Lagrangian two body gravitational conserved quantities I have the Lagrangian for two gravitationally attracting bodies:
$$ L ={\frac{1}{2}}M\dot{R}^2 +\frac{1}{2}{\mu}\dot{r}^2 + \frac{Gm_1m_2}{|r|}$$
Where M is the total mass, mu the reduced mass and r the vector pointing between them and R the centre of mass vector.
I'm trying to prove ${\mu}r\times\dot{r}$ is conserved. 
I've rewritten the vector $\mathbf{\dot{r}^2}$ in polars as $\dot{r}^2 +r^2\dot{\phi}^2$ which creates a cyclic co-ordinate due to the absence of $\phi$ but this results in the conserved quantity ${\mu}r^2\dot{\phi}$, but I am unsure how to prove this is the same as the cross product?
 A: I'm guessing you wanted to say $\mu r^2 \dot{\phi} $ is conserved. You may know by intuition that should be the $z$ component of the angular momentum. If not, you can do this:
$$\vec{L} =\mu \vec{r} \wedge \dot{\vec{r}} =\mu (r \hat{u}_r ) \wedge(\dot{r} \hat{u}_r + r \dot{\phi} \hat{u}_\phi ) =\mu r^2 \dot{\phi} \hat{u}_z $$
And that's it, just analyze it by components. It's ausual result that some component of the angular momentum is conserved, get used to this kind of reasoning. 
A: Actually this can be proved not only for the Newtonian potential energy
$V = -\frac{G m_1 m_2}{|\mathbf r|}$, but for any potential energy $V(|\mathbf r|)$
which depends only on the length $|\mathbf r|$, but not on the direction of $\mathbf r$.
So, let's use the more general Lagrangian
$$ L ={\frac{1}{2}}M\dot{\mathbf R}^2 +\frac{1}{2}{\mu}\dot{\mathbf r}^2 - V(|\mathbf r|) $$
Then the Euler-Langrange equation for $\mathbf r$ gives
$$
\begin{align}
\frac{\text d}{\text d t} \frac{\partial L}{\partial \dot{\mathbf r}} &= \frac{\partial L}{\partial \mathbf r} \\
\frac{\text d}{\text d t} \mu \dot{\mathbf r} &= - \nabla V(|\mathbf r|) \\
\mu \ddot{\mathbf r} &= - \nabla V(|\mathbf r|)  \tag{1}\label{1}
\end{align}
$$
Now let's calculate the time derivative of $\mu \mathbf r \times \dot{\mathbf r}$.
$$
\begin{align}
   \frac{\text d}{\text d t} (\mu \mathbf r \times \dot{\mathbf r})
=& \mu \dot{\mathbf r} \times \dot{\mathbf r} + \mu \mathbf r \times \ddot{\mathbf r} \\
=& \mu \mathbf r \times \ddot{\mathbf r} \\
\stackrel{(1)}=& - \mathbf r \times \nabla V(|\mathbf r|) \\
=& - \mathbf r \times V'(|\mathbf r|) \frac{\mathbf r}{|\mathbf r|} \\
=& \mathbf 0
\end{align}
$$
