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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 0.600c. The pursuit ship is traveling at a speed of 0.800c 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) Away; (b) \(-0.385c\).

Step by step solution

01

Identify the Information

We need to find the velocity of the cruiser relative to the pursuit ship, which requires using relativistic velocity addition as both ships are traveling at significant fractions of the speed of light. We know:- Speed of the cruiser relative to Tatooine: \(v_c = 0.600c\)- Speed of the pursuit ship relative to Tatooine: \(v_p = 0.800c\)
02

Determine Direction of Relative Velocity

Since both the cruiser and the pursuit ship are moving in the same direction as viewed from Tatooine, the cruiser appears to be moving away from the pursuit spacecraft. Therefore, the velocity of the cruiser relative to the pursuit ship will be negative. This means that the cruiser is moving away from the pursuit ship.
03

Use Relativistic Velocity Addition Formula

The formula for the relativistic addition of velocities is:\[v = \frac{v_c - v_p}{1 - \frac{v_c \, v_p}{c^2}}\]where:- \(v\) is the relative velocity of the cruiser as seen by the pursuit ship- \(v_c = 0.600c\)- \(v_p = 0.800c\)
04

Plug in Values

Substitute the known values into the relativistic velocity addition formula:\[v = \frac{0.600c - 0.800c}{1 - \frac{0.600c \, \times \, 0.800c}{c^2}}\]
05

Simplify the Equation

Simplify the numerator and denominator:- Numerator: \(0.600c - 0.800c = -0.200c\)- Denominator: \(1 - \frac{0.480c^2}{c^2} = 1 - 0.480 = 0.520\)Thus, the equation simplifies to:\[v = \frac{-0.200c}{0.520}\]
06

Calculate the Result

Perform the division to find the relative velocity:\[v = \frac{-0.200c}{0.520} = -0.385c\]So, the velocity of the cruiser relative to the pursuit ship is \(-0.385c\). This negative sign indicates the cruiser is moving away from the pursuit ship.

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Key Concepts

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

Pursuit Spacecraft
In the scenario of a pursuit spacecraft, we're looking at a spaceship from Tatooine trying to catch up with another vessel, the Trade Federation cruiser. The role of the pursuit spacecraft is crucial here, as its goal is to close the distance between itself and the cruiser.
The pursuit spacecraft travels at a high velocity, specifically at 0.800c, or 80% of the speed of light. This speed is significant because, according to Einstein’s theory of relativity, as objects move faster, they approach the speed of light, leading to several relativistic effects.
  • Time dilation: Time moves slower for objects traveling near the speed of light relative to a stationary observer.
  • Length contraction: Objects in motion will appear shorter in the direction of the motion to an external observer.
  • Mass increase: As speed increases, so does the relativistic mass of the object.
These effects are essential when calculating velocities like in this scenario. Using relativistic velocity addition, we determine how fast the cruiser moves relative to the pursuit spacecraft.
Trade Federation Cruiser
The Trade Federation cruiser is the lead spaceship being pursued in this space chase. From the observer's perspective on Tatooine, the cruiser moves away from the planet at 0.600c, or 60% of the speed of light.
This spacecraft is central to understanding the relative motion involved in this problem. Its speed forms the baseline against which the pursuit spacecraft's speed is compared. Initially, it seems like the cruiser is fast enough to evade pursuit. However, understanding its velocity relative to another fast-moving object is where relativity plays a pivotal role.
Given that both spacecraft move in the same direction:
  • The relative velocity of the cruiser to the pursuit ship involves subtractive calculation.
  • The sign (negative in this scenario) of the resultant velocity confirms the cruiser moves away.
Overall, this helps illustrate how velocities combine under Einstein's special relativity, contrasting them with our everyday experiences.
Speed of Light
The speed of light plays an instrumental role in relativistic physics. It serves as the ultimate speed limit of the universe, at around 299,792,458 meters per second or approximately 300,000 km/s.In the context of this problem, both spacecraft velocities are expressed as fractions of the speed of light (denoted as "c"). This is necessary to properly apply the relativistic velocity addition formula. The formula is:\[ v = \frac{v_c - v_p}{1 - \frac{v_c \, v_p}{c^2}} \]where "\(v\)" is the velocity of the cruiser relative to the pursuit ship. The constraint "\(c\)" ensures that no object exceeds this universal speed limit. If not properly accounted for, calculations could erroneously suggest impossible speeds greater than \(c\).Relativistic effects significantly impact such high-speed scenarios. They provide critical insights into real-world space travel, even if we are far from reaching such velocities with current technology.

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Most popular questions from this chapter

Calculate the magnitude of the force required to give a 0.145-kg baseball an acceleration \(a =\) 1.00 m/s\(^2\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) 10.0 m/s; (b) 0.900c; (c) 0.990c. (d) Repeat parts (a), (b), and (c) if the force and acceleration are perpendicular to the velocity.

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400c. The enemy ship fires a missile toward you at a speed of 0.700c relative to the enemy ship (Fig. E37.18). (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is 8.00 * 106 km away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

A muon is created 55.0 km above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is 2.20 \(\mu\)s, and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of 0.9860\(c\). (a) In the muon's frame, what is its initial height above the surface of the earth? (b) In the muon's frame, how much closer does the earth get during the lifetime of the muon? What fraction is this of the muon's original height, as measured in the muon's frame? (c) In the earth's frame, what is the lifetime of the muon? In the earth's frame, how far does the muon travel during its lifetime? What fraction is this of the muon's original height in the earth's frame?

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c\) ? (Express the answers in terms of \(mc^2\).) (c) How do your answers in parts (a) and (b) compare?

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