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Spaceship Time dilation problem

  • Thread starter Benzoate
  • Start date
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1. Homework Statement
A spaceship departs from Earth for the star Alpha Centauri whic is 4 light years away. The spaceship travels at .75c. Howlong does it take to get there (a) as measured on Earth and (b) as measured by by a passenger on the spaceship


2. Homework Equations
proper time = delta(t')=2D/v

delta(t)=delta(t')/sqrt(1-v^2/c^2)

3. The Attempt at a Solution

In part a, the observer is going to the measured that the earth travels at the speed of light, so I would used the equation , delta=delta(t')/sqrt(1-v^2/c^2)

In part b, the observer on the space ship is going to measured the proper time and will not notice that the spaceship is traveling closed to the speed of light . so he will used the equation delta(t') =2D/v= 2*(4*light*years)/(.75c)

Are both my observations correct?
 

Doc Al

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You have it a bit backwards. It's the Earth observer who sees the ship move at 0.75c, so the Earth observer simply says the trip takes T = D/V. (Not 2D/V !)

(Note that the distance is 4 light years according to Earth observers; the spaceship observers measure a different distance.)
 
419
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You have it a bit backwards. It's the Earth observer who sees the ship move at 0.75c, so the Earth observer simply says the trip takes T = D/V. (Not 2D/V !)

(Note that the distance is 4 light years according to Earth observers; the spaceship observers measure a different distance.)
How do I have it backwards when the observers on earth who are at rest are gonna see the ship approach the speed of light and therefore , in the inertial frame containing the observers of the earth, time will seem slower. For the person

I didn't think I needed to find the distance the passenger on the spaceship measures from the spaceship to the earth since the time dilation formula contains the proper time interval. delta(t) =delta(t')/gamma
 

Doc Al

Mentor
44,815
1,078
How do I have it backwards when the observers on earth who are at rest are gonna see the ship approach the speed of light and therefore , in the inertial frame containing the observers of the earth, time will seem slower.
It's certainly true that Earth observers will view the moving clocks on the spaceship as running slow. Thus the travel time according to Earth observers will be longer than the travel time according to the spaceship clock.

I didn't think I needed to find the distance the passenger on the spaceship measures from the spaceship to the earth since the time dilation formula contains the proper time interval. delta(t) =delta(t')/gamma
It's also true that you don't need to find the distance according to spaceship observers (that's why it was a parenthetical remark). But in order to use the time dilation formula, you need to start by calculating one of the times. The travel time according to Earth observers is easy to find using D = V*T, since you know the distance and speed as measured by Earth observers.
 

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