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A radar antenna is rotating and makes one revolution every 25 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed \(v\) measure that the antenna makes one revolution every 42 s. What is the ratio o\(/ \mathrm{c}\) of the speed \(v\) to the speed \(c\) of light in a vacuum?

Short Answer

Expert verified
The ratio \(\frac{v}{c}\) is approximately 0.804.

Step by step solution

01

Understanding the problem

We need to find the speed ratio \(\frac{v}{c}\), where \(v\) is the speed of a spaceship relative to Earth, and \(c\) is the speed of light. The radar on Earth takes 25 seconds per revolution, while on the spaceship it appears to take 42 seconds per revolution.
02

Identifying relativistic effect

The change in observed revolution time is due to time dilation, an effect described by the theory of relativity. This occurs because the spaceship is moving at a significant fraction of the speed of light.
03

Setting up the time dilation equation

According to time dilation, the time observed (\(t'\)) is related to the time in the stationary frame (\(t\)) by \(t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}\). Here \(t = 25\) seconds (Earth) and \(t' = 42\) seconds (spaceship).
04

Rearranging the time dilation formula

We need \(\sqrt{1-\frac{v^2}{c^2}}\) to calculate the speed ratio. Rearranging, we have \(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{t'}{t}\), so \(1-\frac{v^2}{c^2} = \left(\frac{t}{t'}\right)^2\).
05

Plugging in the values

Substitute \(t = 25\) s and \(t' = 42\) s into the equation: \(1-\frac{v^2}{c^2} = \left(\frac{25}{42}\right)^2\).
06

Solving for \(\frac{v^2}{c^2}\)

Calculate \(\left(\frac{25}{42}\right)^2 = \left(\frac{25}{42}\right)^2 \approx 0.354\), then \(\frac{v^2}{c^2} = 1 - 0.354 = 0.646\).
07

Finding the ratio \(\frac{v}{c}\)

Take the square root, \(\frac{v}{c} = \sqrt{0.646} \approx 0.804\).

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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, formulated by Albert Einstein, revolutionized our understanding of time and space. It is a cornerstone of modern physics and comprises two main parts: Special Relativity and General Relativity.
Special Relativity focuses on objects moving at constant speeds, particularly speeds close to the speed of light. One of its fundamental insights is that the laws of physics are the same for all observers, regardless of their relative motion. This theory introduced the idea that time and space are interconnected in a four-dimensional continuum known as spacetime.
An important concept arising from this theory is time dilation, where time measured in a moving system differs from time measured in a stationary system. It emphasizes that time is not absolute but can vary depending on an observer's relative velocity. This breakthrough challenges traditional notions of time and asserts that high-speed travel can alter the passage of time itself.
Relativistic Effects
Relativistic effects become significant when objects move at velocities close to the speed of light. They deviate from the classical mechanics we experience in everyday life. Key relativistic effects include time dilation, length contraction, and mass-energy equivalence.
  • Time Dilation: This is the effect observed when time appears to move slower in a high-speed system relative to a stationary observer. As the relative velocity increases, the difference in time perception becomes more pronounced.
  • Length Contraction: This effect means objects appear shorter in the direction of motion when viewed from a stationary frame. It complements time dilation, demonstrating how space and time are interconnected.
  • Mass-Energy Equivalence: Famously encapsulated in Einstein's equation \(E=mc^2\), this principle states that mass can be seen as a form of energy. As objects accelerate, their effective mass increases, which has profound implications for high-speed travel.
Understanding these effects is crucial for grasping the nature of our universe at high speeds. They reveal the limits of classical intuition and open a window into the complex, intriguing world of relativistic physics.
Time Dilation Equation
The Time Dilation Equation is a mathematical representation of how time is perceived differently by observers in different inertial frames. The equation is expressed as \(t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}\), where:
  • \(t'\) is the time interval measured by an observer in motion relative to a stationary observer.
  • \(t\) is the time interval measured by the stationary observer.
  • \(v\) stands for the relative velocity between the two observers.
  • \(c\) denotes the speed of light, which is approximately \(3 \times 10^8\, \mathrm{m/s}\).
When an object moves at high velocity, as described in the radar antenna problem, the time interval \(t'\) observed in the moving frame is longer than \(t\), the time interval measured by a stationary observer.
This equation highlights the concept of time dilation as it directly connects velocity with time distortion. As the spaceship nears the speed of light, \(t'\) becomes significantly larger, illustrating the intriguing effects of high-speed motion on our perception of time.

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Most popular questions from this chapter

Twins who are 19.0 years of age leave the earth and travel to a distant planet 12.0 light-years away. Assume that the planet and earth are at rest with respect to each other. The twins depart at the same time on different spaceships. One twin travels at a speed of \(0.900 c,\) and the other twin travels at \(0.500 c .\) (a) According to the theory of special relativity, what is the difference between their ages when they meet again at the earliest possible time? (b) Which twin is older?

Three particles are listed in the table. The mass and speed of each particle are given as multiples of the variables \(m\) and \(v,\) which have the values \(m=1.20 \times 10^{-8} \mathrm{kg}\) and \(v=0.200 \mathrm{c} .\) The speed of light in a vacuum is \(c=3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) Determine the momentum for each particle according to special relativity. $$ \begin{array}{ccc} \text { Particle } & \text { Mass } & \text { Speed } \\ \hline \mathbf{a} & m & v \\ \hline \mathbf{b} & \frac{1}{2} m & 2 v \\ \hline \mathbf{c} & \frac{1}{4} m & 4 v \\ \hline \end{array} $$

ssm A particle known as a pion lives for a short time before breaking apart into other particles. Suppose that a pion is moving at a speed of \(0.990 c\), and an observer who is stationary in a laboratory measures the pion's lifetime to be \(3.5 \times 10^{-8}\) s. (a) What is the lifetime according to a hypothetical person who is riding along with the pion? (b) According to this hypothetical person, how far does the laboratory move before the pion breaks apart?

A Klingon spacecraft has a speed of \(0.75 c\) with respect to the earth. The Klingons measure \(37.0 \mathrm{h}\) for the time interval between two events on the earth. What value for the time interval would they measure if their ship had a speed of \(0.94 c\) with respect to the earth?

You are driving down a two-lane country road, and a truck in the opposite lane is traveling toward you. Suppose that the speed of light in a vacuum is \(c=65 \mathrm{m} / \mathrm{s} .\) Determine the speed of the truck relative to you when (a) your speed is \(25 \mathrm{m} / \mathrm{s}\) and the truck's speed is \(35 \mathrm{m} / \mathrm{s}\) and \((\mathrm{b})\) your speed is \(5.0 \mathrm{m} / \mathrm{s}\) and the truck's speed is \(55 \mathrm{m} / \mathrm{s} .\) The speeds given in parts (a) and (b) are relative to the ground.

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