Physics Help Forum Relativity- stars and spacecraft.

 Special and General Relativity Special and General Relativity Physics Help Forum

 Apr 3rd 2009, 08:43 AM #1 Member   Join Date: Apr 2009 Posts: 71 Relativity- stars and spacecraft. Hi, I have recently started revising for my relativity exam and don't seem to have grasped it yet. Here is a past exam Q and my attempt at it any, guidance would be greatly appreciated (particularly on part b). 1.(a.) A star is 450 light years from earth. i) what is the minimum time a spacecraft could make a return trip to this star? ii)How fast should the space craft travel to make the journey in 30 astranaught years? iii) What is the distance of eart to this star as measured by the astronauts on their 30 year journey? b.) A star is 4.25 light years from earth (according to earth observations). A spacecraft travels to this star at a uniform velocity and takes 4.25 years to reach this star (according to a clock on the ship). What is the speed of the ship as a fraction of c? c).Spacecraft a travels with a velocity 0.9c w.r.t earth. i).If spacecraft B passes A with a relative velocity of 0.5c what is the velocity of B w.r.t earth? ii). compare this to the result obtained using Gallilean transformations. Here are my attempts: 1.a. (i). 900 years (ii). I don't see how to do this as if the velocity is c in the earth frame then it is also c in every other frame so gamma is 1/0 and hence length contaction etc. do not work. What am I doing wrong? b) using the space time interval (Δx)^-c^2Δt^2=(Δx’)^2 - (Δt’)^2 Δx=Δt’=4.25 Δx’=0 (as is in spacecrafts rest frame) so rearranging: Δt=6.01 therfore v=s/t= 4.25/6.01= 0.707c c(i) (0.5c + 0.9c)/(1+0.9x0.5)=0.966c (ii) using Gallilean transformations we would get 1.4c (impossible). Any glaring errors?
 Apr 4th 2009, 06:50 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 1.(a.) A star is 450 light years from earth. i) what is the minimum time a spacecraft could make a return trip to this star? ii)How fast should the space craft travel to make the journey in 30 astranaught years? iii) What is the distance of eart to this star as measured by the astronauts on their 30 year journey? 1.a. (i). 900 years (ii). I don't see how to do this as if the velocity is c in the earth frame then it is also c in every other frame so gamma is 1/0 and hence length contaction etc. do not work. What am I doing wrong? (1) is OK (2) The problem here is that you assume he has to travel at vel c to cover the journey in 30 years (his time) . Actually because of time dilation 30 years is much more to an earth based observer so the eath based chap doesn't see any contradiction. Now you'll feel how can he travel at < c and still reach as stated. Because he experiences length contraction also .Since it is postulated that the speed of light is the same in all frames of reference, and he experiences a time dilation, he MUST experience a length contraction also for the postulate to hold. So set up the equations for time dilation and length contraction and solve them simultaneously to get both ii) & iii).
 Apr 4th 2009, 07:32 AM #3 Physics Team   Join Date: Feb 2009 Posts: 1,425 b.) A star is 4.25 light years from earth (according to earth observations). A spacecraft travels to this star at a uniform velocity and takes 4.25 years to reach this star (according to a clock on the ship). What is the speed of the ship as a fraction of c? b) using the space time interval (Δx)^-c^2Δt^2=(Δx’)^2 - (Δt’)^2 Δx=Δt’=4.25 Δx’=0 (as is in spacecrafts rest frame) so rearranging: Δt=6.01 therfore v=s/t= 4.25/6.01= 0.707c. I dont think you can take Δx’=0 (as it is the distance to the star as seen by the astronaut) or else he will be inside the star and that wouldn't be very comfortable. Try to set up the eqns for length contraction and time dilation as mentioned in the previous post and you should be able to get the answer in terms of c.
 Apr 4th 2009, 07:45 AM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 c).Spacecraft a travels with a velocity 0.9c w.r.t earth. i).If spacecraft B passes A with a relative velocity of 0.5c what is the velocity of B w.r.t earth? ii). compare this to the result obtained using Gallilean transformations. c(i) (0.5c + 0.9c)/(1+0.9x0.5)=0.966c (ii) using Gallilean transformations we would get 1.4c (impossible). This looks perfect. However, remember impossible to Einstein, but not to Galileo! Though i have tried to address your queries, I MAY BE WRONG (this is relativity!)so please check the answers and also with your prof and post the outcome on the forum.
 Apr 4th 2009, 09:01 AM #5 Member   Join Date: Apr 2009 Posts: 71 Hello again, I am trying to solve the simultanous equations but am having great difficulty. I start with: (where g is gamma and L0=900, to=30) Lg=lo t=tog and whatever I do I can't seem to eliminate enough variables, i.e. I always have a solution in both g and L or t and g or everything disappears and I get 0=0. Are these the right equations to start with? If so, what is going wrong?
 Apr 5th 2009, 05:16 AM #6 Physics Team   Join Date: Feb 2009 Posts: 1,425 1.(a.) A star is 450 light years from earth. i) what is the minimum time a spacecraft could make a return trip to this star? ii)How fast should the space craft travel to make the journey in 30 astranaught years? iii) What is the distance of eart to this star as measured by the astronauts on their 30 year journey? I am going to assume that the journey refers to a one way journey and not a return trip. The astronaut measures 30 years by his clock to complete the journey. He sees a distance of which he has to cover. If this is not clear, think of his spaceship as one frame, and the other frame as a large portion of the universe which includes the star and the earth which are assumed to be stationary with respect to each other for this problem. Think of a rod extending from the earth to the star. Now you will agree that he sees this rod as having contracted due to his motion as mentioned above. Now mentally remove the rod and you will see that the distance too has contracted, (i.e. there is no need for an object like the rod to understand this) He covers this in 30 years according to his clock. Firstly, let us get rid of all this light year business. 450 L.Y = 450 x 3 x 108 x 365 x 24 x 60 x 60 metres. 30 years = 30 x 365 x 24 x 60 x 60 secs So his speed in m/sec is v= d/t = {450 x 3 x 108 x [1-(v/c)^2]^(1/2)}/30 v = 15 c [1-(v/c)^2]^(1/2) or v^2 = 225c^2 - 25v^2 or (v/c)^2 = (225/226) or v = 0.9977c and the distance appears as 1.9908 L.Y.
 Apr 6th 2009, 06:48 AM #7 Member   Join Date: Apr 2009 Posts: 71 I have just worked out the answer to part b using the suggested method and I get 0.707c again, I don't understand why I originally got the right answer via the wrong method, and when I try it for bii it gives 0.989c which is really close, do you think this is just co-incidental? Last edited by C.E; Apr 6th 2009 at 07:43 AM.

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