A spaceship has a proper length of 50 m. As it flies past Earth at 0.95c, what length would an Earth-based observer measure for the ship?

1. Length contraction gives L = L0 divided by gamma, where gamma = 1 divided by the square root of (1 minus v squared over c squared). 2. With v = 0.95c, gamma equals 1 divided by the square root of (1 minus 0.95 squared), which is about 3.20. 3. Divide the proper length by gamma: L = 50 divided by 3.20. 4. Dividing gives L of about 16 m. 5. 50 m is the proper length seen aboard the ship itself, not the length an Earth observer measures. 6. 8 m and 31 m do not match this division. 7. So 16 m is correct, and it is the Earth observer, not the ship's own crew, who sees the shortened length. _Source: OpenStax Physics (CC BY 4.0), Ch 10 "Special Relativity", section 10.2 Consequences of Special Relativity_