Vector addition and Newton's law

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In a static case involving multiple bodies, the net gravitational force on a reference mass can be calculated by vectorially adding the gravitational forces from all other bodies. This requires knowing the positions of the bodies relative to the reference mass, as gravitational forces depend on the difference vectors. While calculating the resultant force is straightforward, the system becomes complex when considering the equilibrium of all bodies, leading to simultaneous non-linear equations. The discussion highlights that even in static scenarios, understanding the dynamics can be intricate, as exemplified by the three-body problem. Thus, while the initial force calculation is simple, the overall analysis can become significantly complicated.
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I know that n-body problem can be complicated, but that's for the dynamics. What about a static case:

e.g. if I have the distances of several bodies A, B and C etc. and their distance to a reference mass m, can I just use the vector addition of the Newton's gravitational force to add up all of the forces from all those bodies to the reference mass and obtain the resulting vector? or is it more complicated than that?
 
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If you want the net force on the reference body with mass m, just add all the gravitational forces vectorially. Yes, it is that simple. You need the position of all other bodies with respect to the reference body, as the forces of gravity depend on the difference vectors ##\vec r_i-\vec r _0##.
The system is static if the net force at each body is zero (not only at the reference body).
 
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xaratustra said:
? or is it more complicated than that?
It doesn't need to be harder to comprehend in principle but you end up with what's effectively a set of simultaneous non linear equations and it gets very complicated if you want to work out the paths of those bodies. Working out the resultant force on each body is only the start. Look at the wiki article on the three body problem (and that's only three.)
 
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