Rocket moving away from the Earth

In summary, the rocket travels 10 light-seconds and the distance between the two mirrors is 8 light-seconds.
  • #36
PeroK said:
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. :wink:
 
  • Like
Likes Frostman
Physics news on Phys.org
  • #37
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:
  • Like
Likes Frostman and PeroK
  • #38
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.
 
  • Like
Likes Frostman
  • #39
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.
 
  • Like
Likes Tom.G, phinds, berkeman and 3 others

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
814
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
741
  • Introductory Physics Homework Help
Replies
4
Views
783
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
953
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top