.$$ (11.24)

Note that $r/r'\sim a/a'=1/389.2$, where $a=384,399\,{\rm km}$ and $a'=149,598,261\,{\rm km}$ are the major radii of the lunar orbit about the Earth, and the orbit of the Earth-Moon barycenter about the Sun, respectively (Yoder 1995). Now,

$$\frac{1}{\sqrt{1-2\,x\,t+t^{\,2}}}=\sum_{n=0,\infty}t^{\,n} \,P_n(x),$$ (11.25)

for $\vert x\vert<1$ and $\vert t\vert<1$, where the $P_n(x)$ are Legendre polynomials (Abramowitz and Stegun 1965b). Thus, expanding Equations (11.23) and (11.24) to third order in the small parameters $[m_2/(m_1+m_2)]\,(r/r)'$ and $[m_1/(m_1+m_2)]\,(r/r)'$, respectively, we obtain

$$\frac{r'}{r_{13}} \simeq 1-\left(\frac{m_2}{m_1+m_2}\right)\left(\frac{r}{r'}\right... ...)+\left(\frac{m_2}{m_1+m_2}\right)^2\left(\frac{r}{r'}\right)^2 P_2(\cos\theta)$$

$$\phantom{\simeq} -\left(\frac{m_2}{m_1+m_2}\right)^3\left(\frac{r}{r'}\right)^3 P_3(\cos\theta),$$ (11.26)

$$\frac{r'}{r_{23}} \simeq 1+\left(\frac{m_1}{m_1+m_2}\right)\left(\frac{r}{r'}\right... ...)+\left(\frac{m_1}{m_1+m_2}\right)^2\left(\frac{r}{r'}\right)^2 P_2(\cos\theta)$$

$$\phantom{\simeq} +\left(\frac{m_1}{m_1+m_2}\right)^3\left(\frac{r}{r'}\right)^3 P_3(\cos\theta),$$ (11.27)

respectively. It follows from Equation (11.2) that

$$U \simeq -G\left\{\frac{m_1\,m_2}{r}+\frac{(m_1+m_2)\,m_3}{r'}+\lef... ...r}\cdot{\bf r}')^{\,2}}{r'^{\,5}}-\frac{(1/2)\,r^{\,2}}{r'^{\,3}}\right]\right.$$

$$\phantom{=}\left.+\left[\frac{m_1\,m_2\,m_3\,(m_1-m_2)}{(m_1+m_2)... ...^{\,7}}-\frac{(3/2)\,r^{\,2}\,({\bf r}\cdot{\bf r}')}{r'^{\,5}}\right]\right\},$$ (11.28)

where use has been made of

$$P_2(x) = \frac{1}{2}\,(3\,x^{\,2}-1),$$ (11.29)

$$P_3(x) = \frac{1}{2}\,(5\,x^{\,3}-3\,x)$$ (11.30)

(Abramowitz and Stegun 1965b), as well as Equation (11.21).

Let (see Chapter 4)

$$G\,(m_1+m_2) = n^{\,2}\,a^{\,3},$$ (11.31)

$$G\,(m_1+m_2+m_3) \simeq G\,m_3=n'^{\,2}\,a'^{\,3},$$ (11.32)

where $n=13.176359^\circ$ per day is the mean orbital angular velocity of the Moon around the Earth, and $n'=0.98560912^\circ$ per day is the mean orbital angular velocity of the Earth–Moon barycenter around the Sun (Yoder 1995). Here, we have neglected the combined mass of the Earth and the Moon, $m_1+m_2$, with respect to the mass of the Sun, $m_3$, because $m_3/(m_1+m_2)=3.29\times 10^{\,5}$ (Yoder 1995). It follows from Equations (11.13), (11.14), and (11.28) that the equations of motion of the Moon and the Earth–Moon barycenter can be written

$$\ddot{\bf r}+n^{\,2}\,a^{\,3}\,\frac{{\bf r}}{r^{\,3}} = {\bf f}({\bf r},{\bf r}')+{\bf g}({\bf r},{\bf r}'),$$ (11.33)

$$\ddot{\bf r}' + n'^{\,2}\,a'^{\,3}\,\frac{{\bf r}'}{r'^{\,3}} = {\bf g}'({\bf r},{\bf r}'),$$ (11.34)

respectively, where

$${\bf f} = \frac{n'^{\,2}\,a'^{\,3}}{r'^{\,3}}\left[\frac{3\,({\bf r}\cdot{\bf r}')\,{\bf r}'}{r'^{\,2}}-{\bf r}\right],$$ (11.35)

$${\bf g} =\frac{n'^{\,2}\,a'^{\,3}}{r'^{\,3}}\left(\frac{m_1-m_2}{m_1+m_2}... ...{\bf r}'}{r'^{\,2}}-\frac{3\,({\bf r}\cdot{\bf r'})\,{\bf r}}{r'^{\,2}}\right],$$ (11.36)

$${\bf g}' =\frac{n'^{\,2}\,a'^{\,3}}{r'^{\,3}}\left[\frac{m_1\,m_2}{(m_1+m_... ...,{\bf r}'}{r'^{\,2}}+\frac{3\,({\bf r}\cdot{\bf r'})\,{\bf r}}{r'^{\,2}}\right]$$ (11.37)

According to Equations (11.33) and (11.35), the relative size of the lowest-order (quadrupole) solar disturbance of the lunar orbit is

$$\frac{f}{n^{\,2}\,a} \sim \left(\frac{n'}{n}\right)^2= \left(7.48\times 10^{-2}\right)^{\,2} = 5.6\times 10^{-3}.$$ (11.38)

On the other hand, according to Equation (11.36), the relative size of the next-order (octupole) solar disturbance of the lunar orbit is

$$\frac{g}{n^{\,2}\,a}\sim \left(\frac{n'}{n}\right)^2\left(\frac{a... ...\right)= \frac{\left(7.48\times 10^{-2}\right)^{\,2}}{389.2}=1.4\times 10^{-5}.$$ (11.39)

Finally, according to Equations (11.34) and (11.37), the relative size of the lowest-order (octupole) disturbance of the orbit of the Earth–Moon barycenter around the Sun, due to the combined effect of the Earth and the Moon, is

$$\frac{g'}{n'^{\,2}\,a'}\sim \left(\frac{m_2}{m_1}\right)\left(\frac{a}{a'}\right)^{\,2} = \frac{1}{81.3\,(389.2)^{\,2}}= 8.1\times 10^{-8}.$$ (11.40)

Here, use has been made of the fact that the ratio of the terrestrial to the lunar mass is $m_1/m_2=81.3$ (Yoder 1995). Clearly, it is an excellent approximation to neglect the octupole disturbance of the orbit of the Earth–Moon barycenter around the Sun, while retaining the quadrupole and octupole disturbances of the orbit of the Moon around the Earth. Adopting this approximation, we conclude that the former orbit is an unperturbed Keplerian ellipse.

One obvious way of proceeding would be to express the right-hand side of the lunar equation of motion, Equation (11.33), as the gradient of a disturbing function (see Section 11.18, Exercise 1), and then to use this function to determine the time evolution of the Moon's osculating orbital elements from Lagrange's planetary equations. (See Chapter 10.) Unfortunately, this approach is fraught with mathematical difficulties. (See Brouwer and Clemence 1961). It is actually more straightforward to directly solve Equation (11.33) in a Cartesian coordinate system. This method of solution is outlined below.