Algebric troubles while trying to get energy of orbit in isochrone (or plummer) potential

Question

I am facing this problem, from a 2016 Astrophysics Tripos past paper:

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The gravitational potential

$$\Phi=-\frac{G M}{b+\sqrt{b^{2}+r^{2}}}$$

where $G$, $M$ and $b$ are constants and $r$ is the distance from the origin, is called the isochrone potential. Show that in the isochrone potential, the energy of a circular orbit is given by $E=-G M /(2 a)$, where $a=\sqrt{b^{2}+r^{2}}$.

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(There is a wikipedia article about the potential they are talking about.)

I know that I can express energy of the orbit by

$$E=\frac{1}{2}r\frac{\partial\Phi}{\partial r}+\Phi$$

and I have done the algebra resulting in:

\begin{equation} \tag{got} \frac{G M\left(\left(-\frac{1}{2}\right)\bbox[5px,border:2px solid green]{\left(r^{2}+2 b^{2}\right)}\bbox[1px,border:2px solid red]{-b \sqrt{b^{2}+r^{2}}}\right)}{\bbox[5px,border:2px solid green]{\left(r^{2}+2 b^{2}\right)} \sqrt{b^{2}+r^{2}}\bbox[1px,border:2px solid red]{+2 b\left(b^{2}+r^{2}\right)}} \end{equation}

which numerically does equal

\begin{equation} \tag{want} \frac{-GM}{2\sqrt{b^2+r^2}} \end{equation}

as wanted (checked here using Desmos).

I could cancel the terms in green but there are the other terms in red, which I cannot get rid of currently.

How can I go algebraicly from the state I got to the state I want?

Answer 1

We factorize $-\frac12GM$ from the numerator and we factorize $\sqrt{r^2+b^2}$ from the denominator:

\begin{align} &\frac{GM \left( -\frac12(r^2+2b^2)-b\sqrt{b^2+r^2}\right)}{(r^2+2b^2)\sqrt{b^2+r^2}+2b(b^2+r^2)}\\ &=\frac{-GM}{2\sqrt{b^2+r^2}}\left(\frac{r^2+2b^2+2b\sqrt{b^2+r^2}}{r^2+2b^2+2b\sqrt{b^2+r^2}} \right)\\ &=\frac{-GM}{2\sqrt{b^2+r^2}} \end{align}