Spaceship moving away from light source

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nick227
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Homework Statement



A spaceship moves radially away from the Earth with an acceleration of 20m/c2. How long does it take the sodium street lamps ([tex]\lambda[/tex]=589 nm) on Earth to be invisible (with a powerful telescope) to the human eye of the astronauts? The visible spectrum is 400-700nm.

Homework Equations



fobs = ((1+(v/c))1/2/(1-(v/c))1/2) fsource

The Attempt at a Solution



I can use the doppler shift equation, to solve for velocity, but what should I use for fobs? Should I use 400nm or 700nm. I'm thinking 400 nm because of length contraction...

After I find a velocity, I can use the acceleration given and solve for time.
 
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Well seeing as how I am moving away from the source wouldn't it be a red shift, so 700nm. But everything i read about red shift is if the source is moving away from the observer. now assuming that doesn't matter, am i right?
finding v=c(f_o^2 - f_s^2)/(f_s^2 + f_o^2)
f_o = f of observer
f_s = f of source

so our time is v/a?
 
But the latter half of the question is right, right? solving for v then saying t=v/a would determine how long it would take to observe the light shift into the non-visible spectrum?
 
You wrote: "I'm thinking 400 nm because of length contraction..."

This is not about length contraction, but about the Doppler shift. Check your textbook for the difference between these concepts!

You wrote: "But everything i read about red shift is if the source is moving away from the observer". You should read and reread what the "principle of relativity" says!
 
borgwal said:
You wrote: "I'm thinking 400 nm because of length contraction..."

This is not about length contraction, but about the Doppler shift. Check your textbook for the difference between these concepts!

You wrote: "But everything i read about red shift is if the source is moving away from the observer". You should read and reread what the "principle of relativity" says!

OK...but we're past that part now and we get it, just curious if utilizing the equations like that seem to be correct
 
Do you think you have the relativistically correct (rather than the classical) equation for the red shift? If so, then of course that equation applies.