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Chapter 37: Problem 7

A spacecraft flies away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) 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 is behind by 0.380 hours; it shows the shorter elapsed time.

Step by step solution

01

Understand time dilation

According to the theory of relativity, a moving clock runs slower than a stationary clock due to time dilation. We need to calculate how much time has passed on the spacecraft's clock compared to the Earth's clock.
02

Identify given data

We know the speed of the spacecraft is \(v = 4.80 \times 10^6 \mathrm{m/s}\). The Earth clock measures \(t_0 = 365\) days, and we need to convert this duration to seconds for further calculations.
03

Convert time to seconds

Since 1 day = 86400 seconds, the total time measured by the Earth clock is \(t_0 = 365 \times 86400 \) seconds.
04

Apply time dilation formula

The time dilation formula is \(t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\), where \(c\) is the speed of light, \(c = 3.00 \times 10^8 \mathrm{m/s}\). Substitute the known values into this equation to find the time interval \(t\) measured by the spacecraft's clock.
05

Simplify the equation

Plug in the values: \(t = \frac{365 \times 86400}{\sqrt{1 - \left(\frac{4.80 \times 10^6}{3.00 \times 10^8}\right)^2}}\) seconds. Calculate the value inside the square root first, then solve for \(t\).
06

Calculate elapsed time difference

After finding \(t\) in seconds, convert it back to days by dividing by 86400. The difference in time \(\Delta t = t_0 - t\) is given in days. Convert \(\Delta t\) to hours (1 day = 24 hours) to find the final answer.
07

Determine the shorter clock

The spacecraft clock will show a shorter elapsed time compared to the Earth clock due to time dilation. Thus, the difference calculated will be the time lag of the spacecraft's clock.

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Key Concepts

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

Special Relativity
To understand time dilation, we first need to dig into Special Relativity. This is a theory developed by Albert Einstein in the early 20th century. Special Relativity primarily deals with how time and space are interconnected when objects move at significant fractions of the speed of light. One of the theory's most revolutionary aspects is that time is not absolute.
  • Time can "slow down" or "dilate" for objects in motion relative to a stationary observer.
  • This effect becomes more noticeable as the object's speed nears the speed of light.
  • Special Relativity is essential to explaining phenomena where high-speed travel is involved.
Understanding these principles allows us to calculate the curious side effects of high-speed space travel. As in the exercise, this is exactly what happens on the spacecraft: the moving clock (spacecraft clock) runs slower than the clock on Earth.
Relativistic Physics
Relativistic Physics is the study of physics that takes into account the effects of Special Relativity. This area of physics explains how objects behave when they move at speeds close to the speed of light. A key factor in relativistic scenarios is the Lorentz Factor, symbolized by \[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\]
    • As velocity \( v \) approaches the speed of light \( c \), the Lorentz Factor increases significantly, resulting in substantial time dilation.
    • The time dilation effect explains why a moving clock ticks slower than one at rest in a high-speed context.
    Calculating these scenarios helps elucidate how differences in time will manifest between two clocks, such as those on a traveling spacecraft versus one stationary on Earth.
    • Atomic Clock Synchronization
    Atomic clocks are highly precise timekeeping devices essential for experiments in Relativistic Physics. They synchronize time to an incredibly accurate degree.
    • Before the spacecraft departs, the clock aboard is synchronized with one on Earth.
    • This synchronization is crucial to accurately measure the effects of time dilation over the journey.
    Atomic clocks operate by measuring the vibrations of atoms, which are at a consistent and high frequency, ensuring remarkable precision.
    When the spacecraft returns, the atomic clock onboard shows less elapsed time compared to the synchronized clock on Earth. This difference is due to time dilation, as indicated by the calculation in the exercise, reflecting one of the most interesting and observable consequences of Special Relativity.

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    Most popular questions from this chapter

    A particle with mass \(m\) accelerated from rest by a constant force \(F\) will, according to Newtonian mechanics, continue to accelerate without bound; that is, as \(t \rightarrow \infty, v \rightarrow \infty .\)Show that according to relativistic mechanics, the particle's speed approaches \(c\) as \(t \rightarrow \infty\) [Note: A useful integral is \(\int\left(1-x^{2}\right)^{-3 / 2} d x=\) \(x / \sqrt{1-x^{2}}\)]

    Measuring Speed by Radar. A baseball coach uses a radar device to measure the speed of an approaching pitched base-ball. This device sends out electromagnetic waves with frequency \(f_{0}\) and then measures the shift in frequency \(\Delta f\) of the waves reflected from the moving baseball. If the fractional frequency shift produced by a baseball is \(\Delta f / f_{0}=2.86 \times 10^{-7}\) , what is the baseball's speed in \(\mathrm{km} / \mathrm{h} ?\) (Hint: Are the waves Doppler- shifted a second time when reflected off the ball?

    (a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-\) vt and \(t^{\prime}=t .\) In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ This has a different form than the wave equation in \(S .\) Hence the Galilean transformation violates the first relativity postulate that all physical laws have the same form in all inertial reference frames. (Hint: Express the derivatives \(\partial / \partial x\) and \(\partial / \partial t\) in terms of \(\partial / \partial x^{\prime}\) and \(\partial / \partial t^{\prime}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Egs. \((37.21),\) and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S :\) $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

    An imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920\(c\) relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360\(c .\) What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

    A spaceship flies past Mars with a speed of 0.985\(c\) relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for 75.0\(\mu\) s. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?
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