# Central forces and conservation problem

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.

• Perhaps this question was more suited to Physics.SE. Apr 26, 2019 at 23:43

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 $$\begin{equation*} \frac{a^2 u^2}{2 - a u^2} \end{equation*}$$

• But this is a lower bound for $r$, and hence a 'lower limit' for this distance AB right? Whereas I am trying to find an upper bound for this 'upper limit', unless my understanding of what this question asks is off Apr 27, 2019 at 16:08
• @AzamatBagatov: I added a clarification in the end of my post. The second value is the upper bound. Apr 27, 2019 at 16:10