Irodov 1.144: Reflect the Trajectory Through the Center

Problem Irodov 1.144

Two particles of masses $m_1$ and $m_2=m_1/2$ interact with each other and with nothing else, so their common center of inertia stays at one fixed point for all time. At a certain moment the positions of both particles are known, and the complete path (past and future) that the particle of mass $m_1$ traces is also known, a curved line through that particle’s marked position. Using only the fixed center of inertia and the mass ratio, construct the trajectory traced by the particle of mass $m_2$ .

Given

$m_2=m_1/2$
the trajectory of $m_1$
both particles’ positions at one instant

Find

the trajectory of

m_2

Solution

Step 1The center of inertia is a fixed point

The center of inertia C located from the one pictured instant: it divides the segment joining the two marked positions (dashed) in the ratio m_2:m_1 = 1:2, so it sits twice as far from m_2's position as from m_1's. — The center of inertia $C$ located from the one pictured instant: it divides the segment joining the two marked positions (dashed) in the ratio $m_2:m_1 = 1:2$ , so it sits twice as far from $m_2$ ’s position as from $m_1$ ’s.

The system is closed (no external forces), so its total momentum is conserved; it is given to be zero.

m_1\vect v_1+m_2\vect v_2 = \vect 0

(1)

Each velocity is the time derivative of its particle’s position vector, so (1) is a statement about the rate of change of one combination of the two positions.

\begin{align*}m_1\frac{d\vect r_1}{dt}+m_2\frac{d\vect r_2}{dt} &= \vect 0 \\ \frac{d}{dt}\bigl(m_1\vect r_1+m_2\vect r_2\bigr) &= \vect 0\end{align*}

Integrating over time from $0$ to $t$ , the combination keeps its initial value, a constant vector; writing that constant as $(m_1+m_2)\vect r_C$ names the point it defines.

\begin{align*}\int_0^t\frac{d}{dt'}\bigl(m_1\vect r_1+m_2\vect r_2\bigr)\,dt' &= \vect 0 \\ m_1\vect r_1(t)+m_2\vect r_2(t) &= m_1\vect r_1(0)+m_2\vect r_2(0)\end{align*}

m_1\vect r_1(t)+m_2\vect r_2(t) = (m_1+m_2)\,\vect r_C

(2)

Being constant, $\vect r_C$ names the same point $C$ for every $t$ , including the pictured instant, so $C$ can be located directly from the one instant shown.

\vect r_C = \frac{m_1\vect r_1+m_2\vect r_2}{m_1+m_2}

Step 2Solve for

m_2

’s position at every instant

Isolating $m_2\vect r_2(t)$ in (2), dividing by $m_2$ , and splitting $(m_1+m_2)/m_2$ as $1+m_1/m_2$ gives $m_2$ ’s position at every instant directly from $m_1$ ’s, at that same instant; inserting $m_2=m_1/2$ (so $m_1/m_2=2$ ) finishes it.

\begin{align*}m_2\vect r_2(t) &= (m_1+m_2)\vect r_C-m_1\vect r_1(t)\\ \vect r_2(t) &= \vect r_C+\frac{m_1}{m_2}\bigl(\vect r_C-\vect r_1(t)\bigr)\\ &= \vect r_C - 2\bigl(\vect r_1(t)-\vect r_C\bigr)\end{align*}

\vect r_2(t) = 3\vect r_C - 2\,\vect r_1(t)

(3)

Step 3Read the construction off the formula

The constructed trajectory of m_2 (blue): the image of m_1's given curve (black) under the point reflection through C with scale factor m_1/m_2=2, so twice as large and on the opposite side of C at every instant. — The constructed trajectory of $m_2$ (blue): the image of $m_1$ ’s given curve (black) under the point reflection through $C$ with scale factor $m_1/m_2=2$ , so twice as large and on the opposite side of $C$ at every instant.

For every point $P$ that lies on $m_1$ ’s given trajectory, the corresponding point on $m_2$ ’s trajectory is obtained by the same map applied to $P$ ,

Q = \vect r_C + \frac{m_1}{m_2}\bigl(\vect r_C-P\bigr) = 3\vect r_C-2P,

so $m_2$ ’s whole path is the point-reflection of $m_1$ ’s path through $C$ , magnified by the factor $m_1/m_2=2$ . That is the blue curve in the figure.

Check

Applied to $m_1$ ’s own pictured position, the map returns exactly $m_2$ ’s pictured position (both are on the same line through $C$ , on opposite sides, with $C$ twice as far from $m_2$ ’s marked position as from $m_1$ ’s, consistent with $m_1=2m_2$ ), which is the one point where the construction can be checked directly against the given data. The mass-weighted sum also closes: $m_1\vect r_1+m_2\vect r_2=m_1\vect r_1+ (m_1/2)\bigl(3\vect r_C-2\vect r_1\bigr) = (3m_1/2)\vect r_C = (m_1+m_2)\vect r_C$ , matching (2).

Remark

Nothing about the shape of $m_1$ ’s path mattered, only that it is a path through points that are each paired, instant by instant, with a point on $m_2$ ’s path by the same fixed linear map. The same construction works for any closed two-body system with its center of inertia at rest, however complicated the interaction force: Kepler’s ellipse for one body and its partner’s ellipse under gravity are related by exactly this point reflection and scaling.