solo

Actually, there is something new here. Examine the following formula: [math]re=\frac{\left(\sqrt[4]{\frac{6}{\pi}}\right)\left({\sqrt[3]10}\right)^{25}}{c^2}[/math]............in cgs units and; [math]re=\frac{\left(\sqrt[4]{\frac{6}{\pi}}\right)\left({\sqrt[3]10}\right)^{7}}{c^2}[/math].............in SI units Notice the involvement of [math]\left(\frac{6}{\pi}\right)[/math] within these formulae.

Welcome mmonaco27, looking forward to hearing more from you friend. Actually momonaco27, the inverse of Planck length would be (1/1.616E-35) and that would be: 6.188E+34, not 1.616E+34 as you formerly stated. Never-the-less, an interesting thought anyway. NO WAY!!, never happen around here my friend..................solo

That formula would be: [math]\left(r=\frac{2GM}{c^2}\right)[/math]

Absolutely Cap'n........., simplified for the sake of convenience.....................solo

The long standing formula for the volume of a sphere is: [math]\left(r^3\right)*\left(\frac{4\pi}{3}\right)[/math] = volume of sphere I have calculated a new and much simpler formula: [math]\left(\frac{dia^3 \pi}{6}\right)[/math] = volume of sphere Anyone have something to add?..............solo

Solving for G: G = 6.67288271516 E-8 cgs.

My own thoughts: [math]\left[\left(\frac{\hbar c}{e^2}\right)^3 \left(\frac{\hbar c}{G^3}\right)^3 \left(\frac{\hbar re}{G^3}\right)^{1.11111....}\right] =\left(\pi\right)^3[/math]

[math]\left[\left(\frac{\hbar c}{e^2}\right)^3 \left(\frac{\hbar c}{G^3}\right)^3 \left(\frac{\hbar re}{G^3}\right)^{1.111111....}\right] = \left(\pi\right)^3[/math]

[MATH]\left(\frac{\sout{h}c}{e^2}\right)^3[/MATH]

Might it be possible that gravity is the zero point field pressure density acting upon matter rather than matter giving rise to gravity itself? [(hbar*c/e^2)^3] X [(hbar*c/G^3)^3] X [(hbar*re/G^3)^1.1111111] = (pi^3)

Most elegant hypothesis yet conceived IMHO..................solo

Many thanks for the warm welcome Spyman. The following speculations are just that, speculations because I have no experimental data to offer regarding any official research. Nevertheless, remembering that even though the Casimir plates produce only a very small gravitational influence, a gravitational effect is none-the-less a factor. Keeping this in mind, one must assume a slight curving or bending of the vacuum field itself. At this small scale however, detection would be extremely difficult............solo

Most assuredly; Because, as this article points out, there is no space empty of field. Understanding that gravity bends and distorts space time, one could rightly assume that the zero-point field would also be affected by the gravitational field.....................solo

Actually, I came across it while looking thru my yahoo search for science forums.................solo

I'm a new member here and thought it only proper to introduce myself. My forum name is solo and for good reason. It's the way I've lived my life so far and most likely the way it will end. At any rate, I hope to be an asset to these forums always keeping in mind the attitude of a peace maker. I am greatful to be able to share in both your thoughts and offer a few of my own..........solo