Deriving equations of motion for restricted 3-body problem

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In the restricted three-body problem, two large masses rotate around their barycentre while a smaller mass is influenced by their gravitational forces. The discussion highlights that the frame of reference where the barycentre is at rest is non-inertial, necessitating the inclusion of angular velocity in the equations of motion. The equation provided in the textbook lacks derivation, prompting a request for guidance on how to derive it. It is clarified that a valid inertial frame exists where the barycentre is at rest, but the focus should be on the rotating frame where the larger bodies remain stationary. To derive the equation, one should consider rotating basis vectors and apply Newton's second law to the entire vector, including these basis vectors.
tomwilliam2
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My textbook is describing a 3-body situation where there are two large masses rotating around their barycentre, and a third much smaller mass experiencing gravitational forces from the two larger masses. If the frame of reference is the one in which the barycentre is at rest, then it is not inertial. For the smaller mass, there is an angular speed ##\Omega## to take into account.
The gravitational force due to the two masses is clear, but when my textbook sets out the equation of motion for this mass, it just gives the result in this frame of reference as:
$$\mathbf{\ddot{r}} + 2(\mathbf{\Omega \times \cdot{r}})+(\mathbf{\Omega \cdot r})\mathbf{\Omega - \Omega^2 r}=-G\left(\frac{M_1}{r_1^3}\mathbf{r_1} + \frac{M_2}{r_2^3}\mathbf{r_2}\right) + \frac{1}{m}\mathbf{F_d}$$
Without any explanation. I'd like to try and derive it for myself. Any idea how to go about doing that?
Obviously I could start with:
$$\mathbf{\ddot{r}} + \mathbf{\alpha(\Omega)}=-G\left(\frac{M_1}{r_1^3}\mathbf{r_1} + \frac{M_2}{r_2^3}\mathbf{r_2}\right) + \frac{1}{m}\mathbf{F_d}$$
Where ##\alpha## represents the contribution to the acceleration caused by the rotating frame of reference, and ##F_d## is the disturbance forces.
 
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tomwilliam2 said:
If the frame of reference is the one in which the barycentre is at rest, then it is not inertial
This is not correct. There is a perfectly valid inertial frame where the barycenter is at rest. The point is that you want to go to the rotating frame in which the larger bodies are also at rest.

You can derive the result directly by considering a set of basis vectors that are rotating with the system. Note that the vector ##\ddot{\bf r}## as probably defined in your text only contains the time derivatives of the components, not the entire vector including the basis. Newton's second law holds for the entire vector - including the basis vector.
 
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