Hello it's my first time posting in here and I hope someone in here can help me as my teachers quite dislike questions that aren't specifically in the curriculum. I'm reading About classical mechanics and this came up in my book: They define $ u=1/r$ and transform the radial energy equation by noticing that $\frac{du}{d\theta} = \frac{1}{r} \frac{dr}{d\theta}$ they inject it in the radial energy equation getting $$\frac{J^{2}}{2m} (\frac{du}{d\theta})^{2} + \frac{J^{2}}{2m} u^{2} + V = E $$ Next the consider explicitly the case of the inverse square law so the $V= \pm|k|u $. Now they introduce $l=\frac{J^2}{m|k|} $ and my multiplying both sides by $\frac{2}{|k|} $ so that they can rewrite the equation as $$ l (\frac{du}{d\theta})^2+l u^2 \pm 2u = \frac{2E}{|k|} $$ Now they set $z=lu \pm 1 $ and multiply the equation by l and then add 1. Thus finally getting: $$ (\frac{dz}{d\theta})^2 + z^2 = \frac{2E l}{|k|} +1 = e^2 $$ where e is dimensionless constant.
My question is how do they solve the latter equation. Because my book only says which are the solutions but never how they got them. Could someone help me with this? Thanks a lot.


1 Answer 1


Just take the square root

\begin{equation} {dz\over d\theta}=\pm\sqrt{e^2-z^2}, \end{equation}


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