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An astronomer on Earth observes a meteoroid in the southern sky approaching the Earth at a speed of 0.800 \(\mathrm{c}\) . At the time of its discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the time interval required for the meteoroid to reach the Earth as measured by the Earth-bound astronomer, (b) this time interval as measured by a tourist on the meteoroid, and (c) the distance to the Earth as measured by the tourist.

Short Answer

Expert verified
The time interval required for the meteoroid to reach Earth as measured by the Earth-bound astronomer is 25.0 years. The time interval as measured by the tourist on the meteoroid is approximately 15.0 years. The contracted length as measured by the tourist is approximately 12.0 ly.

Step by step solution

01

Calculate the time interval for Earth observer

To find the time interval required for the meteoroid to reach Earth as measured by the Earth-bound astronomer, use the formula for time: time = distance / speed. Here, the distance is 20.0 light years (ly) and the speed is 0.800 times the speed of light (c). Thus, time = 20.0 ly / 0.800 c.
02

Apply the time dilation formula

To determine the time interval as measured by a tourist on the meteoroid, apply the time dilation formula from special relativity: \( t' = t / \sqrt{1 - v^2/c^2} \), where t is the time measured by the Earth observer, v is the relative velocity of the meteoroid, and c is the speed of light.
03

Calculate the contracted length for the tourist

Use the length contraction formula to determine the distance to Earth as measured by the tourist on the meteoroid: \( L' = L \sqrt{1 - v^2/c^2} \), where L is the proper length measured by the Earth observer (20.0 ly), and v is the speed of the meteoroid (0.800 c).

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Key Concepts

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

Time Dilation
Imagine you’re watching a clock on a speeding spaceship; you'd notice the clock ticks slower than your own. This phenomenon is known as time dilation, a core principle of Albert Einstein’s theory of special relativity. Time dilation occurs when an object moves at a significant fraction of the speed of light, symbolized as 'c'. Interestingly, the faster an object moves, the more pronounced the time dilation effect.

For instance, let’s break down the example of our meteoroid approaching Earth at 0.800 c. The time interval for the meteoroid to reach Earth as measured by an Earth-bound astronomer is straightforward: just the distance divided by speed. But for a 'tourist' on the meteoroid, this duration is shorter due to time dilation. According to the formula for time dilation, \( t' = \frac{t}{\sqrt{1 - (v/c)^2}} \), where \( t' \) is the 'dilated' time interval, their journey's duration shrinks as speed increases. This time warping shows that time is not absolute but relative to the observer’s speed!
Length Contraction
What about the size of objects moving at these high velocities? They shrink, according to length contraction, another mind-bending aspect of special relativity. Length contraction suggests that the length of an object in the direction of motion, as measured by an observer at rest, is decreased when the object is moving near the speed of light relative to the observer.

For the meteoroid’s tourist, not only do their watches tick differently but the very space between the stars contracts. Based on the length contraction formula, \( L' = L \sqrt{1 - (v/c)^2} \), the meteoroid’s distance to Earth, when viewed from the tourist's perspective, is less than the 20.0 ly measured from Earth. Here, \( L \) represents the 'proper length,' and \( L' \) the contracted length as experienced by the one in motion. So, just as time bends, so does space.
Relativistic Speed
The term relativistic speed refers to velocities that are a significant fraction of the speed of light. In the realms of everyday human experience, traditional Newtonian physics suffices, but as we approach the speed of light, relativistic effects cannot be ignored. At relativistic speeds, objects behave in counterintuitive ways, as illustrated by the time dilation and length contraction already mentioned.

When an object, such as our meteoroid, travels at 0.800 c, we must use Einstein's theory of special relativity to accurately describe its motion and various effects on time and space. It’s essential to understand these relativistic speeds to grasp the full impact on other phenomena, like mass increase and energy requirements, too.
Light Year
A light year is the distance light travels in one Earth year, roughly 5.88 trillion miles (9.46 trillion kilometers). It's used to measure astronomical distances, which are so vast that using kilometers or miles would be impractical. For our discussed meteoroid, the 20.0 light years is not just a measure of space but also a temporal indicator when coupled with relativistic effects.

In the context of special relativity, it’s fascinating that a light year as a measure of distance can contract or 'shorten' for observers moving at high velocities. So while astronomers measure the meteoroid to be 20.0 light years away, our tourist in relative motion would measure a shorter distance, illustrating how special relativity reshapes our perspective on the universe.

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Most popular questions from this chapter

Police radar detects the speed of a car (Fig. P39.19) as follows. Microwaves of a precisely known frequency are broadcast toward the car. The moving car reflects the microwaves with a Doppler shift. The reflected waves are received and combined with an attenuated version of the transmitted wave. Beats occur between the two microwave signals. The beat frequency is measured. (a) For an electromagnetic wave reflected back to its source from a mirror approaching at speed \(v,\) show that the reflected wave has frequency $$ f=f_{\text { source }} \frac{c+v}{c-v} $$ where \(f_{\text { source }}\) is the source frequency. (b) When \(v\) is much less than \(c,\) the beat frequency is much smaller than the transmitted frequency. In this case use the approximation \(f+f_{\text { source }} \approx 2 f_{\text { source }}\) and show that the beat frequency can be written as \(f_{\text { beat }}\) \(=2 v / \lambda .\) (c) What beat frequency is measured for a car speed of 30.0 \(\mathrm{m} / \mathrm{s}\) if the microwaves have frequency 10.0 \(\mathrm{GHz}\) ? (d) If the beat frequency measurement is accurate to \(\pm 5 \mathrm{Hz},\) how accurate is the velocity measurement?

A supertrain (proper length 100 \(\mathrm{m} )\) travels at a speed of 0.950 \(\mathrm{c}\) as it passes through a tunnel (proper length 50.0 \(\mathrm{m}\) ). As seen by a trackside observer, is the train ever completely within the tunnel? If so, with how much space to spare?

A pion at rest \(\left(m_{\pi}=273 m_{e}\right)\) decays to a muon \(\left(m_{\mu}=\right.\) 207\(m_{e} )\) and an antineutrino \(\left(m_{\overline{\nu}} \approx 0\right) .\) The reaction is written \(\pi^{-} \rightarrow \mu^{-}+\overline{\nu} .\) Find the kinetic energy of the muon and the energy of the antineutrino in electron volts.(Suggestion: Conserve both energy and momentum.)

When 1.00 g of hydrogen combines with 8.00 g of oxygen, 9.00 g of water is formed. During this chemical reaction, \(2.86 \times 10^{5} \mathrm{J}\) of energy is released. How much mass do the constituents of this reaction lose? Is the loss of mass likely to be detectable?

An observer in reference frame S sees two events as simultaneous. Event \(A\) occurs at the point \((50.0 \mathrm{m}, 0,0)\) at the instant \(9 : 00 : 00\) Universal time, 15 January 2004 . Event \(B\) occurs at the point \((150 \mathrm{m}, 0,0)\) at the same moment. A second observer, moving past with a velocity of 0.800 \(\mathrm{ci}\) , also observes the two events. In her reference frame \(\mathrm{S}^{\prime},\) which event occurred first and what time interval elapsed between the events?

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