# Hill approximation of the CRTBP (Circular Restricted Three-Body Problem) derivation

Dear fellow space enthusiasts,

Let us first consider the classical CRTBP equations of motion by defining the position vector of the spacecraft

$$\begin{equation} \boldsymbol{r}=[x,y,z]^{T}, \end{equation}$$

which is the distance from the barycentre of let's say the Sun-Earth-spacecraft system to the spacecraft.

Now, using the Transport Theorem

$$\begin{equation} \left(\frac{d\boldsymbol{r}}{dt}\right)_{A}=\left(\frac{d\boldsymbol{r}}{dt}\right)_{B}+\boldsymbol{\omega}_{A/B}\times\boldsymbol{r}, \end{equation}$$

we can transfer $$\boldsymbol{r}$$ in the sidereal (inertial) frame to the synodic (rotating) frame and find the acceleration as

$$\begin{gather}\label{eq:4.5} \boldsymbol{r}'' = \begin{bmatrix} x''-\theta''y-2\theta'y'-\theta'^{2}x \\ y''+\theta''x+2\theta'x'-\theta'^{2}y \\ z'' \end{bmatrix} = \sum\boldsymbol{F}=-\frac{Gm_{1}}{r_{1}^{3}}\boldsymbol{r_{1}}-\frac{Gm_{2}}{r_{2}^{3}}\boldsymbol{r_{2}}, \end{gather}$$

where $$\sum\boldsymbol{F}$$ denotes that the force is gravitational to the acceleration, $$-\frac{Gm_{1}}{r_{1}^{3}}\boldsymbol{r_{1}}$$ is Sun-spacecraft gravitational force and $$-\frac{Gm_{2}}{r_{2}^{3}}\boldsymbol{r_{2}}$$ is the Earth-spacecraft gravitational force.

As CRTBP describes the motion of massless spacecraft under the gravitational influence of the two majors in circular orbits, the eccentricity of the system is $$e=0$$. Therefore, the angular velocity, $$\theta'$$ is equal to the mean motion of the two primaries, $$n$$ as

$$\begin{equation} \theta'=n=\sqrt{\frac{G(m_{S}+m_{E})}{a^{3}}}. \end{equation}$$

Because $$G$$, $$m_{S}$$ and $$m_{E}$$ are constant, the angular acceleration of the system, $$\theta''=0,$$ and the acceleration of the spacecraft becomes

$$\begin{gather}\label{eq:4.7} \boldsymbol{r''} = \begin{bmatrix} x''-2ny'-n^{2}x \\ y''+2nx'-n^{2}y \\ z'' \end{bmatrix} = \sum\boldsymbol{F}=-\frac{Gm_{1}}{r_{1}^{3}}\boldsymbol{r_{1}}-\frac{Gm_{2}}{r_{2}^{3}}\boldsymbol{r_{2}}. \end{gather}$$

Considering the normalised units for the Sun-Earth system:

1. The length unit (LU) and the semi-major $$a$$ (Sun-Earth distance) $$a=1LU,$$
2. The time unit (TU) $$\frac{1}{n}=1TU,$$

let's us write $$G(m_{S}+m_{E})=1,$$

and define the mass ration parameter as

$$\begin{equation}\label{eq:4.8} Gm_{2}=\frac{Gm_{2}}{G(m_{1}+m_{2})}=\frac{m_{2}}{m_{1}+m_{2}}=\mu, \end{equation}$$ but also $$\begin{equation}\label{eq:4.9} Gm_{1}=\frac{Gm_{1}}{G(m_{1}+m_{2})}=\frac{m_{1}}{m_{1}+m_{2}}=1-\mu. \end{equation}$$

Furthermore, expressing $$\boldsymbol{r}_{1}$$ and $$\boldsymbol{r}_{2}$$ in dimensionless coordinates, the CRTBP equations of motion are given by

$$\begin{equation}\ \begin{cases} x''=x-\frac{1-\mu}{r_{1}^{3}}(x+\mu)-\frac{\mu}{r_{2}^{3}}(x+\mu-1)+2y'\\ y''=y-\frac{1-\mu}{r_{1}^{3}}y-\frac{\mu}{r_{2}^{3}}y-2x'\\ z''=-\frac{1-\mu}{r_{1}^{3}}z-\frac{\mu}{r_{2}^{3}}z \end{cases} , \end{equation}$$

where $$\begin{equation} r_{1}=\sqrt{x_{S}^{2}+y^{2}+z^{2}}, \end{equation}$$ is the Sun-spacecraft distance and $$\begin{equation} r_{2}=\sqrt{x_{E}^{2}+y^{2}+z^{2}}, \end{equation}$$ is the Earth-spacecraft distance.

The CRBTP equations of motion can also be expressed in a more compact form as $$\begin{equation} \begin{cases} x''=\Omega_{x}+2y'\\ y''=\Omega_{y}-2x'\\ z''=\Omega_{z} \end{cases} , \end{equation}$$ where $$\begin{equation}\label{eq:4.20} \Omega=\frac{1}{2}(x^{2}+y^{2})+\frac{1-\mu}{r_{1}}+\frac{\mu}{r_{2}}, \end{equation}$$ is the Effective Potential. This concludes the classical CRTBP equations of motion.

Now let's consider that our spacecraft orbits in the vicinity of the Earth (i.e. a considerably lower mass than our primary body, the Sun). From my investigations, this means that the dynamical system can be modelled using the Hill approximation of the CRTBP. Thus, in my mind, we could rewrite the CRTBP as

$$\begin{gather} \boldsymbol{r}'' = \begin{bmatrix} x''-\theta''y-2\theta'y'-\theta'^{2}x \\ y''+\theta''x+2\theta'x'-\theta'^{2}y \\ z'' \end{bmatrix} = \sum\boldsymbol{F}=-\frac{Gm_{E}}{r^{3}}\boldsymbol{r}, \end{gather}$$

because, being now into the Hill region of the Earth, the spacecraft orbits the Earth rather than the Sun and the attraction from the Sun is negligible as long it (i.e. the spacecraft) is in the Hill region of the Earth. Moreover, $$r$$ now represents the distance from the Earth (i.e. the secondary body, which in the Hill Problem is the centre of the system) to the spacecraft.

At this point everything becomes unclear for me with regards to the steps required from the last equation to reach the simplified dimensional formulae of the Hill Problem given by

$$\begin{equation}\ \begin{cases} x''=2n_{S}y'+3n_{S}^{2}x-\frac{\mu_{S}}{r^{3}}x\\ y''=-2n_{S}x'-\frac{\mu_{S}}{r^{3}}y\\ z''=-n_{S}^{2}z-\frac{\mu_{S}}{r^{3}}z \end{cases} , \end{equation}$$

where $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$ is the distance from the secondary (i.e. now origin of the coordinate frame), $$\mu_{S}$$ is the gravitation parameter of the secondary (i.e. the Earth) and $$n_{S}$$ is its mean motion. The last equation then somehow reduces to

$$\begin{equation}\ \begin{cases} x''=2y'+3x-\frac{x}{r^{3}}x\\ y''=-2x'-\frac{y}{r^{3}}y\\ z''=-z-\frac{z}{r^{3}} \end{cases} , \end{equation}$$

Would it be too much of a request to have someone look over the problem and tell if my process thought is correct and maybe explain how to obtain the last two equations? It would be much appreciated.

• The answer to your primary question is embedded in your question, where you wrote $\frac1n = 1\,\text{TU}$. This means that $n$ has a numerical value of one in the system of units you have chosen to use. Also, the $\frac{\mu_s}{r^3}$ terms are dubious. – David Hammen Feb 7 at 16:44
• @DavidHammen Thank you for answering. Yes, that is right. I just modified the last equation. – space_monkey Feb 7 at 16:54
• @DavidHammen What I do not necessarily get is how does one obtain the penultimate equation from the one before (I am not sure that one is correct either) ? – space_monkey Feb 7 at 17:03
• The derivation in the Sun-Earth rotating frame is correct (to the best of my knowledge). However, I fail to understand what you're trying to achieve for Hill frame in the vicinity of the Earth. If the purpose is to represent the motion of the spacecraft as a rotation around the Earth, you'll need to use the orbital motion of the spacecraft. If the purpose is to create a new rotating frame between both the spacecraft and the Earth, then you'll need to "account" for the mass of the spacecraft and repeat the first derivation. – ChrisR Feb 8 at 5:42