# 3 Constants of Motion@

Info

Insights and Conserved Quantities: The general 2-body problem may be simplified greatly by utilizing several special properties admitted by the system. These properties will enable a full solution of the general problem to be determined analytically.

Description

Develop the two-body problem’s constants of motion ($$\mathbf{h}$$, $$\mathbf{A}$$, and $$E$$).

Author

Matt Werner

## 3.1 Angular momentum $$\mathbf{h}$$@

Rewriting the Kepler problem in a convenient form,

$\ddot{\mathbf{r}} + \frac{\mu}{r^3}\mathbf{r} = \mathbf{0},$

lets us immediately show

$\begin{split}\mathbf{0} &= \left(\ddot{\mathbf{r}} + \frac{\mu}{r^3}\mathbf{r}\right) \!\times \mathbf{r} \\ &= \ddot{\mathbf{r}} \times \mathbf{r} + \left(\frac{\mu}{r^3}\mathbf{r}\right) \!\times \mathbf{r} \\ &= \ddot{\mathbf{r}} \times \mathbf{r} + \frac{\mu}{r^3}\left(\mathbf{r} \times \mathbf{r}\right) \\ &= \ddot{\mathbf{r}} \times \mathbf{r} \\ &= \ddot{\mathbf{r}} \times \mathbf{r} + \dot{\mathbf{r}} \times \dot{\mathbf{r}} \\ &= \frac{d}{dt}\underbrace{(\dot{\mathbf{r}} \times \mathbf{r})}_{-\mathbf{h}}.\end{split}$

We conclude that the (specific) angular momentum $$\mathbf{h}$$ is conserved under the dynamics of the 2-body problem.

(3.1.1)@$\mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}} \equiv \mathrm{const.}$

Important

The motion of the two bodies must be planar, and the angular momentum is perpendicular to this plane!

## 3.2 Laplace’s vector $$\mathbf{A}$$@

Knowing that the angular momentum $$\mathbf{h}$$ is conserved, we can do the following calculation.

$\begin{split}\frac{d}{dt}(\dot{\mathbf{r}} \times \mathbf{h}) &= \ddot{\mathbf{r}} \times \mathbf{h} \\ &= \left(-\frac{\mu}{r^3} \mathbf{r}\right) \times (\mathbf{r} \times \dot{\mathbf{r}}) \\ &= -\frac{\mu}{r^3} \big(\mathbf{r} \times (\mathbf{r} \times \dot{\mathbf{r}})\big) \\ &= -\frac{\mu}{r^3} \big((\mathbf{r} \cdot \dot{\mathbf{r}})\mathbf{r} - (\mathbf{r} \cdot \mathbf{r})\dot{\mathbf{r}}\big) \\ &= -\frac{\mu}{r^3} \big(r\dot{r} \mathbf{r} - r^2 \dot{\mathbf{r}}\big) && \quad \left(\mathbf{r} \cdot \dot{\mathbf{r}} = \frac{1}{2}\frac{d}{dt}(\mathbf{r} \cdot \mathbf{r})\right)\\ &= -\mu \left(\frac{\dot{r}}{r^2} \mathbf{r} - \frac{1}{r}\dot{\mathbf{r}}\right) \\ &= \mu \left(\frac{r \dot{\mathbf{r}} - \dot{r} \mathbf{r}}{r^2}\right) \\ &= \frac{d}{dt}\left(\frac{\mu}{r}\mathbf{r}\right).\end{split}$

Finally, we can combine the first and last expressions since the derivative is a linear operation. This provides the conserved quantity $$\mathbf{A}$$, the so-called Laplace-Runge-Lenz (or simply Laplace) vector1.

$\mathbf{A} = \dot{\mathbf{r}} \times \mathbf{h} - \frac{\mu}{r}\mathbf{r} \equiv \mathrm{const.}$

Important

There is a direction in the plane of motion of the two bodies that stays constant as a function of relative position and velocity!

## 3.3 Energy $$\mathcal{E}$$@

Manipulating the relative form of the equations of motion provides

$\begin{split}0 &= \left(\ddot{\mathbf{r}} + \frac{\mu}{r^3}\mathbf{r}\right) \cdot \dot{\mathbf{r}} \\ &= \ddot{\mathbf{r}} \cdot \dot{\mathbf{r}} + \left(\frac{\mu}{r^3}\mathbf{r}\right) \cdot \dot{\mathbf{r}} \\ &= \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}) + \frac{\mu}{r^3}(\mathbf{r} \cdot \dot{\mathbf{r}}) \\ &= \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}) + \frac{\mu}{r^2}\dot{r} && \quad \left(\mathbf{r} \cdot \dot{\mathbf{r}} = r\dot{r}\right) \\ &= \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}) + \frac{d}{dt}\left(-\frac{\mu}{r}\right) \\ &= \frac{d}{dt}\left(\frac{\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}}{2} - \frac{\mu}{r}\right).\end{split}$

This quantity can be easily identified as a sort of total specific mechanical energy — that is, the total mechanical energy per unit mass.

$\mathcal{E} = \frac{v^2}{2} - \frac{\mu}{r} \equiv \mathrm{const.}$

Here, $$v = |\dot{\mathbf{r}}|$$ is the magnitude of the (inertial) velocity. Note that the gravitational potential is appearing to come from a body of mass $$m_1 + m_2$$.

Important

The relative motion of the two bodies must be such that the relative orbital velocity $$v$$ and relative orbital radius $$r$$ interplay with an inverse relationship for a given, fixed energy $$\mathcal{E}$$.

1

Goldstein, Poole, Safko. Classical Mechanics, 3rd Edition. Pgs. 102-103