A rocket traveling at 0.50c sets out for the nearest star, Alp...

Question

A rocket traveling at 0.50c sets out for the nearest star, Alpha Centauri, which is 4.3 ly away from earth. It will return to earth immediately after reaching Alpha Centauri. What distance will the rocket travel and how long will the journey last according to (a) stay-at-home earthlings and (b) the rocket crew? (c) Which answers are the correct ones, those in part a or those in part b?

Accepted Answer

Hey, everyone in this problem, we're told that a spacecraft is moving at a speed of 0.45 C from Mars to a distant star 12 light years away. The spacecraft plans to reach the star and return to Mars in the shortest possible time. And we're asked to calculate the spacecraft's trip distance and time as viewed by in part one, the stationary observers on Mars. And in part two, the crew aboard the spacecraft, we're given four answer choices in this problem A through D each of them containing a different distance and time for part one and part two. Now we're gonna come back to these answer choices as we work through the problem and narrow them down and we'll give some more details about them when we do that. Now, this is a relativity type problem. OK. We're looking for some distance and time in different reference frames. So the first thing we wanna do is go ahead and define what those reference frames are. So we're gonna define our inertial reference frames and we're gonna have frame S OK? And this is going to be our marring. OK. So frame S is gonna be the MARS frame and then we have frame s prime and that's going to be our spacecraft frame. All right. So we have our two frames of reference and we're interested in how this spacecraft is moving. So in the spacecraft frame that spacecraft is stationary, that means we can use our length contraction in time dilation equations when we work through um this problem. OK. So let's start by writing out the information we know. So in frame S hey, what do we know? Well, we know that the speed V OK is going to be 0.45 C and that's the speed of the spacecraft. We know that delta X, the distance it travels relative to Mars is 12 light years. You know, we have to be a little bit careful here. Mars is um 12 light years away from the star that we're moving to, but the spacecraft is going to go to the start and return to Mars. OK. So the distance is actually going to be 12 light years multiplied by two. OK. The distance to get there and then the distance to get back. And so that's going to be 24 light years. OK. All right. So that's what we know about frame S, we don't really know anything about S prime right now. OK. That's what we're asked to calculate. But we do know that frame S prime moves at a speed V equal to 0.45 C relative to frame S OK. So that spacecraft frame is moving that speed that the spacecraft is traveling relative to Mars. OK. Relative to our Mars frame. All right. So let's start with part one here. And on part one, we are looking for the distance and the time um according to that stationary observer on Mars. OK. So for part one, let's start with the easy part here. Now, the distance OK, we've already been told the distance relative to Mars that this um spacecraft is moving, we calculated what the total distance is to get to that distant star and back. And so our distance here is just going to be that 24 light years. OK. So that is the answer for part what for the distance? Now, for the time we know the speed, we know the distance, we can calculate time using our typical equation. OK. So the speed V is equal to the distance delta X divided by the time delta T, we can rearrange to solve for delta T and it's going to be equal to delta X divided by V and we can go ahead and substitute in our values. And so delta T is going to be equal to 24 light years divided by 0.45 C. Now recall that C the speed of light is equivalent to a light year per year. So we can write rewrite this as delta T is equal to 24 light years, divided by 0.45 multiplied by one light year per year. The unit of light years is gonna divide out. We have a unit of per year in the denominator. So that's gonna become a unit of year when we switch that into the numerator. And so we're gonna get that, our time is in years. And when we work this on, on our calculator delta T is going to be about 53.333 repeated years. All right. So this is the distance and the time of the spacecraft trip as viewed from stationary observers on Mars. OK. Standing on Mars looking at this spacecraft traveling. Let's go to our answer choices. Now, for part A and part B we had that the distance was 12 light years, the time was 27 years according to Mars, that's not what we found. So we can eliminate option A and B options, C and D have the distance as 24 light years and the time as 53 years is that is what we found for part one. So either C or D could be correct and now we're gonna move on to part two, we're gonna do the exact same thing, but now we're interested in this frame of reference of our spacecraft. OK. So according to the crew on the spacecraft, what's going on. So for part two, we're actually going to use the length contraction and time dilation equations. And we mentioned being able to use those because our spacecraft is stationary relative to this reference frame that we have here. OK. So we are gonna find the distance using our length contraction formula. And recall that that equation tells us that the length or the distance in frame S prime L prime is going to be equal to the square root of one minus beta squared multiplied by L which is the distance or length in frame S OK. So we know L that's our 24 light years. What about this beta value? Well, recall that beta is a V that speed divided by C, the speed of light, we know that V is 0.45 C. So beta is gonna be 0.45 C divided by C which just gives us 0.45. We can substitute this into a length contraction equation. And we get that the length or distance in frame S prime L prime is going to be the square root of one minus 0.45 squared multiplied by 24 light years. And when we work this out on our calculator, we get a length of about 21.43 27 light years. OK. So we have now the distance of the spacecraft travels according to the crew aboard the spacecraft. Now we're gonna go ahead and do the time that they think it takes or the time that they observe. It to take. And for this one, we're gonna use our time dilation. OK. So we have delta T prime is going to be equal to delta T multiplied by the square root of one minus beta squared. And you may have seen this equation rearrange. Sometimes it's written as delta T is equal to delta T prime divided by the square root of one minus beta squared. We're just multiplying by that square root on both sides to write this in as an equation for delta T prime. Since that's what we're looking for. Now, delta T, we know, OK, that's that time in the mars frame of reference. So 53.33 years and we know beta, we just calculated it. So for delta T prime, we're gonna substitute in these values. We get that it's going to be equal to 53.333 years multiplied by the square root of one minus 0.45 squared. And when we work this out on our calculator, we get a time of 47.6282 years approximately. Now comparing this to our answer choices, what do we have? Ok. We had narrowed it down to option C or D based off of part one and part two A, we found a distance of approximately 21 light years in a time of approximately 48 light years. So this is going to correspond with answer choice C. Hi, thanks. Everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Relativity of Simultaneity

Time Dilation

Length Contraction

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