G-Tron

You are correct, D H, I am not trained to refute Pari Spolter's gravitational theory/math; she is a mathematician, and her published book, "Gravitational Force of the Sun", calculates gravity without the use of mass. I posted the 1st paragraph; Pari & John Lear's statements & math, begins with the 2nd paragraph: "The issue of the Neutral Point, that point at which the moon's gravity is equal to the Earth's gravity is this:" and follows without my making further comment. This is the type of stuff that ignores science by utilizing error math, and is an actual book espousing the theory that graviational calculations are entirely wrong using Newton's & Einstein's laws. To allow these ideas to spread seems like scientific suicide, and I appreciate the feedback you, Cap'n Refsmmat & Atheist have given. Klaynos, I hope to see your efforts posted! The nature of the thread title was to imply some erudite attention to this, as not many care to address the idea. One error in the calculations disqualifies the entire theoretical summation. If you see anything else associated, please post it. Thanks again.

I would appreciate any opinion (with reasons) on the following material by Pari Spolter. Accordingly, using her math forumula, she ascertains mass does not come into the calculations for determining planetary (Sun, etc.) gravity. Also calculates the moon's gravity is at least 68.71% that of Earth's. Any thoughts? See below. Her book is "Gravitational Force of the Sun" by Pari Spolter. Thanks to all who respond! The issue of the Neutral Point, that point at which the moon's gravity is equal to the Earth's gravity is this: Mainstream science says that bsed on Newton's Law of Universal Gravity and its Universal Constant based on the Earth's density of 5.5 gr/cm3 and Moons density of 3.4 gr/cm3 the moon's gravity should be one sixth that of Earth. William Brian II (Moongate: Suppressed Findings of the U.S. Space Program), me and a few others contend that, based on the Bullialdus/Newton Law of Inverse Square the moons gravity should be at least 64% that of Earth's. The Law of Inverse Square states: "The inverse-square law (Bullialdus/Newton) is any physical law stating that some physical quantity or strength is inversely proportional to the square of the distance between them, specifically, the gravitational attraction between two massive objects, in additional to being directly proportional to the product of their masses, is inversely proportional to the square of the distance between them." Using the Bullialdus/Newton Inverse Square for the calculation of the moon's gravity we get: Using the following values: Re = radius of the Earth = 3,960 miles Rm = radius of the Moon = 1,080 miles X = distance from the Earth’s center to the neutral Point = 200,000 miles Y = Distance from the Moon’s center to the neutral point = 43,495 miles Ge = Earth’s surface gravity Gm = Moons surface gravity Since the forces of attraction of the Earth and the Moon are equal at the neutral point, the inverse-square law yields: Ge (Re²/X²) = Gm(Rm²/Y²) Gm/Ge = Re²Y²/Rm²X² = (3,960)2 (43,495)2/(1,080)2 (200,000)2 = .64 Therefore, Gm = .64 Ge Or 64% of the Earth's gravity. Mainstream science says, "Oh but you can't use the Bullialdus/Newton Law of Inverse Square because now we have a 3 body problem, Earth, Moon and Sun. We would have to know for that exact moment of the Neutral Point the perturbation of the Sun, the eccentricity of the Moon and the effects of the particular Moon cycle for that time and date. In the past few months I have exchanged emails with Pari Spolter addressing this problem. Pari agreed with mainstream science that the Inverse Square Law in regards to the Earth-Moon neutral point does constitute a 3 body problem and that the eccentricity of the Moon and the effects of the particular Moon cycle for that time and date would have to be addressed. So I challenged Pari to address the problem and below is our email exchange: ----- Original Message ----- From: John Lear To: Pari Spolter Sent: Friday, May 23, 2008 9:17 AM Subject: Moons gravity May 22, 2008 Pari Spolter Orb Publishing Company 11862 Balboa Blvd. #182 Grenada hills, CA 91334-2753 Dear Pari, Please consider for a moment what the gravity might be on the moon. Also please consider, at least temporarily, these 2 facts: (1) Gravitational force is independent of the nature and the quantity of the attracted body. (2) No Apollo mission ever landed on the moon and took gravity measurements. We are told that the gravity on the Moon is one sixth that of earth. If the gravity of a planet is quantized and if the gravity of a planet has nothing to do with mass/density, then how could the moons gravity be one sixth that of earth? The one sixth gravity of earth is a figure derived based on the alleged density of the Earth and the Moon and an equation using Newton's Law of Universal Gravitation. Now that Newton's Law of Universal Gravitation has been decisively proven wrong, and that there is no relationship between gravity and the mass/density of a planet it is inconceivable that the gravity of the Moon can still be coincidentally, one sixth that of Earth. It is simply not true. And it is not supported by the facts. And the facts are that the gravity on the Moon is F=a.A just as all the other planets. There is no 'exception' for planets orbiting other planets. There is no exception that: "For any moon orbiting a planet F=a.A is suspended." There is no 'third body problem" any more than there is a 9 body problem with F=a.A. In you letter of April 16, 2008 you state that the gravitational force of the Moon is 1.540260256 x 1013 . Then you state that the equatorial radius of the Moon is 1,737,400 M. You also state that the area at the Moon's equator is 9.483082025 x 1012 m2. You then conclude that the gravity on the Moon is one sixth that of earth. The problem is that you have relied on only one source for the gravitational force of the Moon, Apollo 14. If the orbital velocity at semi-major for the earth is 29.771 and the semi-major axis of revolution around the sun for the Earth is 149.57 then why would these numbers be significantly different for the Moon? If the gravitational force of a planet is equal to a.A, then why would the gravitational force of the Earth be any different from the Moon? And why, if indeed the gravitational force of the Moon where less than Earth, would it be 83% less? Surely the fact that the Moon orbits the Earth cannot account for a reduction in gravitational force of 83%. Nowhere else do we find support for the Moon's alleged one sixth gravity of Earth other than the alleged Apollo 14 measurements. Could the FV relationship of the Moon to the Earth of 60% have this kind of effect? There were many failures of probes sent to the Moon in the late 50's and early 60's that appear to have been the result in the assumption of an erroneous figure of the gravitational force of the Moon. In fact, the highly elliptical orbits of the Lunar Orbiter series of 1966 and 1967 from a mean orbital altitude of 60 to 70 miles would indicate a gravitational factor of the moon far greater than one sixth that of Earth. We know that NASA claimed that the 60 to 70 mile lunar orbits for the Lunar Orbiter, Apollo, Clementine were because of the alleged effects of Mascons. I suspect that Mascons are fictitious and introduced to hide the real reason for the 60 to 70 mile high orbits which was because of the high lunar gravity. The alleged movies of the Apollo astronauts on the Moon do not appear to show that the gravity is one sixth that of Earth. The maximum height reached by any jumping Apollo astronaut was barely 18 inches, if that. This is not indicative of one sixth gravity of Earth. What possible set of circumstances exist that would cause the gravity of the Moon determined to be one sixth that of Earth based originally on Newton's Law of Universal Gravitation and hypothesized mass densities of the Earth and the Moon to miraculously coincide after discovering that Newton's was wrong about gravitation being the result of an inert mass. Maybe it happened like this as Mr. Sun talked to Mr. Moon: "Mr. Moon, I realize that you are a planet and that you orbit the sun as well as the Earth. However since you orbit the Earth I am not going to give you the full benefit of F=a.A. I am only going to dole out one sixth of Earths gravity. If at some time you wish to break away from Earth and go into orbit around me on your own, you will then be afforded the full benefit of F=a.A. I hope you understand my feelings on this matter." Mainstream science tell us that we cannot use the Bullialdus/Newton Law of Inverse Square for the determination of the gravity on the Moon because "it creates a third body problem of enormous complexity". In fact, in my opinion, this is nonsense. Since mass/density has nothing to do with gravity and as F=a.A, the alleged 'third body problem of enormous complexity' has no more validity than saying f=a.A doesn't take into account the 8 other planets. So Pari, let us assume that the Bullialdus/Newton Law of Inverse Square is an enormous 'third body' problem. And let us assume that we are going to attempt to see what the effect might be on the 64% of Earths gravity that the Moon has according the Inverse Square Law. What would be the maximum effect on the figure 64% of: (1) Maximum perturbation effect of the Sun for the exact date of the neutral point. (2) Maximum effect of the eccentricity of the orbit of the Moon for that date. (3) Maximum effect of the particular Moon cycle for that date. Please estimate (ballpark) what the sum of the maximum effect of all of these considerations might be and consider if the sum of these affects would account for the difference between the inverse square result of 64% and the alleged Apollo gravity measurement of 16%. Pari, I would respectfully request your careful consideration of the above comments. Thank you and all the best, John Lear continued.... From: Pari Spolter [mailtorbpublishing@msn.com] Sent: Saturday, June 07, 2008 3:39 PM To: John Lear Subject: Re: Moons gravity Dear John Lear, Thank you for your letter. F = a . A is the correct interpretation of Kepler’s third law. We derive the gravitational force of the Sun from the acceleration of planets at their semimajor axis of revolution around the sun. And we derive the gravitational force of a planet from the acceleration at the semimajor axis of revolution of their Moons or of artificial satellites around the planet. If a body does not have a Moon or artificial satellite orbiting it, we do know its gravitational force. The Earth orbits the Sun; the velocity of the Earth v at the semimajor axis of its revolution r gives the gravitational force of the Sun. The Moon orbits the Earth. The v and r of the Moon give the gravitational force of the Earth. The three-body problem is real. When a third body is near the other two, its gravitational force causes perturbation of the motion of the system. Neutral Point According to the references given in Chapter 3 of Moongate by William L, Brian II, the Apollo 11 passed the Neutral Point (N) between Earth and Moon at 43,495 miles (69,998,417.28 m) from the Moon on July 19, 1969. The Horizontal Parallax in The American Ephemeris and Nautical Almanac for the year 1969 on July 19.5 gives the Earth-Moon distance of 395,362,477.4 m. Subtracting the NM distance, we get the EN distance of 325,364,060,1 m. The gravitational force of the Earth is Fe = 1.252240211 × 10 15 m/s2 . m2 At the EN distance, the acceleration is 3.765291596 ´ 10-3 m/s2. This acceleration at the distance NM gives a gravitational force of the Moon Fm = 5.795953977 × 1013 m/s2 . m2. This is 4.628% of the gravitational force of the Earth. The equatorial radius of the Moon is 1,737,400 m. The acceleration at the equatorial surface of the Moon would be 6.1119 m/s2 or 62.38% of the Earth’s acceleration at the equatorial surface. In 1969 The Moon’s Apogee was on July 13.75, and Perigee at July 28.375. The New Moon was on July 14.5917, and First Quarter on July 22.507. The above ratio of Fe / Fm is 21.605. This gives at the above EM distance the center of gravity of the Earth Moon system at 17,489,720 m, and the distance of the Moon from the center of gravity 377,872,757.1 m. The Right Ascension of the Moon on July 19 was 166˚.307975 and on July 20 it was 177˚.264376. The Declination on July 19 was 6˚.5978 and on July 20 it was 0˚.71767. The Cos formula for the arc traversed in one day gives 12˚.42115962 equal to 0.2167901323 radians. At the above distance from the center of gravity it gives a velocity v of 948.1376 m/s, an acceleration of 2.379 ´ 10-3 m/s2, and a gravitational force of the Earth equal to 1.06818 × 10 15 m/s2 . m2. Subtracting (Fe - Fm ) gives - 1.27099496 × 10 14 m/s2 . m2. So on that day the perturbation by the Sun was negative. Adding this to the gravitational force of the Earth, we get 1.3793397 × 10 15 m/s2 . m2. At the EN distance above, we get an acceleration of 4.147 ´ 10-3 m/s2. This acceleration at Neutral Point gives Fm = 6.384 × 1013 m/s2 . m2. This is 5.1% of Fe. The acceleration at the equatorial surface of the Moon would be 6.73 m/s2 or 68.71% of the Earth’s acceleration at the equatorial surface. If Fe = 1.252240211 × 10 15 m/s2 . m2 = ae . BRe2 , and Fm = 1.540260256 × 1013 m/s2 . m2 = am . BRm2 , at Neutral Point ae = am and Fe / Fm = 81.3. Re + Rm = 395,362,477.4 m. Re = 355,892,075.4 m and Rm = 39,470,402.03 m. The ae = am = 3.147 ´ 10-3 m/s2. The gravitational force of the Moon has to be a lot less than the gravitational force of the Earth, because the Moon orbits the Earth. If the gravitational force of the Moon was in the same range, or even 30% of the gravitational force of the Earth, Earth and Moon would be in a binary orbit around their common center of gravity. So the above result would not be inconsistent with observations. But this is the result from just one report of the Neutral Point by Apollo 11 on July 19, 1969. More accurate determinations of the Neutral Point at other times are needed to confirm this and eliminate the possibility of interference from other sources such as the solar wind. In 1969 the Sun was at the peak of its 11-year activity cycle. Several artificial satellites have been placed in orbit around the Moon and the gravitational force of the Moon derived by the use of Kepler’s third law. Also, Apollo 11, 12, and 14 astronauts have measured acceleration on the lunar surface. The gravitational force of the Moon reported from these observations is 1.540260256 × 1013 m/s2 m2 or 1.23% of the gravitational force of the Earth. Acceleration at the equatorial radius of the Moon is 1.624 m/s2 or 16.6% of the acceleration at Earth’s equatorial surface. Regards, Pari Spolter orbpublishing@msn.com ----- Original Message ----- From: John Lear To: 'Pari Spolter' Sent: Tuesday, July 29, 2008 9:19 AM Subject: RE: Moons gravity Hello Pari. I have been talking about your book on various radio programs. I hope you have received some orders. To clarify in my mind what you have said below: “In 1969 The Moon’s Apogee was on July 13.75, and Perigee at July 28.375. The New Moon was on July 14.5917, and First Quarter on July 22.507. The above ratio of Fe / Fm is 21.605. This gives at the above EM distance the center of gravity of the Earth Moon system at 17,489,720 m, and the distance of the Moon from the center of gravity 377,872,757.1 m. The Right Ascension of the Moon on July 19 was 166˚.307975 and on July 20 it was 177˚.264376. The Declination on July 19 was 6˚.5978 and on July 20 it was 0˚.71767. The Cos formula for the arc traversed in one day gives 12˚.42115962 equal to 0.2167901323 radians. At the above distance from the center of gravity it gives a velocity v of 948.1376 m/s, an acceleration of 2.379  103 m/s2, and a gravitational force of the Earth equal to 1.06818 × 10 15 m/s2 . m2. Subtracting (Fe  Fm ) gives  1.27099496 × 10 14 m/s2 . m2. So on that day the perturbation by the Sun was negative. Adding this to the gravitational force of the Earth, we get 1.3793397 × 10 15 m/s2 . m2. At the EN distance above, we get an acceleration of 4.147  103 m/s2. This acceleration at Neutral Point gives Fm = 6.384 × 1013 m/s2 . m2. This is 5.1% of Fe. The acceleration at the equatorial surface of the Moon would be 6.73 m/s2 or 68.71% of the Earth’s acceleration at the equatorial surface.” Taking into consideration for Apollo 11: (1) Maximum perturbation effect of the Sun for the exact date of the neutral point. (2) Maximum effect of the eccentricity of the orbit of the Moon for that date. (3) Maximum effect of the particular Moon cycle for that date. That the gravity on the Moon should have been 68.71% that of earth’s. (We will discuss the other issues you pose in the next email.) All the best, John Lear From: Pari Spolter [mailtorbpublishing@msn.com] Sent: Saturday, July 29, 2008 8:169 PM To: John Lear Subject: Re: Moons gravity Hello John, Thank you for your email. Yes, the reported Neutral Point by the Apollo 11 Astronauts is in conflict with the accepted gravitational force of the Moon by the establishment, and I have not seen any explanation for the discrepancy. Yes, I have sold a lot of books the last three months. Thank you very much. Regards, Pari Spolter orbpublishing@msn.com So, I win. The moon's gravity is at least 68.71% that of Earth's. The importance of the issue of the neutral point is that if the Moon's gravity is 64% or more than that of Earth's the lunar lander could not have undocked from a 60 mile orbit, descended to the Moon's surface, landed, then blasted off, climbed to a 60 mile orbit, and docked with the CSM with 22,000 pounds of fuel. And when I say a 60 mile orbit, the CSM allegedly used an elliptical orbit in which the lander undocked at 10 miles altitude but this issue is in doubt for several reasons. The fuel figures remain the same whether or not they undocked at 60 miles altitude or 10 miles altitude. They could, allegedly, barely do it in one sixth gravity, landing in most cases with only 1 or 2 minutes of fuel remaining.