Some Invalidations of Newton's and Kepler's orbital Mechanics.

Some simple invalidations of Newton’s and Kepler’s orbital mechanics –

1. The orbital mechanics of Newton dictates the earth orbits the sun’s center of mass in an ellipse, yet Newtonian mechanics states the earth also orbits the solar system barycenter. As the solar system barycenter is almost never at the center of mass of the sun, then the earth simply cannot be orbiting the sun in an ellipse. Therefore Newton’s principle of barycentric motion invalidates Kepler’s laws of elliptical motion.

2. Newton’s orbital mechanics dictates the earth orbits the sun’s center of mass in an ellipse. Newton’s orbital mechanics dictates the earth orbits the earth-moon barycenter ever month. Now if the earth is fixed in its elliptical orbit around the sun, then both the earth-moon barycenter and the moon orbit the earth every month, to maintain the earths elliptical orbit shape around the earth every year. Yet if this occurs, Newton’s notion of the barycenter as the center of mass about which the masses orbit as affirmed in the sun-earth motion and then denied in the earth-moon motion. As Newton’s center of mass concludes to a contradiction with more than two bodies in motion, the theory of the center of mass and with it, mass as a cause of gravity, is invalidated.

3. Newton’s orbital mechanics dictates the earth orbits the solar system’s center of mass in an ellipse. Yet, Newton’s orbital mechanics dictates the earth orbits the earth-moon barycenter every month. To properly account for the Earth’s monthly motion around the earth-moon barycenter and the motion of the earth around the solar system barycenter, the earth cannot be orbiting the solar system barycenter in an ellipse as demonstrated above, so the other options available to explain the sun-earth-moon orbits are –

A- The earth-moon barycenter orbits the solar system barycenter in an ellipse. But this option is not in accord with Newton’s notion of a barycenter, where it is a stationary point, relative to the motion of bodies. Furthermore, if the earth-moon barycenter orbits the solar system barycenter, then the earth’s yearly orbit does not conform to Kepler’s orbital laws.

B- The Earth-moon barycentric motion is independent of the earth-solar system barycentric motion. But such movements are never independent in the real, indicating Newton’s notion of the barycenter is not a reflection of the real.

C- The earth does not orbit in the Earth-moon barycentric motion, nor does it move around the solar system barycenter, but is stationary at the barycenter of the universe. This solution gives some respectability to Newton’s notion of the barycenter, but is normally denied by modern science out of blind prejudice.

4. Newton’s orbital mechanics dictates the earth moves around the earth-moon barycenter. If we take the earth’s velocity at the earth center of mass when it orbits the earth-moon barycenter, the velocity of the earth varies between a maximum of v=+-0.012km/s relative to the sun, then the earths orbital velocity relative to a fixed sun is not 30.287 and 29.291 km/s as expected by Kepler’s laws of elliptical orbits, but is 30.229 and 29.279 km/s as expected by Newton’s laws of barycentric motion. As we arrive at two sets of competing velocities for the earth over the Earth’s orbital period, both Newton’s law of barycentric motion and Kepler’s laws of elliptical motion are inconsistent with each other and therefore invalidated.

5. Newton’s orbital mechanics dictates the earth moves around the sun-earth system. Newton’s laws dictate the earth orbits the barycenter of the solar system. As the barycenter of the solar system is rarely (if ever) the same as the sun’s center of mass, then Newton’s laws of barycentric motion as almost always invalidated in the real.

6. Newton’s laws dictate the earth and the sun orbit the barycenter of the solar system. This means the sun orbits the solar system barycenter with a period of approximately 29 years, with a variable velocity of about +-0.0046km/s relative to the earth. Yet the velocity of the earth, relative to the sun, means the earth-sun relative velocity is not 30.229 at the earth’s perihelion and 29.279 km/s at the earth’s aphelion, but is up to approximately 30.234km/s and 29.274km/s respectively depending on the relative directions of the sun’s velocity vector to the earths orbit direction. As these relative velocities are not derivable through Kepler’s laws, the motion of the sun and earth around the solar system barycenter invalidates Kepler’s laws of elliptical orbits.

7. Newton’s law of gravity is claimed to be a universal law of gravity, yet the universal law of gravity is dependent upon the distance r, between the center of mass of bodies orbiting the barycenter. As the barycenter of orbiting masses is arbitrary, depending on the masses included in the calculations, then the universal law is therefore not universal, but merely applicable only at the whim of the man who includes mass or excludes mass. In other words, if the universal law of gravity is really universal, then it would only be applicable when all the masses of the universe are universally taken into account. This means when we exclude masses from the universal law, we are being inconsistent with that law.

One may object to this and say, the universal law of gravitation is applicable to a system of bodies anywhere in the universe and in this respect, the law of gravity is truly said to be universal. Yet when the law is applied anywhere (i.e universally), then it is always applied universally, but arbitrarily with respect to mass included in the calculations. Therefore the so called universal law of gravity is always inconsistent with itself.

Take for example the problem of the earth’s elliptical orbit around the sun. If r1 is the distance between the center of mass of the sun and the earth, then this produces a gravity acceleration of say X, yet, if we include the mass of the other planets in the calcs and use r2 as the distance to the solar system center of mass, r2 is not r1, and the gravity acceleration is not X, but Y caused by the sun and the other planets. So, according to Kepler’s laws, the earth is required to orbit the sun in an ellipse relative to the sun’s center of mass, but according to Newton and Kepler, the earth is also required to orbit the solar systems center of mass in an ellipse.

But as the focus of the two ellipses is not the same, the elliptical orbits are not the same, therefore there are two gravity fields acting on the earth, causing the earth to orbit two elliptical focal points at the same time, depending on what masses are included or excluded from the calculations. This means, as the two ellipses are different, the earths velocities through absolute space are different, meaning there is always more than one gravity field acting on the earth at any one time if both Newton’s and Kepler’s laws are consistently applied to the earth’s orbit. Evidently this is absurd, showing Newton’s law of universal gravity is inconsistent with Kepler’s elliptical laws and is not a universal law at all, but only a mathematical approximation, based upon a selection of local masses, mass attraction and absolute space.

8. Newton’s physics requires the assumption of absolute space, yet relativity denies the existence of absolute space, therefore modern physics invalidates Newtonian physics.

9. Kepler’s laws of elliptical orbits were based upon the assumption of the earth’s motion relative to the stationary sun and other planets. The motion of the other planets are observed and equations are then derived form Newton’s laws with assume the sun is the center of the solar system, because the smaller always orbits the larger mass. As there are several inherent inconsistencies within Newtonian mechanics and contradictions between Newton’s laws and Kepler’s elliptical orbit laws, then it is highly likely that the other planets in the solar system do not orbit the sun in elliptical orbits. In fact, when we take into account the mass of all the planets and the sun, which all are said to orbit the solar system barycenter, the solar system planets definitely do not orbit the sun in ellipses at all, thereby invalidating Kepler’s laws of elliptical orbits.

10. Kepler’s elliptical orbit laws are said to be an accurate approximation of the motion of the planets in the solar system, yet if those laws are derived by assuming the motion of the earth, (which is a false premise in Kepler’s derivation), then Kepler’s laws of elliptical orbits do not hold in the solar system. As many experiments have been performed such as the lunar laser ranging experiment, the Michelson Morley experiment, Michelson Gale experiment, Airy’s experiment and so on, which evidence a stationary earth, then Kepler’s assumed premise of a moving earth is false and the other planets must be orbiting the sun, which in turn orbits a stationary earth every day.

This means the solar system planetary orbits cannot be ellipses with respect to either the sun, solar system barycenter, the earth, or absolute space, but must be accounted for as a complex double motion around the sun and the earth. Therefore Kepler’s laws of orbital ellipses are invalidated and a new physics is required to account for the motion of the solar system planets around the stationary earth (rather than the falsified solar system barycenter).

The more we investigate the status of Newtonian and Kepler’s laws of physics, the more we see how internally and relatively inconsistent those laws are. Evidently the problems with Newton’s physics and Kepler’s laws shown above invalidate those laws and a new physics is needed to account for what modern science tells us about the motion of the sun and the planets relative to a stationary earth.

The Derivation of "the equation of time" Invalidates Heliocentrism and/or Newtonian Mechanics.

It will be shown that the Equation of time derivation of the equation of time ignore the barycentric motions of the Earth-Moon system and the Sun-Jupiter system, thereby demonstrating that the relative motions of the sun and earth are accounted for without the need to use the Newtonian barycenters. In this way, the real success of the equation of time demonstrates the S-J and E-M barycentres do not exist in the real and therefore invalidate the heliocentric model of the solar system.

The following analysis firstly investigates the effect of the relative velocity of the earth to the sun when the earth-moon barycenter motion is included. We then include the motions of the sun caused by the sun-Jupiter barycenter motion to show the variation in earth-sun relative velocities over the 12 year Jupiter orbital period and include in the analysis the combined velocity effects of the S-J and E-M motions at ¼ month intervals at t = 0, 3, 6, 9 and 12 years in the Jupiter period.

This investigation demonstrates that by ignoring the S-J and E-M barycenter motions, the equation of time should change from its standard values by -66s to +66s at times during the 12 year Jupiter period. It also shows the equation of time should vary over the 12 year Jupiter orbit, yet the standard equation of time excludes the Jupiter motion and in practice is assumed to be the same for any year. This is further strong evidence for the non existence of the S-J barycenter and hence an invalidation of the heliocentric system.

Velocity of the Earth center of mass relative to E-M barycenter 

Barycenter of the Earth-moon system

D = M(moon)d(moon)/(M(earth) +M(moon)  = 0.012 x 384405/(1.00 + 0.012) = 4641 kilometers from center of the earth

If the earth orbits the earth – moon barycenter every 27.3 days with respect to the fixed stars and for the sake of simplicity assume the orbit is circular then the earth moves around the E-M barycenter t = 3600 x 24 x 27.3 = 2,358,720 s over a distance of 6.14 x 4641 = 28,495.74 km with an average velocity of 28495.74/2,358,720 = 0.012 km/s = 43.49 km/hr

For the sake of simplicity, let’s assume –

1. The sun is stationary relative to the orbiting earth.

2. The earths orbit velocity is 30 km/s

3. The earths orbit radius to the sun is 1 AU

4. Assume the rotation speed of the earth around its axis is constant.

If we track the motion of the center of mass of the earth, we have the following configurations, with velocities –

T=0

S – E b M

Earth vcm = ↓0.0121 km/s

Earth vorbit = ↑30km/s

Resultant earth’s c of m velocity relative to the sun ↑ 29.988 km/s

T=1/4 month

…..M

…...b

S - E

Earth vcm = → 0.0121 km/s

Earth vorbit = ↑30km/s

Resultant earth’s c of m velocity relative to the sun ↑ 30.00 km/s

T = ½ month

S - M b E 

Earth vcm = ↑ 0.0121 km/s

Earth vorbit = ↑30km/s

Resultant earth’s c of m velocity relative to the sun ↑ 30.012 km/s

T = ¾ month

S – E

……b

……M

Earth vcm = ← 0.0121 km/s

Earth vorbit = ↑30km/s

Resultant earth’s c of m velocity relative to the sun ↑ 30.00 km/s

T = 1 month

S – E b M

Earth vcm = ↓0.0121 km/s

Earth vorbit = ↑30km/s

Resultant earth’s c of m velocity relative to the sun ↑ 29.988 km/s

Therefore over one month, the center of mass orbital velocity of the earth relative to the stationary sun varies according to the following –

Time…………. Earth c of m velocity relative to stationary sun

(month)………(km/s)

0………………29.988

¼……………..30.00

½……………..30.012

¾……………..30.00

1………………29.988

For the sake of simplicity, we can reduce the E-M velocities down to the velocity of the center of the earth’s mass as shown in the table above. When this is done we can calculate the distance travelled according to rotation around the E-M barycenter with respect to the Earth’s constant daily rotation around the axis. Let a day equal = 24 x 3600 = 86,400s.

Earth radius = 6,384 km, v = 6.14 x 6,384/86,400 = 0.454km/s rotation velocity at the Earth-Sun plane (assumed to be along the line of the earths equator- which has no tilt in this example).

Time………….…Earth c of m v rel to sun…..Difference…………Time difference

(month)………..(km/s)……………………...over 24 hrs(km).…………..(s)

0………………29.988…………………………-1,036.8…………...+34.57

¼……………..30.00…………………………..0…………………..…0

½……………..30.012…………………………1,036.8……............-34.55

¾……………..30.00…………………………..0…………………..…0

1………………29.988…………………………-1,036.8………..…..+34.57

The column entitled “Difference” is the derived as follows –

Let the earth’s center of mass velocity be the average velocity of the earth orbiting through space due to the E-M motion. vcom = 0.0121 km/s. The orbital velocity of the Earth around the sun is then the total of the orbital velocity around the sun of 30km +- the E-M barycenter velocity component. The difference is then distance obtained through the addition of the E-M barycenter velocity component over 24 hours.

Time difference is the difference in time between the earth motions without and with the E-M barycenter velocity component for a point on the earth’s earths equator to be facing the same point in space. From the above table we see a sinusoidal variation in the times required for the earth to return to the same point at the sun.

The velocity effect of the 12 year Jupiter orbital period.

We can also add in the 12 year Jupiter-Sun barycenter with the sun orbiting the barycenter at 0.011 km/s. In this way we can arrive at time differences for parts of the month, throughout Jupiter’s orbit period as follows –

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 = 30km/s orbit

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. Therefore, if we assume the earths orbital velocity around the sun is 30km/s, the time difference for a point on the earth’s equator to reach the same point in space, as caused by the S-J motion is 950/30 = -31.66s

Time………….…Time difference……. Time difference……Total time difference

(month)………..E-M(s)…………….. S-J (s)……………..(s)

0………………+34.57………………..-31.66……..………..+2.90

¼……………..0……………………….-31.66………..……..-31.66

½……………..-34.55…………………-31.66…………….+2.90

¾……………..0……………………….-31.66………..…….-31.66

1………………+34.57………………..-31.66………..…….+2.90

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.

↓S _ B _ J ↑...........sun v = 0.011 km/s towards the earth

E -> v = 30km/s orbit

Time………….…Time difference……. Time difference……Total time difference

(month)………..E-M(s)…………….. S-J (s)……………..(s)

0………………+34.57………………..0……………..………+34.57

¼……………..0……………………….0………………..…….0

½……………..-34.55…………………0……………..………-34.55

¾……………..0……………………….0………………..…….0

1………………+34.57………………..0……………..……….+34.57

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 = 30km/s orbit

Time………….…Time difference……. Time difference……Total time difference

(month)………..E-M(s)…………….. S-J (s)……………..(s)

0………………+34.57………………..+31.66……..………..+66.23

¼……………..0……………………….+31.66……..………..+31.66

½……………..-34.55…………………+31.66……..………..-2.89

¾……………..0……………………….+31.66……..………..+31.66

1………………+34.57………………..+31.66……..………..+66.23

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.

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

________ E→ v = 30km/s orbit

Time………….…Time difference……. Time difference……Total time difference

(month)………..E-M(s)…………….. S-J (s)……………..(s)

0………………+34.57………………..0…………..……………+34.57

¼……………..0……………………….0…………..……………0

½……………..-34.55…………………0…………..…………...-34.55

¾……………..0……………………….0…………..….………...0

1………………+34.57………………..0…………..….………...+34.57

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 = 30km/s orbit

J ->

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

Time………….…Time difference……. Time difference……Total time difference

(month)………..E-M(s)…………….. S-J (s)……………..(s)

0………………+34.57………………..-31.66…..…………..+2.90

¼……………..0……………………….-31.66…..…………..-31.66

½……………..-34.55…………………-31.66…..…………..-66.21

¾……………..0……………………….-31.66…..…………..-31.66

1………………+34.57………………..-31.66…..…………..+2.90

From the above tables, it is evident that when incorporating the S-J and E-M motions into the relative velocities of the sun and the earth, the time required for the earth to return to a fixed point in space varies according to the time in the Jupiter period and the E-M orbital period.

However, when calculating the Equation of Time, only the effects of the ellipse and the tilt of the earth are taken into account and the S-J and E-M effects are taken into account.

The mathematical expression of the equation of time can be written as

Δt=(M-v)+( λ- α)/w_E

Δt = t_s − t is the time difference between solar time t_s (essentially time measured by a sundial) and mean solar time t (essentially time measured by a mechanical clock). The first term in parentheses is the time difference due to the eccentricity of the Earth's orbit, and the second term is the time difference due to the inclination of the Earth's rotational axis with respect to its orbital plane. This formulation ignores perturbations due to the moon and other planets, as well as the very small variations in the orbital and rotational parameters with time. http://en.wikipedia.org/wiki/Equation_of_time

Note that the Equation of time is used to determine standard time from the sundial time –

The equation of time expresses the relationship between the sundial and standard time, and the standard time is then available from the sundial by applying the proper value, plus or minus, from the equation of time. But such conversion yields true standard time only if the sundial is on the standard meridian. One must know one's distance east or west of the standard meridian in order to make the remaining correction to the sundial time.

The Earth turns through one time zone in an hour. The time zone is 15 degrees wide (one twenty-fourth of 360 degrees), so each degree of longitude within the time zone is equivalent to four minutes of time (60 min./15^o). This then is the correction to make for each degree of longitude away from the standard meridian: minus if east or plus if west of the standard meridian

As an example, suppose that you are located at longitude 155 degrees west. What is the correction to arrive at standard time for your time zone? The standard meridian is the 150 degree west meridian, so you are located 5 degrees west of that. Every degree is 4 minutes of time, so the sun passes overhead at your longitude 4 x 5 = 20 minutes later than at the standard meridian. Thus, you must add 20 minutes from your sundial time to get the standard zone time. This, of course, is in addition to the time that must be added or subtracted according to the equation of time. See Appendix B for additional examples.

http://www.cso.caltech.edu/outreach/log/NIGHT_DAY/equation.htm

So according to Wicki and Caltech, the equation of time is derived using only the effects of the elliptical orbit and the earth tilt to determine the time lag during the year between the apparent solar time and the mean solar time. Apparently the equation of time is well known and works quite well in practice. Yet the equation has been shown above to not include the Earth-Moon barycenter motion every month or the Sun-Jupiter barycentric motion over 12 years. The Wicki and Caltech sites assume the equation of time does not vary from one month to the next due to the E-M motion, or one year to the next over 12 years, due to the S-J motion.

Evidently the Equation of time, like other astronomical mathematical equations, assumes Newton’s and Keplar’s formulas and physics, yet those same equations, which assume the barycentric motion of the planets to be an accurate model of orbital mechanics, are routinely ignored when the observations of the sun’s location do not match the theory.

Ignoring the Earth-Moon and Sun-Jupiter barycenter motions in the equation of time is no small anomaly in the maths. This indicates that the barycentric motions, which produce time differences from that predicted by the equation of time of between -66s to +66s indicates the E-M and S-J motions simply do not exist in the real. This singular observation indicates the immediate solar system of the sun-earth-moon-Jupiter act in a way that is not compatible with Newtonian mechanics, but is very compatible with geocentrism. According to a geocentric model, the equation of time can be derived assuming an elliptical orbit of the sun, or a circular orbit with a variable sun orbit velocity over the year, without the need for the Earth-moon barycenter motion or the sun Jupiter barycentric motion.

Any way we look at the equation of time, it’s practical success indicates heliocentrism is an invalid model of the solar system. As it is invalid, then either Newtonian barycenter motion is invalidated and a geocentric solar system can be valid. As science points towards a geocentric universe form other experimental failures to find any Earth motion, through space, then it is no surprise to the geocentrist that the equation of time is derived by ignoring the relative velocity changes associated with the E-M and S-J barycenters, simply because –

1. The barycentric motions don’t exist in the real,

2. The universe is really geocentric,

3. Gravity is not caused by mass attraction, but by the aether flow

4. The real cause of planetary orbits is the celestial winds.

Conclusion - The standard derivation of the equation of time is further compelling evidence for a stationary earth in a geocentric universe.

JM

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.