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$$r=\frac{r^2\frac{\mathrm d\theta^2}{\mathrm dt}}{\frac{Gm_2^3}{\left(m_1+m_2\right)^2}\left(1+e\cos\theta\right)}$$

I have this equation for the radial distance of a planet from the barycenter. But I don't understand why there is an $r$ on both sides, the booklet from which this originates states that the $ r^2 \mathrm d\theta^2/\mathrm dt $ is a constant, but what should it represent and how do I obtain this constant?

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It is the angular momentum per unit mass, $L$

$$ L = r^2\dot\theta = r^2 \frac{{\rm d}\theta}{{\rm d}t} $$

In a central potential (e.g., Kepler's potential) this is a conserved quantity. If at any point you know the position (${\bf x}$) and velocity (${\bf v}$) of the test mass then you can calculate it as

$$ {\bf L} = {\bf r}\times {\bf v} $$

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