An Invalidation of Newtonian Mechanics using the Heliocentric model of the Solar System.

Newtonian Mechanics says the barycenter is the center of mass of two or more orbiting bodies. The barycenter is then similar to a pivot point around which bodies rotate. Heliocentrism says the cause of the solar day is the rotating earth relative to the sun.

According to Wicki – If all the planets were aligned on the same side of the Sun, the combined center of mass would lie about 500,000 km above the Sun's surface. This means R = sun radius + 500,000km = 695,500 + 500,000km = 1.3 million km. If Jupiter takes 76% of R = 0.99 million km and the rest of the planets take 24% of R = 0.31 million km.

During the 12 year period in which Jupiter orbits the sun, the other planets orbit the sun according to their respective orbits as well. I intend to include all the planets being on the opposite side of Jupiter, throughout Jupiter’s orbital period as a very conservative calculated estimate of the solar system barycenter location. In this respect the barycenter is then at approximately 0.99 – 0.31 = 0.68 million km from the center of the sun, throughout Jupiter’s orbit around the sun. In reality, the barycenter will be more than this from the sun’s center, due to the other solar system planets not aligning on the opposite side of Jupiter and the sun.

In brief, the solar system barycenter location is very conservatively calculated assuming the following alignment throughout the 12 years –

Jupiter – Barycenter – Sun – Mercury – Venus – Earth –Mars – Saturn – Uranus – Neptune - Pluto

During Jupiter’s 12 year orbit (t = 378,432,000 s) the sun must conservatively move around an assumed circular path of 2 x 3.14 x 0.68 = 4.27 million km. The average velocity of the sun during the 12 year Jupiter orbit is then 0.011 km/s.

For simplicity of calculation I have assumed the earths orbit around the sun to be a circle centered on the solar system barycenter and it is assumed that the sun returns to the same point on the equator each day to highlight the location difference of the sun relative to the earth, if the barycenter notion is used in the heliocentric model.

When the earth moves to the same point in space relative to the stars every year, the barycenter of the solar system has moved every year. If we take J = Jupiter, E = earth, B = barycenter, S= sun and the arrows indicate the motions relative to the barycenter.

If we track the suns motion around the sun-Jupiter barycenter over 12 years, we have these configurations and the corresponding time variation calculations –

At year 0, when the barycenter is between the sun and the earth, the sun moves to the left of the earth.

S <- v = 0.011 km/s

B

E -> v = 0.46 km/s rotation at the equator and 0 km/s at the poles.

J ->

At this time in the Jupiter cycle an observer will see the sun move due to its barycentric motion by 0.011 x 60 x 60 x 24 = 950 km in 24 hours relative to the Earth. During this 24 hour time the observer standing on the equator will travel 6378 x 2 x 3.14 = 40053 km due to the earth’s daily rotation. So the relative distance travelled by the observer on the equator will change when the sun in an opposite motion, relative to the earth, to 40,053 - 950 = 39,103 km, meaning either –

1. The earths rotation velocity of 0.46km/s must change to 0.45 km/s to maintain the regular 24 hour solar day, or

2. The earths rotation velocity of 0.46km/s is constant and the length of the day changes to 39,103/(0.46 x 60 x 60) = 23.61 hrs.

Yet according to the International Rotation Reference Service, the variations in the duration of a day are only up to 4 milli seconds or 4/1000th of 1 second during the 1960-2000 as shown here –

http://www.iers.org/nn_10398/IERS/EN...tml?__nnn=true

This means according to the International Rotation Reference Service, the earth’s rotation velocity is virtually constant. Therefore points 1 and 2 above are not accounted for in the heliocentric model.

At year 3, when the barycenter is between Jupiter and the sun, the sun will seem to not move at all relative to the earth.

\I/S _ B _ J /I\...........sun v = 0.011 km/s towards the earth

E -> v = 0.46 km/s rotation at the equator and 0 km/s at the poles.

The sun will move towards the earth with a velocity of 0.011 km/s and appear to stand still in space due to the sun orbiting the Jupiter barycenter. So according to the barycentric notion in the heliocentric model, there is no change in sola day length.

At year 6, when the barycenter is between Jupiter and the sun, the sun will seem to move to the left from the earth.

J <-

B

S ->…………. sun v = 0.011 km/s

E -> v = 0.46 km/s rotation at the equator.

At year 6 the relative velocity between the sun and the earth’s rotation rate is 0.46-0.011 = 0.45 km/s. For the sun to orbit the earth every 24 hours, the earth is required to –

1. Increase in velocity from 0.46km/s to 0.46 + 0.011 = 0.47km/s. But there is no evidence that the earth’s rotation velocity changes throughout the year by this much.

2. Sun does not move around the solar system barycenter, but this fails to use the barycenter as used in the heliocentric model.

Alternatively the earths rotation velocity is constant at 0.46km/s with an earth-sun relative velocity of 0.45km/s and the solar day increases to 39,103/(0.45 x 60 x 60) = 24.14 hrs. Yet it is well known that the solar day dos not increase by this amount.

At year 9, when the barycenter is between Jupiter and the sun, the sun will move away from the earth and appear to be stationary to the earth bound observer.

\I/ J _ B _ S /I\........... sun v = 0.011 km/s away from the earth

________ E->

At year 12, when the barycenter is between the sun and the earth, the sun moves to the left from the earth.

S <- v = +0.011 km/s

B

E -> v = 0.46 km/s rotation at the equator

J ->

At year 12, we have the same problem discussed at year 0.

So according to the above simplistic model, clearly over 12 years, the motion of the sun through space, as seen from earth varies from about v = +0.011 km/s to an apparent zero to v = -0.011 km/s to an apparent zero velocity again and the back to v = +0.011 km/s at 12 years. In total, the sun would move about 4.27 million km through space over 12 years to return back to its origin in space.

Therefore if the heliocentric model is correct and the notion of the solar system barycenter is also correct, with a constant earth rotation velocity throughout the year, then the solar day must vary over Jupiter’s 12 year orbit according to the following table.

Year of Jupiter orbit -------Solar day length at the Earth (hrs)

0---------------------------------23.61

3---------------------------------24

6---------------------------------24.14

9---------------------------------24

12-------------------------------23.61

The table shows some points as a sinusoidal variation in the length of solar day over 12 years due to the motion of the solar system barycenter.

These calculations are problematic for the helio model. First the sun moves one way, then apparently stops, then the sun moves in the opposite direction and then apparently stops and then moves back again. For the helio model to survive, the earth must –

1. Vary its rotation velocity throughout Jupiter’s 12 year orbital period, but this option is not found in science.

2. The Earth maintains its rotation velocity and the sun disobeys the barycenter rotation as required in Newtonian mechanics. But this is an invalidation of Newton’s physics.

3. The Earth maintains its daily rotation velocity and the sun somehow orbits the solar system and the solar day length does not change. But for all these to be true, there must be a violation of Newtonian mechanics.

Therefore if the heliocentrism model is valid, Newtonian mechanics is invalid. Alternatively, if Newtonian mechanics is valid, then the heliocentric model is invalid.

Alternatively, the solar system is not heliocentric, but geocentric, with the earth at the center of mass of the entire universe. Is the earth at the center of mass of the universe possible?

Estimated mass of the universe is 8 × 10^52 kg

Estimated universe size is at least 13 billion light years

1 light year = 1x10^12 miles

Therefore the size of the universe is = 1x10^12x 13x10^9 =1.3x10^22 miles wide

If the earth is at the center of the universe then it is 6.5x10^21 miles from the edge of the universe.

The mass of the solar system is 1.992 x 10^30kg

We can assume the universes mass is equally distributed or distributed around the universe in shells. Either way the answer is the same regarding this calculation.

Worst case scenario regarding the position of the center of mass of the universe is when all the planets and sun are to align themselves, making the center of mass of the solar system located within the sun. Does this change the center of mass of the universe outside of the earth?

The distance from the earth to the sun is 93 million miles

The center of mass should be the balancing point of the universe within the earth. If we take the center of mass of the universe

Using a = 93 million miles.

M1= 8 × 10^52

M2= 1.992 x 10^30

R1 = 93 x 10^6 x (1.992 x 10^30/(8 × 10^52+1.992 x 10^30)

= 93 x 10^6 x 2.4 x 10^-23

= 2.3 x 10 ^ -15 miles

=1.61 x 10^6 x 2.3 x 10 ^ -15 = 3.7 x 10^-9 mm

This means, when using the center of mass concept as used in Newtonian physics, in the worst case scenario when the sun and planets align, the center of mass of the universe does not move outside the earth. Therefore, if Newton’s physics is valid, only a geostationary universe is valid.