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A spaceship whose rest length is \(350 \mathrm{~m}\) has a speed of \(0.82 c\) with respect to a certain reference frame. \(\mathrm{A}\) micrometeorite, also with a speed of \(0.82 c\) in this frame, passes the spaceship on an antiparallel track. How long does it take this object to pass the ship as measured on the ship?

Short Answer

Expert verified
The micrometeorite takes approximately 1.19 microseconds to pass the spaceship as measured on the ship.

Step by step solution

01

Understand the Scenario

This problem involves relativistic effects since the spaceship and micrometeorite are moving at a significant fraction of the speed of light, denoted as 'c'. The spaceship has a proper length of 350 meters. Both objects move with speeds of 0.82c in opposite directions.
02

Use Velocity Addition to Find Relative Speed

In special relativity, the velocity of one object relative to another moving object is found using the velocity addition formula: \ \( u_{rel} = \frac{v + u}{1 + \frac{v \cdot u}{c^2}} \ \), where \( v = 0.82c \) and \( u = 0.82c \). Plugging these values in: \ \( u_{rel} = \frac{0.82c + 0.82c}{1 + \frac{0.82c \cdot 0.82c}{c^2}} = \frac{1.64c}{1 + 0.6724} = \frac{1.64c}{1.6724} = 0.98c \). Thus, the micrometeorite moves at a relative speed of 0.98c as measured by the spaceship.
03

Relate Time and Distance Using Relative Speed

To determine how long it takes for the micrometeorite to pass by the spaceship, we use the formula: \ \( t = \frac{d}{u_{rel}} \ \), where \(d = 350 \text{ m} \) is the length of the spaceship and \( u_{rel} \) is 0.98c. Thus, \( t = \frac{350 \text{ m}}{0.98c} \).
04

Calculate the Time in the Spaceship's Frame

Substitute \( c = 3 \times 10^8 \text{ m/s}\): \ \( t = \frac{350}{0.98 \times 3 \times 10^8} \approx \frac{350}{2.94 \times 10^8} \approx 1.19 \times 10^{-6} \) seconds.

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Key Concepts

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

Special Relativity
Special Relativity is a fundamental theory proposed by Albert Einstein in 1905. It revolutionized how we understand space, time, and motion, especially at high speeds close to the speed of light, denoted as "c." This theory introduces two main postulates:
  • The laws of physics are the same for all observers, no matter their constant velocity.
  • The speed of light in a vacuum is constant and will always be measured the same by all observers, regardless of their motion relative to the light source.
These principles lead to intriguing phenomena like time dilation and length contraction, significantly differing from our everyday experiences.
Velocity Addition
In the realm of Special Relativity, adding velocities isn't simply a matter of summing them up. Due to the constancy of the speed of light, we use a more complex formula called the relativistic velocity addition formula:
\[ u_{rel} = \frac{v + u}{1 + \frac{v \cdot u}{c^2}} \]
This formula ensures that the resultant speed never exceeds the speed of light. For example, if two objects move toward each other, each at speeds of 0.82c, calculating their relative speed requires this formula, giving a result of 0.98c.
This contrasts with classical mechanics, where you might expect simply adding 0.82c and 0.82c to result in a speed greater than c, which isn't possible in the relativistic context.
Time Dilation
Time Dilation is another cornerstone of Special Relativity. It describes how time is perceived differently for observers in various states of motion.
When an object moves at a significant fraction of the speed of light, time for it seems to slow down relative to a stationary observer. This doesn't imply that one clock ticks slower; rather, both clocks run at the same rate in their own inertial frames.
This concept is mathematically expressed as:
\[ t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}} \]
where \( t' \) is the dilated time, \( t \) is the proper time, and \( v \) is the relative velocity. In our exercise, despite the micrometeorite zipping past the spaceship at 0.98c, the time observed on the spaceship's clock slightly differs from that seen by a stationary observer.
Length Contraction
Length Contraction is the relativistic effect where an object's length appears shorter in the direction of motion when viewed by an observer in a different inertial frame traveling at relativistic speeds. This fascinating effect occurs only along the direction of motion, leaving other dimensions unaltered.
Mathematically, it is given by the formula:
\[ L = L_0 \sqrt{1-\frac{v^2}{c^2}} \]
where \( L \) is the contracted length, \( L_0 \) is the proper length, and \( v \) is the velocity of the object. In our scenario of the spaceship and micrometeorite, you would see the spaceship compress along its direction of travel from an outside observer’s point of view, though passengers aboard would measure its original length of 350 meters.
This underscores one of the key insights of Special Relativity: measurements of space and time depend on the observer's relative motion.

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Most popular questions from this chapter

You wish to make a round trip from Earth in a spaceship, traveling at constant speed in a straight line for exactly 6 months (as you measure the time interval) and then returning at the same constant speed. You wish further, on your return, to find Earth as it will be exactly 1000 years in the future. (a) To eight significant fig. ures, at what speed parameter \(\beta\) must you travel? (b) Does it matter whether you travel in a straight line on your journey?

A pion is created in the higher reaches of Earth's atmosphere when an incoming high-energy cosmic-ray particle collides with an atomic nucleus. A pion so formed descends toward Earth with a speed of \(0.99 c\). In a reference frame in which they are at rest, pions decay with an average life of \(26 \mathrm{~ns}\). As measured in a frame fixed with respect to Earth, how far (on the average) will such a pion move through the atmosphere before it decays?

Certain wavelengths in the light from a galaxy in the constellation Virgo are observed to be \(0.4 \%\) longer than the corresponding light from Earth sources. (a) What is the radial speed of this galaxy with respect to Earth? (b) Is the galaxy approaching or receding from Earth?

To eight significant figures, what is speed parameter \(\beta\) if the Lorentz factor \(\gamma\) is \(\begin{array}{lll}\text { (a) } 1.010 & 000 & 0, \text { (b) } & 10.000000, & \text { (c) } 100.00000 \text { , }\end{array}\) and (d) \(1000.0000\) ?

A spaceship, moving away from Earth at a speed of \(0.900 c\), reports back by transmitting at a frequency (measured in the spaceship frame) of \(100 \mathrm{MHz}\). To what frequency must Earth receivers be tuned to receive the report?

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