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Chapter 33: Problem 8

Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at \(1.0 \mathrm{MW}\). What is the intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system, 4.3 ly away? (An alien civilization at that distance might be able to watch \(X\) Files. ) A light-ycar (ly) is the distance light travels in one year.

Short Answer

Expert verified
The intensity is approximately \(4.81 \times 10^{-29} \text{ W/m}^2\).

Step by step solution

01

Understand the Parameters

We know that the power of the TV station broadcasting is given as 1.0 MW, which is equivalent to \(1.0 \times 10^6 \text{ W}\). We need to find the intensity of the signal at a distance of 4.3 light years. A light year is the distance that light travels in one year, and the speed of light \(c\) is approximately \(3 \times 10^8 \text{ m/s}\).
02

Calculate Distance in Meters

First, we convert 4.3 light years into meters. One light year is the distance traveled by light in one year: \[ 1 \text{ year} = 365.25 \times 24 \times 60 \times 60 \text{ seconds} \] Distance of 1 light year: \[ 1 \text{ light year} = c \times 1 \text{ year} = 3 \times 10^8 \text{ m/s} \times 31,557,600 \text{ s} \approx 9.46 \times 10^{15} \text{ m} \] Thus, 4.3 light years is: \[ 4.3 \times 9.46 \times 10^{15} \text{ m} \approx 4.07 \times 10^{16} \text{ m} \]
03

Use the Intensity Formula

The intensity of a point source is given by the formula: \[ I = \frac{P}{4\pi r^2} \] where \(I\) is intensity, \(P\) is power, and \(r\) is the distance from the source. In this case, \(P = 1.0 \times 10^6 \text{ W}\) and \(r = 4.07 \times 10^{16} \text{ m}\).
04

Calculate Intensity

Substitute the values into the intensity formula: \[ I = \frac{1.0 \times 10^6}{4\pi (4.07 \times 10^{16})^2} \] Calculate the denominator first: \[ 4\pi (4.07 \times 10^{16})^2 \approx 2.08 \times 10^{34} \] Now calculate the intensity: \[ I \approx \frac{1.0 \times 10^6}{2.08 \times 10^{34}} \approx 4.81 \times 10^{-29} \text{ W/m}^2 \]

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Key Concepts

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

Point Source Broadcasting
A TV station can be thought of as a point source broadcasting its signal. In physics, a point source is an idealized source of energy that emits energy equally in all directions. This model simplifies calculations and helps understand how energy spreads over large distances.
  • Point sources are theoretical and don't exist in reality, but they are useful for understanding concepts like wave propagation and energy distribution.
  • The key idea is that all the transmitted energy spreads out uniformly in a spherical shape around the source. This makes it easier to calculate how much energy reaches a certain location.
The point broadcasting concept is crucial when we do calculations of intensity over large distances, like figuring out how much of the TV station's signal might reach a distant star like Proxima Centauri.
Isotropic Emission
Isotropic emission is a type of emission where the energy is spread out evenly in all directions from the source. In the context of our exercise, the TV station emits its signal isotropically. This is important because it means that the signal spreads out over the surface of a sphere.
Getting the concept right for isotropic emission helps to understand the formula used for intensity calculations:
  • Emitted energy forms a spherical wavefront as it expands from the origin.
  • The surface area of the sphere is given by the formula \(4\pi r^2\), where \(r\) is the distance from the source.
  • The intensity drops as you move away from the source because the same amount of energy has to cover a larger area.
Understanding isotropic emission is crucial to calculating how much of the original energy from the source remains when reaching a point in space.
Light Year Conversion
Converting light years into meters is an essential step in understanding astronomical distances. A light year defines the distance that light travels in the vacuum of space in one year. This is an enormous figure often used to describe the vast separations between stars. To convert light years into meters, follow these steps:
  • First, determine the number of seconds in a year: \(365.25\times24\times60\times60=31,557,600 \, ext{s}\).
  • Multiply this by the speed of light: \(3\times10^8 \, ext{m/s}\).
  • So, one light year becomes \(9.46\times10^{15} \, ext{m}\).
Performing this conversion helps us apply the correct physical units when calculating the intensity, giving a measurement that is easier to work with in physics.
Distance in Meters
The concept of distance in meters is crucial in our calculations, especially when dealing with astronomical distances. By converting distances from light years to meters, we can use common physical formulas more effectively.
This particular exercise required converting 4.3 light years into meters:
  • Based on the light year conversion factor (\(9.46\times10^{15}\, \text{m}\) per light year), we multiply this by 4.3 to get the distance to Proxima Centauri.
  • The final distance used in calculations was approximately \(4.07\times10^{16} \, ext{m}\).
  • Using meters standardizes the units and ensures consistency with formulas, all of which typically use the metric system.
This step is invaluable for accurate physics calculations, setting a clear path for further analysis of our broadcasting problem.

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Most popular questions from this chapter

As a comet swings around the Sun, ice on the comet's surface vaporizes, releasing trapped dust particles and ions. The ions, because they are electrically charged, are forced by the electrically charged solar wind into a straight ion tail that points radially away from the Sun (Fig. \(33-39\) ). The (electrically neutral) dust particles are pushed radially outward from the Sun by the radiation force on them from sunlight. Assume that the dust particles are spherical, have density \(3.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) and are totally absorbing. (a) What radius must a particle have in order to follow a straight path, like path 2 in the figure? (b) If its radius is larger, does its path curve away from the Sun (like path 1 ) or toward the Sun (like path 3 )?

A point source of light emits isotropically with a power of \(200 \mathrm{~W}\). What is the force due to the light on a totally absorbing sphere of radius \(2.0 \mathrm{~cm}\) at a distance of \(20 \mathrm{~m}\) from the source?

High-power lasers are used to compress a plasma (a gas of charged particles) by radiation pressure. A laser generating radiation pulses with peak power \(1.5 \times 10^{3} \mathrm{MW}\) is focused onto \(1.0 \mathrm{~mm}^{2}\) of high-electron-density plasma. Find the pressure exerted on the plasma if the plasma reflects all the light beams directly back along their paths.

A beam of intensity \(I\) reflects from a long. totally reflecting cylinder of radius \(R ;\) the beam is perpendicular to the central axis of the cylinder and has a diamcter larger than \(2 R .\) What is the beam's force per unit length on the cylinder?

A certain helium-neon laser emits red light in a narrow band of wavelengths centered at \(632.8 \mathrm{nm}\) and with a "wavelength width" (such as on the scale of Fig. \(33-1\) ) of \(0.0100 \mathrm{nm}\). What is the corresponding "frequency width" for the emission?
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