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

Since the binary separation and mean motion are both one, the Kepler third law givesThe center of mass condition puts the bodies atin the rotating reference frame. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, andIn the inertial sketch this entire configuration rotates uniformly about $O$.

At a Lagrange point, the test particle is stationary in the rotating reference frame, so $F=G=0$. On the $x$ axis, $y=0$ makes $G=0$ automatically. The singularities at the two masses split the axis into three intervals, and the balance between gravity and centrifugal acceleration gives one root of $F=0$ in each interval. These are the three Collinear Lagrange points $L_1,L_2,L_3$.

For an off-axis equilibrium, $y\ne 0$. Put $A_i=1-r_i^{-3}$. ThenwhileHence $A_1=A_2=0$ and therefore $r_1=r_2=1$. The two intersections of unit circles centered at $M_1$ and $M_2$ form equilateral triangles with the binary, giving the Triangular Lagrange pointsTogether with the three collinear points, these are the five equilibria of the circular restricted three-body problem.

At $L_3$, the ordering is $L_3,M_1,O,M_2$. Since $r_1=1+\beta$,The equation $F=0$ becomesWriting $ϵ={\mu }_2/{\mu }_1\ll 1$ and retaining first order in $\beta$ and $ϵ$ givesConsequentlyThe negative sign says that $L_3$ is slightly less than one binary separation from the dominant mass $M_1$.

The required Taylor series giveSeek $\beta =aϵ+b{ϵ}^2+O({ϵ}^3)$. Substitution in the exact equation from part (iii) gives successivelyThus $a=-7/12$ and $b=7/12$, so

Let $X=x-x_{L_3}$ and $Y=y$ be the displacement from the L3 Lagrange point, and put $U=\dot{X}$, $W=\dot{Y}$. The first-order linearization of a dynamical system iswhere every derivative is evaluated at $L_3$. For the state vector $\mathbf{\xi }=(X,Y,U,W{)}^{\mathsf{T}}$, this becomesThe question's reuse of $\mathbf{X}$ for the state vector is only notation; its first two entries are the small displacements $X,Y$, not the absolute coordinates.

Write $d_1=x+{\mu }_2$ and $d_2=x-{\mu }_1$, so $r_i^2=d_i^2+y^2$. Sincedirect differentiation givesEvery Collinear Lagrange point has $y=0$, and in particular