A physicist decides she would like to remain 21 for 10 years. What is the minimum constant speed relative to earth at which she would have to take a rocket trip into outerspace and back in order to achieve this?(adsbygoogle = window.adsbygoogle || []).push({});

Let t=10 and [tex]t_{o}=1[/tex]

[tex]t=\frac{t_{o}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]10=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{10}[/tex]

[tex]1-\frac{v^2}{c^2}=\frac{1}{100}[/tex]

[tex]\frac{v^2}{c^2}=\frac{99}{100}[/tex]

[tex]v^2=8.91\times10^{16}[/tex]

[tex]v=298496231.1=0.9949874371c[/tex]

This was the correct answer when I looked up the solutions.

What Ido not understandis why thenormal timeis considered to be1 yearsinceten yearsofproper timehas actually past. If I do this problem the other way, a negative answer is obtained. Could someone please explain why this occurs, or have I just not grasped the concepts of what is the proper time and what is the relative time?

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# Homework Help: Relativity Question

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