Time Dilation spacecraft watches

AI Thread Summary
The discussion revolves around calculating time dilation between two observers, A on Earth and B in a spacecraft traveling at 2.0 x 10^8 m/s. Participants are confused about the correct application of the time dilation equation, specifically how to derive the time difference of one second between the two observers. One user calculated that for every second that passes for A, 1.34 seconds pass for B, but they are uncertain about the book's answer of 3.92 seconds. Clarification is provided that the goal is to find the time elapsed for A when the difference between their watches is one second. The conversation highlights the complexities of modern physics and the challenges faced in understanding these concepts.
crazynut52
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I have the attached equation
but when I plug it, I don't get the same answer as in the back of the book...

Here is the question

Two observers, A on earth, B in a spacecraft whose speed is 2.0X10^8 m/s both set their watches to the same time, when the ship is abreast the earth.

How much time must elapse by A's reckoning before the watches differ by 1.00 seconds?
 

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just a quick thought

you do know that the speed of light is normalised to 1

hence if you are traveling at 2x10^8 ms^-1 you are traveling 2/3 times the speed of light hence your speed is 2/3 relative to light

does this help in anyway
 
i understand it is 2/3 the speed of light but am I using the right equation?
using this equation I found that one second passes for A as 1.34 seconds passes to person B. But not sure how the book gets 3.92s as the answer
 
I can't see the equation yet because it hasn't been moderated however if you are using the following equation

1/sqrt(1-\v^/c^2)


then I get the same answer as you 1.34s
 
Last edited:
yes I am using the same equation as you, but the answer in the back of the book is 3.93, not 1.34
 
Hey,
For that question, my answer is 2.93 for A. If the question asks B's time, then I guess it's 3.93... I don't know... xD
 
crazynut52 said:
i understand it is 2/3 the speed of light but am I using the right equation?
using this equation I found that one second passes for A as 1.34 seconds passes to person B. But not sure how the book gets 3.92s as the answer

Here's the point, you are looking for how much time has passed for A when the difference between A and B is one sec. In the example you just gave, A reads 1.34 sec and B reads 1 sec, for a difference of 0.34 sec.

Hint: you are looking for the value of T0 when T1= T0-1
 
or in a different notation,

\Delta t_A - \Delta t_B = 1 sec

You are asked to find \Delta t_A
 
I get it now. Thanks guys I appreciate the help. This semster of modern physics is going to be tough.
 
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