# whats the rate of energy increase required for constant acceleration between 0.0c and 0.99c?

I was wondering how much energy would be required to accelerate 1000kg to 0.99c at 1G.

What I don't understand is what the rate of increase of energy is required as velocity increases. I was looking at the Lorentz factor curve but I'm not sure if this affects the energy required or just time dilation. eg. 0.99c = Lorentz factor of 7. So travel only takes 1/7th the time from the point of view of the traveler but does that mean you need 7 times as much energy to get to 0.99c than from say 0.0c to 0.1c.

Does any body have a link to a graph that shows this rate of increase of energy required or is it the same as the Lorentz factor curve?

I'm a programmer not a physics or math person so go easy on me.

Thankyou.

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dave, is this actually a homework question? (Qmechanic I wonder if you might be a tad overzealous about that today ;-P) –  David Z Jul 14 '11 at 22:44
@David Zaslavsky: I understand your general concern, which you and others have explained, e.g., here: meta.physics.stackexchange.com/q/714/2451 In my opinion the 'homework' tag applies to any homework-type problem such as the above. Whether it has actually been assigned by a teacher as homework to OP is, to some extent, irrelevant. –  Qmechanic Jul 15 '11 at 13:04
@Qmechanic: right, the homework tag does apply to all homework-type problems, but I really don't think this is one. Only the OP can tell for sure whether a question should be considered a homework-type problem or not, but in this case the details requested make it seem like something just asked out of curiosity. My issue here is that Edgar's answer, while very good for a non-homework question, might give away a little too much if this is a homework question, and if that were the case it should be edited or deleted - so basically either we remove the homework tag, or remove part of the answer. –  David Z Jul 15 '11 at 18:10
(cont.) In this case, since as I said, it really doesn't strike me as a homework-type question, and since if I were going to edit/delete the answer the appropriate time would have been yesterday, I'm going to remove the homework tag. (For the record, I don't make a habit of reverting other people's edits in general even when I disagree with them) –  David Z Jul 15 '11 at 18:12

The energy of the mass at rest is $$E_0 = m c^2$$ At speed $v = 0.99 c$ it becomes $$E = \frac{m c^2}{\sqrt{1-v^2/c^2}}$$ So, the increase in energy is $$E - E_0 = m c^2\left(\frac{1}{\sqrt{1-v^2/c^2}} - 1\right) \approx 6.09 m c^2.$$ irrespective of the acceleration.
Or about $5.5\times10^{20}$ J. –  Edgar Bonet Jul 14 '11 at 21:16
@Jason Waldrop: The classical formula for the kinetic energy gives $E_K = \frac{1}{2}mv^2 = 4.4\times 10^{19}\;\mathrm{J}$ –  Edgar Bonet Jul 15 '11 at 8:04