The Correct Time on the Clock During the Presentation Ceremony?

Click For Summary

Homework Help Overview

The problem involves relativistic effects on measurements of distance and time as experienced by a spacecraft traveling at significant fractions of the speed of light. It specifically examines how the length of a 100m track appears to an observer in motion and the time experienced on a spaceship compared to Earth time during a presentation ceremony.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of length contraction and time dilation formulas, questioning the correct interpretation of the observer's frame of reference. There are attempts to calculate the apparent length of the track and the time experienced on the spaceship.

Discussion Status

Some participants have provided guidance on the correct application of relativistic equations, while others have expressed confusion over the calculations. There is an acknowledgment of differing results, and clarification is sought regarding the calculations of length and time.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance provided. There is an emphasis on understanding the relativistic effects rather than simply obtaining numerical answers.

prime-factor
Messages
74
Reaction score
0

Homework Statement



1)An ex-student with a new spacecraft 'buzzes' the school oval on sports day at 1.5 * 10^8 m/s. How long does the 100m track appear to him.

2) The sports day began at 9am. The student buzzed the school at an average speed of 0.5c till the presentation ceremony at 3pm. If the clock on the spaceship matched the school time at 9pm, what was the time on the clock when the presentation started.

^^I am not sure I have done these correctly. If someone could check them, that would be great. Thanks very much.

Homework Equations



L = Lo sqrt[1-u2/c2]

T = To/sqrt[1-u2/c2]

The Attempt at a Solution



Let L = 100

Beta Factor:

sqrt[1-u2/c2]

sqrt ( 1 - (1.5^2/3^2))

sqrt ( 1 - 0.25)

sqrt ( 0.75)

So, 100/ (sqrt(0.75))

= 115.47 m

2)

Let u = 0.5c

T(earth) = 6hrs

To(ship) = ?

6 = To / sqrt ( 1 - (0.5^2))

6 = To / sqrt(0.75)

To = 6*Ans

To = 4.5 hrs , therefore time on spaceship clock = 9 + 4.5 = 1:30
 
Physics news on Phys.org
U must consider the observer, or the frame, carefully.

1) "Appear to him", that means the student is the observer. So u must consider he stands still and the school move. So Lo=100m, not L, and u must find L.

L = Lo sqrt[1-u2/c2]= 100.sqrt[1-(1,5/3)^2]=75m.

2)Yep, u r right :D
 
Thank you.

But, I don't get 75m for some reason.

L = sqrt(1 - (1.5^2/3^2)) * 100
L = sqrt(1 - 0.25)*100
L = sqrt(0.75)*100
L = 86.60254038

Where have I gone wrong?
 
No, u r right, I forget the sqrt :D My bad, sorry.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 54 ·
2
Replies
54
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 79 ·
3
Replies
79
Views
7K
  • · Replies 23 ·
Replies
23
Views
5K