# Hill’s Equations: Derivation and Solution Explained

Hill’s equations are equations that describe the relative motion of the chaser object to the target object in the rendezvous of spacecraft. In this article, the meaning and derivation of Hill’s equations and the derivation of the Clohessy-Wiltshire (C-W) solution are explained.

## What is Hill’s equations?

Rendezvous is a maneuver that brings closer a spacecraft to another spacecraft. The object in the goal of rendezvous is called “target”, and the object which conduct rendezvous is called “chaser”. Hill’s equations are equations that describe the relative motion of the chaser to the target in a circular orbit. Hill’s equations are described as below.

$\begin{eqnarray} \ddot{x}-2\omega\dot{y}-3\omega^2x & = & f_x \\ \ddot{y}+2\omega\dot{x} & = & f_y \\ \ddot{z}+\omega^2z & = & f_z \end{eqnarray}$

Hill’s equations are derived first by Hill (1878), and rediscovered by Clohessy and Wiltshire (1960) in research about rendezvous in the orbit. Thus, Hill’s equations are also called Clohessy-Wiltshire equations or C-W equations.

## Derivation of Hill’s equations

Coordinate system is defined supposing that the target is in a circular orbit arount the earth. Placing the origin at center of the target, x axis toward the target from the center of the earth, y axis toward moving direction of the target and z axis to be right-handed.

The equation of motion of the target and the chaser can be obtained by following equations:

$\frac{d^2\boldsymbol{r}_t}{dt^2} = -\mu\frac{\boldsymbol{r}_t}{r^3_t}$

$\frac{d^2\boldsymbol{r}_c}{dt^2} = -\mu\frac{\boldsymbol{r}_c}{r^3_c}+\boldsymbol{f}$

where $$\boldsymbol{r}_t$$ denotes the position vector of the target, $$\boldsymbol{r}_c$$ denotes the position vector of the chaser, $$\mu$$ denotes earth gravity constant and $$f$$ denotes the acceleration caused by the force acting on chaser.

Introducing relative position vector $$\boldsymbol{r} =\boldsymbol{r}_c -\boldsymbol{r}_t$$, the above equations can be written as:

$\frac{d^2\boldsymbol{r}}{dt^2} = -\mu\frac{\boldsymbol{r}_c}{r^3_c}+\mu\frac{\boldsymbol{r}_t}{r^3_t}+\boldsymbol{f}$

Here,

$\begin{eqnarray} r^2_c & = & (\boldsymbol{r}_t + \boldsymbol{r})^2 \\ & = & r^2_t + 2\boldsymbol{r}_t\cdot\boldsymbol{r} + r^2 \\ & = & r^2_t \left[1 + \frac{2\boldsymbol{r}_t\cdot\boldsymbol{r}}{r^2_t} + \left(\frac{r}{r_t}\right)^2\right]\end{eqnarray}$

When raised to the $$-\frac{3}{2}$$ power,

$\boldsymbol{r}^{-3}_c = r^{-3}_t \left[1 + \frac{2\boldsymbol{r}_t\cdot\boldsymbol{r}}{r^2_t} + \left(\frac{r}{r_t}\right)^2\right]^{-\frac{3}{2}}$

considering $$r << r_t$$,

$\boldsymbol{r}^{-3}_c = r^{-3}_t \left[1 + \frac{2\boldsymbol{r}_t\cdot\boldsymbol{r}}{r^2_t}\right]^{-\frac{3}{2}}$

using approximate formula $$(1+x)^p = 1+px$$, the following equation can be obtained:

$\boldsymbol{r}^{-3}_c = r^{-3}_t \left(1 – 3\frac{\boldsymbol{r}_t\cdot\boldsymbol{r}}{r^2_t}\right)$

Substituting this into the equations of motion,

$\frac{d^2\boldsymbol{r}}{dt^2} = -\mu\frac{1}{r^3_t}\left(- 3\frac{\boldsymbol{r}_t\cdot\boldsymbol{r}}{r^2_t}\boldsymbol{r_t} + \boldsymbol{r}\right) +\boldsymbol{f}$

Also, from time derivative of the position vector in the rotating frame,

$\frac{d^2\boldsymbol{r}}{dt^2} = \boldsymbol{a}’ + 2\boldsymbol{\omega} \times \boldsymbol{v’} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r’}) + \dot{\boldsymbol{\omega}} \times \boldsymbol{r’}$

Thus, supposing $$\omega = \sqrt{\frac{\mu}{r^3_t}}$$,

$\left(\begin{array}{c} 2\omega^2x \\ -\omega^2y \\ -\omega^2z \end{array}\right) + \left(\begin{array}{c} f_x \\ f_y \\ f_z \end{array}\right) = \left(\begin{array}{c} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{array}\right) + 2\left(\begin{array}{c} -\omega\dot{y} \\ \omega\dot{x} \\ 0 \end{array}\right) + \left(\begin{array}{c} 0 \\ 0 \\ \omega^2z \end{array}\right) + \left(\begin{array}{c} -\omega^2x \\ -\omega^2y \\ -\omega^2z \end{array}\right)$

Simplifying this,

$\begin{eqnarray} \ddot{x}-2\omega\dot{y}-3\omega^2x & = & f_x \\ \ddot{y}+2\omega\dot{x} & = & f_y \\ \ddot{z}+\omega^2z & = & f_z \end{eqnarray}$

## Derivation of Clohessy-Wiltshire Solution

When considering no force acting on the chaser,

$\begin{eqnarray} \ddot{x}-2\omega\dot{y}-3\omega^2x & = & 0 \tag{1} \\ \ddot{y}+2\omega\dot{x} & = & 0 \tag{2}\\ \ddot{z}+\omega^2z & = & 0 \tag{3} \end{eqnarray}$

$$x$$ and $$y$$ are coupling, these equations can be solved by elimination method. Differentiating Eq. (2),

$\dot{y}+2\omega x = C_1 \tag{4}$

Substituting Eq. (4) to Eq. (1),

$\ddot{x}+\omega^2 x = 2\omega C_1 \tag{5}$

can be obtained. This equation can be solved:

$x = \frac{2}{\omega}C_1+C_2\cos{\omega t}+C_3\sin{\omega t} \tag{6}$

If differentiated,

$\dot{x} = -C_2\omega\sin{\omega t}+C_3\omega\cos{\omega t} \tag{7}$

Substituting Eq. (6) to Eq. (4),

$\dot{y} = -3C_1-2C_2\omega\cos{\omega t}-2C_3\omega\sin{\omega t} \tag{8}$

$y = -3C_1t-2C_2\sin{\omega t}+2C_3\cos{\omega t}+C_4 \tag{9}$

Substituting $$t=0$$ to Eqs. (6)-(9),

$\begin{eqnarray} x_0 & = & \frac{2}{\omega}C_1 + C_2\tag{10} \\ \dot{x}_0 & = & C_3\omega \tag{11}\\ \dot{y}_0 & = & -3C_1-2C_2\omega \tag{12}\\ y_0 & = & 2C_3+C_4 \tag{13} \end{eqnarray}$

This can be solved:

$\begin{eqnarray} C_1 & = & 2\omega x_0+\dot{y}_0\tag{14} \\ C_2 & = & -3x_0-\frac{2}{\omega}\dot{y}_0 \tag{15}\\ C_3 & = & \frac{1}{\omega}\dot{x}_0 \tag{16}\\ C_4 & = & -\frac{2}{\omega}\dot{x}_0+y_0 \tag{17} \end{eqnarray}$

z axis is independent of other axes, by solving Eq. (3),

$z = C_5\cos{\omega t}+C_6\sin{\omega t} \tag{18}$

$\dot{z} = C_5\omega\sin{\omega t}+C_6\omega\cos{\omega t} \tag{19}$

Substituting $$t=0$$ to Eqs. (18)-(19),

$\begin{eqnarray} C_5 & = & z_0 \tag{20} \\ C_6 & = & \frac{\dot{z}_0}{\omega} \tag{21} \end{eqnarray}$

From the above, the following equations can be obtained:

$\begin{eqnarray} x & = & (4-3\cos{\omega t})x_0+\frac{\sin{\omega t}}{\omega}\dot{x}_0+\frac{2}{\omega}(1-\cos{\omega t})\dot{y}_0 \\ y & = & 6(\sin{\omega t} – \omega t)x_0+y_0+\frac{2}{\omega}(\cos{\omega t}-1)\dot{x}_0+\frac{1}{\omega}(4\sin{\omega t}-3\omega t)\dot{y}_0 \\ z & = & (\cos{\omega t})z_0+\frac{1}{\omega}(\sin{\omega t})\dot{z}_0 \\ \dot{x} & = & (3\omega\sin{\omega t})x_0 + (\cos{\omega t})\dot{x}_0+(2\sin{\omega t})\dot{y}_0 \\ \dot{y} & = & 6\omega(\cos{\omega t}-1)x_0-2(\sin{\omega t})\dot{x}_0+(4\cos{\omega t}-3)\dot{y}_0 \\ \dot{z} & = & -\omega(\sin{\omega t})z_0 + (\cos{\omega t})\dot{z}_0 \\ \end{eqnarray}$

This can be written in the form of the matrix:

$\small{\left[\begin{array}{c} x(t) \\ y(t) \\ z(t) \\ \dot{x}(t) \\ \dot{y}(t) \\ \dot{z}(t) \\ \end{array}\right] = \left[\begin{array}{cccccc} 4-3c & 0 & 0 & s/w & 2(1-c)/w & 0 \\ 6(s-\omega t) & 1 & 0 & -2(1-c)/\omega & (4s-3\omega t)/\omega & 0 \\ 0 & 0 & c & 0 & 0 & s/\omega \\ 3\omega s & 0 & 0 & c & 2s & 0 \\ -6\omega(1-c) & 0 & 0 & -2s & -3+4c & 0 \\ 0 & 0 & \omega s & 0 & 0 & c \\ \end{array}\right] \left[\begin{array}{c} x_0 \\ y_0 \\ z_0 \\ \dot{x}_0 \\ \dot{y}_0 \\ \dot{z}_0 \\ \end{array}\right]}$

where $$c=\cos{\omega t}$$, $$s=\sin{\omega t}$$.

This is called Clohessy-Wiltshire solution

## Summary

Derivation of Hill’s equations and Clohessy-Wiltshire solution has done in this article.

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