Sir Isaac Newton, could he be mistaken?
For years now I have been troubled by the possibility that Isaac Newton’s Gravitation formula may be just a little bit from perfect when it gives us the Mass of objects that have moons or satellites. For the Sun the planets can be considered satellites. Because Newton has always been one of my hero’s, and because of the great respect the world holds for him, I have been reluctant to go public with my idea on this subject. Note, and make no mistake, I do not contend that any part of Newton’s Gravitation formulas are in error. This issue deals with a fine point concerning the results his formula provides for the Mass of objects when considered to be the Mass of the object for the value of the object and the result provided being the “working Mass” of the object and all of its orbiting companions.
This may be a little too unique for me to get into it without some up front explanation. Suppose you were the project engineer for a large tract building operation of 1,000 homes. There will be a lot of traffic and the law requires that you water down the dirt and sand so there will be a minimum of airborne particles. The usual way to do this is to have water trucks go over the areas routinely to wet the ground so the dust will not be carried in the wind. You are concerned with ground surfaces and paved surfaces. You cannot risk cracking the paved surfaces or sinking the truck into the dirt. When you consider the weights involved the truck is advertised with their delivered weight. You must know what the “working weight” will be. The truck plus the 500 gallons of water weight will be the working weight when the truck has a full load. I know that is a long-winded hypothetical, but required.
I will be talking about what I contend is the difference between the actual Weight Mass (AWM) for the Sun, for instance, and the “Working Weight Mass” (WWM) of the Sun. I am doing this because I contend that these two conditions are entirely different. I contend that when an Astronomer calculates the Mass of the Sun using Newton’s formulas he is getting the Mass of the Sun and all of the other Mass in the Solar System that is in orbit around the Sun. You could say he is measuring the Mass of the entire Solar System. Why do I believe that, you ask? I have decided years ago that the Solar System is a unit with the focus at the Sun being like an axle for the large wheel that has all the weight of the wheel pulling on the Sun. Because there is only Newton’s work by which to measure the Mass of the Solar Objects there has not been a means to challenge the published results. In the past I came up with another, more simplified method, by which to get a Mass ratio without the use of Newton’s formulas. The results are different scales or units but they do provide an accurate ratio for making comparisons.
I mentioned the Miles-Mass (MM) method and how I came up with it in my book, Surfing the Solar System, published in 2007. A long time before I started to write that book I had worked it out but when I wrote the book I did not have enough confidence in the idea to argue for it in the book. I went back over some of my old notes today and I found some pages of notes dated Feb. 28, 1987, stating ” Workup on Masses correction”. These are handwritten pages with data collected from earlier notes that every so often I would go back over and extract the useful parts for a more logical record. I have lightly discussed my idea for finding the Miles-Mass for the Sun and the planets in other articles that I have written for the American Chronicle , one of which was “the Solar Mechanics II”. Johannes Kepler and Isaac Newton taught us the basic principal that the measure of the Mass of a planet is the ability of a planet to orbit a satellite in a specific period, or at a particular velocity. All things being equal with the orbit radii being the same the more massive the orbited object the greater the velocity of the orbiting object. This is indirectly Included in Newton’s formula as well. While on the topic of orbiting bodies I want to mention something that is important but not normally discussed relating to the Mass of the orbiting object. I contend that the Mass of the object doing the orbiting has nothing to do with the size of the orbit or the velocity of the orbit. The nature of the orbit in all regards is dependent on the Mass of the orbited object with no contribution by the orbiter. No doubt the academics were aware of this but I was not until I got started on my own investigation of the Solar System mechanics.
I want to repeat my contention.
Every Solar System in every Galaxy has a Sun that when measured for WM by Newton’s formula will actually give you the WM for that Sun that will include all of the Mass in orbit around that Sun. The formula doe not segregate the AWM of the Sun from the Mass of the balance of the objects in the Solar System. When measuring planets with Newton’s formula you will get the WM for the planet that will include the WM of all the satellites, if any, in orbit around that planet. These planet systems are units that must be separated someway to see what the Mass of the orbited object would be without any of the satellites. This does not appear to be possible using Newton’s formula but I believe it is possible using my Miles-Mass method. You could think of the Solar System like a large mobile with everything hanging on the Sun and thereafter each planet group identified with its satellites as a group, individually hanging from the Sun, etc. When a scientist wants to measure the effect of planets gravitation on a passing rocket he can use Newton’s formula because he will get the Working WM for the object and that is the measure of the effect he will want to include in his calculations.
I have outlined my contention so now it is time to show the evidence I rely on to support my contention. You may be wondering why this issue is important if in fact we are getting the proper Mass value for our space calculations? The most important factor is that we have been over estimating the Mass of out Solar System and of all the other Solar Systems in this and all other Galaxies. This happens because we use the exaggerated Mass value for the Sun and then add the Mass of the other objects to the Sun’s value exaggerating the final result even more. There is probably less Mass around than thought.
I use Weight-Mass to distinguish Newton’s method from my Miles-Mass approach. There are two Web Sites worth noting.
http://www.nasa.gov/worldbook/sun_worldbook.html wherein they recite that the Sun contains over 99% of the Mass in the Solar System.
http://www.jpl.nasa.gov/solar_system/sun_index.html wherein they recite the Sun contains over 99.8% of the Solar System’s Mass.
There was no mention as to how they discovered that result but the result does not surprise me because I contend they’re measure of the Sun’s Mass includes 100% of the Solar System’s Mass. No one knows, not NASA and not JPL, what Mass remains in the far reaches of the Solar System that has not been discovered. I hope to provide an estimate of that Mass by separating the actual Mass of the Sun so we know what is left is the measure of the System’s remaining Mass.
To appreciate the basis for my contention you must understand what I mean by my Miles-Mass method and how I arrived at it. We know that the very basic attribute of an objects Mass is its ability to orbit other objects at specific velocities depending on the proximity of the orbited object to the object being orbited. Orbits with shortest radius from the central object orbit the fastest. I do not like to use kilometers for any of my measurements. Also when I first started on this project I discovered quickly that miles worked a lot better with Solar System mechanics than kilometers. My first curious discovery was by taking the circumference of the Sun (2,714,342.4 miles) divided by the equatorial diameter of the Earth (7,926.6 miles) providing 342.43. The square root of 342.43 turns out to be 18.505 and that turned out to be the Earths orbital velocity in miles per second as it makes its trip around the Sun. You can make the calculations in kilometers but the final result in Km is worthless.
As a result of a lot of trial and error investigation I eventually moved on to the Sun/Earth relationship thinking that the square of the true miles per second orbit velocity might work backwards to give me a value for the Sun’s ability to accelerate orbiting objects. The Earth’s mean orbit radius is about 92,961,440 miles. That times 342.43 is 31,832,785,899.2. Take note that I had been checking the various orbits and orbit velocities and came to see quite clearly that there was a common number that applied to all of the planets consistent with the Earth derived number. Getting the actual orbital velocity of each planet squared when the mean orbit radius was divided into my new Solar System number mathematically proved this. This was far too accurate to be coincidence. I eventually settled on a minimum Solar System circumference of 200,000,000,000.00 miles for a minimum diameter of 63,661,828,367.7 and a minimum Solar System radius of 31,830,914,183.8 miles. This would be the point in space where an object in orbit around the Sun will travel at a velocity of 1 mile per second. Beyond this point the velocities will be a fraction of one. I will provide a graph using my old data and calculations for the listing but pointing out that I have improved the numbers by careful review with no important changes. I will use 31,830,914,183.8 as my Solar #.
Planet====Mean orbit radius===vel^2=== Sun MM #========Vel mps
The reason I contend that the Miles-Mass results do not include the satellites can be found above. Note that Mars and Jupiter provide a very similar result even though their actual Mass difference is very large. I also found this to be true when doing the individual planet Miles-Mass calculations. I double-checked Neptune with no change.
The total Sun MM of 286,380,261,990 / 9 equals 31,820,029,110. It is my view that the S# of 31,830,914,183.8 is valid even though the average is a little less. I say that because it will always provide an accurate ovmps when divided by the mean orbit radius. This S# is my contention of the Miles-Mass of the Sun when expressed in Miles as opposed to WM in Kg. The importance here is that I contend the S# represents the Mass of the Sun in Miles radius without including all of the remainder of the Mass in the Solar System. Following up on this approach I decided to try the same method on each of the planets to obtain what I hoped would be the Miles-Mass value for each planet without including the Mass of the satellites orbiting those planets. To do this I had to translate the current published data for the satellite periods to the appropriate orbit velocity expressed in miles per second like I did for the planets. When I started this I only used the satellites that were listed with data at the time. That was years ago so it was only the largest known to us then. After all this time many more have been listed with some useful data. Saturn has about 60 satellites listed now just to give you one example.
To double-check my earlier results I went back and included all the currently known satellites for all of the planets. I included the basic data I required for orbit radius in miles and my adjustment of the period for each in orbit velocity miles per second. This was run in a BASIC computer program I wrote to run the calculations and proved to be revealing in that the results were consistent with my original figures. I will not set forth all of the many objects tested. I must add that for Saturn’s satellites, for one example, the results were highly consistent for most of the satellites and even more consistent when the orbit velocity was adjusted no more then about .001 miles a second. Which renews the need for accurate data to get the best results.
So far I have provided an alternative means to get a Mass ratio that is not created by the Newton formulas. This means I can now compare Newton’s results, as published by academics, to an alternative result. I want to stress that I do not see any conflict with Newton’s WM and my MM based upon his expression as weight and mine expressed as miles because I am using the appropriate ratios and not the values.
For completeness I will provide a listing of the planets MM values. The first time I did this I used the number of satellites published at that time. The details are in my book and there is no reason to repeat the details here. I will explain the Earth/Moon result as an example and you will know how it works. First I use the published data for the Moon’s orbital period and reduce it to the miles per second velocity. It turns out the Moon travels about .63679 miles per second in its orbit around the Earth. The mean orbit radius is about 239,265 miles so I square the velocity to get .405503. That .405503 times the 239,265 miles radius give me 97022 as what I contend is the Miles-Mass value for the Earth. The process is repeated for each planet/satellite unit and the only difference was that if a planet had more than one satellite I did it for all of them and averaged the result. Mercury and Venus have no satellites so I used the WM ratio as it compared to the Earth. This is the conclusion.
Planet====MM value===/ Earth====WM value===WM/Earth=#2- #1
Mercury===5,413.8==== .0558====3.3e10+23==== .0058=== 0
Venus=====79,073==== .8150====4.87e10+24=== .8150=== 0
Earth======97,022====1.0======5.97e10+24=== 1.0===== 0
Mars======10,139===== .1063===6.42e10+23=== .1074=== .0011
For Mercury and Venus I am simply using the published values as they relate to Earth related ratios for MM, MM%, WM and WM%.
In every instance the WM planet values result in a ratio suggesting greater mass then the ratios for MM. The only value of this chart is to show the prospect that each WM for each planet when divided into the WM for the Sun will show the Sun as an exaggerated Mass when compared to the MM values between the same objects. The last number in each line of the listing is not intended to reflect the actual difference between the MM and the WM. It is simply a guideline. There are so many other potential considerations I can only claim here that the chart shows a measurable variance but not necessarily the degree of the variance between MM and WM. The chart is showing the relative value of Earth’s WM divided by Mars WM is more than the MM that I am contending represents the values for the planets without the inclusion of the satellites and the WM values represent the Unit values of planet and satellites. The Sun WM is used as 1.99D+30.
This time I will compare the WM value of the Sun divided by the WM value of the planets and the MM values on the same basis. Lastly the chart will show the value for each planet after deducting the MM values from the WM values. How does the planet value differ?
Planet==SWM val./planet WM ==SMM val./planet MM val. = Result ============#1=====================#2================
Venus=====408,624.2299794 402,551.3568576= 6,072.87312178
Earth======333,333.3333333 328,079.3558390= 5,253.97749432
Mars======3,099,688.473520 3,084,689.91784= 14,998.55567818
Jupiter====1,047.9199578185 1,047.82715931= .0292804740625
Saturn=====3,499.824129440 3.491.36670739= 8.457422049683
Uranus=====22,979.2147806 23,196.94842621= -217.733645614
Neptune====19,320.3883495 19,131.75377000= 188.634579506
I know that some academics dislike the use of long numbers but for charts like this one I like to see as much of the result as I can and I am not an academic. There is something mysterious about using miles in these measurements even though I am unable to reveal it. In general the last value is always larger for the WM comparison than for the MM and I contend that supports my viewpoint. For some reason the largest planets show the least change and I contend that is because the satellites of the largest planets are a very small part of the total WM of Jupiter and Saturn. The fact that Uranus shows a negative result is not a surprise to me because in my original efforts over the years I came to believe that the WM calculations as well as the MM calculations for Uranus were not accurate and I mentioned that point in my book. I will not discuss that issue here because it will not help to resolve the primary issue.
Interpretation of data can be more difficult than devising ways of breaking it down. To me, this last chart suggests that each planets WM value is exaggerated to the extent the excess is shown in the results. Mercury has the largest disparity and that is consistent as compared to the least disparity of the largest planets.
It is not possible to go through all of the satellites and total up their Mass because Newton’s formula is essentially for a two-body system as is my formula. This means that most of the satellites around the larger planets have no calculated Mass. In some cases you will find a source on the Web where in a Mass value is offered for a satellite in the 2nd or 3rd orbit position. I have a possible rough method of doing an estimate based on size but that would not satisfy anyone.
The published WM for the Sun is 1.99e10+30 Kg and if we seek the relative Mass of the Sun divided by the Earth’s Mass we will find that not all the published data agrees. The one published figure is 330,000 and another is 332,948. Because 332,948 are close to my double-checking effort I will use that result. This is a case of simple subtraction. If we have the correct Mass values there should be no issue.
Now we want to get the MM relative values. I prefer to work backwards from the minimum circumference at times while getting relative value of the Earth’s MM compared to the Sun’s MM. This is a lot easier due to there being whole numbers. ((200000000000/6.2832)/97022) will show the relative value of the Sun’s MM to Earths MM is 328,079. From my standpoint this supports exaggeration in the WM values now in use. Think of it as the Sun as 332,948 times Earth WM while the same Sun is 328,079 MM times the Earth. 332.948 328,079 = 4,869 Earth Mass size planets! That is less than the chart showing 5,253 above. It may help to look at this a different way using the planet Jupiter for a comparison. Earth WM of ((5.97e+24 * 4.869)/1.899e+27) = 15.31. We can reduce that to 14 by eliminating what we know about the other stuff and settle for that. Consider the prospect of 14 Jupiter Mass size objects out there that we never considered because we have not tried to separate the Mass values we have been working with.
I contend that the Solar System has a lot of Mass out there that we have not considered and that we have yet to discover. Assuming for discussion that my contention has merit I can go on changing the WM values for the Sun and the planets as if they had no satellites. For this purpose I will use the published Mass for the Sun at 1.99D+30 = SM.
I was surprised to see a uniform progression for WM divided by MM because it looked like I was dividing oranges by apples. Here it is.
Planet======WM value==== / MM value==== Result
Average of the above results = 6.200511977D+19, with out the Sun.
This could be a meaningless collection of numbers unless there is a way to make some useful interpretation about the current issue. I think the fairly uniform result obtained by dividing the Newton WM figures by my MM figures demonstrates that there is value in the use of the MM approach. I never would have expected these results. The fact that Venus and Earth come in very close and that Venus has no satellite concerned me at first because if the WM value includes the satellite I would have expected the Earth result to be larger than the result I got for Venus because Earth’s Moon would otherwise increase the WM value for the Earth. Uranus shows the largest result suggesting that either my MM value is too small or the WM value is too large. In my book I explained the trouble I had evaluating Uranus, It does not mean anything for this article, While this part of my effort may seem to you to be unorthodox I contend it is worthwhile supporting the MM method under circumstances wherein I have the burden of proof and there are limited means by which I can present my proof. I have carefully considered the relationship between the values recited above without finding anything to help my contention. However I am not skilled with numbers and many people are so I included this chart for the unique surprise it offered and in the hope that some person more skilled than I will find some significance in the results, even if it serves to challenge my contentions. If all of the data relied upon were perfect the result should be identical for each object. To compare results I must eliminate the +19.
Exponential numbers are not required for my comparison. By doing away with the D+19 I will shorten the effort while still providing some comparisons. I will try an example of dividing the Sun’s derived result by each of the results for the planets.
Planet=====Derived Sun value===/planet derived value==== Result
This is mainly redundant confirming what we already know that the small planet Mercury provides the largest comparison while Jupiter provides the smallest, all of which I contend is as expected.
Uranus is consistently providing an unlikely and improbable result. I have no other ideas for attempting to prove my basic contention. It is my conclusion that the weight mass values we get using Isaac Newton’s formula for the determination exaggerates the mass of the Sun by including the mass of the entire Solar System.
This idea will not be received with favor but I felt compelled to get it published. Possibly some knowledgeable person will show why my contention is incorrect. James J. Wood, Sr. all rights reserved.
Surface Gravity; Isaac Newton, follow up, was he mistaken?
You may not be familiar with the use of the surface gravity of a Star, a planet or a satellite, but it is one of the measurements that the academics decipher for us when they are working up the data. Surface gravity, like the words imply, relate to the effect of gravitation at the surface of a spherical body. It is different from the basic method used to measure the overall gravitation of a planet. You might think that the surface gravity of a planet would simply be a value directly related to the measured gravitation of the object as discovered by Sir Isaac Newton; Nope.
I have been concerned from the start that my Miles-Mass formula, the one I used to argue that Newton’s formula may slightly mistaken was itself not set in concrete and questionable. I did try my scheme on the determination of surface gravity using my Miles-Mass values.
I had wondered about the relationship between the surface gravity of a planet and the way we measure the extent of a planet’s gravitation. Having already created what I call my Miles-Mass measure of a planets gravitation based on the ability of a planet to orbit a satellite at a specific velocity I wanted to see if I could find a way to determine surface gravity using the Miles-Mass values. This was mainly a trial and error project that I had no reason to expect would be in any way productive. If you look up the formula for making this calculation normally you will find that it is a little mathematically complicated. Check out Wikipedia:
I will show the formula I worked out for finding a planets surface gravity by using my Miles-Mass method as opposed to using Newton’s laws of gravitation. First of all I had to find the Miles-Mass value for each planet, something I had already done for other comparisons. The method is set forth in my article” Isaac Newton, was he slightly mistaken?” so I will not repeat it here. I rely on the published data for the values of surface gravity and apply my method for comparison. The data as published tends to focus on the Earth and relates every thing else to the Earth values. To get individual results the method must first be interpreted to show the actual values for the other planets. The published Earth value is 32.16 feet per second per second, meaning for each second of fall another 32.16 feet velocity per second is added.
Planet-published-Feet per second-Planet Radius
Mercury-.38 of Earth-times 32.16 = 12.22 -1,515.5 miles
Venus-.91 of Earth-times 32.16 = 29.26 -3,759.5 miles
Earth-1. -times 32.16 = 32.16 -3,963.3
Moon-.17 of Earth-times 32.16 = 5.46 -1,080.0 miles
Mars-.38 of Earth-times 32.16 = 12.22 -2,111.5 miles
Jupiter-2.5 of Earth-times 32.16 = 80.40 -44,678.7 miles
Saturn-1.07 of Earth-times 32.16 = 34.41 -37,284.0 miles
Uranus-.88 of Earth-times 32.16 = 28.30 -16,246.5 miles
Neptune-1.14 of Earth-times 32.16 = 36.66 -15,379.6 miles
The Sun-28. of Earth-times 32,16 = 900.48-432,000 miles
None of this data is of my creation except the simple conversion.
Very often the acceleration due to gravity, using Earth as an example, will be shown as 32.16 ft./s^2. This question was on my mind at the time because I was a little concerned that my Miles-Mass values may not have been realistic and I was looking for ways to seek confirmation of the accuracy. One means was provided in the comparison to the Weight-Mass per Newton. If I could come up with a new way to get an accurate result that would be encouraging. By a long series of trial and error efforts I settled on a formula that used my Miles-Mass value times the feet in a mile at 5,280 divided by the planet radius in miles squared. A preliminary trial for Earth was so close I ran it out on all the planets and the Sun.
Planet-MM-X-Mile feet-result-/-Radius^2-FPS-Pub FPS
Mercury-5,413.82-X-5,280-= 28,584,969 / 1,515.5^2-12.44-12.22
Venus-79.072.94-X-5,280-=417,505,070 / 3,759.5^2-29.12-29.26
Earth-97,022.0-X-5,280-=512,276,160 / 3,963.3^2-32.61-32.16
Moon-1,193.4-X-5,280-=6,301,152.0 / 1,080.0^2-5.402-5.46
Mars-10,319.-X-5,280-=54,484,320. / 2.111.5^2-12.22-12.22
Jupiter-30,378,021-X5,280-=160,395,950,080 / 44.678^2-80.35-80.40
Saturn-9,118,897X5,280-=4,814,816,160 / 37,284^2-34.63-34.41
Uranus-1,372,202.7-X-5.280-=7,245,230,256 / 16,246.5^2-27.45-28.30
Neptune-1,663,774.0-X-5,280-=8,784,726,720 / 15,379.6^2-37.13-36.66
Sun-31830914183.8 X 5280-=168,067,232,480 / 432000^2-900.56-900.48
Because Mercury and Venus have no satellites and I did not have the orbit details for the spacecraft we sent there I used the Weight-Mass ratios for Earth to arrive at a Miles-Mass value for each of them by multiplying the Earth’s MM by their percent of Earth that is published as .0558 for Mercury and .8159 for Venus.
You may be quick to point out that my method does not produce the very same results that the published results show. However you should also note that the overall results are remarkable when considered in the light of the simple formula that was used to obtain them, and that the method serves to prove that my use of the Miles-Mass approach to measuring Mass is valid. You should also keep in mind that my method gives no consideration to the potential density of any object included in the chart. For my personal education I ran out the numbers in reverse to see how much different my Miles-Mass number must be to get the results like the published figures. For the Sun dividing my MM by the newly worked MM will give you a result that the new relates to the old like 0.99990451814. I did not feel it was worth the creation of another chart but if you are curious then go for it. I am aware that some fine-tuning could improve the MM results.
One last point, on the issue of whether my Miles-Mass value does actually exclude the value of the planets satellites, I think this is another step in the proof that I seek. It is also remotely possible that the published values are not the best rendition of the facts. In a program like this I must take small strides and try to keep all my details in line logically. For now this particular article was published solely on the comparative ease we can now determine Surface Gravity my way. I have a series of articles to share using my own unique methods of investigating the mechanics of the System, including the measure of the speed of light and stuff you may not be aware of.
Keep an open mind and expand you horizon. www.egpok.com
James J. Wood, Sr. http://www.SurfingtheSolarSystem.com