Length contraction: Two space stations are at rest relative to each other and are 6.0 × 10⁷ m apart, as measured by observers on the stations. A spaceship traveling from one station to the other at 0.90c relative to the stations passes both of them, one after the other. As measured by an observer in the spaceship, how long does it take to travel from one station to the other? (c = 3.00 × 10⁸ m/s) A) 97 ms B) 220 ms C) 510 ms D) 58 ms E) 39 ms

Question

Quizplus · Accepted Answer

Explanation:The length contraction formula is:$$L = L_0/\gamma$$where:L is the length of the object as measured by an observer in the moving frameL_0 is the length of the object as measured by an observer in the rest frameγ is the Lorentz factor, which is given by:$$\gamma = 1/\sqrt{1 - v^2/c^2}$$In this problem, the length of the object is 6.0 × 10^7 m, and the speed of the spaceship is 0.90c. Therefore, the Lorentz factor is:$$\gamma = 1/\sqrt{1 - 0.90^2} = 2.29$$The length of the object as measured by an observer in the spaceship is:$$L = 6.0 × 10^7 m/2.29 = 2.62 × 10^7 m$$The time it takes for the spaceship to travel from one station to the other is:$$t = L/v = (2.62 × 10^7 m)/(0.90 × 3.00 × 10^8 m/s) = 97 ms$$Therefore, the correct answer is A.

Length Contraction: Two Space Stations Are at Rest Relative to Each