What is the second $r$ in this equation for the Two Body Problem? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-16T10:54:40Z https://physics.stackexchange.com/feeds/question/449132 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/449132 1 What is the second $r$ in this equation for the Two Body Problem? Ian Ronk https://physics.stackexchange.com/users/215782 2018-12-18T18:39:22Z 2018-12-18T19:06:02Z <p><span class="math-container">$$r=\frac{r^2\frac{\mathrm d\theta^2}{\mathrm dt}}{\frac{Gm_2^3}{\left(m_1+m_2\right)^2}\left(1+e\cos\theta\right)}$$</span></p> <p>I have this equation for the radial distance of a planet from the barycenter. But I don't understand why there is an <span class="math-container">$r$</span> on both sides, the booklet from which this originates states that the <span class="math-container">$r^2 \mathrm d\theta^2/\mathrm dt$</span> is a constant, but what should it represent and how do I obtain this constant?</p> https://physics.stackexchange.com/questions/449132/-/449138#449138 5 Answer by caverac for What is the second $r$ in this equation for the Two Body Problem? caverac https://physics.stackexchange.com/users/135145 2018-12-18T19:06:02Z 2018-12-18T19:06:02Z <p>It is the angular momentum per unit mass, <span class="math-container">$L$</span></p> <p><span class="math-container">$$L = r^2\dot\theta = r^2 \frac{{\rm d}\theta}{{\rm d}t}$$</span></p> <p>In a central potential (e.g., Kepler's potential) this is a conserved quantity. If at any point you know the position (<span class="math-container">${\bf x}$</span>) and velocity (<span class="math-container">${\bf v}$</span>) of the test mass then you can calculate it as</p> <p><span class="math-container">$${\bf L} = {\bf r}\times {\bf v}$$</span></p>