Physics+ Newton Laws, Gravity Two Objects: UNIZOR.COM - Physics+ - Laws ...

Notes to a video lecture on UNIZOR.COM

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 Problem 1 in this chapter of this course presented a proof that gravitation field of a point-mass M located at some point Q in three-dimensional space is conservative.

The work performed by this field when it moves some test object of mass m from point A to point B depends only on the location of these endpoints of its trajectory.

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.

Earlier in this course the gravitational force F(P) at some point P of the field was presented as
F(P) = F(r) = −G·M·m·r/r³
where r=QP is the relative position vector from the source of gravity Q to point P,
r is the magnitude (scalar) of a relative position vector r,
M is a mass of an object that is the source of gravitation,
m is a mass of a test object and
G is the Universal Gravitational Constant.

Of course, r²=x²+y²+z²
where (x,y,z) are Cartesian coordinates of vector QP (point Q is the source of gravity).

In coordinate form, components of the gravitational force vector F(P)=F(r) are:
F_x(P)=−G·M·m·x/(x²+y²+z²)^3/2
F_y(P)=−G·M·m·y/(x²+y²+z²)^3/2
F_z(P)=−G·M·m·z/(x²+y²+z²)^3/2

The potential of a gravitational field was determined as a scalar function defined at any point P of the field as
U(P) = U(r) = G·M/r =
= G·M/(x²+y²+z²)^1/2 =
= G·M·(x²+y²+z²)^−1/2

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).

Indeed,
∂U(P)/∂x =
= −½G·M·(x²+y²+z²)^−3/2·2x =
= −G·M·x/(x²+y²+z²)^3/2 =
= (1/m)·F_x(P)

Similarly, partial derivatives by y and z produce the corresponding components of the gravitational field intensity - the force acting on a unit of mass of a test object.
∂U(P)/∂y =
= −G·M·y/(x²+y²+z²)^3/2 =
= (1/m)·F_y(P)
∂U(P)/∂z =
= −G·M·z/(x²+y²+z²)^3/2 =
= (1/m)·F_z(P)

We came to conclusion that
∇U(P)={∂U/∂x,∂U/∂y,∂U/∂z}=
= (1/m)·(F_x(P),F_y(P),F_z(P)) =
= E(P)
where E(P) is a vector of field intensity (the force per unit of test object's mass) at point P.

Let's determine a potential of the gravitational field produced by two point-masses M_a and M_b 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:
r_a=AP originated at A (first source of gravitation) and ending at point P and
r_b=BP originated at B (second source of gravitation) and ending at the same point P.

Assume, coordinates of all points involved are
A(x_a,y_a,z_a)
B(x_b,y_b,z_b)
P(x_p,y_p,z_p)
Then the coordinates of vectors r_a and r_b are
r_a = (x_p−x_a, y_p−y_a, z_p−z_a)
r_b = (x_p−x_b, y_p−y_b, z_p−z_b)

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·M_a·m)·(r_a/r_a³) −
− (G·M_b·m)·(r_b/r_b³)

The field intensity (the force per unit of mass of a test object) at point P, therefore, is
E(P) = −(G·M_a)·(r_a/r_a³) −
− (G·M_b)·(r_b/r_b³)
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.

More rigorously, work is an integral of a scalar product of force by differential of distance
W = ∫[P,∞](F·dl)

If F=F_a+F_b, the work of a combined force is the sum of work of its components
W = ∫[P,∞]((F_a+F_b)·dl) =
= ∫_[P,∞](F_a·dl) + ∫_[P,∞](F_b·dl) =
= W_a + W_b
which proves additive characters of work and, consequently, of field potential in case a field is generated by more than one source.

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 M_a and M_b are not fixed at particular points A and B in space, but can move freely, and their corresponding position vectors are r_a and r_b.
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 M_a is the gravitation force sourced at b-object with mass M_b.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ab(t) = G·M_a·M_b/d²(t).
where d=|r_a − r_b| is a time-dependent distance between these objects.

In a more appropriate vector form it looks like
F_ab(t) =
= G·M_a·M_b·[r_a(t)−r_b(t)]/d³(t).

Similarly, the only force acting on the b-object with mass M_b is the gravitation force sourced at a-object with mass M_a.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ba(t) = G·M_b·M_a/d²(t).

In a more appropriate vector form it looks like
F_ba(t) =
= G·M_a·M_b·[r_b(t)−r_a(t)]/d³(t).

So, the magnitudes of these forces, F_ab(t) and F_ba(t), are the same, but directions are opposite to each other, F_ab(t) is directed from a-object to b-object, while F_ba(t) is directed from b-object to a-object:
F_ab(t) = −F_ba(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 r_a" and r_b".
r_a" · M_a = F_ab
r_b" · M_b = F_ba
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 M_a and M_b are positioned at r_a and r_b, the center of mass is at position
r = (M_a·r_a + M_b·r_b)/(M_a+M_b)
The acceleration of this point is it's second derivative of position
r" = (M_a·r_a" + M_b·r_b)"/(M_a+M_b) =
= (F_ab + F_ba)/(M_a+M_b) = 0
because, as we stated above, F_ab(t)=−F_ba(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.