# Laplace Runge Lenz (LRL) Vector

I have never encountered the LRL vector before (except the physics cup 2021). Please help in solving this. Any guidance will be helpful!

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I am assuming cycle means a single rotation period.

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At first, it actually seems like Laplace-Runge-Lenz vector just comes out of nowhere. So in order to give you a better understanding of this quantity, I’m going to show you the derivation from scratch:

Consider the Newton’s 2nd Law for a planet:

\dot{\vec p} = -\frac{GMm}{r^2}\hat r.

The rate of momentum’s change is pointed along the orbit’s plane. Motivated by the reason that \vec L is pointed perpendicular to the plane, let’s consider the following quantity:

\dot{\vec p}\times\vec L = -\frac{GMm}{r^2}\cdot(\hat r\times(\vec p\times\vec r)) = -\frac{GMm}{r^2}\cdot(\vec p(\hat r\cdot\vec r)-\vec r(\hat r\cdot\vec p))=
=-\frac{GMm}{r^2}(\vec p r -(\vec p\cdot\hat r)\vec r).

Here I have used the double cross product rule \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\times\vec b). Now, as you remember from my overkilling Baseball problem solution,

\dot{\vec r} = \vec r\hat r + r\dot\theta\hat\theta,

So \vec p=m\vec r , and its projection onto \hat r is simply m\dot r, i.e. \vec p\cdot\hat r= m\dot r. Hence,

\dot{\vec p}\times\vec L = -\frac{GMm^2}{r^2} (\dot{\vec r}r-\dot r\vec r).

At this point, we can rewrite \dot{\vec p}\times\vec L as \displaystyle\frac{d}{dt}\left(\vec p\times\vec L\right), because \vec L is conserved. Also note that \displaystyle\frac{r\dot{\vec r}-\dot r\vec r}{r^2} can be reduced to -\displaystyle\frac{d}{dt} \left(\frac{\vec r}{r}\right). Finally,