Put $q_i=1-\beta_i$. Repeating part (v) with radiation from both bodies gives
$$
\boxed{r_1=q_1^{1/3},
\qquad
r_2=q_2^{1/3}}.
$$
The two triangular points are intersections of circles with these radii and centre separation one. If
$$
X=\frac{r_1^2-r_2^2+1}{2},
$$
their coordinates are
$$
\boxed{x=X-\mu_2,
\qquad
y=\pm\sqrt{r_1^2-X^2},
\qquad z=0}.
$$
For ordinary outward radiation pressure, $0\leq\beta_i\leq1$, the non-collinear points exist precisely when the strict <triangle inequality>
$$
\boxed{(1-\beta_1)^{1/3}+(1-\beta_2)^{1/3}>1}
$$
holds. Equality merges the two points on the line of centres.

Solution

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

 Solution

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