The Cost of Speed
To accelerate a mass to a speed requires energy and energy costs money to buy.
To accelerate a mass m, to a speed v requires adding kinetic energy, E
E = mc^2(1/(1-v^2/c^2)^0.5 -1)
where c is the speed of light.
The term 1/(1-v^2/c^2)^0.5 occurs so often in relativity that it is given the name gamma, g.
E = mc^2 (g - 1)
Particle accelerators all over the world verify this equation every day llook at their electric bill. e.g. Stanford Linear Accelerator Center.
A set of three 250 KV power ines feeds power to SLAC.
Explore The graph
The cost goes up asymptotically as the speed of light is approached.
For example, at a speed of v = 0.866, g = 2.
Sample calculation: How many Joules does it take to accelerate 1 kilogram to 0.866 c ?
E = mc^2(g -1) = 1 * (3*10^8)^2 * 1 = 10^17 joules.
The cost of energy can be estimated from the cost of electricity as about $0.3 per kilowatt-hour where a kilowatt-hour is about 3.6 * 10^6 joules.
So the cost is about $ 10^-7 per joule. ( a tenth of a microdollar per joule)
So the cost for 10^17 joules will be Cost = 10^17 J * 10^-7 $/J = 10^10 $
About ten billion dollars to accelerate a kilogram to 0,866 times the speed of light.
(It actually costs twice this much for a trip because you have to slow down at the far end.)
The real cost will be greater than this since I assumed perfect conversion of electric energy to kinetic energy of motion.
Notice that the relativistic equation for kinetic energy is not the same as the Newtonian version E = 1/2 mv^2
However, the relativistic equation becomes the Newtonian version at low speeds.
Scientific Explorations with Paul Doherty
8 May 2005