How to Calculate Time Required for 11 Light Years @ 0.9c

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Homework Help Overview

The discussion revolves around calculating the time required for a spacecraft traveling at 0.9c to reach a planet 11 light years away, focusing on the perspective of the passenger onboard the spacecraft.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of different frames of reference, questioning whether the distance of 11 light years is measured from the passenger's perspective or a stationary observer's perspective. There is discussion about the validity of using the formula time = distance / speed in this context.

Discussion Status

Participants are actively engaging with the problem, raising questions about the clarity of the original question and the assumptions involved. Some guidance has been offered regarding the need to consider the frame of reference when calculating time, but no consensus has been reached on the correct approach.

Contextual Notes

There is uncertainty regarding whether the distance of 11 light years is in the passenger's frame of reference or a stationary observer's frame, which affects the calculations being discussed.

doc.madani
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You are in a spacecraft that is traveling at 0.9c (according to the passenger), and you want to get to a planet that is 11 light years away. How long will it take to get to the planet ?

This may seem like a simple problem but it's been bugging me. I answered this question by using the simple time = distance / speed since the person is in the same frame of reference as the space craft, is this correct?
 
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Hi doc.madani! :smile:
doc.madani said:
You are in a spacecraft that is traveling at 0.9c (according to the passenger), and you want to get to a planet that is 11 light years away. How long will it take to get to the planet ?

The question isn't clear …

according to the passenger, the speed of the spacecraft is zero. :confused:

I'll guess it means, if the speed and distance as measured by a stationary observer are 09.c and 11 l-y, then how long on the passenger's clock does it take?
 
I was just quoting the exact question that I was given In a test :s however since the passenger is in the same frame of reference (inertial frame if reference) to the spacecraft you can simply use the time = distance over speed formula ?
 
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Yes, but if he got the 11 light-years from a standard Galactic Maritime Federation astro-chart (sorry, I don't have a link :redface:), that'll be the distance in a stationary frame, and you need the distance in his frame. :wink:
 
Ok for arguments sake let's say it was 11 light years in his frame of reference :) your starting to scare me that there's more to the question than I anticipated :s lol that's ok
 
doc.madani said:
Ok for arguments sake let's say it was 11 light years in his frame of reference :) your starting to scare me that there's more to the question than I anticipated :s lol that's ok

If it was 11 light years in his frame of reference, then yes, he can divide by the speed to get the time on his clock. :smile:
 
ohh :) good thank you very much tiny-tim for your help :D
 

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