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Homework Help: Time taken to travel to the andromeda galaxy, oserved from earth

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A traveller with mass m = 66,5 kg travels with a velocity of 0,82 c to the Andromeda galaxy which is a distance of 2,55·106 lightyears away. Light with the wavelength 459,4 nm is emitted from the galaxy. assume the galaxy doesnt move and use the exact speed of light.

    How long time (in million years, year=365,25 days) will the journey take for an observer on earth?

    2. Relevant equations
    i dont really know...
    time dialation T = To / √( 1 - (v2/c2)

    3. The attempt at a solution
    first of all, id just like to confess that the whole relativity concept is, unsurprisingly, doing my head in a little. i can pretty much grasp the way the speed of light is a constant and that time slows down when things approach the speed of light, and im usually ok with questions on the doppler effect, red/blue shift and relativistic energy, but im fairly certain all of the pieces have yet to fall into place.

    also, i know a lot of the info in the question isnt relevant (im guessing the velocity and the distance are the ones that actually matter) nonetheless, im giving the entire question just in case.

    Ok, so the part that's confusing me about this specific problem is the matter of what effect time dialation would have on the observer on earth. i understand that the observer on earth would observe clocks moving slowly on the spaceship, but wouldnt that not affect the observer since he is outside the travelling object?

    intuitively, i'd say that the answer is just 2,55·106 lightyears divided by 0.82 c, but this is obviously too simple.

    my next instinct is to plug the values into the time dialation equation, but i would love to understand why im supposed to do that. even then, i imagine i'm missing some aspects of the question....

    obviously, its a little amateurish to try to do physics on instinct and intuition, but reading up on the subject has only made me more confused so far. would love any help on the subject, even if its just a link to yet another explanation on special relativity.
  2. jcsd
  3. Apr 11, 2013 #2


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    Your instincts are rather more accurate than you give them credit for. It's not a complicated question. You have the right procedure.
  4. Apr 11, 2013 #3
    Your intuitive answer is not only simple, it's right (congratulations!). To see this, consider the following:
    1. The observer on Earth sees the Andromeda galaxy as 2 550 000 light-years away.
    2. The observer on Earth sees the traveler moving at a speed 0,82 c towards the Andromeda galaxy.
    3. The speed of something, as measured by any observer, is just the distance travelled (as measured by that observer) divided by the time interval (as measured by the same observer).

    Different observers measure different distances, time intervals, and speeds (for that you need relativity), but we're only talking about one observer (and one who isn't changing motion by firing rocket engines to turn around or something similar).

    Therefore: the observer on Earth measures a time (for the trip out to the Andromeda galaxy, not the round trip) of 2 550 000 c year/0,82 c = 2 550 000/0,82. For a round trip (to Andromeda and back), the Earth observer would measure a time twice as long, assuming the traveler turned around as soon as she or he got to Andromeda and returned at the same speed.

    The traveler would measure a different time interval (for the traveler, it's the galaxy that moves toward the traveler, but because of the Lorentz constraction, the traveler measures a shorter distance between Earth and Andromeda than we (on Earth) do.
  5. Apr 11, 2013 #4

    well, now that i've typed out my thoughts i realised more specifically what my question is: i'm guessing i'd get the time from the velocity times distance, but which side of the equation would that time go to? would that be the T or the To ?

    i imagine the To, bbecause the To in the equation refers to the time one would experience on the spaceship, and thats the way these questions are usually formulated. this would imply that the time experienced on the spaceship would be the time one would expect the journey to take if we were dealing with normal speeds and distances, and the time experienced on earth would be longer. why is it not the other way round?

    oh, and thanks for the help so far. i reckon ill be able to do the question now, however, if you have the time i would love some explanation as to why this is so. :)
  6. Apr 11, 2013 #5
    to clarify: i'd have thought the time one would expect to experience from earth would be the distance times velocity, and the time experienced on the spaceship would be shorter. why is this not so (phew, its hard to express oneself accurately in words about this sort of thing!)
  7. Apr 11, 2013 #6
    ooops, missed you there, sir/mam! thank you, that confirmed my suspicions and made everything a whole lot clearer! sneaky, sneaky question....
  8. Apr 11, 2013 #7


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    umm... divided by
    What will be the distance as perceived from the spaceship?
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