/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A spaceship flies past Mars with... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 37: Problem 4

A spaceship flies past Mars with a speed of \(0.985 c\) relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for \(75.0 \mu\) s. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?

Short Answer

Expert verified
(a) The observer on Mars measures the proper time because the event (blinking light) is happening at the same position for him. (b) Use the time dilation formula to find the duration of the light pulse as measured by the spaceship pilot.

Step by step solution

01

Determine the observer measuring proper time

The proper time is the time interval that is measured by an observer who perceives the start and end of an event happening in the exact same position. Here, the event is the signal light turning on and off. This event occurs at a fixed location on Mars surface. Hence, the observer on Mars measures the proper time.
02

Apply the time dilation formula

The formula for time dilation is given by \( \Delta t = \frac{\Delta t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \), where \( \Delta t \) is the dilated time (time measured by the moving observer), \( \Delta t_0 \) is the proper time, \( v \) is the relative speed and \( c \) the speed of light. In this case, \( \Delta t_0 \) is the time measured by Mars observer (75.0 碌s), \( v \) is the spaceship speed (0.985 c), and \( c \) is the speed of light. Now, we just need to insert these values into the equation and solve it.
03

Compute the time as measured by spaceship pilot

By substituting the given values into the time dilation formula, we calculate the time interval (duration of the light pulse) as measured by the spaceship pilot.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Time Dilation
Time dilation is a fascinating concept in the realm of special relativity, introduced by Albert Einstein. It's all about how time appears to stretch or dilate when observed from a moving frame of reference. This might sound odd, but imagine you're on a spaceship zipping past a planet. Time for you, relative to someone standing still on the planet, would actually be experienced differently.
An important aspect of time dilation is the time dilation formula:\[ \Delta t = \frac{\Delta t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \]
Here, \( \Delta t \) is the dilated time鈥攚hat the person on the spaceship would measure. \( \Delta t_0 \) is the proper time,\( v \) is the relative speed of the moving object, and \( c \) is the speed of light. In simpler terms, this equation tells us how much time stretches by when something is moving incredibly fast. So, as the speed of the spaceship increases, the time dilation effect becomes more pronounced.
Proper Time
Proper time is the time measured by an observer for whom the events (such as a light blinking on and off) occur at the same point in space. This is crucial because it provides a baseline measurement, unaffected by high speeds or other factors.
In our exercise, the observer standing on Mars is the one who measures the proper time, since the signal light from Mars blinks at a fixed location relative to them. This observer is stationary with respect to the event, so their measurement of time (75.0 microseconds) is the proper time in this context.
Understanding proper time helps us unravel complex situations where multiple frames of reference are involved. It gives us a grounded perspective on the event without the influence of motion.
Relative Speed
Relative speed is the speed of one object as observed from another moving object. It鈥檚 a crucial element in understanding phenomena in special relativity, because many effects like time dilation depend on this relative motion.
In our example, the spaceship travels at a speed of \(0.985 c\) relative to Mars. This high relative speed is significant because it puts the theory of special relativity into action. The closer the speed of the spaceship approaches the speed of light, the more pronounced the relativistic effects will become.
Relative speed allows us to compare the motions of two objects effectively. It's this relative nature of speeds that determines how much time dilation we observe. When solving problems, understanding this concept helps clarify how different observers, moving at different speeds, perceive time and space differently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of \(2.2 \mu \mathrm{s}\) They are produced when cosmic rays bombard the upper atmosphere about \(10 \mathrm{~km}\) above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its \(2.2 \mu\) s lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the \(2.2 \mu \mathrm{s}\) lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of \(0.999 c,\) what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only \(2.2 \mu \mathrm{s},\) so how does it make it to the ground? What is the thickness of the \(10 \mathrm{~km}\) of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

Electrons are accelerated through a potential difference of \(750 \mathrm{kV},\) so that their kinetic energy is \(7.50 \times 10^{5} \mathrm{eV}\). (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c ?\) (b) What would the speed be if it were computed from the principles of classical mechanics?

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?

Quarks and gluons are fundamental particles that will be discussed in Chapter \(44 .\) A proton, which is a bound state of two up quarks and a down quark, has a rest mass of \(m_{\mathrm{p}}=1.67 \times 10^{-27} \mathrm{~kg}\). This is significantly greater than the sum of the rest mass of the up quarks, which is \(m_{\mathrm{u}}=4.12 \times 10^{-30} \mathrm{~kg}\) each, and the rest mass of the down quark, which is \(m_{\mathrm{d}}=8.59 \times 10^{-30} \mathrm{~kg} .\) Suppose we (incorrectly) model the rest energy of the proton \(m_{\mathrm{p}} c^{2}\) as derived from the kinetic energy of the three quarks, and we split that energy equally among them. (a) Estimate the Lorentz factor \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) for each of the up quarks using Eq. \((37.36) .\) (b) Similarly estimate the Lorentz factor \(\gamma\) for the down quark. (c) Are the corresponding speeds \(v_{\mathrm{u}}\) and \(v_{\mathrm{d}}\) greater than \(99 \%\) of the speed of light? (d) More realistically, the quarks are held together by massless gluons, which mediate the strong nuclear interaction. Suppose we model the proton as the three quarks, each with a speed of \(0.90 c,\) with the remainder of the proton rest energy supplied by gluons. In this case, estimate the percentage of the proton rest energy associated with gluons. (e) Model a quark as oscillating with an average speed of \(0.90 c\) across the diameter of a proton, \(1.7 \times 10^{-15} \mathrm{~m}\). Estimate the frequency of that motion.

(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 \(\left.m c^{2} .\right)\) (c) How do your answers in parts (a) and (b) compare?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Oscillations

Read Explanation

Electrostatics

Read Explanation

Medical Physics

Read Explanation

Waves Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.