A cargo ship of proper length leaves a station in deep space

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · The equivalence principle and gravity as geometry · The equivalence principle and gravity as geometry

Relativity and Gravitation, Ch Accelerated observers, hyperbolic motion and the equivalence principle

Problem

A cargo ship of proper length 150 m150\ \mathrm{m} leaves a station in deep space, where no gravitational field acts, and holds a constant proper acceleration a=20.0 m s−2a = 20.0\ \mathrm{m\,s^{-2}} along the +x+x axis of the station frame SS. It starts from rest at x=0x = 0 at t=τ=0t = \tau = 0, with τ\tau the proper time of the ship's clock. The nose is the forward end. Take c=2.998×108 m s−1c = 2.998\times 10^{8}\ \mathrm{m\,s^{-1}} and 1 day=86400 s1\ \mathrm{day} = 86400\ \mathrm{s}. ((a)) At what station-frame time, in days after launch, is the last moment at which a radio pulse sent from the station at x=0x = 0 can still reach the ship? A pulse is sent 150150 days after launch: by how many days does it beat that deadline? ((b)) Working to first order in ah/c2ah/c^2, by what fraction does the nose clock run fast compared with the tail clock? ((c)) How far from the station, in metres, is the ship when its own clock reads τ=200\tau = 200 days?

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