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A certain light source sends out \(2 \times 10^{15}\) pulses each second. As a spaceship travels parallel to the Earth's surface with a speed of \(0.90 \mathrm{c}\), it uses this source to send pulses to the Earth. The pulses are sent perpendicular to the path of the ship. How many pulses are recorded on Earth each second?

Short Answer

Expert verified
Approximately \(8.7 \times 10^{14}\) pulses per second are recorded on Earth.

Step by step solution

01

Understand the Problem

We are asked to find out how many pulses per second are recorded on Earth when the light source is on a spaceship traveling at 0.9 times the speed of light, while the light pulses are sent perpendicularly to the motion.
02

Identify Known Values

The spaceship moves at a velocity of 0.90c relative to Earth. The light source on the spaceship sends out \(2 \times 10^{15}\) pulses per second in its rest frame.
03

Consider Effects of Relativity

Since the pulses are sent perpendicular to the ship's motion, we should use the transverse Doppler effect, which alters the frequency of pulses due to the relativistic motion of the source.
04

Apply Transverse Doppler Effect

The transverse Doppler effect formula is \(f = f_0 \sqrt{1 - \frac{v^2}{c^2}}\), where \(f\) is the observed frequency and \(f_0\) is the source frequency. Use the given speed of the spaceship to calculate.
05

Plug in Values and Compute

The given speed is 0.90c, so substitute \(v = 0.90c\) and \(f_0 = 2 \times 10^{15} \text{ pulses/sec}\) into the formula: \[ f = 2 \times 10^{15} \times \sqrt{1 - (0.90)^2} \]. Simplifying inside the square root: \[ 1 - 0.81 = 0.19 \], so, \[ f = 2 \times 10^{15} \times \sqrt{0.19} \].
06

Calculate Final Result

Calculate \(\sqrt{0.19} \approx 0.435\) and multiply: \[ f \approx 2 \times 10^{15} \times 0.435 = 8.7 \times 10^{14} \]. Thus, approximately \(8.7 \times 10^{14}\) pulses are recorded on Earth each second.

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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 that was revolutionized by Albert Einstein's Theory of Relativity. It essentially describes how the laws of physics are the same for all non-accelerating observers, and it introduces the idea that space and time are interwoven into a single continuum known as spacetime.
One of the key ideas in relativity is that the speed of light is constant for all observers, regardless of their motion relative to the light source. This leads to several surprising effects, such as time dilation and length contraction. In particular, when objects move at speeds close to the speed of light, their observed behavior changes significantly from what we would expect in classical mechanics.
In the problem involving the spaceship, relativity comes into play because the spaceship is traveling at a significant fraction of the speed of light (0.90c). This high speed means we must use the principles of relativity, rather than classical physics, to accurately describe how the pulses of light are perceived on Earth.
Light Pulses
Light pulses in the realm of physics refer to brief bursts of light emitted from a source. These pulses can be used to transmit information or to perform various measurements. Because light is composed of photons, these pulses essentially consist of numerous photons moving through space.
A critical aspect of light is its electromagnetic nature, which means it can travel through a vacuum and its speed is the universal speed limit, noted as "c." The speed of light in a vacuum is approximately 299,792,458 meters per second. In this context, the light source on the spaceship emits an impressive number of pulses each second, specifically, \(2 \times 10^{15}\) pulses, while moving almost as fast as light itself.
Understanding how these light pulses travel and are perceived is crucial when looking at scenarios involving high velocities, such as those described by special relativity. When the spaceship travels at such high speeds compared to the stationary observer on Earth, peculiar effects on the light pulses, due to relativistic principles, must be taken into account.
Frequency Calculation
Frequency calculation in physics involves determining the rate at which waves repeat over a specific time period. It is often expressed in Hertz (Hz), indicating cycles per second. In this exercise, frequency calculation deals with determining how many light pulses are observed on Earth each second, given the relativistic effects at play.
Due to the transverse Doppler effect, the observed frequency of light changes when the source is moving relative to the observer. This effect considers not the path of travel directly towards or away from the observer, but perpendicular to the travel direction, as is the case here. The formula used for this frequency shift is:
\[ f = f_0 \sqrt{1 - \frac{v^2}{c^2}} \]
where \( f \) is the frequency observed on Earth, \( f_0 \) is the rest frame frequency (\(2 \times 10^{15} \) pulses/sec), \( v \) is the velocity of the spaceship, and \( c \) is the speed of light.
By substituting the given values into the formula, we calculate the observed frequency as \( 8.7 \times 10^{14} \) pulses each second. This reduction in observed frequency illustrates how relativistic speeds affect the perception of time and frequency, fundamental to understanding the finer points of relativity and its implications.

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Most popular questions from this chapter

A spacecraft moving at \(0.95 \mathrm{c}\) travels from the Earth to the star Alpha Centauri, which is \(4.5\) light years away. How long will the trip take according to \((a)\) Earth clocks and \((b)\) spacecraft clocks? (c) How far is it from Earth to the star according to spacecraft occupants? ( \(d\) ) What do they compute their speed to be? A light year is the distance light travels in 1 year, namely 1 light year \(=\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\left(3.16 \times 10^{7} \mathrm{~s}\right)=9.47 \times 10^{15} \mathrm{~m}\) Hence the distance to the star (according to earthlings) is $$ \begin{array}{l} d_{e}=(4.5)\left(9.47 \times 10^{15} \mathrm{~m}\right)=4.3 \times 10^{16} \mathrm{~m} \\ \text { (a) } \Delta t_{e}=\frac{d_{e}}{v}=\frac{4.3 \times 10^{16} \mathrm{~m}}{(0.95)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=1.5 \times 10^{8} \mathrm{~s} \end{array} $$ (b) Because clocks on the moving spacecraft run slower, $$ \Delta t_{\text {cnth }}=\Delta t_{\varepsilon} \sqrt{1-(v / \mathrm{c})^{2}}=\left(1.51 \times 10^{8} \mathrm{~s}\right)(0.312)=4.7 \times 10^{7} \mathrm{~s} $$ (c) For the spacecraft occupants, the Earth-star distance is moving past them with speed 0.95c. Therefore, that distance is shortened for them; they find it to be $$ d_{\text {cath }}=\left(4.3 \times 10^{16} \mathrm{~m}\right) \sqrt{1-(0.95)^{2}}=1.3 \times 10^{16} \mathrm{~m} $$ (d) For the spacecraft occupants, their relative speed is $$ v=\frac{d_{\text {cnft }}}{\Delta t_{\text {caff }}}=\frac{1.34 \times 10^{16} \mathrm{~m}}{4.71 \times 10^{7} \mathrm{~s}}=2.8 \times 10^{8} \mathrm{~m} / \mathrm{s} $$ which is 0.95c. Both Earth and spacecraft observers measure the same relative speed.

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.

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 \(v=0.14 \mathrm{c}=4.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

The insignia painted on the side of a spaceship is a circle with a line across it at \(45^{\circ}\) to the vertical. As the ship shoots past another ship in space, with a relative speed of \(0.95 \mathrm{c}\), the second ship observes the insignia. What angle does the observed line make to the vertical?

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

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