Rotational mechanics $u$-substitution

In classical physics, when we try to get the force as a function of $$r$$; Why do we substitue $$r=\frac{1}{u}?$$ I don't get it, what is the point of it? Is the substitution arbitrary?

I am looking for a mathematical explanation of it. I understand it makes the equation solvable, otherwise we would have some messy chain differentiations. But I want it explained in Mathematical term. Perhaps like a proof. That way I wouldn't feel like I am putting in a formula without understanding.

• You should go into more detail about the problem you have in mind, but assuming this is a two-body problem thing, the answer is "it makes the differential equations easier to solve". In scattering, there's also the fact that annoying boundary conditions at $r=\infty$ become more tractable boundary conditions at $u=0$. – eyeballfrog Oct 25 '18 at 18:40
• @eyeballfrog Hi, first off, thank you for your answer. However, I am looking for a mathematical explanation of it. I understand it makes the equation solvable, otherwise we would have some messy chain differentiations. But I want it explained in Mathematical term. Perhaps like a proof. – Bertrand Wittgenstein's Ghost Oct 25 '18 at 18:48
• @eyeballfrog Thank you for that explanation; however, I don't understand how that is relevant in 2-particle systems, with $r$ being a finite number. We don't deal with $r\rightarrow \infty$. – Bertrand Wittgenstein's Ghost Oct 25 '18 at 18:55

It has to do with orbit trajectories and angular momentum. The conservation of angular momentum says $$mr^2\frac{d\theta}{dt} = \ell.$$ Suppose we want to find $$r(\theta)$$ instead of $$r(t)$$ in our orbit equations. $$dr/dt = (dr/d\theta)(d\theta/dt)$$, which means $$\frac{dr}{dt} = \frac{\ell}{mr^2}\frac{dr}{d\theta} = -\frac{\ell}{m}\frac{d}{d\theta}\left(\frac{1}{r}\right)$$ $$\begin{multline} \frac{d^2 r}{dt^2} = \frac{d^2 r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 + \frac{dr}{d\theta}\frac{d}{dt}\left(\frac{d\theta}{dt}\right) = \frac{\ell^2}{m^2r^4}\frac{d^2r}{d\theta^2}-\frac{2\ell}{mr^3}\frac{dr}{d\theta}\frac{dr}{dt} = \\ = \frac{\ell^2}{m^2r^4}\frac{d^2r}{d\theta^2}-\frac{2\ell^2}{m^2r^5}\left(\frac{dr}{d\theta}\right)^2 = -\frac{\ell^2}{m^2 r^2}\frac{d^2}{d\theta^2}\left(\frac{1}{r}\right) \end{multline}$$ So we see that the orbit trajectories are more easily expressed in terms of the $$\theta$$ derivatives of $$u= 1/r$$. In particular, the radial force equation $$m\ddot{r}-\ell^2/(mr^3) = F(r)$$ becomes $$\frac{d^2u}{d\theta^2} + u = -\frac{m}{\ell^2u^2}F\left(\frac{1}{u}\right),$$ and the conservation of energy becomes $$\left(\frac{du}{d\theta}\right)^2 + u^2 - \frac{2m}{\ell^2}V\left(\frac{1}{u}\right) = \frac{2mE}{\ell^2},$$ where $$-dV/dr = F(r)$$. In the common case of an inverse-square law force, the force term in the force equation is constant and the potential term in the energy equation is linear, which are easily solved exactly.
Using $$u = 1/r$$ has the added advantage that it is always bounded, as the bodies never have zero separation, while $$r$$ can become unbounded in scattering-type problems where the bodies do not form bound orbits.