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Chapter 37: Problem 17

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 \(\mathrm{c}\) . The pursuit ship is traveling at a speed of 0.800 \(\mathrm{c}\) relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the speed of the cruiser relative to the pursuit ship be positive or negative? (b) What is the speed of the cruiser relative to the pursuit ship?

Short Answer

Expert verified
(a) Negative (b) Approximately -0.385c.

Step by step solution

01

Understanding Relative Motion

We want to find the speed of the cruiser as observed from the pursuit spacecraft. For part (a), we should consider that if the pursuit ship is faster, then relative to the pursuit ship, the cruiser is moving in the opposite direction. Therefore, in such a case, the relative speed would be negative. Conversely, if the pursuit ship were slower, which is not the case here, the relative speed would be positive.
02

Using the Relativistic Velocity Addition Formula

To solve part (b), we use the relativistic velocity addition formula:\[ v' = \frac{v - u}{1 - \frac{vu}{c^2}} \]where \( v = 0.800c \) is the speed of the pursuit ship, \( u = 0.600c \) is the speed of the cruiser, and \( c \) is the speed of light.
03

Substituting the Values

Substitute the known values into the formula:\[ v' = \frac{0.800c - 0.600c}{1 - \frac{(0.800c)(0.600c)}{c^2}} \]
04

Calculating the Numerator

Calculate the numerator:\[ 0.800c - 0.600c = 0.200c \]
05

Calculating the Denominator

Calculate the denominator:\[ 1 - \frac{(0.800)(0.600)c^2}{c^2} = 1 - 0.480 = 0.520 \]
06

Final Calculation

Now perform the final calculation:\[ v' = \frac{0.200c}{0.520} \approx 0.385c \]So, the speed of the cruiser relative to the pursuit ship is approximately \(-0.385c\), indicating that from the perspective of the pursuit ship, the cruiser appears to move backwards at this speed.

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Key Concepts

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

Relative Motion
When we talk about relative motion, we are focusing on how the speed of one object appears to another observer. This isn't as straightforward as it seems, especially when dealing with high velocities close to the speed of light. In the context of high speeds, such as a spacecraft racing after another, we need to consider how different observers view their relative velocities. Here, since the pursuit ship is faster, it perceives the cruiser as moving in the opposite direction. Therefore, for the pursuit spacecraft, the relative velocity of the cruiser is negative. This is simply because the pursuit is closing the gap as it moves faster than the cruiser.
Special Relativity
Special relativity is a theory coined by Albert Einstein, which radically transformed our understanding of motion, especially when objects move near the speed of light. One of its most mind-boggling concepts is that time and space are not absolute, and they vary depending on the observer's speed. When objects travel at a significant fraction of the speed of light, relativistic effects become essential. These effects include how distances, time intervals, and even the order of events can appear different to observers in different inertial frames.
Velocity Transformation
Velocity transformation is a method within special relativity to calculate how quickly an object appears to move relative to another moving reference frame. The key tool for this is the relativistic velocity addition formula:
  • It ensures that no object exceeds the speed of light irrespective of the observers' frame of reference.
  • Under this formula, velocities do not add up the same way they do at lower velocities.
The formula used here, \[ v' = \frac{v - u}{1 - \frac{vu}{c^2}} \]is specifically designed for combining velocities that are along the same line. It accounts for the relativistic effects by adjusting the sum of velocities as perceived by a stationary observer.
Speed of Light
The speed of light, often denoted as \( c \), is a fundamental constant of nature, approximately valued at \( 3 \times 10^8 \) meters per second. It's the ultimate speed limit in the universe, dictating not only the flow of information, but also shaping the very fabric of time and space as we perceive it. No material object can reach, let alone surpass, this speed.
  • In relativity, \( c \) serves as a natural speed boundary.
  • Its constancy ensures equivalence in the laws of physics across all frames of reference.
The concept of light's speed touches on deep principles about the universe's structure, influencing how we calculate velocities and ensure nothing violates this cosmic speed limit.

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Most popular questions from this chapter

Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (Hint: Since \(u \ll c,\) you can simplify \(\sqrt{1-u^{2} / c^{2} \text { by a binomial expansion. }}\))

A spaceship moving at constant speed \(u\) relative to us broadcasts 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 \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the time 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, \(f-f_{0} ?(\mathrm{c})\) 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)?

(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 \(m c^{2}-)(c)\) How do your answers in parts \((a)\) and \((b)\) compare?

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earth-based observers? Do present-day jet planes approach such speeds?

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?
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