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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 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 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 99.68% of the speed of light.

Step by step solution

01

Identify Proper Time

Proper time is the time measured by an observer who is at rest relative to the event being measured. In this case, the proper time is the time measured by the first officer on the spacecraft, as the officer is at rest relative to the searchlight.
02

Use Time Dilation Formula

The time dilation formula is given by \( t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \), where \( t \) is the time measured by an observer moving relative to the event, \( t_0 \) is the proper time, \( v \) is the velocity of the moving observer, and \( c \) is the speed of light.
03

Convert Measurements to Same Units

Convert the proper time \( t_0 = 12.0 \) ms to seconds: \( 12.0 \text{ ms} = 0.012 \text{ s} \). The observer on Earth measures \( t = 0.150 \text{ s} \).
04

Solve for Velocity Fraction

Rearrange the time dilation formula to solve for \( \frac{v}{c} \): 1. \( t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \) 2. Solve for \( \frac{v^2}{c^2} \), giving \( 1-\frac{v^2}{c^2} = \left(\frac{t_0}{t}\right)^2 \)3. \( \frac{v^2}{c^2} = 1-\left(\frac{t_0}{t}\right)^2 \)4. \( \frac{v}{c} = \sqrt{1-\left(\frac{0.012}{0.150}\right)^2} \)
05

Calculate and Simplify

Calculate \( \left(\frac{0.012}{0.150}\right)^2 = 0.0064 \). Therefore, \( \frac{v}{c} = \sqrt{1 - 0.0064} = \sqrt{0.9936} \). This gives \( \frac{v}{c} \approx 0.9968 \).

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Key Concepts

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

Proper Time
In the realm of physics, specifically in the context of special relativity, **proper time** becomes an essential concept. Proper time is the time interval measured by an observer who is at rest relative to the process being observed. It is the true duration of an event from the perspective of someone or something that is not moving relative to the event.

This concept is crucial when dealing with time dilation because the proper time represents the "true" time for events. In our exercise, the officer on the spacecraft measures the searchlight blink duration as 12.0 ms. This is considered the proper time because the officer and the light event are both in the same reference frame鈥攖he spacecraft itself. In contrast, the observer on Earth measures a different time, 0.150 s, due to the relativistic effects of time dilation since the spacecraft is moving relative to Earth.
Velocity of Spacecraft
In understanding how fast the spacecraft is moving relative to Earth, we need to delve into the **velocity of the spacecraft** in terms of the speed of light. The formula that governs this relationship is derived from time dilation, which takes into account the effects of an object's speed as it approaches the speed of light.

The important thing to remember is:
  • The observer on Earth perceives a longer time for the searchlight blink due to the relative motion of the spacecraft.
  • This time dilation occurs because the spacecraft is moving at a significant fraction of the speed of light (\( c \)).
  • By rearranging the time dilation formula, we solved for the ratio of the spacecraft's velocity to the speed of light.
In our example, after using the time dilation formula and substituting the known values, we calculated the velocity of the spacecraft as a fraction of the speed of light, approximately 0.9968. This indicates the spacecraft is traveling at almost the speed of light itself, leading to the significant time difference measured by the observers.
Speed of Light
The **speed of light** in a vacuum is a fundamental constant of nature, symbolized by \( c \). It holds a value of approximately 299,792,458 meters per second. This isn't just any random measurement鈥攊t defines the ultimate speed limit of the universe.

Key points about the speed of light:
  • Nothing can travel faster than this limit.
  • In our exercise, the speed of light serves as a reference to determine how close the spacecraft's velocity gets to this universal ceiling.
  • The time dilation effects become significant as velocities approach the speed of light, as seen with the spacecraft.
Understanding the speed of light helps in imagining how time can "stretch" and "shrink" depending on the observer's frame of reference and velocity. As the velocity approaches \( c \), time dilation becomes noticeable, which is exactly what we observe in the time difference between the spacecraft and Earth in the exercise.

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Most popular questions from this chapter

(a) How fast must you be approaching a red traffic (\(\lambda=\) 675 nm) for it to appear yellow (\(\lambda=\) 575 nm)? Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is $1.00 for each kilometer per hour that your speed exceeds the posted limit of 90 km/h.

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.600c. A scientist on Coruscant measures the length of the moving spacecraft to be 74.0 m. The spacecraft later lands on Coruscant, and the same scientist measures the length of the now stationary spacecraft. What value does she get?

After being produced in a collision between elementary particles, a positive pion (\(\pi^+\)) must travel down a 1.90-km-long tube to reach an experimental area. A \(\pi^+\) particle has an average lifetime (measured in its rest frame) of 2.60 \(\times\) 10\(^{-8}\) s; the \(\pi^+\) we are considering has this lifetime. (a) How fast must the \(\pi^+\) travel if it is not to decay before it reaches the end of the tube? (Since \(u\) will be very close to \(c\), write \(u\) = (1 - \(\Delta\))c and give your answer in terms of \(\Delta\) rather than \(u\).) (b) The \(\pi^+\) has a rest energy of 139.6 MeV. What is the total energy of the \(\pi^+\) at the speed calculated in part (a)?

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 observer in frame \(S'\) is moving to the right (+\(x\)-direction) at speed \(u\) = 0.600c away from a stationary observer in frame S. The observer in \(S'\) measures the speed \(v'\) of a particle moving to the right away from her. What speed \(v'\) does the observer in S measure for the particle if (a) \(v'\) = 0.400c; (b) \(v'\) = 0.900c; (c) \(v'\) = 0.990c?
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