A probe leaves Earth 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 probe leaves Earth in deep space, where no gravitational field acts, and thereafter has a constant proper acceleration

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

along the

+x

axis of the Earth frame

S

. It starts from rest at

x = 0

t = \tau = 0

, where

\tau

is the proper time read by the probe's own clock. The probe's hull has proper length

80\ \mathrm{m}

, with the nose the forward end. Take

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

1\ \mathrm{yr} = 3.156\times 10^{7}\ \mathrm{s}

and

1\ \mathrm{ly} = 9.461\times 10^{15}\ \mathrm{m}

. ((a)) When the probe's clock reads

\tau = 1.50\ \mathrm{yr}

, find its speed in

S

(as a multiple of

c

), the Earth-frame time

t

in years, and its distance

x

from Earth in light-years. ((b)) How far behind the launch point, in light-years, is the probe's Rindler horizon, and by what Earth-frame time must a radio message leave Earth at

x = 0

if it is ever to reach the probe? ((c)) Working to first order in

ah/c^2

, by what fraction does the nose clock run fast compared with the tail clock?

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