LAW OF GRAVITATION FOR METIS MOON OF JUPITER:

The Metis! Which is an best moon of Jupiter, because his distance contain on mass, if the distance is long then mass will be large, and here mass and distance are better condition, And from this reason we shall use double day of Metis. Such Metis days of rotation are 0.29 around of Jupiter, then we shall use 0.29 + 0.29 = 0.58 days for find out outer equator of Metis by real size, or Metis size containing on 0.58 days, 

THE SECRET LAW OF METIS, 

(1) The earth period is 365 day * 14.4 charge of a electron = 5256 km distance of earth, because 5256 km / 24 hours / 60 minute = 3.65 * 100 cm = 365 yearly days of earth. 

(2) The Metis period 0.29 day * 14.4 charge of one electron = 4.176 * 1000 meter = 4176 km distance of Metis from Jupiter, Now 4176 km / 24 hours / 60 minute = 2.9 / 10 cm = 0.29 day period of Metis, now for outer equator of Metis, we shall do double of year, such 0.29 + 0.29 = 0.58 days for outer equator of Metis, 

DETAILS OF METIS, 

Distance is 4176 km from Jupiter, 0.29 days period of Metis, Radius 10 km, Outer equator 60.63552 km, Diameter 20 km, and the Metis size is contain on 0.58 days. And Metis is 1.392 light seconds away from Jupiter. Because distance 4176 km / 3 one triangle = 1392 / 1000 meter = 1.392 / 4.8 neutron energy = 0.29 days period of Metis.   

LAW IN KILOMETERS OF METIS:

Gravitation law of King-S is 104.544 * 0.58 day size of Metis = 60.63552 kilometers outer equator of Metis. So 60.63552 / 6 radius = 10.10592 kilometers radius of Metis,               

LAW OF AGAINST GRAVITATION OF METIS. 

So what would be gravitation or value of “hg” at the Metis. That we are doing solve this gravitation through the Math. So the Metis outer equator is 60.63552 kilometer, and radius is 10.10592 kilometers, and 0.015168 m/s “hg” of Metis. or 0.015168 m/s is against gravitation or “hg” of Metis. Because Metis outer equator is 60.63552 km / 10.89 earth “g” = 5.568 / 0.58 day size of Metis = 9.6 It was first step of gravitation. So Metis radius is 10.10592 km / 6359.76 radius of earth = 0.00158-Sa and it was second step of gravitation. So after multiply of both, 9.6 * 0.00158-Sa = 0.015168 m/s against gravitation of Metis. And this Metis “gh” gravitation is attached with gravitation of Earth. So 0.015168 meter per seconds against gravitation of Metis, which is 1.392 light seconds away from Jupiter, 

PROOF FOR EQUATOR OF METIS: 

Then 0.015168 “hg” * 10.89 “g” = 0.16517952 * 0.58 days size of Metis = 0.0958041216 km / 0.00158-Sa = 60.63552 km outer equator of Metis,     

PROOF FOR EQUATOR OF EARTH: 

Then 0.015168 “hg” * 10.89 “g” = 0.16517952 * 365 days size of Earth = 60.2905248 km / 0.00158-Sa = 38158.56 km outer equator of earth, 

PROOF FOR EQUATOR OF SUN: 

Then 0.015168 “hg” * 10.89 “g” = 0.16517952 * 399443296.60 days of sun = 65979851.999605632 km / 0.00158-Sa = 41759399999.7504 km outer equator of sun, because 41759399999.7504 / 6 radius = 6959899999.9584 + 0.0416 = 6959900000 radius of sun.  

PROOF BY YOUR LAW EQUATOR OF METIS: 

Then 0.015168 “hg” * 9.8 = 0.1486464 * 0.58 days size of Metis = 0.086214912 km / 0.00158-Sa = 54.5664 km outer equator of Metis, 

LAW OF FALLING GRAVITATION OF METIS: 

So what would be gravitation or value of “g” at the Metis. That we are doing solve this gravitation through the Math. So the Metis outer equator is 60.63552 kilometer, and radius is 10.10592 kilometers, and “g” is 0.0172062 m/s of Metis. Or 0.0172062 m/s is a falling gravitation or “g” of Metis Because Metis outer equator is 60.63552 km / 9.6 earth “hg” = 6.3162 / 0.58 days size of Metis = 10.89, It was first step of gravitation. So Metis radius is 10.10592 / 6359.76 radius of earth = 0.00158-Sa and it was second step of gravitation. So after multiply of both, 10.89 * 0.00158-Sa = 0.0172062 m/s falling gravitation of Metis. And this Metis “g” gravitation is attached with gravitation of earth. So 0.0172062 meter per seconds falling gravitation of Metis or “g” and which is 1.392 light seconds away from Jupiter,

PROOF FOR EQUATOR OF METIS: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 0.58 days size of Metis = 0.0958041216 km / 0.00158-Sa = 60.63552 km outer equator of Metis, 

PROOF FOR EQUATOR OF EARTH: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 365 days size of Earth = 41.5928304 km / 0.00158-Sa = 38158.56 km outer equator of earth, 

PROOF FOR EQUATOR OF MARS: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 321 days size of Mars = 53.02262592 km / 0.00158-Sa = 33558.624 km outer equator of Mars, 

PROOF FOR EQUATOR OF JUPITER: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 4337.5 days size of Jupiter = 716.466168 km / 0.00158-Sa = 453459.6 km outer equator of Jupiter, 

PROOF FOR EQUATOR OF VENUS: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 223.75 days size of Venus = 36.9589176 km / 0.00158-Sa = 23391.72 km outer equator of Venus, 

PROOF FOR EQUATOR OF MERCURY: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 86.25 days size of Mercury = 14.2467336 km / 0.00158-Sa = 9016.92 km outer equator of Mercury, 

PROOF FOR EQUATOR OF SUN: 

Then 0.0172062 “g” * 9.6 “hg” = 0.16517952 * 399443296.60 days of Sun = 65979851.999605632 / 0.00158-Sa = 41759399999.7504 km outer equator of Sun, because 41759399999.7504 / 6 radius = 6959899999.9584 + 0.0416 = 6959900000 radius of Sun.  

PROOF BY YOUR LAW EQUATOR OF EARTH, 

Then 0.0172062 “g” * 9.8 = 0.16862076 * 365 days size of Earth = 61.5465774 km / 0.00158-Sa = 38953.53 km outer equator of Earth, so by your law the Earth is 794.97 km large than my measurement of Earth.

So the Metis “hg” = 0.015168 meter per second, and the “g” = 0.0172062 meter per second. And 0.00158-Sa is a code of both gravitation, and the gravitation is main function of universe. And all system of universe is contains on gravitation of particle. And you all are very different from each other by face and body. Your gravitation of cells is very different from each other. And it is a long chapter about genetic gravitation. Which will be started in 2017, (1000 / 10.89 = 91.82!! 1000 / 1.9602 = 510.15 / 91.82 = 5.55!! 1000 / 5.55, 180.18),