A comet, traveling with speed $V$, is at a very great distance from a star. If the path of the comet were unaffected by the presence of the star it's closest distance of approach would be $d$. However, the comet experiences an attractive force $\frac{\gamma}{r^2}$ per unit mass directed towards the star. Calculate the actual distance of the closest approach.
The first part of this exercise was to prove that $$\frac{1}{2}mv^2+\int{F(r)dr}=\ const.$$ which I proved.
May I use this instead of Binet's equation?
I've done this:
$\vec r=r\vec{e_r}$ - position vector
$\dot{\vec r}=\dot r\vec{e_r}+r\dot\theta\vec{e_\theta}$ - velocity
$F(r)=\frac{\gamma m}{r^2}$
$\int F(r)dr=\int \frac{\gamma m}{r^2}dr=-\frac{\gamma m}{r}$
Hence,
$$\frac{1}{2}mv^2-\frac{\gamma m}{r}=\frac{1}{2}mV^2-\frac{\gamma m}{d}$$
Since $v=0$ at an apse, I have $$-\frac{\gamma m}{r}=\frac{1}{2}mV^2-\frac{\gamma m}{d}$$ at the closest approach.
From this I have that a distance of the closest approach is $$r=\frac{2\gamma d}{dV^2+2\gamma}$$
In effect you can bypass Binet.
Start from the conservation laws $$r^2 \dot \theta=dV$$$$\dot r^2+r^2 \dot \theta^2=V^2+2 \gamma \left (\frac 1{r}- \frac 1{d} \right)$$Obtain $\dot \theta$ from the first and substitute in the second.
Then set $\,\dot r=0$ and solve with respect to $r$.
The solutions are $$r=d \quad ,\quad r= \frac {d^2V^2}{2\gamma-dV^2}$$If $\,V<\sqrt \frac {\gamma}{d}$, then $$\frac {d^2V^2}{2\gamma-dV^2}<d$$