Question

An astronaut travels to a star system 3.9 lyly away at a speed of 0.90 cc . Assume that the time needed to accelerate and decelerate is negligible.

Answer 1

Total time elapsed is =8.2y

The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:

Δt=3.9ly/0.9c

Δt=4.3y

For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,

Δτ

$= \sqrt{1 - \frac{v {}^{2} }{c {}^{2} } } \times Δt$

Δτ

$= \sqrt{1 - ( \frac{0.9c}{c} ) {}^{2} } \times (4.3y)$

Δτ=1.8ly

The total elapsed time is:

T elapsed=Δt+3.9

T elapsed=4.3+3.9

T elapsed=8.2y

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