/*! 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 6 You see a sudden eruption on the... [FREE SOLUTION] | 91影视

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Chapter 38: Problem 6

You see a sudden eruption on the surface of the Sun. From solar theory you predict that the eruption emitted a pulse of particles that is moving toward the Earth at oneeighth the speed of light. How long do you have to seek shelter from the radiation that will be emitted when the particle pulse hits the Earth? Take the light-travel time from the Sun to the Earth to be 8 minutes.

Short Answer

Expert verified
64 minutes.

Step by step solution

01

Understand the Problem

The goal is to determine the time it will take for a pulse of particles, moving at one-eighth the speed of light, to travel from the Sun to the Earth.
02

Determine the Speed of Particles

Given that the particles are moving at one-eighth the speed of light, denote the speed of light as \(c\). Then, the speed of the particles is \(\frac{c}{8}\).
03

Calculate the Light-Travel Time

The light-travel time from the Sun to the Earth is given as 8 minutes.
04

Write the Formula

Use the relationship between speed, distance, and time: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]. Since the speed of the particles is \(\frac{c}{8}\) and the distance is the same as the light, we use light-travel time to find the distance.
05

Calculate the Time for Particle Travel

If the light takes 8 minutes to travel from the Sun to the Earth, the time for the particles is: \[\text{Time} = \frac{8 \text{ minutes}}{\frac{1}{8}} = 8 \times 8 = 64 \text{ minutes} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Particle Speed Calculation
Understanding the speed of particles emitted from solar eruptions is essential. In the given exercise, we know the particles travel at one-eighth the speed of light. To find their exact speed, we use the speed of light, denoted by the symbol \( c \). The speed of light is approximately \( 300,000 \ \text{km/s} \). If particles move at one-eighth of this speed, their speed is calculated as follows: \[ \frac{c}{8} = \frac{300,000 \ \text{km/s}}{8} = 37,500 \ \text{km/s} \]. Thus, the particles travel at a speed of \ 37,500 \ \text{km/s} \.
Light-Speed Relationship
The relationship between the speed of light and other speeds is crucial in understanding their interactions. Light speed, denoted by \( c \), is always constant in a vacuum. For the mentioned problem, light takes 8 minutes to travel from the Sun to Earth. Knowing the speed of light is essential in better understanding some key concepts like relativity and electromagnetic radiation. Here, we leverage this constant speed to compute other relationships. When particles travel at a fraction of the speed of light, we can use this information for various calculations, such as predicting travel time for different types of particles with specific speeds relative to light speed.
Time Calculation in Physics
Time calculation is a fundamental aspect of physics. We often need to relate time, speed, and distance in our calculations. From the exercise, we know the light-travel time from the Sun to Earth is given as 8 minutes. This is our constant for distance. To find the travel time for particles moving at one-eighth the speed of light, we use the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \].Given that particles move at a speed of \(\frac{c}{8} \)and the distance is the same as the light-travel distance, we perform the calculation like this: \[ \text{Time} = \frac{8 \text{minutes}}{\frac{1}{8}} = 8 \times 8 = 64 \text{minutes} \]
Solar Radiation Safety
It鈥檚 important to understand how long it takes for potentially harmful radiation to reach Earth in case of a solar eruption. Once you know the particle speed and travel time, you can better prepare and seek shelter. Solar radiation can pose various risks, such as increased cancer risk from high-energy particles and damage to electronic devices. By calculating the travel time, 64 minutes in our example, you gain a crucial window to take protective measures. Understanding these calculations and maintaining awareness about solar eruptions can enhance overall safety and preparedness.

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Most popular questions from this chapter

The famous twin paradox is often introduced as follows: Two identical twins grow up together on Earth. When they reach adulthood, one twin zooms to a distant star and returns to find her stay-at-home sister much older than she is. Thus far no paradox. But Alexis Allen formulates the Twin Paradox for us: "The theory of special relativity tells us that all motion is relative. With respect to the traveling twin, the Earth-bound twin moves away and then returns. Therefore it is the Earthbound twin who should be younger than the 'traveling' twin. But when they meet again at the same place, it cannot possibly be that each twin is younger than the other twin. This Twin Paradox disproves relativity." The paradox is usually resolved by realizing that the traveling twin turns around. Everyone agrees which twin turns around, since the reversal of direction slams the poor traveler against the bulkhead of the decelerating starship, breaking her collarbone. The turnaround, evidenced by the broken collarbone, destroys the symmetry required for the paradox to hold. Good-bye Twin Paradox! Still, Alexis's father Cyril Allen has his doubts about this resolution of the paradox. "Your solution is extremely unsatisfying. It forces me to ask: What if the retro-rockets malfunction and will not fire at all to slow me down as I approach a distant star a thousand light-years from Earth? Then I cannot even stop at that star, much less turn around and head back to Earth. Instead, I continue moving away from Earth forever at the original constant speed. Does this mean that as I pass the distant star, one thousand light-years from Earth, it is no longer possible to say that I have aged less than my Earth-bound twin? But if not, then I would never have even gotten to the distant star at all during my hundred-year lifetime! Your resolution of the Twin Paradox is insufficient and unsatisfying." Write a half-page response to Cyril Allen, answering his objections politely but decisively.

A particle moves with speed \(v^{\prime}\) in the \(x^{\prime} y^{\prime}\) plane of the rocket frame and in a direction that makes an angle \(\phi^{\prime}\) with the \(x^{\prime}\) axis. Find the angle \(\phi\) that the velocity vector of this particle makes with the \(x\) axis of the laboratory frame. (Hint: Transform space and time displacements rather than velocities.)

A train moves at \(10 \mathrm{~km} / \mathrm{h}\) along the track. A passenger sprints toward the rear of the train at \(10 \mathrm{~km} / \mathrm{h}\) with respect to the train. Our knee-jerk motto says that the train clocks "run slow" with respect to clocks on the track, and the runner's watch "runs slow" with respect to train clocks. Therefore the runner's watch should "run doubly slow" with respect to clocks on the track. But the runner is at rest with respect to the track. What gives? (This example illustrates the danger of the simple knee-jerk motto "Moving clocks run slow.")

A spaceship moving away from Earth at a speed \(0.900 c\) radios its reports back to Earth using a frequency of 100 MHz measured in the spaceship frame. To what frequency must Earth's receivers be tuned in order to receive the reports?

(a) Two events occur at the same time in the laboratory frame and at the laboratory coordinates \(\left(x_{1}=\right.\) \(\left.10 \mathrm{~km}, y_{1}=4 \mathrm{~km}, z_{1}=6 \mathrm{~km}\right)\) and \(\left(x_{2}=10 \mathrm{~km}, y_{2}=7 \mathrm{~km}, z_{2}=\right.\) \(-10 \mathrm{~km})\). Will these two events be simultaneous in a rocket frame moving with speed \(v^{\text {rel }}=0.8 c\) in the \(x\) direction in the laboratory frame? Explain your answer. (b) Three events occur at the same time in the laboratory frame and at the laboratory coordinates \(\left(x_{0}, y_{1}, z_{1}\right),\left(x_{0}, y_{2}, z_{2}\right)\), and \(\left(x_{0}, y_{3}, z_{3}\right)\), where \(x_{0}\) has the same value for all three events. Will these three events be simultaneous in a rocket frame moving with speed \(v^{\text {rel }}\) in the laboratory \(x\) direction? Explain your answer. (c) Use your results of parts (a) and (b) to make a general statement about simultaneity of events in laboratory and rocket frames.
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