A satellite in a circular orbit intersects the $x$-axis and is inclined to the $xy$-plane at an angle of $30{{}^\circ}$. Suppose the orbit has a radius $a,$ and the motion has angular speed $\omega$. Then the position of the satellite at time $t$ is $$\begin{equation*} \mathbf{r}(t)=a\!\left( \cos ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{ 2}\sin ( \omega t) \mathbf{j}+\frac{1}{2}\sin ( \omega t) \mathbf{k}\right) \end{equation*}$$

The velocity vector $\mathbf{v}$ is $$\begin{equation*} \mathbf{v}(t)=\mathbf{r}^{\prime} (t)=a\omega \!\left( -\!\sin ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{2}\cos ( \omega t) \mathbf{j}+ \frac{1}{2}\cos ( \omega t) \mathbf{k}\right) \end{equation*}$$

In this case, using (7), the UVW system of axes is $$\begin{array}{rcl@{\qquad}rcl} \mathbf{i}_{\mathbf{U}}&=&\cos ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{2}\sin ( \omega t) \mathbf{j}+\frac{1}{2}\sin ( \omega t) \mathbf{k} & \color{#0066A7}{\mathbf{i}_{\mathbf{U}}}&{\color{#0066A7}=}&\color{#0066A7}{\tfrac{\mathbf{r}}{\left\Vert \mathbf{r}\right\Vert }}\\ \mathbf{i}_{\mathbf{W}}&=&-\frac{1}{2}\mathbf{j}+\frac{\sqrt{3}}{2}\mathbf{k}& \color{#0066A7}{\mathbf{i}_{\mathbf{W}}}&{\color{#0066A7}=}& \color{#0066A7}{\tfrac{\mathbf{r}\times \mathbf{v}}{\left\Vert \mathbf{r}\times \mathbf{v}\right\Vert }}\\ \mathbf{i}_{\mathbf{V}}&\;=\;&-\sin ( \omega t) \mathbf{i}+\frac{ \sqrt{3}}{2}\cos ( \omega t) \mathbf{j}+\frac{1}{2}\cos ( \omega t) \mathbf{k} & \color{#0066A7}{\mathbf{i}_{\mathbf{V}}}&{\color{#0066A7}=}&\color{#0066A7}{\mathbf{i}_{\mathbf{W}}\times \mathbf{i}_{\mathbf{U}}} \end{array}$$

The directions of the onboard gyroscope axes UVW are now expressed in terms of $\mathbf{i},$ $\mathbf{j},$ and $\mathbf{k}.$