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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

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.80 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control (on earth) who is watching the experiment? (b) If each swing takes 1.80 s as measured by a person at mission control, how long will it take as measured by the astronaut in the spaceship?

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9930c. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory?

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?

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earthbased observers? Do present- day jet planes approach such speeds?

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