Theory of zero

Gravity

Figures 1 & 2 shows a lever in balance, i.e. the input and output forces are equal. The fulcrum is the point along the lever where the acting force is zero. In the second diagram, the forces have moved further along the lever, but the fulcrum remains at the same point. For the lever to remain in balance, the product of force and distance must remain the same for both. Figure 3 shows the generalised relationship between forces, where a.f is the applied force, i.f the input force and u.f the unit force, or force per unit of distance. The relationships will follow the trigonometric ratios because the applied force will act at a right angle to the input force. The angle θ = arctangent(applied force / input force). The ratios are as follows:

The last ratio is especially important because it represents the distance between the application point and the input point of the applied force. As any point can be considered an application point or input force, this formula can be used to (absolutely) describe the distance between any two points. Figure 4 shows the situation where the applied point is at an angle (φ) to the horizontal; the applied and input forces are still at right angles to each other, and the distance from the application point to input point remains a.f(tan θ).
Your browser does not support HTML 5
Your browser does not support HTML 5
Your browser does not support HTML 5
Your browser does not support HTML 5


Input force as potential energy.

The previous examples, the lever connected the applied, input and output forces. But what if there was no lever? In that case force could not be transferred, and so it will exist as in potential form only. To see this, consider a man digging a hole with a spade. He will need to apply a certain force to move the dirt. If he puts the spade down he won't be able to dig of course, so his energy becomes potential. When he takes the spade up again, he will have to apply the same force as before to move the dirt - the ratio of applied force to input force is the same regardless of whether there is a mechanism to transmit it or not. If the man uses a longer handled spade, then the force he will have to apply will be less, but the amount of dirt he can move will be less also. Again, the ratio of applied force to input/output will depend on the distance between the two, thus distance can be described as this ratio.

Applied force as kinetic energy.

In the scenario where there is no connection, the force that cannot be transferred is equivalent to kinetic energy.
With applied force being equivalent to kinetic energy and potential energy equivalent to input force, then θ = arctangent(kinetic energy / potential energy).

Your browser does not support HTML 5
Your browser does not support HTML 5

When a force moves along a lever (and is kept the same), the ratio of applied force to input force changes - but the nett force at the fulcrum remains zero. The same principle applies to energy; as the distance from the barycentre changes, the ratio of a bodys' potential to kinetic energy, changes, with the nett energy of the system remaining zero at the barycentre. As both nett force and nett energy must be zero at the same point, this implies that there must be a universal constant linking the two - let's call it G for traditions' sake.


Your browser does not support HTML 5
Your browser does not support HTML 5

Your browser does not support HTML 5


Mass, distance and conservation of energy.

For every action, there is an equal and opposite reaction. The recoil of a gun is a good examples of this; the gun travels in one direction at a certain velocity - the bullet travels in the opposite with a velocity that ensures that the mass multiplied by the distance in both directions cancel. So, what then is the mass of a body? If energy has to be conserved, and the point at which it conserved to remain constant, then the mass of a body must be the change in energy over a distance. The mass of a body is given by the formula m=d.θ, where d is the distance from the barycentre, and θ is as already described. The distance is given by d=ke/(tanθ). The reason for this is conservation of energy. If energy is to be conserved at a particular point, then the position of each body must be determined by its mass.


Your browser does not support HTML 5
Frames of reference.

The standard way of creating a frame of reference is to consider an observer to be stationary relative to a body they are observing moving. This is impossible, of course; any real observer will be moving relative to the body they are observing. To an observer in their own frame of reference, the origin of the co-ordinate system is their own centre of mass. In actual reality, the origin of the co-ordinate system is the barycentre of the observer and the mass point of the body they are observing. Any real observer has to have mass, and thus can never be at the barycentre of themselves and the body they are observing moving. Every observer in the system will agree that the nett energy of the system is conserved, but no two observers will agree what the energy of the system is. This is because they will not take their own energy levels into account. In figure 14, the movement away from the barycentre represents an action/reaction event; movement towards represents a collision. At all times, the nett work done is zero at the barycentre.


Your browser does not support HTML 5
Your browser does not support HTML 5
Your browser does not support HTML 5

Acceleration

When a body accelerates, its energy levels are changing. This change in a body's energy changes the position of its barycentre with other bodies, and thus the point which it is moving relative to. The example below focuses on the change in the position of the barycentre between body A and B when body B accelerates. When body B's energy levels increase, the barycentre moves closer to body B. This increases the distance travelled by body A. When body B's energy levels decrease, the barycentre moves closer to body A. Neither observer is aware of the change in the position of the barycentre, but observer B will feel the acceleration as his energy levels change, whereas observer A's do not. After Body B has finished accelerating, the barycentre will again be stationary, and inertial motion will resume relative to it. The energy level of body B has changed so the amount of work it does over a given distance (its mass) will also have changed.
Your browser does not support HTML 5
Your browser does not support HTML 5

Any co-ordinate system needs an origin. The obvious point to use in a two-body system is the barycentre (point zero). Using a polar system, the barycentre becomes the pole, and the polar axis is a horizontal line drawn through the pole. The position of the mass point of a body in the system can be given by its distance from the barycentre and angle (φ) to the polar axis. The distance of the body from the barycentre will depend on its mass - with the mass times the distance of both bodies being equal at all times (zero nett energy). As stated previously, when angle and distance change simultaneously, this is a spiral.

Your browser does not support HTML 5

Figure 18 shows two bodies moving relative to their barycentre. Once they come into contact with each other, they no longer be able to move relative to their barycentre (which is now their common centre of mass). They will continue to attempt to move relative to their centre of mass because this is the point where the nett energy of the system is zero. The fact that they cannot move relative to this point is what gives them weight. So there is no real need to introduce gravity as a separate concept - the principle of conservation of energy is enough.

Your browser does not support HTML 5
Summary.

Every object in existence has both potential and kinetic energy, and the motion of a body can be described as the change in the profile of this energy. The nett energy of a body remains unchanged unless it comes into contact with another body. The centre of mass of one body moves relative to the centre of mass of another, in such a way that the nett energy of the system is zero at their barycentre.

Formula


Ends.