A rocket of proper length leaves Earth from rest and makes a round trip in four legs

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 rocket of proper length

h = 120\ \mathrm{m}

leaves Earth from rest and makes a round trip in four legs, each of proper time

\tau_{\rm leg} = 2.50\ \mathrm{yr}

on the ship's clock and each at constant proper acceleration

a = 9.80\ \mathrm{m\,s^{-2}}

: accelerate away, brake to rest, accelerate back, brake to arrive at rest at Earth. Earth is treated as an inertial frame, the rocket is rigid, all clocks are ideal, and gravity is ignored throughout. Take

c = 2.998 \times 10^{8}\ \mathrm{m\,s^{-1}}

1\ \mathrm{yr} = 3.156\times10^{7}\ \mathrm{s}

1\ \mathrm{light\text{-}year} = 9.461\times10^{15}\ \mathrm{m}

. ((a)) Using

ct = (c^2/a)\sinh(a\tau/c)

and

x = (c^2/a)[\cosh(a\tau/c) - 1]

for one leg, and the symmetry of the four legs, derive

T = (4c/a)\sinh(a\tau_{\rm leg}/c)

for the total Earth-frame time and evaluate it in years. ((b)) Derive the Earth-frame distance

D = (2c^2/a)[\cosh(a\tau_{\rm leg}/c) - 1]

of the turn-round point and evaluate it in light-years. ((c)) During the first leg only, the thrust points from tail to nose. Working to first order in

ah/c^2

, by how much does the nose clock gain on the tail clock over that leg?

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