Gravity as a Force related to the Energy Density of the Vacuum and the Critical Mass Density

We know that the force of gravity is proportional to the mass of the object in question, and inversely proportional to the square of the distance between the center of the mass-carrying objects, but the problem with gravitational force is that, in absence of a second mass carrying body, the force does not exist except as a potential force. An analysis of G needs to take into account the context within which the gravitational constant is used, so we will first look at the Newtonian equation used to calculate the gravitational force between two mass carrying objects:

F = GM₁M₂ / r²

My first observation here is that the gravitational constant is only used once; it does not need to be used a second time in conjunction with the second mass (M₂). This would seem to imply that the gravitational constant, G, must be a proportional constant that describes the properties of both gravitational fields (of both masses) simultaneously, since when we add the second mass into the equation the constant does not change or need adjusting. What I would like to suggest here is that we substitute G and replace it with a constant that purports only to explain the properties of one gravitational field.

Given that the units of G are Nm² kg², it may be better to represent the equation by stating that F = (√G)² M₁M₂ / r² . This gives a value of -8.168^-06 N(1/2) m kg for the square root of G (√G). This also allows us to split the equation into two parts and allows us to show exactly how the gravitational constant should be related to each mass. For the ease of notation, we shall say that AF = √G (thus AF = -8.168^-06 N(1/2) m kg), where AF is the constant of proportionality of the force of the vacuum (F_V) exerted upon any mass, and can be accurately represented by stating that the force of the vacuum (F_V) is proportional to the mass of the object, and inversely proportional to the distance from the center of the mass:

F_V = A_FM / r

To show how this forms the basis of Newton's equation for the force of gravity, we take the F_V for two masses, and represent them as:

F = (A_FM₁ / r) x (A_FM₂ / r)

or

F = (A_FM₁) x (A_FM₂)
r²

If we substitute the values into the above equation and use two familiar masses for M₁ and M₂, then we should get exactly the same result as when using G. We shall say M₁ is the Earth, and M₂ is a 1kg object at the earth's surface. Thus:

(A_FM₁) = (8.168^-06 N(1/2) m kg x 5.98²⁴ kg) = 4.884464¹⁹ N(1/2) m

(A^FM²) = (8.168^-06 N(1/2) m kg x 1kg) = 8.168^-06 N(1/2) m

The force of gravity is essentially the product of the force exerted by the vacuum by each of the two masses. In this respect, the force exerted upon one mass by the vacuum is multiplied by the force exerted upon the second mass. This results in an attractive force (gravity) between the two masses, which is strong enough to give rise to motion where one of the masses is substantially large enough.

Therefore, the force of gravity between the two masses must be:

F = (4.884464¹⁹ N(1/2) m) x (8.168^-06 N(1/2) m kg)
r²

or

F = (4.884464¹⁹ N(1/2) m / r) x (8.168^-06 N(1/2) m / r)

The distance, r, between the center of the earth and the 1kg mass (essentially the radius of the earth) is 6376500m, so if we input this final information we should get:

Certainly using the value A_F demonstrates why the dimensions of the gravitational constant G are Nm² kg², but its primary advantage is that it shows us from a classical mechanics point of view how gravity is propagated, and indeed what it is; gravity is a resultant force (F_R) of the product of the two partial forces exerted on two mass carrying bodies by the vacuum, thus: