C= the speed of light. M= mass in kg.
First take the speed of light in meters per second and then take a mass value; lets say 5 kg. And multiply the 5 by 5 to represent the square value of the 5 kg mass, now multiply the square value of the mass with the speed of light and you will get value of the (potential) force in Joules that exists within the said quantity of mass, then you need to divide that quantity by the original mass quantity; 5 in this case, that will give you the amount of kinetic energy in joules that exists within the said quantity of mass, by multiplying the result by the speed of light you will get the total amount of energy in Joules that the said mass quantity can have.
In other words: E = cm^2 divided by m and multiplied by c. So, to illustrate and use “c” as 300000000 m/s and the 5 kg as the mass, then m^2 = 25 multiplied by 300000000 = 7,500,000,000, Then take the 7.5*10^9 and divide it by 5 = 1,500,000,000 then multiply that result by 300000000 which gives you the final result of 4.7*10^17 Joules. As a comparison; C * C = 9*10^16 * 5 = 4.5*10 17 Joules.
Previously I have used the formula E= (U/k)c, with the same results, that formula was deciphered from a real old book, a book that many people consider useless and not worth saving.
Thanks for the link, it has another way of looking at the problem. Mass being accelerated to (almost) the speed of light by using external force.
Off Topic » in reply to The world's quickest derivation of E=mc^2
Perhaps I did not use the proper way of representing the formula.
The whole thread (11 posts)
- Einstein got the outcome of the calculation E= mc^2 figured out correctly but (originally from Mysteries of the Multiverse)
