# Spaceship moving away from light source

1. Oct 15, 2008

### nick227

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Oct 15, 2008

### D H

Staff Emeritus
You're on the right track, but you are solving it backwards. You know the frequency of the light source as observed in the light source's rest frame, and you know what happens to the frequency. Hint: Does the moving away from the Earth make the street lamp redshift or blueshift?

3. Oct 15, 2008

### nick227

Well seeing as how im 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?

4. Oct 15, 2008

### D H

Staff Emeritus
The observer is moving away from the source when the source is moving away from the observer. There is no such thing as absolute position.

5. Oct 15, 2008

### phyguy321

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?

6. Oct 15, 2008

### borgwal

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!

7. Oct 16, 2008

### phyguy321

OK....but we're past that part now and we get it, just curious if utilizing the equations like that seem to be correct

8. Oct 16, 2008

### borgwal

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.