/*! 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 50 The distance to a particular sta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 37: Problem 50

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

Short Answer

Expert verified
The trip takes approximately 4.82 years as observed from earth and the passengers on the spaceship measure a distance of approximately 3.35 light-years for the trip.

Step by step solution

01

Calculate Earth Time

The spaceship's time-travel is \(3.35 \, \mathrm{y}\) and the distance in light years is \(7.11 \, \mathrm{light-years}\). This indicates that the ship traveled at a speed of \(\frac{7.11}{3.35} = 2.12 \, \text{light-years per year}\) as measured from the ship. Since the speed of light is 1 light-year per year, we can say that the velocity \(v\) is \(2.12c\). Now, to find the time as observed on Earth, we'll need to make use of the time dilation principle from the theory of relativity, which states that time observed in a moving frame of reference (ship's frame) is shorter than the time observed in a stationary frame (Earth's frame). The formula for time dilation is \(t=\frac{t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\). Substitute \(t_0 = 3.35 \, \mathrm{y}\) and \(v = 2.12c\) into this formula to get the time observed on earth.
02

Calculate Spacecraft Distance

For finding the distance observed by passengers, it's known that the length observed in a moving frame (length in the spaceship) is shorter than the length observed in a stationary frame (length observed on earth), a result of length contraction in relativity. The formula for this phenomenon is \(L = L_0 \sqrt{1-\left(\frac{v}{c}\right)^2}\). \(L_0 = 7.11 \, \mathrm{light-years}\) is the distance observed on earth. Substituting \(v = 2.12c\) into this formula gives the distance observed in the spaceship.

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 from Einstein鈥檚 theory of special relativity. It explains how time can appear to pass at different rates in different frames of reference.
For example, if you are traveling at a very high speed in a spaceship, time for you (as observed from the ship) appears to move slower compared to an observer on Earth.
The formula for time dilation is \( t = \frac{t_0}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \), where:
  • \( t \) is the time observed in the stationary frame (Earth).
  • \( t_0 \) is the time observed in the moving frame (spaceship).
  • \( v \) is the velocity of the moving frame.
  • \( c \) is the speed of light.
When the speed \( v \) approaches that of light \( c \), time dilation becomes more pronounced.
Length Contraction
Length contraction is another intriguing result from special relativity, revealing that objects traveling at high speeds appear shorter in the direction of motion to a stationary observer. This effect is a direct consequence of the same relativistic principles that cause time dilation.
The formula to calculate the contracted length is \( L = L_0 \sqrt{1 - \left(\frac{v}{c}\right)^2} \), where:
  • \( L \) is the length measured in the moving frame (spaceship).
  • \( L_0 \) is the original length measured in the stationary frame (Earth).
  • \( v \) is the velocity of the moving object.
  • \( c \) is the speed of light.
As the velocity \( v \) approaches the speed of light \( c \), the length \( L \) becomes significantly shorter.
Speed of Light
The speed of light, denoted as \( c \), is a fundamental constant in the universe and is approximately \( 299,792,458 \) meters per second, or \( 1 \) light-year per year when considering astronomical distances. According to the theory of relativity, nothing can travel faster than the speed of light.
This principle shapes the very framework of how time and space interact.
When objects move close to the speed of light, unusual effects like time dilation and length contraction become significant. The significance of the speed of light is its role in nearly all modern physics equations related to relativity.
Frame of Reference
A frame of reference in physics is a set of coordinates or a viewpoint used to measure or observe physical phenomena. A crucial aspect of relativity is understanding that the observations can differ depending on the observer's frame of reference.
For instance, an observer on Earth perceives time and distance differently compared to an observer traveling in a high-speed spaceship.
In the realm of special relativity:
  • The stationary frame of reference often makes measurements from an "at rest" perspective, like Earth's frame.
  • The moving frame of reference measures from within the moving object, such as a spaceship.
Every scenario is relative to the speed and direction of the observer's frame, making relativity the universal rule for such high-speed interactions.

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

(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?

If a muon is traveling at \(0.999 c,\) what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

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?

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?

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes \(1.80 \mathrm{~s}\) as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control (on earth) who is watching the experiment? (b) If each swing takes \(1.80 \mathrm{~s}\) as measured by a person at mission control, how long will it take as measured by the astronaut in the spaceship?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Mechanics and Materials

Read Explanation

Waves Physics

Read Explanation

Quantum Physics

Read Explanation

Classical Mechanics

Read Explanation

Space 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.