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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

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

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800c relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled 1.20 \(\times\) 10\(^8\) m past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 m. (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of 4.00 \(\times\) 10\(^7\) m/s relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9930c. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame \(S'\) moving relative to \(S\), the second event is observed to occur 2.15 s after the first. What is the difference between the positions of the two events as measured in \(S'\)?

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