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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—the 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—it 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

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?

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c\) ? (Express the answers in terms of \(mc^2\).) (c) How do your answers in parts (a) and (b) compare?

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 0.800c relative to you. At the instant the spaceracer 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\(^8\) m 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?

A spaceship moving at constant speed u relative to us broadcasts a radio signal at constant frequency \(f_0\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_0\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_0\)? (\(Hint\): In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1 /T. Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_0\) = 345 MHz as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f - f_0\)? (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift (\(f - f_0\)) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.890c. Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

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