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The question isn' clear at all. However you can consider the two masses a part of a inertial system . Then , Physics, with some rule, tell us we can see them as a lonely mass (mass sum) that in a case of dynamic action, have a behaviour run according its centrer of mass.
Determinated a system of riferiment, let be `M_1` and `M_2` the two mass. As far as I could understand the mass laying on the same plane, then called `G_1` and `G_2` their relative centers of gravity, and let it be `O` the original point. According rules:
`(M_1 xx OG_1 +M_2 xx OG_2)/(M_1+M_2)= OG` where `G` stands for the gravity center of the system.
Then you can think system as a lonely gravitiy and study its motion as all the mass is tookl all around `G` .
Of course the sysyem has to be hard, (its clear that if masses moved according force effetcs, changing their relative positions `G ` changes too and no more system has to be commpared to an lonely mass)
If you consider a point of riferiment `O` between `M_1` and `M_2` at 5mts from the two mass, for the rule above we have:
So `G ` is exactly in the midle.
Then you can consider the two masses of 10 Kgs as one of 20 Kg in the midest of, far `+- 25` meters away .
This is the graph of the value of `OG` left the mass `M_1=` `10 Kgs` e changing `M_2` . Note , for a mass `M_2=10 Kgs` the center `G ` is in the middle ( `y=0` ), as we have calculated.
Newton's law of universal gravitation states that
Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them .
We can write this as
`F = G ((m_1m_2)/r^2)`
where `G` is the universal gravitational constant, equal to
6.674 x 10^(-11) Nm^2 kg^(-2) (Cavendish, 1798).
Therefore, if we have two particles mass 10kg, 50cm apart, the gravitational force between them is
`F = G(((10)(10))/50^2) = G(100/2500) = G/25` Newtons
`= ` 2.670 x 10^(-12) N
Since `F = ma` (Newton's 2nd law of motion) each mass is accelerating
toward the other at the same rate equal to
` ``a = F/m_1 = ` 2.670 x 10^(-13) ms^(-2)
This is of course very small relative to the distance between them :-
5 x 10^(-1) m
The gravitational force between them is F = 2.670 x 10^(-12) N.
Their centre of mass, considering them as a unit, is at the midpoint
 Page 297 in H W Turnbull (ed.), Correspondence of Isaac Newton, Vol 2 (1676-1687), (Cambridge University Press, 1960), document #235, 24 November 1679.
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