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Relativistic Mechanics

Problem 1.366

An imaginary space rocket launched from the Earth moves with an acceleration w′=10gw^{\prime}=10 g which is the same in every instantaneous co-moving inertial reference frame. The boost stage lasted τ=1.0\tau=1.0 year of terrestrial time. Find how much (in per cent) does the rocket velocity differ from the velocity of light at the end of the boost stage. What distance does the rocket cover by that moment?

Reveal Answer
1.366. Let us make use of the relation between the acceleration w′w^{\prime} and the acceleration ww in the reference frame fixed to the Earth: w′=(1−v2/c2)−3/2dvdt w^{\prime}=\left(1-v^{2} / c^{2}\right)^{-3 / 2} \frac{d v}{d t} This formula is given in the solution of the foregoing problem (item (a)) where it is necessary to assume V=vV=v. Integrating the given equation (for w′=w^{\prime}= const), we obtain v=w′t/1+(w′t/c)2v=w^{\prime} t / \sqrt{1+\left(w^{\prime} t / c\right)^{2}}. The sought distance is l=(1+(w′t/c)2−1)c2/w′=0.91l=\left(\sqrt{1+\left(w^{\prime} t / c\right)^{2}}-1\right) c^{2} / w^{\prime}=0.91 lightyear; (c−v)/c=1/2(c/w′t)2=0.47%(c-v) / c=1 / 2\left(c / w^{\prime} t\right)^{2}=0.47 \%
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