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?

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    <h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.3.40.$ 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 $E$?
    </p>

    <h3>Solution</h3>
    <p>

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 $r_2$
$$l\frac VE\left(-1-\sqrt{1+E/V}\right)\leq r\leq l\frac VE\left(-1+\sqrt{1+E/V}\right)$$

    <h4>Answer</h4>
    <p>
        With $E>0$ area of motion $r\geqslant l\frac VE\left(-1+\sqrt{1+E/V}\right)$</p><p>

if $E<0$ area of motion, $r$ between $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$

Almaskhan Arsen

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