‘‘THIRTIETH ORBIT OF SUN’’

SOME COLD ORBIT.  

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 Proton = 3000000 x 100 Centimetres = 300000000 light speed meter per seconds. 

Nearest Thirtieth Orbit OF Sun.  

Sun`s Thirtieth orbit which belongs with living Particles. this orbit is Thirty minute far from sun. and Thirty minute are equal to 1800 seconds. So 1800 x 300000000 = 5.40000000000 meters distance from sun. and to the know days 5.40000000000 / 24 hour / 60 minute / 1000000 light measure = 375 days of distance of Thirty minute from the sun. 

Now to the orbit 5.40000000000 x 6 = 3.240000000000 meter orbit. to the know days 3.240000000000 / 24 hours / 360 degree / 1000000 light measurement = 375 days of Thirty minute orbit of the sun.   

Here is not present any satellite.

But I am doing solve value of each orbit. even which orbit is away from the sun. So it is Thirtieth orbit of sun. If in future you made a humans colony into space. then you can acquire the energy from even any orbit of sun. 

Diameter of Satellite. 

So orbit radius or distance is 1800 light seconds between Orbit and sun. Now for get the diameter, If any satellite happens there. So 1800 x 21.78 code of diameter = 39204 kilometres diameter of the satellite. and by radius that 39204 / 6 = 6534 kilometres radius. 

and diameter of satellit is 1045.44 kilometre large from our Earth. because much distance contains on over diameter. but the Mars is small than our Earth, because Mars not separated alone from the sun in beginning. that the mars was attached with the earth. and some time later, the Mars again separated from the earth. from this reason, Mars`s size is small from our earth. 

So the Mars was separated two times in own life. Such as` first times separated from the sun with Earth. and second times from the earth to live alone. Now satellite total seconds are 1800 x 3.63kg length weight code = 6534 / 17.424 Code of days = 375 days of Thirtieth orbit of sun.  

Diameter of Mars. 

The diameter find out of mars from the distance of sun is very complicated situation. because the mars and earth was couple in beginning. therefore we find out the diameter of Mars, through the distance of earth. because mars diameter is some less from the earth. 

As a trial, 

If we do the solve diameter of mars from the distance of sun. then diameter of mars would be double large from the earth. Such as` mars distance is 3292.8 light seconds. So for diamete 3292.8 x 21.78 = 71717.184 diameter. 

and this diameter is double from the diameter of earth. or 71717.184 - 38158.56km = 33558.625 kilometre difference of diameter between the earth and mars. as a trial we make the days of figures of difference. So 33558.625 / 6 = 5593.104 radius of diameter. 

then 5593.104 / 17.424 Code = 321 days difference of distance between the earth and mars. because 321 + 365 = 686 days of the Mars. So the mars is small 44 days from the earth. therefore 321 x 4.8 = 1540.8 light seconds distance of mars from the earth. and 1540.8 x 21.78 Code for diameter = 33558.624 kilometre diameter of mars.

Temperature of Mars. 

We are doing compear of temperature with the Mars for each distance. So the Mars temperature is 63.26 Centigrade, and Mars distance is 54.88 Minute far from the sun. 

satellite. perhaps would have remember you that law of temperature. ( that light speed / distance in seconds / 24 hours / 60 minute = temperature of planet). So satellite or orbit`s radius is 1680 light seconds. Then 300000000 / 1800 =  1666666667  / 24 = 6944.444444 / 60 = 115.7407407 Centigrade temperature will happen of satellite or of the Thirtieth orbit. and it is right and perfect temperature. because when we ll’ divide with Venus temperature with the satellite or for the temperature of orbit. 

then our proof will be strong. So satellite temperature is 115.7407407 / 63.26 = 1.8 x 30 minute = 54.88 minute distance of Mars  from the sun. here we shall say. if your basic maths would have to be right. then you can solve each matter of universe. Now we have 115.74 Centigrade temperature to the satellite or of the orbit. 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 About space.  

Earth temperature: 

300000000 / 1752 seconds distance = 171232.8767 / 24 hours = 7134.703196 / 60 minute = 118.91 temperature of earth. So water temperature 59.455 - 118.91 = 59.455 Centigrade temperature of earth 

The velocity of satellite in own orbit. 

The satellite`s velocity will happen 32.22 kiometres per seconds in own orbit. and it velocity is 5 percent excessive from the sun diameter. So it is finally accurate law of velocity. Now to the make law we ll’ use the sun diameter and atom two conditions. So satellite days are 375 in own orbit. next 1.25 is warm condition of Hydrogen atom and 1.25 is cool condition of Hydrogen. after collect of both. 1.25 + 1.25 = 2.5.

Law of velocity

Now satellite total days are 375 / 2.5 = 150 and it is the code of speed. The sun diameter is 4833.26 days kilometres. So 4833.26 / 150 speed code = 32.22 kilometres per seconds speed of satellite. I had explained about velocity of planet 2.5 years ago. which 

was sixth percent accurate. and it law is hundred percent accurate . Actually! I am explaining about many subject in same time. So can be mistake but I itself shall do accurate own each mistake. And in the beginning the Internet and english  were new things for my. 

Velocity of Mars. 

The Earth days are 686 / 2.5 = 274.4 and sun diameter 4833.26 / 274.4 = 17.61 kilometres per seconds velocity of Mars around the sun.   

The size of sun from Thirtieth minute distance. 

It is also very important situation. you always see the sun in 1.5 feet size from the earth. but if you want imagine size of sun from Thirtieth minute away. then this law will help us. many events rises up into the Universe. sometimes a star suddenly very quickly  melt to be the small size. fact about a star. which was against across of our solar system in

the milky-way. and around this star the life were present in two planets. there in only four centuries. their star was fifty percent melt. from this reason. there life was suddenly finished into the two planets. why it happened, actually you do not know about the StarQuake which happens very dangerous. tell later to which details. So we are doing probe orbit of sun 

from Thirty minute away. So from Thirty minute away the sun size will happen 2.5 feet. Now by maths that the sun diameter is 4833.26 days kilometres. So 4833.26 / 1800 seconds = 2.685144444 and we shall only use 2 here. So 2 x 10 atom = 20 inch size of sun from on Thirty minute away. and by the feet that 20 inch / 12 = 1.66 feet height of sun from Thirty minute away. or, size of sun.