
deriving Jupiter's 15 pulsar pulses per 10 hours Chapt16.15 Gravity Cells #1485 ATOM TOTALITY 5th ed
Posted:
Apr 10, 2013 4:47 PM


On Apr 6, 12:33 am, Archimedes Plutonium <plutonium.archime...@gmail.com> wrote in #1473: > This math is bothering me, so let me try to nail it. > > We have the problem of the force of gravity. There is no problem if > the Sun had 0 speed, but the Sun is 220 km/sec, far faster than any of > its planets and so there is a huge problem. > > To solve the problem, we need the force of gravity to be EMgravity > with Space itself having been formed by gravity and then, that newly > formed space to have a spin. > > So, now, how to work out the spin. > > We go to the planet most close to the Sun, which is Mercury. > > Now there would be no problem if the planets had their orbital speeds > plus tacked on a 220km/sec. > > Thus, Mercury would be alright if its true speed were 267km/sec rather > than 47km/sec. (sic) > > So that if Mercury and the Sun were moving in a straight line and at a > distance apart of 0.4AU in Euclidean Space that they would be the > orbits we know of today. However, Mercury is not moving at a 267km/sec > speed in Space. > > So, we have elliptical geometry Space and this space has a spin. So > what measure of a spin of Space so that 47km/sec is required to go > around the Sun moving at 220km/sec. Now one complete revolution of > Mercury is 88 days. > > So how much of a spin is required of this Elliptical geometry Space to > carry the Mercury planet around at 47km/sec, around the Sun moving at > 220km/sec. > > So, now, if the Space was spinning at 88 days for one complete > revolution, then the planet Mercury need go 0km/sec and it would be > making 1 complete orbit. But Mercury is not going at 0 km/sec but > rather at 47km/sec. So, I think the ratio I need to build on is this > ratio: > > 88 267 > X 220 > > And now, cross multiplying I get 88x220 = 267X > > And solving for X, I get 19360/267 = 72.5. > > So the rotation for one complete revolution of the Solar gravitycell > is that of 72.5 days. > > So when I did this last night and came up with 70 days, I was more > correct than in the next day thinking it had to be 110 days. > > Now the other worry was how would a spinning of the Solar gravity cell > fix the revolution of Earth at 29km/sec or the revolution of Jupiter > at 13 km/sec? > It is solid body rotation of the Solar Gravity Cell, but it takes 1 > complete turn of 72.5 days to fix Mercury at 47km/sec and 88 days. It > takes 3 or more complete turns of the Solar Gravity Cell to render > Venus and 5 or more complete turns to render Earth of its 29km and 1 > year full revolution and takes about 60 or more Cell complete spins > to render Jupiter of its 13km/sec and its 11.8 years. > > Now the spinning of the Solar Gravity Cell is solid body rotation, but > the orbits of the planets is not solid body rotation. The orbits of > the planets are "aided or helped out along their path, by the spinning > of the Sun's gravity cell." > > So if Michael Wright were to be so kind as to build a Antikythera > Mechanism that is heliocentric and which has the Sun depicted as a > linear forward motion of 220km/sec, then to counterbalance that motion > a gear wheel has to be engineered so that > it makes the entire Solar Ecliptic revolve one complete circuit in > 72.5 days. > > Hopefully I got it correct now. On issues so complex as astro body > motion, we often mistake one side for the other side, we get 70 and > then think it wrong and get 110, and finally we go back to the 70 > (72.5). > > Now Gravity Cells would solve many other problems and questions in > astronomy. Earlier today I wrote that pulsars are likely to be two > nearby gravity cells pushing on one another as a Faraday law and > producing a pulsed radio signal. > > Gravity Cells also solve the question of redshift of stars and > galaxies, because of the high degree of bent space, that light > traveling through the bent space is shifted by diffraction or > refraction into the red wavelengths. So that the Shapley concentrate > of galaxies may actually be a close by neighbor of the Milky Way. > > Gravity Cells also allow us to unravel the mystery of the morphology > and evolution of galaxies, that they are all ball shaped once we > include the gravity cell with the galactic plane of the galaxy. > > And Gravity Cells solves the question of where galaxies are locating > in the mapping of the Universe. > And I am delighted to say that the galaxies are spaced very evenly and > uniformly, much like a fruit tree orchard of evenly spaced trees in > rows. >
So now, the gravity cell of the Sun spins around its axis and augments the orbit of Mercury so that it can orbit with 47km/sec whilst the Sun is moving in space at 220km/sec.
So now I am wanting to get 15 Jupiter pulsar pulses in 10 hours time rotation of Jupiter from this 2002 NASA report:
 quoting from ?http://science.nasa.gov/sciencenews/scienceatnasa/2002/07mar_jupit... "The pulses are coming from the north pole of Jupiter," says Randy Gladstone, a scientist at the Southwest Research Institute and leader of the team that made the discovery using NASA's orbiting Chandra X ray Observatory. ?Above: Every 45 minutes an xray source blinks near Jupiter's north ?magnetic pole. This animation, based on data from the Chandra Xray ?Observatory, shows the hot spot pulsing 15 times during one complete ?10hour rotation of the giant planet. ? end quote 
I want to apply that Solar gravity cell of 72.5 days of rotation per every single complete revolution.

Approximately 90 percent of AP's posts are missing in the Google newsgroups author search starting May 2012.
Only Drexel's Math Forum has done a excellent, simple and fair author archiving of AP sci.math posts since May 2012 as seen here:
http://mathforum.org/kb/profile.jspa?userID=499986
Archimedes Plutonium http://www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electrondotcloud are galaxies

