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As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 \(\mathrm{m} .\) (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of \(4.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) 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?

Short Answer

Expert verified
(a) 3575.6 m, (b) 9.0×10^{-5} s, (c) 9.06×10^{-5} s

Step by step solution

01

Identify the Problem

The problem relates to relativistic effects, specifically length contraction and time dilation, as predicted by Einstein's theory of special relativity.
02

Given Quantities

We have: 1. Rest length of the runway on Earth, \(L_0 = 3600 \; \text{m}\).2. Relative speed of the spacecraft, \(v = 4.00 \times 10^7 \; \text{m/s}\).3. Speed of light, \(c = 3.00 \times 10^8 \; \text{m/s}\).
03

Calculate Lorentz Factor

The Lorentz factor (\(\gamma\)) is defined as \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\).Substitute the given values, \(v = 4.00 \times 10^7 \; \text{m/s}\) and \(c = 3.00 \times 10^8 \; \text{m/s}\) into the formula: \[\gamma = \frac{1}{\sqrt{1 - \left(\frac{4.00 \times 10^7}{3.00 \times 10^8}\right)^2}} \approx 1.0067.\]
04

Calculate Length Contraction

The length of the runway as measured by the pilot of the spacecraft is given by:\[L = \frac{L_0}{\gamma}.\]Substitute \(L_0 = 3600 \; \text{m}\) and \(\gamma \approx 1.0067\) into the formula:\[L \approx \frac{3600}{1.0067} \approx 3575.6 \; \text{m}.\]
05

Calculate Earth's Measured Time

The time interval as measured by an observer on Earth is given by:\[t = \frac{L_0}{v}.\]Substitute \(L_0 = 3600 \; \text{m}\) and \(v = 4.00 \times 10^7 \; \text{m/s}\) into the formula:\[t = \frac{3600}{4.00 \times 10^7} = 9.0 \times 10^{-5} \; \text{s}.\]
06

Calculate Spacecraft Pilot's Time

The time interval as measured by the pilot on the spacecraft is dilated because of time dilation:\[t' = \gamma \cdot t.\]Substitute \(\gamma \approx 1.0067\) and \(t = 9.0 \times 10^{-5} \; \text{s}\) into the formula:\[t' = 1.0067 \times 9.0 \times 10^{-5} \approx 9.06 \times 10^{-5} \; \text{s}.\]
07

Conclusion

(a) The length of the runway as measured by the pilot is approximately 3575.6 m.(b) The Earth observer measures the time interval as \(9.0 \times 10^{-5}\) s.(c) The pilot measures the time interval as \(9.06 \times 10^{-5}\) s.

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Key Concepts

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

Length Contraction
Length contraction is a phenomenon predicted by Einstein's theory of special relativity. It describes how objects appear shorter when they are moving close to the speed of light, relative to an observer.Let's take an example of a spacecraft moving at a significant fraction of the speed of light (denoted as \(c\)), as it flies past a runway on Earth. An observer on the runway, under normal conditions, measures its length as \(3600 \, \text{m}\). However, due to the spacecraft's high speed, the pilot sees the runway as shorter.This contraction can be calculated using the Lorentz factor (\(\gamma\)), which accounts for the effects of high speeds:- Formula: \(L = \frac{L_0}{\gamma}\)- Here, \(L_0\) is the rest length (\(3600 \, \text{m}\)), and \(L\) is the contracted length.In this problem, the calculated contracted length seen by the pilot is approximately \(3575.6 \, \text{m}\). This is less than the rest length, demonstrating that high-speed travel affects spatial measurements.
Time Dilation
Time dilation is another fascinating consequence of special relativity, showing that time does not pass at the same rate for everyone.When you have an observer and an object moving at relativistic speeds, time experienced by each will differ. This is captured through an experiment like measuring how long a spacecraft takes to travel the length of a runway from the perspective of Earth vs. the spacecraft itself.Consider these insights:- From Earth's perspective, the event takes a certain duration (\(t\)).- Like with length contraction, we use the Lorentz factor to relate this to the pilot's time measurement.For our example:- Earth's measurement is computed as \(9.0 \times 10^{-5} \, \, \text{s}\).- The pilot’s view, accounting for time dilation, results in a slightly longer time, \(9.06 \times 10^{-5} \, \, \text{s}\).This shows that time runs slower for the spacecraft relative to Earth, reflecting time dilation effects.
Lorentz Factor
The Lorentz factor (\(\gamma\)) plays a critical role in understanding relativistic effects like length contraction and time dilation.It is defined by the formula:- \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\)Where:- \(v\) is the velocity of the moving object- \(c\) is the speed of lightIn our scenario, the spacecraft moves at \(4.00 \times 10^7 \, \text{m/s}\), much less than the speed of light \(3.00 \times 10^8 \, \text{m/s}\). Substituting these values gives us a Lorentz factor of approximately \(1.0067\).The factor determines how much physical dimensions and time intervals change due to high speed:- If \(\gamma = 1\), there's negligible relativistic effect.- A \(\gamma\) greater than 1 suggests measurable changes, as seen in this problem.Thus, understanding \(\gamma\) is key to applying special relativity accurately in practical scenarios.
Relativistic Effects
Relativistic effects occur when objects travel at speeds close to the speed of light, leading to unique physical phenomena established by Einstein's theory of special relativity.These include:
  • Length Contraction: Moving objects appear shorter along the direction of motion.
  • Time Dilation: Time between events is experienced differently depending on the observer's relative speed.
  • Mass Increase: As speeds increase, so does the object's effective mass. (While not explored in this exercise, it's another core concept.)
These effects become significant only at velocities nearing a substantial fraction of \(c\). Our example with the spacecraft and runway clearly illustrates length contraction and time dilation, highlighting how these effects become practical challenges in high-speed travel.By understanding these principles, we can better grasp not only the theoretical aspects but also practical applications, such as in particle physics and future space exploration scenarios.

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Most popular questions from this chapter

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

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?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800\(c .\) Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate \(x\) and \(t\) as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. \((37.6),\) to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-\) vt and \(t^{\prime}=t .\) In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ 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 \(\partial / \partial x^{\prime}\) and \(\partial / \partial t^{\prime}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Egs. \((37.21),\) and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S :\) $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

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