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#### MermaidWonders

##### Active member

- Feb 20, 2018

- 113

Here, I know that I'm supposed to find $v$, but I'm having a hard time setting up my equation(s) in order to reach the final answer.

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- Thread starter
- #1

- Feb 20, 2018

- 113

Here, I know that I'm supposed to find $v$, but I'm having a hard time setting up my equation(s) in order to reach the final answer.

Let's start with a question. You do know that you can go 85 light years in 85 years at 1x Speed of Light, right?

Here, I know that I'm supposed to find $v$, but I'm having a hard time setting up my equation(s) in order to reach the final answer.

- Aug 30, 2012

- 1,123

There is a wrinkle here. If we are talking about an observer on Earth then as tkhunny points out there must be some kind of misprint... It can't be done.

Here, I know that I'm supposed to find $v$, but I'm having a hard time setting up my equation(s) in order to reach the final answer.

If you are talking about an observer on the flight we can do it. What do you know about time dilation? (See the section "Simple Inference of Velocity Time Dilation.")

Can you finish from here? If not just let us know.

-Dan

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- #4

- Feb 20, 2018

- 113

Yes.Let's start with a question. You do know that you can go 85 light years in 85 years at 1x Speed of Light, right?

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- #5

- Feb 20, 2018

- 113

I know that the time measured in the frame in which the clock is at rest is called the proper time, and so a moving clock runs slower (hence dilated).There is a wrinkle here. If we are talking about an observer on Earth then as tkhunny points out there must be some kind of misprint... It can't be done.

If you are talking about an observer on the flight we can do it. What do you know about time dilation? (See the section "Simple Inference of Velocity Time Dilation.")

Can you finish from here? If not just let us know.

-Dan

- Aug 30, 2012

- 1,123

So what we have here isI know that the time measured in the frame in which the clock is at rest is called the proper time, and so a moving clock runs slower (hence dilated).

\(\displaystyle \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}\)

where \(\displaystyle \Delta t'\) is measured by the observer's clock and \(\displaystyle \Delta t\) is the time in the moving frame, aka "proper time."

So one possibility is if we wish the moving frame to experience 85 years and the observer's frame to be 240 years, then \(\displaystyle \Delta t' = 240\) and \(\displaystyle \Delta t = 85\). Solve for v.

-Dan

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- #7

- Feb 20, 2018

- 113

But isn't the light year a unit of distance? For instance, the 240 light years in this question would represent a certain distance?So what we have here is

\(\displaystyle \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}\)

where \(\displaystyle \Delta t'\) is measured by the observer's clock and \(\displaystyle \Delta t\) is the time in the moving frame, aka "proper time."

So one possibility is if we wish the moving frame to experience 85 years and the observer's frame to be 240 years, then \(\displaystyle \Delta t' = 240\) and \(\displaystyle \Delta t = 85\). Solve for v.

-Dan

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- #8

- Feb 20, 2018

- 113

What do you mean? It says "240 light years away".... I'm confused.You were talking about times in your original post, not distances.

-Dan

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- #9

- Mar 5, 2012

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Indeed, the distance travelled is $v\Delta t' = 240\text{ lightyear}$.What do you mean? It says "240 light years away".... I'm confused.

So we have $\Delta t' = \frac{240\text{ lightyear}}{v}$ and $\Delta t = 85\text{ year}$.

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- #10

- Feb 20, 2018

- 113

Ah, OK, that makes sense. So is $\Delta t'$ the time measured by the person on board the spaceship, and 85 years measured by a person at rest on Earth?Indeed, the distance travelled is $v\Delta t' = 240\text{ lightyear}$.

So we have $\Delta t' = \frac{240\text{ lightyear}}{v}$ and $\Delta t = 85\text{ year}$.

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- #11

- Mar 5, 2012

- 8,780

It's the other way around.Ah, OK, that makes sense. So is $\Delta t'$ the time measured by the person on board the spaceship, and 85 years measured by a person at rest on Earth?

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- #12

- Feb 20, 2018

- 113

Oops, my bad. Thanks!It's the other way around.