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A starship voyages to a distant planet 10 ly away. The explorers stay 1 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?

$$

0.8 \mathrm{C}, 16 \mathrm{y}

$$

Gravitation

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Numerade Educator

University of Washington

Simon Fraser University

University of Sheffield

do? Mm. The total time taken by the spaceship in travel. So the total time, the total time is equal to 26 minus one, which is equal to 25 years and the total distance covered by the spaceship total distance covered by especially people's to multiply by 10 eso 20 light years. The velocity of the spaceship is so you lost steam. So the spaceship is 20 light years divided by 25 years, which is equal to 0.8 times the speed of light. Ah, since speed of light C equals one light here for a year and no, we do part B well, in my family preference, the time taken different travel is 35 years. So in my frame of reference in my frame off reference in my primal preference, our time taken these, uh, 35 years since for one year. Uh, since for one year, the spaceship waas stash Nery in arts frame of reference. So the time will be seen, uh, for the data. One year I didn't travel the time in spaceships. Krone meter is, uh we have a delta T prime is equal to delta t times The gamma which is equal to 25. Multiply by route one minus R 0.8 square. Right, which is equal to 15 years. Thank you. And so the total time they will be shown in spaceships. Criminal, uh, problem eater is 15 plus 1 16 years. So the total time. So the total time, uh, shown in a Cron a meter krone meter is equal to 15 plus one is equal to 16 years.