Time Dilation Problem; Find the proper time, ts.

In summary, the conversation discusses the time it would take for a spaceship to reach Alpha Centauri, the closest star to Earth, if it were traveling at 0.999c. The solution involves using the distance-time formula to find the proper time (ts) and the time interval for an observer moving with a speed v (tm). The confusion arises from the use of the variable "a" which is not clearly defined. The answer key suggests further calculations, but it is unclear how they relate to the solution. It is recommended to consult with a teacher for clarification.
  • #1
Ammar SH
19
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
  • #2
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.
 
  • #3
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?
 
  • #4
"= 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.
 
  • #5
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:

1. What is time dilation?

Time dilation is a phenomenon in which time appears to pass at different rates for objects in motion relative to one another. It is a prediction of Einstein's theory of relativity and can be observed when comparing clocks on objects moving at different speeds.

2. How does time dilation affect the measurement of time?

Time dilation causes time to pass at a slower rate for objects that are moving at high speeds or in strong gravitational fields. This means that measurements of time may differ between two objects that are in motion relative to each other.

3. What is the proper time?

The proper time, also known as the proper time interval, is the time measured by a clock that is at rest with respect to the observer. It is the time experienced by an object or observer in its own frame of reference and is considered the "true" time for that object.

4. How is the proper time, ts, calculated?

The proper time can be calculated using the formula ts = t√(1 - v^2/c^2), where t is the time measured by a clock in motion, v is the velocity of the object, and c is the speed of light. This formula takes into account the effects of time dilation and allows for the calculation of the proper time for objects in motion.

5. What are some real-life examples of time dilation?

Some real-life examples of time dilation include the time difference experienced by GPS satellites in orbit compared to clocks on Earth, the time dilation experienced by astronauts in space, and the time dilation observed in particle accelerators. These examples all demonstrate the effects of high speeds and strong gravitational fields on the measurement of time.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
931
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Special and General Relativity
Replies
4
Views
744
  • Introductory Physics Homework Help
Replies
4
Views
796
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
945
Back
Top