Wednesday, November 19, 2008

Disproof of Gravity

The standard particle model has yet to find any sign of Gravity. This is most concerning since gravity is used by most modern equations. Great particle accelerators have hunted for any signs of gravity. None found!

But one simple experiment shows there is no gravity. The Helium Balloon. It rises. How is this possible? Classical Mechanics shows that Force equals the inverse of the Constant of Gravity multiplied by the Mass of Object 1 multiplied by the Mass of Object 2 divided by the Distance between the two masses raised to the second power. F=-GM1M2/r^2

In this equation the mass of the earth is so great that the helium balloon would have no choice but to be attracted to the earth. It would not rise.

We see in the experiment that the helium is rising to meet its level of density.

Density is the most important function in determining the position of an object. Density is the vibration intensity within a volume in relation to the density of the surrounding medium.

Addendum 5/27/09: From the General Science Journal forum

Anonymous, (who ever you are?)
In standard conditions of temperature and pressure (SCTP), 1 kg of He fills a volume of 5.6 m³.
This also means that the density of He in SCTP is 0.1786 kg/m³.

This is an elementary secondary school calculation based of the atomic mass of Helium (4.0026 gr/mol) and the volume occupied by a mole in SCTP (0.022414 m³/mol).

Air composition is roughly 20.9% O2 and 79.1% N2.
This implies a volume of 0.777 m³ for 1kg of air, and a density of 1.287 kg/m³.

+++++++++++

So what you are saying is that lifting force is the differential between differing densities. Why does the balloon accelerate? Is there a terminal velocity for the balloon in SCTP?

If we were to create a box 500m in the y, 10m in both the x and z in reference to the ground. This box has door with a 500m on one side. We can open the door to introduce our SCTP. But we don't want other influences on our experiment. So we close the door. At the bottom of the box we have a mechanical device that will release the helium balloon as described above but including the density of the mylar balloon in our calculations.

What set of equations would accurately describe the balloon's voyage? Gravity suggests that the massive Earth would pull the balloon or any mass towards it. The balloon would never leave the ground. So any equation based on the axiom of gravity would not work. Mass does not accurately describe anything. So any equations based on mass will not work. Only density equations will work.


The Disproof of Gravity
Post a Comment