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 \(E\)).

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