Time Dilation Problem; Find the proper time, ts.

AI Thread Summary
The discussion revolves around calculating the time it takes for a spaceship traveling at 0.999c to reach Alpha Centauri, which is 4.3 x 10^16 meters away. The user calculated the proper time (ts) as 1.43 x 10^8 seconds but is confused about the variable 'a' and how the textbook derives the final answer of 0.2a, equating it to two months. They also express uncertainty about the distinction between proper time (ts) and the dilated time (tm), which they calculated as 3.19 x 10^9 seconds. The conversation highlights the need for clarity on reference frames and the definitions of the variables involved in the calculations. The user plans to seek further clarification from their teacher regarding the discrepancies in the textbook's answers.
Ammar SH
Messages
18
Reaction score
0

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:
Physics news on Phys.org
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top