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(Kindly read the details before answering) How much energy would be actually required for a 1 kg object to reach the speed of light in a vacuum?

Asked by krrazypassions (1355points) May 12th, 2011

Given:
1) Lets say we have a an object A with a mass of 1kg (at rest)-(mass of 1 kg is assumed so that we can calculate a definite amount of required energy)
2) we want it to travel at the speed of light ( c ) in empty space (vacuum).
3) In this experiment, we will be increasing our speed in equal increments, so that after n equal increments, our speed will be c.

Part A
According to relativity theory, as we tend closer to c, for an earthly observer,
a) the mass of A will increase- and hence the energy required to increase the speed of this heavier A will also have increased.
b) But then, the mass of the fuel being used will also have increased. Therefore (by E=mc^2), wouldn’t the energy that can be generated from that fuel also be larger, and therefore can possibly satisfy the increased energy requirement!

Part B
Also, for an observer in frame of reference of A,
a) the mass of A will still be 1 kg only
b) the energy required to increment its speed in equal amounts will be constant for all increments. And hence, the mass of fuel required to do this will also be some constant value.

Thus, if we need M mass of fuel for one increment in the speed of A, and if we need a total of n equal increments to reach the speed of light c, the total mass of fuel required will be nM. So we can carry n fuel-packs* of mass M and use up one fuel-pack for every increment.

Gist of the question:
Hence, wont this actual required mass and energy be a finite and not-very-large value after all,
because even though energy required to accelerate the increasingly massive object A will be increasing from an earthly observer’s perspective,
the mass of the fuel and energy produced from it will also be increasing from an earthly observer’s perspective to be a very large value, and would therefore succeed at attaining speed of light using finite energy!

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