Irodov 1.205: Only the Total Mass Matters – Amit & Deepali Sinha Foundation

Problem Irodov 1.205

Two stars, of individual masses that are not separately specified but sum to a total mass $M$ , orbit their common center of mass under their mutual gravitational attraction alone, each completing one revolution in the same period $T$ . Find the distance $l$ separating the two stars.

The two stars m_1 and m_2 on opposite sides of the center of mass C, each on its own dashed circle; l is their separation, and v_1, v_2 are their velocities, tangent to the circles and oppositely directed. — The two stars $m_1$ and $m_2$ on opposite sides of the center of mass $C$ , each on its own dashed circle; $l$ is their separation, and $\vect v_1$ , $\vect v_2$ are their velocities, tangent to the circles and oppositely directed.

Given

Find

l

Solution

Step 1Locate the common center of mass

Gravity is the only force on either star and it acts along the line joining them, so both stars share the same orbital period and stay diametrically opposite each other about the center of mass $C$ at every instant; this much follows from the force law alone. The problem asks for “the distance” between the stars as a single number, which exists only if that separation is constant, so we take the orbits to be circular, both centered on $C$ . Writing $r_1$ , $r_2$ for each star’s distance from $C$ and $l=r_1+r_2$ for their (now constant) separation, the center of mass condition fixes the split.

\begin{align*}m_1 r_1 &= m_2 r_2 \\ r_1+r_2 &= l\end{align*}

Eliminating $r_2$ between the two and collecting the terms in $r_1$ ,

\begin{align*}m_1 r_1 &= m_2(l-r_1) \\ (m_1+m_2)\,r_1 &= m_2 l\end{align*}

so that, with $M=m_1+m_2$ the total mass,

r_1 = \dfrac{m_2}{M}\,l

(1)

Step 2Newton’s second law for star 1

m_1\omega^2 r_1 = \frac{\gamma m_1 m_2}{l^2}

Step 3Eliminate the individual masses

Dividing by $m_1$ , substituting (1) for $r_1$ , and dividing by $m_2$ removes both individual masses in turn, leaving a relation in $l$ and the total mass $M$ alone.

\begin{align*}\omega^2 r_1 &= \frac{\gamma m_2}{l^2} \\ \omega^2\left(\frac{m_2}{M}l\right) &= \frac{\gamma m_2}{l^2} \\ \omega^2\frac{l}{M} &= \frac{\gamma}{l^2}\end{align*}

\omega^2 l^3 = \gamma M

(2)

Step 4Solve for the separation

With $\omega=2\pi/T$ , (2) rearranges directly for $l$ .

l^3 = \frac{\gamma M}{\omega^2} = \gamma M\left(\frac{T}{2\pi}\right)^2

l = \sqrt[3]{\gamma M\left(\dfrac{T}{2\pi}\right)^2}

(3)

Check

Dimensionally $[\gamma M][T^2] = (\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg})(\mathrm{s^2}) = \mathrm{m^3}$ , and its cube root is a length, as required. For equal masses $m_1=m_2=M/2$ , (1) gives $r_1=r_2=l/2$ : each star simply orbits the midpoint of the segment joining them, the familiar equal-mass double star.

Remark

The individual masses $m_1$ , $m_2$ never appear separately in (2), only their sum: two double stars with the same total mass $M$ and period $T$ have exactly the same separation $l$ , however unevenly that mass is split between the pair. This is the same content as Kepler’s third law for a planet, generalized to a total mass shared between two bodies rather than concentrated in one.