# Kepler's laws: show the planet always stays in the same plane？

I know angular momentum $L=q \times p$ is conserved, where $p=L_{\dot q}$ is linear momentum. How to apply this to a planet orbiting the star, described by the position vector $q$ relative to the star. Here $L=\frac{1}{2}M|\dot q ^2|+\frac{GM}{|q|}$.

To deduce the Kepler's law

A planet always stays in one fixed plane and the radius vector $q$ sweeps equal areas in equal time.

My attempt:

I compute $L_{\dot q}=M\dot q$, so the given condition means $q \times \dot q$ is conserved. I think the cross product can denote the area the planet sweeps in $\triangle t$.

But I don't know how to show the planet always stays in the same plane.

Angular momentum $\mathbf{L}=\mathbf{q}\times m\mathbf{\dot{q}}$ is a vector which is always perpendicular to $\mathbf{q}$ (as it comes from a cross product). So, since angular momentum is conserved, $\mathbf{q}$ is always perpendicular to the same unit vector $\hat{\mathbf{L}}$ which can be taken as the unit normal of a plane.
In general, the magnitude of the cross product $|\mathbf{a}\times\mathbf{b}|$ gives the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$. Here you want the triangle with sides $\mathbf{q}$ and $d\mathbf{q}$, i.e. $\frac{1}{2}|\mathbf{q}\times d\mathbf{q}|\approx \frac{1}{2}|\mathbf{q}\times \mathbf{\dot{q}}|\,dt$.