# Inducing circular motion problem

Suppose a planet is moving under the influence of the central force of a star. When it reaches its closest distance to the star (i.e. where $$\dot{r}=0$$) its velocity $$\textbf{v}=\dot{r}\textbf{e}_r + r\dot{\theta}\textbf{e}_\theta$$ is reduced by some factor $$\beta$$ so that it then moves in a circular orbit. How am I able to find an expression for $$\beta$$ from this information?

You need additional assumptions, such as the force $$\mathbf{F}(\mathbf{r})=F(\mathbf{r})\,\mathbf{e}_r$$ is only dependent on the instantaneous $$\mathbf{r}=\mathbf{r}(t)$$ and not its derivatives $$\dot{\mathbf{r}}, \ddot{\mathbf{r}},\dots$$. We do not assume $$F(\mathbf{r})=F(r)$$ or the inverse square law, for example. However, we will need to assume $$F(a\mathbf{e}_r)=F(a)$$ for the circle $$r=a$$ that the planet will end up orbiting on.
Before the reduction in speed, the angular momentum $$\ell=r^2\dot\theta$$ is constant (since it is a central force motion), and we know the radial part of acceleration is $$\ddot r-r\dot\theta^2=\ddot{r}-r^{-3}\ell^2$$.
At $$\dot r=0$$, $$r=a$$, we reduce the speed, hence also angular momentum, by a factor $$\beta$$. Equating the radial part of acceleration before and after: $$\left.\bigg(\ddot{r}-r\dot\theta^2\bigg)\right\rvert_{r\to a+}=-\frac{\beta^2\ell^2}{a^3}$$ i.e., $$\left.\ddot{r}\right\rvert_{r\to a+}-\frac{\ell^2}{a^3}=-\frac{\beta^2\ell^2}{a^3}$$ rearranging for $$\beta$$: $$\beta=\sqrt{1-\frac{a^3}{\ell^2}\ddot{r}\Big\rvert_{r\to a+}}=\sqrt{1-\left.\frac{\ddot r}{r\dot\theta^2}\right\rvert_{r\to a+}}.$$