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Chapter 37: Problem 8

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.190 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 \(\mathrm{ms}\) . (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth expressed as a fraction of the speed of light \(c\) ?

Short Answer

Expert verified
(a) The proper time is 12.0 ms. (b) The speed is approximately 0.998c.

Step by step solution

01

Understanding Proper Time

The proper time is the time interval measured by an observer who sees two events occurring at the same place. In this case, the proper time is measured by the first officer on the spacecraft, as they measure the time duration of the light being on in their frame of reference. Therefore, the proper time is 12.0 ms.
02

Apply Time Dilation Formula

To find the speed of the spacecraft relative to the earth, we use the time dilation formula \[ t = \gamma t_0 = \frac{t_0}{\sqrt{1-v^2/c^2}} \]where \( t \) is the time observed on Earth (0.190 s) and \( t_0 \) is the proper time (12.0 ms or 0.012 s). Rearrange to solve for \( v \).
03

Solve for Relative Speed

Rearrange the time dilation formula to solve for the velocity \( v \):\[ \frac{t_0}{t} = \sqrt{1-\frac{v^2}{c^2}} \]Square both sides:\[ \left( \frac{t_0}{t} \right)^2 = 1 - \frac{v^2}{c^2} \]Rearrange further to isolate \( v^2/c^2 \):\[ \frac{v^2}{c^2} = 1 - \left( \frac{t_0}{t} \right)^2 \]
04

Calculation

Substitute \( t_0 = 0.012 \) s and \( t = 0.190 \) s into the equation:\[ \frac{v^2}{c^2} = 1 - \left( \frac{0.012}{0.190} \right)^2 \]Calculate the numerical value:\[ \frac{v^2}{c^2} = 1 - \left( 0.0632 \right)^2 \]\[ \frac{v^2}{c^2} = 1 - 0.00399 \]\[ \frac{v^2}{c^2} = 0.996 \]Solve for \( v/c \):\[ v/c = \sqrt{0.996} \approx 0.998 \]
05

Conclusion

The proper time is the one measured by the spacecraft officer, 12.0 ms. The spacecraft's speed relative to Earth is approximately 0.998 times the speed of light.

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Key Concepts

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

Proper Time
Proper time is a crucial concept in the theory of relativity, particularly when dealing with time dilation. It refers to the time measured by an observer who perceives the start and end of an event occurring at the same place in their frame of reference. For instance, if you are riding in a train and see two events happen within the train, you measure the proper time between them. In the original exercise, the first officer on the alien spacecraft measures the proper time for how long the searchlight is on because both events (the searchlight turning on and off) are happening at the same location in their reference frame, which is the spacecraft itself.
  • Proper time is always the shortest time interval measured for an event.
  • It provides a basis to compare time intervals observed in different frames.
Understanding proper time helps us appreciate how motion affects the perception of time. In the spacecraft's frame, the searchlight is on for 12.0 ms, which is the proper time. Meanwhile, observers in a different reference frame, like someone standing on Earth, would measure a different, longer time interval due to time dilation.
Relative Speed
Relative speed is how fast one object is moving in relation to another. In the context of the exercise, it refers to the speed of the spacecraft as seen from Earth. When objects move at speeds approaching the speed of light, time dilation becomes significant, and the actual speed can be calculated using the time dilation formula.
  • Relative speed affects the time experienced by observers in different frames.
  • As speed approaches that of light, relativistic effects become prominent.
In our solution, we used the time dilation formula to find the spacecraft's speed relative to Earth. By rearranging the formula and substituting the given time values, we solved for the spacecraft's velocity as a fraction of the speed of light. The calculated relative speed was approximately 0.998 times the speed of light. This calculation shows the immense speeds involved when approaching light's velocity, which emphasizes how crucial understanding relativity is for space travel.
Speed of Light
The speed of light, denoted by the symbol \(c\), is a fundamental constant in physics. It represents the fastest speed at which information can be transmitted through the universe. In vacuum, the speed of light is approximately 299,792,458 meters per second (or about 300,000 km/s). This speed forms the upper limit for how fast anything can travel, as postulated by Einstein's theory of special relativity.
  • Nothing can travel faster than the speed of light in vacuum.
  • Light's speed remains constant, regardless of the motion of the observer or the light source.
In the exercise at hand, the spacecraft's speed is expressed as a fraction of the speed of light. This approach is common in dealing with relativistic speeds because it provides a clear perspective on how close an object's velocity is to this natural limit. Understanding the boundaries set by light's speed helps in grasping the broader implications of relativity, such as time dilation experienced at high velocities, as demonstrated when calculating the spacecraft's speed relative to Earth.

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Most popular questions from this chapter

A \(0.100-\mu g\) speck of dust is accelerated from rest to a speed of 0.900\(c\) by a constant \(1.00 \times 10^{6} \mathrm{N}\) force. (a) If the nonrelativistic form of Newton's second law \((\Sigma F=m a)\) is used, how far does-the object travel to reach its final speed? (b) Using the correct relativistic treatment of Section 37.8 , how far does the object travel to reach its final speed? (c) Which distance is greater? Why?

Find the speed of a particle whose relativistic kinetic energy is 50\(\%\) greater than the Newtonian value for the same speed.

Show that when the source of electromagnetic waves moves away from us at 0.600 \(\mathrm{c}\) , the frequency we measure is half the value measured in the rest frame of the source.

A photon with energy \(E\) is emitted by an atom with mass \(m\) which recoils in the opposite direction. (a) Assuming that the motion of the atom can be treated nonrelativistically, compute the recoil speed of the atom. (b) From the result of part (a), show that the recoil speed is much less than \(c\) whenever \(E\) is much less than the rest energy \(m c^{2}\) of the atom.

When a particle meets its antiparticle, they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{19} \mathrm{J}\) of energy per year (a) If all this energy came from a futuristic ant-matter reactor, how much mass of matter and antimatter fuel would be consumed yearly? (b) If this fuel had the density of iron \(\left(7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) and were stacked in bricks to form a cubical pile, how high would it be? (Before you get your hopes up, antimatter reactors are a long way in the future\(- \)if they ever will be feasible.)
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