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A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of \(0.600 c\). The pursuit ship is traveling at a speed of \(0.800 c\) relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

Short Answer

Expert verified
a) The velocity of the cruiser, relative to the spacecraft, should be directed towards the spacecraft. b) The speed of the cruiser relative to the spacecraft is \(0.429c\), moving towards the spacecraft.

Step by step solution

01

Analyze the speed and direction

To begin with, let's recapitulate the given problem. The spacecraft and the cruiser are moving in the same direction but at different speeds. As both velocities are positive, considering the direction from Tatooine, the cruiser's velocity, relative to the spacecraft, should be directed away from the pursuit ship if the cruiser's actual speed is to be considered.
02

Applying Einstein's velocity addition formula

Now using the velocity addition formula prescribed by Einstein, calculate the speed of the cruiser as observed from the spacecraft. This formula can be applicable when the objects involved are moving at speeds nearly approaching the speed of light. Here the formula is written as \[w=\frac{u+v}{1+uv/c^2}\]. In this case \(u\) will represent the speed of the cruiser which is \(0.600c\) and \(v\) will be the speed of the spacecraft which is \(-0.800c\). Here, the speed is taken as negative since it is in a direction opposite to what was considered positive (away from Tatooine).
03

Computations and result

Substituting the given values into the formula, we get \[w=\frac{0.600c - 0.800c}{1 - (0.600c)(-0.800c)/ c^2}\]. On simplification, this will yield the result \(w = -0.429c\). Here, the negative sign indicates that the cruiser, seen from the spacecraft, is moving towards the spacecraft. The magnitude of the speed is \(0.429c\) (in terms of the speed of light).

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Key Concepts

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

Einstein's velocity addition formula
In the wild realm of high-speed motion, especially when dealing with velocities approaching the speed of light, Einstein's velocity addition formula becomes an essential tool. This formula helps us understand how velocities combine in a relativistic context. Unlike classical physics, where simply adding velocities is sufficient, relativity requires a more intricate approach. This is crucial because, according to Einstein's theory of relativity, the speed of light is the ultimate speed limit and cannot be exceeded. Thus, the traditional addition of velocities would conflict with this principle. The formula can be written as:\[w = \frac{u+v}{1+\frac{uv}{c^2}}\]In this equation:
  • \(u\) represents the velocity of one object relative to a stationary observer.
  • \(v\) is the velocity of the second object relative to the same stationary observer.
  • \(c\) denotes the constant speed of light (approximately 299,792,458 meters per second).
  • \(w\) stands for the resultant velocity of one object as seen from the other.
Einstein's formula ensures that even when adding two high-speed velocities, the resultant speed doesn't exceed the speed of light. This fascinating principle showcases how our universe operates differently at extreme speeds.
Pursuit problem
A pursuit problem, like the scenario of the Tatooine spacecraft and the Trade Federation cruiser, involves one object trying to catch up to another. In these problems, it's important to analyze both the relative velocities and the direction of motion.

In the given example, the spacecraft is moving at a higher speed than the cruiser relative to the planet Tatooine. With a pursuit in play, a key step is determining whether the relative velocity is directed toward or away from the pursuing object, which is essential in leading to a successful catch-up.
In our scenario, when applying Einstein's velocity addition formula, we computed that the velocity of the cruiser relative to the spacecraft ended up being negative, \(-0.429c\). The negative sign here is quite important as it communicates the cruiser is approaching the spacecraft in its frame of reference. This deduction aligns with the expectation for the spacecraft to eventually reach the cruiser.A typical pursuit problem illustrates the fascinating dance of velocities and directions, showcasing how relativity plays a crucial role in scenarios involving speeds nearing that of light.
Speed of light
The speed of light, denoted as \(c\), is not just a number but a universal constant in all physical laws. It's approximately 299,792,458 meters per second. In the context of relativistic physics, it sets an ultimate limit for speed within our universe.

This speed limit is integral to how we understand motion at extreme velocities and underpins the rationale for Einstein's velocity addition formula.

The world of relativistic speeds challenges the classical idea of simply adding up speeds. Instead, any velocity calculation must respect the limit imposed by the speed of light. This results in the unique relativistic effects we discussed earlier.In our scenario, where a spacecraft and cruiser are traveling at such high speeds, these principles of relativistic physics safeguard that their combined velocities do not surpass the speed of light. This enforces a fundamental rule of our universe and assures that the laws of physics about speed hold true consistently, regardless of the observers or their state of motion.

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Most popular questions from this chapter

In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\operatorname{kaon}\left(\mathrm{K}^{+}\right)\) $$ p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+} $$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is \(493.7 \mathrm{MeV},\) and the rest energy of each proton is \(938.3 \mathrm{MeV}\). (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of \(0.890 c .\) Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

A spaceship moving at constant speed \(u\) relative to us broad. casts a radio signal at constant frequency \(f_{0}\). As the spaceship approaches us, we receive a higher frequency \(f\); after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0},\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed \(0.758 c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?\) (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

An observer in frame \(S^{\prime}\) is moving to the right \((+x\) -direction \()\) at speed \(u=0.600 c\) away from a stationary observer in frame \(S .\) The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in \(S\) measure for the particle if (a) \(v^{\prime}=0.400 c ;\) (b) \(v^{\prime}=0.900 c ;\) (c) \(v^{\prime}=0.990 c ?\)

In a particle accelerator a proton moves at constant speed \(0.750 c\) in a circle of radius \(628 \mathrm{~m} .\) What is the net force on the proton?

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