Does the $e$ in the following formula for the two-body problem represent the eccentricity?

$$r = \frac{k^2}{cu} = \frac{k^2}{c(1+e\cos{(\theta - \theta_0))}} $$ with $$c = \frac{Gm_2^3}{(m_1 + m_2)^2}$$ and $$k = r^2\frac{\mathrm d\theta}{\mathrm d t} $$

  • 1
    $\begingroup$ Yes, it does represent the eccentricity. $\endgroup$ – PM 2Ring Jan 27 at 8:09

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