Perifocal Frame

In this section, we define the perifocal reference frame, a convenient reference frame for describing the planar orbits that we’ve been discussing. The perifocal frame is a Cartesian reference centered at the occupied focus of the orbit, the location where \(m_1\) exists. Since all the orbits we’ve looked at so far are planar, the \(\pf{x}\pf{y}\) plane of the perifocal frame lies in the same plane as the orbit. This means that the \(\pf{z}\) direction is pointing in the same direction as the angular momentum.

The unit vectors that define this coordinate system are:

  1. \(\uvec{p}\) pointing along the \(\pf{x}\) axis

  2. \(\uvec{q}\) pointing along the \(\pf{y}\) axis

  3. \(\uvec{w}\) pointing along the \(\pf{z}\) axis

../_images/definition-of-perifocal-frame.svg

Fig. 41 The definition of the perifocal frame. The \(\uvec{w}\) direction is pointing direction towards the viewer.

As shown in Fig. 41, \(\uvec{p}\) points along the apse line to the right of the focus, towards periapsis. \(\uvec{q}\) points 90° true anomaly from \(\uvec{p}\). Finally, \(\uvec{w}\) points in the same direction as the angular momentum, and can be defined by:

(164)\[\uvec{w} = \frac{\vector{h}}{h}\]

Position Vector

In the perifocal frame, the position vector \(\vector{r}\) may be written in terms of the \(\pf{x}\) and \(\pf{y}\) Cartesian coordinates. Since the orbit is planar, the \(\uvec{w}\) component is zero.

(165)\[\vector{r} = \pf{x}\uvec{p} + \pf{y}\uvec{q}\]

\(\pf{x}\) and \(\pf{y}\) can be transformed into the radial-true anomaly polar coordinate system by:

(166)\[\begin{aligned} \pf{x} &= r\cos\nu & \pf{y} &= r\sin\nu \end{aligned}\]

By plugging in the orbit equation, Eq. (113), \(\vector{r}\) can be written as:

(167)\[\vector{r} = \frac{h^2}{\mu}\frac{1}{1 + e\cos\nu}\left(\cos\nu\uvec{p} + \sin\nu\uvec{q}\right)\]

Velocity Vector

The velocity is found by taking the time derivative of the position:

(168)\[\vector{v} = \dot{\vector{r}} = \dot{\pf{x}}\uvec{p} + \dot{\pf{y}}\uvec{q}\]

Then we need to apply the product rule, because both \(r\) and \(\nu\) are functions of time:

(169)\[\begin{split}\begin{aligned} \pf{\dot{x}} &= \dot{r}\cos\nu - r\dot{\nu}\sin\nu \\ \pf{\dot{y}} &= \dot{r}\sin\nu + r\dot{\nu}\cos\nu \end{aligned}\end{split}\]

Substituting some of the relationships from previously, we can simplify the velocity vector as:

(170)\[\vector{v} = \frac{\mu}{h}\left[-\sin\nu\uvec{p} + \left(e + \cos\nu\right)\uvec{q}\right]\]