A starship voyages to a distant planet 10 ly away. The explorers stay 3.0 yr , return at the same speed, and arrive back on earth 26 yr after they left. Assume that the time needed to accelerate and decelerate is negligible.a What is the speed of the starship?b. How much time has elapsed on the astronauts' chronometers? answer in years.

Respuesta :

Answer:

14.357 years

Explanation:

Total distance travelled is 10+10 ly = 20 ly

Time they stayed on the planet is 3 years

Total time from leaving to returning is 26 years

So, time they travelled is 26-3 = 23 years

Speed of light = c

Average speed

[tex]v=\frac{20}{23}c[/tex]

[tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \gamma=\frac{1}{\sqrt{1-\frac{\left (\frac{20}{23} \right)^2c^2}{c^2}}}\\\Rightarrow \gamma=\frac{1}{\sqrt{1-0.83^2}}\\\Rightarrow \gamma=2.025\ years[/tex]

So, elapsed time

[tex]\frac{23}{2.025}=11.357\ years[/tex]

Including 3 years stationary time we get

14.357 years

∴ Time elapsed on the astronauts' chronometers is 14.357 years

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