Time Dilation of Starship

1. Apr 13, 2010

iaberrant

1. The problem statement, all variables and given/known data

On a Starship, which travels at 0.93c, dinner is served between 8:15pm and 8:45pm, according to the clock on the ship. How long is dinner served according to stationary observes on Earth watching the ship on a video monitor?

2. Relevant equations

I believe the equation for time dilation to be

t'= t √ 1 - v^2/ c^2

I am sure as i do not fully understand time dilation

3. The attempt at a solution

t'= t √ 1 - v^2/ c^2

t' = 8:15 pm
v=0.93c
c=300 m/s

is this correct so far?
i need help continuing on with solving the question

Thankyou!

2. Apr 13, 2010

Staff: Mentor

Remember that moving clocks are observed to run slow. According to Earth observers, the spaceship's clocks are moving. How much time elapses according to the moving clock? What does the time dilation formula tell you about the time according to Earth clocks?

3. Apr 14, 2010

iaberrant

ok so would that mean that that t would be the time in the space ship ? 30 minutes which converts to 1800 seconds?
then velocity is simply 0.93c and substituting these values into the equation we get:
t'= t / (√ 1 - v^2/ c^2)
t'= 1800 / [√ 1 - (0.93c^2)^2/ c^2]
t'= 1800 / [√ 1 - 0.93c^2]
t'=4897.166056 seconds
and converted to minutes is 81.61 minutes?

is this correct?

thank you soo much for your help !!~

4. Apr 14, 2010

Staff: Mentor

Looks good! You don't really have to convert to seconds, but no harm done.

The Lorentz factor (λ) = 2.72, so the observed time is 2.72 X 30 minutes.

5. Apr 15, 2010

iaberrant

thankyou!!

also how do you calculate the lorentz factor ?

6. Apr 15, 2010

Matterwave

It's just 1/sqrt(1-v^2/c^2)

7. Apr 15, 2010

Staff: Mentor

The "Lorentz factor" is just a name for that factor that appears in many relativistic equations; it's also called "gamma" or γ. (See Matterwave's post.) You've been calculating it all along.