= Solution
For the attractive <inverse-square force>, choose zero <potential energy> at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the <kinetic energy> is purely tangential, with conserved <angular momentum> $J$. Thus
$$
E=\frac{J^2}{2mr^2}-\frac Ar,\qquad
Er^2+Ar-\frac{J^2}{2m}=0.
$$
Its two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the <ellipse> has $r_p+r_a=2a$. Therefore \b[$a=-A/(2E)$], with $E<0$ for a nondegenerate bound <ellipse>. The circular case follows by the double-root limit.
For the <two-body problem>, define $\mathbf r=\mathbf r_B-\mathbf r_A$, total <mass> $M=m_A+m_B$, <reduced mass> $\mu=m_Am_B/M$, and <center of mass> $\mathbf R$. The relative equation is $\mu\ddot{\mathbf r}=\mathbf F$, while the isolated <center of mass> has $\ddot{\mathbf R}=0$. Once the relative <orbit> is known,
$$
\mathbf r_A=\mathbf R-\frac{m_B}{M}\mathbf r,\qquad
\mathbf r_B=\mathbf R+\frac{m_A}{M}\mathbf r.
$$
The relative <energy> uses $\mu$ in its <kinetic energy> and $A=Gm_Am_B$ in the gravitational <potential energy>.
In the equal-mass case the initial relative <speed> is $u$, $\mu=m/2$, and the center-of-mass <velocity> is half B's initial <velocity>. Thus
$$
E_{\rm rel}=\frac14mu^2-\frac{Gm^2}{a_1}.
$$
Negative relative <energy> requires $u^2<4Gm/a_1$. The \b[escape threshold is $u_E=2\sqrt{Gm/a_1}$]; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.
For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\rm rel}=-3Gm^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse <speed> is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. Hence
$$
\boxed{r_p=2a-a_1=\frac{a_1}{3},\qquad e=\frac12.}
$$
Choose the initial separation along $+x$ from A to B, with B's initial <velocity> $+u\widehat{\mathbf y}$. <Conservation of angular momentum> gives relative periapsis <speed> $3u$. At the opposite end of the <ellipse> the direction is $-\widehat{\mathbf y}$, so
$$
\mathbf V_{\rm CM}=\frac u2\widehat{\mathbf y},\qquad
\mathbf v_{B,p}=\mathbf V_{\rm CM}+\frac12\mathbf v_{{\rm rel},p}
=\boxed{-u\widehat{\mathbf y}}.
$$
Thus B has the same <speed> as initially but the opposite <velocity>. A has $2u\widehat{\mathbf y}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame <speed>, not the requested initial laboratory frame result.
The relative <orbital period> follows from <Kepler's third law> with total gravitating <mass> $2m$:
$$
\boxed{T=2\pi\sqrt{\frac{(2a_1/3)^3}{2Gm}}
=\frac{4\pi}{3\sqrt3}\sqrt{\frac{a_1^3}{Gm}}
=\frac{4\pi a_1}{3\sqrt3\,u}.}
$$
In the center-of-mass frame each star follows an <ellipse> of semimajor axis $a_1/3$ and <eccentricity> $1/2$, with the <center of mass> at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The <laboratory paths of an equal-mass binary> are therefore not closed <ellipses>, even though their relative motion repeats.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-1-binary-orbits.png]
{title=Equal-mass binary trajectories in the center-of-mass frame and the original laboratory frame}
Back to article page
