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Chapter 35: Problem 12

Suppose you are explaining the theory of relativity to a friend, and you have told him that nothing can go faster than \(300,000 \mathrm{~km} / \mathrm{s}\). He says that is obviously false: Suppose a spaceship traveling past you at \(200,000 \mathrm{~km} / \mathrm{s}\), which is perfectly possible according to what you are saying, fires a torpedo straight ahead whose speed is \(200,000 \mathrm{~km} / \mathrm{s}\) relative to the spaceship, which is also perfectly possible; then, he says, the torpedo's speed is \(400,000 \mathrm{~km} / \mathrm{s}\). How would you answer him?

Short Answer

Expert verified
In the given scenario of a spaceship and a torpedo, the theory of relativity takes into account the combination of their velocities and ensures that the total velocity does not exceed the speed of light (\(300,000 \mathrm{~km} / \mathrm{s}\)). According to special relativity's velocity addition formula, the relative velocity of the torpedo with respect to the observer is approximately \(327,273 \mathrm{~km} / \mathrm{s}\), which is less than the speed of light. Therefore, there is no contradiction, and the student's misconception is addressed.

Step by step solution

01

Introduce the Principle of Relativity

The theory of relativity states that the laws of physics are the same for all inertial observers, regardless of their relative velocities. In this case, we have a spaceship and a torpedo that are moving relative to an observer.
02

Explain Velocity Addition in Classical Mechanics

In classical mechanics, velocities simply add together. According to this framework, if the spaceship is traveling at \(200,000 \mathrm{~km} / \mathrm{s}\) and the torpedo is traveling at \(200,000 \mathrm{~km} / \mathrm{s}\) relative to the spaceship, the student is correct that the total speed would be \(400,000 \mathrm{~km} / \mathrm{s}\), which appears to contradict the theory of relativity.
03

Explain Velocity Addition in Special Relativity

However, special relativity provides a different equation for combining velocities, accounting for the fact that nothing can travel faster than the speed of light (\(300,000 \mathrm{~km} / \mathrm{s}\)). The relativistic velocity addition formula is given by: \(v = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\), where \(v\) is the relative velocity between the two objects, \(v_1\) and \(v_2\) are their individual velocities, and \(c\) is the speed of light.
04

Calculate the Relative Velocity of the Torpedo

Using the given example, let's calculate the relative velocity of the torpedo (in km/s) with respect to the observer: \(v_1 = 200,000 \mathrm{~km} / \mathrm{s}\) (spaceship velocity) \(v_2 = 200,000 \mathrm{~km} / \mathrm{s}\) (torpedo velocity relative to the spaceship) \(c = 300,000 \mathrm{~km} / \mathrm{s}\) (speed of light) Plug these values into the relativistic velocity addition formula: \(v = \frac{200,000 + 200,000}{1 + \frac{200,000 \cdot 200,000}{300,000^2}}\) \(v = \frac{400,000}{1 + \frac{40,000,000,000}{90,000,000,000}}\) \(v = \frac{400,000}{1 + \frac{2}{9}}\) \(v = \frac{400,000}{\frac{11}{9}}\) \(v = \frac{400,000 \cdot 9}{11}\) \(v \approx 327,273 \mathrm{~km} / \mathrm{s}\)
05

Conclusion

The relative velocity of the torpedo with respect to the observer is approximately \(327,273 \mathrm{~km} / \mathrm{s}\), which is less than the speed of light, \(300,000 \mathrm{~km} / \mathrm{s}\). Thus, the theory of relativity holds true and the student's misconception is cleared. There is no contradiction, and no object can travel faster than the speed of light according to the theory of relativity.

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Most popular questions from this chapter

Michelson and Morley used an interferometer to show that the speed of light is constant, regardless of Earth's motion through any perceived luminiferous aether. An analogy can be understood from the different times it takes for a rowboat to travel two different round-trip paths in a river that flows at a constant velocity \((u)\) downstream. Let one path be for a distance \(D\) directly across the river, then back again; and let the other path be the same distance \(D\) directly upstream, then back again. Assume that the rowboat travels at constant speed, \(v\) (with respect to the water), for both trips. Neglect the time it takes for the rowboat to turn around. Find the ratio of the cross-stream time divided by the upstream-downstream time, as a function of the given constants.

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