[Bored chemist]

To get a more general relationship you would need to have the masses equal, as has been mentioned before.

And why so?

Obviously if you wanted to compare the forces since the charges are 1 to 1 you would want the masses to be 1 to 1 for a valid comparison.  Doesn't that make sense to you?

It's still arbitrary.
The ratio of the gravitational to electrostatic forces is not the same for a couple of protons as it is for a couple of electrons (By nearly 4 million fold).
So all you can really say is that the electrostatic force is enormously much bigger.
(or, of course, zero if you use uncharged particles).

Since the ratio is a moveable feast, it can't  have any numerical significance.
So the OP is still doing numerology.

[Origin]

The ratio of the gravitational to electrostatic forces is not the same for a couple of protons as it is for a couple of electrons (By nearly 4 million fold).

Dooh!  You are right of course.

[PaulTalbot (OP)]

How can it be out of your comfort zone if it is something you put in your own equation?  It sounds like you are saying you combined a bunch of constants and then put a factor of 2 in the equation to get the answer you want.  Assuming that is what you did, that is precisely why I am calling your technique numerology. 

My comment was about the physical meaning of the factor of 2 (or ½), not where it comes from.
As mentioned in a previous post and in the text of the preview, this factor of 2 results from the inclusion of the dimensionless constant α in the equation:

r_p = e² / ε₀ E_ep
r_p = e² / ε₀ (β H h)
r_p = (e² / ε₀ h) / βH
since α = e² / 2 ε₀ hc, we get
r_p = 2 αc / βH

This is pure algebra, not numerology.

[Origin]

rp = e2 / ε0 Eep
rp = e2 / ε0 (β H h)

This is pure algebra, not numerology.

In the above you are saying:


Where

This is pure numerology.  The Hubble constant is not a frequency.  This has been pointed out to you several times.  Even if it was a frequency it would still be meaningless.  If I swing a ball on a string around at a frequency of 5 cycles per second, do you think multiplying that by 'h' will give me anything meaningful?

You have presented an idea and it has been shown to be wrong, not accepting this means you are not doing science you are doing pseudoscience and are entering the land of crankdom.  That's too bad, but it happens all the time.

[PaulTalbot (OP)]

This section of the forum is for new theories and intended to be on the lighter side. I appreciate the opportunity to debate new ideas and, in my opinion, some comments and questions were useful in my approach. So, I intend to continue answering questions and reply to constructive criticism.

As for your last post, sorry Origin, but all you have proven so far is that you are close-minded and that you don't understand the arguments provided by Edington, Dirac, Wesson, Valev and others.

[Bored chemist]

The Hubble constant is not a frequency.

It's not a constant either.

If I swing a ball on a string around at a frequency of 5 cycles per second, do you think multiplying that by 'h' will give me anything meaningful?

Well... if it's the resonant frequency of the pendulum then multiplying 5 by h/2 gives you the zero point energy of the system.

But the universe isn't an oscillation; so the point's moot.
The Hubble constant currently has a value near 1/(14 billion years).
But tomorrow it will be slightly different because the universe isn't going to turn round and go back.

If I live for 100 years than the reciprocal of my lifetime will be once per century.
But it's not meaningful to say that I live once per century because that would imply that I live 10 times per millennium and I didn't and  won't.
I will live once in the whole of infinite time, so my frequency is 1 in essentially infinity.
If you want to put a number on that it should be zero.

You can do the same thing with the universe.
It's either a one off, or infinitely rare. It has no "frequency" to multiply by h.
And that's the root of the OP's problem.