# Gravity at 5 nanoseconds

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Announcement

The Primitive Equation for the Force of Gravity by Alan Folmsbee, Master of Science

force = f = N/D = numerator / denominator

N = pi^(4/3) square meters (cube root 4/3) Mmr

D = (R^2)(seconds^(3/2)) (sqrt(tau)) kilogram

M is mass of Earth for example m is a test mass r is proton radius R is distance from center of M to m

tau is The Universal Constant for Conservation of Hadronic Continuum = 5.5 nanoseconds _____________________________________

Discussion

Newton wrote in 1684

f = GMm/(R^2)

G = 6.67 x 10^-11 meters cubed per kilogram second squared

____________________________________________

Therefore, you, the lay person, will be persuaded if G is equal to the new primitive coefficient : G(N)/G(D)

G(N) = pi^(4/3) square meters (cube root 4/3) r and G(D) = (seconds^(3/2)) (sqrt(tau)) kilogram

\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$

G(N) = 3.14^(4/3) sq meters cube root(4/3) 10^-15 meters and G(D) = sqrt(5.5 nanoseconds) kilogram seconds^(3/2)

\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$

G(N) = 4.60113 x 1.1006 x 10^-15 cubic meters

G(D) = 7.41 x 10^-5 kilogram seconds

\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$

G = (4.6 x 1.1/7.41 ) x 10^(-15+5) meters cubed per kg second^2

G = 6.8 x 10^-11 cubic meters per kg sec^2

It is shown!

"It is so simple it must be true".

Announcing the Folmsbee-Christian Theory of Gravity December 26, 2014 Wailuku Hawaii Protons shrink space and grow time. This is gravity. Diffusion under a conservation of continuum rules the force of interplanetary gravity. see sciphysicsforums dot com muon200

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Given two 1 kg spherical masses at zero relative velocity in space and no external influences and separated by 1 meter, how long will it take for them to make contact?

Or, is this too easy?

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No, I don't think it is. Without knowing the density of the masses (or equivalently their diameters) there is no unique solution.

John

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The unique solution is they will never make contact. The op specified that they have zero relative velocity.

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Not necessary, at least until their surfaces touch: Gauss's law.

The calculation works just fine for point masses, in which case the time taken is for the 1 meter separation without worrying about their dimensions. Any other dimension (up to 1m) can of course be found once the solution is known.

Tim

```--
Seven Transistor Labs
Electrical Engineering Consultation ```
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I omitted 1kg mass each. My apologies.

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For those playing along at home, something like,

*pi*sqrt%281+m%5E3%2F%28G*1kg%29%29%2F4

Tim

```--
Seven Transistor Labs
Electrical Engineering Consultation ```
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Again, my apologies for the oversight.

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Actually, if they're point masses they'll have evaporated long before they would have had a chance to draw together.

```--
www.wescottdesign.com```
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;-)

Technically still not point masses, as that still has a well defined (if rather small) dimension.

Point masses, like point charges, are a mathematical abstraction; real potential wells do not have singularities (notwithstanding the above exception :) ), but we approximate them as such, as long as Gauss' law holds, which is the important / handy part.

Tim

```--
Seven Transistor Labs
Electrical Engineering Consultation ```
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5 nanoseconds?
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Would they not attract each other the same, were they two one ton spheres where one is a half meter of whatever makes a ton at a half meter, and the other is a 1 mm ball of neutron star material or other dense medium where a 1mm ball weighs in at a ton?

Is not the absolute mass the only factor, given a diameter smaller than the space between them for the test?

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Yes.

But if they have a large diameter they will touch sooner than if they have a small diameter.

John

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So, gravity does not exist before 5nSec, pops into existence AT 5nSec and then s l o w l y disperses...

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The mass is not a factor at all. Two steel hammers will fall at the same rate, and if you weld them together they won't fall any faster. In a vacuum, a feather falls to earth as quickly as a stone.

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It is not about falling idiot. They attract each other.

The entire question here has nothing to do with Earth or any planetary level gravity nor any object falling toward one.

It has to do with minute gravitational attraction between two free masses in space. They 'fall' toward each other, idiot.

Get a clue. Nothing is falling in the manner you state.

Don't be so Dip Daffy, Shift Shafty.

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LOL! What do you think "falling" is?

"Planetary level" gravity huh? LOL! Pay attention, dimwit: The planet earth is a collection of stones. It, THEY, attract. and are attracted to, other stones just the same either collectively or singly.

They FALL. DUH!

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Thanks, Tim. I think I came up with about 48 hours when I calculated it a few months ago. At least I was in the ball park.

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They are orbiting each other! No contact will ever occur.

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"Zero relative velocity"

However, the motion in one direction (aka, the in-plane side view) will be identical, so the period is still correct (assuming they don't collide, of course, but even the first 1/4 period before they hit is still valid). That's the formula I posted earlier.

Tim

```--
Seven Transistor Labs
Electrical Engineering Consultation ```

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