/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 When electromagnetic radiation s... [FREE SOLUTION] | 91Ó°ÊÓ

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When electromagnetic radiation strikes perpendicular to a flat surface, a totally absorbing surface feels radiation pressure \(I_{0} / c\) where \(I_{0}\) is the intensity of incident electromagnetic radiation. A totally reflecting surface feels twice that pressure. More generally, a surface absorbs a proportion \(e\) of the incident radiation and reflects a complementary proportion, \(1-e,\) where \(e\) is the emissivity of the surface, as introduced in Chapter \(17 .\) Note that \(0 \leq e \leq 1 .\) (a) Determine the radiation pressure \(p_{\text {rad }}\) in terms of \(I_{0}\) and \(e .\) (b) Consider cosmic dust particles in outer space at a distance of \(1.5 \times 10^{11} \mathrm{~m}\) from the sun, where \(I_{\text {sun }}=1.4 \mathrm{~kW} / \mathrm{m}^{2}\). We can model these particles as tiny disks with \(e=0.61,\) diameter \(8.0 \mu \mathrm{m}\) and mass \(1.0 \times 10^{-10} \mathrm{grams},\) all oriented perpendicular to the sun's rays. What is the force on one of these particles that is exerted by the radiation from the sun? (c) What is the ratio of this force to the attractive force of gravity exerted by the sun on the particle?

Short Answer

Expert verified
The radiation pressure \(p_{\text {rad }}\) is found to be \(\frac{I_{0}}{c}(2-e)\), and the ratio of the radiation force to the gravitational force is calculated from the corresponding forces.

Step by step solution

01

Derive the Radiation Pressure Formula

We know that for a fully absorbing surface, the radiation pressure \(p_{\text {rad }}\) is \(I_{0}/c\), and for a fully reflecting surface, it's \(2I_{0}/c\). If a surface absorbs a proportion \(e\), it means it reflects \(1-e\). Therefore, it feels a radiation pressure that is a linear combination of both: \n\n\[p_{\text {rad }}=e \frac{I_{0}}{c}+(1-e) \frac{2 I_{0}}{c} = \frac{I_{0}}{c}(2-e)\]
02

Calculate the Radiation Force

Using the derivation of the previous step, we can use the density of solar radiation \(I_{\text {sun }}=1.4 \mathrm{~kW/m}^{2}\) and the emissivity of a cosmic dust particle \(e=0.61\) to calculate the radiation pressure upon the particle. From that the radiation force F exerted upon the particle is obtained using: \n\n\[F = p_{\text {rad }} \times A, \]\n\nwhere \(A\) is the area of the particle, which can be computed as the area of a circle, \(A = \pi \times (\text {diameter}/2)^2\].
03

Find the Gravitational Force

The gravitational force \(F_{\text {grav }}\) between the sun and the particle is given by the universal law of gravitation: \n\n\[F_{\text {grav }} = G \frac{m_{\text {sun }} \times m_{\text {particle}}}{r^2},\]\n\nwhere \(m_{\text {particle}}\) is the mass of the dust particle, \(r\) is the distance between the dust and the sun, and \(m_{\text {sun }}\) and \(G\) are the mass of the sun and the gravitational constant respectively.
04

Compute the Ratio of Forces

Finally, calculate the ratio of the radiation force \(F\) over the gravitational force \(F_{\text {grav }}\). This will yield the relation of the two forces acting on the particle.

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation refers to waves of the electromagnetic field, propagating through space, carrying electromagnetic energy. These waves include visible light, radio waves, X-rays, and gamma rays.
In the context of radiation pressure, which is an intriguing concept when considering electromagnetic radiation, the pressure is exerted by the force of the photons as they strike a surface.
The interaction between electromagnetic radiation and surfaces leads to pressure due to the momentum change upon striking. This pressure can differ based on whether the surface absorbs or reflects the radiation:
  • Totally absorbing surfaces feel a pressure of \( I_{0} / c \).
  • Fully reflective surfaces experience twice this pressure.
The beauty of this phenomenon lies in its linearity and how it adjusts for mixed surfaces that absorb and reflect in specific proportions. For example, a surface with an emissivity of \( e \) absorbs \( e \) portion and reflects \( 1-e \) portion of the radiation, leading to a combined pressure expressed as  \[ p_{\text{rad}} = \frac{I_{0}}{c}(2-e) \]
Cosmic Dust Particles
Cosmic dust particles, scattered throughout the universe, play a vital role in various cosmic phenomena. These tiny particles, often made up of silicate and carbon materials, range from a few molecules large to several micrometers in size.
In our exercise, we're analyzing cosmic dust particles at a distance of \(1.5 \times 10^{11} \;\text{m}\) from the Sun.
These particles encounter electromagnetic radiation from the Sun, experiencing radiation pressure that exerts a force on them.
This force calculation is done by modeling the particle as a disk with defined diameter and emissivity:
  • Diameter: \(8.0 \; \mu \text{m}\)
  • Emissivity \( e \): 0.61
The area \( A \) of these disks can be computed using \( A = \pi \times (\text{diameter}/2)^2 \).
Given the intensity of solar radiation \( I_{\text{sun}} = 1.4 \;\text{kW/m}^2 \), we can determine the pressure and subsequently the radiation force using \( F = p_{\text{rad}} \times A \). This interaction not only introduces radiation pressure but significantly affects the dynamics of cosmic dusty regions, influencing their movements and interactions with other cosmic bodies.
Gravitational Force
Gravitational force, a fundamental force of nature, is responsible for the attraction between masses. It governs astronomical movements and interactions across the universe.
When calculating gravitational attraction, especially with tiny cosmic dust particles near massive objects like the Sun, one utilizes Newton's Law of Universal Gravitation.
The gravitational force \( F_{\text{grav}} \) between an object and the Sun is defined by:
  • \( F_{\text{grav}} = G \frac{m_{\text{sun}} \times m_{\text{particle}}}{r^2} \)
Here, \( G \) is the gravitational constant, \( m_{\text{sun}} \) the solar mass, and \( m_{\text{particle}} \) the mass of the dust particle, while \( r \) denotes the distance between them.
This force is incredibly small in scale for dust particles due to their minuscule mass and vast distance from the solar body. Interestingly, when these particles are compared against the radiation force exerted by electromagnetic radiation, we often find a fascinating balance or even a dominance of one force over the other. This balance can result in various outcomes, such as causing particles to move away from the Sun or remain in stable orbits.

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Most popular questions from this chapter

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from \(\mathrm{WCCO}\) is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

An air-filled cavity for producing electromagnetic standing waves has two parallel, highly conducting walls separated by a distance \(L\). One standing- wave pattern in the cavity produces nodal planes of the electric field with a spacing of \(1.50 \mathrm{~cm}\). The next-higher-frequency standing wave in the cavity produces nodal planes with a spacing of \(1.25 \mathrm{~cm} .\) What is the distance \(L\) between the walls of the cavity?

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 vacuum about an axle perpendicular to it through its center (Fig. \(\mathbf{P 3 2 . 3 7}\) ). These reflectors 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?

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area \(0.500 \mathrm{~m}^{2}\). At the window, the electric field of the wave has rms value \(0.0400 \mathrm{~V} / \mathrm{m}\). How much energy does this wave carry through the window during a 30.0 s commercial?

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the \(+x\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(+y\) -direction; (b) \(\overrightarrow{\boldsymbol{E}}\) in the \(-y\) -direction, \(\vec{B}\) in the \(+x\) -direction; (c) \(\vec{E}\) in the \(+z\) -direction, \(\vec{B}\) in the \(-x\) -direction; (d) \(\overrightarrow{\boldsymbol{E}}\) in the \(+y\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-z\) -direction.

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