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Question
A spaceship is heading directly toward the Earth at a velocity of $0.800c$ . The astronaut on board claims that he can send a canister toward the Earth at $1.20c$ relative to the Earth. (a) Calculate the velocity the canister must have relative to the spaceship. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
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Final Answer
  1. $10c$
  2. A velocity of $10c$ is impossibly fast.
  3. It's unreasonable to presume a velocity greater than $c$.
Solution Video

OpenStax College Physics Solution, Chapter 28, Problem 19 (Problems & Exercises) (3:15)

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This is College Physics Answers with Shaun Dychko. A spaceship is moving towards the Earth with a speed of 0.800<i>c</i>. And the astronaut on the ship claims that they can fire a canister at some speed that would be measured at 1.20<i>c</i> by somebody on the Earth. So let's suppose that we are in the reference frame of the Earth and we see this canister coming towards us with a speed of 1.20<i>c</i> and the ship is moving towards us with this speed and so our question is, what speed will the astronaut on the ship measure for this canister? And they are going to be measuring <i>u prime</i> and so in this relativistic addition of velocities, we need to solve for <i>u prime</i> because <i>u prime</i> is measured by the observer that is moving with speed <i>v</i>, with respect to some other reference frame. OK. So, first, we'll multiply both sides by this denominator— 1 plus <i>v u prime</i> over <i>c</i>. And then we end up with <i>u</i> plus <i>vu u prime</i> over <i>c</i> squared; there's a squared on the bottom there and that equals <i>v</i> plus <i>u prime</i>, on the right hand side because these two cancel. And then we want to collect the <i>u prime</i> terms together so we'll subtract this term from both sides so we are moving it to the right hand side, in other words. And we'll subtract the <i>v</i> from both sides and then we'll switch the sides around. And we have <i>u prime</i> minus <i>v u u prime</i> over <i>c</i> squared equals <i>u</i> minus <i>v</i> and this is useful because we have the <i>u prime</i> factor in both of these terms and then we can factor it out. So you have <i>u prime</i> times 1 minus <i>vu</i> over <i>c</i> squared equals <i>u</i> minus <i>v</i>. And then we'll divide both sides by this bracket is 1 minus <i>vu</i> over <i>c</i> squared and then we solve for <i>u prime</i>. So <i>u prime</i> then is <i>u</i> minus <i>v</i> over 1 minus <i>vu</i> over <i>c</i> squared. So, <i>u</i> is the velocity measured by the Earth based observer; <i>v</i> is the velocity with which the spacecraft is approaching the Earth, and <i>u prime</i> is the velocity of the canister measured by the astronaut which is in this moving reference frame. So we have 1.20<i>c</i> is the alleged speed of the canister, measured by somebody on Earth minus 0.80<i>c</i>, speed that the ship is approaching the Earth divided by 1 minus 0.80<i>c</i> times 1.20<i>c</i> over <i>c</i> squared and this gives 10 times <i>c</i>. Now, that is very impossible because nothing can go faster than the speed of <i>c</i> and it's unreasonable to presume that this canister will approach the Earth with a velocity of 1.20<i>c</i>. It's unreasonable to think that it will approach the Earth at any speed greater than <i>c</i>.