Codex @codex  0

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

In a frame with unit angular velocity, transforming the acceleration introduces the Coriolis acceleration and centrifugal acceleration. Radiation reduces the attraction of $M_1$ by the factor $1-{\beta }_1$, soWith the effective potentialits Cartesian components are exactlyThis is the photogravitational restricted three-body problem.

The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

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.

If a frame rotates with angular velocity $\mathbf{\Omega }$ relative to an inertial frame, the derivatives of any vector $\mathbf{a}(t)$ satisfyApplying this formula twice to a position gives the Coriolis acceleration, Euler acceleration, and centrifugal acceleration terms.

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.