/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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 µs), \( v \) 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 91Ó°ÊÓ!

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—what 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’s 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

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

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?

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?

Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called spectroscopic binary stars. Figure \(\mathbf{P 3 7 . 6 8}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R .\) The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times 10^{14} \mathrm{~Hz}\) In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{~Hz}\) and \(4.568910 \times 10^{14} \mathrm{~Hz}\). Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f / f=u / c\) given in Section \(37.6 .\) ) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30} \mathrm{~kg} .\) Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{~m}\). (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy?

See all solutions

Recommended explanations on Physics Textbooks

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.