Speed of rocket as viewed from Earth

  • Thread starter Rococo
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  • #1
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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.
 

Answers and Replies

  • #2
TSny
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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##.
 
  • #3
Andrew Mason
Science Advisor
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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
 
  • #4
67
9
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?
 
  • #5
TSny
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That looks correct to me.

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