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Chapter 38: Problem 41

An unpowered rocket moves past you in the positive \(x\) direction at speed \(v^{\text {rel }}=0.9 c\). This rocket fires a bullet out the back that you measure to be moving at speed \(v_{\text {bullet }}=0.3 c\) in the positive \(x\) direction. With what speed relative to the rocket did the rocket observer fire the bullet out the back of her ship?

Short Answer

Expert verified
The speed of the bullet relative to the rocket is \(0.82c\)

Step by step solution

01

- Identify Variables

Let the speed of the rocket relative to the observer be denoted as \(v_{\text{rel}} = 0.9c\). Let the speed of the bullet relative to the observer be denoted as \(v_{\text{bullet}} = 0.3c\). We need to find the speed of the bullet relative to the rocket.
02

- Use the Relativistic Velocity Addition Formula

The relativistic velocity addition formula is \[u' = \frac{u - v}{1 - \frac{uv}{c^2}}\], where \(u'\) is the velocity of the bullet relative to the rocket, \(u\) is the velocity of the bullet relative to the observer, and \(v\) is the velocity of the rocket relative to the observer.
03

- Substitute Values into the Formula

Substitute \(u = 0.3c\) and \(v = 0.9c\) into the formula: \[u' = \frac{0.3c - 0.9c}{1 - \frac{(0.3c)(0.9c)}{c^2}} = \frac{-0.6c}{1 - 0.27}\].
04

- Simplify the Denominator

Simplify the denominator: \[u' = \frac{-0.6c}{0.73}\].
05

- Divide to Find the Resulting Speed

Calculate the division: \[u' = -0.82c\].
06

- Interpret the Result

The negative sign indicates that the bullet was fired in the direction opposite to the motion of the rocket from the rocket鈥檚 frame of reference.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

relativistic velocity addition
Relativistic velocity addition is crucial when dealing with speeds close to the speed of light. Unlike classical mechanics, where velocities simply add or subtract, relativity requires a more complex formula. This is due to the fact that time and space are intertwined in special relativity.
The formula used to combine velocities in special relativity is: \[u' = \frac{u - v}{1 - \frac{uv}{c^2}} \], where:
  • \(u'\) is the velocity of the bullet relative to the rocket
  • \(u\) is the bullet's speed from the observer's viewpoint
  • \(v\) is the rocket's speed from the observer's viewpoint
This formula ensures that no object's speed surpasses the speed of light \(c\). When combining speeds, the denominator corrects for relativistic effects, maintaining the principles of special relativity.
In our exercise example, the observer sees the bullet moving at \(0.3c\) and the rocket at \(0.9c\). Using the relativistic addition formula, we find the speed of the bullet as seen from the rocket.
speed of light
The speed of light \(c\) is a cornerstone of modern physics, constantly about \(3 \times 10^8 \) meters per second. It's not only a high-speed limit in the universe but also integral to Einstein's Theory of Relativity.
One interesting thing about the speed of light is that it's invariant, meaning all observers, regardless of their motion, measure it the same.
Why is this important in our exercise? The relativistic velocity addition formula ensures that computed velocities respect this universal limit. Despite the high speeds of the rocket and bullet, neither exceeds \(c\).
Understanding \(c\) underpins our grasp of relativity and ensures the accurate calculation of relativistic effects. So, next time you think about fast-moving objects, remember that the speed of light sets an unbreakable limit on how fast anything can go.
rocket motion
Rocket motion in the context of relativity introduces some intriguing aspects. When rockets approach significant fractions of the speed of light, we can't rely on Newtonian mechanics alone; Einstein's special relativity comes into play.
In our exercise, we observed a rocket moving at \(0.9c\). At such high velocities, time dilation and length contraction should be considered, though they don't directly affect the speed calculation here.
What makes the scenario unique is the interaction of the rocket speed with the bullet speed. The rocket's high velocity dramatically alters the perception of the bullet's speed depending on the observer's frame of reference. From the rocket's perspective, the bullet seems to travel almost as fast backward relative to it, despite both moving forward relative to a stationary observer.
Thus, mastering rocket motion in relativity involves understanding how different reference frames interact and how movements are perceived differently based on these frames. Combine this with the relativistic velocity addition formula, and you get a comprehensive picture of high-speed motion in the universe.

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Most popular questions from this chapter

A train moves at \(10 \mathrm{~km} / \mathrm{h}\) along the track. A passenger sprints toward the rear of the train at \(10 \mathrm{~km} / \mathrm{h}\) with respect to the train. Our knee-jerk motto says that the train clocks "run slow" with respect to clocks on the track, and the runner's watch "runs slow" with respect to train clocks. Therefore the runner's watch should "run doubly slow" with respect to clocks on the track. But the runner is at rest with respect to the track. What gives? (This example illustrates the danger of the simple knee-jerk motto "Moving clocks run slow.")

A spaceship moving away from Earth at a speed \(0.900 c\) radios its reports back to Earth using a frequency of 100 MHz measured in the spaceship frame. To what frequency must Earth's receivers be tuned in order to receive the reports?

One kilogram of hydrogen combines chemically with 8 kilograms of oxygen to form water; about \(10^{8} \mathrm{~J}\) of energy is released. Ten metric tons \(\left(10^{4} \mathrm{~kg}\right)\) of hydrogen combines with oxygen to produce water. (a) Does the resulting water have a greater or less mass than the original hydrogen plus oxygen? (b) What is the numerical magnitude of this difference in mass? (c) A smaller amount of hydrogen and oxygen is weighed, then combined to form water, which is weighed again. A very good chemical balance is able to detect a fractional change in mass of 1 part in \(10^{8}\). By what factor is this sensitivity more than enough -or insufficient - to detect the fractional change in mass in this reaction?

The Giant Shower Array detector, spread over 100 square kilometers in Japan, detects pulses of particles from cosmic rays. Each detected pulse is assumed to originate in a single high-energy cosmic proton that strikes the top of the Earth's atmosphere. The highest energy of a single cosmic ray proton inferred from the data is \(10^{20} \mathrm{eV}\). How long would it take that proton to cross our galaxy \(\left(10^{5}\right.\) light-years in diameter) as recorded on the wristwatch of the proton? (The answer is not zero!)

Evelvn Brown does not approve of our latticework of rods and clocks and the use of a light flash to synchronize them. (a) "I can synchronize my clocks in any way I choose!" she exclaims. Is she right? (b) Evelyn wants to synchronize two identical clocks, called Big Ben and Little Ben, which are at rest with respect to one another and separated by one million kilometers in their rest frame. She uses a third clock, identical in construction with the first two, that travels with constant velocity between them. As her moving clock passes Big Ben, it is set to read the same time as Big Ben. When the moving clock passes Little Ben, that outpost clock is set to read the same time as the traveling clock. "Now Big Ben and Little Ben are synchronized," says Evelyn Brown. Is Evelyn's method correct? (c) After Evelyn completes her synchronization of Little Ben by her method, how does the reading of Little Ben compare with the reading of a nearby clock on a latticework at rest with respect to Big Ben (and Little Ben) and synchronized by our standard method using a light flash? Evaluate in milliseconds any difference between the reading on Little Ben and the nearby lattice clock in the case that Evelyn's traveling clock moved at a constant velocity of 500000 kilometers per hour from Big Ben to Little Ben. (d) Evaluate the difference in the reading between the EvelynBrown- synchronized Little Ben and the nearby lattice clock when Evelyn's synchronizing traveling clock moves 1000 times as fast as the speed given in part (c).
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