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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?

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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 $\beta$: $$ \beta=\sqrt{1-\frac{a^3}{\ell^2}\ddot{r}\Big\rvert_{r\to a+}}=\sqrt{1-\left.\frac{\ddot r}{r\dot\theta^2}\right\rvert_{r\to a+}}. $$

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