00:01
Hello, everyone.
00:02
In this problem, we're given a situation where we have a spaceship that is flying towards a distant star that is measured in our frame to be nine light years away.
00:13
So the length of the trip that is going to be taken by the spaceship is measured to be nine light years.
00:20
Now, nine light years, we can estimate that distance, or we can say that that distance is the distance that light travels in nine years.
00:27
So we can actually write the distance as 9 .0 times the speed of flight, times the time interval of one year.
00:38
Then we're going to be asked to find how long this trip is going to take in our frame, how long this trip is going to take in the frame of someone on the spaceship.
00:48
What is the distance to the star according to someone on the spaceship? and then finally, what is the velocity of the star as seen? by someone on the spaceship.
01:00
So the first thing we can do is, since we're given the speed of the spaceship, we can calculate the gamma factor, the gamma factor for 0 .90c, so 90 % of the speed of flight, is equal to 2 .294.
01:16
All right, so then we can just kind of use our usual kinematic equations for calculating how long this time takes in our frame, how long this trip takes in our frame, we're given a distance in our frame and we also know the velocity of the spaceship in our frame.
01:33
So that all we have to do is divide the distance by the speed and since we've written nine light years as 9 .0 times c times one year we can write the distance as 9 .0 times c and that's going to be divided by the velocity which is 0 .90c and so that works out to be 10 years.
01:53
So in our frame this trip is going to take 10 years to complete.
01:57
Now, the question is what is this time as measured or how does this time change as measured by someone on the spaceship due to algorithmistic effects? so someone on the spaceship is someone who's actually completing this trip.
02:16
So they are kind of the inertial frame for this trip, meaning that they are the ones actually measuring their proper time for this trip.
02:23
So t prime, i .e.
02:26
The time of the trip as measured by someone in the spaceship or in the moving frame is the proper time for this event, meaning that the 10 years is the dilated time.
02:37
So in that case, we can write t as being gamma times t prime.
02:43
And so t prime is t over gamma, which means that t prime we can find just divided by gamma...