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Suppose that I have a system of two particles. They have the same mass $m$. At the origin O is something with a very large mass $M$ which is attracting the two particles. The first particle has distance 1 to O. The second particle has distance 2 to O. Now, what is the total gravitational force on the system?

If I first calculate the two forces separately, I see that the force on particle 1 is $\frac{C}{1^2}$, where $C=GMm$ is a constant. The force on particle 2 is $\frac{C}{2^2}$. Therefore, the total force is $\frac54 C$.

However, we could also calculate the force on the entire system using the center of mass. The center of mass is at distance $\frac32$ from $O$. The total mass of the system is $2m$. Therefore, the total force is $\frac{2C}{(\frac32)^2}=\frac89 C$.

So is the total force equal to $\frac54 C$ or $\frac89 C$?

Update: the origin and the two particles are on a straight line. Also, to view the two particles as one ‘system’, we can say that the two particles are attached to the ends of a massless rope with a length larger than 1. Finally, the force between the particles is negligible.

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    $\begingroup$ "total gravitational force on the system" What does that even mean? You should calculate the gravitational forces between each pair of objects. $\endgroup$
    – PM 2Ring
    2 days ago
  • $\begingroup$ What should I clarify for the question to reopen? $\endgroup$
    – Riemann
    16 hours ago

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When calculating the force due to gravity, you can replace a system of objects with a single object at the system's centre of mass in a uniform gravitational field. However, in this case the gravitational field is not uniform. In other words, the centre of gravity of the system of two particles does not coincide with their centre of mass.

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  • $\begingroup$ The center of gravity of the system would be $\left(x_1 m_1 g(x_1) + x_2 m_2 g(x_2)\right)/\left(m_1 g(x_1) + m_2 g(x_2)\right) = \left(1 \frac{1}{1^2} + 2 \frac{1}{2^2}\right)/\left(\frac{1}{1^2} + \frac{1}{2^2}\right) = \frac{6}{5}$. And the force of gravity on an object of mass $2m$ located there would be $\frac{25}{18}C$. But I'm not sure what that tells you. I think the only answer is just the net (external) gravitational force on the system of smaller masses: $\frac{5}{4}C$. $\endgroup$
    – Ben H
    2 days ago
  • $\begingroup$ @Ben H, I think that your formula for the center of gravity is incorrect, because it should give $\frac54C$ for the total force $\endgroup$
    – Riemann
    2 days ago
  • $\begingroup$ @Riemann I'm happy to adjust or delete. Can you explain what you mean? In the case of three masses along a line, I was applying the equation for "parallel fields" found at the wikipedia page for center of gravity in nonuniform fields, i.e., the weighted sum of weights. $\endgroup$
    – Ben H
    2 days ago
  • $\begingroup$ @BenH Since the distance between the objects has the same order of magnitude as their distance from the primary, the gravitational field between them is not even approximately parallel. $\endgroup$
    – gandalf61
    2 days ago
  • $\begingroup$ @gandalf61 The gravitational field vectors of M along a line through M are certainly parallel, no? I assumed all masses were along a line (otherwise Riemann's statement of center of mass distance being 3/2 would not have made sense). $\endgroup$
    – Ben H
    2 days ago

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