# Basic Special Relativity

## Homework Statement

A spaceship travels from Earth to the vicinity of the star that is measured
by astronomers on Earth to be six light-years away. The spaceship
and its occupants have a total rest mass of 32 000 kg. Assume that the
spaceship travels at constant velocity. The time taken as measured by
clocks on the spaceship is 2.5 years

i) Calculate the velocity of the spaceship.

There are other parts but I think once I find the velocity I should be able to do them

## Homework Equations

I'm not sure exactly if it's using Lorentz transformations or just time dilation/length contraction. In any case here are those formulae:

ΔT=γΔT 0
L=L 0 /γ

Δx ′ =γ(Δx−vΔt)
Δt ′ =γ(Δt−vΔx/c 2 )

## The Attempt at a Solution

So I set the distance from Earth to the star as the standard length, L 0, and the time taken as measured by the spaceship to be ΔT. To find the velocity I then tried to find either ΔT 0 or L to substitute into v =Δx/Δt but to compute both I needed the velocity for γ.

Of course I tried it algebraically, solving for v but ended up with a polynomial I couldn't solve. I can post the working if necessary but I have a feeling I've made a more fundamental error with my reference frames which are still confusing me a bit.

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Doc Al
Mentor
Hint: Start by relating distance speed and time, all in the Earth frame.

Chestermiller
Mentor

## Homework Statement

A spaceship travels from Earth to the vicinity of the star that is measured
by astronomers on Earth to be six light-years away. The spaceship
and its occupants have a total rest mass of 32 000 kg. Assume that the
spaceship travels at constant velocity. The time taken as measured by
clocks on the spaceship is 2.5 years

i) Calculate the velocity of the spaceship.

There are other parts but I think once I find the velocity I should be able to do them

## Homework Equations

I'm not sure exactly if it's using Lorentz transformations or just time dilation/length contraction. In any case here are those formulae:

ΔT=γΔT 0
L=L 0 /γ

Δx ′ =γ(Δx−vΔt)
Δt ′ =γ(Δt−vΔx/c 2 )

## The Attempt at a Solution

So I set the distance from Earth to the star as the standard length, L 0, and the time taken as measured by the spaceship to be ΔT. To find the velocity I then tried to find either ΔT 0 or L to substitute into v =Δx/Δt but to compute both I needed the velocity for γ.

Of course I tried it algebraically, solving for v but ended up with a polynomial I couldn't solve. I can post the working if necessary but I have a feeling I've made a more fundamental error with my reference frames which are still confusing me a bit.

If you know Δt and Δt', then you know γ. Or, using the Lorentz Transformation, in the spaceship frame S', Δx' is zero. This, of course, gives Δt' = Δt / γ.

Thanks so much guys for the speedy replies.

Hint: Start by relating distance speed and time, all in the Earth frame.
Is that not just v=L/T where the L and T were both measured in the earth frame?
(forgive the notation I'm trying to be consistent with those in the other equations)

@chestermiller
But I don't have Δt do I? Only the time as measured on the spaceship.

Doc Al
Mentor
Is that not just v=L/T where the L and T were both measured in the earth frame?
Yes, that's all it is. Now express T in terms of the spaceship time (which was given).

yes, that's all it is. Now express t in terms of the spaceship time (which was given).
t= t'γ

Edit: and then I subbed that into v=L/t giving v=L/(t'γ)

but then γ=1/ √(1−(v^2 /c^2))

Last edited:
Doc Al
Mentor
t= t'γ
Good. Now combine those into a single equation. The only unknown will be v.

Good. Now combine those into a single equation. The only unknown will be v.
Yep sorry as I was tryng to edit on my last post I did that giving v=L/(t'γ)

but then γ=1/ √(1−(v^2 /c^2)) which is dependant on v

Thanks again for the help. I hope this isn't just me being incredibly dense.

Doc Al
Mentor
Yep sorry as I was tryng to edit on my last post I did that giving v=L/(t'γ)
Good.
but then γ=1/ √(1−(v^2 /c^2)) which is dependant on v
Sure. Good thing v is what you want to find.

Hint: Square both sides of your equation.

Good.

Sure. Good thing v is what you want to find.

Hint: Square both sides of your equation.

Got it now great! For some reason when I tried this earlier I got t and t' muddled in the time dilation equation so then when I progressed I got (v^4)/(c^2) + v^2 etc and couldn't solve it. My final answer came to be 0.92c which seems reasonable and checks out fine.
Thanks for being patient and helping out.

ps. I'm still kicking myself

can i ask a question,if i did all the calculations in the spaceship's frame,i had v=0.70c,why is it wrong?

Doc Al
Mentor
can i ask a question,if i did all the calculations in the spaceship's frame,i had v=0.70c,why is it wrong?
What exactly did you do?

What exactly did you do?
i assumed v=kc,therefore distance as measured by spaceship=6√(1-k2) light yrs, time=2.5 light yrs and velocity=6√(1-k2) /2.5 =k

solving the above equation,i have... WOW, k=0.923
ok,i think i dont have a problem anymore now that i can get v=0.923c

maybe previously i got the calculation wrong,i actually believe the velocity of spaceship should be the same no matter which reference frame i use to calculate,because velocity is usually the only thing that doesn't change across reference frames(?)

Thank you still for trying to help^^

Doc Al
Mentor
i assumed v=kc,therefore distance as measured by spaceship=6√(1-k2) light yrs, time=2.5 light yrs and velocity=6√(1-k2) /2.5 =k

solving the above equation,i have... WOW, k=0.923
ok,i think i dont have a problem anymore now that i can get v=0.923c
Good.

maybe previously i got the calculation wrong,i actually believe the velocity of spaceship should be the same no matter which reference frame i use to calculate,because velocity is usually the only thing that doesn't change across reference frames(?)
You are correct: The relative velocity between two frames is the same in both frames. (Except for sign, of course.) That's why I knew you must have made an error, since you can do this calculation starting from either frame.

You are correct: The relative velocity between two frames is the same in both frames. (Except for sign, of course.) That's why I knew you must have made an error, since you can do this calculation starting from either frame.