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Homework Help: Why am I getting this relativity velocity addition problem wrong?

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A spacecraft S2 is capable of firing a missile which can travel 0.98c. S2 is escaping from S1 at a speed of 0.95c when it fires a missile towards S1.

    part A) According to the pilot of S2, what speed does the missile approach S1?
    Part B) according to pilot of S1, what speed does the missile approach it?

    2. Relevant equations
    Call the S1 frame x and S2 frame y and speed of missile U

    Velocity addition V[itex]_{x}[/itex] = [itex]\frac{v_{y} + U}{1 + \frac{v_{y} U}{C^{2}}}[/itex]

    3. The attempt at a solution

    My problem lies with part A. The answer is just a simple 0.98c - 0.95c = 0.03c. However I cant get this result with the velocity addition formula, why is it in this case the velocity addition formula does not work/ does not apply?

    I tried imagining S2 moving to right (positive) and firing the missile backwards towards S1 (left direction which is negative). Taking the frame of reference of S2, the spaceship S2 is stationary and S1 is moving to the left at a velocity of -0.95c, the missile is also moving to left with speed -0.98c

    if I try use the velocity addition formula Velocity addition V[itex]_{x}[/itex] = [itex]\frac{-0.98 - 0.95}{1 + \frac{0.98 x 0.95}{C^{2}}}[/itex] I get -0.9994C, which is wrong. The answer is just 0.98c - 0.95C but I cannot see what I am doing wrong with the velocity addition formula or why it is not needed in this case.

    I solved part B) using the formula V[itex]_{x}[/itex] = [itex]\frac{0.98 - 0.95}{1 - \frac{0.98 x 0.95}{C^{2}}}[/itex]. The signs are as they are as in the frame of S1, the ship S2 is moving in + direction with speed 0.98C and the missile is moving with -0.95C. It seems to work for part B but not for part A and I cannot see why.

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

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 26, 2012 #2


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    For part (A) you don't need to use the velocity addition formula. You already know how fast the missile travels relative to S2 and also how fast S1 moves relative to S2. You just want to know how fast the missile is "closing" on S1 as measured by S2.

    It's the same as asking if S2 rolls a ball at 5 m/s along her x-axis and then rolls a ball at 7 m/s along her x-axis, how fast is the second ball closing on the first ball according to S2? No relativity needed since all measurements are in one inertial frame. You are not switching frames of reference.
  4. Dec 26, 2012 #3


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    You use the relativistic formula when you are working between two frames.

    For instance, in part B) you add adding two velocities from S2's frame but want an answer for S1 frame.

    When you are working just in one frame, it is is not used. For example in part A) you are adding two velocities according to S2 and want an answer for that frame.

    Think of it is this way. in part (A, you have a missile traveling away from S2 at .98c

    After 1 sec, the missile will be 0.98 light secs further away.

    S1 is traveling away a 0.95c, so after 1 sec, it will be 0.98 light sec away.

    This means that, according to S2, after 1 sec the missile will be 0.03 light sec closer to each other. which works out to a difference of 0.03 c between S1 and the missile according to S2.
  5. Dec 26, 2012 #4
    Ahhh I see thats much clearer now. So its basically because for part A the measurements are given from the frame of S2 and you are working in the same frame of reference because you want an answer for S2 so it is not needed. However in part B you are trying to take the position from S1 and so you are using measurements which were given from the frame in S2 and therefore need to use the velocity addition because you are switching frames of reference
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