Hi all please read Part I first.
In Part I it was shown by me how the AVERAGE speed of Neptune around The Sun was in the ratio of "the cube root of 1/2 the diagonal of The Great Pyramid (622.00 / 2) divided by 100 or 3.11 when we use the perfect measurements to show Pi of 439.82 for the base and 280.00 for the height . So I had a thought maybe ALL the planets were at a cube root of a number that we would find significance with and so I tried another planet and so I tried the one furthest from The Sun and that would be Pluto at about 5.874,000,000 kilometers from The Sun on it's semi major axis. I wondered could this give me a number that was important so I started trying "things" and came up with this marvelous solution. In books and online the suggested average speed of Pluto is anywhere from 4.67 to 4.7 ( I would guess that the latter was rounded) and that would give Pluto a ratio of Earth's speed of 29.78 / 4.67 or 6.37687366167 to 1. So in keeping with our theme I decided to cube this value to see what I would get and arrived at this number: 6.37687366167 cubed is 259.31249186591 but I have learned not to trust measurements online and so going to my elliptical measuring site I found out that the actual distance traveled by Pluto in a year is 36,348,870,300 kilometers. Now to calculate average speed we simply need to know how long a year is and we find two values for the year on Pluto. The first is "the sidereal orbital period" and this is 90,465 days while the second is "the tropical orbital period" and this is 90,588 days/ so now all we do is divide 36,348,870,300 kilometers by 90,588 / 24 (hours) / 60 (minutes) / 60 (seconds) and we get 4.6441528388 if we use 90,465 we get 4.65046722336. If we use the latter we get 29.78 / 4.65046722336 or 6.4036576476490935125430844545394 and this number cubed is 262.59371 (very close to the height of The Great Pyramid now and many feel that this is as high as it got) but the real numbers I am looking at are 6.3972269796157684960801651350896 and 4.6551420005717308801570872085312 for 29.78 / 6.3972269796157684960801651350896 = 4.6551420005717308801570872085312 okay be why 6.3972269796157684960801651350896 well this number cubed is Phi squared x 100 or 261.803398875.
Using these values we get 4.6551420005717308801570872085312 x 90,465 x 24 x 60 x 60 = 36,385,409,181.5 and the accepted value of 36,348,870,300 divided by 36,385,409,181.5 = 0.998996
So is the ratio of Pluto's average orbital speed to that of Earth's in the ratio of 1 to the cube root of Phi squared x 100 ?
Well what do you think ?
I was about to close this posting when I decided to see what the actual radius would be with our values for Pluto and got these values. For 36,348,870,300 if this was a perfect circle the radius would be 5,785,102,384 and if we used the number we got when we used Phi squared x 100 cubed then cube rooted we get 36,385,409,181.5 as the perfect circle and radius would be ... 5,790,917,728 and if we divide by 100 we get 57,909,177.28 and we have to be amazed that it is almost precisely the semi major axis of Mercury of 57,909,050 or an accuracy rating this time of 0.9999978 or about 1 in 22/10 millionths ! (22/10,000,000). So in other words we now have this solution for the orbital distance of Pluto. Actually quite easy to remember and quite brilliant as well. We simply use the semi major axis distance of Mercury (and how many times have I tried to tell you that Mercury is the messenger of the information of the gods ?) as a radius and simply calculate for circumference and we get 57,909,050 x 2 x Pi x 100 to get 36,385,329,211 and Phi squared x 100 cube rooted yields 36,385,409,182 simply amazing ... don't you think ?
But there is still a bit more and I will continue with a solution for Eris and an explanation (maybe) for the outer reaches of our solar system and The Oort Cloud in Part III of Cubes and roots and squares, oh my !