Since the binary separation and <mean motion> are both one, the <Kepler third law> gives
$$
\mu_1+\mu_2=G(M_1+M_2)=1.
$$
The <center of mass> condition puts the bodies at
$$
\mathbf r_{M_1}=(-\mu_2,0),
\qquad
\mathbf r_{M_2}=(\mu_1,0)
$$
in the <rotating reference frame>. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, and
$$
\boxed{OM_1=\mu_2,\qquad OM_2=\mu_1},
$$
$$
\boxed{M_1P=r_1=\sqrt{(x+\mu_2)^2+y^2}},
$$
$$
\boxed{M_2P=r_2=\sqrt{(x-\mu_1)^2+y^2}}.
$$
In the inertial sketch this entire configuration rotates uniformly about $O$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 3 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/3/i/solution