Define
$$
\boxed{A=\frac{\mu_1(1-\beta_1)}{r_1^3}+\frac{\mu_2}{r_2^3}}.
$$
At an <equilibrium point>, the velocity and acceleration vanish. Direct differentiation gives
$$
U_z=-Az,
\qquad
U_y=(1-A)y,
$$
and
$$
U_x=(1-A)x-\mu_1\mu_2\left[(1-\beta_1)r_1^{-3}-r_2^{-3}\right].
$$
Thus the equilibrium conditions are
$$
\boxed{Az=0,
\qquad
(1-A)y=0,
\qquad
(1-A)x=\mu_1\mu_2\left[(1-\beta_1)r_1^{-3}-r_2^{-3}\right]}.
$$

Solution

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

 Solution

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