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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

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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Many of the stars in the sky are actually \(binary\space stars\), in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called \(spectroscopic\space binary\space stars. \textbf{Figure P37.68}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m\), orbiting their center of mass in a circle of radius \(R\). The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of 4.568110 \(\times\) 10\(^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) Hz. Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (\(Hint\): The speeds involved are much less than \(c\), so you may use the approximate result \(\Delta f/f = u/c\) given in Section 37.6.) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, 1.99 \(\times\) 10\(^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, 1.50 \(\times\) 10\(^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

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