Velocity of relativistic spaceship

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Homework Help Overview

The problem involves a spaceship traveling from Earth to a star 6 light years away, taking 2.5 years in the spaceship's frame. The objective is to calculate the spaceship's velocity, considering relativistic effects.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss relating the equations of distance and time in the context of special relativity. There are attempts to express velocity in terms of the given distance and time, with some questioning the validity of their calculations and assumptions regarding the equations used.

Discussion Status

Some participants have provided feedback on the calculations presented, suggesting that there may be errors in the application of the equations. There is an ongoing exploration of the relationships between the variables involved, with no explicit consensus reached on the correct approach or solution.

Contextual Notes

Participants note the importance of unit consistency and the implications of using relativistic equations, indicating potential confusion regarding the interpretation of the results.

Rct33
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Homework Statement


A spaceship travels from Earth to a star that is 6 light years away. The spaceship takes 2.5 years to reach the star in its frame. Calculate the velocity of the spaceship.


Homework Equations


x=\frac{x_0}{γ}, t=γt_0


The Attempt at a Solution


I guess I have to relate the two equations to work out velocity somehow. Previous attempts where I considered the distance, 6ly divided by the velocity of the spaceship was equal to the time it takes to travel to the star as seen on Earth. I then substituted t for γt_0 where t_0=2.5y and rearranged to find v, but this was unsuccessful. I don't have any other ideas to try so hints would be appreciated.
 
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Your approach looks good. Please show your work so we can see where the calculations went wrong.
 
\frac{6}{v}=t=2.5γ where the velocity is a fraction of c, 6 is in light years and t, 2.5 are in years.

Implies:

\frac{6}{2.5}=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}

∴v=\frac{12c}{\sqrt{25c^2 + 144}}=2.4 solved with wolfram because tired

Can't understand why I get 2.4c as an answer?
 
Rct33 said:
\frac{6}{v}=t=2.5γ where the velocity is a fraction of c, 6 is in light years and t, 2.5 are in years.

That should be: \frac{6c}{v}=t=2.5γ

Chet
 
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Chestermiller said:
That should be: \frac{6c}{v}=t=2.5γ

Chet

Cheers
 
$$\frac{6}{2.5}=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}$$
This has a solution with v smaller than 1.

$$∴v=\frac{12c}{\sqrt{25c^2 + 144}}=2.4$$ Don't use c here (or plug in 1), as you worked with years=speed of light = 1 anyway.

A proper calculation with units would not have this issue...
 

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