Time Dilation Problem; Find the proper time, ts.

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Homework Help Overview

The problem involves calculating the proper time for a spaceship traveling to Alpha Centauri at a relativistic speed of 0.999c. The original poster expresses confusion over the relationship between their calculations and the textbook's answers, particularly regarding the variable 'a' and the conversion of time into months.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations for proper time (ts) and time in the moving frame (tm), questioning the definitions of variables and the implications of time dilation. There is uncertainty about how the textbook arrives at its conclusions, particularly regarding the variable 'a' and its relation to time in months.

Discussion Status

The discussion is ongoing, with participants seeking clarification on variable definitions and the steps leading to the final answer. Some participants suggest that the original poster may have misinterpreted the reference frames involved in the problem.

Contextual Notes

There is a lack of explicit information about reference frames in the problem statement, leading to confusion about the calculations and the meaning of certain variables. Participants are encouraged to clarify their understanding of the problem setup and the definitions of the variables used.

Ammar SH
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Homework Statement


Alpha Centauri, the closest star to Earth, is 4.3x10^6 m away. How long would it take a spaceship to reach the star if it were traveling at 0.999c?


Homework Equations



I did get the answer as 0.2a...and the textbook also said it would take about 2 months,
I do not understand what the a represents/stands for and how the textbook got 2 months from 2a.
Also, I do not why I had to find tm and then to?

The Attempt at a Solution



ts = d/v = 4.3 x 10^16/0.999c = 1.43x10^8 s

tm = ts/√(1-v^2/c^2
= 1.43x10^8/√(1-(0.999c)^2/c^2
= 3.19 x 10^9 s
tm = 4.5a

THE ANSWER KEy gets the same answer till the part above but then the answer key starts doing more calculations afterwards(below) which I do not get

ts = tm√1-v^2/c^2

= 4.5a√1-(0.999c)^2/c^2
ts = 0.2a

the textbook gave 0.2a but also said it will take two months to reach the star and that is what I do not understand.


Thanks for your help.
 
Last edited:
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Quick question, could you define all of your variables? a, tm, ts to etc.

Also, is that really the full question? It doesn't state anything about reference frames.
 
yes that is the whole question and you are right it does not tell you the frames...Im guessing you are suppose to figure out the frames yourself..


d from Earth to star= 4.3x10^16 m
v= 0.999c=3x10^8x0.999=2.9997X10^8

I used the distance time formula to find the ts(well I think the answer I found is ts)
ts = 1.43x10^8 s

In case if you do not know: ts is the proper time, the time interval between two events measured by an observer who sees the events occur at one position
and tm is the time interval for an observer moving with a speed v relative to the sequence of events.


And that is what confuses me... I found the right answer but i don't understand how and why

tm=3.19x10^9s is also equal to 4.5a

i mean what is a??! I am going crazy...

Uptill 4.5a I got but then the answer key does the next steps which makes no sense and
says the final answer is 0.2a which is two months.

how is 0.2a = two months?? what does a stand for and what# does it equal?
 
"= 3.19 x 10^9 s
tm = 4.5a"

Super confused about what happened here.

Am I missing some unit of time?

Btw, you should probably try to put units into your solutions throughout.

I don't think you found ts, I think you found the time in alpha centuri's reference frame. You can then use time dilation to find out how long it took in the reference frame of the ship. Or you could use length contraction to find the distance in the reference frame of the ship and solve it that way.
 
yea sure I will do that but the book literally states in answers(unless they are wrong)

4.5a=3.19x10^9 secondsi will try one more time

thanks for your help I think I will speak to teacher about this question, maybe something is wrong with it
 
Last edited:

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