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An astronaut travels from the Earth to the nearest star system, Alpha Centauri, 4.3 light years (ly) away as measured by an Earth-bound observer. If the astronaut travels at 2.88e+8 m/s, how far apart are the Earth and Alpha Centauri as measured by the astronaut?

User Ulluoink
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Final answer:

The distance between Earth and Alpha Centauri as measured by the astronaut traveling at a high speed is contracted to 0.1433 light years due to the effects of length contraction in special relativity. The Lorentz factor, given as 30.00, is used to calculate this new distance.

Step-by-step explanation:

The question revolves around the concept of length contraction in the realm of special relativity, as experienced by an astronaut traveling to Alpha Centauri. According to an Earth-bound observer, Alpha Centauri is 4.300 light years (ly) away, which is the proper distance L0. However, for the astronaut moving at a velocity v, the observed distance will be contracted. We use the lorentz factor y, also expressed as γ, which is given as 30.00 to calculate this contracted length L. Applying the length contraction formula L = L0 / y, we find:

L = 4.300 ly / 30.00 = 0.1433 ly

The far-apart distance between the Earth and Alpha Centauri as measured by the astronaut is therefore 0.1433 light years.

For (b), the astronaut's velocity v in terms of the speed of light c can be deduced by the Lorentz factor, where y = 1 / sqrt(1 - (v2 / c2)). Since y is given as 30.00, v can be found via the inverse process. However, without the detailed calculations included in the question, we do not provide a specific value for v here.

User Mr Beardsley
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