Is the ratio of the two times in the right direction?

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Homework Help Overview

The problem involves calculating the speed of a spacecraft traveling from the Sun to the Earth, a distance of 8.3 light-minutes, as measured in the rest frame. The spacecraft's onboard clocks measure the trip duration as 5.0 minutes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the distance, time, and the Lorentz factor (gamma). There is uncertainty regarding the correct application of time dilation and the equations involved. Some participants question the original poster's calculation of gamma and the interpretation of the time measurements.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to relate the distance and time using the appropriate equations. There is exploration of different interpretations of the time dilation relationship and its implications for calculating velocity.

Contextual Notes

Participants note the need to clarify the time experienced by an observer at rest versus the onboard time of the spacecraft. There is also mention of confusion regarding the equations used and the assumptions made about the scenario.

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Homework Statement



Sun and Earth are 8.3 lightminutes apart, as measured in their rest frame. What is the speed of a spacecraft that makes the trip in 5.0 min according to its onboard clocks?

Homework Equations





The Attempt at a Solution


I'm not really sure how to proceed. I figured out that 8.3 lightmins = 1.5 x 10^11 m. I was thinking that since the ship traveled in a time span of 5 minutes that gamma = 1.66 (8.3/5) but apparently that was incorrect.

Is the ratio of the two times in the right direction? I don't see any other way to find the velocity
 
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I'm confused by this equation you gave...

gamma = 1.66 (8.3/5)

this isn't right. remember 8.3lightminutes is a distance as you calculated...

What is the time the trip takes according to an observer at rest? The distance divided by that time is the velocity you need... you'll have both a gamma and a v in your equation... but you can rewrite gamma in terms of v... then you can solve the equation for v.
 
Last edited:
t = x / v = 8.3 lightmins / v

What equation should I be using so far I've tried x' = gamma(x - vt) and t' = gamma(t -vx/c^2) but I haven't been able to come up with anything useable with them yet.

I would think the former, but I tried that and I actually got c for the answer.
 
jesuslovesu said:
t = x / v = 8.3 lightmins / v

Yes, so t = 1.5*10^11/v

That's the time seen by an observer at rest... an observer inside the ship sees 5 min = 300s.

So use the time dilation relationship to relate 1.5*10^11/v and 300
 
Last edited:
Time dilation: [tex]\Delta{t} = \gamma\Delta{t_0}[/tex]

where [tex]\Delta{t_0}[/tex] is the proper time (time elapsed in the ship)
 

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