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How fast must an object be moving if its corresponding value of \(y\) is to be \(1.0\) percent larger than \(y\) is when the object is at rest? Give your answer to two significant figures. Use the definition \(\gamma=1 / \sqrt{1-(v / \mathrm{c})^{2}}\) to find that at \(v=0, y=1.0\). Hence, the new value of \(y=1.01(1.0)\), and so $$ 1-\left(\frac{v}{\mathrm{c}}\right)^{2}=\left(\frac{1}{1.01}\right)^{2}=0.980 $$ Solving yields \(v=0.14 \mathrm{c}=4.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The object must move at \( 4.2 \times 10^7 \text{ m/s} \).

Step by step solution

01

Understanding the Problem

We need to find the speed of an object such that its Lorentz factor, \( y = \gamma \), is 1.0% larger than when the object is at rest. For an object at rest, \( y = 1.0 \). Therefore, the new \( y \) should be \( 1.01 \times 1.0 = 1.01 \).
02

Using the Lorentz Factor Formula

The Lorentz factor is given by \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \). We need \( \gamma = 1.01 \). Thus:\[ \frac{1}{\sqrt{1 - (v/c)^2}} = 1.01 \]
03

Isolating \(v/c\)

We rearrange the equation to solve for \( (v/c)^2 \).First, invert both sides of the equation:\[ \sqrt{1 - (v/c)^2} = \frac{1}{1.01} \]Then square both sides to remove the square root:\[ 1 - (v/c)^2 = \left(\frac{1}{1.01}\right)^2 \]
04

Calculating Numerical Values

Calculate \( \left(\frac{1}{1.01}\right)^2 \):\( \left(\frac{1}{1.01}\right)^2 = 0.9801 \).Substitute back into the equation:\[ 1 - (v/c)^2 = 0.9801 \]
05

Solving for \((v/c)^2\)

Subtract \(0.9801\) from 1:\[ (v/c)^2 = 1 - 0.9801 = 0.0199 \]
06

Finding \(v/c\) and \(v\)

Take the square root to find \(v/c\):\[ v/c = \sqrt{0.0199} \approx 0.141 \]To find \(v\), multiply by \(c\) (speed of light \(3 \times 10^8\) m/s):\[ v = 0.141 \times 3 \times 10^8 \text{ m/s} = 4.23 \times 10^7 \text{ m/s} \]
07

Rounding Off

Round \(4.23 \times 10^7 \) m/s to two significant figures, resulting in:\( v = 4.2 \times 10^7 \) m/s.

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Key Concepts

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

Relativity
Relativity is a fundamental concept in physics, introduced by Albert Einstein, that explains how the laws of physics are the same for all observers, regardless of their relative motion. This is known as the principle of relativity. The most famous theory stemming from this concept is the Theory of Special Relativity, which deals with objects moving at a constant speed, particularly speeds close to the speed of light.
Relativity has transformed our understanding of time and space, showing that they are interconnected dimensions rather than separate entities. This concept leads to phenomena such as time dilation and length contraction, where time can slow down or lengths can shrink for objects moving at significant fractions of the speed of light, as observed from a stationary frame.
Understanding relativity is crucial for working with the Lorentz factor, as it arises directly from the need to reconcile the measurements of time and space between different inertial frames. It allows us to predict how objects behave at high speeds and provides the framework for the Lorentz transformation, which mathematically describes how coordinates change between moving observers.
Lorentz Transformation
The Lorentz transformation is a set of equations that describe how the coordinates of an event change when viewed from two different inertial frames moving relative to each other. It forms the backbone of the Theory of Special Relativity and ensures that the speed of light is constant in all inertial frames, a core tenet of relativity.
It can be expressed using the Lorentz factor, \[ \gamma = \frac{1}{\sqrt{1 - (v/c)^2}}. \]This factor accounts for the differences in the observations of space and time between the two frames. When working with the Lorentz transformation:
  • If you know the speed of an object relative to light speed, you can determine how time and distance measurements differ between two observers.
  • It highlights how time can "slow down" (time dilation) or distances can "shorten" (length contraction) relative to the speed of the moving object.
Using these equations enables scientists and engineers to handle scenarios involving high-velocity motion, such as in particle physics or in calculating GPS satellite positions relative to Earth, which must account for relativistic effects to maintain accuracy.
Speed of Light
The speed of light, often denoted as "c," is a universal constant with a value of approximately \(3 \times 10^8\) meters per second in a vacuum. It is not only the speed at which light travels but also the ultimate speed limit for any form of information or matter, according to the principles of relativity.
This speed is crucial in physics because it defines the relationship between space and time. It's the constant that ensures the equations of special relativity hold, such as those involving the Lorentz factor and transformation.
  • In calculations, such as finding the Lorentz factor, knowing the value of "c" lets us compute how close an object's velocity is compared to light.
  • The invariance of "c" in all inertial frames is what causes strange relativistic effects, like time dilation and length contraction, which don't occur when dealing with everyday speeds that are insignificant compared to "c."
The constancy and finite nature of the speed of light also lead to the notion that nothing can travel faster than light, preserving causality within the universe and serving as a basis for modern physics theories.

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Most popular questions from this chapter

Determine the energy required to give an electron a speed equal to \(0.90\) that of light, starting from rest.

Two cells that subdivide on Earth every \(10.0 \mathrm{~s}\) start from the Earth on a journey to the Sun \(\left(1.50 \times 10^{11} \mathrm{~m}\right.\) away) in \(\mathrm{a}\) spacecraft moving at \(0.850 \mathrm{c}\). How many cells will exist when the spacecraft crashes into the Sun? According to Earth observers, with respect to whom the cells are moving, the time taken for the trip to the Sun is the distance traveled \((x)\) over the speed \((v)\) $$ \Delta t_{\mathrm{M}}=\frac{x}{v}=\frac{1.50 \times 10^{11} \mathrm{~m}}{(0.850)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=588 \mathrm{~s} $$ Because spacecraft clocks are moving with respect to the planet, they appear from Earth to run more slowly. The time these clocks read is $$ \Delta t_{\mathrm{s}}=\Delta f_{\mathrm{M}} / \gamma=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}} $$ and so $$ \Delta l_{\mathrm{s}}=310 \mathrm{~s} $$ The cells divide according to the spacecraft clock, a clock that is at rest relative to them. They therefore undergo 31 divisions in this time, since they divide each \(10.0 \mathrm{~s}\). Therefore, the total number of cells present on crashing is $$ (2)^{31}=2.1 \times 10^{9} \text { cells } $$

A proton has a mass of \(1.6726 \times 10^{-27} \mathrm{~kg}\) and is traveling at \(\mathrm{a}\) speed where \(y=2.294157\); that's \(90 \%\) the speed of light. Determine its total energy. Four significant figures will do. Discuss your answer as it compares with classical energy.

Two twins are \(25.0\) years old when one of them sets out on a journey through space at nearly constant speed. The twin in the spaceship measures time with an accurate watch. When he returns to Earth, he claims to be \(31.0\) years old, while the twin left on Earth knows that she is \(43.0\) years old. What was the speed of the spaceship? The spaceship clock as seen by the space-twin reads the trip time to be \(\Delta t_{\mathrm{S}}\), which is \(6.0\) years long. The Earth-bound twin sees her brother age \(6.0\) years, but her clocks tell her that a time \(\Delta t_{\mathrm{M}}=18.0\) years has actually passed. Hence, \(\Delta t_{\mathrm{M}}=y \Delta t_{\mathrm{S}}\) becomes \(\Delta t_{\mathrm{S}}=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}}\) and so $$ 6=18 \sqrt{1-(v / \mathrm{c})^{2}} $$ from which \((\mathrm{u} / \mathrm{c})^{2}=1-0.111\) or \(v=0.943 \mathrm{c}=2.83 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

A person in a spaceship holds a meterstick as the ship shoots past the Earth with a speed \(u\) parallel to the Earth's surface. What does the person in the ship notice as the stick is rotated from parallel to perpendicular to the ship's motion? The stick behaves normally; it does not change its length, because it has no translational motion relative to the observer in the spaceship. However, an observer on Earth would measure the stick to be \((1 \mathrm{~m}) \sqrt{1-(v / \mathrm{c})^{2}}\) long when it is parallel to the ship's motion, and \(1 \mathrm{~m}\) long when it is perpendicular to the ship's motion.

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