Irodov 1.191: The Mass Hiding in Two Turning Points – Amit & Deepali Sinha Foundation

Problem Irodov 1.191

A particle of unknown mass $m$ moves on a closed orbit under a central force directed toward a fixed point $O$ , with potential energy $U(r)=kr^2$ , where $k$ is a given positive constant and $r$ is the distance from $O$ . The orbit’s closest approach to $O$ is $r_1$ , and the particle’s speed at its farthest point from $O$ is $v_2$ . Determine the particle’s mass $m$ in terms of $k$ , $r_1$ , and $v_2$ .

The orbit under U=kr^2 is a closed curve centered on the force center O itself, not on a focus. Its closest and farthest distances are r_1 and r_2; at the farthest point the speed is the given v_2. — The orbit under $U=kr^2$ is a closed curve centered on the force center $O$ itself, not on a focus. Its closest and farthest distances are $r_1$ and $r_2$ ; at the farthest point the speed is the given $v_2$ .

Given

$U(r) = kr^2$
$r_1$ (closest distance)
$v_2$ (speed at farthest point)

Find

m

Solution

Step 1Angular momentum conservation on the orbit

The angular momentum of the particle about the force center $O$ is the moment of its momentum $\vect p=m\vect v$ about that point.

\vect L = \vect r\times\vect p = m\,\vect r\times\vect v \tag{1}

What changes it is the torque about the same point. Differentiating (1) by the product rule shows that nothing but the moment of the force survives.

\begin{align*} \vect N = \frac{d\vect L}{dt} &= m\,\frac{d\vect r}{dt}\times\vect v + m\,\vect r\times\frac{d\vect v}{dt} \\ &= m\,\vect v\times\vect v + \vect r\times\left(m\,\frac{d\vect v}{dt}\right) \\ &= \vect r\times\vect F \tag{2}\end{align*}

The field is central, which is exactly the statement that $\vect F$ lies along the line joining the particle to $O$ at every instant. Two parallel vectors have a vanishing cross product, so the torque (2) is identically zero.

\vect F \parallel \vect r \quad\Longrightarrow\quad \vect N = \vect r\times\vect F = \vect 0 \tag{3}

A vector whose rate of change is zero at every instant does not change at all, in size or in direction.

\vect L = \text{constant}

(4)

Step 2Both turning points have purely tangential velocity

Let $r_2$ (unknown) be the farthest distance and $v_1$ the (unknown) speed at $r_1$ . Taking the magnitude of (1) at a point where $\vect v\perp\vect r$ gives $L=mrv$ ; writing that at each of the two turning points and equating the two, since (4) says the value is the same,

L = mv_1r_1 = mv_2r_2

(5)

Step 3Energy conservation between the two turning points

The total mechanical energy $E=mv^2/2+kr^2$ is also conserved, so equating its value at $r_1$ and at $r_2$ ,

\frac{1}{2}mv_1^2+kr_1^2 = \frac{1}{2}mv_2^2+kr_2^2. \tag{6}

Step 4Eliminate the unknowns

v_1

and

r_2

From (5), $v_1=v_2r_2/r_1$ . Substituting into the energy balance (6) and multiplying through by 2,

\begin{align*}m\left(\frac{v_2r_2}{r_1}\right)^2+2kr_1^2 &= mv_2^2+2kr_2^2 \\ mv_2^2\left(\frac{r_2^2}{r_1^2}-1\right) &= 2k\left(r_2^2-r_1^2\right) \\ \frac{mv_2^2}{r_1^2}\left(r_2^2-r_1^2\right) &= 2k\left(r_2^2-r_1^2\right).\end{align*}

Since the orbit is genuinely closed with two distinct turning points, $r_2\neq r_1$ , so the common factor $(r_2^2-r_1^2)$ cancels from both sides,

\frac{mv_2^2}{r_1^2} = 2k.

m = \dfrac{2kr_1^2}{v_2^2}

(7)

Check

Dimensionally, $U=kr^2$ requires $[k]=\text{kg/s}^2$ , and $[k][r_1]^2/[v_2]^2$ then reduces to kilograms, a mass, as required. As a limiting case, a circular orbit has $r_2=r_1$ and $v_2=v_1$ ; the two conservation laws (5) and (6) are then satisfied identically for any $m$ , consistent with the canceled factor $(r_2^2-r_1^2)$ vanishing exactly there: it is this specific route, combining only $L$ and $E$ , that needs a genuinely closed, non-circular orbit to pin down $m$ . A circular orbit still fixes $m$ uniquely, to the same value, but by the separate centripetal-force balance $mv_1^2/r_1=2kr_1$ , not from $L$ and $E$ alone.

Remark

The farthest distance $r_2$ never had to be found: it canceled out of the combination of the two conservation laws, so the mass is fixed by $r_1$ and $v_2$ alone. The cancellation relied on both points being turning points, where the full speed is tangential and $L=mvr$ holds without a separate radial piece; at a generic point on the orbit the same two conservation laws still apply, but $v$ there is not simply $L/(mr)$ , so this shortcut does not carry over unless the second point is also a turning point.