# Non-relativistic Kepler orbits

Consider the Newtonian gravitational potential at a distance of Sun:

$$\varphi \left ( r \right )~=~-\frac{GM}{r}.$$

I write the classical Lagrangian in spherical coordinates for a planet with mass $m$:

$$L ~=~ \frac{1}{2}m (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{GM}{r},$$

and find that the canonical momentum $p_{\phi }$ is a constant of motion, because:

$$\dot{p_{\phi }}~=~ \frac{\partial L}{\partial \phi} ~=~ 0.$$

1. What is the physical interpretation of the canonical momentum?

2. How can we from the Lagrangian see that it is a constant of motion?

• The Lagrangian for the classical two body problem is: $$L = \frac{M m}{M\!+\!m} \frac{1}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{GMm}{r}$$ the Lagrangian per unit mass is: $$L =\frac{1}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{G(M\!+\!m)}{r}$$ For motion about a fixed mass M, the Lagrangian is: $$L = \frac{m}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{G M m}{r}$$
– Nick
May 12, 2013 at 22:24

$$p_{\phi}~:=~\frac{\partial L}{\partial \dot{\phi}}$$
is the (polar) $z$-component of the angular momentum $L_z$ of the point mass $m$ relative to the heliocentric reference frame. It is a constant of motion because the azimuthal angle $\phi$ is a cyclic coordinate.