GRAVITY – the Maths

A ball is dropped from the surface of the earth down a shaft

r and R represent the radius of the earth
a and A represent the distance the ball is from its surface
b and B represent the distance the ball is from its centre

The lower case is the actual distance and
The higher case is the apparent distance to an observer

So a + b = r and A + B = R where R is a constant    ............................(1)

Assume matter and the above mentioned observer are expanding in a geometric progression by a constant factor K ( = 1 + k ) every time unit t.

So r = R .K ^t , a = A .K ^t , and b = B.K ^t        .......................................(2)

Applying simple Newtonian kinetics, we can now show that this expansion of matter produces a constant apparent acceleration near the surface of the earth, proportional to its radius.
As ‘time’ is just the state of expansion of matter, it is unlike a physical dimension and we only consider one type of time - apparent time or better-named - real time.

Initially, when the ball is dropped:

    t = 0 and a = A = 0 and b = B = r = R       .........................................(3)

Also at t = 0, as the ball is dropped, it has the same velocity away from the centre of the earth as the surface has and continues at this speed, if unhindered.

Thus    db/dt = dr/dt   at  t = 0.

From 2.   r = R . K^t     then    dr/dt = R K^t . log K
    and at t = 0   dr/dt = R . log K = db/dt (a constant) for all t
    and therefore b = R . logk . t + c  where c is a constant.......................(4)

When t = 0   b = c and also b = R from 3
therefore c = R and, substituting in 4,    b = R.logK . t + R = R(1 + logK . t)

From 1,2 and above.     a = r - b = R.K^t - R(1 + logK . t)
Also from 2                  A = a/K^t = R - R(1+ logK . t)/K^t
                                         = R{1 - (1+ logK . t)/K^t}

K = 1+ k   where k is a small constant

So logK can be represented by the series:  k - k²/2 + k³/3  - .................
and 1/K^t  by the series...........................:  1 - t.k + t(t+1)k²/2 - t(t+1)(t+2)k³ + ........

As k is small and t is not a very large value, when the ball is still close to the surface, ignore k to the power of 3 and above.

Then:  A = R.[1 - (1+ k.t - k²t/2)(1 - k.t + k².t/2 + k².t²/2)]
             = R.[1 - (1 - k.t + k².t/2 + k².t²/2 + k.t - k².t² - k².t/2)] ......ignoring k³ and above
             = R.k².t²/2 .........after cancelling like terms.

Then dA/dt = R.k².t and the acceleration of the ball towards the centre of the earth
is   d²A/dt² = R.k²  a constant.

So, given a geometric expansion of matter, an object released near the surface of a planetary body, appears to move towards it's centre with an acceleration proportional to its distance from the centre. If this was the only component of gravity, it obviously wouldn't agree with scientific observation (the moon's radius is a quarter of that of the earth, yet it's surface gravity is only a sixth of the earth's, also gravity doesn't keep increasing with distance from the earth or a planet). The other component of gravity, as suggested in "the Four Forces", is the interaction of the flux envelopes of the bodies involved. When the bodies are far apart they tend to repel each other and sometimes this is enough as to counteract the effect of the "expansion" factor. But if this doesn't happen, and they get "apparently" closer together, their flux-envelopes start to lose their identities and merge together. In so doing, they repel each other less. This doesn't necessarily mean they impact. In their merged identity, they could revolve around each other, either stably or for a short time and then fly apart - it all depends on their relative masses and original trajectories.