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An imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920\(c\) relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360\(c .\) What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

Short Answer

Expert verified
The spaceship's speed is -0.646c, moving away from Arrakis.

Step by step solution

01

Understand the Problem

We have three objects involved: the spaceship, the rocket, and the observer on Arrakis. The rocket's speed relative to the spaceship is 0.920c, and its speed relative to Arrakis is 0.360c. We need to find the speed of the spaceship relative to Arrakis.
02

Identify the Variables

Let \( v \) be the speed of the spaceship relative to Arrakis, \( u' = 0.920c \) be the speed of the rocket relative to the spaceship, and \( u = 0.360c \) be the speed of the rocket relative to Arrakis.
03

Use the Relativistic Velocity Addition Formula

In special relativity, the velocity of an object relative to another is not simply the sum of the velocities. We use the formula: \[ u = \frac{u' + v}{1 + \frac{u'v}{c^2}} \] where \( u' \) is the velocity of the rocket relative to the spaceship, and \( v \) is the velocity of the spaceship relative to Arrakis.
04

Plug in Given Values

Substitute the given values into the equation: \[ 0.360c = \frac{0.920c + v}{1 + \frac{0.920c \cdot v}{c^2}} \].
05

Simplify the Equation

Let \( x \) be \( \frac{v}{c} \). This gives us the equation: \[ 0.360 = \frac{0.920 + x}{1 + 0.920x} \].
06

Solve for x

Multiply both sides by \( 1 + 0.920x \) to clear the denominator: \[ 0.360(1 + 0.920x) = 0.920 + x \]. Expand and simplify: \[ 0.360 + 0.3312x = 0.920 + x \]. Rearrange: \[ 0.360 - 0.920 = x - 0.3312x \], which simplifies to \[ x = -0.646 \].
07

Conclude the Calculation

Since \( x = \frac{v}{c} \), we have \( v = -0.646c \). The negative sign indicates that the spaceship is moving away from Arrakis.

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Key Concepts

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

Special Relativity
Special relativity is a theory developed by Albert Einstein that revolutionized our understanding of space and time. It emphasizes how different observers perceive the same event when they are moving at high speeds relative to one another. One of the key ideas in special relativity is that the speed of light is the same for all observers, regardless of their relative motion. This contrasts with classical mechanics and leads to phenomena like time dilation and length contraction. In the context of our problem, special relativity affects how velocities are added. If we used the classical approach, we might expect to simply add velocities. However, at high speeds, particularly those close to the speed of light, special relativity requires us to use a more complex formula to calculate the relative velocity between moving objects.
Velocity Calculation
Calculating velocity in the framework of special relativity involves more than simple addition or subtraction. We rely on the relativistic velocity addition formula. This formula accounts for the fact that as objects approach the speed of light, their velocities don't just add up in a straightforward way.
In the exercise, we use the formula: \[ u = \frac{u' + v}{1 + \frac{u'v}{c^2}} \]where:
  • \(u'\) is the velocity of the rocket relative to the spaceship,
  • \(v\) is the velocity of the spaceship relative to Arrakis,
  • \(u\) is the velocity of the rocket relative to Arrakis.
Simply putting the speeds into this equation gives us the tools to determine the spaceship's speed relative to Arrakis. The complexity comes from the need to consider how velocities interact at these high speeds.
Reference Frames
The concept of reference frames is crucial in physics, especially in special relativity. A reference frame is essentially a perspective from which an observer or a set of observers measure and interpret physical events. Each reference frame comes with its own view of space and time.
In our problem, we have two main reference frames: one attached to the spaceship and the other to the planet Arrakis. The observer on Arrakis sees the rocket's speed as 0.360\(c\), while the spaceship sees it as 0.920\(c\). Understanding how these differing perspectives relate is the essence of relativistic calculations. In practice, this means that the motions observed in one frame need adjustments using the relativistic formulas to be interpreted correctly in another. This principle ensures consistency among different observers, even when they are in relative motion to each other at significant fractions of the speed of light.

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Most popular questions from this chapter

The negative pion \(\left(\pi^{-}\right)\) is an unstable particle with an average lifetime of \(2.60 \times 10^{-8}\) s (measured in the rest frame of the pion). (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be \(4.20 \times 10^{-7} \mathrm{s}\) . Calculate the speed of the pion expressed as a fraction of \(c .\) (b) What distance, measured in the laboratory, does the pion travel during its average lifetime?

(a) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(m v ?\) Express your answer in terms of the speed of light. (b) A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

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A spacecraft flies away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days \((1\) year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shorter elapsed time?

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