/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 \(\mathrm{As}\) you pilot your s... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{As}\) 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 spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{~m}\) 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
The calculated results would be (a) the race pilot's timer reading, (b) the measured distance from the race pilot to you and (c) your timer reading when the race pilot reads a calculated value from part (a).

Step by step solution

01

Apply Lorentz Transformation to Find the Race Pilot's Time

We first calculate the time read by the race pilot when the space utility vehicle has travelled a distance of \(1.20 \times 10^{8} m\). We use the time dilation formula, \( t = \frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \), where \( t_{0} \) is the proper time (time in moving frame), \( v \) is the velocity of the moving frame relative to the observer and \( c \) is the speed of light. In this case, \( t_{0} = \frac{d}{v} = \frac{1.20 \times 10^{8} m}{0.800c} \) and then substitute into the time dilation formula.
02

Calculate the Distance from the Race Pilot to the Utility Vehicle

Next, we calculate the distance from the race pilot to the utility vehicle when her timer reads the value calculated in part (a). This is simply her speed multiplied by the time, \( d = vt \). Now substitute the time from step 1 and the given speed of the race pilot into the formula.
03

Calculate the Utility Vehicle Pilot's Time Reading

Finally, we find the time the pilot of the utility vehicle reads on his timer when the race pilot's timer reads the value from part (a). Here again we use the Lorentz transformation and time dilation formula. The proper time \( t_{0} \) is the time calculated in part (a) and we put this back to the formula of time dilation to find \( t \).

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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 can appear to move at different rates for different observers, depending on their relative motion. The formula used to calculate time dilation is:\[ t = \frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \]Here:
  • \( t \) is the time interval for the observer in a stationary reference frame.
  • \( t_{0} \) is the proper time, or the time interval for the observer moving at velocity \( v \).
  • \( c \) represents the speed of light, which is approximately \( 3 \times 10^{8} \, \text{m/s} \).
Time dilation occurs because as you approach the speed of light, time appears to slow down for the observer moving relative to a stationary observer. This was beautifully illustrated in the exercise, where the race pilot traveling at a high speed experienced time differently compared to the stationary pilot on the utility vehicle.
Understanding this concept helps explain why astronauts returning from space may age slightly less than people who remain on Earth, a phenomenon known as the "twin paradox." The time dilation formula is key to calculations involving fast-moving objects, like the space racer in our exercise.
Special Relativity
Special relativity is one of the two pillars of modern physics, alongside quantum mechanics. Developed by Albert Einstein in 1905, it revolutionized how we understand space and time. This theory introduced the idea that the laws of physics are the same for all observers, regardless of their relative motion, provided they are moving at constant speeds.Two key postulates of special relativity are:
  • The laws of physics are identical for every observer not undergoing acceleration.
  • The speed of light in a vacuum is constant for all observers, regardless of their motion or the source's motion.
In our scenario with the space racer, special relativity helps explain the discrepancy in time and distance measurements between the two pilots. Cracking these complex ideas open showcases the non-intuitive reality that distances contract, and clocks tick differently when velocities approach that of light. Einstein's equation, \( E = mc^2 \), emerged from this theory, linking mass and energy as interchangeable.Special relativity does not address accelerations—this is where general relativity comes in—but it provides insights into the behavior of particles at high speeds, such as those in particle accelerators. It sets the scene for our understanding of the universe's fabric.
Speed of Light
The speed of light, denoted by \( c \), is a fundamental constant in the universe and plays a central role in the theory of relativity. Its value is approximately \( 3 \times 10^{8} \, \text{m/s} \), and it is the ultimate speed limit for the transmission of information and mass.The invariance of this speed in all inertial frames is a cornerstone of special relativity. Regardless of how fast you move, you will always measure the speed of light in a vacuum to be \( c \). This leads to some mind-boggling consequences, such as time dilation and length contraction.In our exercise, both the utility vehicle and the space racer deal with measurements that hinge on the speed of light. The speed sets the stage for time dilation calculations and dictates how the distance each pilot perceives changes as they move. As an object approaches this speed, relativistic effects become pronounced, affecting measurements of time and space.By understanding the speed of light as a cosmic speed limit, students can grasp why no massive object can reach or exceed it. This insight is pivotal not only in theoretical physics but also in practical applications like GPS systems that rely on relativistic corrections to function accurately.

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Most popular questions from this chapter

Spaceship \(A\) moves past the earth at \(0.80 c\) to the west. Spaceship \(B\) approaches \(A,\) moving to the east. Both spaceship crews measure their relative speed of approach to be \(0.98 c .\) What mass would the crews of both spaceships measure for the standard kilogram, kept at rest on the earth, (a) according to classical physics and (b) according to the special theory of relativity?

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

Electrons are accelerated through a potential difference of \(750 \mathrm{kV},\) so that their kinetic energy is \(7.50 \times 10^{5} \mathrm{eV}\). (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c ?\) (b) What would the speed be if it were computed from the principles of classical mechanics?

As measured by an observer on the earth, a spacecraft runway on earth has a length of \(3600 \mathrm{~m}\). (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of \(4.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of \(0.520 c\). If the radiation has a frequency of \(8.64 \times 10^{14} \mathrm{~Hz}\) in the rest frame of the star, what is the frequency measured by an observer on earth?

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