$2.3.40.$ Potential energy of interaction of a particle with a fixed point source $U = V (\frac{l^2}{r^2} - \frac{2l}{r})$, where $r$ is the distance between the particle and the source, $V$ and $l$ are positive constants having the dimension of energy and distance, respectively. In what region does the rectilinear motion of a particle occur, if the total energy of the system is equal to $E$?
Let us consider the case where the system (the second particle) starts moving at a distance $r$ from the first particle.
Since this is the beginning of the movement, the kinetic energy of the first and second particles tends to zero. $$E_{k1}\to 0;\quad E_{k2}\to 0$$ From where the total energy would be equal to the potential energy $$E=E_p$$ Substituting expression from the statement $$\frac{E}{V}=\frac{l^2}{r^2}-\frac{2l}{r}$$ Transforming it into a quadratic equation $$Er^2+2Vlr-Vl^2=0$$ Let's solve it $$r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$$ Since the distance is not less than zero, we choose a positive sign for $E>0$ $$r\geq l\frac VE\left(\sqrt{1+E/V}-1\right)$$ For $E<0$ it will be in the range between $r_1$ and $r_2$ $$l\frac VE\left(-1-\sqrt{1+E/V}\right)\leq r\leq l\frac VE\left(-1+\sqrt{1+E/V}\right)$$
With $E>0$ area of motion $r\geqslant l\frac VE\left(-1+\sqrt{1+E/V}\right)$
if $E<0$ area of motion, $r$ between $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$
Almaskhan Arsen