/*! 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 12 An unstable particle is created ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 0.99540 c relative to the earth. A scientist at rest on the carth's surface measures that the particle is created at an altitude of 45.0 \(\mathrm{km}\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) 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) 1.505 x 10^-4 s; (b) 4473.94 m; (c) 4.56 x 10^-6 s (both methods agree).

Step by step solution

01

Calculate Time in Earth's Frame

To find the time it takes for the particle to reach the Earth's surface as measured by the Earth scientist, we use the formula for time: \( t = \frac{d}{v} \), where \( d = 45.0 \text{ km} = 45000 \text{ meters} \) and \( v = 0.99540c \). Converting speed into meters per second, \( c = 3 \times 10^8 \text{ m/s} \), so \( v = 0.99540 \times 3 \times 10^8 \text{ m/s} \). Thus, \( t = \frac{45000}{0.99540 \times 3 \times 10^8} \). Calculate to find \( t = 1.505 \times 10^{-4} \text{ seconds} \).
02

Calculate Length in Particle's Frame using Length Contraction

To find the contracted length in the particle's frame, we use the length contraction formula: \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \), where \( L_0 = 45.0 \text{ km} = 45000 \text{ meters} \) is the proper length, and \( v = 0.99540c \). Therefore, \( L = 45000 \sqrt{1 - (0.99540)^2} \). Solving gives \( L = 4473.94 \text{ meters} \).
03

Calculate Time in Particle's Frame using Time Dilation

Time dilation can be calculated using \( t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \). From Step 1, \( t = 1.505 \times 10^{-4} \text{ seconds} \). Therefore, \( t' = \frac{1.505 \times 10^{-4}}{\sqrt{1 - (0.99540)^2}} \). Solving gives \( t' = 4.56 \times 10^{-6} \text{ seconds} \).
04

Calculate Time in Particle's Frame from Contracted Length

Using the contracted length from Step 2, the time taken in the particle's frame can also be calculated by \( t' = \frac{L}{v} \), where \( L = 4473.94 \text{ meters} \) and \( v = 0.99540 \times 3 \times 10^8 \text{ m/s} \). Therefore, \( t' = \frac{4473.94}{0.99540 \times 3 \times 10^8} \). Solving gives \( t' = 4.56 \times 10^{-6} \text{ seconds} \).
05

Compare Time Calculations

The time calculated in the particle's frame using both methods (Step 3 and Step 4) is approximately the same: \( 4.56 \times 10^{-6} \text{ seconds} \). This agreement confirms that both the time dilation and length contraction calculations are consistent with the theory of special relativity.

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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, which was formulated by Albert Einstein. It describes how time experienced by an object is observed to be different depending on the relative velocity between the observer and the object in motion. In essence, time flows differently for observers moving relative to each other.
In the given exercise, time dilation is evident when calculating how much time it takes for a particle, moving at a relativistic speed, to reach Earth's surface. In an observer's frame on Earth, the journey takes a certain amount of time. However, in the particle's frame, time is perceived to move more slowly. This effect can be quantified by the time dilation formula:
  • ' = rac{t}{ ext{sqrt}{1 - rac{v^2}{c^2} }}
This formula shows how time 't' in a stationary observer’s frame (Earth) is different from 't'' in the moving particle's frame. The more significant the velocity (v) is—close to the speed of light (c)—the more pronounced the time dilation effect. In the problem, this was used to show that even though an Earth observer measures 1.505 imes 10^{-4} seconds, the particle perceives time as a mere 4.56 imes 10^{-6} seconds for the same journey.
Length Contraction
Length contraction is another cornerstone concept of special relativity. It reveals that the length of an object in motion is observed to be shorter when measured from a stationary frame compared to the object's rest length. This occurs only in the direction of motion and is negligible unless speeds approach that of light.
Using the length contraction formula, the contracted length of the path taken by the particle is reduced in its own frame:
  • L = L_0 ext{sqrt}(1 - rac{v^2}{c^2})
Where L is the length observed in the moving frame, and L0 is the proper length. With a velocity of 0.99540c, the length contracted effect is significant, reducing the distance from 45000 meters to 4473.94 meters in the particle's frame.
This contracted length helps explain the shorter travel time experienced by the cosmic particle itself. It needs to cross less distance in its own reference frame, taking only a fraction of the time compared to a stationary observer measuring the same journey from Earth.
Cosmic Rays
Cosmic rays are high-energy particles originating from outer space. They enter Earth's atmosphere with tremendous velocities, often close to the speed of light. Their study provides a compelling context for exploring special relativity.
As with the exercise, cosmic rays often create secondary particles high up in Earth's atmosphere during collisions with atmospheric molecules. These secondary particles, like the one in the problem, travel rapidly towards the Earth's surface.
  • Their high velocities make relativistic effects—such as time dilation and length contraction—noticeable and significant.
  • Understanding cosmic rays helps physicists study not only particle physics but also astrophysical phenomena.
The analysis of these particles and associated relativistic effects further confirms the consistency of Einstein’s theory of relativity, as it accurately predicts behaviors otherwise difficult to intuit at non-relativistic speeds. This aspect of studying cosmic rays adds a practical framework for applying abstract physical theories.
Relativistic Speed
Relativistic speeds are velocities that are a significant fraction of the speed of light. When objects reach these speeds, familial notions of space and time transform significantly, as predicted by Einstein's theory of special relativity.
In the original exercise, the particle moves at 0.99540 times the speed of light. Such a velocity clearly illustrates relativistic effects:
  • Time dilation becomes profound, as the time experienced by the particle is far less than the time measured by a stationary Earth observer.
  • Length contraction is evident, with considerable reduction in travel distance observed in the particle's own frame.
  • These effects are measurable only when velocities approach the speed of light, showing that classical mechanics gives way to relativistic concepts.
The implications of understanding relativistic speeds extend far beyond academic exercises, influencing technologies like GPS and deepening our grasp of the universe's fundamental laws. By studying particles moving at such speeds, scientists explore the boundaries of physics and validate theories that unravel the fabric of space-time.

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Most popular questions from this chapter

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800 c relative to you. At the instant the space-racer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{in}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) 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) \(v^{\prime}=0.990 c ?\)

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