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I am having difficulty solving the following problem:

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earth-based observers?


My approach to the question:

$ \Delta t $ is the time measured by the earth-based "rest frame" observers. $ \Delta t_0 $ is the time measured by the rocket in the "moving frame" relative to earth.

If the rocket "slows down" to half its rate as measured by earth-based observers, does that mean $ \Delta t= \frac{\Delta t_0}{2} $ ?

If so, then $ \Delta t= \frac{\Delta t_0}{2} = \gamma \Delta t_0 $ , Where $ \gamma $= $\frac{1}{\sqrt[]{1-u^2/c^2}} $

Rearranging becomes $ \frac{1}{2} = \frac{1}{\sqrt[]{1-u^2/c^2}}$

so $ 1- \frac{u^2}{c^2} =2^2$

then $ c^2 (1-4)=u^2$

then $u^2=-3c^2$

However, this cannot be correct, since I get a negative square root by solving for u.


All help is greatly appreciated!

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The rocket's clock appears to be ticking slowly. This means that a time which I measure as, say, ten seconds, is only measured as, say, five seconds by the slow ticks of the rocket clock.

Hence, $\Delta t = 2 \Delta t_0$, not $1/2$.

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