Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Short Answer

Expert verified
(a) 1.508 × 10^-4 s; (b) 4307 m; (c) 1.574 × 10^-3 s, results agree.

Step by step solution

01

Calculate time as measured by the scientist

To find the time taken for the particle to travel 45.0 km at speed 0.99540c, use the formula \( t = \frac{d}{v} \), where \( d = 45000 \text{ m} \) and \( v = 0.99540c \). First calculate the velocity: \[ v = 0.99540 \times 3 \times 10^8 \text{ m/s} = 2.9862 \times 10^8 \text{ m/s} \]Next, use the formula to calculate time:\[ t = \frac{45000}{2.9862 \times 10^8} = 1.508 \times 10^{-4} \text{ s} \]
02

Calculate contracted distance in particle's frame

Using the length-contraction formula \( L = L_0 \sqrt{1 - v^2/c^2} \), where \( L_0 = 45000 \text{ m} \) and \( v = 0.99540c \), find \( L \):\[ L = 45000 \times \sqrt{1 - (0.99540)^2} \]Calculate the contraction factor:\[ \sqrt{1 - (0.99540)^2} = \sqrt{1 - 0.99082} = \sqrt{0.00918} = 0.0958 \]So, contracted distance:\[ L = 45000 \times 0.0958 = 4307 \text{ m} \]
03

Calculate time in particle's frame using time dilation

Using the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \), substitute \( t = 1.508 \times 10^{-4} \text{ s} \) and solve for \( t' \):\[ t' = \frac{1.508 \times 10^{-4}}{0.0958} = 1.574 \times 10^{-3} \text{ s} \]
04

Calculate time in particle's frame using distance

For the particle's frame, use \( t' = \frac{L}{v} \) with \( L = 4307 \text{ m} \) and \( v = 0.99540c = 2.9862 \times 10^8 \text{ m/s} \):\[ t' = \frac{4307}{2.9862 \times 10^8} = 1.574 \times 10^{-3} \text{ s} \]
05

Compare results

Both methods to find time in the particle's frame give \( t' = 1.574 \times 10^{-3} \text{ s} \). The results agree, confirming the calculations.

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 Vaia!

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 physics that comes from Einstein's theory of special relativity. It tells us that time passes differently for two observers who are moving relative to each other. To understand this, imagine a scientist and a fast-moving particle.
For the scientist, who is stationary on Earth, time ticks at a normal pace. However, for the particle whizzing toward Earth at nearly the speed of light (0.99540c, where c is the speed of light), time slows down. This slowing of time means that the particle, in its own "frame of reference," experiences less time passing compared to the scientist who is observing from a stationary position.
To calculate the time experienced by the particle, we use the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \). Here, \( t \) is the time measured by the observer (the scientist), and \( t' \) is the dilated time experienced by the particle. Hence, even though the scientist measures the journey to take \( 1.508 \times 10^{-4} \) seconds, the particle experiences a longer time: \( 1.574 \times 10^{-3} \) seconds.
This example shows how high velocities can significantly affect time perception, which is only observable at speeds close to the speed of light. It’s exciting to think about how flexible time becomes in such extreme conditions.
Length Contraction
Length contraction is another intriguing result of special relativity. It explains how the length of an object is perceived to be shorter when it is moving at high speeds relative to an observer. Imagine the path the particle takes from the atmosphere to the Earth's surface.
From the scientist's perspective on Earth, the distance is a standard \( 45.0 \) kilometers or \( 45000 \) meters. But for the particle moving at the speed of \( 0.99540c \), the distance appears shorter. According to the formula \( L = L_0 \sqrt{1 - v^2/c^2} \), the length \( L \) becomes \( 4307 \) meters in the particle's frame.
This occurs because, at such high speeds, the dimensions along the direction of motion contract. This effect doesn’t affect motions or objects we encounter in our daily lives but becomes significant when dealing with particles or objects moving at speeds close to that of light.
Length contraction is a key piece of understanding how interstellar travel or high-speed journeys would alter human perceptions and measurements, a captivating point for future explorations.
Special Relativity
Special relativity is a theory introduced by Albert Einstein in 1905. It reshaped our understanding of space and time. One of its core ideas is that the laws of physics are the same for all non-accelerated observers, meaning there are no "special" or "preferred" reference frames.
One of the most famous outcomes of special relativity is the equation \( E=mc^2 \), linking mass (m) and energy (E). However, the theory also leads to phenomena like time dilation and length contraction, as discussed earlier. These phenomena reveal how different observers, moving relative to one another, can perceive time and space in contrasting ways.
Special relativity becomes especially crucial when understanding cosmic rays or particles traveling at light-like speeds because it gives us the tools to calculate how time and space change in such extraordinary conditions. Its implications for high-speed travel, particle physics, and even the fabric of the universe continue to be areas of active research and wonder.
Whether it's calculating times or distances in unique scenarios, special relativity helps us comprehend the universe's intricate dance as particles flit across it, governed by these timeless laws.
Cosmic Rays
Cosmic rays are high-energy particles that travel through space and sometimes enter Earth's atmosphere. They can be electrons, protons, and other atomic nuclei. When these cosmic rays strike the Earth's atmosphere, they often produce a shower of secondary particles.
These particles collide with atmospheric atoms and create more particles, a phenomenon scientists observe at high altitudes. In the provided problem, the creation of an unstable particle from a cosmic ray in the upper atmosphere is being studied.
The particle's high speed (0.99540c) as it descends toward Earth brings special relativity into play. Time dilation and length contraction reflect the incredible speeds involved. Studying cosmic rays isn't just an exercise in theory but an exploration of nature’s powerful, energetic phenomena that might give us clues about the universe's origins.
Cosmic rays offer a fascinating avenue for both experimental and theoretical physics, allowing scientists to test theories like special relativity under extreme conditions and providing insights into high-energy processes far beyond Earth’s boundaries.

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 imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920c relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360c. What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400c. The enemy ship fires a missile toward you at a speed of 0.700c relative to the enemy ship (Fig. E37.18). (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is 8.00 * 106 km away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

As you have seen, relativistic calculations usually involve the quantity \(\gamma\). When \(\gamma\) is appreciably greater than 1, we must use relativistic formulas instead of Newtonian ones. For what speed \(v\) (in terms of \(c\)) is the value of \(\gamma\) (a) 1.0% greater than 1; (b) 10% greater than 1; (c) 100% greater than 1?

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.

Sign-up for free