# 4 The Orbital Plane@

Info

A Change of Perspective: The solution of the general 2-body problem is developed several different ways utilizing the now known special properties admitted by the system.

Description

Switch into the orbital frame and rewrite the constants of motion in polar coordinates.

## 4.1 Reorienting the coordinate system@

Knowing that the motion must be 2-dimensional, we can choose a more convenient reference frame using a coordinate rotation. Let this rotation be denoted $$\mathbf{R} \in \mathrm{SO}(3)$$ such that

$\begin{split}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \mathbf{R} \begin{pmatrix}X \\ Y \\ Z\end{pmatrix}.\end{split}$

Although they look different, $$(x, y, z)$$ and $$(X, Y, Z)$$ both represent the same object.

It’s very convenient to utilize the conservation of angular momentum and the Laplace vector to define the precise action of $$\mathbf{R}$$ on the two coordinate systems. We choose to orient the $$z$$ axis in the direction of the angular momentum and the $$x$$ axis in the direction of the Laplace vector.

This rotation is not time-dependent — that is, $$\mathbf{R}$$ is a constant matrix representing a rotation between the two coordinate systems.

The standard form of the equations of motion is consequently unchanged.

## 4.2 Switching into polar coordinates@

By choosing this new coordinate system, we have reduced the problem to determining just 2 unknown quantities — the parameterizations of polar coordinates,

(4.2.1)@$\begin{split}x &= r \cos\theta \\ y &= r \sin\theta \\ z &= 0,\end{split}$

where $$r = |\mathbf{r}|$$ as usual and $$\theta$$ is the angle made between the reduced mass particle and the $$x$$ axis.

Important

The relative motion of the two bodies ($$\mathbf{r}$$) in the orbital plane is fully determined by the set of polar coordinates $$(r, \theta)$$. The 2-body problem has been reduced from determining all 6 variables to only 2.

### 4.2.1 Developing new equations of motion@

With polar coordinates being used to describe position, it’s extremely convenient to change our reference frame again by switching into the (noninertial) cylindrical coordinate system.

This transformation is especially useful as the expression for the relative position simplifies dramatically to

$\mathbf{r} = r\,\mathbf{e}_r.$

From this, we need to calculate the inertial velocity and acceleration before continuing. These expressions are given by

$\begin{split}\dot{\mathbf{r}} &= \dot{r}\,\mathbf{e}_r + r \dot{\theta}\,\mathbf{e}_\theta \\ \ddot{\mathbf{r}} &= (\ddot{r} - r\dot{\theta}^2)\mathbf{e}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\mathbf{e}_\theta.\end{split}$

Simply inserting these back into the standard form of the 2-body problem gives us a new set of differential equations in terms of $$r$$ and $$\theta$$.

(4.2.2)@$\begin{split}\ddot{r} - r\dot{\theta}^2 &= -\frac{\mu}{r^2} \\ r\ddot{\theta} + 2\dot{r}\dot{\theta} &= 0\end{split}$

### 4.2.2 Rewriting the conserved quantities@

#### 4.2.2.1 Angular momentum $$h(r,\dot{\theta})$$@

The angular momentum $$\mathbf{h}$$ expressed in polar coordinates is

$\begin{split}\mathbf{h} &= \mathbf{r} \times \dot{\mathbf{r}} \\ &= \left(r\,\mathbf{e}_r\right) \times \left(\dot{r}\,\mathbf{e}_r + r \dot{\theta}\,\mathbf{e}_\theta\right) \\ &= r\dot{r} (\mathbf{e}_r \times \mathbf{e}_r) + r^2\dot{\theta} (\mathbf{e}_r \times \mathbf{e}_\theta) \\ &= r^2\dot{\theta}\,\mathbf{e}_z.\end{split}$

Since $$\mathbf{h} = h \mathbf{e}_z$$ too by construction of $$\mathbf{R}$$ (where $$h = |\mathbf{h}|$$ is a constant of motion), the conservation of angular momentum provides

(4.2.3)@$r^2\dot{\theta} = h.$

This simple equation proves to be very useful, and we’re even able to get some information out of it (since we know that $$h \geqslant 0$$ and $$r > 0$$).

Important

1. The trajectory always goes in the counterclockwise direction! ($$\dot{\theta} \geqslant 0$$)

2. A collision will occur if there’s no angular momentum.

Proof

Set $$\dot{\theta} = 0$$. Then $$\ddot{r} < 0$$. This behavior eventually drives $$r \to 0$$ (called a collision) in finite time.

#### 4.2.2.2 Laplace vector $$A(r,\theta,\dot{r},\dot{\theta})$$@

The Laplace vector $$\mathbf{A}$$ expressed in polar coordinates is

$\begin{split}\mathbf{A} &= \dot{\mathbf{r}} \times \mathbf{h} - \frac{\mu}{r}\mathbf{r} \\ &= (\dot{r}\,\mathbf{e}_r + r\dot{\theta}\,\mathbf{e}_\theta) \times (r^2\dot{\theta}\,\mathbf{e}_z) - \frac{\mu}{r} (r \,\mathbf{e}_r) \\ &= r^2\dot{r}\dot{\theta}(\mathbf{e}_r \times \mathbf{e}_z) + r^3\dot{\theta}^2(\mathbf{e}_\theta \times \mathbf{e}_z) - \mu\,\mathbf{e}_r \\ &= r^2\dot{r}\dot{\theta}(-\mathbf{e}_\theta) + r^3\dot{\theta}^2(\mathbf{e}_r) - \mu\,\mathbf{e}_r \\ &= (r^3\dot{\theta}^2 - \mu)\mathbf{e}_r - r^2\dot{r}\dot{\theta}\,\mathbf{e}_\theta \\ &= A \underbrace{(\cos\theta\,\mathbf{e}_r - \sin\theta\,\mathbf{e}_\theta)}_{\mathbf{e}_x}.\end{split}$

(The last equality holds by construction of $$\mathbf{R}$$.)

To summarize, we get two scalar equations from the invariance of the Laplace vector.

(4.2.4)@$\begin{split}r^2\dot{r}\dot{\theta} &= A\sin\theta \\ r^3\dot{\theta}^2 - \mu &= A\cos\theta\end{split}$

Important

Combined with angular momentum, we can get expressions of $$r(\theta)$$ and $$\dot{r}(\theta)$$ very quickly!

#### 4.2.2.3 Energy $$\mathcal{E}(r,\theta,\dot{r},\dot{\theta})$$@

The energy $$\mathcal{E}$$ expressed in polar coordinates is

(4.2.5)@$\mathcal{E} = \frac{1}{2}\underbrace{(\dot{r}^2 + r^2\dot{\theta}^2)}_{v^2 = \dot{\mathbf{r}} \cdot \dot{\mathbf{r}}} - \frac{\mu}{r}.$

Important

• Paired with angular momentum, this expression “integrates” the $$\ddot{r}$$ equation of motion with integration constant $$2\mathcal{E}$$.

• Similarly, this relation provides another expression for $$\dot{\theta}$$ in addition to that from the conservation of angular momentum using the knowledge that $$\dot{\theta} > 0$$ for interesting motion.