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If the intensity of direct sunlight at a point on the earth's surface is \(0.78 \mathrm{kW} / \mathrm{m}^{2},\) find \((\mathrm{a})\) the average momentum density (momentum per unit volume) in the sunlight and (b) the average momentum flow rate in the sunlight.

Short Answer

Expert verified
(a) Average momentum density is \( \frac{13}{3 \times 10^7} \, \text{kg/m}^2 \cdot \text{s} \). (b) Average momentum flow rate is \( 1.3 \times 10^{-5} \, \text{kg/s}^2 \).

Step by step solution

01

Understand Relationship between Intensity and Energy Density

The intensity of light is related to the energy density (u) of the light by the equation: \( I = c u \), where \( I \) is the intensity and \( c \) is the speed of light in vacuum. We can rearrange this equation to find the energy density: \( u = \frac{I}{c} \).
02

Calculate Energy Density

Given that the intensity \( I = 0.78 \, \text{kW/m}^2 = 780 \, \text{W/m}^2 \), and knowing \( c = 3 \times 10^8 \, \text{m/s} \), substitute these values into the formula: \( u = \frac{780}{3 \times 10^8} \, \text{J/m}^3 \).
03

Calculate Average Momentum Density

The average momentum density \( p \) is related to the energy density via: \( p = \frac{u}{c} \). Use the energy density calculated in the previous step: \( p = \frac{780}{(3 \times 10^8)^2} \). This simplifies to \( p = \frac{780}{9 \times 10^{16}} = \frac{13}{3 \times 10^7} \, \text{kg/m}^2 \cdot \text{s} \).
04

Calculate Average Momentum Flow Rate

The average momentum flow rate (momentum per unit time) is equivalent to the radiation pressure times the speed of light. For electromagnetic radiation, this can be written as \( P = \frac{I}{c} \). Use the given intensity \( I = 780 \, \text{W/m}^2 \): \( P = \frac{780}{3 \times 10^8} = \frac{13}{5 \times 10^7} \, \text{N/m}^2 = 1.3 \times 10^{-5} \, \text{N/m}^2 \). Since this is already in units of momentum flow rate, it represents \( 1.3 \times 10^{-5} \, \text{kg/s}^2 \) through each square meter.

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Key Concepts

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

Energy Density
In discussing electromagnetic radiation, energy density is a core component that allows us to understand how much energy is contained within a specific volume. We use the relationship between intensity and energy density, which is described by the equation: \( I = c u \). Here, \( I \) represents the intensity of the light in watts per square meter, \( c \) is the speed of light, and \( u \) is the energy density in joules per cubic meter.

To find the energy density, the equation can be rearranged to become \( u = \frac{I}{c} \). Given sunlight intensity at \( 780 \, \text{W/m}^2 \) and the speed of light as \( 3 \times 10^8 \, \text{m/s} \), we substitute these values into the formula to get \( u = \frac{780}{3 \times 10^8} \, \text{J/m}^3 \).
  • Energy density provides insight into how intense the light is, effectively measuring the energy available within that space.
  • A low energy density implies less energy in the radiation, impacting how much can be absorbed or used for processes like photosynthesis.
Intensity of Light
The intensity of light is central to understanding how much power is transferred through a given area. Defined as power per unit area, it is measured in watts per square meter \( \text{W/m}^2 \).

Consider the intensity of sunlight calculated in the original problem at \( 780 \, \text{W/m}^2 \). This value indicates the concentration of energy hitting the earth's surface directly from the sun.
  • Higher intensity means more energy per unit area and can significantly affect temperature and energy dynamics in the environment.
  • Intensity is also essential in determining luminosity and overall energy output in astrological settings.
Knowing the intensity helps in calculating the energy density and ultimately the momentum carried by photons in light, key in understanding how light interacts with matter and contributes to solar power devices.
Radiation Pressure
Radiation pressure is exerted upon a surface by light or other electromagnetic radiation. It arises because photons, which make up light, possess momentum. When light strikes a surface, it can transfer this momentum, resulting in a small but measurable force.
  • In the case of sunlight, the radiation pressure is calculated by \( P = \frac{I}{c} \), where \( I \) is the intensity and \( c \) is the speed of light.
  • For the intensity of \( 780 \, \text{W/m}^2 \), the radiation pressure is \( 1.3 \times 10^{-5} \, \text{N/m}^2 \).
This pressure is minute, yet it is critical in understanding solar sails in spacecraft technologies, where radiation pressure from the sun's light is used to propel spacecraft without fuel. Such a concept leverages the multiplication of this tiny force over an extended area to generate forward momentum and illustrates the practical applications of radiation pressure in technology.

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Most popular questions from this chapter

An intense light source radiates uniformly in all directions. At a distance of 5.0 \(\mathrm{m}\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10^{-6} \mathrm{Pa}\) . What is the total average power output of the source?

Two square reflectors, each 1.50 \(\mathrm{cm}\) on a side and of mass 4.00 \(\mathrm{g}\) , are located at opposite ends of a thin, extremely light, \(1.00-\mathrm{m}\) rod that can rotate without friction and in a vacuum about an axle perpendicular to it through its center (Fig. 32.24\()\) . These reficctors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude 1.25 \(\mathrm{N} / \mathrm{C}\) that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating. What is the angular acceleration of this device?

In a TV picture, ghost images are formed when the signal from the transmitter travels to the receiver both directly and indirectly after reflection from a building or other large metallic mass. In a 25 -inch set, the ghost is about 1.0 \(\mathrm{cm}\) to the right of the principal image if the reflected signal arrives 0.60\(\mu\) s after the principal signal. In this case, what is the difference in path lengths for the two signals?

In the \(25-\) fil Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 \(\mathrm{W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

The plane of a flat surface is perpendicular to the propagation direction of an electromagnetic wave of intensity \(1 .\) The surface absorbs a fraction \(w\) of the incident intensity, where \(0 \leq w \leq 1,\) and reflects the rest. (a) Show that the radiation pressure on the surface equals \((2-w) I / c .\) (b) Show that this expression gives the correct results for a surface that is (i) totally absorbing and (ii) totally reflective. (c) For an incident intensity of \(1.40 \mathrm{kW} / \mathrm{m}^{2},\) what is the radiation pressure for 90\(\%\) absorption? For 90\(\%\) reflection?

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