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Chapter 37: Problem 6

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 0.800 c relative to you. At the instant the space-racer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{in}\) 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) 0.833 seconds; (b) 2.00 x 10^8 meters; (c) 0.5 seconds.

Step by step solution

01

Understanding the Problem

This exercise involves the relative motion of two observers in space, with speeds approaching the speed of light. We're dealing with special relativity due to high velocities, given as 0.800c (where c is the speed of light). We need to calculate time and distance measurements from two different frames of reference.
02

Calculations for Part (a)

The velocity of the race pilot relative to the space utility vehicle is 0.800c. We need to find the time on the race pilot's clock when the space utility vehicle measures that she has traveled 1.20 x 10^8 meters.Using the formula for time dilation: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( t = \frac{d}{v} \), \( v = 0.800c \), and \( d = 1.20 \times 10^8 \) meters.Calculate: \[ t = \frac{1.20 \times 10^8}{0.800 \times 3.00 \times 10^8} \approx 0.5 \, \text{seconds} \]Thus, \[ t' = \frac{0.5}{\sqrt{1 - 0.6400}} \approx \frac{0.5}{0.6} \approx 0.833 \text{ seconds} \]
03

Calculations for Part (b)

Now, we calculate the distance the race pilot measures from her perspective when her timer reads 0.833 seconds. To her, the space utility moves in the opposite direction but at the same speed (0.800c). Using the formula \[ d' = v \cdot t' \],we find:\[ d' = 0.800 \times 3.00 \times 10^8 \times 0.833 \approx 2.00 \times 10^8 \, \text{meters}\]
04

Calculations for Part (c)

Finally, find the time you will read on your timer when the race pilot reads 0.833 seconds on her timer. We have to account for time dilation from the race pilot's frame:Since the time is already given for the race pilot as 0.833 seconds, due to the symmetry of time dilation,\[ t = 0.833 \times \sqrt{1 - 0.6400} \approx 0.5 \, \text{seconds}\] for your measurement.

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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 a fascinating concept from Einstein's theory of special relativity. It describes how time experienced by an observer moving at a significant fraction of the speed of light differs from time experienced by a stationary observer. This discrepancy occurs because, at high velocities, time seems to slow down.

In the exercise scenario, the race pilot is moving at 0.800 times the speed of light or 0.800c. When you, as the pilot of the space utility vehicle, measure the movement of the spaceracer, the time on the race pilot's clock will be less than what your clock measures.

To find the time on the race pilot's clock, you use the formula for time dilation, which is: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where:
  • \(t'\) is the time observed in the space racer鈥檚 frame,
  • \(t\) is the time observed in your frame,
  • \(v\) is the velocity relative to you,
  • \(c\) is the speed of light.


Calculating step-by-step helps visualize how relative motion affects time. It's an excellent demonstration of how everyday intuitions about time can change dramatically at cosmic speeds.
Relative Motion
Relative motion is key in understanding scenarios involving two moving objects. In this case, both the space utility vehicle and the space racer are moving, but the focus is on the space racer's speed relative to your vehicle.

Each observer measures time and distance differently based on their relative motion. Due to the symmetry in their speeds (both consider the other moving at 0.800c away from them), their observations are tied directly to their respective frames of reference.

From your perspective, when the race pilot has moved a certain distance, time passes differently due to your high-speed difference. Interestingly, from the race pilot鈥檚 perspective, while she reads 0.833 seconds on her timer, she perceives different distances for your movement. This difference encapsulates the core idea where each observer sees the other in motion, yet differently from themselves.
Speed of Light
In special relativity, the speed of light (\(c\)) is a universal constant, approximately \(3.00 \times 10^8\) meters per second. It sets the ultimate speed limit in the universe and greatly influences the nature of time and space.

Since neither object in the exercise can exceed this speed, their relative motion calculations are based on fractions of light speed, hence the notation \(0.800c\).

Understanding light's role here involves recognizing how it shapes the measurements from different frames of reference. This constant speed ensures that regardless of the observers' motions, both the space racer and the utility vehicle perceive the limits of their speed relative to light unchanged. It highlights the revolutionary impact of relativity theories on physics, demonstrating the principle of invariant speed regardless of relative movement, a centerpiece in modern physics discussions.

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Most popular questions from this chapter

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?

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)

(a) By what percentage does your rest mass increase when you climb 30 \(\mathrm{m}\) to the top of a ten-story building? Are you aware of this increase? Explain. (b) By how many grams does the mass of a \(120-\mathrm{g}\) spring with force constant 200 \(\mathrm{N} / \mathrm{cm}\) change when you compress it by 6.0 \(\mathrm{cm} \%\) Does the mass increase or decrease? Would you notice the change in mass if you were holding the spring? Explain.

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts \((a)\) and \((b)\) , find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Two events observed in a frame of reference Shave positions and times given by \(\left(x_{1}, t_{1}\right)\) and \(\left(x_{2}, t_{2}\right),\) respectively. (a) Frame \(S^{\prime}\) moves along the \(x\) -axis just fast enough that the two events occur at the same position in \(S^{\prime} .\) Show that in \(S^{\prime},\) the time interval \(\Delta t^{\prime}\) between the two events is given by $$\Delta t^{\prime}=\sqrt{(\Delta t)^{2}-\left(\frac{\Delta x}{c}\right)^{2}}$$ where \(\Delta x=x_{2}-x_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Hence show that if \(\Delta x>c \Delta t,\) there is \(n o\) frame \(S^{\prime}\) in which the two events occur at the same point. The interval \(\Delta t^{\prime}\) is sometimes called the proper time interval for the events. Is this term appropriate? (b) Show that if \(\Delta x>c \Delta t,\) there is a different frame of reference \(S\) in which the two events occur simultaneously. Find the distance between the two events in \(S^{\prime} ;\) express your answer in terms of \(\Delta x, \Delta t,\) and \(c\). This distance is sometimes called a proper length. Is this term appropriate? (c) Two events are observed in a frame of reference \(S^{\prime}\) to occur simultancously at points separated by a distance of 2.50 \(\mathrm{m}\) . In a second frame \(S\) moving relative to \(S^{\prime}\) along the line joining the two points in \(S^{\prime},\) the two events appear to be separated by 5.00 \(\mathrm{m}\) . What is the time interval between the events as measured in \(S ?[\text { Hint: Apply the result obtained in part (b).1 }\)
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