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