A satellites around the sun to acquire the heavy Proton

Magnetic Gas of Sun or Stars`s. 

If you want heavy pure proton. then you should send a satellites around the sun to acquire it. to which details I shall tell continuous in this chapter. and this proton will too much powerful from Earth proton. and if you want faster travel in the space. then this heavy pure proton will solve your this faster journey. and which distance I am telling you about a orbit of  60 seconds far from sun. there are too much quantity of heavy proton. and through this heavy proton you will easily and faster reach at  others planet. or if you takes three year to travel of Mars. then through this heavy pure proton nearly you will take four or five days. which is very less number of days. and to the Pluto I think only some week. because light travels in open process. and we shall use it. into covered process. and to which will be results. conform the speed of ray 1800000000 meter per seconds. 

Rate of Orbit. 

we shall solve this matter with light speed. but you should know about light speed. and it is not very difficult. Number one, A star`s temperature of surface always happens 15000000 Centigrade. and always into one proton are happens five photons. and the light speed begins from one meter, one seconds. when we shall divide the sun surface temperature with five photons. then one heated particle would be divided. and when we multiply with 100 centimetres. then will be begun the travel of ray. 

First Basic Law. 

Temperature of sun`s surface is ‘‘15000000 Centigrade’’ / 300000000 light speed per meter seconds = 0.05 x 100 centimetre = 5 photons of one proton. 

Second Basic Law. 

I know it that 0.05 is very cheap value of any mass. because it is the heat value of mass. and photons happens very small. to which you cannot see. only can be feel them. So one photon mass equal to .008. So then 0.05 / .008 = 6.25 / 1.25 Earth hydrogen mass unit = 5 photons of one proton. 

Third Basic Law. 

Temperature of sun`s surface is ‘‘15000000 Centigrade’’ / 12.5 mass unit of sun hydrogen atom = 1200000 x .008 hot photon = 9600 x .008 cold photon = 76.8 / 60 minute = 1.28 - 0.03 = 1.25 mass unit of earth`s hydrogen.  

Fourth Basic law.  About light speed

Stars`s surface temperature 15000000 / 5 Photons or one Neutron = 3000000 x 100 Centimetres   = 300000000 light speed meter per seconds. 

First nearest orbit of Sun.  

Sun`s first orbit which belongs with magnetic particle. this orbit is one minute far from sun. and one minute equal to 60 seconds. So 60 x 300000000 = 18000000000 meters distance from sun.. and to the know days. 18000000000 / 24 hour / 60 minute / 1000000 light  measure = 12.5 days of Magnetic distance. Now to the orbit. 18000000000 x 6 = 10800000000 meter orbit. to the know days. 108000000000 / 24 hours / 360 degrees / 1000000 light measure = 12.5 days of orbit. 

Diameter of Satellites. 

If the satellites diameter proper perfect. then they will right rotate around the sun. and it is first important matter. because which you make the satellites to rotate around of sun. their size will nearly equal to a city. So orbit radius is 60 seconds from sun. in which satellites will be sent. So 60 x 21.78 code of diameter = 1306.8 kilometres diameter of the satellites. and by radius that 1306.8 / 6 = 217.8 kilometres. and it is some big from normal city. 

TEMPERATURE OF SATELLITE. 

Who metal will use you to the satellite. which could bear the sun temperature. because orbit of one seconds far from sun. would have very warm, you will have to use special metal. Although Mercury`s temperature is 503 Centigrade. and Mercury is 6.9 minute away from sun. Now we are going to solve of temperature of  satellite. perhaps would have remember you temperature`s law. ( that light speed / distance in seconds / 24 hours / 60 minute = temperature of planet). So satellite distance is 60 seconds. Then 300000000 / 60 =  5000000 / 24 = 20833333333 / 60 = 3472.22 Centigrade would have temperature of satellite. and it is right and perfect temperature. because when we ll’ divide Mercury temperature with the satellite temperature. then our proof will be strong. So satellite temperature is 3472.22 / 503 = 6.9 minute distance of Mercury  from the sun. here we shall say. if your basic maths would have right. then you can solve each matter of universe. Now we have 3472.22 Centigrade temperature to the satellite. If you are thinking that it is too much  temperature. then you are wrong. because which maths I am going to tell you. that our problem will be solved.