At a <Triangular Lagrange point>, $y\ne0$, so $A=1$. The $x$ equation then gives
$$
\frac{1-\beta_1}{r_1^3}=\frac1{r_2^3}.
$$
Substitution into $A=1$ and $\mu_1+\mu_2=1$ yields
$$
r_2=1,
\qquad
r_1=(1-\beta_1)^{1/3}\equiv s.
$$
Intersecting these two circles gives
$$
\boxed{x=\frac{s^2}{2}-\mu_2,
\qquad
y=\pm s\sqrt{1-\frac{s^2}{4}},
\qquad z=0}.
$$
As $\beta_1$ rises from zero to one, $s$ falls from one to zero. The two points move along the unit circle about $M_2$, from the classical equilateral positions toward $M_1$, where they coalesce when the attraction of $M_1$ is fully cancelled.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 316 / 2 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-316/2/v/solution