Speed of rocket as viewed from Earth

[Rococo]

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.

 

[TSny]

Your formulas for $z$ and $t$ must be assuming units where $c = 1$. It would be a good idea to insert factors of $c$ so that you can work with meters and seconds.

The z component of the 4-velocity is $dz/d\tau$. But this is not the velocity you want. In the Earth frame, a time increment would be $dt$, not $d \tau$.

 

[Andrew Mason]

There seems to be an error in your calculation of $\tau$. You have $\tau$ = 2.4 seconds (23.93/g). But for the first several months $t \approx \tau$ because the rocket speed << c.

AM

 

[Rococo]

Your formulas for $z$ and $t$ must be assuming units where $c = 1$. It would be a good idea to insert factors of $c$ so that you can work with meters and seconds.

The z component of the 4-velocity is $dz/d\tau$. But this is not the velocity you want. In the Earth frame, a time increment would be $dt$, not $d \tau$.

Yes, I forgot to say the question mentions that the equations $z(\tau) = \frac{cosh(g\tau) - 1}{g}$ and $t = \frac{sinh(g\tau)}{g}$ are in units where $c=1$. So reinserting factors of $c$, would these equations become:

$gt/c = sinh(g\tau /c)$
$z(\tau) = \frac{cosh(g\tau /c) - 1}{g/c^2}$?

Then for $t=40 years=1.26 \times 10^9 s$, we have

$g\tau /c = sinh^{-1}(gt/c) = sinh^{-1}[\frac{(9.8)(1.26 \times 10^9)}{3 \times 10^8}]= 4.41$

Giving a distance measured in the Earth frame:

$z= \frac{cosh(g\tau /c) - 1}{g/c^2} = \frac{(3 \times 10^8)^2}{9.8}[cosh(4.41) -1] = 3.69 \times 10^{17} m$.

The speed in the Earth frame would then be:

$U^z = \frac{dz}{d\tau}\frac{d\tau}{dt} = c sinh(g\tau /c) \frac{1}{cosh(gt/c)} = c tanh(gt/c) = c tanh(4.41) =0.9997c$

Is this the right way to do it?

 

[TSny]

That looks correct to me.

It might be nice to express the distance in light-years.