An object A moves under the influence of a planet B's gravity, a central force of magnitude $\frac{1}{r^2}$. At some point, its velocity is $u$e$_{\theta}$ and $r=a$. It is given that $au^2 < 2$. I am looking to find an upper limit to the distance A gets from B.
Integrating the force I found the potential energy to be V = $-\frac{1}{r}$.
From conservation of angular momentum I found that: L = $au$k = $r^2\dot{\theta}$k $\implies$ $\dot{\theta} = \frac{au}{r^2}$.
From conservation of energy I found that: \begin{equation*} \frac{1}{2}\Big(\dot{r}^2 + r^2\dot{\theta}^2\Big) - \frac{1}{r} = \frac{\dot{r}^2}{2} + \frac{a^2u^2}{2r^2} - \frac{1}{r} = \frac{u^2}{2} - \frac{1}{a}. \end{equation*}
If I use that $\frac{\dot{r}^2}{2} \geq 0$, simplify everything and use $au^2 < 2$ then I end up with: \begin{equation*} au^2(r^2 -a^2) -2r^2 +2ar \geq 0 \implies 2a(r-a) \geq 0. \end{equation*}
So at this point I only have $r \geq a$, but that only gives me a lower bound for this distance, right? If someone could point out where I'm going wrong that would be great.
Note that the condition that at $r=a$ the velocity is $u$e$_{\theta}$ means that when $r=a$ we are either at a minimum distance or at a maximum distance.
This is because the condition that marks that we are at an extreme distance is $\dot{r} = 0$, i.e., no radial velocity.
So, from the equation \begin{equation*} \frac{\dot{r}^2}{2} + \frac{a^2u^2}{2r^2} - \frac{1}{r} = \frac{u^2}{2} - \frac{1}{a} \end{equation*}
Using $\dot{r} = 0$ and after multiplying by $2 a r^2$ gives \begin{equation*} a^3u^2 - 2 a r = (a u^2 - 2 ) r^2 \end{equation*}
Which has as solutions \begin{equation*} r = a \frac{1 \pm \sqrt{1 - (2 - a u^2)a u^2}}{2 - a u^2} \end{equation*}
Hence \begin{equation*} r = a \frac{1 \pm \sqrt{(1 - a u^2)^2}}{2 - a u^2} \end{equation*}
And, finally \begin{equation*} r = a \vee r = \frac{a^2 u^2}{2 - a u^2} > a \end{equation*}
So, the lower bound for $r$ is $a$ and the upper bound for $r$ is \begin{equation*} \frac{a^2 u^2}{2 - a u^2} \end{equation*}