How to Calculate Time Dilation with Spaceship Speeds | Expert Help

AI Thread Summary
To calculate time dilation for a spaceship moving at 0.92 c, the observer on the spaceship measures a time interval of 38 hours for events on Earth. Using the formula Δt = Δtp/sqrt(1 - v^2/c^2), the speed of the spaceship is first converted to meters per second. The calculations show that the time interval for the faster speed of 0.92 c results in approximately 30.97 hours. However, the initial attempt at the solution was incorrect, indicating a need for careful re-evaluation of the calculations. The discussion emphasizes the importance of accurate application of relativistic equations in time dilation scenarios.
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Time Dilation help please

Homework Statement


A spaceship is moving at 0.75 c (0.75 times the speed of light) with respect to the earth. An observer on the spaceship measures the time interval between two events on Earth as 38 hrs. If the spaceship had been moving with a speed of 0.92 c with respect to the earth, what would the time interval between the events have been?

Δt0.92 c = _____hrs

Homework Equations




Δt=Δtp/sqrt1-v^2/c^2



The Attempt at a Solution



v=.75(3e8)=2.25e8m/s
t=1/368e5sec
v=d/t
d=3.078e13m

Δtp=3.078e13/v

v=.92(3e8)=2.76e8


Δtp=3.078e13/2.76e8
=1.115e5sec
=30.97hr

but this is not the correct ans
 
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