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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?

Short Answer

Expert verified
(a) 0.9s, (b) 7.2m, (c) 1.5s.

Step by step solution

01

Identify Given Data

The problem involves special relativity and relative motion. The key information provided is: * Spaceracer's speed = 0.800c (where c is the speed of light) relative to you. * The distance measured by you as 1.20 \( \times \) 10\(^8\) meters past you.
02

Calculate Time on Race Pilot’s Timer

To find the time the race pilot reads on her timer when you measure the distance, use the formula for time dilation: \( \Delta t = \gamma \times \Delta t' \), where \( \Delta t' \) is the proper time (time from your frame), \( \Delta t = \frac{L}{v} \) is the time from your perspective, \( L = 1.20 \times 10^8 \) meters is the distance, and \( v = 0.800c \) is the relative velocity. First, calculate the time you measure: \( \Delta t = \frac{1.20 \times 10^8}{0.800c} \). Then, calculate the Lorentz factor: \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} = \frac{1}{\sqrt{1-0.64}} \approx 1.6667 \). Substitute into \( \Delta t \): \( \Delta t' = \Delta t / \gamma \).
03

Calculate Distance Measured by Race Pilot

Once we have \( \Delta t' \), multiply by her velocity \( v = 0.800c \) to calculate the distance she measures: \( d' = v \times \Delta t' = 0.800c \times \Delta t' \). This is the distance the race pilot reads when her timer shows the time calculated earlier.
04

Calculate Your Timer Reading When Race Pilot’s Timer Matches

At the instant when the race pilot’s timer reads the value we calculated, use the same gamma factor to find the corresponding reading on your timer. We've already calculated that from the earlier time calculations.

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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 an intriguing aspect of Einstein's theory of special relativity. It describes how time can pass at different rates for observers who are moving relative to one another.
In our exercise, you and the race pilot both start your timers simultaneously. However, because her speed is 0.800c (where \(c\) is the speed of light), she experiences time differently than you do. This is because the faster you move through space, the slower you move through time.
To find the time read on the race pilot's timer, we use the time dilation formula:
  • \( \Delta t = \gamma \times \Delta t' \)
Where \( \Delta t' \) is her proper time (the time as measured in her own frame) and \( \Delta t \) is the time as you measure it.
This tells us how time is 'stretched' or dilated when objects move at high velocities. So, although you and the race pilot start timing together, she'll read less time on her clock compared to yours because of this unique feature of time dilation.
Lorentz Factor
The Lorentz factor, denoted by \( \gamma \), plays a crucial role in calculating time dilation and other effects of special relativity. It essentially scales the time or distance observed in a moving frame relative to a stationary one.
In our situation, the Lorentz factor helps in adjusting times and lengths measured by stationary observers to the moving observer's frame. It is given by the formula:
  • \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \)
Where \( v \) is the velocity of the moving object, and \( c \) is the speed of light.
In the given exercise, since the race pilot flies at 0.800c, we calculate:
  • \( \gamma = \frac{1}{\sqrt{1-0.64}} \approx 1.6667 \)
This factor quantifies how much time, length, and mass appear to change with velocity. It's a fundamental piece of understanding how our universe handles high-speed travel.
Relative Velocity
Relative velocity is an essential concept when dealing with objects moving at high speeds. It's the velocity of one object as observed from another object that is also moving. In the exercise, the relative velocity is between your space vehicle and the race pilot's spaceracer.
Since you're moving towards the moon and the race pilot passes by you at 0.800c, this is her speed relative to you. Understanding relative velocity helps us solve how time and distance are perceived differently by each observer. Consider these points to grasp the concept of relative velocity better:
  • Relative velocity determines how two observers moving with respect to one another perceive time and space.
  • It is crucial for applying the Lorentz factor to adjust time and distance measurements.
In high-speed scenarios like this one, calculations based on relative velocity reveal that time and space are not absolute but depend on the observer's motion.
Proper Time
Proper time, represented often as \( \Delta t' \), is the time interval measured by an observer at rest relative to the event being timed. In the context of special relativity, it's the time recorded by a clock that moves along with the object being measured.
For the race pilot, her timer measures the proper time since it's moving with her as she flies past you. This proper time differs from the time you measure because of the effects of time dilation.To further clarify proper time:
  • It is the time measured in the rest frame of the object involved.
  • Time dilation causes the proper time to be the shortest time interval observed, compared to the time in a moving observer's frame.
In essence, proper time gives us insight into the specific timing of events from the viewpoint of someone 'traveling with the clock' so to speak. It's a foundational concept for accurately understanding the discrepancies in time recordings noted by different observers.

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Most popular questions from this chapter

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of 0.520c. If the radiation has a frequency of 8.64 \(\times\) 10\(^{14}\) Hz in the rest frame of the star, what is the frequency measured by an observer on earth?

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light in the laboratory reference frame, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 h. By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (\(Hint\): Since \(u \ll c\), you can simplify \(\sqrt{1 - u^2/c^2}\) by a binomial expansion.)

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 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 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?

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)?

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