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Suppose that you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon your return, the people on earth will have advanced exactly one hundred years into the future. According to special relativity, how fast must you travel? Express your answer to five significant figures as a multiple of \(c-\) for example, \(0.95585 \mathrm{c}\)

Short Answer

Expert verified
The spacecraft must travel at approximately 0.99995c.

Step by step solution

01

Understand Time Dilation in Special Relativity

Time dilation is a concept from Einstein's theory of special relativity, where time measured in a moving system (such as the spacecraft) is slower compared to time measured in a stationary system (like Earth's). The time dilation formula is given by:\[ T = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}} \]Where \(T\) is the time measured on Earth, \(T_0\) is the time measured on the spacecraft, \(v\) is the velocity of the spacecraft, and \(c\) is the speed of light.
02

Identify the Given Time Periods

- The time elapsed on Earth \(T\) is 100 years.- The time elapsed on the spacecraft \(T_0\) is 1 year (6 months outbound, 6 months inbound).
03

Set Up the Time Dilation Equation

Using the time dilation formula, we have:\[ 100 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
04

Solve for \(v\)

Rearrange to solve for velocity \(v\):1. Take the reciprocal of both sides: \[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{100} \]2. Square both sides: \[ 1 - \frac{v^2}{c^2} = \frac{1}{10000} \]3. Solve for \(v^2/c^2\): \[ \frac{v^2}{c^2} = 1 - \frac{1}{10000} \] \[ \frac{v^2}{c^2} = \frac{9999}{10000} \]4. Solve for \(v\): \[ v = c \times \sqrt{\frac{9999}{10000}} \]
05

Calculate the Result

Compute the value:\[ v = c \times \sqrt{0.9999} \]\[ v \approx c \times 0.99995 \]Thus, the spacecraft must travel at approximately \(0.99995c\).

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Key Concepts

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

Time Dilation
In the world of special relativity, time dilation is one of the most fascinating concepts. It describes how the passage of time is relative and can vary for observers in different frames of motion. Imagine you are on a high-speed spacecraft cruising away from Earth. For you inside the spacecraft, time ticks normally. However, for someone observing your journey from Earth, time appears to pass at a different rate.
This idea is encapsulated in the time dilation formula:
  • \(T\) represents the time interval measured in the stationary frame (Earth).
  • \(T_0\) is the time interval measured in the moving frame (spacecraft).
  • \(v\) stands for the velocity at which the spacecraft travels.
  • \(c\) is the constant speed of light.
By using this formula, we can understand how much time will pass differently between two observers, as was shown in the exercise where the spacecraft's time was just one year, compared to Earth's one hundred years.
Velocity
Velocity is a key player in the realm of special relativity. It defines the speed of an object in a given direction. In our spacecraft scenario, velocity plays a crucial role in determining the extent of time dilation.
Consider this: The faster your spacecraft travels, the more pronounced the effects of time dilation become. To achieve the extreme outcome of returning to Earth 100 years into the future while experiencing only 1 year on the spacecraft, you must be traveling at a speed close to that of light.
This exercise showed us the importance of solving for velocity in the context of Einstein's equations. By calculating the velocity using the time dilation formula, we found the spacecraft needed to travel at approximately \(0.99995c\), meaning 99.995% of the speed of light.
Speed of Light
The speed of light, denoted as \(c\), is one of the most fundamental constants in the universe. It is approximately equal to 299,792,458 meters per second. In the context of special relativity, nothing can surpass this speed. It is the universal speed limit.
The speed of light influences how we view both space and time. In our exercise, the spacecraft's velocity approaches this cosmic speed limit. Because the speed of light is the maximum velocity possible, the effects of time dilation become significant only when nearing this speed.
This concept reveals that as you approach the speed of light, not only does time slow down for you, but your perception of distance, energy, and mass will alter as well. That's why understanding \(c\) is crucial when exploring the depths of Einstein's special relativity.
Einstein's Theory
Einstein's theory of special relativity is a groundbreaking framework that reshaped our understanding of physics. It introduced revolutionary ideas about how time and space interact. One of its key predictions is the relativity of simultaneity—events that are simultaneous for one observer may not be so for another.
Special relativity hinges on two core postulates:
  • The laws of physics are invariant in all inertial frames of reference.
  • The speed of light in a vacuum is the same for all observers, regardless of their motion relative to the light source.
This theory reshuffles our notions of time, space, and energy, showing how they are intertwined. In the context of the exercise, Einstein's insights provided the framework to calculate the velocity required to experience the different passage of time between Earth and the spacecraft. This demonstration of time dilation is just one of the many fascinating implications of Einstein's revolutionary theory.

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Most popular questions from this chapter

A \(6.00-\mathrm{kg}\) object oscillates back and forth at the end of a spring whose spring constant is \(76.0 \mathrm{N} / \mathrm{m}\). An observer is traveling at a speed of \(1.90 \times 10^{8} \mathrm{m} / \mathrm{s}\) relative to the fixed end of the spring. What does this observer measure for the period of oscillation?

Two kilograms of water are changed (a) from ice at \(0^{\circ} \mathrm{C}\) into liquid water at \(0^{\circ} \mathrm{C}\) and \((\mathbf{b})\) from liquid water at \(100^{\circ} \mathrm{C}\) into steam at \(100^{\circ} \mathrm{C}\). For each situation, determine the change in mass of the water.

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