I am facing this problem, from a 2016 Astrophysics Tripos past paper:
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 $a=\sqrt{b^{2}+r^{2}}$.
(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?
Best Answer
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}