Pedestrian Physics 10

Newton 2
Universal Gravitation

Newton’s dynamics has adapted to Einstein’s new definition of mass. His system of universal gravitation has not. Force is the gradient of an energy field, and the field must have a mass. Due to the singularity at the core of the definition we cannot assign a definite value to this mass.

What we can do, however, is to arbitrarily assign a minimum value.

Consider two bodies, each with a mass “m” and a diameter ‘r’ cm. When they abut, we stipulate the gravitational field to be zero. When they are separated by an infinite distance, the field will have a value

e(1) = Gm²/r

Now introduce “n” identical bodies. The total body mass is n*m, with the energy equivalent

e(2) = nmc² .

The number of permutations, taken two at a time, is n²/2. The total value of the energy fields is then

e(3) ≈ n²m² G/r

The ratio
e(2)/e(3) ≈ c²/(mnG/r).

Now plug in the value of the universal constant, G; the number of stars in the universe (8*10²²) and the mass of the sun (6*10²⁷ g):

e(2)/e(3) ≈ r*10²⁹/10⁴⁹ ≈r/10²⁰

The calculation is certainly way off the mark but no matter how one tweaks the numbers, even if one substitutes the Solar diameter (r = 10¹¹ cm) as the minimum, the stellar mass is dwarfed by the gravitational mass in the interstellar fields. This contrasts with the estimate that the stellar-to-interstellar ratio is about 1/25 (or 1/200, depending on what paper you read).

Discussion

There is no gravitational energy field. The concept must go the way of the luminous aether.

I do not pretend to have the answer or even an answer. What follows is an exercise. It seems inevitable that a solution must be found outside the three-dimensional box. One model may be based on the four-dimensional equivalent to the circle and the sphere.