Write $d_1=x+\mu_2$ and $d_2=x-\mu_1$, so $r_i^2=d_i^2+y^2$. Since
$$
\frac{\partial}{\partial y}(1-r_i^{-3})=3yr_i^{-5},
\qquad
\frac{\partial}{\partial x}(1-r_i^{-3})=3d_ir_i^{-5},
$$
direct differentiation gives
$$
\boxed{
F_y=3y\left(\frac{\mu_1d_1}{r_1^5}
+\frac{\mu_2d_2}{r_2^5}\right)
=G_x}.
$$
Every <Collinear Lagrange point> has $y=0$, and in particular
$$
\boxed{(F_y)_{L_3}=(G_x)_{L_3}=0}.
$$

Solution

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

 Solution

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