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A spacecraft flies away from the earth with a speed of 4.80 \(\times\) 10\(^6\) m/s relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days (1 year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shorter elapsed time?

Short Answer

Expert verified
The spacecraft clock shows a shorter elapsed time by 0.040 hours.

Step by step solution

01

Understand the situation

This is a problem about time dilation in special relativity. The concept means that a clock moving at high speed relative to an observer will appear to tick more slowly than a stationary clock to that observer. We need to determine the time difference between clocks on Earth and on the spacecraft after a year of travel.
02

Apply Time Dilation Formula

In special relativity, the time dilation formula is given by \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \(t'\) is the time experienced by the stationary observer (Earth clock), \(t\) is the time experienced by the moving observer (spacecraft clock), \(v\) is the velocity of the spacecraft, and \(c\) is the speed of light (approximately \(3 \times 10^8\) m/s).
03

Calculate the Lorentz Factor

Calculate the Lorentz factor, \(\gamma\), using the spacecraft's speed:\[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{4.80 \times 10^6}{3 \times 10^8}\right)^2}} \]Compute the value inside the square root first, and then find the reciprocal to determine \(\gamma\).
04

Calculate Time Experienced by the Spacecraft Clock

Using the Lorentz factor, compute the time \(t\) experienced by the spacecraft:\[ t = \frac{t'}{\gamma} = \frac{365}{\gamma} \]Here, \(t' = 365 \text{ days}\) is the time measured on Earth. Convert this into hours after calculating \(t\).
05

Find the Time Difference

The time difference \(\Delta t\) between the two clocks is given by \[ \Delta t = t' - t \]Compute \(\Delta t\) in hours by converting days into hours.
06

Conclude the Shorter Time

Based on the calculations, the time \(t\) on the spacecraft clock, is less than \(t'\). Therefore, the spacecraft clock shows a shorter elapsed time compared to the Earth clock.

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Key Concepts

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

Special Relativity
Special relativity is a crucial concept when we study objects moving at very high speeds, close to the speed of light. Developed by Albert Einstein, this theory revolutionized our understanding of space and time. One of its core principles is that the laws of physics are the same for all observers, regardless of their relative motion.
This theory introduces the idea that time is not an absolute constant but can vary for different observers. As a result, time seems to pass slower for objects moving at high velocities compared to stationary observers. This phenomenon is known as time dilation.
  • Time dilation affects how we perceive time when traveling close to the speed of light.
  • This principle is key in understanding the difference in clock readings for the moving spaceship versus the stationary observer on Earth.
This is the underlying reason why the astronaut in our exercise perceives time differently compared to people on Earth.
Lorentz Factor
The Lorentz factor, denoted as \( \gamma \), helps us quantify the effects of time dilation in special relativity. It is a mathematical expression that describes how much time dilation or length contraction occurs at a given velocity relative to the speed of light.
The formula for calculating the Lorentz factor is:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where:
  • \( v \) is the velocity of the moving object (e.g., the spacecraft, in this case).
  • \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s.

The Lorentz factor becomes significant when the object's velocity approaches the speed of light. In our exercise, it is used to determine how much slower time passes on the spacecraft compared to Earth. Understanding and calculating \( \gamma \) is essential when dealing with high-speed travel.
Atomic Clock Synchronization
Atomic clocks are incredibly precise timekeeping devices used to maintain accurate time measurement. In our exercise, an atomic clock is placed on the spacecraft, and another remains on Earth. Initially, these clocks are synchronized to ensure they start counting time simultaneously.
Synchronization is crucial as it allows us to observe the effects of time dilation. As the spacecraft travels at high speeds, its atomic clock experiences time differently compared to the one on Earth. When the spacecraft returns, we can compare the two clocks to quantify how much time dilation has occurred.
  • Synchronized atomic clocks are fundamental in testing and observing real-world scenarios of special relativity.
  • They help scientists verify theoretical predictions about the differences in time perception at varying speeds.
In essence, synchronization helps emphasize the importance of timekeeping in understanding special relativity.
Spacecraft Velocity
Spacecraft velocity plays a vital role in this exercise, particularly its impact on time dilation. In our scenario, the spacecraft moves at a velocity of \( 4.80 \times 10^6 \) m/s relative to Earth. This speed, while considerably lower than the speed of light, is nonetheless significant enough to cause detectable time differences on its onboard clock due to time dilation.
The concept of relative velocity is central to special relativity. The faster an object travels compared to the constant speed of light, the more pronounced the effects of time dilation.
  • This exercise highlights how even a fraction of the speed of light can result in measurable differences in time due to high velocities.
  • The spacecraft returns to Earth showing that its clock lagged behind the Earth's clock, illustrating the real implications of special relativity.
Understanding how velocity affects time perception is crucial for both theoretical physics and real-world applications like space travel.

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Most popular questions from this chapter

The negative pion (\(\pi^-\)) is an unstable particle with an average lifetime of 2.60 \(\times\) 10\(^{-8}\)s (measured in the rest frame of the pion). (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be 4.20 \(\times\) 10\(^{-7}\) s. Calculate the speed of the pion expressed as a fraction of c. (b) What distance, measured in the laboratory, does the pion travel during its average lifetime?

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame \(S'\) moving relative to \(S\), the second event is observed to occur 2.15 s after the first. What is the difference between the positions of the two events as measured in \(S'\)?

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9380c as measured in the laboratory. What is the magnitude of the velocity of one particle relative to the other?

The positive muon (\(\mu^+\)), an unstable particle, lives on average 2.20 \(\times\) 10\(^{-6}\) s (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of 0.900c, what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

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