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Laws of Newton -
Gravitation of Two Objects

We recommend to refresh your knowledge of the gravitational field in three-dimensional space and the concept of field potential (scalar defined at each point of a field) using the material presented in this chapter in lectures
Field & Potential,
Work Lemmas,
Potential Theorem and
Problem 1.

The potential of the field at point P is an amount of work needed by a field force to move a test object of a unit mass from point P to infinity.

Also, the gradient of a potential equals to the field intensity - the force (vector) acting on a test object of unit mass at any point.

Its gradient, a vector of partial derivatives by coordinates, is
∇U(r)={∂U/∂x, ∂U/∂y, ∂U/∂z}
which should be equal to a vector of gravitational field intensity (the force per unit of test object's mass).

Let's determine a potential of the gravitational field produced by two point-masses Ma and Mb fixed at positions A and B correspondingly within some inertial frame of reference.

At any point P in space we can consider two positional vectors:
ra=AP originated at A (first source of gravitation) and ending at point P and
rb=BP originated at B (second source of gravitation) and ending at the same point P.

Assume, coordinates of all points involved are
A(xa,ya,za)
B(xb,yb,zb)
P(xp,yp,zp)
Then the coordinates of vectors ra and rb are
ra = (xp−xa, yp−ya, zp−za)
rb = (xp−xb, yp−yb, zp−zb)

According to a principle of superposition, the combined gravitational force of our two sources acting on a test object of mass m positioned at point P should be a vector sum of two separate forces, that is
F(P) = −(G·Ma·m)·(ra/ra³) − (G·Mb·m)·(rb/rb³)

The field intensity (the force per unit of mass of a test object) at point P, therefore, is
E(P) = −(G·Ma)·(ra/ra³) − (G·Mb)·(rb/rb³)
Notice, the above formula involves the addition of vectors.

From the fact that field potential is an amount of work needed to move a test object of a unit mass from its initial location P to infinity and that the work is, generally speaking, an additive quantity intuitively follows that potential of two fields should be equal to a sum of potentials.

The sum of potentials involves the addition of scalars, which is easier than addition of vectors of forces, so dealing with potentials is a preferable way.

Now consider a different scenario.

Assume, our two objects of mass Ma and Mb are not fixed at particular points A and B in space, but can move freely, and their corresponding position vectors are ra and rb.
Assume further that there are no external forces that can influence the movement of our system of two objects, and the only force that somehow affects their movement is the force of gravitation of one to another.
We will prove that in this case a center of mass of these two objects moves uniformly in space along a straight line with constant velocity.

The only force acting on the a-object with mass Ma is the gravitation force sourced at b-object with mass Mb.
This force is directed along a straight line connecting these two objects and its magnitude is
Fab(t) = G·Ma·Mb/r²(t).
where r=|ra − rb| is a time-dependent distance between these objects.

Similarly, the only force acting on the b-object with mass Mb is the gravitation force sourced at a-object with mass Ma.
This force is directed along a straight line connecting these two objects and its magnitude is
Fba(t) = G·Mb·Ma/r²(t).

So, the magnitudes of these forces, Fab(t) and Fba(t), are the same, but directions are opposite to each other, Fab(t) is directed from a-object to b-object, while Fba(t) is directed from b-object to a-object:
Fab(t) = −Fba(t)

Using the Newton's Second Law, knowing the forces and masses, we obtain equations for accelerations of these two objects, second derivatives of position vectors ra" and rb".
ra" · Ma = Fab
rb" · Mb = Fba
In the equations above and following it is assumed that all variables except masses are time-dependent functions. We just skip (t) for brevity.

If masses Ma and Mb are positioned at ra and rb, the center of mass is at position
r = (Ma·ra + Mb·rb)/(Ma+Mb)
The acceleration of this point is it's second derivative of position
r" = (Ma·ra" + Mb·rb)"/(Ma+Mb) =
= (Fab + Fba)/(Ma+Mb) = 0
because, as we stated above, Fab(t)=−Fba(t)

Since acceleration of the center of mass is zero, it velocity is a constant vector, which means that the center of mass in the absence of external forces moves along straight line with the same velocity.