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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 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . 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 \(\sqrt{1-u^{2} / c^{2} \text { by a binomial expansion. }}\))

Short Answer

Expert verified
The plane's clock reads about 0.36 milliseconds less, showing shorter time.

Step by step solution

01

Understanding Time Dilation

We need to calculate the difference in time due to the effects of time dilation, a relativistic effect predicted by Einstein's theory of relativity. When an object moves at a high speed relative to a stationary observer, the time experienced by the moving object is less than that experienced by the stationary observer. This phenomenon is described by the equation: \[\Delta t' = \Delta t \sqrt{1 - \frac{v^2}{c^2}}\]where \(\Delta t\) is the time interval for the stationary observer, \(\Delta t'\) is the time interval for the moving clock, \(v\) is the velocity of the moving clock, and \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\).
02

Simplifying with Binomial Expansion

Given that \(v \ll c\), we can use a binomial expansion to simplify the term \(\sqrt{1 - \frac{v^2}{c^2}}\). The binomial expansion is given by:\[(1 - x)^n \approx 1 - nx \,\text{for small } x.\]Applying this to our equation, we get:\[\sqrt{1 - \frac{v^2}{c^2}} \approx 1 - \frac{1}{2} \cdot \frac{v^2}{c^2}.\]
03

Calculating the Time Difference

We substitute this approximation back into the original time dilation equation:\[\Delta t' \approx \Delta t \left(1 - \frac{1}{2} \cdot \frac{v^2}{c^2}\right).\]Given that \(\Delta t = 4.00 \text{ hours} = 4.00 \times 3600 \text{ seconds}\) and \(v = 250 \text{ m/s}\), the time difference becomes:\[\Delta t' = 14400 \times \left(1 - \frac{1}{2} \cdot \frac{(250)^2}{(3 \times 10^8)^2}\right).\]
04

Solving for the Time Experienced by the Plane

Calculate the right-hand side to find the time according to the airplane clock:\[\Delta t' = 14400 \times \left(1 - \frac{1}{2} \cdot \frac{62500}{9 \times 10^{16}}\right)= 14400 \times \left(1 - \frac{31250}{9 \times 10^{16}}\right).\]Evaluating this, we find:\[\Delta t' \approx 14400 - 1.0 \times 10^{-11} (14400) \approx 14399.99999964 \text{ seconds}.\]
05

Calculating the Difference in Time

The difference in elapsed time between the two clocks is:\[\Delta \tau = \Delta t - \Delta t' = 14400 - 14399.99999964 \approx 3.6 \times 10^{-4} \text{ seconds}.\]This indicates that the clock aboard the plane shows slightly less time elapsed compared to the clock that remained in New York.
06

Conclusion on Which Clock Shows Shorter Time

Since \(\Delta t' < \Delta t\), the clock that was on the airplane shows less elapsed time. This is consistent with the predictions of special relativity, where time dilation causes the moving clock to experience a shorter duration.

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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 fundamental theory in physics introduced by Albert Einstein in 1905. It describes how time and space are intertwined and behave differently depending on the observer's state of motion. This theory introduces the concept of time dilation, which explains that time passes at different rates for observers in different inertial frames of motion. When an object moves at a fast speed, such as an airplane traveling across the world, time for that object slows down relative to a stationary observer. This means that less time passes for someone on the plane compared to someone on the ground. The resulting effect is time dilation, a key aspect of special relativity. Time dilation is represented by the equation: \[\Delta t' = \Delta t \sqrt{1 - \frac{v^2}{c^2}}\] where \( \Delta t \) is the time interval for a stationary observer, \( \Delta t' \) is the time interval for the moving clock, \( v \) is the velocity of the moving object, and \( c \) is the speed of light, a constant value approximately \( 3 \times 10^8 \text{ m/s} \). This theory becomes particularly noticeable only at speeds close to the speed of light, but even at lower speeds, small effects can be detected with precise measurements.
Einstein's Theory of Relativity
Einstein's Theory of Relativity consists of two main parts: special relativity and general relativity. Einstein proposed these groundbreaking theories to describe the interactions of motion, time, and gravity. In special relativity, introduced in 1905, Einstein established that the laws of physics are the same for all observers no matter their speed, and the speed of light is constant in a vacuum. This means that as an object approaches the speed of light, time, as experienced by anyone on that object, slows down compared to someone stationary. General relativity, introduced in 1915, expands on this by describing gravity not as a force, but as the curvature of spacetime caused by mass. While general relativity includes more complex factors, it works harmoniously with special relativity's basic principles. These theories revolutionized our understanding of physics, leading to new insights and technologies, such as GPS, which requires adjustments for time dilation to provide accurate location data.
Atomic Clocks
Atomic clocks are the most accurate timekeeping devices available today. They use the vibrations of atoms as their timekeeping element. Unlike traditional clocks, which might use a pendulum or a crystal, atomic clocks measure the frequency of microwave radiation that causes electrons in atoms to transition between energy levels. The reason atomic clocks are so precise is that the frequency of these atomic transitions remains consistent, leading to enormous accuracy. These clocks can measure time down to billionths of a second, making them essential in fields where precision is crucial, such as in global positioning systems (GPS) and scientific research that involves time-sensitive data. In the concept of time dilation, atomic clocks are used to measure the small yet significant differences in time experienced by moving and stationary observers, as they are sensitive enough to detect changes resulting from the effects of special relativity even at relatively low velocities.
Binomial Expansion
The binomial expansion is a mathematical method used to simplify expressions involving powers. When dealing with time dilation, if the velocity of the object \( v \) is much less than the speed of light \( c \) (\( v \ll c \)), the term \( \sqrt{1 - \frac{v^2}{c^2}} \) can be approximated using this expansion.The binomial theorem states that for any real number \( x \) close to zero, and \( n \) as a power, the expression \( (1 - x)^n \) can be approximated as:\[ (1 - x)^n \approx 1 - nx \]Applying this to the time dilation equation, we substitute \( x = \frac{v^2}{c^2} \) and \( n = 0.5 \) resulting in:\[ \sqrt{1 - \frac{v^2}{c^2}} \approx 1 - \frac{1}{2} \cdot \frac{v^2}{c^2} \]This simplification is valuable because it allows us to calculate time dilation effects efficiently in scenarios involving relatively low speeds like an airplane's flight. Such approximations make the calculations more manageable while still providing accurate results.

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Most popular questions from this chapter

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c ?\) (Express the answers in terms of \(m c^{2}-)(c)\) How do your answers in parts \((a)\) and \((b)\) compare?

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400 \(\mathrm{c}\) . The enemy ship fires a missile toward you at a speed of 0.700\(c\) relative to the enemy ship (Fig. 37.28\()\) . (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is \(8.00 \times 10^{6} \mathrm{km}\) away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

Calculate the magnitude of the force required to give a \(0.145-\mathrm{kg}\) baseball an acceleration \(a=1.00 \mathrm{m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{m} / \mathrm{s} ;\) (b) 0.900 \(\mathrm{c}\) (c) \(0.990 c .(\mathrm{d})\) Repeat parts \((\mathrm{a}),\) and \((\mathrm{c})\) if the force and acceleration are perpendicular to the velocity.

Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light-year is the distance that light travels in a year.) You plan to travel at constant speed in a \(1000-\mathrm{kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{19} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }},\) \(t_{\text { earth }},\) \(\mathbf{t}_{\text { rouket }} \boldsymbol{E}\) \((\text { in } \mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) (c) \(0.9999 \mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

When a particle meets its antiparticle, they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{19} \mathrm{J}\) of energy per year (a) If all this energy came from a futuristic ant-matter reactor, how much mass of matter and antimatter fuel would be consumed yearly? (b) If this fuel had the density of iron \(\left(7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) and were stacked in bricks to form a cubical pile, how high would it be? (Before you get your hopes up, antimatter reactors are a long way in the future\(- \)if they ever will be feasible.)

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