A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth. How much time for this trip would elapse on a clock on Earth?
A) 3.23 y
B) 5.38 y
C) 4.40 y
D) 5.15 y

Question

Quizplus · Accepted Answer

Explanation:The time dilation formula is:$$t' = t/\gamma$$where:$$t'$$ is the time measured in the moving frame$$t$$ is the time measured in the stationary frame$$\gamma$$ is the Lorentz factor, which is given by:$$\gamma = 1/\sqrt{1 - v^2/c^2}$$where:$$v$$ is the speed of the moving object$$c$$ is the speed of lightIn this problem, the spaceship is moving at a speed of 0.800c relative to Earth, and the distance to the star is 4.30 light-years. The time for this trip would elapse on a clock on Earth is:$$t = t'/\gamma = (4.30 	ext{ light-years})/(1/\sqrt{1 - 0.800^2/c^2}) = 5.38 	ext{ years}$$Therefore, the correct answer is A.

A Spaceship with a Constant Speed of 0