G-Tron Posted August 11, 2008 Posted August 11, 2008 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 2.379 103 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 4.147 103 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.
D H Posted August 11, 2008 Posted August 11, 2008 I would appreciate any opinion (with reasons) on the following material by Pari Spolter. Pure unadulterated garbage, and that is putting it nicely.
timo Posted August 11, 2008 Posted August 11, 2008 I'm afraid I didn't read all of the long text but only up to the calculation. I wonder where the value for Y comes from?
Klaynos Posted August 11, 2008 Posted August 11, 2008 I'm afraid I didn't read all of the long text but only up to the calculation. I wonder where the value for Y comes from? I got to teh calculation as well and I'm going to work threw it in a min, I don't like using "gravity at planet surface" very much in this type of equation...
D H Posted August 11, 2008 Posted August 11, 2008 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. Mainstream science says that the Moon's mass is about 0.0123 Earth masses based on vehicles that have been placed in orbit around the Moon since the 1960s. Based on assessments of the Moon's size, this means the Moon has a density of about 3.4 gm/cm3 and a surface gravity of about 1/6 that of Earth's surface gravity. In short, you are bass-ackwards here. 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 What's wrong here is that you are using the location of the Earth-Moon L1 point, not the "neutral point". The latter, the point where the gravitational acceleration toward the Earth is equal to the gravitational acceleration toward the Moon, is rather meaningless. The former, the L1 point, is quite important but it is not the same as what are calling the neutral point. (2) No Apollo mission ever landed on the moon and took gravity measurements. The Lunar Surface Gravimeter was flown on Apollo 17 but failed. Are you claiming this is part of some vast conspiracy? 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? What is this bunch of non-sequiturs? Mass is quantized, but the size of the individual particles is so incredibly tiny that quantum mechanics simply doesn't come into play on a planetary scale. 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. Wrong. We can determine the Moon's gravity field by observing how the Moon affects vehicles placed in orbit about the Moon behave. For example, see http://www.sciencemag.org/cgi/content/full/281/5382/1476. Now that Newton's Law of Universal Gravitation has been decisively proven wrong All you have done is to demonstrate a lack of understanding. Newton's law of gravity, at least so far as the Earth and Moon are concerned, is quite fine. We have of course discovered that Newton's law of gravity is wrong in the sense that it does not account for the precession of Mercury or behavior around incredibly large, small masses (e.g., black holes). One needs general relativity to describe gravity for fast moving objects and very large masses. General relativity is equivalent to Newtonian gravity for smallish velocities and smallish masses.
Cap'n Refsmmat Posted August 11, 2008 Posted August 11, 2008 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. Could you explain this right here? Which "inverse-square law" are you using? It corresponds to none I've ever learned. Perhaps you could explain what exactly you are doing in these equations.
timo Posted August 11, 2008 Posted August 11, 2008 Could you explain this right here? Which "inverse-square law" are you using? It corresponds to none I've ever learned. Perhaps you could explain what exactly you are doing in these equations. [math]F_e® = Gm_e/r^2, G_e := Gm_e/r_e^2 \Rightarrow F_e® = G_e \frac{r_e^2}{r^2}[/math] (expand the former equation by [math]\frac{r_e^2}{r_e^2}[/math] to see it). Same for the force by the moon. The only and supposedly (see also D_H's comment) striking problem is the choice of Y.
D H Posted August 11, 2008 Posted August 11, 2008 Could you explain this right here? Which "inverse-square law" are you using? It corresponds to none I've ever learned. Perhaps you could explain what exactly you are doing in these equations. Since I doubt the OP can explain what he is talking about, I will do so. First, the "neutral point". This is the point in space at which the gravitational acceleration toward the Earth is exactly equal but opposite to the gravitatinal acceleration toward the Moon. This point is obviously located along the line between the Earth and Moon. Denoting the distance from the Earth to the Moon as [math]r_m[/math] and the distance from the Earth to this neutral point as [math]r_p[/math], [math]-\,\frac{GM_e}{{r_p}^2} + \frac{GM_m}{(r_m-r_p)^2} = 0[/math] or [math]r_p = r_m\,\frac 1 {1 + \sqrt{\frac{M_m}{M_e}}}[/math] or about 90% of the way to the Moon. Second, the Earth-Moon L1 point. This is the point in space along the Earth-Moon line that is stationary in the rotating Earth-Moon frame. Denoting this point as [math]r_1[/math], [math]-\,\frac{GM_e}{{r_1}^2} + \frac{GM_m}{(r_m-r_1)^2} + r_1\,\frac{G(M_e+M_m)}{{r_m}^3}= 0[/math] or [math]M_e(r_m-r_1)^2({r_m}^3-{r_1}^3) = M_m{r_1}^2({r_m}^3+r_1(r_m-r_1)^2))[/math] This is a fifth order equation. The Earth-Moon L1 point lies about 80% of the way to the Moon, or about 200,000 miles from the Earth. The OP conflated the L1 point with the neutral point and as a result obtained goofy results.
G-Tron Posted August 12, 2008 Author Posted August 12, 2008 Since I doubt the OP can explain what he is talking about, I will do so. First, the "neutral point". This is the point in space at which the gravitational acceleration toward the Earth is exactly equal but opposite to the gravitatinal acceleration toward the Moon. This point is obviously located along the line between the Earth and Moon. Denoting the distance from the Earth to the Moon as [math]r_m[/math] and the distance from the Earth to this neutral point as [math]r_p[/math], [math]-\,\frac{GM_e}{{r_p}^2} + \frac{GM_m}{(r_m-r_p)^2} = 0[/math] or [math]r_p = r_m\,\frac 1 {1 + \sqrt{\frac{M_m}{M_e}}}[/math] or about 90% of the way to the Moon. Second, the Earth-Moon L1 point. This is the point in space along the Earth-Moon line that is stationary in the rotating Earth-Moon frame. Denoting this point as [math]r_1[/math], [math]-\,\frac{GM_e}{{r_1}^2} + \frac{GM_m}{(r_m-r_1)^2} + r_1\,\frac{G(M_e+M_m)}{{r_m}^3}= 0[/math] or [math]M_e(r_m-r_1)^2({r_m}^3-{r_1}^3) = M_m{r_1}^2({r_m}^3+r_1(r_m-r_1)^2))[/math] This is a fifth order equation. The Earth-Moon L1 point lies about 80% of the way to the Moon, or about 200,000 miles from the Earth. The OP conflated the L1 point with the neutral point and as a result obtained goofy results. 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.
MrGamma Posted August 12, 2008 Posted August 12, 2008 (edited) I would appreciate any opinion (with reasons). I am sorry I cannot join you on the math side of things ( still learning ) but I can offer you a place to look in history. Maybe it will give you ideas? The Michelson–Morley experiment was an attempt to find the Aether. This same interference pattern went on to create a laser interferometer gyroscope which can innately detect spin and orientation. Properties which might be gravitational in nature. Perhaps the interference pattern is worth investigating? Edited August 12, 2008 by MrGamma
Pari Spolter Posted August 13, 2008 Posted August 13, 2008 The distance of 43,495 miles from the Moon reported by the Apollo 11 astronauts is the Neutral Point where lunar gravity exerted a force equal to the gravity of the earth, then some 200,000 miles distant. IT IS NOT THE L1 POINT. This is quoted from July 25, 1969 issue of Time Magazine and from page 238 of the History of Rocketry & Space Travel by Wernher von Braun and Frederick I. Ordway III.
D H Posted August 14, 2008 Posted August 14, 2008 Pari, You have based your work on two false hypotheses. One is your interpretation of the meaning of the term "neutral point" in non-scientific magazines such as Time and lay literature such as the History of Rocketry & Space Travel. The other is your misunderstanding of how we have assessed the mass of the Moon. Your so-called "neutral point", the point where the gravitational force toward the Moon is exactly equal to the gravitational force toward the Earth, has little or no physical meaning. A body at this point moving at a speed less than lunar escape velocity will orbit the Moon because the Moon's Hill sphere is about 60,000 km in radius. As another example, consider the "neutral point" between the Earth and the Sun. The Moon itself is well beyond this point (do the math). The Moon still orbits the Earth because the Moon is well inside the Earth's Hill sphere. The location of the Sun-Earth "neutral point" has little physical meaning. The same goes for the Earth-Moon "neutral point". Because this point has little meaning, nobody uses the term "neutral point" to describe this point. On the other hand, the L1 point in a very real sense is the "neutral point" between the Earth and the Moon. What happens to a vehicle with the same orbital angular velocity as the Moon placed near this point depends very much on the vehicle's location. If it is just a bit closer to the Earth than the L1 point it will fall Earthward; just a bit closer to the Moon and it will fall toward the Moon. Scientists and engineers try not use technical jargon when they write for the general public. Because the L1 point is in a very real sense the neutral point between the Earth and the Moon, lay writing refers to this point as the "neutral point" rather than the first collinear libration point. Your other misconception is that this point forms the basis for our assessment of the Moon's mass. It is exactly the other way around. We can calculate the location of this point on the basis of the already known values of the mass of the Moon and the separation between the Earth and the Moon. We had reasonably good estimates of the Moon's mass well before the start of the space age. Columbus not withstanding, we have known the size of the Earth for thousands of years. We determined the distance between the Earth and the Sun on the basis of transits of Venus in 1761 and 1769 and later transits in 1874 and 1882. These values coupled with multiple observations of solar and lunar eclipses gave us estimates of the distance to the Moon. The Moon's orbital period leads to estimates of the mass of the Earth-Moon system. Finally, observations of the apparent monthly nutation of the Sun and planets gives an estimate of the extent of the Earth's orbit around the Earth-Moon center of mass. Put all of this together and you get estimates of both the Earth's and the Moon's mass. We had a good estimate of the Moon's mass over one hundred years ago. By sending vehicles to the Moon we obtained even more refined estimates of the Moon's mass. We never used estimates of the density of the Moon (or of the location of the L1 point or the neutral point) to determine the Moon's mass. It is exactly the other way around. Knowing the Moon's mass let us make estimates of the Moon's density and the location of the L1 point. 1
Pari Spolter Posted August 15, 2008 Posted August 15, 2008 "At a point 43,495 miles from the moon, lunar gravity exerted a force equal to the gravity of the earth, then some 200,000 miles distant." This is quoted from the article "the Moon-A Giant for Mankind," published in Time Magazine, July 25, 1969, page 14. "At a distance of 43,495 miles from the Moon, Apollo 11 passed the so-called 'neutral' point, beyond which the Lunar gravitational field dominated that of Earth." This is quoted from page 238 of the book History of Rocketry & Space Travel by Wernher von Braun and Frederick I. Ordway III, published by Crowell Company, New York in 1969. These are quoted on pages 45 and 46 of the Brian book mentioned above by John Lear. There is no more authoritative person than Wernher von Braun to quote on this subject. On the other hand, the D.H. (Physics expert) does not give any reference, from any public or any scientific source, to support his own contention that the distance of 43,495 miles from the Moon is the L1 point and not the Neutral Point. F = a . A or gravitational force is equal to the product of the acceleration times the area of a circle with radius equal to the semimajor axis of revolution is the correct interpretation of Kepler's third law. Kepler's laws were based on observation. Newton's inclusion of the term product of the masses (m1m2) in his Universal Law was arbitrary and as shown in my book GRAVITATIONAL FORCE OF THE SUN and in the article "New Concepts in Gravitation," published in PHYSICS ESSAYS volume 18, Number 1, pages 37-49, is incorrect. Please also see "Problems with the Gravitationa Constant," published in INFINITE ENERGY, Issue 59, 2005, page 39. Science should be a search for truth based on observations and experimental data. It should not be indoctrination in dogma based on faith. P.S. Please sign your full name and address, so we know who we are communicating with.
insane_alien Posted August 15, 2008 Posted August 15, 2008 Science should be a search for truth based on observations and experimental data all spaceflights so far are experiments AND observations in favour of newton(einstein too if they were carrying a reasonably accurate clock). we win!.
Royston Posted August 15, 2008 Posted August 15, 2008 F = a . A or gravitational force is equal to the product of the acceleration times the area of a circle with radius equal to the semi major axis of revolution is the correct interpretation of Kepler's third law. Kepler's laws were based on observation. Newton's inclusion of the term product of the masses (m1m2) in his Universal Law was arbitrary There's nothing arbitrary about the inclusion of mass, it's an extension of and refinement of Kepler's third law, for example... If you take a body (the moon) of mass [math]m[/math], with a uniform angular speed [math]\omega[/math] and obviously the radius of the orbit [math]r[/math] it will be subject to a force [math]F = mr\omega^2[/math]. So the force from the gravitational pull of the Earth (E= Earth) is... [math]mr\omega^2 = \frac {Gm_E m}{r^2}[/math] So if the moon has an orbital period of [math]T = \frac{2\pi}{\omega}[/math] then... [math]\omega = \frac {2\pi}{T}[/math] Now substituting [math]\omega[/math] into the first equation with [math]\frac {2\pi}{T}[/math], and a bit of algebra... [math]T^2 = \frac{4\pi^2}{Gm_E}r^3[/math] Now compare that with Kepler's third law, so that [math]K[/math] is... [math]K = \frac{4\pi^2}{Gm_E}[/math] Now [math]K[/math] can be determined from observations, and provides the mass of the Earth, plus this can apply to other systems e.g a planet around the Sun. Kepler's laws are successful, because of universal gravitation...i.e an inverse square law. Newton bolstered and refined the equations, and they certainly hold up to observation and experiment. I really don't understand why you're arguing the point. Plus we're not going to advertise our names and addresses on a public forum, for obvious reasons.
Pari Spolter Posted August 15, 2008 Posted August 15, 2008 Kepler's third law does not have any mass in it, and has been verified to very high precision. In the 1960s several artificial satellites were placed in heliocentric orbits by NASA and by the former Soviet Union. We know their exact Mass in kg. Please see Tables III and IV of the paper "New Concepts in Gravitation" in PHYSICS ESSAYS volume 18, Number 1, pages 37-49 for experimental refutation of Newton's second law and Universl Law. There are more Tables on Earth's artificial satellites in my book. You have my name and address and John Lear's name and address on this Forum. So we are not hiding and are not afraid to stand behind our statements.
Royston Posted August 15, 2008 Posted August 15, 2008 Kepler's third law does not have any mass in it, and has been verified to very high precision. That's correct, it has been verified that Kepler's laws don't have mass included in the equations, it has also been verified that the inclusion of mass has provided more precise approximations of orbital motion. I'm a first year (part time) physics student, and even I can pick holes in your argument.
D H Posted August 16, 2008 Posted August 16, 2008 (edited) "At a point 43,495 miles from the moon, lunar gravity exerted a force equal to the gravity of the earth, then some 200,000 miles distant."This is quoted from the article "the Moon-A Giant for Mankind," published in Time Magazine, July 25, 1969, page 14. "At a distance of 43,495 miles from the Moon, Apollo 11 passed the so-called 'neutral' point, beyond which the Lunar gravitational field dominated that of Earth." This is quoted from page 238 of the book History of Rocketry & Space Travel by Wernher von Braun and Frederick I. Ordway III, published by Crowell Company, New York in 1969. These are quoted on pages 45 and 46 of the Brian book mentioned above by John Lear. There is no more authoritative person than Wernher von Braun to quote on this subject. On the other hand, the D.H. (Physics expert) does not give any reference, from any public or any scientific source, to support his own contention that the distance of 43,495 miles from the Moon is the L1 point and not the Neutral Point. First off, this is a red herring because the location of either of these points, the L1 point or your "neutral point", is derived, not measured. Second, your so-called "neutral point" has negligible physical meaning. In comparson, the L1 point has deep physical meaning. For example, NASA currently has a satellite operating at the Sun_Earth L1 point and has contemplated putting a space station at the Earth-Moon L1 point. In a very real sense, the Earth-Moon L1 point is the neutral point between the Earth and the Moon. In the Earth-Moon synodic frame, the second derivative at the L1 point is zero acceleration. Many people refer to the L1 point as the neutral point because in a very real sense it is the neutral point between the Earth and the Moon. Examples (emphasis mine): http://umbra.nascom.nasa.gov/soho/ssu/ssu_stills.html "3. THE UNQUIET SUN--- This sequence of images of the the Sun in ultraviolet light was taken by the Solar and Heliospheric Observatory (SOHO) spacecraft on Feb. 11, 1996 from its unique vantage point at the "L1" gravity neutral point 1 million miles sunward from the Earth." http://www.gsfc.nasa.gov/news-release/releases/1999/99-079.htm "The Triana mission will carry three independent instrument groups mounted on the SMEX-Lite spacecraft. The spacecraft will be launched from the Space Shuttle (as a secondary payload) to provide continuous scientific measurements of the Earth's atmosphere and surface from the Lagrange (L1) point between the Earth and the Sun, the neutral gravity point between the Earth and the Sun." What has happened is that you have misinterpreted the meaning of the term "neutral point" in that lay article in Time magazine and that lay book by Von Braun and Ordway. What is the mass of the Moon? Valado, "Fundamentals of Astrodynamics and Applications", McGraw-Hill, 1997 "Table D-4 Mean Planetary Constants for Epoch J2000 Moon mass = 0.01230 earth mass = 7.3483*1022 kg" NASA determines the moon's mass by observing satellites sent to orbit the Moon. Thus they are measuring the product of the univesal gravitation constant and the Moon's mass. While the uncertainty in G isn't all that great (one part in 105), the uncertainty in the product GMMoon is much reduced: better than one part in 107. Lunar geodesy (sic) typically reports GMMoon rather than MMoon. Ralph B. Roncoli, "Lunar Constants and Models Document", JPL Technical Document D-32296, California Institute of Technology, 2005 Available online at http://ssd.jpl.nasa.gov/dat/lunar_cmd_2005_jpl_d32296.pdf "GMMoon = 4902.801076 km3/s2 (Gravitational parameter from LP150Q) Note: While the LP150Q gravity field represents the current best (complete) lunar gravity field, the GMMoon value from LP150Q does not represent the single best estimate of the gravitational parameter (GM) of the Moon. Reference 2 provides a more precise estimate of the lunar GM -- it is 4902.8000 ± 0.0003 km3/s2. For calculations that require a lunar GM value (as opposed to a complete gravity field), the value from Reference 2 should be used." The 2006 CODATA value for G is 6.67428*10-11 ± 0.00067 m3/kg/s2, or 6.67428*10-20 km3/kg/s2. With this value and the above estimate of GMMoon, the Moon's mass is 7.3458*1022 kg, more-or-less the same as that published by Valado. F = a . A[/i'] or gravitational force is equal to the product of the acceleration times the area of a circle with radius equal to the semimajor axis of revolution is the correct interpretation of Kepler's third law. [math]F=ma[/math] is the definition of force. Your equation reduces to [math]F=ma = aA[/math], or [math]m=A[/math]. That doesn't make a bit of sense and completely disagrees with observation. Your "equation" doesn't even have the right units. Force has units of mass times acceleration, or mass times length per time squared. Acceleration times area has units of length cubed per time squared. You published this nonsense in a book?? Kepler's laws were based on observation. Newton's inclusion of the term product of the masses (m1m2) in his Universal Law was arbitrary Kepler's Laws are purely empirical laws and apply to the orbit of planets around the Sun only. Newton's law of gravitation is much deeper and much more general. It applies to the orbit of the Moon about the Earth, the orbit of a pair of binary stars about each other, and of course to an apple falling to the surface of the Earth. Since Kepler's laws describe an acceleration, the force is obviously proportional to the smaller mass. (F=ma). Newton realized that there is nothing special about the larger mass -- the force should also be proportional to its mass as well as to the mass of the smaller object. That the gravitational force should exhibit some symmetry makes even more sense when you look at it in light of Newton's third law. Forces come in equal but opposite pairs. Note well: Kepler's laws are only approximately valid (the same goes for Newton's law of gravitation; general relativity is an even better model of gravity than Newton's). Kepler's laws work because the planets are many orders of magnitude smaller than is the Sun. The Moon is 1/81 of the Earth's mass. A tiny satellite at the Earth-Moon distance would orbit the Earth a bit slower than does the Moon. Edited August 16, 2008 by swansont Distinguish quotes of postings from quotes from outside sources. swan: fix superscript tag
Pari Spolter Posted August 16, 2008 Posted August 16, 2008 I wish you would first read my book and the article "New Concepts in Gravitation" before criticizing it. In many tables using recent data I have provided evidence that Newton's Second Law and Universal Law are incorrect. I have defined consistent units for Force, for Energy, and for Weight. The Weight of a body is the product of its mass and the acceleration: W = ma. Weight is not equal to Force. Newton himself arrived at the inverse square of the distance by the sole consideration of the orbital speed of the moon, the acceleration due to gravity at the surface of the earth, and the respective distances of the moon and of an object falling near the earth's surface from the center of gravity. Newton did not estimate the masses of the earth and the moon to deduce the reciprocal square of the distance from the center of attraction. Please see the references and quotes in Chapter one of my book. F = a . A is the correct interpretation of Kepler's third law. Kepler's third law has been verified to very high precision, it is not an approximate law. Please see references in my book. I think that you are confusing the Neutral Point with the L1 Point. The references cited in my previous posting makes this clear. Please sign your full name and addresss so we know who we are comminicating with.
Pangloss Posted August 16, 2008 Posted August 16, 2008 Please sign your full name and addresss so we know who we are comminicating with. Negative sir, that is not a requirement of this web site. You are welcome to communicate real personal information about yourself if that is your wish, but nobody is required to do so. Thanks.
swansont Posted August 16, 2008 Posted August 16, 2008 "At a point 43,495 miles from the moon, lunar gravity exerted a force equal to the gravity of the earth, then some 200,000 miles distant."This is quoted from the article "the Moon-A Giant for Mankind," published in Time Magazine, July 25, 1969, page 14. "At a distance of 43,495 miles from the Moon, Apollo 11 passed the so-called 'neutral' point, beyond which the Lunar gravitational field dominated that of Earth." This is quoted from page 238 of the book History of Rocketry & Space Travel by Wernher von Braun and Frederick I. Ordway III, published by Crowell Company, New York in 1969. These are quoted on pages 45 and 46 of the Brian book mentioned above by John Lear. There is no more authoritative person than Wernher von Braun to quote on this subject. On the other hand, the D.H. (Physics expert) does not give any reference, from any public or any scientific source, to support his own contention that the distance of 43,495 miles from the Moon is the L1 point and not the Neutral Point. Time and History of Rocketry & Space Travel aren't scientific sources, either. D H gave an equation. Stating Newton's gravitation law and doing a little math doesn't require a citation, so complaining about the lack of one is a rather hollow complaint. If you think the stated work is wrong, then rebut it. And scientists typically work in SI units. F = a . A or gravitational force is equal to the product of the acceleration times the area of a circle with radius equal to the semimajor axis of revolution is the correct interpretation of Kepler's third law. Kepler's laws were based on observation. Newton's inclusion of the term product of the masses (m1m2) in his Universal Law was arbitrary and as shown in my book GRAVITATIONAL FORCE OF THE SUN and in the article "New Concepts in Gravitation," published in PHYSICS ESSAYS volume 18, Number 1, pages 37-49, is incorrect. Please also see "Problems with the Gravitationa Constant," published in INFINITE ENERGY, Issue 59, 2005, page 39. Not really scientific sources here, either. Science should be a search for truth based on observations and experimental data. It should not be indoctrination in dogma based on faith. A fallacy of distraction. D H has been providing the former, and not the latter. P.S. Please sign your full name and address, so we know who we are communicating with. Focus on the subject matter.
D H Posted August 16, 2008 Posted August 16, 2008 I wish you would first read my book and the article "New Concepts in Gravitation" before criticizing it. You have provided more than enough evidence right here in this thread to convince me that there is no reason to read your book. In many tables using recent data I have provided evidence that Newton's Second Law and Universal Law are incorrect. You have done nothing of the sort. What you have completely misunderstood the meaning of the phrase "neutral point" as used in lay literature and you have completely misunderstood how we assess the mass of the Moon. Your entire analysis is based upon these misunderstandings. I have shown that NASA itself describes the L1 point as the "neutral gravity point" in lay writing. I have shown how scientific organizations assess the mass of the Moon agencies. These scientific organizations do not assess the mass of the Moon by in terms of location of either the L1 point or your so-called "neutral point" because no instrument can be constructed to assess the location of either point directly. Doing so would require constructing an instrument that could directly assess the force of gravity. No instrument that we know of can do this and there are very good reasons to believe that no such instrument could be constructed, period. F = a . A is the correct interpretation of Kepler's third law. No, it is not. Kepler's third law has been verified to very high precision, it is not an approximate law. Yes, it is, just as is every other physical law, including Newton's laws. Kepler's laws are only approximately correct even in our own solar system. Kepler's laws explicitly state that the planets have elliptical orbits. This is not quite true (google the phrase anomalistic precession). Accounting form the gravitational interactions among the planets explains most of the anomalistic precession. Kepler's laws give an incorrect answer for Jupiter's orbital period because Kepler's laws implicitly assume that the planets have masses many orders of magnitude smaller than the Sun's mass. Jupiter's mass is 1/1048 solar masses, large enough to make Jupiter's orbit deviate from that predicted by Kepler's laws. Newton's laws themselves have been proven to be only approximations that fail in regimes far removed from day-to-day experience. Newton's laws yield incorrect results when applied in very small domains (quantum mechanics provides more accurate models), very fast domains (special relativity), or very large domains (general relativity). You, on the other hand, are arguing that Newton's laws are wrong, period -- that they don't even apply in the day-to-day world. This goes counter to an incredible number of experiments performed from before Newton's time to the present day. It goes counter to almost all of modern day engineering, which are largely applied Newtonian mechanics. We use Newton's laws when we build bridges, cars, buildings, and spacecraft, just to name a few things. I think that you are confusing the Neutral Point with the L1 Point. The references cited in my previous posting makes this clear. The references you cited simply talked about a "neutral point". It is you who has misinterpreted this as the point where the gravitational force toward the Moon equals the force toward the Earth. In lay terms, the L1 point is the neutral point. I have provided references that specifically describe the L1 point as the neutral point. The distances you cited are exactly the distances between the Earth (or Moon) and L1 point. Nobody but you thinks of the point where the gravitational forces are equal but opposite as the neutral point because this point is physically meaningless. Please sign your full name and addresss so we know who we are comminicating with. Thanks, but no thanks. Please stick to the topic of discussion. My full name and address are not part of that discussion. These, along with my phone number and social security number, are things I do not give out freely on the internet.
Pari Spolter Posted August 19, 2008 Posted August 19, 2008 Dear Snail, There is no experimental veification for your statement that it has also been verified that the inclusion of mass has provided more precise approximations of orbital motion. Please give me a reference to support your statement. Dear D.H. (Physics Expert), Gravitational Force is equal to the product of the acceleration times the area of a circle with radius equal to the semimajor axis of revolution (F = a . A) is the correct interpretation of Kepler's third law. Kepler's third law is the law of gravitation and has been verified to very high precision. Please see the references in my book. What you call deviations from Kepler's law are due to perturbations by other bodies. In many tables using recent data, I have presented the experimental refutation of Newton's Second Law and Universal Law. What works in Celestial Mechanics is Kepler's third law. And what work in modern day engineering are the equations for Energy (not Newton's Force.) The references you have given show that NASA has used the term "gravity neutral point" for Legendre point. You have not provided any reference to state that the distace of 43,495 miles from the moon reported by the Apollo 11 astronauts was the Legendre point . On the other hand, the references cited in my previous posting on this Forum, state in no uncertain term that on July 19, 1969, at a distance of 43,495 miles from the moon, Apollo 11 passed the point where the acceleration due to the gravitational force of the earth became equal to the acceleration due to the gravitational force of the moon. The attachment is from page 371 of ORBITAL MOTION by A E Roy, Professor of Astronomy at the University of Glasgow, published by the Institute of Physics in 1988. Please note that the italic for Neutral Point is Roy's not mine. So obviously the Neutral Point is not the radius of the Moon's sphere of influence. When I was reading all these articles in the scientific journals, with most of them now available on the internet, I did not realize that the authors names and addresses printed on the papers were putting them in danger of Identity Theft. You call yourself 'Physics Expert,' so obviously people are interested in your biography. I will be happy to send you a complimentary copy of my book and the articles, if you give me your address (or any address.) My email is: orbpublishing@msn.com My address is: Orb Publishing Company 11862 Balboa Boulevard, # 182 Granada Hills, CA 91344-2753 My biography is available in CONTEMPORARY AUTHORS, Volume 163. Roy.pdf
swansont Posted August 19, 2008 Posted August 19, 2008 I thought Kepler's third law was that the square of the orbital period is proportional to the cube of the semimajor axis. 43,495 miles from the moon Your source mentions an equation (6.11) that is not in your attachment. How about including the derivation of that, so we can see exactly what the author is talking about and how r'A and rM are being used. (it also gives 66190 km = 41100 miles)
D H Posted August 20, 2008 Posted August 20, 2008 Dear D.H. (Physics Expert) Stop that! "Physics Expert" is not a title I gave myself. It is a title conferred upon me by the moderators of this forum. Gravitational Force is equal to the product of the acceleration times the area of a circle with radius equal to the semimajor axis of revolution (F = a . A) is the correct interpretation of Kepler's third law. Politicians can turn a complete nonsense into the "truth" by spouting the same nonsense over and over. Fortunately, that is not how science works. Spouting scientific nonsense over and over is just scientific nonsense spouted over and over. This equation of yours doesn't make one bit of sense because it doesn't even have the right units. Acceleration times area has units of length3/time. Force has units of mass*length/time2by definition. Kepler's third law is the law of gravitation and has been verified to very high precision. Let's see. If Kepler's Laws are true the ratio [math]P^2/a^3[/math] should be constant. Let's see: Mercury.[math]P=0.240846\,\text{yr}\,\, a=0.387098\,\text{AU}\;\Rightarrow P^2/a^3=1.00004\,\text{yr}^2/\text{AU}^3[/math] Earth.[math]P=1.0000175\,\text{yr}\,\, a=1.0000001124\,\text{AU}\;\Rightarrow P^2/a^3=1.00004\,\text{yr}^2/\text{AU}^3[/math] Mars.[math]P=1.8808\,\text{yr}\,\, a=1.523679\,\text{AU}\;\Rightarrow P^2/a^3=1.00001\,\text{yr}^2/\text{AU}^3[/math] Saturn.[math]P=29.657296\,\text{yr}\,\, a=9.58201720\,\text{AU}\;\Rightarrow P^2/a^3=0.999751\,\text{yr}^2/\text{AU}^3[/math] Jupiter.[math]P=11.85920\,\text{yr}\,\, a=5.204267\,\text{AU}\;\Rightarrow P^2/a^3=0.99777\,\text{yr}^2/\text{AU}^3[/math] Whoa! The supposedly constant ratio is only approximately constant. Moreover, the value for Jupiter differs significantly with respect to the less massive planets. A bit more than two decimal places of accuracy was extremely good four hundred years ago. It is not so good nowadays. These measurements of orbital period and semimajor axis falsify Kepler's Laws as the ratio P2/a3 is not a constant. Please see the references in my book. What you call deviations from Kepler's law are due to perturbations by other bodies. You are contradicting yourself. There are no perturbations by other bodies according to Kepler's Laws. Perturbations do however arise in Newton's law of gravitation (and in general relativity, which surpassed Newtonian theory almost 100 years ago). The references you have given show that NASA has used the term "gravity neutral point" for Legendre point. You have not provided any reference to state that the distace of 43,495 miles from the moon reported by the Apollo 11 astronauts was the Legendre point. I stand corrected. That distance is not the L1 point. It is the point when the Apollo 11 entered the Moon's "sphere of influence". Let's take a look at those numbers in Time magazine: From http://www.time.com/time/magazine/article/0,9171,901102-1,00.html (page 6), At a point 43,495 miles from the moon, lunar gravity exerted a force equal to the gravity of the earth, then some 200,000 miles distant. From this article, it appears the Earth and Moon are 243,495 miles apart. This differs from the 238,855 miles distance cited in literature. Did Times make a typo? No. The Moon isn't in a circular orbit. The Moon was near apogee on July 19, 1969. From http://history.nasa.gov/ap11-35ann/apollo11_log/log.htm 4:40 p.m.- One of the clearest television transmissions ever sent from space is begun, with the spacecraft 175,000 nautical miles from Earth and 48,000 from the Moon. It lasts an hour and 36 minutes. 11:12 p.m.- Velocity of spacecraft has slowed to 2,990 ft. per second just before entering the Moon's sphere of influence at a point 33,823 nautical miles away from it. 223,000 nautical miles is 257,000 miles or 413,000 km (to three digits). Using that number and Roy's sphere of influence equation, the sphere of influence on that day was 44,000 miles. End of conundrum. You call yourself 'Physics Expert,' so obviously people are interested in your biography. As I mentioned before, I don't call myself Physics Expert. This board endowed me with that title. My biography is available in CONTEMPORARY AUTHORS, Volume 163. You've already listed where you have published articles on physics: A borderline journal (Physics Essays) that is no longer indexed by the Web of Science abstract and citation database, and a completely quack journal (Infinite Energy) that has never been indexed. Think about it this way: If your conjecture is correct, you have seen something that completely flew over the heads of Newton, Euler, Lagrange, Hamilton, Poincaré, Einstein, and a host of lesser scientists. If your conjecture is correct, the US never landed people and the USSR never landed automated spacecraft on the Moon, and every space agency in the world is lying about the Moon's gravity. If your conjecture is correct there are thousands of people alive today who know that NASA, Rosaviakosmos, ESA, JAXA, CNSA, and others have created the greatest fiction ever written. Think about it.
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