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Spaceship travels at .83 c, relativity |
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| Jun2-12, 07:44 PM | #1 |
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Spaceship travels at .83 c, relativity
1. The problem statement, all variables and given/known data
Suppose our sun is about to explode and we escape in a spaceship toward the star Tau Ceti. When we reach the midpoint of our journey, which takes place at v=.83 C we see our sun explode and, unfortunately, we see Tau Ceti explode as well (we observe the light arriving from each explosion. (there's a part a and b but i understand those parts, so i'll go straight to c and d) c)In the spaceship frame of reference, how long before we saw the Sun explode did it actually explode? (enter a positive value for times in the past.) d) In the spaceship frame of reference, how long before we saw Tau Ceta explode did it actually explode? 2. Relevant equations Δt = [itex]\gamma[/itex](Δt' - v/c2Δx') } S'-->S 3. The attempt at a solution c) I have worked out an answer using the equation I gave but I'm not sure if its correct. I said that S is my frame for the spaceship and S' is my frame for the sun. Δt = [itex]\gamma[/itex](Δt' - v/c2Δx') so Δt = 1/(sqrt(1-(.83c)2/c2))*((6 years)+(.83c*6c)/c2) Δt=19.686 years Δt' = 6 years So it actually exploded 13.686 years ago? for part d), i used the same equation but instead used -v so Δt = 1/(sqrt√(1-(-.83c)2/c2))*((6 years)+(-.83c*6c)/c2) = 1.823 years so it actually happened 1.823-6 = -4.171 4 years into the future? This is the main reason why I'm not believing my answer. |
| Jun3-12, 01:17 AM | #2 |
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hi harts! welcome to pf!
![]() ![]() (and what do you think "∆" means? )
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| Jun3-12, 05:31 PM | #3 |
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Thanks for the reply!
OK, I thought I was finding the difference in time between the two frames, which is why I subtracted the change in time from one frame from the change in time in the other frame. Am I misinterpreting the question? "In the spaceship frame of reference, how long before we saw the Sun explode did it actually explode?" |
| Jun4-12, 03:09 AM | #4 |
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Spaceship travels at .83 c, relativity
hi harts!
![]() (just got up )both measurements must be made in the spaceship frame … the time the sun exploded, and the time the light reached the spaceship
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| Jun4-12, 12:37 PM | #5 |
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OK I think I understand it now. I know that the sun actually exploded 6 years ago, but because of relativity I have to use a lorentz transformation to figure out how much time it took for us to see it.
For part c, I use my equation t'= ((-6 years)-(.83c/c^2)(-6ly))/ sqrt(1-(.83c)^2/c^2)= -1.83 For part d, I use the same equation but use 6 light years for my x value and I got -19.686 years. I guess it just took me a while to understand those lorentz transformation equations. Thanks tiny tim |
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