Irodov 1.131: Not at Equilibrium, the Force Peaks – Amit & Deepali Sinha Foundation

Problem Irodov 1.131

A particle sits in a central force field whose potential energy depends only on the distance $r$ from the center of the field, according to $U(r)=a/r^2-b/r$ , where $a$ and $b$ are positive constants.

Find the radius $r_0$ at which the particle is in equilibrium, and determine whether that equilibrium is stable.
Find the greatest possible magnitude of the attractive force the particle can feel, and sketch how $U(r)$ and the radial force $F_r(r)$ (the force’s projection onto the radius vector) vary with $r$ .

Given

$U(r)=\dfrac{a}{r^2}-\dfrac{b}{r}$
$a,b>0$ constants

Find

(a) the equilibrium radius

r_0

and whether it is steady; (b) the maximum magnitude of the attractive force, and the shape of

U(r)

F_r(r)

Solution

Step 1The radial force from the potential

For a central field the radial component of the force is minus the derivative of the potential energy along $r$ . Differentiating the given $U(r)$ first,

\begin{align*}\frac{dU}{dr} = \frac{d}{dr}\left(\frac{a}{r^2}-\frac{b}{r}\right) = -\frac{2a}{r^3}+\frac{b}{r^2}.\end{align*}

and then flipping the sign gives the radial force itself:

F_r = -\frac{dU}{dr} = \frac{2a}{r^3}-\frac{b}{r^2}.

F_r=\dfrac{2a}{r^3}-\dfrac{b}{r^2}

(1)

Step 2Equilibrium is where the radial force vanishes

Setting (1) to zero and clearing denominators,

\frac{2a}{r_0^3}=\frac{b}{r_0^2} \quad\Longrightarrow\quad 2a = b\,r_0.

r_0=\dfrac{2a}{b}

(2)

Step 3Stability from the curvature of

U

An equilibrium is steady when $U$ has a genuine minimum there, so the second derivative must be positive at $r_0$ :

\begin{align*}\frac{d^2U}{dr^2} = \frac{6a}{r^4}-\frac{2b}{r^3}, \qquad \left.\frac{d^2U}{dr^2}\right|_{r_0=2a/b} = \frac{6a}{(2a/b)^4}-\frac{2b}{(2a/b)^3} = \frac{b^4}{8a^3}.\end{align*}

Since $a,b>0$ this is positive, so $r_0$ is a genuine minimum of $U$ and the equilibrium is steady.

Step 4Locating the point of strongest pull

For $r>r_0$ , (1) is negative, so the force is attractive there; its magnitude is

\begin{align*}|F_r| = \frac{b}{r^2}-\frac{2a}{r^3}.\end{align*}

Maximizing over $r$ :

\begin{align*}\frac{d}{dr}|F_r| = -\frac{2b}{r^3}+\frac{6a}{r^4} = 0 \quad\Longrightarrow\quad r_{\max}=\frac{3a}{b}.\end{align*}

Substituting back:

\begin{align*}|F_r|_{\max} = \frac{b}{(3a/b)^2}-\frac{2a}{(3a/b)^3} = \frac{b^3}{9a^2}-\frac{2b^3}{27a^2} = \frac{3b^3-2b^3}{27a^2}.\end{align*}

F_{\max}=\dfrac{b^3}{27a^2}

(3)

Step 5The shape of

U(r)

and

F_r(r)

U(r): a well with its minimum at r_0=2a/b, crossing the r axis at r=a/b and approaching 0 from below as r->∞. Illustrative curve for a=b=1, shape only. — $U(r)$ : a well with its minimum at $r_0=2a/b$ , crossing the $r$ axis at $r=a/b$ and approaching $0$ from below as $r\to\infty$ . Illustrative curve for $a=b=1$ , shape only.

F_r(r): repulsive (positive) for r<r_0, crossing the r axis at r_0=2a/b, then attractive (negative) with its deepest point, the answer to (b), at r_=3a/b, and approaching 0 from below as r->∞. Illustrative curve for a=b=1, shape only. — $F_r(r)$ : repulsive (positive) for $r<r_0$ , crossing the $r$ axis at $r_0=2a/b$ , then attractive (negative) with its deepest point, the answer to (b), at $r_{\max}=3a/b$ , and approaching $0$ from below as $r\to\infty$ . Illustrative curve for $a=b=1$ , shape only.

Both curves follow directly from (1): a shared zero at $r_0$ , $U$ turning from falling to rising there, and $F_r$ turning from positive to negative.

Check

Dimensionally, $U=a/r^2$ forces $[a]=[U][r]^2$ and $U=b/r$ forces $[b]=[U][r]$ , so $b^3/a^2$ carries units $([U][r])^3/([U][r]^2)^2=[U]/[r]$ , energy per length, which is force, as an $F_{\max}$ must be.

Remark

This $U(r)$ is a toy version of the interatomic potentials used in molecular physics: a steep repulsive term at short range ( $a/r^2$ ) against a gentler attractive tail ( $-b/r$ ), the same qualitative shape as the Lennard-Jones potential, just with different powers. Note also that $r_{\max}=3a/b=(3/2)r_0$ sits beyond the equilibrium radius, never at it, since $r_0$ is by definition where the force is zero and the pull only builds up past that point.