Question

Suppose an astronaut travels to a star 10 lightyears away at a significant fraction of the speed of light (for example, v=0) and returns. Assume the negligible amount of time is spent at the destination.

U-0.99C

1. Does this round-trip journey take less than, equal to, or more than 20 years as observed by an Earth-based observer?
2. Does this round-trip journey take less than, equal to, or more than 20 years as observed by the astronaut?
3. Describe the journey as the astronaut would observe it. In particular, describe what the astronaut would observe as happening on Earth during the journey.

Answer 1

To Asteconaut, length or distance will appear to be shortened L - L_0 squareroot 1 minus v^2/c^2 L - 10 squareroot 1 minus (0.99 c)^2/c^2 L = 10 squareroot 0.0199 L = 1.410673598 light years. Time taken = 1.410673598/0.99 = 1.4249 years. While coming back also, he will take same time. Total time taken to cover 20 light years = 1.4249 + 1.4249 = 2.8498 years (a) It takes 20/0.99 = 20.2020 years according to the observer on earth. Time = Distance/speed \r\n (c) To astronaut, earth is in motion relative to him at a speed of 0.99 c. so, all processes on earth should be slowed down by a factor of squareroot 1 minus v^2/c^2. But, this is not true. As Astronaut changed from one inertial frame to a different one when he started out, when he reversed the direction to head home, and when he landed on the earth.