I've added the alternative approach, which results in the same correct prediction of G, to my pages http://nigelcook0.tripod.com/ and http://members.lycos.co.uk/nigelbryancook/ :

Another way of proving the same result as above uses the same illustration but a different mathematical approach, which requires inserting the shield area given by mass against the light-velocity radiation which causes gravity. There is strong evidence from electromagnetic theory (April 2003 Electronics World article, reprinted toward the end of this page) that every fundamental particle has black-hole properties. The effective shielding radius of a black hole of mass M is equal to 2GM/c^2. A shield, like the planet earth, is composed of very small, sub-atomic particles. The very small shielding area per particle means that there will be an insignificant chance of the fundamental particles within the earth ‘overlapping’ one another by being directly behind each other. The total shield area is therefore directly proportional to the total mass: the total shield area is the area of shielding by 1 fundamental particle multiplied by the total number of particles. Notice that the earth’s mass in the standard model is due to the virtual particles associated with up and down quarks: the Higgs field. If indeed all mass is due to the Higgs boson, there is then only one type of shielding particle for this gravity mechanism, and no differing shielding areas for different quarks. From the illustration, the total outward force of the big bang, (total outward force) = ma = (mass of universe).(Hubble acceleration), while the gravity force is the shielded inward reaction (by Newton’s 3rd law the outward force has an equal and opposite reaction): F = (total outward force).(cross-sectional area of shield projected to radius R) / (total spherical area with radius R). The cross-sectional area of shield projected to radius R is equal to the area of the fundamental particle (pi multiplied by the square of the radius of the black hole of similar mass), multiplied by the (R/r)^2 which is the inverse-square law for the geometry of the implosion. ...... (see above mentioned pages for the remainder, which involves equations which will not turn out properly here)

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