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Chapter 36: Problem 13

(II) Suppose a news report stated that starship Enterprise had just returned from a 5-year voyage while traveling at \(0.74 c .(a)\) If the report meant 5.0 years of Earth time, how much time elapsed on the ship? (b) If the report meant 5.0 years of ship time, how much time passed on Earth?

Short Answer

Expert verified
(a) 7.43 years elapsed on the ship. (b) 3.36 years passed on Earth.

Step by step solution

01

Understanding Time Dilation

According to the theory of relativity, time measured in a moving system (like the ship) is different from the time measured in a stationary system (like Earth). This is referred to as time dilation. The relationship is given by the formula: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \(t'\) is the time elapsed for the moving object, \(t\) is the time elapsed for the stationary observer, \(v\) is the velocity of the moving object, and \(c\) is the speed of light.
02

Solving for Time Elapsed on the Ship (Part a)

Given that \(v = 0.74c\) and \(t = 5.0\) years (Earth time), we plug these into the time dilation formula for \(t'\):\[ t' = \frac{5.0}{\sqrt{1 - (0.74)^2}} \]Calculate \(1 - (0.74)^2\), which gives 0.4524. Taking the square root of 0.4524 gives approximately 0.6726. Thus,\[ t' = \frac{5.0}{0.6726} \approx 7.43 \text{ years} \]The ship's time is approximately 7.43 years.
03

Solving for Time Passed on Earth (Part b)

Given that \(v = 0.74c\) and \(t' = 5.0\) years (ship time), we rearrange the dilation formula to solve for \(t\):\[ t = t' \times \sqrt{1 - \frac{v^2}{c^2}} \]Substitute the values:\[ t = 5.0 \times \sqrt{0.4524} \approx 5.0 \times 0.6726 \approx 3.36 \text{ years} \]Thus, Earth time is approximately 3.36 years.

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Key Concepts

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

Theory of Relativity
The theory of relativity, introduced by Albert Einstein, revolutionized our understanding of space and time. At its core, this theory suggests that the measurements of time and space are not absolute but vary depending on the frame of reference.
In simple terms, the faster you move through space, the slower you move through time, relative to a stationary observer.There are two main parts: special relativity and general relativity. This exercise focuses on special relativity, which deals with objects moving at significant fractions of the speed of light. One key outcome of special relativity is the concept of time dilation, where time for a moving object appears to pass differently compared to a stationary observer.
This difference, although tiny at everyday speeds, becomes significant at speeds close to the speed of light, represented by the symbol \(c\).
Velocity and Speed of Light
Velocity refers to an object's speed in a particular direction, whereas speed is just how fast something is moving, regardless of direction. In relativity, the speed of light, or \(c\), holds a special place.
It's the cosmic speed limit, meaning nothing can go faster than light in a vacuum, which is approximately 299,792,458 meters per second.When we talk about velocities close to this speed, as in the original exercise with \(0.74c\), we mean 74% of the speed of light. At such high speeds:
  • Motion affects the flow of time.
  • Time dilation comes into play, altering how time is experienced by moving objects.
This incredible aspect of physics illustrates why journeys at high speed, like those imagined for the starship Enterprise, involve complex time calculations to determine how time passage differs between Earth and a high-speed spacecraft.
Dilation Formula
The dilation formula is a mathematical representation of how time is experienced differently by someone moving relative to someone stationary, as explained by special relativity.
This formula helps calculate the perceived difference in time between the two observers.The formula is:\[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]Here's what each symbol means:
  • \(t'\): Time elapsed for the moving observer (e.g., on a spaceship).
  • \(t\): Time elapsed for the stationary observer (e.g., on Earth).
  • \(v\): The velocity of the moving object.
  • \(c\): Speed of light in a vacuum.
This formula illustrates how increased speed \(v\) approaches the speed of light \(c\), significantly increasing the time observed by the stationary observer.
The scenario outlined in the original problem demonstrates this principle: as the spacecraft travels at \(0.74c\) and alters the duration of the journey between ship and Earth perspectives.

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Most popular questions from this chapter

(II) Two identical particles of mass \(m\) approach each other at equal and opposite speeds, \(v\). The collision is completely inelastic and results in a single particle at rest. What is the mass of the new particle? How much energy was lost in the collision? How much kinetic energy was lost in this collision?

(II) A stick of length \(\ell_{0},\) at rest in reference frame S, makes an angle \(\theta\) with the \(x\) axis. In reference frame \(\mathrm{S}^{\prime}\), which moves to the right with velocity \(\overrightarrow{\mathbf{v}}=v \hat{\mathbf{i}}\) with respect to \(\mathbf{S}\), determine \((a)\) the length \(\ell\) of the stick, and \((b)\) the angle \(\theta^{\prime}\) it makes with the \(x^{\prime}\) axis.

(II) A spaceship leaves Earth traveling at \(0.61 c .\) A second spaceship leaves the first at a speed of \(0.87 c\) with respect to the first. Calculate the speed of the second ship with respect to Earth if it is fired \((a)\) in the same direction the first spaceship is already moving, \((b)\) directly backward toward Earth.

(II) Calculate the speed of a proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) whose kinetic energy is exactly half ( \(a\) ) its total energy, (b) its rest energy.

(II) In its own reference frame, a box has the shape of a cube \(2.0 \mathrm{~m}\) on a side. This box is loaded onto the flat floor of a spaceship and the spaceship then flies past us with a horizontal speed of \(0.80 c\). What is the volume of the box as we observe it?
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