/*! 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 43 A space traveler moving at a spe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A space traveler moving at a speed of 0.70\(c\) with respect to the earth makes a trip to a distant star that is stationary relative to the earth. He measures the length of this trip to be 6.5 light-years. What would be the length of this same trip (in light-years) as measured by a traveler moving at a speed of 0.90\(c\) with respect to the earth?

Short Answer

Expert verified
The length is 3.96 light-years for the traveler moving at 0.90\(c\).

Step by step solution

01

Understanding the Problem

We are given a speed and a measured length for a traveler moving at 0.70\(c\) and need to find the length as perceived by another traveler moving at 0.90\(c\). This is a problem involving length contraction, a key idea from the theory of relativity.
02

Length Contraction Formula

The formula for length contraction is \( L = L_0 \sqrt{1 - v^2/c^2} \), where \( L \) is the contracted length, \( L_0 \) is the proper length (the length measured by an observer at rest relative to the object), and \( v \) is the object's speed relative to the observer. We need to use this formula twice: once for each traveler.
03

Calculate Proper Length

We first find the proper length \( L_0 \). The traveler at 0.70\(c\) measures the length as 6.5 light-years. Therefore, we solve for \( L_0 \) using the formula: \( L_0 = \frac{6.5}{\sqrt{1 - (0.70)^2}} \).
04

Compute Proper Length

Calculating the above expression, we find \( L_0 = \frac{6.5}{\sqrt{1 - 0.49}} = \frac{6.5}{0.714} \approx 9.1 \) light-years. This is the length of the trip as seen from Earth.
05

Calculate Contracted Length for Second Traveler

Now, calculate the contracted length for the traveler moving at 0.90\(c\) using the proper length \( L_0 = 9.1 \) light-years: \( L = 9.1 \sqrt{1 - (0.90)^2} \).
06

Compute Length for Second Traveler

Evaluating the above expression gives \( L = 9.1 \sqrt{1 - 0.81} = 9.1 \times 0.435 = 3.96 \) light-years.

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.

Theory of Relativity
The theory of relativity is a revolutionary scientific concept developed by Albert Einstein. It fundamentally changed our understanding of how space and time work. This theory comprises two main parts: the special theory of relativity and the general theory of relativity. Let's focus on the special theory in this context.
The key idea of relativity is that laws of physics are the same for all observers, regardless of their position or speed. An especially mind-bending consequence of this is that time and space are not absolute and can change depending on the observer's relative motion. One fascinating effect of relativity is length contraction, where objects moving at high speeds will appear shorter in the direction of motion from the point of view of a stationary observer.
  • Imagine you are on a spaceship moving close to the speed of light. The distance to your destination seems shorter compared to how someone at rest would measure it.
  • This concept helps explain why space travelers perceive distances differently than observers on Earth.
Special Relativity
Special relativity is an integral part of Einstein's theory that addresses the physics of objects moving at constant high speeds, specifically near the speed of light. It builds up our understanding from relying on the constants of time and distance to learning about their variability.
For example, when addressing the problem of a space traveler moving at various speeds, special relativity gives us the tools to calculate how length contractions occur. The contraction effect is expressed in a mathematical formula:
  • The formula for length contraction is \( L = L_0 \sqrt{1 - v^2/c^2} \),
  • where \( L \) is the contracted length seen by a moving observer, \( L_0 \) is the proper length or the length at rest, and \( v \) is the velocity of the moving observer relative to the speed of light \( c \).

These predictions, although they sound strange, have been verified through experiments involving rapidly moving particles. Understanding these principles highlights the significant differences between intuitive, everyday physics and the high-speed regimes where relativity rules.
Speed of Light
The speed of light, denoted by \( c \), is a fundamental constant of nature, standing at approximately 299,792,458 meters per second or about 186,282 miles per second. It serves as the ultimate speed limit in the universe. Nothing can travel faster than light in vacuum.
This constant is crucial for the calculations and concepts within relativity, including length contraction. The invariant speed of light implies that even at high velocities, the laws of physics, including the speed of light itself, remain consistent.
  • In our exercise, light-years, a unit measuring distance as light travels in one year, helps us relate distances in cosmic scales.
  • The constant \( c \) appears in equations to calculate time dilation and length contraction effects experienced by objects traveling at significant fractions of this speed.
  • This ensures that even as objects move close to light speed, the mechanics of light and its role in physics stay unwavering.

This fundamental limit reshapes the relationship we experience with time and space, underpinning the many remarkable predictions of relativity theory.

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 woman is 1.6 \(\mathrm{m}\) tall and has a mass of 55 kg. She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of \(2.0 \times 10^{10} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) s. What does the observer measure for her height?

Two atomic particles approach each other in a head-on collision. Each particle has a mass of \(2.16 \times 10^{-25} \mathrm{kg}\) . The speed of each particle is \(2.10 \times 10^{8} \mathrm{m} / \mathrm{s}\) when measured by an observer standing in the laboratory. (a) What is the speed of one particle as seen by the other particle? (b) Determine the magnitude of the relativistic momentum of one particle, as it would be observed by the other.

Refer to Conceptual Example 11 as an aid in solving this problem. An intergalactic cruiser has two types of guns: a photon cannon that fires a beam of laser light and an ion gun that shoots ions at a velocity of 0.950\(c\) relative to the cruiser. The cruiser closes in on an alien space- craft at a velocity of 0.800\(c\) relative to this spacecraft. The captain fires both types of guns. At what velocity do the aliens see \((a)\) the laser light and \((b)\) the ions approach them? At what velocity do the aliens see (c) the laser light and (d) the ions move away from the cruiser?

At what speed is the magnitude of the relativistic momentum of a particle three times the magnitude of the nonrelativistic momentum?

Two spaceships \(A\) and \(B\) are exploring a new planet. Relative to this planet, spaceship \(A\) has a speed of 0.60\(c\) , and spaceship \(B\) has a speed of 0.80\(c\) . What is the ratio \(D_{A} / D_{B}\) of the values for the planet's diameter that each spaceship measures in a direction that is parallel to its motion?

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.