- #1

- 67

- 9

## Homework Statement

A rocket starts at rest from the Earth, moving in the z direction so that in its own instantaneous rest frame the acceleration is always g = 9.8 m/s^2. To an observer on Earth, its position is given by

## z(\tau) = \frac{cosh(g\tau) - 1}{g} ##

## t = \frac{sinh(g\tau)}{g} ##

where ##\tau## is the rocket's proper time. Measured in a reference frame at rest on the Earth, how far from the Earth will the rocket be in 40 years? How fast will it be going?

## Homework Equations

Four-velocity:

## U^\mu = (\gamma, \frac{d\underline{x}}{d \tau}) = (\gamma, 0, 0, sinh(g\tau)) ##

## The Attempt at a Solution

t = 40 years = ##1.26\times10^9 s##

g = ##9.8m/s^2##

##g\tau = sinh^{-1}(gt) = sinh^{-1}((9.8)(1.26\times10^9)) = 23.93##

##z = \frac{cosh(g\tau)-1}{g} = \frac{cosh(23.93)-1}{9.8} = 1.26\times10^9 m##

## U^z = sinh(g\tau) = sinh(23.93) = 1.24\times10^{10} m/s##

But this is greater than the speed of light so it can't be right.