# PlanetPhysics/Conservation of Angular Momentum

The angular momentum, ${\displaystyle \mathbf {L} }$ of a particle with position vector, ${\displaystyle \mathbf {r} }$, and total linear momentum, ${\displaystyle \mathbf {p} }$ is given by ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$. If some force, ${\displaystyle \mathbf {F} }$, acts on that particles, then the torque is defined similarily as ${\displaystyle \mathbf {N} =\mathbf {r} \times \mathbf {F} =\mathbf {r} \times d\mathbf {p} /dt}$.
${\displaystyle {\begin{matrix}{\frac {d\mathbf {L} }{dt}}&=&{\frac {d}{dt}}\left(\mathbf {r} \times \mathbf {p} \right)\\&=&\left({\frac {d\mathbf {r} }{dt}}\times \mathbf {p} \right)+\left(\mathbf {r} \times {\frac {d\mathbf {p} }{dt}}\right).\end{matrix}}}$
Consider the term, ${\displaystyle d\mathbf {r} /dt\times \mathbf {p} }$. Since ${\displaystyle \mathbf {p} =md\mathbf {r} /dt}$, it follows that ${\displaystyle {\frac {d\mathbf {r} }{dt}}\times \mathbf {p} =m\left({\frac {d\mathbf {r} }{dt}}\times {\frac {d\mathbf {r} }{dt}}\right).}$ But, given an arbitrary vector, ${\displaystyle \mathbf {A} }$, ${\displaystyle \mathbf {A} \times \mathbf {A} =\mathbf {0} }$ (the zero vector), so the expression for the time derivative of the angular momentum becomes, ${\displaystyle {\frac {d\mathbf {L} }{dt}}=\left(\mathbf {r} \times {\frac {d\mathbf {p} }{dt}}\right)=\mathbf {N} .}$ Writing the above simplistically as ${\displaystyle d\mathbf {L} /dt=\mathbf {N} }$ is is clear that when the torque is zero, then the angular momentum is constant in time; it is conserved.