Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Urgent Help! Time Dilation

A team of astronauts set off from a stationary space station at one quarter of the speed of light. They are carrying a highly accurate atomic clock. They travel to another space station, stationary with respect to the first one , and which is one quarter of a light year away from the first one. They immediately turn round and travel back to the original space station, at half the speed of light. How much time has elapsed on the atomic clock ? How does this compare to the time elapsed on the space station ?

Therefore I know for the first journey v=0.25c x=0.25 light years.
The return journey is v=0.5c and x=0.25 light years.

I know I have to find the time but Im confused as to which formula to use please help guys exam in afternoon! :frown:

:frown:

OP

Anyone?

:frown:

Special Relativity states:

t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

Now then for the first part, the time it takes is exactly one year, so:

t_{actual} = 3.15 \times 10^7

Put the numbers into the equation:

t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

For the second part of the journey, the time it takes is half a year, so:

t_{actual} = 1.58 \times 10^7

Again into the equation:

t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

Overall time =

4.42 \times 10^7 \text{seconds}

(edited 10 years ago)

OP

Is that overtime for the atomic clock?

Original post by Piguy

Special Relativity states:

t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

Now then for the first part, the time it takes is exactly one year, so:

t_{actual} = 3.15 \times 10^7

Put the numbers into the equation:

t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

For the second part of the journey, the time it takes is half a year, so:

t_{actual} = 1.58 \times 10^7

Again into the equation:

t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

Overall time =

4.42 \times 10^7 \text{seconds}

OP

Also how did you the the T actual value?

Original post by Piguy

Special Relativity states:

t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

Now then for the first part, the time it takes is exactly one year, so:

t_{actual} = 3.15 \times 10^7

Put the numbers into the equation:

t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

For the second part of the journey, the time it takes is half a year, so:

t_{actual} = 1.58 \times 10^7

Again into the equation:

t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

Overall time =

4.42 \times 10^7 \text{seconds}

Original post by mathslover786

Also how did you the the T actual value?

Speed = Distance * Time,
you can just cancel out the speed of light parts, so you end up with a year, and half a year respectively...

Quick Reply

Latest

Trending

Trending

Articles for you

Everyday worries about starting university (and what to do about them)

Everyday worries about starting university (and what to do about them)

What is an LLM and how could it benefit your career?

What is an LLM and how could it benefit your career?

Powerful questions business students forget to ask their career advisers

Powerful questions business students forget to ask their career advisers

Why I chose a law degree and what studying law was like

Why I chose a law degree and what studying law was like