Metaphysics (Alternate View)/Creation/One-Planet System

(An alternative view by Calgea 20:52, 7 June 2007 (UTC))

History
Before Copernicus (1473-1543), people believed that God made the Earth and man. God was perfect; the Earth was the center of the universe, and all heavenly bodies moved in perfect circles about the Earth. After, Copernicus, these bodies moved in perfect circles about the sun. This heliocentric view gain acceptance and was believed by Kepler (1571-1630). However, after reviewing Tycho Brahe’s (1546-1601) data, Kepler changed his mind and formulated his three famous laws. This article provides an alternative view of a one-planet system and help explains the why of Kepler’s laws.

Ancient
After the Big Bang, clouds of Hydrogen moved outward in a radial fashion. For this article, one cloud is moving in a straight line and is a zillion light-years from all other clouds. As it moves along its straight-line path, gravity compresses the Hydrogen atoms. Under this compressive influence of gravity, these atoms formed Hydrogen molecules. As the molecules and atoms move inward, they collided with other atoms and molecules. These collisions allowed the conversion of the inward Linear-motion energy to Spin or angular motion energy. These atoms and molecules spin about their own axes and about the center of mass of the Hydrogen cloud. It is this center of mass that is moving in a straight line.

As the compression continues, some atoms convert to other forms of energy. The cloud emits light and starts its existence as a sun. Later, there is an explosion under the surface of this sun, which sends a large mass into space to become a planet.

Planet-mass
At this surface and before the explosion, this planet-mass had an angular energy about the center of mass of the Hydrogen cloud. This energy remains with the planet-mass as it moves outward from the sun. However, as its radius to the center of mass increases, its angular velocity slows to keep its angular energy constant.

As the force of the explosion moves the planet-mass outward, the force of gravity acts in opposition to slow the outward movement. These opposing Linear-motion energies convert to Spin, and the planet-mass rotates about its own axis.

Gravity also acts on this planet-mass to compress its atoms into a spherical shape like it did for the sun. This third energy of rotation applies to the internal rotation the mass had while it was a part of the sun.

Eventually, the planet-mass stops its outward movement under the influence of gravity. Since energy cannot be destroyed, all the planet-mass’s explosive energy is now in Spin.

However, gravity does not stop. Gravity now acts to return the planet-mass to the sun. But at this point, the planet-mass acts as a gyroscope. The force of gravity is inward toward the sun, and the planet-mass moves perpendicular to this force in the direction of its original rotation while it was in the sun. If all parameters are correct, the mass goes into orbit about the sun and becomes a planet. Otherwise, it crashes into the sun.

Gyroscope (AV)

According to Newton’s (1642-1727) third law of motion, when an explosion occurs in the sun that sends a mass into space, there is an equal and opposite reaction that moves the sun into space. Therefore, the same energy that places one mass, e.g., planet, into orbit also places the other mass, e.g., sun, into orbit. The center of mass of this one-planet system continues in a straight line.

Energy Circles


The movement of a planet is a combination of two energy circles. The radii of both circles turn at a constant angular velocity $$ (\omega_s) $$ with the same period (T) as the planet. The first circle of radius {c} shows energy oscillating, i.e., converting, between Spin and Linear- motion. The second circle of radius {b} shows energy oscillating, i.e., transitioning, between two Linear-motion dimensions.

Simple Pendulum (AV)   Conversion Conical Pendulum (AV)   Transition

Parametric Equations
$$ X_c = c (cos (\theta)) $$_____Conversion

$$ S_n = c (sin(\theta)) $$

$$ X_T = b (cos(\theta)) $$ ____Transition

$$ Y_T = b (sin(\theta)) $$

Consider the two amounts of energy in the Linear-motion fields $$ (X_c) and (X_T) $$ which lie along the same axis. They are independent and add by the sum of squares.

$$ X^2 = (X_T)^2 + (X_c)^2  $$

$$ X^2 = b^2 (cos^2 (\theta)) + c^2 (cos^2 (\theta)) $$

$$ X^2 = (b^2 + c^2) cos^2 (\theta) $$

$$ X^2 = a^2 (cos^2 (\theta)) $$



$$ X = a cos (\theta) $$

$$ Y = b sin (\theta) $$

$$ S_n = c sin (\theta) $$

These last three equations are the parametric equations for an ellipse.

KEPLER’S FIRST LAW
The orbit of each planet is an ellipse with the sun at a focus.

Returning to Newton’s third law about action and reaction, one says the sun is in an elliptical orbit. That is, the same laws that apply to the planet apply to the sun. If the sun is in an elliptical orbit, it cannot be at its focus and cannot be at the focus of the planet’s ellipse. Kepler did not know about energy circles and placing the sun at a focus was his way of trying to explain why planets moved in an elliptical orbit and not in a circle.

KEPLER’S SECOND LAW


The line joining the planet to the sun sweeps through equal areas in equal times.

Again, Kepler believed the sun was at a focus. The following diagram uses the center of mass of the system.

To understand this second law, let the planet sweep out areas while traveling at different angular velocities $$ (\omega_s) $$. Also, let the interval be small so the areas swept out approximate triangles. Each triangle has an altitude (R) and a base (S) and an area (A).

$$ A_1 = A_2 $$ ____Second Law

$$ (1/2)R_1S_1 = (1/2)R_2S_2 $$ ____Triangles

$$ S = R (\theta) = R(\omega)t $$

$$R_1^2 (\theta_1) = R_2^2 (\theta_2) $$

$$ R_1^2 (\omega_1) t = R_2^2 (\omega_2) t $$

$$ mR_1^2 (\omega_1) = m R_2^2 (\omega_2) $$

Restatement
The angular momentum of a planet in its elliptical orbit is a constant.

One can easily reason backwards from the conservation law to equal areas. But for Kepler to formulate his laws four hundred years ago was outstanding. It was also amazing he even recognized the equal area idea was important. In his time, it took a brave man to formulate and publish heretical ideas.

KEPLER'S THIRD LAW




The square of the period (T) of a planet is proportional to the cube of its mean distance from the sun.

To understand this law, one needs both a system ellipsoid and a system circle for this one-planet system. The system circle is the system clock and has the same period (T) as the planet. Its radius $$ (R_s) $$ is the mean radius for the planet. The ellipsoid shows the constant conservation of the angular momentum and energy.

From the conservation law, the planet’s angular momentum and angular energy are constants. This means the volume of the system ellipsoid is also a constant. This volume depends on its three semi-axes, a, b, & c, that are respectively $$ (R_s (\omega_s)), (R_s (\omega_s)), and (R_s) $$.

This in turn leads to an equation for the period of the system ellipsoid.

$$ V_s = 4 (\pi) abc/3 $$     ____Volume

$$ V_s = K_1 (abc) $$
 * where $$ K_1  =  4 (\pi)/3  $$          ____a constant

$$ V_s = K_1 (R_s(\omega_s)) (R_s(\omega_s)) (R_s) $$ $$ V_s = K_1 (R_s)^3 (\omega_s)^2  $$
 * where $$ (\omega_s) = 2 (\pi)/T $$

$$ T^2 = K_2 (R_s)^3   $$        ____3rd Law
 * where $$ K_2 = 16 (pi)^3 /(3V_s)  $$

Note: The above is not an accepted view.

Creation (Alternate View)