# Rocket moving away from the Earth

Mentor
When the signal is received back on Earth. Originally, I didn't think light bouncing off a mirror was a signal being received by the ship.

Ah, I see. Well, one can always calculate both to be covered either way.

Frostman
Homework Helper
Gold Member
2022 Award
The OP seems to have put effort in. Much of the explanation has been given, but is 'scattered'. So I’m taking a bit of a liberty and posting this, hoping it will help and that it's OK with @PeroK and @PeterDonis.

Q1.

It takes time ##T## for the light to return from the rear mirror: ##\frac T 2## traveling outwards and ##\frac T 2## returning.
If the rear mirror is a distance ##d## from the source at the moment of reflection, this means: ##d = speed * time = c\frac T 2##

This is the same calculation as used in elementary (sound) echo problems. It has nothing to do with relativity.

Q2.

Let P be the position of the rear mirror at the moment it reflects.

##\Delta T## is the time for the light to travel from P to the front mirror and back to P: ##\frac {\Delta T} 2## traveling outwards and ##\frac {\Delta T} 2## returning.

Distance moved by front mirror in time ##\frac {\Delta T} 2## is ##v \frac {\Delta T} 2##.

If ##l## is the (contracted) length (distance between mirrors as observed from earth) then in time ##\frac {\Delta T} 2## the light covers a total distance ##l + v\frac {\Delta T} 2##

Also in time ##\frac {\Delta T} 2## we know light covers a distance ##c\frac {\Delta T} 2##, giving the equation:

##c\frac {\Delta T} 2 = l + v\frac {\Delta T} 2##

No relativity needed up to this point! If we replace ##l## with ##l_0 \sqrt{1-{(\frac v c})^2}## we now have an equation from which v can be found.

Edit - unclear wording improved.

Last edited:
Frostman and PeroK
Staff Emeritus
Comment 1: you didn't put in equations in the "relevant equations" part of the template. That was a mistake.

Comment 2: you don't seem to have a plan. You seem to be calculating random things hoping one will be the answer. This strategy will not get you successfully through a PhD program.

I suggest you start over. Write down the plan on how you will solve this. Write down the equations you need to execute this plan. Then execute the plan.

Frostman
Frostman
I understand the problem, I find it hard to frame the situation.
For example, I didn't really get to the fact that yes, ##c \frac{\Delta T} {2}## is the distance traveled by the light, but it's not the length of the rocket, it's the length of the rocket plus the distance traveled by the rocket in that time interval.
The exercise itself is as simple as it is banal. In the end it was a question of remembering the equations of uniform motion and then understanding what those times actually represented.
I apologize to all of you, especially @PeroK for wasting your time.
In the next few times I will try to focus more on the knowledge I have available and to try, as far as I can, to understand the text better.

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